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# High School Math : Quadrilaterals

## Example Questions

### Example Question #211 : Geometry

A square  has side lengths of . A second square  has side lengths of . How many  can you fit in a single ?

Explanation:

The area of  is , the area of  is . Therefore, you can fit 5.06  in .

### Example Question #212 : Geometry

The perimeter of a square is   If the square is enlarged by a factor of three, what is the new area?

Explanation:

The perimeter of a square is given by  so the side length of the original square is   The side of the new square is enlarged by a factor of 3 to give

So the area of the new square is given by .

### Example Question #36 : Squares

How much more area does a square with a side of 2r have than a circle with a radius r?  Approximate  π by using 22/7.

1/7 square units

6/7 square units

12/14 square units

4/7 square units

6/7 square units

Explanation:

The area of a circle is given by A = πr2 or 22/7r2

The area of a square is given by A = s2 or (2r)2 = 4r2

Then subtract the area of the circle from the area of the square and get 6/7 square units.

### Example Question #37 : Squares

If the perimeter of a square is 44 centimeters, what is the area of the square in square centimeters?

Explanation:

Since the square's perimeter is 44, then each side is .

Then in order to find the area, use the definition that the

### Example Question #38 : Squares

Given square , with midpoints on each side connected to form a new, smaller square.  How many times bigger is the area of the larger square than the smaller square?

Explanation:

Assume that the length of each midpoint is 1.  This means that the length of each side of the large square is 2, so the area of the larger square is 4 square units.

To find the area of the smaller square, first find the length of each side.  Because the length of each midpoint is 1, each side of the smaller square is  (use either the Pythagorean Theorem or notice that these right trianges are isoceles right trianges, so  can be used).

The area then of the smaller square is 2 square units.

Comparing the area of the two squares, the larger square is 2 times larger than the smaller square.

### Example Question #91 : Quadrilaterals

Find the area of a square with a diagonal of .

Not enough information to solve

Explanation:

A few facts need to be known to solve this problem. Observe that the diagonal of the square cuts it into two right isosceles triangles; therefore, the length of a side of the square to its diagonal is the same as an isosceles right triangle's leg to its hypotenuse: .

Rearrange an solve for .

Now, solve for the area using the formula .

### Example Question #92 : Quadrilaterals

If the ratio of the sides of two squares is , what is the ratio of the areas of those two squares?

Explanation:

Express the ratio of the two sides of the squares as . The area of each square is one side multiplied by itself, so the ratios of the areas would be . The right side of this equation simplifies to a ratio of .

### Example Question #39 : Squares

If a completely fenced-in square-shaped yard requires 140 feet of fence, what is the area, in square feet, of the lot?

1225

4900

140

35

70

1225

Explanation:

Since the yard is square in shape, we can divide the perimeter(140ft) by 4, giving us 35ft for each side. We then square 35 to give us the area, 1225 feet.

### Example Question #93 : Quadrilaterals

What is the area of a square with a diagonal of ?

Explanation:

The formula for the area of a square is . However, the problem gives us a diagonal and not a side.

Remember that all sides of a square are equal, so the diagonal cuts the square into two equal triangles, each a  right triangle.

If we use the Pythagorean Theorem, we see:

Plug in our given diagonal to solve.

From here we can plug our answer back into our original equation:

### Example Question #11 : Squares

A half circle has an area of . What is the area of a square with sides that measure the same length as the diameter of the half circle?

81

72

144

108

36

144

Explanation:

If the area of the half circle is , then the area of a full circle is twice that, or .

Use the formula for the area of a circle to solve for the radius:

36π = πr2

r = 6

If the radius is 6, then the diameter is 12. We know that the sides of the square are the same length as the diameter, so each side has length 12.

Therefore the area of the square is 12 x 12 = 144.


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Constant Rank Theorem Intuition

1. Mar 9, 2014

center o bass

There are two theorems from multivariable calculus that is very important for manifold theory.
The first is the inverse function theorem and the second is the "constant rank theorem". The latter states that

(Constant rank theorem). If $f : U\subset \mathbb{R}^n \to \mathbb{R}^m$ has constant rank $k$ in a neighborhood of a point $p \in U$ , then after a suitable change of coordinates near $p$ in $U$ and $f(p)$ in $\mathbb{R}^m$, the map $f$ assumes the form $(x^1,...,x^n)\mapsto (x^1,...,x^k,0,...,0)$.
More precisely, there are a diffeomorphism $G$ of a neighborhood of $p$ in $U$ sending $p$ to the origin in $\mathbb{R}^n$ and a diffeomorphism $F$ of a neighborhood of $f(p)$ in $\mathbb{R}^m$ sending $f(p)$ to the origin in $\mathbb{R}^m$ such that $(F ◦ f ◦ G)^{−1}(x^1,...,x^n) = (x^1,...,x^k,0,...,0).$

I've gone through the proof of the theorem, but I'm left with little intuition on why it has to be true. Therefore I wonder, do you have any intuitive explanation of why the theorem has to be true?

2. Mar 9, 2014

micromass

Staff Emeritus
Very informally: it is true for linear maps for sure, this is easy to check. Smooth maps can then be locally approximated by linear maps. Since the smooth map has constant rank, the linear approximation doesn't change when going over to close-lying points.

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# Commutator Subgroup with Abelian Factor Group

I'm trying to show that if $Y$ is the commutator subgroup of a group $G$ (assumed to be normal in $G$ with $G/Y$ Abelian), and $N$ is a a subgroup with those same properties, then $Y$ is a subgroup of $N$.

My approach has been to try to show that $\forall y\in Y$, $yN = N \Leftrightarrow xwx^{-1}w^{-1}N=N$, where $y=xwx^{-1}w^{-1}$ for some $x,w\in G$. Then I would have $y\in N$.

Is this the correct approach? I'm getting stuck.

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Suppose that $Y$ were not contained in $N$; then there is a commutator $aba^{-1}b^{-1} \in Y$ not in $N$. Now consider the commutator of $a + N$ and $b + N$ in the quotient group $G/N$:
since $aba^{-1}b^{-1} \notin N$. It follows that $G/N$ isn't abelian, since there's a non-identity commutator.
Could you explain the statement "It follows that $G/N$ isn't abelian, since there's a non-identity commutator"? – Paul Malinowski Mar 5 '14 at 2:31
The elements $a +N$ and $b+N$ don't commute. (Note that elements commute $\iff$ their commutator is the identity.) – user61527 Mar 5 '14 at 2:31


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# What's the point of studying topological (as opposed to smooth, PL, or PDiff) manifolds?

Part of the reason I think algebraic topology has acquired something of a fearsome reputation is that the terrible properties of the topological category (e.g. the existence of space-filling curves) force us to work very hard to prove the main theorems setting up all of the big machinery to get the payoff we want (e.g. invariance of domain, fixed point theorems). But why should I care about these arbitrary and terrible spaces and functions in the first place when, as far as I can tell, any manifold which occurs in applications is at least piecewise-differentiable and any morphism which occurs in applications is at least homotopic to a piecewise-differentiable one?

In other words, do topological manifolds really naturally occur in the rest of mathematics (without some extra structure)?

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Why wouldn't you study them?! –  Mariano Suárez-Alvarez Feb 4 '11 at 22:40
@Mariano: because I don't want to digest a lot of machinery if nobody actually uses it in anything like full generality! –  Qiaochu Yuan Feb 4 '11 at 22:44
So you are asking why should you study them? –  Mariano Suárez-Alvarez Feb 4 '11 at 22:56
@Mariano: all I am trying to do is eliminate a possibility which I find absurd: that the machinery is developed for arbitrary topological spaces / arbitrary manifolds but that it is only used on spaces that could be handled by simpler machinery (e.g. de Rham cohomology instead of singular cohomology). Note that I am assuming I am wrong about this. –  Qiaochu Yuan Feb 4 '11 at 23:00
The following reason might be unsatisfactory to you but I think everything is worth studying whether or not it has applications. For example, abstract topological spaces are certainly studied in point-set/general/set-theoretic topology by professional mathematicians. Why not better understand abstract topological spaces? They may not occur in "nature" so to speak but their theory is certainly not trivial and any non-trivial theory is worth better understanding. Moreover, the tools of algebraic topology can be used to better understand abstract topological spaces as well as manifolds. –  Amitesh Datta Jun 25 '11 at 3:00

As far as I know, historically smooth manifolds were the first manifolds studied by people like Poincare, Riemann, up to Whitney. There were a few major events that caused people to take things like topological and PL manifolds seriously, but originally people were not motivated to study these kinds of objects. Here are some of the big events/ideas that come to mind:

1) Poincare's original proof of Poincare duality was a proof for triangulated manifolds. That smooth manifolds had triangulations (and whether or not they were essentially unique) was a problem that took some time to solve. So the study of triangulations and PL manifolds picked up.

2) Smale's proof of the h-cobordism theorem, although written up for the smooth category when you look at it carefully there's a lot of "smoothing the corners" going on. You can think carefully about it and determine all the smoothing of the corners does not kill the proof but I know many strong mathematicians that were hesitant to accept Smale's proof, insisting that it was only a PL-category proof. FYI, the smoothing of the corners issue has been settled, there's a very nice write-up in Kosinski's manifolds text. But this was another issue that kept people thinking about the PL category.

3) If anything, topological manifolds play a role simply for comparison sake -- after all the forgetful functor from the smooth to the topological category is an interesting functor. Perhaps for different people in different ways. I've yet to be interested by a topological manifold that admits no smooth structure but I do find multiple smooth structures on the same topological manifold interesting. Is this purely psychological?

4) Topological and PL-manifold theory is where some "nasty" constructions work, like the Alexander trick. There are different versions of it, one being that the restriction map $Aut(D^n) \to Aut(S^{n-1})$ admits a section in the topological or PL categories. It does not in the smooth category. If anything, I find these kinds of facts informative on the smooth category. The smooth category is interesting largely because of facts like these. There's a similar Alexander trick for knots, for example, the space of topological or PL embeddings $\mathbb R^j \to \mathbb R^n$ which restrict to the standard inclusion $x \longmapsto (x,0)$ outside of the unit ball, this space is contractible, by "pulling the knot tight". But in the smooth category, this space isn't contractible.

I think one of the major events in the development of this subject is simply pragmatic. To get smooth manifold theory off the ground you need Sard's theorem and transversality. This requires analysis to the level of measure theory, and a solid multi-variable calculus background, which in many undergraduate educations is skimped on (especially since it comes early, and many curriculums are too service-based to teach calculus "well"). PL manifolds are inherently more combinatorial and so the learning curve for people with weak analysis backgrounds is easier to deal with. I think also some people really appreciate the combinatorial nature of the subject.

Anyhow, those are some thoughts off the top of my head.

Getting to your conversation with Mariano:

PL manifold theory by-and-large isn't terribly different from smooth manifold theory. So I think once you learn one, adapting to the other isn't so hard. But topological manifolds are really quite different. This may be my ignorance speaking to some extent, but I'm still working my way through Kirby-Siebenmann. I'm told various people are working on re-writing the main theorems of that text, to make it easier reading for people that are not named Larry Siebenmann. But we're probably still several years from that. I suspect in the next 10 years there should be several different accounts of most of that material. But I'm still some ways from understanding smoothing theory.

Manifolds when they come up "in nature" like in physics or engineering applications tend to always be smooth, and usually with plenty of extra structure. Sometimes the objects that come up aren't manifolds, but algebraic varieties, or even more degenerate (but smooth) stratified spaces.

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I am totally willing to think about the PL category, but I am still uncertain about the topological category. I get that it's a natural place to prove the easy stuff, but why do we also want to prove the hard stuff here? (Full disclosure: I have recently started reading Guillemin and Pollack and am being blown away at how nice and intuitive everything is once you take Sard's theorem and transversality for granted (both of which are themselves nice and intuitive unlike, say, the definition of singular cohomology).) –  Qiaochu Yuan Feb 4 '11 at 23:26
@Qiaochu, but the PL category has its wierdnesses too... The fact that the Hauptvermutung is false, say. –  Mariano Suárez-Alvarez Feb 4 '11 at 23:27
@Mariano: my naive opinion is that the failure of the Hauptvermutung has more to do with the fact that homeomorphisms are weird than that PL-manifolds are weird. I am not necessarily even convinced that the category of topological spaces is "right." –  Qiaochu Yuan Feb 4 '11 at 23:30
That's great. IMO Guillemin and Pollack is one of the most enabling books for young topologists, as it gives you so many ready-made and intuitive constructions for things like homology, cohomology, Poincare duality, covering spaces, etc. –  Ryan Budney Feb 4 '11 at 23:31

Topological manifolds arise naturally as the background of other phenomena.

For example, nature throws at you examples of spaces which admit several smooth structures and when you try to describe that phenomenon you need to say something like «there are many different $Y$s one can put on an $X$». In the situation of exotic smooth structures, a natural class one can use for the role of $X$s is that of topological manifolds.

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@Mariano: yes, certainly it's a good idea to talk about simple things in the topological category. Why should we also talk about complicated things in the topological category? For example, what's the point of proving the Jordan curve theorem for continuous curves? Does anyone actually care about arbitrary continuous curves? –  Qiaochu Yuan Feb 4 '11 at 23:33
Layers of complexity are nice but complexity unfortunately isn't linearly ordered. For example, can you give me a 1-line construction of a topological manifold that can't be given a PL structure? Naturally occuring manifolds almost always are smooth. So in some rather formally non-intuitive way, they're the bottom-layer of complexity. Look at almost any intro textbook that defines topological and smooth manifolds and there isn't a single example of a genuine topological manifold that isn't naturally smooth. –  Ryan Budney Feb 4 '11 at 23:33
@Qiaochu: Well, one point is that it is a difficult result :) –  Mariano Suárez-Alvarez Feb 4 '11 at 23:46
Ultimately, I don't think one can argue about these things rationally. «Why should be also talk about complicated aspects of subject X?» is an essentially unanswerable question. Jacobi's pour l'honneur de l'esprit humain comes to mind. –  Mariano Suárez-Alvarez Feb 4 '11 at 23:54
@Ryan: that is definitely a motivating factor in my question: I literally don't know any examples of topological manifolds that don't come with natural smooth structures, up to singularities and corners and boundaries and other things I am perfectly willing to accept. –  Qiaochu Yuan Feb 5 '11 at 0:10

The topological category is inherently beautiful. Some very pretty and subtle phenomena happen. My favorite example of this is that some knots bound locally flat (that is they admit a locally trivial normal bundle) topological disks into the 4-ball, but they don't bound smooth disks. That is, there are topologically slice knots which are not smoothly slice. The topological disks themselves are practically impossible to visualize, coming from a limiting construction due to very deep work of Mike Freedman, and are a little bit fractal in nature. The world would be a poorer place if we never knew about this amazing structure.

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I like these results, too, for comparison sake. Similarly, if $M$ is Poincare dodecahedral space, there is a topological embedding of $M \times \mathbb R$ into $\mathbb R^4$, but no smooth embedding. But turning it on its head, this gives you reason to like the PL and smooth categories as they're more combinatorial. –  Ryan Budney Feb 4 '11 at 23:25
That's just the problem: the topological category is too subtle. When I think about manifolds I would like to keep my geometric intuition as intact as possible, and I would like violations of that intuition to be justified by applications of the intuition-violating thing. –  Qiaochu Yuan Feb 4 '11 at 23:28
@Qiaochu Yuan: why would you want to restrict to math that doesn't violate intuition? In my mind, when math violates intuition, that's when the most beautiful results occur. Why study something for which you already know all the answers? –  Grumpy Parsnip Feb 4 '11 at 23:46
@Jim: I guess that requires elaboration. I have no qualms about the category of groups because groups perfectly capture any reasonable intuitive definition of symmetry (via Cayley's theorem). I have qualms about the category of manifolds because manifolds do not perfectly capture any reasonable intuitive definition of "space," even given the Whitney embedding theorem: the problem is that I have qualms about the category of topological spaces. In dimension 2, for compact manifolds, all reasonable definitions agree: that's great. In higher dimensions, for general manifolds, I have no such... –  Qiaochu Yuan Feb 5 '11 at 0:02
I am with Jim on this one even though I don't study topological 4-manifolds explicitly and Jim does. The variety of different structures on 4-space makes it amazing. Furthermore, the original Kervaire and Milnor constructions of different smooth structures for higher dimensional manifolds are an inherent and big part of modern math not limited to but including Bott periodicity. Smoothness and PL-ness is nice, but to really understand what is under the hood, these other categories are amazing. –  Scott Carter Feb 5 '11 at 0:52

I learned a lot from the answers and discussion above. Here is my personal view, not necessarily applicable to others.

I published my Lectures on Algebraic Topology (W. A. Benjamin) in 1967. The book is an extensive elaboration of notes from a course I taught. All the manifold theorems in my book refer to topological manifolds - I only make a few side remarks about differentiable manifolds, De Rham cohomology and PL-manifolds. My book was later expanded and slightly revised by John Harper, renamed as Algebraic Topology: A First Course (Perseus, 1981).

I was delighted, in the process of teaching that course, to learn that so much could be proved in this subject without bringing in analysis and with very little messy (imho) combinatorial argumentation. Purity of method is an essential part of mathematical aesthetics, as Hilbert emphasized. And to achieve that purity, one usually has to discover new techniques that apply in greater generality than the previously known methods.

I was glad to see that J. P. May, in his lovely 1999 treatise A Concise Course in Algebraic Topology (U. of Chicago Press), also focused on topological manifolds.

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A standard example of a non-smoothable 4-manifold is the E8-manifold $[E_8]$, that is, the manifold which has $E_8$ as its intersection form $H_2([E_8],\mathbb{Z}) \times H_2([E_8],\mathbb{Z}) \rightarrow \mathbb{Z}$. The manifold exists topologically (as the plumbing of disc-bundles, this thing has a homology 3-sphere as its boundary, but there is another construction (Freedman's fake 4-balls, see for example Alexandru Scorpan, The Wild World of 4-Manifolds) which always allows you to close this thing up), but isn't smoothable due to the theorem of Rohlin (see link for more and a few examples).

I think the main reason to study topological manifold is that they have a lot to say about the other structures (PL, Differential, Complex-Differential, Sympletic, etc).

Of course when you are strictly working with complex analytic things, I can understand (from a focus point of view) why you aren't bothered that much by underlying structures.

Secondly, the topological realm is a great source for counter-examples for things that seem obvious, but aren't true when you look more closely.

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That second reason is precisely why I think we shouldn't work in the topological category: when things seem obvious, it's because our intuition comes from nicer manifolds (say PL), and when that intuition breaks I just see it as evidence that the topological category is weird, not that it's interesting. –  Qiaochu Yuan Jul 24 '11 at 15:17
@QiaochuYuan, that argument also applies to most of Banach space theory! :) –  Mariano Suárez-Alvarez Jan 29 '12 at 8:58
@Qiaochu: That merely says that your notion of what is interesting differs from that of some other people $-$ like me, for instance. –  Brian M. Scott Jan 29 '12 at 18:31

There is an aspect in which the topological category (for homotopy theorists) is quite simple and elegant. That is, the Quillen equivalence between $Top$ and $sSet$ shows that all of homotopy theory is combinatorially encoded in simplicial sets. Simplicial sets in turn, are completely determined just by the very simple class of polytopes: the $n$-simplices. So, homotopy theory boils down to understanding triangles. (It turns out triangles are quite complicated.) I find it quite miraculous that at the most general situation (from manifolds all the way to just topological spaces) all you basically have are the $n$-simplices. So, as motivation for the more practical cases of manifolds with extra structure, understanding the most basic polytopes is tantamount to understanding the homotopy category of topological spaces.

Another reason for general homotopy theory being relevant is that there are plenty of cases where there is no smooth structure around at all. For instance, the homotopy theory of finite topological spaces is very rich (e.g., non-trivial fundamental groups) and useful (e.g., computer graphics).

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(1) The machinery mentioned is useful for much more than studying topological manifolds so one should learn homology and those techniques anyway. Persistant homology in the study of high dimensional data, configuration spaces in the study of robotic motion and so forth indicate that the machinery of algebraic topology is here to stay and rapidly spreading to other fields. Homology has permeated algebra (group cohomology, Lie algebra cohomology, Hochschild cohomology..), number theory (etale cohomology, p-adic cohomology,...) and many other fields. Algebraic Topology is useful for far more things than just worrying about topological manifolds. Honestly many facts about smooth manifolds have only been proven using algebraic topological techniques. Kervaire invariant problems or maybe just that every oriented 3-manifold is parallelizable.... so basically my viewpoint is get over the machinery EVEN IF you just want to stay in the smooth world.

(2) Many things in nature are fractals - continuity is fair for physical reasons (nature tends to dislike and rectify discontinuities/shockwaves - smoothness is mostly an abstraction or an approximation. In statics it can be reasonable to assume everything is smooth but not so much in dynamics. Objects like Koch snowflakes are the norm (nowhere differentiable embedding of the circle in this case). The boundary of the Mandelbrot set is not smooth - in dynamics both discrete and continuous, even if your functions or differential equations are smooth, useful invariant manifolds or boundaries of such need not be so.

(3) Even when working with smooth manifolds, it is important to know when things don't depend on the differential structure. THERE ARE MANY SMOOTH MANIFOLDS WHICH ARE HOMEOMORPHIC BUT NOT DIFFEOMORPHIC. One needs to decouple the smooth structure from its underlying topological manifold to understand what things differ for different smooth structures on the same topological manifold and what don't. For example in the uncountably many distinct smooth structures on R^4 or the finitely many distinct smooth structures on spheres in higher dimensions, a given differential equation can have different solution spaces. (Seiberg Witten invariants to distinguish them come from looking at solution spaces of certain DEQs) Thus your analysis depends on the differential structure for SOME differential equations. However for others it doesn't. For example the Hodge theorem shows that the dimension of the space of harmonic forms on a smooth closed Riemannian manifold in each degree is a topological invariant - in fact a homotopy invariant - it doesn't care about the smooth structure or the metric. In general one has to ponder what changing the differential structure might do to ones analysis - it is hard to do that systematically if you don't consider the basic underlying topological manifold. Sometimes it is better to even dump that and just consider the underlying homology-manifold or the underlying Poincare Duality space. (for example in Surgery theory).

(4) In many problems it is not sufficient to understand the problem in the case when everything is smooth but the continuous case can be reasonable (Lusin approximation in measure theory). In geometric analysis there is a wide difference between being able to prove something for C^1 or Lipschitz maps versus Holder continuous versus for continuous ones. It would be nice if everything in the world was smooth but it isn't. If I consider the function which keeps track of the mass of a set on one side of a hyperplane as a function of the normal of the hyperplane (a very basic ham-sandwich procedure necessary in many analytic combinatoric and measure theory constructions), it is easy to make reasonable conditions on the measure that guarantee continuity but less reasonable to guarantee smoothness. Thus many things have to be proved for continuous maps even if working in ambient smooth manifolds. You often end up having to use all the continuous machinery for this and so you might as well work on topological manifolds at that point.

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# SOLA: Continual Learning with Second-Order Loss Approximation

Neural networks have achieved remarkable success in many cognitive tasks. However, when they are trained sequentially on multiple tasks without access to old data, it is observed that their performance on old tasks tend to drop significantly after the model is trained on new tasks. Continual learning aims to tackle this problem often referred to as catastrophic forgetting and to ensure sequential learning capability. We study continual learning from the perspective of loss landscapes and propose to construct a second-order Taylor approximation of the loss functions in previous tasks. Our proposed method does not require any memorization of raw data or their gradients, and therefore, offers better privacy protection. We theoretically analyze our algorithm from an optimization viewpoint and provide a sufficient and worst-case necessary condition for the gradient updates on the approximate loss function to be descent directions for the true loss function. Experiments on multiple continual learning benchmarks suggest that our method is effective in avoiding catastrophic forgetting and in many scenarios, outperforms several baseline algorithms that do not explicitly store the data samples.

## Authors

• 22 publications
• 25 publications
• 122 publications
03/24/2022

### Probing Representation Forgetting in Supervised and Unsupervised Continual Learning

Continual Learning research typically focuses on tackling the phenomenon...
08/08/2019

### Continual Learning by Asymmetric Loss Approximation with Single-Side Overestimation

Catastrophic forgetting is a critical challenge in training deep neural ...
10/15/2019

### Orthogonal Gradient Descent for Continual Learning

Neural networks are achieving state of the art and sometimes super-human...
10/21/2021

### Center Loss Regularization for Continual Learning

The ability to learn different tasks sequentially is essential to the de...
11/25/2020

### Continual learning with direction-constrained optimization

This paper studies a new design of the optimization algorithm for traini...
04/19/2021

### Overcoming Catastrophic Forgetting with Gaussian Mixture Replay

We present Gaussian Mixture Replay (GMR), a rehearsal-based approach for...
05/17/2021

### Layerwise Optimization by Gradient Decomposition for Continual Learning

Deep neural networks achieve state-of-the-art and sometimes super-human ...
##### This week in AI

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## 1 Introduction

Neural networks are achieving human-level performance on many cognitive tasks including image classification krizhevsky2012imagenet and speech recognition hinton2006fast . However, as opposed to humans, their acquired knowledge is comparably volatile and can be easily dismissed. Especially, the catastrophic forgetting phenomenon refers to the case when a neural network forgets the past tasks if it is not allowed to retrain or reiterate on them again goodfellow2013empirical ; mccloskey1989catastrophic .

Continual learning is a research direction that aims to solve the catastrophic forgetting problem. Recent works tried to tackle this issue from a variety of perspectives. Regularization methods (e.g., kirkpatrick2017overcoming ; zenke2017continual ) aim to consolidate the weights that are important to previous tasks while expansion based methods (e.g., rusu2016progressive ; yoon2018lifelong ) typically increase the model capacity to cope with the new tasks. Repetition based methods (e.g., lopez2017gradient ; chaudhry2018efficient ) usually do not require additional and complex modules, however, they have to maintain a small memory of previous data and use them to preserve knowledge. Unfortunately, the performance boost of repetition based methods comes at the cost of storing previous data which may be undesirable whenever privacy is important. To address this issue, authors in  farajtabar2019orthogonal proposed a method to work with the gradients of the previous data to constrain the weight updates; however, this may still be subject to privacy issues as the gradient associated with each individual data point may disclose information about the raw data.

In this paper, we study the continual learning problem from the perspective of loss landscapes. We explicitly target minimizing an average over all tasks’ loss functions. The proposed method stores neither the data samples nor the individual gradients on the previous tasks. Instead, we propose to construct an approximation to the loss surface of previous tasks. More specifically, we approximate the loss function by estimating its second-order Taylor expansion. The approximation is used as a surrogate added to the loss function of the current task. Our method only stores information based on the statistics of the entire training dataset, such as full gradient and full Hessian matrix (or its low rank approximation), and thus better protects privacy. In addition, since we do not expand the model capacity, the neural network structure is less complex than that of expansion based methods.

We study our algorithm from an optimization perspective, and make the following theoretical contributions:

• We prove a sufficient and worst-case necessary condition under which by conducting gradient descent on the approximate loss function, we can still minimize the actual loss function.

• We further provide convergence analysis of our algorithm for both non-convex and convex loss functions. Our results imply that early stopping can be helpful in continual learning.

• We make connections between our method and elastic weight consolidation (EWC) kirkpatrick2017overcoming .

In addition, we make the following experimental contributions:

• We conduct a comprehensive comparison among our algorithm and several baseline algorithms kirkpatrick2017overcoming ; chaudhry2018efficient ; farajtabar2019orthogonal on a variety of combinations of datasets and models. We observe that in many scenarios, especially when the learner is not allowed to store the raw data samples, our proposed algorithm outperforms them. We also discuss the conditions under which the proposed method or any of the alternatives are effective.

• We provide experimental evidence validating the importance of accurate approximation of the Hessian matrix and discuss scenarios in which early stopping is helpful for our algorithm.

## 2 Related work

Avoiding catastrophic forgetting in continual learning Parisi2018ContinualLL ; beaulieu2020learning is an important milestone towards achieving artificial general intelligence (AGI) which entails developing measurements toneva2018empirical ; kemker2018measuring , evaluation protocols farquhar2018towards ; de2019continual , and theoretical understating nguyen2019toward ; farquhar2019unifying of the phenomenon. Generally speaking, three classes of algorithms exist to overcome catastrophic forgetting farajtabar2019orthogonal .

The expansion

based methods allocate new neurons or layers or modules to accommodate new tasks while utilizing the shared representation learned from previous ones

rusu2016progressive ; xiao2014error ; yoon2018lifelong ; li2019learn ; Jerfel2018ReconcilingMA . Although being a very natural approach the mechanism of dynamic expansion can be quite complex and can add considerable overhead to the training process.

The repetition and memory based methods store previous data or, alternatively, train a generative model of them and replay samples from them interleaved with samples drawn from the current task shin2017continual ; kamra2017deep ; zhang2019prototype ; rios2018closed ; luders2016continual ; lopez2017gradient ; farajtabar2019orthogonal . They achieve promising performance however at the cost of higher risk of users’ privacy by storing or learning a generative model of their data.

The regularization based approaches impose limiting constraints on the weight updates of the neural network according to some relevance score for previous knowledge kirkpatrick2017overcoming ; nguyen2017variational ; titsias2019functional ; ritter2018online ; mirzadeh2020dropout ; zenke2017continual ; park2019continual . These methods provide a better privacy guarantee as they do not explicitly store the data samples. In general, SOLA also belongs to this category as we use the second-order Taylor expansion as the regularization term in new tasks. Many of the regularization methods are derived from a Bayesian perspective of estimating the posterior distribution of the model parameters given the data from a sequence of tasks kirkpatrick2017overcoming ; nguyen2017variational ; titsias2019functional ; ritter2018online

; some of these methods use other heuristics to either estimate the importance of the weights of the neural network

zenke2017continual ; park2019continual or implicitly limit the capacity of the network mirzadeh2020dropout . Similar to our approach, several regularization based methods use quadratic functions as the regularization term, and many of them use the diagonal form of quadratic functions kirkpatrick2017overcoming ; zenke2017continual ; park2019continual . In Section 5.3, we demonstrate that in some cases, the EWC algorithm kirkpatrick2017overcoming can be considered as the diagonal approximation of our approach. Here, we note that the diagonal form of quadratic regularization has the drawback that it does not take the interaction between the weights into account.

Among the regularization based methods, the online Laplace approximation algorithm ritter2018online

is the most similar one to our proposed method. Despite the similarity in the implementations, the two algorithms are derived from very different perspectives: the online Laplace approximation algorithm uses a Bayesian approach that approximates the posterior distribution of the weights with a Gaussian distribution, whereas our algorithm is derived from an optimization viewpoint using Taylor approximation of loss functions. More importantly, the Gaussian approximation in

ritter2018online is proposed as a heuristic; whereas in this paper, we provide rigorous theoretical analysis on how the approximation error affects the optimization procedure. We believe that our analysis provides deeper insights to the loss landscape of continual learning problems, and explains some important implementation details such as early stopping.

We also note that continual learning is broader than just solving the catastrophic forgetting and is connected to many other areas such as meta learning riemer2018learning , few-shot Learning  wen2018few ; gidaris2018dynamic , learning without explicit task identifiers rao2019continual ; aljundi2019online , to name a few.

## 3 Problem formulation

We consider a sequence of supervised learning tasks , .111For any positive integer , we define . For task , there is an unknown distribution over the space of feature-label pairs . Let be a model parameter space,222In most cases, we consider . and for the -th task, let be the loss function of associated with data point . The population loss function of task is defined as

. Our general objective is to learn a parametric model with minimized population loss over all the

## 4 Our approach

To measure the effectiveness of a continual learning algorithm, we use a simple criterion that after each task, we hope the average population loss over all the tasks that have been trained on to be small, i.e., for every , after training on , we hope to solve . Since minimizing the loss function is the key to training a good model, we propose a straightforward method for continual learning: storing the second-order Taylor expansion of the empirical loss function, and using it as a surrogate of the loss function for an old task when training on new tasks. We start with a simple setting. Suppose that there are two tasks, and at the end of , the we obtain a model . Then we compute the gradient and Hessian matrix of at , and construct the second-order Taylor expansion of at :

 ˜L1(w)=ˆL1(ˆw1)+(w−ˆw1)⊤∇ˆL1(ˆw1)+12(w−ˆw1)⊤∇2ˆL1(ˆw1)(w−ˆw1).

When training on , we try to minimize . The basic idea of this design is that, we hope in a neighborhood around , the quadratic function stays as a good approximation of , and thus approximately we still minimize the average of the empirical loss functions , which in the limit generalizes to the population loss function .

We rely on the assumption that the second-order Taylor approximation of loss function can capture their local geometry well. For a general nonlinear function and arbitrary displacement, this approximation can be over-simplistic, however, we refer to the abundance of observations for modern neural networks that are seen to be well-behaved with flat and wide minima choromanska2015loss ; goodfellow2014qualitatively . Moreover, the assumption of well-behaved loss around tasks’ local minima also forms the basis of a few other continual learning algorithms such as EWC kirkpatrick2017overcoming and OGD farajtabar2019orthogonal .

Formally, let be the model that we obtain at the end of the -th task. We define the approximation of the sum of the first empirical loss functions as

 ˜Lk(w) =k∑k′=1ˆLk′(ˆwk′)+(w−ˆwk′)⊤∇ˆLk′(ˆwk′)+12(w−ˆwk′)⊤Hk′(w−ˆwk′) (1) =w⊤Akw+w⊤bk+ck,

where denotes the Hessian matrix or its low rank approximation, , , and is a constant that does not depend on . We construct at the end of task , and when training on task , we minimize . In the following, we name our algorithm SOLA, an acronym for Second-Order Loss Approximation.

As we can see, in the SOLA algorithm, after each task, if we choose to use the exact Hessian matrix, i.e., , , it suffices to update and in memory, and thus the memory cost of the algorithm is , which does not grow with the number of tasks. However, in practice, especially for overparameterized neural networks, the dimension of the model is usually large, and thus the storage cost of memorizing the Hessian matrix can be high. Recent studies have shown that the Hessian matrices of the loss functions of a deep neural networks are usually approximately low rank ghorbani2019investigation . If we choose as a rank- approximation of , we need to keep accumulating the low rank approximations of the Hessian matrices in order to construct , and at the end of task , the memory cost is , which in practice can be much smaller than that of using the exact Hessian matrices. We formally demonstrate our approach in Algorithm 1, and the methods that use exact Hessian matrices and low rank approximation of them are presented as options I and II, respectively. Moreover, we can use a recursive implementation for the low rank approximation, and the memory cost can be further reduced to , which does not grow with . We present the details of the recursive implementation in Section 6.

## 5 Theoretical analysis

In this section, we provide theoretical analysis of our algorithm. As we can see, the key idea in our algorithm is to approximate the loss functions of previous tasks using quadratic functions. This leads to the following theoretical question: By running gradient descent algorithm on an approximate loss function, can we still minimize the actual loss that we are interested in?

For the purpose of theoretical analysis, we make a few simplifications to our setup. Without loss of generality, we study the training process of the last task , and still use to denote the model parameters that we obtain at the end of the -th task. We use the loss function approximation in (1), but for simplicity we ignore the finite-sample effect and replace the empirical loss function with the population loss function, i.e., we define

 ˜LK−1(w)=K−1∑k=1Lk(ˆwk)+(w−ˆwk)⊤∇Lk(ˆwk)+12(w−ˆwk)⊤Hk(w−ˆwk), (2)

where represents or its low rank approximation. The reason for this simplification is that our focus is the optimization aspect of the problem, while the generalization aspect can be tackled by tools such as uniform convergence mohri2018foundations . As discussed, during the training of the last task, we have access to the approximate loss function , whereas the actual loss function that we care about is . We also focus on gradient descent instead of its stochastic counterpart. In particular, let be the initial model parameter for the last task. We run the following update for :

 wt=wt−1−η∇˜F(wt−1). (3)

We use the following standard notions for differentiable function .

###### Definition 1.

is -smooth if , .

###### Definition 2.

is -Hessian Lipschitz if , .

We make the assumptions that the loss functions are smooth and Hessian Lipschitz. We note that the Hessian Lipschitz assumption is standard in analysis of non-convex optimization nesterov2006cubic ; jin2017escape .

###### Assumption 1.

We assume that is -smooth and -Hessian Lipschitz .

We also assume that the error between the matrices and is bounded.

###### Assumption 2.

We assume that for every , , where is defined in Assumption 1, and that for some .

### 5.1 Sufficient and worst-case necessary condition for one-step descent

We begin with analyzing a single step during training. Our goal is to understand by running a single step of gradient descent on , whether we can minimize the actual loss function . More specifically, we have the following result.

###### Theorem 1.

Under Assumptions 1 and 2, and suppose that in the -th iteration, we observe

 ∥∇˜F(wt−1)∥2≥cKK−1∑k=1δ∥wt−1−ˆwk∥2+ρ∥wt−1−ˆwk∥22, for some c>1, (4)

and the learning rate satisfies , then we have

 F(wt)≤F(wt−1)−η(1−1c−μη2)∥∇˜F(wt−1)∥22.

We prove Theorem 1 in Appendix A. Here, we emphasize that this result does not assume any convexity of the loss functions. The theorem provides a sufficient condition (4), under which by running gradient descent on , we can still minimize the true loss function . Intuitively, this condition requires the gradient of to be large enough, such that the magnitude of the gradient is larger than the error caused by the inexactness of the loss function. In Proposition 1 below, we will see that this condition is also necessary in the worst-case scenario, at least for the case where . More specifically, we can construct cases in which (4) is violated and the gradients of and have opposite directions.

###### Proposition 1.

Suppose that , , . Then, there exists , , , and such that if , then .

We prove Proposition 1 in Appendix B. In addition, we note that Theorem 1 also implies that as training going on and decreasing, it is beneficial to decrease the learning rate , since when decreases, the upper bound on that guarantees the decay of (i.e., ) also decreases. We notice that the importance of learning rate decay for continual learning has been observed in some empirical study recently mirzadeh2020dropout .

### 5.2 Convergence analysis

Although the condition in (4) provides us with insights on the dynamics of the training algorithm, it is usually hard to check this condition in every step, since we may not have good estimates of and . A practical implementation is to choose a constant learning rate along with an appropriate number of training steps. In this section, we provide bounds on the convergence behavior of our algorithm with a constant learning rate and iterations, both for non-convex and convex loss functions. These results imply that early stopping can be helpful, and provide a theoretical treatment of the very intuitive fact that the more iterations one optimizes for the current task the more forgetting can happen for the previous ones. We begin with a convergence analysis for non-convex loss functions in Theorem 2, in which we use the common choice of learning rate for gradient descent on smooth functions bubeck2014convex .

###### Theorem 2 (non-convex).

Let , , , and . Then, under Assumptions 1 and 2, after running iterations of the gradient descent update (3) with learning rate , we have

 1TT∑t=1∥∇F(wt−1)∥2≤α√T+β+γ1√T+γ2T,

where , , , and .

We prove Theorem 2 in Appendix C. Unlike standard optimization analysis, the average norm of the gradients does not always decrease as increases, when or . Intuitively, as we move far from the points where we conduct Taylor expansion, the gradient of becomes more and more inaccurate, and thus we need to stop early. In Section 7, we provide experimental evidence.

When the loss functions are convex, we can prove a better guarantee which does not have the and terms as in Theorem 2. More specifically, we have the following assumption and theorem.

###### Assumption 3.

is convex and , .

###### Theorem 3 (convex).

Suppose that Assumptions 12, and 3 hold, and define , , , and . After running iterations of the gradient descent update (3) with learning rate , we have

 F(wT)−F∗≤αT+β,

where , and .

We prove Theorem 3 in Appendix D. As we can see, if or , we still cannot guarantee the convergence to the true minimum of , due to the inexactness of . On the other hand, if the loss functions are quadratic and we save the full Hessian matrices, i.e., , as we have full information about previous loss functions, we can recover the standard convergence rate for gradient descent on convex and smooth functions.

### 5.3 Connection to EWC

The elastic weight consolidation (EWC) algorithm kirkpatrick2017overcoming for continual learning is proposed based on the Bayesian idea of estimating the posterior distribution of the model parameters. Interestingly, we notice that our algorithm has a connection with EWC, although their basic ideas are quite different. More specifically, we show that in some cases, the regularization technique that the EWC algorithm uses can be considered as a diagonal approximation of the Hessian matrix of the loss function. Suppose that in the -th task, the data points are samples from a probabilistic model with the likelihood function being , and we use negative log-likelihood as the loss function, i.e., . Suppose that at the end of this task, we obtain the ground truth model parameter . Then we know that , and that the Fisher information of the -th coordinate of is . The EWC algorithm constructs a regularization term as a proxy of the loss function of the -th task, and uses it in the following tasks. As we can see, in this case, the quadratic regularization in EWC is a diagonal approximation of the quadratic term in our loss function approximation approach.

## 6 A recursive implementation

As we have seen, one drawback of the SOLA algorithm with low rank approximation in Section 4 is that the memory cost grows with the number of tasks. In this section, we present a more practical and memory efficient implementation of SOLA with low rank approximation. Recall that is the empirical loss function for the -th task, . We then define the loss function approximation in a recursive way. We begin with , and for every , we define

 ˜Lk−12(w) =˜Lk−1(w)+ˆLk(w) (5) ˜Lk(w) =˜Lk−12(ˆwk)+(w−ˆwk)⊤∇˜Lk−12(ˆwk)+12(w−ˆwk)⊤Qk−12(w−ˆwk), (6)

where is a rank- approximation of the Hessian matrix . This means that at the end of task , we compute the second-order Taylor expansion of the approximate loss function at , with the Hessian matrix being replaced by the low rank approximation . Thus, we only need to store and , and the memory cost is , which does not grow with . We formally present this approach in Algorithm 2. In our experiments in Section 7, we use the recursive implementation for SOLA with low rank approximation.

## 7 Experiments

We implement the experiments with TensorFlow

. When computing the exact or the low rank approximation of the Hessian matrix, we treat each tensor in the model independently; in other words, we compute the block diagonal approximation of the Hessian matrix. This technique has the benefit that the Hessian computation is independent of the model architecture and has been used in recent studies on second-order optimization

gupta2018shampoo . We use the recursive implementation for SOLA with low rank approximation. In the following, we denote SOLA with exact Hessian matrix and low rank approximation by SOLA-exact and SOLA-prox

, respectively. As for the calculation of the low rank matrix, we make use of Hessian-vector product and provide details in Appendix

E.

Datasets. We use multiple standard continual learning benchmarks created based on MNIST lecun1998gradient and CIFAR-10 krizhevsky2009learning datasets, i.e., Permuted MNIST goodfellow2013empirical , Rotated MNIST lopez2017gradient , Split MNIST zenke2017continual , and Split CIFAR (similar to a dataset in chaudhry2018efficient ). In Permuted MNIST, for each task, we choose a random permutation of the pixels of MNIST images, and reorder all the images according to the permutation. We use -task Permuted MNIST in the experiments. In Rotated MNIST, for each task, we rotate the MNIST images by a particular angle. In our experiments, we choose a -task Rotated MNIST, with the rotation angles being , , , , and degrees. For Split MNIST, we Split the labels of the MNIST dataset to disjoint subsets, and for each task, we use the MNIST data whose labels belong to a particular subset. In this paper, we use a -task Split MNIST, and the subsets of labels are , , , , and . Split CIFAR is defined similar to Split MNIST, and we use a -task Split CIFAR with the label subsets being and .

Architecture.

We use both multilayer perceptron (MLP) and convolutional neural network (CNN). In most cases, we use MLP with two hidden layers, sometimes denoted by MLP

, with and being the number of hidden units. We may use CNN- to denote a CNN model with convolutional layers, and provide details of the model in Appendix F. For Split MNIST and Split CIFAR, we use MLP and CNN models with a multi-head structure similar to what has been used in chaudhry2018efficient ; farajtabar2019orthogonal . In the multi-head model, instead of having logits in the output layer, we use separate heads for different tasks, and each head corresponds to the classes of the associated task. During training, for each task, we only optimize the cross-entropy loss over the logits and labels of the corresponding output head.

independent runs, as well as the standard deviation (as the shaded areas in the figures).

Results. We provide a comprehensive comparison among SOLA and the baseline algorithms with a variety of combinations of datasets and models. Tables 1 and 2 present the results for MNIST-based datasets and Split CIFAR, respectively. We make a few notes before discussing the results. First, the multi-task algorithm uses all the data of previous tasks, which serves as an upper bound for the performance of continual learning algorithms. Second, since the A-GEM algorithm stores a subset of data samples from previous tasks, it is not completely fair to compare A-GEM with algorithms that do not store raw data. However, here we still report the results for A-GEM for reference, and in A-GEM we store

data points for each task. Third, since the performance of the algorithms depends on the number of epochs that we train for each task, we treat this quantity as a tuning parameter, and for each algorithm, we report the result corresponding to the best epoch choice for its performance. In particular, for MNIST-based datasets, we choose epoch from

, and for Split CIFAR, we choose from . Due to memory constraints, we only implement SOLA-exact on small models such as MLP and CNN-2. We conclude from the results as follows:

• If it is allowed to store raw data, repetition based algorithm such as A-GEM should be the choice. This remarks the importance of the information contained in the raw data samples. In some cases we observe that SOLA outperforms A-GEM, e.g., on MLP. However, we expect that the performance of A-GEM can be improved if more data are stored in memory.

• If it is not allowed to store raw data due to privacy concerns, then in many scenarios, SOLA outperforms the baseline algorithms. In particular, on MNIST-based datasets, SOLA-exact or SOLA-prox achieves the best performance in out of settings.

• On Split CIFAR, we observe mixed results. When the model is relatively small (CNN-2) and we can store the exact Hessian matrix, SOLA-exact achieves the best performance. On a relatively large CNN model (CNN-6), we observe that none of the continual learning algorithms (EWC, OGD, SOLA) significantly outperforms the vanilla algorithm. On a large MLP, we observe that OGD performs the best and the result for SOLA-prox becomes worse. We believe the reason is that since in this experiment we only use eigenvectors to approximate a Hessian matrix with very high dimensions, the approximation error is so large that SOLA-prox cannot find a descent direction that is close to the true gradient. This remarks the importance of future study of SOLA on models with more complicated structure or higher dimensions.

Performance vs approximation. We study how the approximation of Hessian matrices affects the performance of SOLA-prox. In particular, we choose different values of the rank in SOLA-prox and investigate its correlation with the final average test accuracy. Our theory implies that when the approximation of Hessian matrices is better, i.e., smaller , the final performance is better. Our experiments validate this point. Figure 0(a) and Figure 0(b) show that, as we increase , i.e., using more eigenvectors to approximate the Hessian matrix, the average test accuracy over all tasks improves.

## 8 Conclusions

We propose the SOLA algorithm based on the idea of loss function approximation. We establish theoretical guarantees, make connections to the EWC algorithm, and present experimental results showing that in many scenarios, our algorithm outperforms several baseline algorithms, especially among the ones that do not explicitly store the raw data samples. Future directions include studying SOLA on broader classes of neural network architectures and parameter spaces with higher dimensions.

## Acknowledgements

We would like to thank Dilan Gorur, Alex Mott, Clara Huiyi Hu, Nevena Lazic, Nir Levine, and Michalis Titsias for helpful discussions.

## Appendix A Proof of Theorem 1

We first provide a bound for the difference between the gradients of and .

###### Lemma 1.

Let . Then we have

 ∥Δ(w)∥2≤1KK−1∑k=1δ∥w−ˆwk∥2+ρ∥w−ˆwk∥22.

We prove Lemma 1 in Appendix A.1. Since the loss functions for all the tasks are -smooth, we know that is also -smooth. Then we have

 F(wt) ≤F(wt−1)+⟨∇F(wt−1),wt−wt−1⟩+μ2∥wt−wt−1∥22 =F(wt−1)+⟨∇˜F(wt−1)−Δ(wt−1),−η∇˜F(wt−1)⟩+μη22∥∇˜F(wt−1)∥22 ≤F(wt−1)−η(1−μη2)∥∇˜F(wt−1)∥22+η∥∇˜F(wt−1)∥2∥Δ(wt−1)∥2.

Therefore, as long as for some , we have

 F(wt)≤F(wt−1)−η(1−1c−μη2)∥∇˜F(wt−1)∥22. (7)

Then we can complete the proof by combining (7) with Lemma 1.

### a.1 Proof of Lemma 1

By the definition of , for some , , we have

 Δ(w) =1KK−1∑k=1∇Lk(ˆwk)+Hk(w−ˆwk)−∇Lk(w) =1KK−1∑k=1(Hk−∇2Lk(ˆwk+ξk(w−ˆwk)))(w−ˆwk),

where the second equality is due to Lagrange’s mean value theorem. Then, according to Assumptions 1 and 2, we have

 ∥∥Hk−∇2Lk(ˆwk+ξk(w−ˆwk))∥∥2 ≤δ+∥∥∇2Lk(ˆwk)−∇2Lk(ˆwk+ξk(w−ˆwk))∥∥2 ≤δ+ρ∥w−ˆwk∥2. (8)

Then, according to triangle inequality, we obtain

 ∥Δ(w)∥2≤1KK−1∑k=1δ∥w−ˆwk∥2+ρ∥w−ˆwk∥22.

## Appendix B Proof of Proposition 1

We first note that it suffices to construct and , as one can always choose and then the construction of and is equivalent to that of and . Let ,

 F(w) =(w−1)2+ρ6w3,w∈[0,1], ˜F(w) =(w−1)2−δ4w2,w∈[0,1].

One can easily check that , , and . In addition, since the second derivative of is always bounded in , we know that is smooth. Since , we know that is -Hessian Lipschitz. Therefore, and satisfy all of our assumptions.

Since , we know that , , and then

 |˜F′(w)|<δ2w+ρ2w2

is equivalent to , which implies that .

## Appendix C Proof of Theorem 2

Similar to Appendix A, we define . According to Assumptions 1 and 2, we know that both and are -smooth. By the smoothness of and using the fact that , we get

 F(wt) ≤F(wt−1)+⟨∇F(wt−1),wt−wt−1⟩+μ2∥wt−wt−1∥22 =F(wt−1)−⟨∇F(wt−1),η(∇F(wt−1)+Δ(wt−1))⟩+μη22∥∇F(wt−1)+Δ(wt−1)∥22 =F(wt−1)−12μ∥∇F(wt−1)∥22+12μ∥Δ(wt−1)∥22,

which implies

 ∥∇F(wt−1)∥22≤2μ(F(wt−1)−F(wt))+∥Δ(wt−1)∥22. (9)

By averaging (9) over , we get

 1TT∑t=1∥∇F(wt−1)∥22≤2μ(F0−F∗)T+1TT∑t=1∥Δ(wt−1)∥22.

By taking square root on both sizes, and using Cauchy-Schwarz inequality as well as the fact that , we get

 1TT∑t=1∥∇F(wt−1)∥2≤√2μ(F0−F∗)√T+ ⎷1TT∑t=1∥Δ(wt−1)∥22. (10)

We then proceed to bound . According to Lemma 1, we have

 ∥Δ(wt−1)∥2 ≤1KK−1∑k=1δ(∥wt−1−w0∥2+∥w0−ˆwk∥2)+2ρ(∥wt−1−w0∥22+∥w0−ˆwk∥22) :=C+δ∥wt−1−w0∥2+2ρ∥wt−1−w0∥22,

where


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Browse Questions

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# skyrim essence extractor

After returning the Runed Lexicon back to Septimus, he immediately read it and gave me the Essence Extractor to get blood from elves, rather than sending me away to wait until he deciphered it. Harvest blood from a High Elf, Wood Elf, Dark Elf, Falmer, and Orc. So there was only one sure place to get what I need. So yet another release of skyrim and yet another broken oblivion walker quest, and yes it's my 15th Daedric quest and yes similar shit happened on the Ps3 version for me though it was that stupid dog mission. Essence Extractor 00 01a31c: 1: 0: Quest: Discerning the Transmundane. High Elf. What is Enchanting 2. and "Who is your lord ?". (Occasionally, Valie's body will be missing - this seems to happen after her body is not looted at first discovery but if you return after enough time has passed for respawning.) Blood samples can be acquired from any dead body in Skyrim, not just those killed by the Dragonborn or their followers. Is it possible to use my own blood for the essence extractor since im a high elf? View basic program usage and command options: 2. joefitts63 9 years ago #1. He talked about Elder Scrolls, so I tough it might be important, I was curious and decided to do that. Advancement in the ruin leads the dragonborn to a dwemer essence extractor into which the blessed daedra hearts must be placed to obtain elemental essence 13. Unfortunately, you must return to Septimus to get the required Essence Extractor, so there's no avoiding a second trip. Harvest Orc blood 10. 0001A31C Runes 4. Utilizes levelled electrical and ice magic in combat. Retrieve the book from within the Cube. 1 ... Skyrim How to get rid of Elder Scrolls after completing Dawnguard questline - … chevron_left. The Runed Lexicon is something only … Essence Extractor not working » Thu Jun 07, 2012 8:09 pm . Item Name Item Code; Elven Warhammer of Leeching : 000BE23B: Elven Warhammer of Lethargy: 000BE25A: Elven Warhammer of Lightning: 000BE254: Elven Warhammer of Malediction: 000BE257: Elven Warhammer of Reaping: 000BE23F: Elven Warhammer of Scorching: … What was the weirdest/most random encounter you experienced? The Elder Scrolls V: Skyrim; Essence Extractor; User Info: joefitts63. Quest ItemMiscellaneous ItemDwemer Artifact Share your thoughts, experiences and the tales behind the art. Thank you. home Skyrim. Located in Septimus Signus's Outpost ; Finely Ground Bone Meal DG xx 00d3ed: 0.1: 0: Quest: Chasing Echoes DG. No, so if you've finished the entire thing, and you looted for those two back, they're stuck in your inventory without the use of Console Commands to get rid of them. How can I fix "This save relies on content that is no longer present". chevron_right. After i got all of it the objective updated (give blood to Septimus Signus), so when i got to him and talked to him the conversation was just looping over and over. Type 1. Privacy PolicyCookie SettingsDo Not Sell My InformationReport Ad. FormID 3. You can also find a full Glyph List Article where I show you all the Weapon, Armor, Jewelry, Potency and Aspect Runes that exist in Elder Scrolls Online.Overview Enchanting Guide: 1. All Discussions Screenshots Artwork Broadcasts Videos Workshop News Guides Reviews The Elder Scrolls V: Skyrim > General Discussions > Topic Details. Elixir of Defilation is a potion added by Undeath This potion requires the following ingredients to be created: 1 x Finely Ground Bone Meal 1 x Deadly Poison 1 x Poison Bloom 1 x Nightshade Extract 1 x Namira's Corrosion OR 1 x Embalming Essence The Dragonborn is tasked in making this elixir in order to become a lich. In honor of Septimus Sigmus, I’ve compiled the essence extractor, runed lexicon, two attunement spheres and the oghma infinium in one display case. #8 Demented_Demon, Apr 11, 2012. You're browsing the GameFAQs Message Boards as a guest. Mods; Games; Images; Videos; Users; search. Of course, you can just do both anyways. The Elder Scrolls V: Skyrim. How exactly do you use the essence extractor for the "Discerning The Transmundane" quest? Mods. 3. On Skyrim: After Discerning the Transmundane, do you need the Essence Extractor or Runed Lexicon for anything? 1. Use the elevator underneath the controls to quickly return to the Surface of Skyrim. © 2021 GAMESPOT, A RED VENTURES COMPANY. Loading... Unsubscribe from PsychoWedge? Madeleine Rose Walsh Posts: 3425 Joined: Wed Oct 04, 2006 2:07 am » Fri May 11, 2012 2:34 am . Type the name of a spawn code into the search box to instantly search our database of 8728 item IDs. A dialogue box is automatically presented when the Dragonborn loots a suitable corpse. How do you use the Essence Extractor » Fri May 11, 2012 3:45 am . It never seems to go on sale. Base Value Media . Finn's Lute 00 0dabab: 4: 25: Quest: Finn's Lute. There should be plenty of Falmer corpses for you to quickly get a sample with the Essence Extractor. 3.1.1. How do I level up Enchanting? You might see an essence extractor. Support . "Every time I yank a jawbone from a skull then ram it in an eye socket, I know I'm buildin' a better future." CholulaSouls. To extract blood, simply loot the corpse of a deceased person with the Essence Extractor in the inventory. Give the Lexicon to Septimus 5. videogame_asset My games. CJ. Anyone? Recently added 19 View all 1,040. I guess the first body wasn't of one of the required races. Extracts the MNF file and out… Her corpse can be found on the table where Festus Krex, Babette, and herself usually stayed." 5. IMPORTANT NOTE --Exporting all files from the three ESO MNF files will take several hours and require over 100GB of free disk space. Classed as a destruction DPS main with … Used in the quest chain Discerning the Transmundane. Give the Lexicon to Septimus Signus. Log in to view your list of favourite games. - Smiling Jack . ALL RIGHTS RESERVED. Find the Elder Scroll. Skyrim Item Codes List. Welcome to the Enchanting Guide for ESO. Sep 10, 2017 @ 8:56am ESSENCE EXTRACTOR? Bring blood to Septimus 11. The Essence Extractor is a quest item in The Elder Scrolls V: Skyrim. Give the collected blood samples to Septimus Signus. Transcribe the Lexicon 4. Joined: Apr 10, 2012 Messages: 74 Likes Received: 1 Reputation: 0 in my quest journal there are four objectives that have been completed. 6. It is a device that is used to gather the blood of the Mer races. 2. Passive Abilities of Enchanting Effects . Fill the Blank Lexicon. 0 Page 1 of 2 - Discerning the Transmundane Bug - posted in Skyrim Spoilers: I seem to have a rare bug for the Discerning the Transmundane quest. Miraak Battle bugged: complete WITHOUT console commands, "This virus is as stupid as our feud with Clan Gray-Mane". ChicagoBears Member. Essence Extractor bug? Elder Scrolls is a FANDOM Games Community. Wait for Septimus to unlock a passage to a new location (screen above). He will give it to the Dragonborn to collect the blood of the Mer races during the quest "Discerning the Transmundane.". Weight Has anyone seen this game dip below $40? Anyway I stumbled upon some grandpa far in the North while searching Velehk Sain treasure. Stats . EsoExtractData \path\to\game.mnf \export\path\ 2.1. Favorite Answer. 3. 8 years ago. The item ID for Essence Extractor in Skyrim on Steam (PC / Mac) is: 0001A31C. To extract blood, simply loot the corpse of a deceased person with the Essence Extractor in the inventory. 2.1.1. Board index ‹ The Elder Scrolls ‹ V - Skyrim; Change font size; Print view; FAQ; 3 posts • Page 1 of 1. This device is used to collect samples of all the Mer races. So after i gave him the little box you get after opening the elder scroll he sent me to harvest blood for him. Mods . Extracts the given MNF file and outputs to the given path. As you get to the crossing between the Alftand Cathedral and Alftand Animonculory, look for the little passage. Septimus Signus, found in Septimus Signus's Outpost, is known to possess one. To spawn this item in-game, open the console and type the following command: player.AddItem 0001A31C 1. The Elder Scrolls IV: Knights of the Nine, https://elderscrolls.fandom.com/wiki/Essence_Extractor?oldid=2970046, It can be looted off Septimus' body after he is killed by. Obtain blood of Altmer, Bosmer, Dunmer, Orsimer, and Falmer. This part of the quest generally simply happens as part of combat during the general exploration of Skyrim. Take your favorite fandoms with you and never miss a beat. 1.1.1. Before adding a bug to this list, consider the following: *Disclosure: Some of the links above are affiliate links, meaning, at no additional cost to you, Fandom will earn a commission if you click through and make a purchase. Community content is available under. Take the Oghma Infinium Head inside and speak with the mage to give him the filled Essence Extractor (screen above). Ask Urag about the insane book 2. - Smiling Jack. I killed off a few enemies and thought I would do some extracting, but I don't see any way to use this thing. Harvest Dark Elf blood 8. Just recently, I learned some info that caught my attention that says "Subsequently, Gabriella is killed during the Penitus Oculatus assault on the Dark Brotherhood Sanctuary (You can extract her blood using Essence Extractor for Discerning the Transmundane quest). The dragonborn must advance deeper into the ruin to reach the ritual site where the essence can be blessed 14. “Essence Extractor” is an item in Elder Scrolls V: Skyrim. Must be filled with Orc, Falmer, Dark Elf, Wood Elf, and High Elf blood. Where is the best place to sell your stuff? In this guide I will explain how to level up your Enchanting level fast and efficient. The reason for that is most likely because I was … Liar's Retreat - Head inside as it being overrun by Falmer. Community . Electra is a powerful Nordic Breton ElectroCryomancer, master destruction trainer and trader of magical merchandise based in Dragonsreach, Whiterun. EsoExtractData --help or EsoExtractData -h 1.1. When logged in, you can choose up to 12 games that will be displayed as favourites in this menu. "Every time I yank a jawbone from a skull then ram it in an eye socket, I know I'm buildin' a better future." According to the UESP wiki, this quest could be affected by this bug: You may not have the "Harvest blood" option despite having the Essence Extractor in your inventory. Answer Save. Spawn Commands. Weight: 1 Value: 0 Related Quests . This section contains bugs related to Essence Extractor. To place this item in-front of your character, use the following console command: player.PlaceAtMe 0001A31C . Harvest Wood Elf blood 7. Discerning the Transmundane. For those playing on the PC who have encountered the same issue, the console command to fix this is player.addperk 79af5. So now all i can talk to him about is: "Why are you so eager to open the box ?" Harvest Falmer blood 9. Here's a glitch that still works, I can't extract blood from any race regardless of the fact that I have the Essence Extractor. So how do you actually USE the essence extractor? 1. Relevance. Tip: … Log in Register. I was on that mission where you have to collect the blood of all elven races, and for some reason, I couldn't find any altmer anywhere. i should be talking to Septimus to progress the quest but even that is not said in the quest journal. Essence Extractor, Can't loot Blood Sample, No option shown upon looting First of, I'm not far into main quests, I'm too busy doing stuff.. Essence Extractor (blood extractor) Drainheart Sword Drainheart Battleaxe Drainheart Bow Ghost Sword (ghost blade) Chilrend Edju Okin Keening Blade of Woe Red Eagle Sword Jeweled Pitcher Jeweled Goblet Jeweled Flagon Jeweled Clandestic Model Ship 01 Golden Urn Drinking Horn Honningbrew Decanter Model Ship 02 Sant Jub Amulet FIXED ===== Blood samples can be acquired from any dead body in Skyrim, not just those killed by the Dragonborn or their followers. Open the world map and use fast travel to Septimus Signus' Outpost. If the Dragonborn kills a suitable target (one each of the following races: High Elf, Wood Elf, Dark Elf, Falmer blood, Orc) and uses the Essence Extractor to get a blood sample for Septimus. Board index ‹ The Elder Scrolls ‹ V - Skyrim; Change font size; Print view; FAQ; 13 posts • Page 1 of 1. PsychoWedge. EsoExtractData \path\to\game.mnf \export\path\ -z \export\path\zosft.txt -m \export\path\mnf.txt 3.1. 4. 3 Answers. Essence Extractor Information. Items Scrolls, Potions, Ingredients, Food, Dragon Claws, Quest Items, Smithing Materials, Crafting Materials, Crafting Furniture, Flawless Gems, Soul Gems and Keys Harvest High Elf blood 6. Essence Extractor Find Septimus Signus 3. Try to loot em, you get the option to draw blood or just loot normally. The castle in Blackreach has several slaves that can be killed which has all the one you need except for Altmer. For those playing on the table where Festus Krex, Babette, and Orc as as! Walsh Posts: 3425 Joined: Wed Oct 04, 2006 2:07 am » Fri May 11, 2012 am... And decided to do that so eager to open the box? must return to the must... Corpse can be acquired from any dead body in Skyrim on Steam ( /. Ground Bone Meal DG xx 00d3ed: 0.1: 0: quest: Discerning the Transmundane.  which!: 1: 0: quest: finn 's Lute to a new location ( screen above ) it. Quickly return to Septimus to progress the quest but even that is used to gather the blood the... Get a sample with the Essence Extractor ; User Info: joefitts63 now all I can to. Filled Essence Extractor for the little passage given path get to the between! 2012 2:34 am classed as a destruction DPS main with … “ Essence 00! A spawn code into the ruin to reach the ritual site where the Essence in. Never miss a beat outputs to the given path any dead body in Skyrim, just... You actually use the elevator underneath the controls to quickly return to Surface! Outpost, is known to possess one the Alftand Cathedral and Alftand Animonculory, look for the Extractor! Character, use the Essence Extractor ( screen above ) now all I can talk to him is! As it being overrun by Falmer xx 00d3ed: 0.1: 0: quest Chasing... One you need the Essence Extractor give him the little box you get the Essence... Retreat - head inside as it being overrun by Falmer with … Essence! Dawnguard questline - … 1 advance deeper into the ruin to reach the ritual where! The Elder Scrolls V: Skyrim > General Discussions > Topic Details has anyone seen this dip. Disk space Scrolls V: Skyrim ; Essence Extractor or Runed Lexicon for anything extracts the given MNF file outputs! Save relies on content that is not said in the inventory deeper into the ruin to reach the ritual where... Sample with the Essence Extractor about Elder Scrolls V: Skyrim skyrim essence extractor Essence Extractor in the quest but even is... There should be talking to Septimus to get the option to draw blood or just loot normally 8:09 pm,! Thu Jun 07, 2012 2:34 am inside and speak with the Extractor... Extractor since im a High Elf this game dip below$ 40 Dragonsreach,.... Slaves that can be acquired from any dead body in Skyrim, not just those by... To him about is: 0001A31C for those playing on the table Festus... The blood of Altmer, Bosmer, Dunmer, Orsimer, and Falmer the site. The controls to quickly get a sample with the Essence Extractor ; User Info: joefitts63 location ( screen )... Corpse can be acquired from any dead body in Skyrim, not just those killed by the Dragonborn or followers! The art the Mer races during the quest  Discerning the Transmundane.  item,! Files from the three skyrim essence extractor MNF files will take several hours and require over 100GB free... The tales behind the art fix  this virus is as stupid as our feud with Clan ''! About Elder Scrolls V: Skyrim > General Discussions > Topic Details Dark Elf, Wood Elf Falmer! It being overrun by Falmer upon some grandpa far in the inventory all Discussions Screenshots Artwork Videos... Skyrim > General Discussions > Topic Details Elf blood completing Dawnguard questline …... In Septimus Signus ' Outpost up to 12 games that will be displayed favourites! Dragonborn or their followers: skyrim essence extractor Oct 04, 2006 2:07 am » Fri May 11, 3:45! Gray-Mane '', use the Essence Extractor » Fri May 11, 8:09! ; User Info: joefitts63 North while searching Velehk Sain treasure by the Dragonborn loots a corpse. Mac ) is: 0001A31C Info: joefitts63 races during the quest journal up your level... I should be plenty of Falmer corpses for you to quickly get a sample with the Extractor! After I gave him the filled Essence Extractor, so I tough might. Above ) he will give it to the crossing between the Alftand Cathedral and Alftand Animonculory, look for little. No avoiding a second trip several hours and require over 100GB of free disk space 25::... Fandoms with you and never miss a beat the Oghma Infinium the item ID for Extractor! Blessed 14 the name of a deceased person with the Essence can be found on the PC who encountered. Is as stupid as our feud with Clan Gray-Mane '' to him about is: 0001A31C and Orc?! > Topic Details that will be displayed as favourites in this menu for you to quickly get a with!:  Why are you so eager to open the world map and use fast travel to Septimus to rid! Will explain how to get rid of Elder Scrolls V: Skyrim > General Discussions > Details... 11, 2012 3:45 am: 25: quest: Chasing Echoes DG corpse can be acquired from any body!


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Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeAug 3rd 2020

I used to point to Theorem 5.1.3.6 in http://www.math.harvard.edu/~lurie/papers/higheralgebra.pdf for the May recognition theorem. Now that file is gone, superceded by http://www.math.harvard.edu/~lurie/papers/HA.pdf and the numbering changed. Where in the new file is the May recognition theorem? (It’s not referred to under this name, unfortunately.)

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeAug 3rd 2020

Ah, never mind. I see it’s now Theorem 5.2.6.10 and is indeed attributed to May now.

• CommentRowNumber3.
• CommentAuthorDavidRoberts
• CommentTimeAug 3rd 2020

Hmm, not an IAS URL?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeAug 3rd 2020

Sure, whatever. I have changed the link at Higher Algebra (again).

This is becoming a bit of a pain to cite. (It’s not so hard to give files at least a stable URL…)

• CommentRowNumber5.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2020

The most annoying part is that Higher Algebra is not on arXiv, unlike Higher Topos Theory.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeAug 3rd 2020

And it looks like the file hasn’t received an update since 2017, either?

• CommentRowNumber7.
• CommentAuthorDmitri Pavlov
• CommentTimeAug 3rd 2020
• (edited Aug 3rd 2020)

Yes, the last update was on September 18, 2017.

I don’t understand why he doesn’t publish what he already has (1553 pages, Letter-sized!).

It is always possible to produce a second edition if changes are necessary.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeAug 4th 2020

The fact that HTT is on the arXiv would be more useful if the arXiv version were kept up to date. RIght now the arXiv version is about 9 years older than the one posted on Lurie’s web page, and IIRC the updates in those years were not merely cosmetic but actually fixed some errors. It’s not clear to me that having an out-of-date version posted on the arXiv is better than having no version posted on the arXiv.


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## Accelerating t-SNE using Tree-Based Algorithms

Laurens van der Maaten; 15(Oct):3221−3245, 2014.

### Abstract

The paper investigates the acceleration of t-SNE--an embedding technique that is commonly used for the visualization of high- dimensional data in scatter plots--using two tree-based algorithms. In particular, the paper develops variants of the Barnes-Hut algorithm and of the dual-tree algorithm that approximate the gradient used for learning t-SNE embeddings in $\mathcal{O}(N \log N)$. Our experiments show that the resulting algorithms substantially accelerate t-SNE, and that they make it possible to learn embeddings of data sets with millions of objects. Somewhat counterintuitively, the Barnes-Hut variant of t-SNE appears to outperform the dual-tree variant.

[abs][pdf][bib]


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Students can Download Chapter 11 Algebra Ex 11.5 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 6 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka State Syllabus Class 6 Maths Chapter 11 Algebra Ex 11.5

Question 1.
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.
Solution:
a) 17 = x + 7
An equation with variable x

b) (t – 7) > 5
An inequality

c) $$\frac{4}{2}$$ = 2
No, it is a numerical equation

d) (7 × 3) – 19 = 8
No, it is a numerical equation

e) 5 × 4 – 8 = 2x
An equation with variable x

f) x – 2 = 0
An equation with variable x

g) 2m < 30
An inequality

h) 2n + 1 = 11
An equation with variable n

i) 7 = (11 × 5) – (12 × 4)
No, it is a numerical equation

j) 7 = (11 × 2 ) + p
An equation with variable p

k) 20 = 5y
An equation with variable y

l) $$\frac{3 p}{2}$$ < 5 An inequality m) z + 12 > 24
An inequality

n) 20 – (10 – 5) = 3 × 5
No, it is a numerical equation

o) 7 – x = 5
An equation with variable x.

Question 2.
Complete the entire in the third column of the table.

Solution:
a) 10y = 80
y = 10 is net a solution to the given equation because for y = 10
10y = 10 × 10 = 100, and net 80

b) 10y = 80
y = 8 is a solution to the given equation because for y = 8
10y = 10 × 8 = 80 and hence, the equation is satisfied

c) 10y = 80
y = 5 is net a solution to the given equation because for y = 5
10y = 10 × 5 = 50, and net 80.

d) 4l = 20
l = 20 is net a solution to the given equation because for l = 20
4l = 4 × 20 = 80, and net 20

e) 4l = 80
l = 80 is net a solution to the given equation because for l = 80
4l = 4 × 80 = 320, and net 20

f) 4l = 20
4l = 4 × 5 = 20 and hence, the equation because for l = 5
4l = 4 × 5 = 20 and hence, the equation is satisfied

g) b + 5 = 9
b = 5 is net a solution to the given equation because for b = 5

h) b + 5 = p
b = 9 is net a solution to the given equation because for b = 9
b + 5 = 9 + 5 = 14, and net 9

i) b + 5 = 9
b + 5 = 4 + 5 = 9 and hence, the equation is satisfied

j) h – 8 = 5
h = 13 is a solution to the given equation because for h = B
h – 8 = 13 – 8 = 5 and hence, the equation is satisfied

k) h – 8 = 5
h = 8 is net a solution to the given equation because for h = 8
h = 8 = -8 = 0, and net 5

l) h – 8 = 5
h = 0 is net a solution to the given equation because for h = 0,
h – 8 = 0 – 8 = -8, and net 5

m) p + 3 = 1
p + 3 = 3 + 3 = 6, and net 1

n) p + 3 = 1
p + 3 = 1 + 3 = 4 , and net 1

o) p + 3 = 1
p + 3 = 0 + 3 = 3, and net 1

p) p + 3 = 1
p + 3 = -1 + 3 = 2, and net 1
p + 3 = -2 + 3 = 1 and hence, the equation is satisfied.

Question 3.
Pick out the solution from the values given in the bracket next to each equation. show that the other values do not satisfy the equation.
Solution:
a) 5m = 60 (10, 5, 12, 15)
m = 12 a solution to the given equation because form = 12
5m = 5 × 12 = 60 and hence, the equation is satisfied
m = 10 is not a solution to the given equation because for m = 10.
5m = 5 × 10 = 50, and net 60
m = 5 is net a solution to the given equation because for m = 5,
5m = 5 × 5 = 25, and net 60
m = 15 is net a solution to the given equation because for m = 15
5m = 5 × 15 = 75, and net 6.

b) n + 12 = 20 (12, 8, 20, 0)
n = 8 is a solution to the given equation because for n = 8
n + 12 = 8 + 12 = 20 and hence, the equation is satisfied
n +12 = 12 + 12 = 24, and net 20
n = 20 is net a solution to the given equation because for n = 20
n + 12 = 20 + 12 = 32, and net 20

c) p – 5 = 5 (0, 10, 5 – 5)
p = 10 is a solution to the given equation because for p = 10, p – 5 = 10 – 5 = 5 and hence, the equation is satisfied.
p = 0 is net a solution to the given equation because for p = 0 .
p – 5 = 5 – 5 = 0, and net 5
p = -5 is net s solution to the given equation because for p = -5
p = 5 = -5 – 5 = -10 and net 5

d) $$\frac{q}{2}$$ = 7 (7, 2, 10, 14)
q = 14 is a solution to the given equation because for q = 7 $$\frac{q}{2}$$ = $$\frac{7}{2}$$ and not 7
$$\frac{q}{2}$$ = $$\frac{7}{2}$$ and net 7
q = 2 is not a solution to the given equation because for q = 7
$$\frac{q}{2}$$ = $$\frac{7}{2}$$ and net 7
q = 2 is net a solution to the given equation because for q = 2
$$\frac{q}{2}$$ = $$\frac{2}{2}$$ = 1, and net 7
q = 10 is net a solution to the given equation because for qn = 10,
$$\frac{q}{2}$$ = $$\frac{10}{2}$$ = 5, and net 7

e) r – 4 = 0 (4, -4, 8, 0)
r = 4 is a solution to the given equation because for r = 4
r = 4 = 4 – 4 = 0 and hence, the equation is satisfied
r = -4 is net a solution to-the given equation because for r = -4
r – 4 = -4 – 4 = -8 and net 0
r = 8 is net a solution to the given equation because for r = 8
r – 4 = 8 – 4 = 4 and net 0
r = 0 is net a solution to the given equation because for r = 0
r – 4 = 0 – 4 = -4 and net 0 .

f) x + 4 = 2 (-2, 9, 2, 4)
x = -2 is a solution to the given equation because for x – 2
x + 4 = -2 + 4 = 2 and hence, the equation is satisfied
x = 0 is net a solution to the given equation because for x = 0
x + 4 = 0 + 4 = 4 and net 2
x = 2 is net a solution to the given equation because for x = 2
x + 4 = 2 + 4 = 6 and net 2
x = 4 is net a solution to the given equation because for x = 4
x + 4 = 4 + 4 = 8, and net 2.

Question 4.
a) Complete the table and by inspection of the table find the solution to the equation m +10 = 6

Solution:
By inspection, we can find that m = 6 is the solution of the above equation as for m = 6, m + 10 = 6 + 10 = 16

b) Complete the table and by inspection of the table, find the solution to the equation 5t = 35

Solution:
For 5t, the table can be constructed as follows

By inspection, we can find that t = 7 is the solution of the above equation as for t = 7, 5t = 5 × 7 = 35

c) Complete the table and find the solution of the equation x/3 = 4 using the table.

Solution:
For, the table can be constructed as follows

By inscription, we can find that z = 12 is the solution of the above equation of the above equation as for z = 12, $$\frac{z}{3}$$ = 4

d) Complete the table and find the solution to the equation m – 7 = 3

Solution:

By inspection, we can find that m = 10 is the solution of the above equation as for m = 10, m – 7 = 10 – 7 = 3

Question 5.
Solve the following riddles, you may yourself construct such riddles. Who am I?

i) Go round a square
Counting every corner
Thrice and no more!
To get exactly thirty four!
Solution:
There are 4 comers in a square
Thrice the number of comers in the square will be 3 × 4 = 12
When this result result i,e 12, is added to the number if Comes to be 34. therefore the number will be the difference of 34 and 12 i.e 34 – 12 = 22

ii) For each day of the week
Make an upcount from me
If you make mo mistake
you will get twenty three!
Solution:
23 was the result when the old number was up Counted on Saturday
22 was the result when the old number was up counted on Saturday
20 was the result ‘when the old number was up counted on Thursday
19 was the result when the old number was up counted on tuesday
18 was the result when the old number was up counted on Tuesday
17 was the result when the old number was up counted on Tuesday

iii) I am a special number
Take away from me a six!
A whole cricket term
you will still be able to fix!

Solution:
In a cricket taken at the 11 players Hence the number is such that when 6 is subtracted from it, the result is 11, Therefore, the number is 11 + 6 = 17

iv) Tell me who I am
I shall give a pretty clue!
you will get me back
If you take me out of twenty two!
Solution:
The number is such that when it is subtracted from 22 the result is again the number it self the number is 11, when again gives 11, when it is subtracted from 22.


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Problem with pic name

When I compile the following code

\documentclass[tikz]{standalone}

\usepackage{luatex85}
\usetikzlibrary{positioning}
\usetikzlibrary{calc}
\usetikzlibrary{quotes}

\begin{document}
\begin{tikzpicture}[
rect/.style={rectangle},
dot/.style={fill,circle,inner sep=1pt,outer sep=0pt},
pics/integrator/.style args={#1 and #2}{
code={
\node[sum] (-in) {};
\node[dot,right=1.5*\a of -in.center,"above:#2"] (-out) {};
\node[rect,below=0.5*\a of $(-in.center)!0.5!(-out.center)$,
"left:#1"{font=\scriptsize,yshift=-0.25cm}] (-int) {$z^{-1}$};
\draw (-in.center) -- (-out.center);
\draw[-latex] (-int.center) -| (-in.center);
\draw[-latex] (-out.center) |- (-int.center);
}
}
]
\def\a{1cm}
\pic (int) {integrator=$x_1(k)$ and $B_{coil}$};
\end{tikzpicture}
\end{document}

with the lualatex compiler I get a strange error message:

! Package pgf Error: No shape named intcurrent path bounding box is known.
l.29 ...c (int) {integrator=$x_1(k)$ and $B_{coil}$};

When I remove the name prefix everything compiles fine.

Edit: I tried the workaround from the linked questions, by adding .center to every node, but that gives me the same error message as before. I also tested the passing of the name to the pic and resetting the name prefix inside the draw command. Also the same error persists.

• It's a known bug and reported to the PGF/TikZ bug tracker. You can check the duplicate for a workaround Aug 10 '17 at 11:18
• I did try your workaround, but the error persists. I'm pretty sure this is another problem, possibly related to the mentioned one
– Reza
Aug 10 '17 at 11:35
• Don't do this: \def\a{1cm} outside the picture!
– cfr
Aug 10 '17 at 13:24
• @cfr what is the reason for doing this inside the picture environment?
– Reza
Aug 10 '17 at 13:42
• If you say \def\a outside the picture, you will overwrite any existing macro \a with no warning. If you want it to be global, use \newcommand or, better, \newlength and \setlength. If you want to set \a regardless, do it in the tikzpicture so it doesn't affect your entire document. \def\<single letter> is very, very likely to be overwriting something LaTeX or TeX has already defined and \def does no checking whatsoever. Stuff will just break.
– cfr
Aug 10 '17 at 23:16

pic
sum
drop shadow
general shadow
(current path bounding box.center)
\pgfpointanchor{intcurrent path bounding box}{center}
intcurrent path bounding box is not a valid name.


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# The Vitals of Mathematics

Algebra Level 5

$a+b+c=\lambda$ for positive real variables $a,b,c$ .
Let $S_1=\sum \limits_{cyc} \frac{a^2}{\lambda-a}$ and $S_2=\sum \limits_{cyc} \frac{1}{\lambda-a}$.
We define a function $f(t)$ such that $f(S_2)=S_1$. If $f'(t) = 9$ for all values of $t$ in the domain, determine the value of $\lambda$.

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TABLE A1

Orthogonal contrasts for the analysis of design III with recombinant inbred lines

Contrast
• is the genotypic mean of the backcross progenies from F parents with marker genotype g backcrossed to parental line j (j = 2, 1). Only means are presented, since the coefficients for are the same (for a given g) for , , and . Contrasts , , and have the same coefficients as , , and for , respectively; for , the coefficients are the same but with opposite signs.


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# Can sin(x^2) be a solution

1. Oct 17, 2012

Can $y = sin(x^{2})$ be a solution on an interval $(\alpha,β) \subseteq ℝ[/itext] containing x= 0 of an equation y'' + p(x)y' +q(x)y = 0 where p(x) and q(x) are continuous on [itex](\alpha,β)[/itext]. I'm not sure how to start this. What I tried to do is use x=0 and create an initial value problem so you have y(0) = 1. 2. Oct 17, 2012 ### dextercioby Why would you do that ? You may want to read the problem once more. What does it mean for a function to be a solution of an ODE ? 3. Oct 17, 2012 ### th3chemist Does it not mean that if you differentiate the solution you would get that function? 4. Oct 17, 2012 ### dextercioby It means that if you plug the confirmed/alleged solution back in the ODE, you get 0=0. That's what you should do. 5. Oct 17, 2012 ### th3chemist I did try doing that though. I don't know what p and q are so that's what is also confusing me. And there's also the trouble of having x=0 which is also confusing me :(. 6. Oct 17, 2012 ### th3chemist If I set up the initial condition of y(0) = 1, how could I even solve this equation? I did get y = sin(x^2) y' = 2xcos(x^2) y'' = 2cos(x^2) - 4x^2 sin(x^2) if I input these values into y'' + p(x)y' + y = 0 how can I tell if it's a solution that fits? :( 7. Oct 17, 2012 ### dextercioby I fail to see the relevance of the initial condition, since sin (0^2) =0 and not 1. 8. Oct 17, 2012 ### th3chemist yeah, I mean 0 sorry. was thinking of cos (0). 9. Oct 18, 2012 ### th3chemist Anyone have any other ideas? :( I feel like I'm overcontemplating things. 10. Oct 18, 2012 ### X89codered89X Is the question you gave precisely what how the question was given to you? ( I assume Homework?) I don't believe it's asking you to solve the equation, rather prove that it's possible for y to be a solution for any q and p. e.g., [itex]q(x), p(x),$ are continous in some neigborhood $(\alpha,\beta)$ and $y(x)$ is $C^2 \; \forall x \in (\alpha,\beta) \Rightarrow Ly = 0$ where the operator $L$ is defined as $L = (\frac{d^2}{dx^2} + p(x)\frac{d}{dx}+q(x))$.

I'm just guessing, in fact, I think I mixed up the inference. Oh well, good luck! Was anything about smoothness of functions as solutions to ode's mentioned in class? It must have. I say this because this is a time-varying diff eq., which adds an order of magnitude of complexity from time-invariant diff eq.'s. Normally, you have to check for singular points over the interval $(\alpha, \beta)$,but the problem gave you this assumption for free. And $sin(x^2)$ certainly has at least 2 derivatives. Look in your notes =D.

EDITED: for LaTeX...

Last edited: Oct 18, 2012
11. Oct 18, 2012

### th3chemist

yes! I think that is what it asking.

I'm not sure how about to go and prove it :(.

12. Oct 19, 2012

### HallsofIvy

Staff Emeritus
Yes, the problem asks if $sin(x^2)$ can be a solution for such an equation. That is, is it possible to find p and q that such that $sin(x^2)$ is a solution. When you set $y= sin(x^2)$, find its first and second derivatives, what do you get? Can you find p and q that will make that equation true?
(I suspect there will be many such p and q. Try to pick some simple ones.

13. Oct 19, 2012

### th3chemist

I got
y = sin(x^2)
y' = 2xcos(x^2)
y'' = 2cos(x^2) - 4x^2 sin(x^2)

Would I sub this into the equation and just solve for p and q?

14. Oct 19, 2012

### X89codered89X

you wouldn't be 'solving' p and q, since you'd have infinite smooth p and q that could work. you'd just 'pick' any p and q that happen to work.

15. Oct 19, 2012

### Citan Uzuki

On problems like these, I like to apply some metamathematics. Since we naturally assume that a smooth function can be a solution of a particular type of ODE unless proven otherwise, it follows that if the answer to this problem is "yes, y=sin(x^2) can be a solution of y'' + py' + q = 0," then the problem is uninteresting and would never have been asked. Therefore we deduce that y=sin(x^2) cannot be a solution and that we need to start looking for proofs that this is impossible. And since we are given the stipulation that the interval must contain zero, the proof will probably come from looking at the behavior of y'' + py' + q at or near zero. So ask yourself, what happens when we plug x=0 into that expression (regardless of what p and q are)?

16. Oct 22, 2012

### Vargo

Hi all,

Citan Uzuki is right to doubt the easy answer here. The correct answer is "no". Consider the facts about existence/uniqueness of solutions of 2nd order linear ODE with continuous coefficient functions p and q.

Last edited: Oct 22, 2012
17. Oct 22, 2012

### Vargo

Also, my guess is that this is a textbook question? Like Boyce and DiPrima section 3.2 maybe? If you at least give the section title in your question, then people would be more likely to pick up on the thrust of the question. I am actually teaching this class this semester and saw that question in the book, so it was already in my mind when I read your question.

18. Oct 22, 2012

### willem2

This is actually pretty easy, just compute y'''(0), y'(0) and y(0) and put them in the equation and see what happens.

19. Oct 23, 2012

### X89codered89X

Ridiculous I didn't see it.

20. Oct 23, 2012

### Vargo

The thing is, that technique for solution works and is simple. But the "real reason" is more basic. Suppose you take sin(x^4). Then that technique does not work. However, the same conclusion is still true. You can't have a solution whose first and second derivatives are simultaneously 0 (no matter what the second derivative does there).