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# If 4^(4x) = 1600, what is the value of [4^(x–1)]^2?

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Director
Joined: 10 Feb 2006
Posts: 657
If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? [#permalink]

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23 Nov 2007, 08:56
2
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Difficulty:

55% (hard)

Question Stats:

69% (02:54) correct 31% (02:00) wrong based on 208 sessions

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If 4^(4x) = 1600, what is the value of [4^(x–1)]^2?

A. 40
B. 20
C. 10
D. 5/2
E. 5/4

Solution :
[Reveal] Spoiler:
2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-4-4x-1600-what-is-the-value-of-4-x-161823.html
[Reveal] Spoiler: OA

_________________

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Joined: 05 Jul 2006
Posts: 1747

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23 Nov 2007, 09:53
If 4 ^4x = 1600 ..........impossible
CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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23 Nov 2007, 13:46
1
KUDOS
Expert's post
yezz wrote:
If 4 ^4x = 1600 ..........impossible

possible

4^(4*1,330482)=1600
Manager
Joined: 26 Sep 2007
Posts: 65

### Show Tags

23 Nov 2007, 14:11
2
KUDOS
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2
VP
Joined: 08 Jun 2005
Posts: 1145

### Show Tags

23 Nov 2007, 14:15
sevenplus wrote:
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2

4^(4*5/2) = 4^10 and not 1,600

Manager
Joined: 26 Sep 2007
Posts: 65

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23 Nov 2007, 14:56
KillerSquirrel wrote:
sevenplus wrote:
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2

4^(4*5/2) = 4^10 and not 1,600

Question is asking to find out the value of (4^x-1)^2
So 5/2 is the value of (4^x-1)^2 and not of x
So you can't plug in x=5/2
Intern
Joined: 11 Nov 2007
Posts: 3

### Show Tags

23 Nov 2007, 18:36
sevenplus wrote:
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2

Why (4^x - 1)^2 = 4^2x-2 ????

i thought (4^x - 1)^2 = 4^2x - 2 * 4^x + 1 ?
SVP
Joined: 29 Aug 2007
Posts: 2473

### Show Tags

23 Nov 2007, 21:54
sevenplus wrote:
KillerSquirrel wrote:
sevenplus wrote:
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2

4^(4*5/2) = 4^10 and not 1,600

Question is asking to find out the value of (4^x-1)^2
So 5/2 is the value of (4^x-1)^2 and not of x
So you can't plug in x=5/2

why only 40 and not -40?

(4)^(2x)^2 = (40)^2
4^2x = +/- 40
Manager
Joined: 26 Sep 2007
Posts: 65

### Show Tags

24 Nov 2007, 15:01
GMAT TIGER wrote:
sevenplus wrote:
KillerSquirrel wrote:
sevenplus wrote:
If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ?

40
20
10
5/2
5/4

solution :

2^8x = 1600 (2^2x-2) ^2 = 2^4x-4
2^8x = 2^6 * 5 ^2

2^8x-6 = 5^2

(4^x-1)^2 = 4^2x-2 = 4^2x/4^2 = 4^2x/16 ____________(1)

Now from 4^4x=1600, we can get a value of 4^2x
(4)^(2x)^2 = (40)^2
4^2x=40

Now plug in 4^2x value in eqn 1
40/16=5/2

4^(4*5/2) = 4^10 and not 1,600

Question is asking to find out the value of (4^x-1)^2
So 5/2 is the value of (4^x-1)^2 and not of x
So you can't plug in x=5/2

why only 40 and not -40?
(4)^(2x)^2 = (40)^2
4^2x = +/- 40

Because no answer choice is -ve
SVP
Joined: 05 Jul 2006
Posts: 1747

### Show Tags

24 Nov 2007, 22:33
walker wrote:
yezz wrote:
If 4 ^4x = 1600 ..........impossible

possible

4^(4*1,330482)=1600

walker still dont get it??

4^y is a repitition of the factors 2 multiplied by each other

to get 1600 ( we need a 5 factor ) ... what am i doing wrong here??
CEO
Joined: 17 Nov 2007
Posts: 3583
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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24 Nov 2007, 23:43
Maybe (4^x-1)^2 means (4^(x-1))^2

otherwise
((4^x)-1)^2~28

yezz wrote:
walker wrote:
yezz wrote:
If 4 ^4x = 1600 ..........impossible

possible

4^(4*1,330482)=1600

walker still dont get it??

4^y is a repitition of the factors 2 multiplied by each other

to get 1600 ( we need a 5 factor ) ... what am i doing wrong here??

you are right in the case of x-integer. But finding x from 4^(4*x)=1600 is out of scope of GMAT. (I use MS Excel to calculate x)
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Posts: 15908
Re: If 4 ^4x = 1600 , what is the value of (4^x-1)^2 ? 40 20 10 [#permalink]

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14 Dec 2013, 13:18
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Re: If 4^(4x) = 1600, what is the value of [4^(x–1)]^2? [#permalink]

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15 Dec 2013, 04:32
4
KUDOS
Expert's post
If 4^(4x) = 1600, what is the value of [4^(x–1)]^2?
A. 40
B. 20
C. 10
D. 5/2
E. 5/4

$$4^{4x} = 1600$$ --> $$4^{2x} = 40$$ -

$$4^{(x-1)^2}=4^{2(x-1)}=4^{2x-2}=\frac{4^{2x}}{4^2}=\frac{40}{16}=\frac{5}{2}$$.

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-4-4x-1600-what-is-the-value-of-4-x-161823.html[/quote]
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Re: If 4^(4x) = 1600, what is the value of [4^(x–1)]^2?   [#permalink] 15 Dec 2013, 04:32
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text:
## Preprints of Dave Witte Morris

These are only the papers that have not yet appeared in print.
For a complete publication list, see Lie theory and Graph Theory.

with Kirsten Wilk:
Cayley graphs of order kp are hamiltonian for k < 48
(2018)
(arXiv:1805.00149)

with Indira Chatterji and Riddhi Shah:
Relative Property (T) for nilpotent subgroups (2017)
to appear in Documenta Mathematica
(arxiv:1702.01801)

with Dallan McCarthy:
Hamiltonian paths in $$m \times n$$ projective checkerboards (2016)
to appear in Contributions to Discrete Mathematics
(arxiv:1607.04001)

format: plain


text:
# B What is resistance?

Tags:
1. Apr 2, 2016

### Sho Kano

My professor says resistance depends on,
$R\quad =\quad \frac { \rho L }{ A }$

And is defined as,
$R\quad =\quad \frac { V }{ I }$

What does she mean? What is the difference?
A definition is denoted by $\equiv$, what is the difference between that and an ordinary equal sign?

2. Apr 2, 2016

### jfizzix

Electrical resistance is a physical concept like force, and not a mathematical one (so we wouldn't necessarily use the triple equals sign here).
In a related example, kinetic energy has been at times defined as half the mass times the square of the velocity, but this an equation that doesn't work when you're going near the speed of light. There are other definitions of kinetic energy that work well with special relativity, and reduce to the usual formula at low velocities.
The amount of resistance (as far as I know) is always equal to voltage over current, so you could define electric resistance by that ratio, but it's important to consider that that resistance is not just one constant for a given material. It depends on temperature, pressure, and applied voltage, among other things.

Last edited: Apr 2, 2016
3. Apr 2, 2016

### David Lewis

R=V/I is not the definition of resistance. It's a formula to calculate its value.

Resistance is the property of a circuit or circuit element that impedes current by converting the current's kinetic energy to heat.

Last edited: Apr 2, 2016
4. Apr 2, 2016

### Staff: Mentor

I disagree. Resistance is defined by $R=V/I$, and the KE is negligibe, it is the PE in the field which is converted to heat.

5. Apr 2, 2016

### Staff: Mentor

A definition in physics serves the same purpose as in ordinary language, it establishes a convention for the meaning of a given word or symbol. So $R\equiv V/I$ says that whenever we say $R$ we mean $V/I$.

We can then use other information and definitions to deduce the relationship between $R$ and other quantities. Those deduced relationships are identified using the usual equals sign.

6. Apr 2, 2016

### David Lewis

Dale wrote: "Resistance is defined by R=V/I."

David Lewis wrote: You apply 1V to a capacitor, the initial current is 1A, so the resistance of the capacitor at that moment is 1 ohm?

7. Apr 2, 2016

8. Apr 3, 2016

### davenn

the apparent resistance of the capacitor is 1 Ohm, Yes
the current into and out of the cap is initially 1A and falls rapidly as the capacitor is energised ( reaches equilibrium)
NOTE: NO current flows through the capacitor tho.

and another note ... please learn to use the reply button so as to quote people ...
doing the "he wrote, she wrote" etc thing can get very confusing to read

Regards
Dave

9. Apr 3, 2016

### Staff: Mentor

That isn't the way that capacitors work, but in principle, yes. In a capacitor I is not proportional to V so it's resistance is not constant. Materials for which I is proportional to V are called "Ohmic".

10. Apr 3, 2016

### Staff: Mentor

Thanks, that is an excellent paper!

11. Apr 3, 2016

### anorlunda

Dale is right, resistance is merely the relationship between voltage V and current I. Components do not need to be linear, nor passive, so that many relationships are possible. The picture attached illustrates a few. (The one labeled battery represents a battery with internal resistance.).

The arbitrary curve is not impossible. R could be defined as the slope of a line connecting any two points on the arbitrary curve. The value could be positive, negative or zero.

It is only when we have approximately linear materials, that R becomes nearly constant (see the black line in the chart). That is what your professor had in mind, but it is not the only possibility.

The international standard defines the ohm, not resistance. ohm is to resistance as a meter is to distance.

12. Apr 3, 2016

### David Lewis

It would be more accurate to say the reactance of the capacitor is 1 ohm. Resistance is not the only physical quantity that can impede current or influence voltage/current ratios.

13. Apr 3, 2016

### davenn

ONLY when dealing with AC, not DC

Capacitive reactance XC and Inductive reactance XL
are very different beasts to plain resistance. Yes they are both measured in Ohms
but that is where the similarity ends

You cannot use Ohms Law, R = V/I, to determine resistance in an AC circuit

D

14. Apr 3, 2016

### anorlunda

That's true davenn in that form, but Ohms Law comes in several forms. See the Insights Article, AC Power Analysis: Part 1, Basics for the complex form.

15. Apr 3, 2016

### davenn

which is what I said

if capacitive and inductive reactances are followed up with some research by David Lewis
he's going to come across the term impedance pretty quickly

16. Apr 3, 2016

### SteamKing

Staff Emeritus
This is a graphical illustration of resistance:

Last edited by a moderator: May 7, 2017
17. Apr 3, 2016

### David Lewis

It may be AC or DC. The only requirement is that the voltage is changing.

18. Apr 3, 2016

### Staff: Mentor

If it is DC then the voltage is not changing, by definition.

19. Apr 3, 2016

### davenn

this is incorrect ... so you just contradicted yourself
see Dale's response

20. Apr 4, 2016

### houlahound

DC voltage can change as much as it wants it just can't change polarity from pos/neg...or neg/pos.

Last edited: Apr 4, 2016


text:
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text:
# Reversing the Ricci flow

Suppose $S$ is a closed, oriented surface (2-manifold) embedded in $\mathbb{R}^3$, which inherits the metric from $\mathbb{R}^3$, so that distances are measured by shortest paths on the surface. If it is at least crudely accurate to say that Ricci flow smooths out the metric/curvature so that the surface (eventually) evolves to a sphere, is there a sense in which "reverse Ricci flow" concentrates curvature, in some sense sharpens the curvature, and perhaps partitions $S$ into distinct regions?

This is a naive question, for which I apologize in advance; it could be complete nonsense. What I mean by reverse Ricci flow (my own terminology; perhaps there is standard terminology?), could be simply changing the sign in Hamilton's equation:

$$\frac{\partial g}{\partial t} \ = \ 2 \ {\bf Ric}(g) \;.$$

What I mean by "sharpens the curvature" is something akin to image-processing operators which enhance the boundaries between regions to segment an image (edge detection). I am imagining segmenting a surface $S$ by reverse Ricci flow.

One problem I can foresee is that is that reversing the heat equation is inherently unstable, and maybe the same is true here. But perhaps in the specific situation of $S \subset \mathbb{R}^3$ inheriting the Euclidean metric, the instabilities are not as severe as they might be for arbitrary Riemannian manifolds.

All this is speculation on my part. Reality checks, references, or further speculation—all welcomed!

-

What you are probably after is something like mean curvature flow. This evolution equation does concentrate curvature -- a good reference for this happening with convex initial hypersurfaces is Huisken's seminal 1984 paper in JDG. More recently, people have worked on surgery for the mean curvature flow, which does break the hypersurface into distinct regions where the curvature concentrates into singularities.

For your specific question, it isn't possible. Uniqueness in backward time is rubbish: the PDE is not parabolic. There is a clear and intuitive reason for this. The Ricci flow evolves geometries toward a model state: on a sphere with positive curvature, toward the geometry of a sphere. Now think about this for a second. All positive curvature geometries evolve toward one geometry -- as you wrote, the effect of the Ricci flow is to `smooth out' regions of curvature. Now if I begin at the goal model geometry, and flow in the backward time direction, how can I possibly know which less-ideal-but-positively-curved geometry to pick?

This is the basic problem with equations which aren't parabolic. In general, to reverse time is a very special and tricky thing and you need a lot more information to do it with any meaning. In this sense, I'd venture that the Ricci flow isn't the tool you're looking for.

-
This is very useful intuition, Glen---Thanks! –  Joseph O'Rourke Oct 7 '12 at 2:01

In http://en.wikipedia.org/wiki/Ricci_flow#Relation_to_diffusion you will find an explicit computation of the form of the equation on a 2-dimensional manifold. The equation is, up to trivial manipulations, the usual heat equation in the plane. This is a non-reversible evolution equation.

-
I thought that the heat equation, roughly, had unique backward time solutions? I.e. if two initial conditions give the same time evolution then they are the same. (I'm not an expert.) –  Grumpy Parsnip Oct 15 '11 at 2:33
I know little about this, but there is an Osher-Rudin approach to solving a time-reversed heat equation to effect image deblurring... –  Joseph O'Rourke Oct 15 '11 at 2:46
@Jim, if I remember correctly, there is uniqueness but not existence in general, no? –  Mariano Suárez-Alvarez Oct 15 '11 at 2:48
Hmm. I probably shouldn't stick my nose into an area of which I have so little knowledge, but I agree with that. So I guess the question includes: "Under what conditions can the flow be reversed, and how far can it be reversed before hitting some kind of singularity?" –  Grumpy Parsnip Oct 15 '11 at 2:53
@GrumpyParsnip: time-reversing heat flow equation is impossible, provided that heat diffusion process by definition generate entropy and thus the information is lost. This is not true for most of the differential equations, notably in solid dynamics, where intial conditions and equation uniquely define all time positions. The reason it is true for heat diffusion equations is that it not with a set of point in canonical corrdinates but a volume. However, because of convergence of solutions of diffusion equation, this volume diverges if time is reversed. –  Andrei Kucharavy Apr 19 '14 at 20:51


text:
•   Wikis

# Doubly-periodic function: Wikis

Note: Many of our articles have direct quotes from sources you can cite, within the Wikipedia article! This article doesn't yet, but we're working on it! See more info or our list of citable articles.

# Encyclopedia

In mathematics, a doubly periodic function is a function defined at all points on the complex plane and having two "periods", which are complex numbers u and v that are linearly independent as vectors over the field of real numbers. That u and v are periods of a function ƒ means that

$f(z) = f(z + u) = f(z + v) \,$

for all values of the complex number z.

The doubly periodic function is thus a two-dimensional extension of the simpler singly periodic function, which repeats itself in a single dimension. Familiar examples of functions with a single period on the real number line include the trigonometric functions like cosine and sine. In the complex plane the exponential function ez is a singly periodic function, with period 2πi.

As an arbitrary mapping from pairs of reals (or complex numbers) to reals, a doubly periodic function can be constructed with little effort. For example, assume that the periods are 1 and i, so that the repeating lattice is the set of unit squares with vertices at the Gaussian integers. Values in the prototype square (i.e. x + iy where 0 ≤ x < 1 and 0 ≤ y < 1) can be assigned rather arbitrarily and then 'copied' to adjacent squares. This function will then be necessarily doubly periodic.

If the vectors 1 and i in this example are replaced by linearly independent vectors u and v the prototype square becomes a prototype parallelogram, which still tiles the plane. And the "origin" of the lattice of parallelograms does not have to be the point 0; the lattice can start from any point. In other words, we can think of the plane and its associated functional values as remaining fixed, and mentally translate the lattice to gain insight into the function's characteristics.

However, what is usually meant is a "smooth" complex function, a mapping satisfying the Cauchy–Riemann equations and providing an analytic function (away from some reasonable number of singularities), in other words, a meromorphic function. Quite a bit of information about such a function can be obtained by applying some basic theorems from complex analysis.

• To be of much interest, the "smooth" doubly periodic function cannot be bounded on the prototype parallelogram. For if it were it would be bounded everywhere, and therefore constant by Liouville's theorem.
• If the function has no essential singularities (i.e. it is meromorphic), then its poles are isolated and a translated lattice that does not pass through any pole can be constructed. But then the contour integral around any parallelogram in the lattice must vanish, because the values assumed by the doubly periodic function along the two pairs of parallel sides are identical, and the two pairs of sides are traversed in opposite directions as we move around the contour. By the residue theorem the function cannot have a single simple pole inside each parallelogram – it must have at least two simple poles within each parallelogram (Jacobian case), or it must have one or more poles of order greater than one (Weierstrassian case).
• A similar argument can be applied to the function g = 1/ƒ where ƒ is doubly periodic. Clearly the zeroes of ƒ become the poles of g, and vice versa, under this inversion. So the meromorphic doubly periodic function ƒ cannot have one simple zero lying within each parallelogram on the lattice—it must have at least two simple zeroes, or it must have at least one zero of multiplicity greater than one. It follows that ƒ cannot attain any value just once, since ƒ minus that value would itself be a meromorphic doubly periodic function with just one zero.

See elliptic function for a more complete account of doubly periodic functions that are meromorphic on the complex plane, and fundamental pair of periods for an account of the lattices involved. Also see Jacobi's elliptic functions and Weierstrass's elliptic functions.


text:
## Saturday, August 20, 2011

### SICP 2.44 & 2.45: A Picture Language

From SICP section 2.2.4 Example: A Picture Language

The next section of SICP is an extended example that illustrates the features of the picture language introduced in SICP Lecture 3A: Henderson Escher Example. I'm going to be using the PLT Scheme SICP Picture Language package to run the examples and exercises. You can load the picture package by putting the following (require...) expression at the beginning of your Scheme file.
(require (planet "sicp.ss" ("soegaard" "sicp.plt" 2 1)))

The painter primitive in the PLT package behaves slightly differently than the one presented in the text, in that it doesn't display itself. To display a painter you must call the paint procedure, which takes a painter as its argument.
> (paint einstein)

> (paint diagonal-shading)

The einstein and diagonal-shading painter values are built in to the PLT Scheme Picture Language package. The package also includes procedures for flipping, rotating and combining painter values, similar to those discussed in the lecture.
> (paint (flip-horiz einstein))

> (paint (flip-vert einstein))

> (paint (rotate90 einstein))

> (paint (beside einstein einstein))

> (paint (below einstein einstein))

Now that we can work with painters, we can start to combine them in ways that form patterns. For example, the text shows us how we can combine an image beside a flipped representation of itself, then draw the resulting painter below itself.
(define (flipped-pairs painter)
(let ((painter2 (beside painter (flip-vert painter))))
(below painter2 painter2)))

> (paint (flipped-pairs einstein))

We can also define recursive operations, like right-split, which makes painters split and branch towards the right as many times as we specify.
(define (right-split painter n)
(if (= n 0)
painter
(let ((smaller (right-split painter (- n 1))))
(beside painter (below smaller smaller)))))

> (paint (right-split einstein 3))

Exercise 2.44 asks us to define the procedure up-split. It's similar to right-split, except that it switches the roles of below and beside.
(define (up-split painter n)
(if (= n 0)
painter
(let ((smaller (up-split painter (- n 1))))
(below painter (beside smaller smaller)))))

> (paint (up-split einstein 3))

Once we have both right-split and up-split defined, we can create balanced patterns by branching in both directions.
(define (corner-split painter n)
(if (= n 0)
painter
(let ((up (up-split painter (- n 1)))
(right (right-split painter (- n 1))))
(let ((top-left (beside up up))
(bottom-right (below right right))
(corner (corner-split painter (- n 1))))
(beside (below painter top-left)
(below bottom-right corner))))))

> (paint (corner-split einstein 1))

> (paint (corner-split einstein 2))

By combining four copies of the corner-split pattern, we can create the square-limit pattern from the Escher drawing that we saw in the lecture.
(define (square-limit painter n)
(let ((quarter (corner-split painter n)))
(let ((half (beside (flip-horiz quarter) quarter)))
(below (flip-vert half) half))))

> (paint (square-limit einstein 4))

Exercise 2.45 points out that right-split and up-split can be expressed as instances of a general splitting operation. We are to define a split procedure, and then rewrite right-split and up-split in terms of split as follows:
(define right-split (split beside below))
(define up-split (split below beside))

We already noted the similarity between the two procedures right-split and up-split when we solved exercise 2.44 above. The only real difference between them is in which direction they split the painter first. Given the amount of duplication between the two, we can use either one of these procedures as a template for split.
(define (split dir1 dir2)
(lambda (painter n)
(if (= n 0)
painter
(let ((smaller ((split dir1 dir2) painter (- n 1))))
(dir1 painter (dir2 smaller smaller))))))

The biggest change is that split uses a lambda to return the procedure that it creates. Once you're done substituting the new definitions for right-split and up-split, you can run the square-limit procedure again to verify that the output hasn't changed.

Related:

For links to all of the SICP lecture notes and exercises that I've done so far, see The SICP Challenge.


text:
# Problem Statement

R’s great strengths as a facilitator of reproducible analysis are due to packages that provide access to literate programming (e.g rmarkdown, knitr). To solve other aspects of the reproducibility problem, the R community has built a culture of best practice that encourages the use of version control, automation, and software testing. Packages that support these have also developed (e.g. devtools. remake, testthat, rhub,githug).

If we consider the facets of reproducibility outlined in the rOpenSci Guide, R is currently weak in the area of Provenance. By that I mean, creating provable lineages for data, code, model, and results objects that ensure none can exist in invalid or unexpected states.

This weakness manifests itself in several reproducibility traps one can encounter, particularly when working with teams of analysts. The subject of this proposal is for a data, code, and results journalling tool that can be used to identify and avoid these reproducibility pitfalls.

## Reproducibility Traps for Teams with the Best of Intentions

Local copies. They are a the crux of the issues described here. Despite the best of intentions, a team using literate programming with version control best practice can still be fall afoul of them. In my personal experience this is for two main reasons, driven separately by data and code.

### Data Copies

Data of moderately large size cannot usually be stored in git repositories. This is because repository size can easily blow out to uneconomical levels when tracking changes of large files. Best practice is to store data inputs for analysis in a remote data store (perhaps with data version control). All members of the team can access the same data file which can act as a ‘single source of truth’ for that data.

Unfortunately when working with moderately large data files (e.g. > 100MB) local copies become necessary due to latency induced by copying data over a network. As soon as a local copy is made, the same latency issue will mean any changes or updates to the data are slow to propagate back to the data store.

### Code Copies

Local copies of code are the norm when using distributed version control. Even for teams following strict versioning practices, confusion can arise about the code lineage of results due to the existence of working branches where experimental updates are tested before being pulled to the master branch.

When it comes time to disseminate results of the analysis via a literate programming document (word, pdf, html), the document provides no information (verifiable or otherwise) about the version of the code used to produce the analysis. Code inconsistencies are extremely difficult to detect looking at results in this form. There is also no way to verify the output was indeed generated by a particular source file even if that file is sitting beside it in a repository.

## Problem Manifestation

Although the examples make reference to teams, they can just as easily arise in solo work. The problems caused by the necessary use of local copies will manifest themselves as inability to reproduce results leading to uncertainty and debate about the validity of all aspects of the analysis. For analysis that is non-deterministic, for example involving random samples, the possibility can never be fully discounted that analysis discrepancies are due to sample variation. In fact this possibility creates an ‘easy-out’ that has in my own experience caused discrepancies due to underlying code and data to go unnoticed for too long.

# Proposed Solution

Here I propose a package intended to provide lightweight, flexible support for data, code and result provenance with minimum change to typical workflows. Design is high level at this stage, no implementation has begun.

## Journalling

The proposed package would allow teams to instantiate a journal file object which lives in the root level of a project directory. The journal would have an R API that would allow registration and verification of data and results. Some fictional usage is shown below. The verbs used in the API are to be decided upon and are very much up for discussion.

### Datasets

For example:

register(source = "//datastore/project/datafile.csv",
name = "data_file",
immutable = TRUE)

Would source a file from the data store, tabulating summary statistics for later verification. The record would be stored with the name in code, in this case data_file. The journal record could be made immutable, forcing analysts to register new names for updated data sets.

Other things that could be registered in the journal:

• Data objects representing useful constants
• Package functions/namespaces (to later verify functions are not masked)
• S3/S4 R objects.

Later when a local copy is read in or about to be analysed, it can be verified against the journal using stored metadata with:

verify(data_file)

You could also perform more expensive verification of the remote source, journal record, and local copy:

verify_remote(data_file)

#### Interaction with Dat and data version control

The Dat protocol could be used as a remote sourcing method for registration. In that case verify() may be able to get away with very little work, instead leveraging the Dat API.

So far I have not imagined that data version control is a requirement for the journalling workflow. Certainly it would be advisable. The journalling workflow and data version control are complementary. If DVC is in place it will make resolving issues faster after the journal verification tools have raised the alarm. Due to this, you can view the journal as a link source and data version control.

#### Data Diffing

The ability to deterine diff data is not a prerequisite for verifying it. Diffing is a desireable validation failure diagnostic.

### Code

Code execution journalling could be supported using an expression wrapper:

journal("Modelling A", {
model_fit <- lm(data = data_file,
formula = A ~ B + c)
record(formula = model_fit\$call, fit = glance(model_fit), score = RMSE(model_fit))
})

Before execution the code can be examined for validity using R’s non-standard evaluation. For example journal could:

• Check objects registered in the journal are not redefined anywhere
• verify() all objects that are registered in the journal, optionally verify_remote().
• Check no uncommitted/unpushed changes to analysis repository.

During execution the record method collects data to store in the journal entry. You could imagine it employing functions from broom and modelr to do so. Other things one might want to record:

• plot images.
• Serialised S3/S4 objects

Post execution an entry would be added to the journal containing the information placed on record() along with useful metadata like:

• A session or execution identifier to identify code executed as blocks
• To allow filtering of journal entries that are related.
• The time executed and time taken to execute.
• The registered objects that were verified.
• The repository branch and version.
• Warnings encountered.
• The code executed (optionally)
• Contents of local environment (optionally).

### Literate Code

The creation of journaled chunks is naturally suggestive of some type of special type of knitr chunk for R markdown files. Relevant journal metadata could be rendered along with the code chunk. Example syntax:

  {r, journal = "Modelling A"}
model_fit <- lm(data = data_file,
formula = A ~ B + c)
...

Thanks to Nick Tierney for the suggestion.

## Opportunities for Journal Analysis

With such a rich amount of metadata available, analysis of the data could lead to useful project insights. For example the trend of model scores over time might be suggestive of when the analysis has reached the a point of diminishing returns.

## Opportunities for Journal Usage

Records in the journal could be read by other methods to provide reproducibility utility. For example:

• The session identifier could be printed in the document output. Mapping the report (even in hard copy) to a specific set of results, generated by a specific code version, by a specific author, using verified data inputs.
• Tests that generate alerts or errors if certain journaled items trend in an undesirable direction (e.g. model scores getting worse)

### As a record of analyis

With larger projects that go through many iterations of analysis the journalling method can provide a record that is provenance for the current analytical approach. For example a new collaborator or superisor may make a suggestion which you believe has already been sufficiently explored. If you have been journalling all modelling efforts you can easily interrorgate the journal to verify to what extent the proposed analysis was already run.

Analysis can be a difficult thing to discuss in words due to the ‘garden of forking paths’ it presents. The ability to recall analysis details from the journal to build a frame of reference for a conversation about new approaches has the potential to save a lot of time and confusion.

Applied to this use-case the journal is acting much like a traditional labratory notebook.

# Feature Breakdown

Looking toward developing this kind of capability there are many ways to break it down. One idea would be in chrnonlogical order:

• Data registration and verification ‘gate’.
• Exeuction journalling with expression wrapper.
• Automated verification by execution wrapper.
• Journal inspection and modification tools.
• Incremental Enhancments:


text:
# How do you find the indefinite integral of int (12/x^4+8/x^5) dx?

Mar 3, 2018

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$

#### Explanation:

We can split the integral up, as $\int \left(f \left(x\right) \pm g \left(x\right)\right) \mathrm{dx} = \int f \left(x\right) \mathrm{dx} \pm \int g \left(x\right) \mathrm{dx} :$

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = \int \left(\frac{12}{x} ^ 4\right) \mathrm{dx} + \int \left(\frac{8}{x} ^ 5\right) \mathrm{dx}$

Let's factor the constants $12$ and $8$ out of their respective integrals:

$\int \left(\frac{12}{x} ^ 4\right) \mathrm{dx} + \int \left(\frac{8}{x} ^ 5\right) \mathrm{dx} = 12 \int \frac{\mathrm{dx}}{x} ^ 4 + 8 \int \frac{\mathrm{dx}}{x} ^ 5$

Let's rewrite our integrals using negative exponents. Recall that $\frac{1}{x} ^ a = {x}^{-} a .$ In other words, bringing a positive exponent from the denominator into the numerator yields the negative form of that exponent.

$12 \int \frac{\mathrm{dx}}{x} ^ 4 + 8 \int \frac{\mathrm{dx}}{x} ^ 5 = 12 \int {x}^{-} 4 \mathrm{dx} + 8 \int {x}^{-} 5 \mathrm{dx}$

Now we can integrate each of these. Recall that $\int {x}^{a} \mathrm{dx} = {x}^{a + 1} / \left(a + 1\right) + C$ where $C$ is the constant of integration.

$12 \int {x}^{-} 4 \mathrm{dx} + 8 \int {x}^{-} 5 \mathrm{dx} = 12 \left({x}^{- 4 + 1} / \left(- 4 + 1\right)\right) + 8 \left({x}^{- 5 + 1} / \left(- 5 + 1\right)\right) + C = \frac{12 {x}^{-} 3}{-} 3 + \frac{8 {x}^{-} 4}{-} 4 + C = - 4 {x}^{-} 3 - 2 {x}^{-} 2 + C$

It may appear confusing that we only have one constant of integration $C$ despite taking two integrals. Yes, each of these integrals did originally produce its own distinct constant of integration; however, we combined them both into a single constant $C .$ Adding two constants yields one constant; therefore, we just have one constant at the end.

We can revert to positive exponents:

$- 4 {x}^{-} 3 - 2 {x}^{-} 2 + C = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$

Thus,

$\int \left(\frac{12}{x} ^ 4 + \frac{8}{x} ^ 5\right) \mathrm{dx} = - \frac{4}{x} ^ 3 - \frac{2}{x} ^ 2 + C$


text:
# Does $g$ map $\mathbb{R}$ onto the Cantor set?

For $x\in\mathbb{R}$ define

$$g(x)=1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}x\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}x\right\rfloor}{2^{2k+1}}\right)$$

Desmos link to first four convergents of $g(x)$

Question: Does $g$ map $\mathbb{R}$ onto the middle-thirds Cantor set of $[0,1]$?

One need only consider the question whether $g$ maps the interval $[0,1)$ onto the Cantor set since for $x\in\mathbb{R}$,

$$g(x)=g(x-\lfloor x\rfloor)$$

\begin{align*} g(x) &= g(\lfloor x\rfloor+(x-\lfloor x\rfloor))\\ &= 1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}(\lfloor x\rfloor+(x-\lfloor x\rfloor))\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}(\lfloor x\rfloor+(x-\lfloor x\rfloor))\right\rfloor}{2^{2k+1}}\right)\\ &= 1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{2^{2k}\lfloor x\rfloor+\left\lfloor2^{2k}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k}}-\frac{2^{2k+1}\lfloor x\rfloor+\left\lfloor2^{2k+1}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k+1}}\right)\\ &=1+\tfrac{3}{2}\sum_{k=0}^{\infty}\left(\frac{\left\lfloor2^{2k}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k}}-\frac{\left\lfloor2^{2k+1}(x-\lfloor x\rfloor)\right\rfloor}{2^{2k+1}}\right)\\ \end{align*}

• It seems that $g$ sends each number $x$ in $(0,1)$ such that $$x=\sum_{n=1}^\infty\frac{b_n(x)}{2^n}$$ with $b_n(x)=0,1$ to $$g(x)=1-\frac32\sum_{k=0}^\infty\frac{b_{2k+1}(x)}{2^{2k+1}}=\sum_{k=1}^\infty\frac{1}{2^k}-\sum_{k=0}^\infty\frac{b_{2k+1}(x)}{2^{2k+1}}-\sum_{k=0}^\infty\frac{b_{2k+1}(x)}{2^{2k+2}}=\sum_{k=1}^\infty\frac{c_k(x)}{2^k},$$ where, for every $k\geqslant0$, $$c_{2k+1}(x)=c_{2k+2}(x)=1-b_{2k+1}(x).$$ Thus $S=g(\mathbb R)$ solves $$S=\frac14S+\frac14\left(\frac34+S\right),$$ and $S$ is a Cantor set, not the standard one, but the one where one drops the middle-halves. – Did May 28 '16 at 5:58
• In other words, the recursive step to build $S$ is $$[0,1]\to[0,\tfrac14]\cup[\tfrac34,1]$$ rather than $$[0,1]\to[0,\tfrac13]\cup[\tfrac23,1]$$ in the standard case. – Did May 28 '16 at 5:59
• Of course, I should have seen that. On the Desmos graph there is a wide-open space on the interval $(0.25,0.75)$. – John Wayland Bales May 28 '16 at 6:04


text:
# HMAC_SHA1

Chandra Bergmann

#### The HMAC_SHA1 function takes two strings as input, a base string and a key, and returns the hash of the base string as a Hex- or Base64-encoded string, arrived at via the HMAC-SHA1 algorithm.

This function requires two strings as input: a base string and key. As an option, it also accepts another string that defines if Hex or Base64 encoding will to be used to calculate the output: a string that gives the hash of the first input, as calculated by the HMAC-SHA-1 algorithm

### Declaration

HMAC_SHA1(base_string, key, encode) -> hash

### Parameters

base_string (required, type: string)
Any string.
key (required, type: string)
The key used in the calculation of the HMAC.
encode (optional, type: string, default: "Hex")

The string "Hex" or "Base64". This defines if the hash of the base string will be Hex- or Base64-encoded, respectively.

If no value is given for encode, HMAC_SHA! will return the Hex-encoded hash. If a value other than "Hex" or "Base64" is given for encode, HMAC_SHA1 will throw an error.

### Return Values

hash (type: string)
The hash of the input as calculated by the HMAC-SHA-1 algorithm.

### Examples

The following example uses the key "111" to calculate Hex-encoded hash of the string "This is a string." according to the HMAC-SHA1 algorithm:

HMAC_SHA1("This is a string.", "111") => "78a6c59e2712dcee89d25e1ccb1884c316f2ea11"

The following example also uses the key "111" to return the Hex-encoded hash of the string "This is a string." as calculated by the HMAC-SHA1 algorithm. Note that it returns the same output as the above example; there is no difference in output between explicitly defining "Hex" as the value for encode and not defining encode at all:

HMAC_SHA1("This is a string.", "111", "Hex") => "78a6c59e2712dcee89d25e1ccb1884c316f2ea11"

The following example uses the key "111" to calculate Base64-encoded hash of the string "This is a string." according to the HMAC-SHA1 algorithm. Note how this differs from the first two examples:

HMAC_SHA1("This is a string.", "111", "Base64") => "eKbFnicS3O6J0l4cyxiEwxby6hE="

The following example uses the key "ai4nga0" to calculate Hex-encoded hash of the string "This is a string." according to the HMAC-SHA1 algorithm. Note how the output is impacted by the key:

HMAC_SHA1("This is a string.", "ai4nga0") => "c998cce59510feaea598bbc4272aa7768201c7b3"


text:
# dark-hammer / 3rdparty / vs2010 / glfw.vcxproj

The branch '0.4.7' does not exist.
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text:
# ThinPlateSplines¶

class menpo.transform.ThinPlateSplines(source, target, kernel=None, min_singular_val=0.0001)[source]

Bases: Alignment, Transform, Invertible

The thin plate splines (TPS) alignment between 2D source and target landmarks.

kernel can be used to specify an alternative kernel function. If None is supplied, the R2LogR2RBF kernel will be used.

Parameters: source ((N, 2) ndarray) – The source points to apply the tps from target ((N, 2) ndarray) – The target points to apply the tps to kernel (RadialBasisFunction, optional) – The kernel to apply. min_singular_val (float, optional) – If the target has points that are nearly coincident, the coefficients matrix is rank deficient, and therefore not invertible. Therefore, we only take the inverse on the full-rank matrix and drop any singular values that are less than this value (close to zero). ValueError – TPS is only with on 2-dimensional data
aligned_source()

The result of applying self to source

Type: PointCloud
alignment_error()

The Frobenius Norm of the difference between the target and the aligned source.

Type: float
apply(x, batch_size=None, **kwargs)

Applies this transform to x.

If x is Transformable, x will be handed this transform object to transform itself non-destructively (a transformed copy of the object will be returned).

If not, x is assumed to be an ndarray. The transformation will be non-destructive, returning the transformed version.

Any kwargs will be passed to the specific transform _apply() method.

Parameters: x (Transformable or (n_points, n_dims) ndarray) – The array or object to be transformed. batch_size (int, optional) – If not None, this determines how many items from the numpy array will be passed through the transform at a time. This is useful for operations that require large intermediate matrices to be computed. kwargs (dict) – Passed through to _apply(). transformed (type(x)) – The transformed object or array
apply_inplace(*args, **kwargs)

Deprecated as public supported API, use the non-mutating apply() instead.

For internal performance-specific uses, see _apply_inplace().

compose_after(transform)

Returns a TransformChain that represents this transform composed after the given transform:

c = a.compose_after(b)
c.apply(p) == a.apply(b.apply(p))

a and b are left unchanged.

This corresponds to the usual mathematical formalism for the compose operator, o.

Parameters: transform (Transform) – Transform to be applied before self transform (TransformChain) – The resulting transform chain.
compose_before(transform)

Returns a TransformChain that represents this transform composed before the given transform:

c = a.compose_before(b)
c.apply(p) == b.apply(a.apply(p))

a and b are left unchanged.

Parameters: transform (Transform) – Transform to be applied after self transform (TransformChain) – The resulting transform chain.
copy()

Generate an efficient copy of this object.

Note that Numpy arrays and other Copyable objects on self will be deeply copied. Dictionaries and sets will be shallow copied, and everything else will be assigned (no copy will be made).

Classes that store state other than numpy arrays and immutable types should overwrite this method to ensure all state is copied.

Returns: type(self) – A copy of this object
pseudoinverse()[source]

The pseudoinverse of the transform - that is, the transform that results from swapping source and target, or more formally, negating the transforms parameters. If the transform has a true inverse this is returned instead.

Type: type(self)
set_target(new_target)

Update this object so that it attempts to recreate the new_target.

Parameters: new_target (PointCloud) – The new target that this object should try and regenerate.
has_true_inverse

typeFalse

n_dims

The number of dimensions of the target.

Type: int
n_dims_output

The output of the data from the transform.

None if the output of the transform is not dimension specific.

Type: int or None
n_points

The number of points on the target.

Type: int
source

The source PointCloud that is used in the alignment.

The source is not mutable.

Type: PointCloud
target

The current PointCloud that this object produces.

To change the target, use set_target().

Type: PointCloud


text:
# Drafts and diet

Bad habits are insidious and hard to shake. As my wife went through cancer treatment, I began to eat poorly and too much. The modest discipline that had allowed me to lose 15 pounds a couple of years earlier seemed unimportant, and I adopted the attitude of “at least I’m not drinking.” Now it’s time to put that year of self-indulgence and rationalization behind me and redevelop the good habits I abandoned.

The only way I’ve ever been able to lose weight is by weighing myself every day and keeping track of the progress (and backsliding). There is, I know, a school of thought that says this approach is wrong-headed, that daily weight fluctuations will drive you crazy. Luckily, I know enough about statistical process control that I expect certain variations and don’t let them mask the overall trend.

My assistant in the daily record keeping is Drafts, the fast and flexible note-taking app, and a simple Dropbox Action I’ve created for it. After I weigh myself, I open Drafts, type in the weight, and tap the “Weight” action. A second later, I hear the bi-doop sound that tells me the weight, with a datestamp, has been appended to a file, also named “Weight,” in my Dropbox folder. Here’s how the Action is defined.

The line appended to the file starts with a simple mm/dd/yy datestamp, followed by two spaces (I like the extra separation) and the weight.

This is, I think, the fastest way to do this. Drafts launches quickly and always opens in edit mode, ready for me to type.1 Its timestamping tag language means I don’t even have to go through the minimal effort of typing a TextExpander abbreviation to add the date.

When I want to track my progress, I can open the “Weight” file in Notesy on my phone or in BBEdit on my Mac. The format I’ve chosen makes plotting the data easy, too, although I’ve never found that especially helpful.

1. I have Drafts set to start a new note whenever I’ve been away from it for more than a minute, so it always launches with a clean new note in the morning when I weigh myself.


text:
# All Questions

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Code Genius was started at MIT to connect programmers in the scientific community to experts of advanced languages for efficient and cost effective support. Our platform helps programmers find the ...
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### Min/Max Depth of a binary tree [on hold]

Is the Min Depth of this tree 2, because of the distance of leaf node "9"? Is the Max Depth if this tree 4, because of the distance of leaf node "2"?
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### Dissipative time-stepping scheme for first order in time system

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### Generating random numbers for long molecular dynamics simulations

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### Relation between Time dependent problem and advection diffusion

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### Calculating theoretical order of accuracy of least squares fit advection scheme

I'm familiar with finding the order of accuracy using von Neumann analysis for finite difference schemes formulated using Taylor series expansions. But is there a similar technique for finding the ...
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### fixed point iteration to find out second order non-linear diff equations

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### Efficient method to multiply floating point matrix with binary matrix and get double precision results

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(First off, a disclaimer: I'm new to the site, and not quite sure this is the proper forum for this question. My apologies if it's out of place). I'm working on a computational algebra project and ...
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### GAMS AND C++ CONNECTION [on hold]

I wrote a optimization model in GAMS and I want to solve the same model with about 100 different data. Instead of inputting the data one by one by hand, I need to do it with a for loop on C++. Is it ...
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### minimization of normalized constrained quadratic function

I'm a computer science student. Please I need a help in solving a constrained normalized quadratic function. I'm familiar with solving quadratic constrained optimization function with matlab by ...
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### Creating a 3D spatial density map from simulation results

I want so visualize the spatial density for my simulation. The result of my simulation is a (time-dependent) system of large number (~100k) of moving particles in a confined space. Each particle has ...
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### How to calculate $det(X^TX)$ efficiently, update one column of X each time

$X_{1} = (A, b)$, where $X_{1}$ is a $n\times p$ matrix, $A$ is a $n\times (p-1)$ and $b$ is $n\times1$. First calculate $\det(X_{1}^T X_{1})$, then update $b$ with $c$, st. $X_{2} = (A, c)$ and ...
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### Numerical Solution of the Advection Dispersion equation

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### Modern C++ in scientific computing?

I am looking for books or articles, or blog-posts, or any published material in general, that address specifically the uses of C++ modern features (move semantics, the STL, iterators, lazy evaluation, ...
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### Solving quasilinear/nonlinear equations obtained from the discretization of partial differential equations

When you solve numerically a (system of) linear partial differential equation (PDE) as for example Lapace's equation $\nabla^2\varphi = 0$ or Poisson's equation $\nabla^2\varphi = f$ you obtain a ...
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### Fortran solver for the Sparse LSE problem

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### How to represent a binary number in a matrix in Matlab?

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### Is there an efficient $O(n^2)$ way to get the eigen decomposition given a LDL factorization?

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text:
Q

# Using the property of determinants and without expanding prove that determinant x a x+a y b y+b z z+c = 0 Ex 4.2 Q : 1

Q : 1        Using the property of determinants and without expanding, prove that

$\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}=0$

Views

We can split it in manner like;

$\begin{vmatrix}x &a &x+a \\y &b &y+b \\z &c &z+c \end{vmatrix}= \begin{vmatrix} x &a &x \\ y & b &y \\ z &c &z \end{vmatrix} + \begin{vmatrix} x &a & a\\ y &b &b \\ z&c & c \end{vmatrix}$

So, we know the identity that If any two rows (or columns) of a determinant are identical (all corresponding elements are same), then the value of the determinant is zero.

Clearly, expanded determinants have identical columns.

$\therefore 0 + 0 = 0$

Hence the sum is zero.

Exams
Articles
Questions


text:
# Corecursion

## Idea

Corecursion exploits the existence of a morphism from a coalgebra for an endofunctor to a terminal coalgebra for the same endofunctor to define an operation. It is dual to recursion. See also coinduction.

## Examples

1. For the endofunctor $H(X) = 1 + X$ on Set, the terminal coalgebra is $\bar{\mathbb{N}}$, the extended natural number system. Define a function $add\colon \bar{\mathbb{N}} \times \bar{\mathbb{N}} \to 1 + \bar{\mathbb{N}} \times \bar{\mathbb{N}}$:

$add(n, m) = \begin{cases} (pred(n), m) & if\; n \gt 0; \\ (0, pred(m)) & if\; n = 0,\; m \gt 0; \\ * & if\; m = n = 0, \end{cases}$

where $pred(x)$ is as defined at extended natural number.

Then $(\bar{\mathbb{N}} \times \bar{\mathbb{N}}, add)$ is an $H$-coalgebra. The unique coalgebra morphism ${+}\colon \bar{\mathbb{N}} \times \bar{\mathbb{N}} \to \bar{\mathbb{N}}$ (to the terminal coalgebra $\bar{\mathbb{N}}$) is addition on the extended natural numbers.

## Reference

• Jiří Adámek, Introduction to Coalgebra (pdf)

• Lawrence Moss, Norman Danner, On the Foundations of Corecursion (journal, pdf)

Revised on September 28, 2012 16:18:36 by Urs Schreiber (82.169.65.155)


text: If a vector space had bases of two different finite sizes $m < n$ say, then expressing one in terms of the other gives $m$ by $n$ and $n$ by $m$ matrices $A$ and $B$ such that $BA=I_n$. Now use Gaussian elimination (quote their first-year course) to "prove" (with an appropriate amount of hand-waving) that $Av=0$ has a nonzero solution, and thus derive a contradiction. Surely less than a lecture :-)


text:
# Learning crystal field parameters using convolutional neural networks

### Submission summary

 As Contributors: Peter P. Orth Preprint link: scipost_202012_00008v1 Code repository: https://github.com/Orth-Research/ml_cr Data repository: https://doi.org/10.6084/m9.figshare.13321985.v2 Date submitted: 2020-12-11 11:43 Submitted by: Orth, Peter P. Submitted to: SciPost Physics Academic field: Physics Specialties: Condensed Matter Physics - Theory Condensed Matter Physics - Computational Approach: Computational

### Abstract

We present a deep machine learning algorithm to extract crystal field (CF) Stevens parameters from thermodynamic data of rare-earth magnetic materials. The algorithm employs a two-dimensional convolutional neural network (CNN) that is trained on magnetization, magnetic susceptibility and specific heat data that is calculated theoretically within the single-ion approximation and further processed using a standard wavelet transformation. We apply the method to crystal fields of cubic, hexagonal and tetragonal symmetry and for both integer and half-integer total angular momentum values $J$ of the ground state multiplet. We evaluate its performance on both theoretically generated synthetic and previously published experimental data on PrAgSb$_2$ and PrMg$_2$Cu$_9$, and find that it can reliably and accurately extract the CF parameters for all site symmetries and values of $J$ considered. This demonstrates that CNNs provide an unbiased approach to extracting CF parameters that avoids tedious multi-parameter fitting procedures.

###### Current status:
Editor-in-charge assigned

### Submission & Refereeing History

Submission scipost_202012_00008v1 on 11 December 2020

## Reports on this Submission

### Report

The manuscript entitled “Learning crystal field parameters using convolutional neural networks” by Berthusen et al presents a machine learning algorithm to extract crystal field parameters from experimental thermodynamic data obtained in rare-earth-based materials. The authors apply the method to cubic, tetragonal, and hexagonal site symmetries as well as integer and half-integer total angular momentum, J, ions. Finally, the authors evaluate their algorithm against experimental data on PrAgSb2 and PrMg2Cu9.
Using convolutional neural networks to solve this important inverse problem in condensed-matter physics is a worthwhile idea, which is presented in a very pedagogical way that will benefit early-career researchers in the field. Extracting crystal field parameters is known to be time consuming, and the fact that the algorithm is provided as an open source will be useful to the community. Overall, this manuscript meets SciPost’s criteria for publication. A few issues need to be addressed, however, before the manuscript is published.
1. One general limitation of performing fits of thermodynamic data to a crystal field Hamiltonian is the possibility of reaching a local minimum that reproduces the data reasonably well but that does not necessarily contain the correct Steven parameters. In particular, this is more likely to happen when a large number of parameters is taken into account. The authors address the issue of over-parametrization, but could the authors comment on whether their algorithm has encountered local minima (of e.g. MSE) during the development of this work? In particular, one would naively think that local minima may be why “the CNN can accurately predict the Stevens parameters for the majority of the data points that it was tested on”, but not all data points.
2. To address the issue above, it might be very enlightening to apply this algorithm to data sets of materials in which neutron scattering measurements have been performed. Two examples that come to mind are CeAgSb2 (from the same family of PrAgSb2) and CeRhIn5 (at high temperatures T>T_K, the Kondo temperature). One interesting aspect here is that CEF fits and neutron data agree well for CeAgSb2, whereas the agreement is worse for CeRhIn5. This could obviously be due to the Kondo effect, but nonetheless it would be a very informative comparison.
3. I further encourage the authors to consider Cerium instead of Praseodymium because the former has only one f-electron, whereas the latter has two, which could give rise to J mixing. This has been shown, for instance, in the case of Pr2Sn2O7 via neutron spectroscopy [PRB 88, 104421 (2013)]. Even if the authors are absolutely convinced there is no J mixing in the Pr compounds investigated in their work, it is worth mentioning this general possibility in the manuscript to inform the reader of this potential issue.
4. On the Kondo effect, the authors correctly point out in the Introduction that the crystal field scheme can also have important ramifications for the nature of the Kondo effect in the system. In a simple local picture, this can be understood by considering that the shape/anisotropy of the ground state wavefunction is related to the overlap between f electrons and conduction electrons, which in turn determines the Kondo hybridization. However, the citations provided [11-14] are not particularly general, and they mostly focus on U-based materials, for which the LS coupling may not be valid as mentioned in point #3. To address SciPost’s criterion to “Provide citations to relevant literature in a way that is as representative and complete as possible” – and the fact that the community is considering crystal field effects more strongly in the recent past – the authors are encouraged to provide more citations to the relevant literature.
5. It could be useful to mention that the leading term, B20, is proportional to the difference in Weiss temperatures along different axes. Even though the method introduced by the authors is supposed to be unbiased, there are tricks that could be useful for training data.
6. Regarding the outlook of this work, including magnetic interactions will be very valuable. The easiest way of doing this is by adding a molecular field term as employed in JPSJ 70, 877 (2001), for example. The next step would be to include a mean-field Hamiltonian with one exchange constant and so on.
7. This is a minor point, but it could stimulate readers to actually use the open software in Ref. [47]: one of the main motivations of this work is to circumvent a time-consuming fitting procedure, but if one is using a conventional minimization method, a significant portion of the overall time is also spent i) preparing/converting the data and ii) waiting for iterations. Could the authors comment on the overall time spent during their approach (machine time and researcher time)?
8. A couple of final points regarding input data: i) similar to the subtraction of the phonon contribution to obtain the magnetic part of the specific heat, one also has to subtract an enhanced Pauli susceptibility from magnetic susceptibility data (if applicable); ii) trying to fit experimental data at low temperatures (eg M(H) at 1.7K) may be very hard, even if there is no magnetic order because of the effects of magnetic exchange. I wonder whether this is causing the disagreement in Fig 8d. This could also be an indication that, although the CF splitting is predicted correctly, the ground state wavefunctions are not. Therefore, it might be more useful to only fit magnetic susceptibility and magnetization data at higher temperatures, e.g., T>25K or so.

• validity: high
• significance: high
• originality: high
• clarity: high
• formatting: excellent
• grammar: excellent

### Strengths

- Well-posed physical and realistic inverse problem adapted to machine learning
- Useful for experiments
- Open source code associated
- Well-written paper

- No weaknesses

### Report

This manuscript presents a neural network analysis to infer crystal field properties (here in terms of Steven parameters) for thermodynamic measurements in f-electrons materials . This is a typical well-posed inverse problem in experimental physics, where the ‘deduction’ of crystal field parameters can become quite cumbersome.
This problem is well suited to a machine learning approach as the number of outputs (Steven parameters) to find is not too large (of the order O(1) to O(10)) and it is easy to generate training data in the single-ion approximation.
The approach chosen is through a 2d convolutional neural network of the wavelet-transformed thermodynamic data. At first glance, this seems a bit complicated and it is not clear why this approach is taken; later in the paper we understand that other simpler approaches (simpler feed-forward neural networks, 1d CNN) do not give good results.

The efficiency of solving this inverse problem is measured through a metrics in the Steven parameters space (distance between measured and true parameters) and in the thermodynamic data. Convincing results are presented for standard symmetries for synthetic data, as well as by using input thermodynamical data from experiments on two compounds to obtain Steven parameters and then recompute thermodynamics (which are found to reproduce the main features of the experimentally obtained one).

While not revolutionary, this is a simple yet useful example of how the machine learning approach can reduce a somewhat tedious task often performed manually in experimental labs. The paper is very clearly written, and is accompanied by an open source repository where the corresponding code can be directly used to compute Steven parameters from thermodynamic measurements. This clearly will be useful for future experiments.

This paper can be accepted as it is, and I have only basic questions / suggestions.

- To improve quality of the results for the hexagonal and tetragonal point group, it is suggested to add other input data such as a second magnetisation curve (obtained from a field aligned in a different direction). I guess this is quite easy to add in the training data and approach, and I would suggest to try it to see if it indeed improves the results. Same question for a larger temperature range.

- When comparing to experimental data, it would be interesting to see how good/bad is this approach when one of the experimental input is missing (as could happen in a lab, if e.g. specific heat measurements are not available).

- Typo : the from (2nd column, page 4)

### Requested changes

- No changes requested, but the authors could consider the two suggestions above

• validity: top
• significance: good
• originality: ok
• clarity: top
• formatting: perfect
• grammar: perfect


text:
# Left and right derivative

This function does not have a derivative at the marked point, as the function is not continuous there. However, it has a right derivative at all points, with $\partial_+f(a)$ constantly equal to 0.

In mathematics, the terms left derivative and right derivative are sometimes used to describe certain kinds of derivatives.

## Derivatives arising from one-sided limits

The most common use is to describe the one-sided limit of a function defined on a subset of the real line,[1] (see semi-differentiability). Under this notation, the right derivative is defined as

$\partial_+f(a):=\lim_{{\scriptstyle x\to a+\atop\scriptstyle x\in I}}\frac{f(x)-f(a)}{x-a}$

and the left derivative as

$\partial_-f(a):=\lim_{{\scriptstyle x\to a-\atop\scriptstyle x\in I}}\frac{f(x)-f(a)}{x-a}.$

### Application

If a real-valued, differentiable function f, defined on an interval I of the real line, has zero derivative everywhere, then it is constant, as an application of the mean value theorem shows. The assumption of differentiability can be weakened to continuity and one-sided differentiability of f. The version for right differentiable functions is given below, the version for left differentiable functions is analogous.

Theorem: Let f be a real-valued, continuous function, defined on an arbitrary interval I of the real line. If f is right differentiable at every point a ∈ I, which is not the supremum of the interval, and if this right derivative is always zero, then f is constant.

Proof: For a proof by contradiction, assume there exist a < b in I such that f(a) ≠ f(b). Then

$\varepsilon:=\frac{|f(b)-f(a)|}{2(b-a)}>0.$

Define c as the infimum of all those x in the interval (a,b] for which the difference quotient of f exceeds ε in absolute value, i.e.

$c=\inf\{\,x\in(a,b]\mid |f(x)-f(a)|>\varepsilon(x-a)\,\}.$

Due to the continuity of f, it follows that c < b and |f(c) – f(a)| = ε(c – a). At c the right derivative of f is zero by assumption, hence there exists d in the interval (c,b] with |f(x) – f(c)| ≤ ε(x – c) for all x in (c,d]. Hence, by the triangle inequality,

$|f(x)-f(a)|\le|f(x)-f(c)|+|f(c)-f(a)|\le\varepsilon(x-a)$

for all x in [c,d], which contradicts the definition of c.

## Differential operators acting to the left or the right

Another common use is to describe derivatives treated as binary operators in infix notation, in which the derivatives is to be applied either to the left or right operands. This is useful, for example, when defining generalizations of the Poisson bracket. For a pair of functions f and g, the left and right derivatives are respectively defined as

$f \stackrel{\leftarrow }{\partial }_x g = \frac{\partial f}{\partial x} \cdot g$
$f \stackrel{\rightarrow }{\partial }_x g = f \cdot \frac{\partial g}{\partial x}.$

In bra–ket notation, the derivative operator can act on the right operand as the regular derivative or on the left as the negative derivative.[2]