text:
## Nonisotopic symplectic tori in the fiber class of elliptic surfaces.(English)Zbl 1078.57025

For a complex surface, there are at most finitely many smooth complex curves (up to smooth isotopy) in a given homology class. However, there is no analogous finiteness for symplectic two-dimensional surfaces. Namely, R. Fintushel and R. Stern [J. Differ. Geom. 52, No. 2, 203–222 (1999; Zbl 0981.53085)] gave examples of 4-dimensional simply connected symplectic manifolds $$M$$ and homology classes $$\alpha\in H_2(M)$$ that can be realized by infinitely many mutually non-isotopic symplectic tori. However, it was not clear whether this can happen for primitive classes $$\alpha$$.
The author constructs an infinite family of mutually non-isotopic symplectic tori realizing an arbitary multiple $$q[F]$$ of the homology class $$[F]$$ of the fiber $$F$$ of an elliptic surface $$E(n), n\geq 3$$. Compare also with the author’s paper [Proc. Am. Math. Soc. 133, No. 8, 2477–2481 (2005; Zbl 1073.57016)].

### MSC:

 57R17 Symplectic and contact topology in high or arbitrary dimension 53D35 Global theory of symplectic and contact manifolds 57N13 Topology of the Euclidean $$4$$-space, $$4$$-manifolds (MSC2010) 57R57 Applications of global analysis to structures on manifolds 57R95 Realizing cycles by submanifolds

### Keywords:

symplectic tori; elliptic surfaces

### Citations:

Zbl 0981.53085; Zbl 1073.57016
Full Text:


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Comparison between treatment timepoints in a single timeseries (control is first timepoint)
1
0
Entering edit mode
Jigyasa • 0
@8f90b644
Last seen 6 days ago
Japan

Hey!

Thank you for a great resource! I am learning how to use DESEQ2 on time-series and I would like to get advice if my analysis is statistically "correct".

I have a single time-series i.e. the first time point is the control (t=-1) and the treatment starts from the second timepoint onwards (t=0). I have total of 8 timepoints of treatment (t=0, 1, 2, 3, 4, 5, 6, 7).

I want to compare each treatment timepoint with the control, and each treatment timepoint with the other. I am able to extract out DESeq results using the following code.

ddsMat <- DESeqDataSetFromMatrix(countData = countdata2,
design = ~ Time)
ddsMat <- DESeq(ddsMat)

#output of each timepoint with "-1"
res_1 <- as.data.frame(results(ddsMat, name="Time_0_vs_.1", test="Wald",alpha=0.05,pAdjustMethod = "BH",format="DataFrame"))
res_2 <- as.data.frame(results(ddsMat, name="Time_1_vs_.1", test="Wald",alpha=0.05,pAdjustMethod = "BH",format="DataFrame"))

#-----------------------------------
#condition2-output of each timepoint with the one before-
res2 <- as.data.frame(results(ddsMat, contrast=c("Time","1","0"),test="Wald",alpha=0.05,pAdjustMethod = "BH"))

I am interested in comparing res_1 and res_2, and found search results suggesting using ANOVA to compare these time points. When I examine res2 (called as Time_0_vs_1) for certain genes, I find that their log2FoldChange value lie between res_1 (called as Time_0_vs_-1) and res_2 (called as Time_1_vs_-1). Any suggestions on how to proceed?

DESeq2 single-timeseries TimeSeriesExperiment • 106 views
0
Entering edit mode
@mikelove
Last seen 4 days ago
United States

I don't follow the problem here. The LFCs make sense to me.

format: plain


text: Special issue on Selected papers from the Twenty-sixth International Symposium on Graph Drawing and Network Visualization, GD 2018 Pole Dancing: 3D Morphs for Tree Drawings Vol. 23, no. 3, pp. 579-602, 2019. Regular paper. Abstract We study the question whether a crossing-free 3D morph between two straight-line drawings of an $n$-vertex tree $T$ can be constructed consisting of a small number of linear morphing steps. We look both at the case in which the two given drawings are two-dimensional and at the one in which they are three-dimensional. In the former setting we prove that a crossing-free 3D morph always exists with $O({rpw}(T))\subseteq O(\log n)$ steps, where ${rpw}(T)$ is the rooted pathwidth or Strahler number of $T$, while for the latter setting $\Theta(n)$ steps are always sufficient and sometimes necessary. Submitted: November 2018. Reviewed: January 2019. Revised: May 2019. Accepted: June 2019. Final: July 2019. Published: September 2019. Communicated by Therese Biedl and Andreas Kerren article (PDF) BibTeX


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# What is the slope-intercept form of 6x+3y=8 ?

Feb 15, 2016

$y = - 2 x + \frac{8}{3}$

#### Explanation:

The slope-intercept form is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{b l u e}{m} x + \textcolor{red}{b}$
for a linear equation with a slope of $\textcolor{b l u e}{m}$ and a y-intercept of $\textcolor{red}{b}$

Given
$\textcolor{w h i t e}{\text{XXX}} 6 x + 3 y = 8$

Subtract $6 x$ from both sides
$\textcolor{w h i t e}{\text{XXX}} 3 y = - 6 x + 8$

Divide both sides by $3$
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{b l u e}{- 2} x = \textcolor{red}{\frac{8}{3}}$


text:
Sunday, June 22, 2014

GMAT test1

Q1. If $3x-21y=15$, then $7y-x$ is
1. $-5$
2. $5$
3. $3$
4. $7$
5. Cannot be determined
Q2.

Tuesday, July 2, 2013

Question 02/07/2013 3.12

How many different 7-place license plates are possible if the first 2 places are for letters and the other 5 for numbers?

Ans : 67 600 000

Monday, December 31, 2012

Question 12/31/2012 10:06

What is the probability that the arrangement has both consecutive W's and consecutive Y's in WYSIWYG?

Ans. 0.0952

Tuesday, December 25, 2012

Question 12/25/2012 2012 7:23

In how many ways may fifty-two cards be divided amongst tour players, so that each may have thirteen ?

Ans. $\frac{52!}{(13!)^4}$


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# Tighter bound on a spectrum of a matrix obtained by partial diagonal shift

A matrix $V\in\mathbb{R}^{n\times n}$ is a full symmetric matrix with negative off-diagonal elements summing (in absolute value) to diagonal elements. Suppose that $V$ has all entries, and let $D=diag(V)$. It is known that the spectrum of $D^{-1}V$ is in the range $[0,2)$.

Let matrix $R$ have non-negative elements (so, some of them might be zero), and let $K=V+diag(R)$. If $L=diag(K)$, what can be said about the spectrum of $L^{-1}K$?

According to Gershgorin, if the diagonal entries of $R$ were all positive, the bound is $(0, 2)$. With non-negative diagonal entries of $R$, the best bound according to Gershgoring is $[0, 2]$. Is there a way to obtain a bound $[0, 2)$ for $L^{-1}K$, with non-negative diagonal entries of $R$ (so, some diagonal entries of $R$ may be zero)?

-
I'm a bit confused. Do you want to show that the spectrum of $L^{-1}K$ lies inside a half-closed, half-open interval $[0,2)$ when $R$ is entrywise nonnegative? –  user1551 Nov 15 '12 at 10:10
@user1551 I would like to show that the spectrum of $L^{-1}K$ is within $[0,2)$ when entries of $diag(R)$ are non-negative (so, either positive or zero). –  user506901 Nov 15 '12 at 12:50

Note that $L^{-1}K = (\textrm{diag}(K))^{-1/2}(\textrm{diag}(K))^{-1/2}K(\textrm{diag}(K))^{-1/2}(\textrm{diag}(K))^{1/2}$. As the spectrum of a matrix is invariant under similarity transform, and the positive semi-definiteness of a real symmetric matrix is preserved by matrix congruence, the above equality shows that $L^{-1}K$ has nonnegative eigenvalues, i.e. the spectrum of $L^{-1}K$ is bounded below by $0$.

If $R=0$ and $V=\begin{pmatrix}1&-1\\-1&1\end{pmatrix}$, the eigenvalues of $L^{-1}K=K=V$ are exactly $0$ and $2$. So, $2$ is a sharp upper bound for the spectral radius of $L^{-1}K$ and there is no room of improvement.

If $R$ contains some positive diagonal entries, assume WLOG that the last one is positive. Let $A=2\ \textrm{diag}(K)$ and $B=K$. Then $A-B=2\ \textrm{diag}(K)-K=2\ \textrm{diag}(V)-V+\textrm{diag}(R)$. Let $V_{n-1}$ be the $(n-1)\times(n-1)$ submatrix obtained by deleting the last row and last column of $V$. Since

• $2\ \textrm{diag}(V)-V$ is positive semidefinte (as it is diagonally dominant),
• $\textrm{diag}(R)$ is positive semidefinite,
• $2\ \textrm{diag}(V_{n-1})-V_{n-1}$ is positive definite (as it is strictly diagonally dominant) and
• the last diagonal entry of $R$ is positive,

we see that $A-B$ is positive definite. Now, one result in matrix theory says that if $A$ is a positive definite matrix and $B$ is positive semidefinite, then $A-B$ is positive definite if and only if $\rho(A^{-1}B)<1$ (here $\rho(\cdot)$ denotes spectral radius). See, e.g., Theorem 7.7.3 (p.471) of Horn and Johnson (1985), Matrix Analysis, Cambridge University Press, New York. So we have $\rho(A^{-1}B) = \rho\left(\frac12L^{-1}K\right) < 1$. Hence the spectrum of $L^{-1}K$ lies inside $[0,2)$ in this case.

Edit: By a similar argument to the four bullet points in the above, we see that when $R$ has at least one positive diagonal entry, $K = V+\textrm{diag}(R)$ is actually positive definite. Hence the first paragraph of this answer shows that $L^{-1}K$ has positive eigenvalues. In other words, the spectrum of $L^{-1}K$ lies inside $(0,2)$, not just $[0,2)$.

-


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# Multiplicity-one theorem

Multiplicity-one theorem In the mathematical theory of automorphic representations, a multiplicity-one theorem is a result about the representation theory of an adelic reductive algebraic group. The multiplicity in question is the number of times a given abstract group representation is realised in a certain space, of square-integrable functions, given in a concrete way.

A multiplicity one theorem may also refer to a result about the restriction of a representation of a group G to a subgroup H. In that context, the pair (G, H) is called a strong Gelfand pair.

Contents 1 Definition 2 Results 3 Strong multiplicity one theorem 4 References Definition Let G be a reductive algebraic group over a number field K and let A denote the adeles of K. Let Z denote the centre of G and let ω be a continuous unitary character from Z(K)Z(A)× to C×. Let L20(G(K)/G(A), ω) denote the space of cusp forms with central character ω on G(A). This space decomposes into a direct sum of Hilbert spaces {displaystyle L_{0}^{2}(G(K)backslash G(mathbf {A} ),omega )={widehat {bigoplus }}_{(pi ,V_{pi })}m_{pi }V_{pi }} where the sum is over irreducible subrepresentations and mπ are non-negative integers.

The group of adelic points of G, G(A), is said to satisfy the multiplicity-one property if any smooth irreducible admissible representation of G(A) occurs with multiplicity at most one in the space of cusp forms of central character ω, i.e. mπ is 0 or 1 for all such π.

Results The fact that the general linear group, GL(n), has the multiplicity-one property was proved by Jacquet & Langlands (1970) for n = 2 and independently by Piatetski-Shapiro (1979) and Shalika (1974) for n > 2 using the uniqueness of the Whittaker model. Multiplicity-one also holds for SL(2), but not for SL(n) for n > 2 (Blasius 1994).

Strong multiplicity one theorem The strong multiplicity one theorem of Piatetski-Shapiro (1979) and Jacquet and Shalika (1981a, 1981b) states that two cuspidal automorphic representations of the general linear group are isomorphic if their local components are isomorphic for all but a finite number of places.

References Blasius, Don (1994), "On multiplicities for SL(n)", Israel Journal of Mathematics, 88 (1): 237–251, doi:10.1007/BF02937513, ISSN 0021-2172, MR 1303497 Cogdell, James W. (2004), "Lectures on L-functions, converse theorems, and functoriality for GLn", in Cogdell, James W.; Kim, Henry H.; Murty, Maruti Ram (eds.), Lectures on automorphic L-functions, Fields Inst. Monogr., vol. 20, Providence, R.I.: American Mathematical Society, pp. 1–96, ISBN 978-0-8218-3516-6, MR 2071506 Jacquet, Hervé; Langlands, Robert (1970), Automorphic forms on GL(2), Lecture Notes in Mathematics, vol. 114, Springer-Verlag Jacquet, H.; Shalika, J. A. (1981a), "On Euler products and the classification of automorphic representations. I" (PDF), American Journal of Mathematics, 103 (3): 499–558, doi:10.2307/2374103, ISSN 0002-9327, JSTOR 2374103, MR 0618323, retrieved 2021-08-06 Jacquet, H.; Shalika, J. A. (1981b), "On Euler products and the classification of automorphic forms. II" (PDF), American Journal of Mathematics, 103 (4): 777–815, doi:10.2307/2374050, ISSN 0002-9327, JSTOR 2374050, MR 0618323, retrieved 2021-08-06 Piatetski-Shapiro, I. I. (1979), "Multiplicity one theorems", in Borel, Armand; Casselman., W. (eds.), Automorphic forms, representations and L-functions (Proc. Sympos. Pure Math., Oregon State Univ., Corvallis, Ore., 1977), Part 1, Proc. Sympos. Pure Math., XXXIII, Providence, R.I.: American Mathematical Society, pp. 209–212, ISBN 978-0-8218-1435-2, MR 0546599 Shalika, J. A. (1974), "The multiplicity one theorem for GLn", Annals of Mathematics, Second Series, 100: 171–193, doi:10.2307/1971071, ISSN 0003-486X, JSTOR 1971071, MR 0348047 Categories: Representation theory of groupsAutomorphic formsTheorems in number theoryTheorems in representation theory

Si quieres conocer otros artículos parecidos a Multiplicity-one theorem puedes visitar la categoría Automorphic forms.

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# The probabilistic method to find out a matrix is MDS

Suppose that $M$ is a matrix of order $n$ such that entries of matrix $M$ com from finite field $GF(2^n)$. The matrix $M$ is called MDS (Maximum Distance Separable) matrix if and only if every sub-matrix of $M$ is non-singular over $GF(2^n)$.

For a matrix of order $n$, we should obtain $\sum_{i=1}^n \, {n \choose i }^2$ determinant over $GF(2^n)$, to find out that a matrix is MDS or not. So, when the order of a matrix is less than $10$, we can use the mentioned definition to check that is it MDS matrix or not. But for large order of matrix, this definition is not applicable.

My question:

Is there a probabilistic method to find out a matrix is MDS or not MDS.

I would be appreciate for any suggestions.

Edition1:

With searching, I found one class of MDS matrices over field of real numbers. Consider the following companion matrix in the following form

$$C_n=\left( \begin{array}{cccccc} 0 &1 &0 &\cdots &\cdots &0 \\ 0 &0 &1 &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &\vdots \\ \vdots &\ddots &\ddots &\ddots &\ddots &0 \\ 0 &\cdots &\cdots &0&0 &1 \\ 1 &1 &1 &\cdots &1 &1 \end{array} \right)_{n\times n}$$

The $n\times (n-1)$ power of matrix $C_n$ is a MDS matrix and the matrices $C_n^i$, $1\leq i \leq n^2-n-1$ are not MDS matrices. For example, assume the matrix $C_3$ as shown

$$C_3= \left( \begin {array}{ccc} 0&1&0\\ 0&0&1\\ 1&1&1 \end {array} \right)$$

The matrices $C_3^i$, $1\leq i \leq 5$ are not MDS matrix but the matrix $C_3^6$ as follows

$$C_3^6= \left( \begin {array}{ccc} 4&6&7\\ 7&11&13\\ 13&20&24 \end {array} \right)$$ is a MDS matrix. It means, the entries of $C_3^6$ are non-zero and the determinant of $C_3^6$ is $1$. In addition all square sub-matrices of order $2$ of matrix $C_3^6$ have non-zero determinant.

Edition2:

As an answer to @Daniel McLaury comment, you can check that the matrices $C_3^j$, $15 \leq j \leq 18$, are not MDS. For example, the matrix $C_3^{17}$ is as follows

$$C_3^{17}= \left[ \begin {array}{ccc} 3136&4841&5768\\ 5768& 8904&10609\\ 10609&16377&19513 \end {array} \right]$$

the determinant of the following two sub-matrices of $C_3^{17}$ are zero

$$\begin{array}{ccc} \left[ \begin {array}{cc} 5768&8904\\ 10609&16377 \end {array} \right] &, & \left[ \begin {array}{cc} 3136&5768\\ 5768&10609 \end {array} \right] \end{array}$$

In fact, design of MDS matrix is on the finite field. I just made this example in the field of real numbers to clarify my question. You can see the original form of my question with example, in my Mathover.

• Given that 100% of matrices have this property, I don't know if probabilistic algorithms are likely to give you any further information. – Daniel McLaury Jan 1 '17 at 23:18
• @DanielMcLaury Please see the new edition. – Amin235 Jan 2 '17 at 7:49
• I see. I think it just confuses things to do everything over $\mathbb{R}$ and then mention that you're actually interested in finite fields at the end. I don't specifically know if it matters, but are you trying to do this over every finite field, or only $\mathbb{F}_p$, or only $\mathbb{F}_{2^n}$ or something? – Daniel McLaury Jan 2 '17 at 14:53
• @DanielMcLaury In cryptography, just $GF(2^n)$ can be used. I made an example in the field of real numbers because your first comment was absolutely wrong. I do not know why do you said "100% of matrices have this property:! – Amin235 Jan 2 '17 at 16:44
• If you want the short answer: because the set of non-MDS $m \times n$ matrices is a union of finitely many hypersurfaces, thus it has dimension $m n - 1$, and a reasonable probability measure on an $mn$-dimensional space shouldn't assign a positive probability to anything with dimension less than $mn$. – Daniel McLaury Jan 2 '17 at 17:21


text:
## Saturday, June 28, 2014

### ICPC Problem L: Wires

While this problem wasn't too conceptually difficult, it requires a lot of code (my solution is about 400 lines), and careful implementation of a number of geometric algorithms. A good chunk of the code comes from implementing a rational number class in order to precisely represent the intersection points of the wires. It is also very easy to suffer from overflow: I spent a long time banging my head against an assertion failure on the server until I upgraded my rational number class to use 128-bit integers everywhere, instead of just for comparisons.

The wires will divide the space up into connected regions. The regions can be represented in a planar graph, with edges between regions that share an edge. The problem is then to find the shortest path between the regions containing the two new end-points.

My solution works in a number of steps:
1. Find all the intersection points between wires, and the segments between intersection points. This just tests every wire against every other wire. The case of two parallel wires sharing an endpoint needs to be handled carefully. For each line, I sort the intersection points along the line. I used a dot produce for this, which is where my rational number class overflowed, but would probably have been safer to just sort lexicographically. More than two lines can meet at an intersection point, so I used a std::map to assign a unique ID to each intersection point (I'll call them vertices from here on).
2. Once the intersection points along a line have been sorted, one can identify the segments connecting them. I create two copies of each segment, one in each direction. With each vertex A I store a list of all segments A->B. Each pair is stored contiguously so that it is trivial to find its partner. Each segment is considered to belong to the region to its left as one travels A->B.
3. The segments emanating from each vertex are sorted by angle. These comparisons could easily cause overflows again, but one can use a handy trick: instead of using the vector for the segment in an angle comparison, one can use the vector for the entire wire. It has identical direction but has small integer coordinates.
4. Using the sorted lists from the previous step, each segment is given a pointer to its following segment from the same region. In other words, if one is tracing the boundary of the region and one has just traced A->B, the pointer will point to B->C.
5. I extract the contours of the regions. A region typically consists of an outer contour and optionally some holes. The outermost region lacks an outer contour (one could add a big square if one needed to, but I didn't). A contour is found by following the next pointers. A case that turns out to be inconvenient later is that some segments might be part of the contour but not enclose any area. This can make a contour disappear completely, in which case it is discarded. Any remaining contours have the property that two adjacent segments are not dual to each other, although it is still possible to both sides of an edge to belong to the same contour.
6. Each contour is identified as an outer contour or a hole. With integer coordinates I could just measure the signed area of the polygon, but that gets nasty with rational coordinates. Instead, I pick the lexicographically smallest vertex in the contour and examine the angle between the two incident segments (this is why it is important that there is a non-trivial angle between them). I also sort the contours by this lexicographically smallest vertex, which causes any contour to sort before any other contours it encloses.
7. For each segment I add an edge of weight 1 from its containing region to the containing region of its dual.
8. For each hole, I search backwards through the other contours to find the smallest non-hole that contains it. I take one vertex of the hole and do a point-in-polygon test. Once again, some care is needed to avoid overflows, and using the vectors for the original wires proves useful. One could then associate the outer contour and the holes it contains into a single region object, but instead I just added an edge to the graph to join them with weight 0. In other words, one can travel from the boundary of a region to the outside of a hole at no cost.
9. Finally, I identify the regions containing the endpoints of the new wire, using the same search as in the previous step.
After all this, we still need to implement a shortest path search - but by this point that seems almost trivial in comparison.

What is the complexity? There can be $$O(M^2)$$ intersections and hence also $$O(M^2)$$ contours, but only $$O(M)$$ of them can be holes (because two holes cannot be part of the same connected component). The slowest part is the fairly straightforward point-in-polygon test which tests each hole against each non-hole segment, giving $$O(M^3)$$ time. There are faster algorithms for doing point location queries, so it is probably theoretically possible to reduce this to $$O(M^2\log N)$$ or even $$O(M^2)$$, but certainly not necessary for this problem.


text:
# Rigidity of Newton dynamics

Abstract : We study rigidity of rational maps that come from Newton's root finding method for polynomials of arbitrary degrees. We establish dynamical rigidity of these maps: each point in the Julia set of a Newton map is either rigid (i.e. its orbit can be distinguished in combinatorial terms from all other orbits), or the orbit of this point eventually lands in the filled-in Julia set of a polynomial-like restriction of the original map. As a corollary, we show that the Julia sets of Newton maps in many non-trivial cases are locally connected; in particular, every cubic Newton map without Siegel points has locally connected Julia set. In the parameter space of Newton maps of arbitrary degree we obtain the following rigidity result: any two combinatorially equivalent Newton maps are quasiconformally conjugate in a neighborhood of their Julia sets provided that they either non-renormalizable, or they are both renormalizable in the same way''. Our main tool is the concept of complex box mappings due to Kozlovski, Shen, van Strien; we also extend a dynamical rigidity result for such mappings so as to include irrationally indifferent or renormalizable situations.
Document type :
Preprints, Working Papers, ...
Complete list of metadata

https://hal.archives-ouvertes.fr/hal-03285952
Contributor : Kostiantyn Drach <>
Submitted on : Tuesday, July 13, 2021 - 4:55:37 PM
Last modification on : Wednesday, July 14, 2021 - 3:35:37 AM

### Identifiers

• HAL Id : hal-03285952, version 1
• ARXIV : 1812.11919

### Citation

Kostiantyn Drach, Dierk Schleicher. Rigidity of Newton dynamics. 2021. ⟨hal-03285952⟩

Record views


text:
# How to Create Diagonal Lines with CSS

A few days ago, I received my invite to Google Music. While browsing the app, I noticed a tiny, but neat trick they use to create tabs and diagonal borders with plain-old CSS. I'll show you how to do the same in your projects today!

## Step 1: The Markup

We begin with some simple markup.

We'll keep this demo as simple as we possibly can.

## Step 2

Next, only for the layout, let's apply a background color and a bit of spacing to the body element.

## Step 3

Now, we'll style the anchor tag somewhat -- apply a new color, make it bold, add some borders, etc.

Notice how, when you set large border widths, at the point the two intersect, it creates a diagonal line? I wonder if we can use that intersection as a tab edge? Let's try reducing the height to zero.

Getting closer! Maybe if we now adjust the line-height of the anchor tag, we can make the text fit into that box!

## Step 4

We're getting there! But, certainly, we don't need that ugly red border on the right. Is there a way to specify that it should receive the body's background color without restating the hex value? Yep! We use the transparent keyword for this purpose.

## Step 5

The obvious issue, at this point, is that it looks a bit odd, when the background stretches for the entire width of the container. The natural instinct might be to update the display to inline. However, as the image below shows, that won't work.

We need the element to retain its block-like nature in order for it to honor the 0 height. The answer to this dilemma is to use display: inline-block;, or to float the anchor tag, and then clear the elements that follow it. We'll use the former.

Much better!

## Step 6

Lastly, let's style the div below it a bit, and we'll be just about finished!

In Safari
In Firefox

Uh oh. We have a problem. It looks like Firefox and Webkit don't exactly agree on how to render elements with zero heights and displays of inline-block.

On a casual note, this one had me a bit stumped. Without resorting to hacks, I couldn't get Firefox and Chrome to render the layout exactly. If you find a way, let me know in the comments! Update - refer to the comments for a work-around.

## Step 7

There are ways to target Webkit browsers with CSS, but it's important for me to note that tricks like this should only be used in situations of last resort. In other words, don't try this at home kiddos (unless you have to). In our situation, it appears that it would be smarter to use floats to accomplish this layout. Nonetheless, let's be dangerous and experiment with some alternate techniques.

Many aren't aware that media queries can be used to target Webkit browsers. Watch what happens when we use a webkit-specific property as the condition for the media query...

Ta da! It works. Remember though - only reach for this trick if you have no other option!

## Final Product

It's a simple enough technique, but, nonetheless, one we should all have in our tool belts. Have you used any neat tricks like this in your projects? Let me know in the comments!


text:
In a previous article, we saw how easy it was to get up and running with Rust on the TI Tiva/Stellaris launchpad development boards. In this post, we will write a small Rust program which generates a random stream of 1’s and 0’s and use it to randomly blink an LED.

## The Linear Feedback Shift Register

An LFSR is a shift register whose input bit is a function of some of the bits already present in the register. This wikipedia article describes the idea in detail. A common use of the LFSR is to generate pseudo-random bit sequences.

Here is a C program for a 16 bit Fibonacci LFSR (taken from the Wikipedia article):

# include <stdint.h>
int main(void)
{
uint16_t start_state = 0xACE1u;  /* Any nonzero start state will work. */
uint16_t lfsr = start_state;
uint16_t bit;                    /* Must be 16bit to allow bit<<15 later in the code */
unsigned period = 0;

do
{
/* taps: 16 14 13 11; feedback polynomial: x^16 + x^14 + x^13 + x^11 + 1 */
bit  = ((lfsr >> 0) ^ (lfsr >> 2) ^ (lfsr >> 3) ^ (lfsr >> 5) ) & 1;
lfsr =  (lfsr >> 1) | (bit << 15);
++period;
} while (lfsr != start_state);

return 0;
}

Let’s convert this to an iterator in Rust:

struct Lfsr {
start: u16,
}

impl Iterator for Lfsr {
type Item = u16;

fn next(&mut self) -> Option<Self::Item> {
let bit  = ((self.start >> 0) ^
(self.start >> 2) ^
(self.start >> 3) ^
(self.start >> 5)) & 1;

self.start =  (self.start >> 1) | (bit << 15);
Some(bit)
}
}

fn new_lfsr(n: u16) -> Lfsr {
Lfsr { start: n }
}

It’s easy to plug this into a main function:

fn main() {
portf_init();
let led = red_led();

let l = new_lfsr(0x1234);

for bit in l {
if bit == 0 {
led.off();
} else {
led.on();
}
led::delay(10000);
}

}

$git clone http://github.com/pcein/rust-lfsr$ cd rust-lfsr
$make debug; make flash$


text:
# Derivation of magnetic field of a Solenoid: Biot savart law

## Main Question or Discussion Point

Hello,

I have seen that biot savart's law works for infinitely narrow wires:
"The formulations given above work well when the current can be approximated as running through an infinitely-narrow wire."

When I wanted to derive the magnetic field of a solenoid, I had to do this substitution:
$n_o = N/L$

$k = n_o dx$
Where k is the number of turns per dx.. But shouldn't K be an integer? so I can substitute it in the formula for circular coils. That means I have infinite number of turns and turn density of something like $\frac{a}{dx}$ where a is an integer.

Is there is something wrong or that this is the idealization that we do to the solenoid? Wouldn't it be way off the correct value?

Page 8, Example 9.

Related Other Physics Topics News on Phys.org
BvU
Homework Helper
2019 Award
But shouldn't K be an integer?
No reason. If you have an 11 cm coil with 10 turns, you have 90.91 turns/m.

Note: never, never ever write something like "$\frac{a}{dx}$ with $a$ finite".
$dx$ is universally seen as an infinitesimal (something that goes to zero). So $\frac{a}{dx}$ does not exist.

No reason. If you have an 11 cm coil with 10 turns, you have 90.91 turns/m.
Don't we idealize a solenoid as a number of circular coils?

https://i.imgur.com/RCO3qcQ.png

If we take a dx piece of this solenoid and treat at as k number of coils,

The equation for a single coil is:
$B = \frac{ u_o i R^2}{2 ( R^2 + x^2)^{\frac{3}{2}}}$
If dx has 2 turns then we multiply by 2, If it has k turns then we multiply by k. But a dx piece cant have a 2.5 or a fraction of a coil (It doesn't make sense), Can it?
That is why I said $k = n_o dx$ has to be an integer.

BvU
Again: do NOT use infinitesimals this way -- it will get you into trouble. A infinitesimal $dz$ piece has zero turns and nevertheless contributes to B with an infinitesimal contribution $dB$ proportional to ${N\over L}$. It is only when you integrate $dB$ over a finite range in $z$ that you get a finite result. The derivation (it's not a proof) in the pdf is just fine.


text:
## Stream: new members

### Topic: maximal finset in finset of finsets

#### Bryan Gin-ge Chen (Sep 13 2018 at 18:36):

Excuse the weird title. I'm trying to prove that a finset of subsets of a finset must contain an element of largest size, without relying on classical choice:

lemma not_empty_has_max_size {E : finset α} {F : finset (finset α)} :
F ⊆ powerset E → F ≠ ∅ → ∃ x ∈ F, ∀ g ∈ F, card g ≤ card x := sorry

Here's my attempt so far, which seems to have taken me down a strange path:

import data.finset

variable α:Type*
variable [decidable_eq α]

open finset

lemma finset.card_one_eq {E : finset α} : card E = 1 ↔ ∃ a, E = finset.singleton a :=
begin
have h : _ := @multiset.card_eq_one α E.val,
unfold card,
rw [iff_congr h, exists_congr],
intro a,
rw [←singleton_val, val_inj],
refl
end

lemma powerset_empty : powerset (∅ : finset α) = finset.singleton ∅ := rfl

lemma subset_singleton {F : finset (finset α)} :
F ⊆ finset.singleton (∅:finset α) ↔
F = ∅ ∨ F = finset.singleton ∅ :=
begin
split,
intro h,
by_cases h1 : F = ∅,
exact or.inl h1,
have : F = finset.singleton ∅ := begin
sorry
end,
sorry,
intro h,
exact or.elim h (begin
intro h1,
rw h1,
simp
end) (begin
intro h1,
rw h1,
simp
end)
end

lemma not_empty_has_max_size {E : finset α} {F : finset (finset α)} :
F ⊆ powerset E → F ≠ ∅ → ∃ x ∈ F, ∀ g ∈ F, card g ≤ card x :=
begin
intros FPE Fne,
induction E using finset.induction with a E ha hE,
rw [powerset_empty, subset_singleton] at FPE,
exact or.elim FPE (by contradiction) begin
intro h,
exact exists.intro ∅ begin
rw h,
exact exists.intro (mem_singleton_self ∅) (begin
intros g hg,
rw mem_singleton at hg,
rw hg
end)
end
end,
by_cases hh1 : F ⊆ powerset E,
exact hE hh1,
sorry
end

I'm hoping someone can point me in a smarter direction. Thanks as always!

#### Reid Barton (Sep 13 2018 at 18:39):

Doesn't every finset of finsets contain an element of largest size, regardless of E and the associated condition?

#### Bryan Gin-ge Chen (Sep 13 2018 at 18:41):

Maybe, I got worried that without having some kind of bound on the size of the finsets I wouldn't be able to prove it without choice.

#### Kevin Buzzard (Sep 13 2018 at 18:44):

just use finset.map (I'm assuming this exists!) on finset.card to get a finset of nats and then take the max and work your way back?

#### Reid Barton (Sep 13 2018 at 18:45):

A finset is basically a list, and you could write a program that takes a list of lists and returns one of the maximum length, so there should be no trouble proving it constructively.

#### Reid Barton (Sep 13 2018 at 18:47):

Yeah, I was looking for a function on finset which is max after applying a function, but I guess since nat has decidable equality you can use finset.map in this case.

#### Reid Barton (Sep 13 2018 at 18:47):

Sorry, not finset.map but finset.image.

#### Chris Hughes (Sep 13 2018 at 18:48):

lemma not_empty_has_max_size {E : finset α} {F : finset (finset α)} :
F ⊆ powerset E → F ≠ ∅ → ∃ x ∈ F, ∀ g ∈ F, card g ≤ card x :=
λ _ h, let ⟨n, hn⟩ := (max_of_ne_empty (mt image_eq_empty.1 h) : ∃ a, a ∈ finset.max (F.image card)) in
let ⟨x, hx₁, hx₂⟩ := mem_image.1 (mem_of_max hn) in
⟨x, hx₁, hx₂.symm ▸ λ g hg, le_max_of_mem (mem_image.2 ⟨g, hg, rfl⟩) hn⟩

Wow, thanks!

#### Mario Carneiro (Sep 13 2018 at 18:49):

That uses decidable_eq, doesn't it?

rofl

#### Mario Carneiro (Sep 13 2018 at 18:50):

My suggestion:

import data.finset

section
variables {α : Type*} (r : α → α → Prop) [is_preorder α r]
local infix  ≼ :50 := r

theorem list_zorn : ∀ l : list α, l ≠ [] → ∃ a ∈ l, ∀ b, a ≼ b → b ≼ a := sorry
theorem finset_zorn : ∀ s : finset α, s ≠ ∅ → ∃ a ∈ s, ∀ b, a ≼ b → b ≼ a := sorry

end

#### Bryan Gin-ge Chen (Sep 13 2018 at 18:50):

I would have never thought of using finset.image. And yes, decidable_eq is allowed.

#### Mario Carneiro (Sep 13 2018 at 18:50):

here the relation is card s <= card t on finsets

#### Kenny Lau (Sep 13 2018 at 18:51):

lemma not_empty_has_max_size {E : finset α} {F : finset (finset α)} :
F ⊆ powerset E → F ≠ ∅ → ∃ x ∈ F, ∀ g ∈ F, card g ≤ card x :=
λ _ h, let ⟨n, hn⟩ := (max_of_ne_empty (mt image_eq_empty.1 h) : ∃ a, a ∈ finset.max (F.image card)) in
let ⟨x, hx₁, hx₂⟩ := mem_image.1 (mem_of_max hn) in
⟨x, hx₁, hx₂.symm ▸ λ g hg, le_max_of_mem (mem_image.2 ⟨g, hg, rfl⟩) hn⟩

what is the role of E in your theorem?

#### Kevin Buzzard (Sep 13 2018 at 18:51):

you can just prove this using the zorn in mathlib

#### Mario Carneiro (Sep 13 2018 at 18:51):

of course, but this (1) doesn't use nonconstructive axioms and (2) doesn't need the chain condition

#### Bryan Gin-ge Chen (Sep 13 2018 at 18:52):

If E is unnecessary here all the better. In the "application", it's the ground set of a matroid.

#### Reid Barton (Sep 13 2018 at 18:52):

decidable_eq nat doesn't use axioms either

#### Mario Carneiro (Sep 13 2018 at 18:53):

yes, but the theorem can be proven constructively without that assumption

#### Reid Barton (Sep 13 2018 at 18:53):

Yeah, the max/min stuff should be on multisets.

#### Mario Carneiro (Sep 13 2018 at 18:54):

Unfortunately, here max/min don't work directly since it's not a poset

#### Kenny Lau (Sep 13 2018 at 18:54):

you can have zorn on a set (i.e. type) S as long as you have a choice function on P(P(S))\{{}}, right

#### Mario Carneiro (Sep 13 2018 at 18:54):

yes, but we don't have that here and I claim it's still provable

sure

#### Reid Barton (Sep 13 2018 at 18:55):

I mean we should just define max for a set in the same style as prod, in terms of max on multiset, and nat is already a decidable linear whatever so there would be no problem there.

protected def prod [comm_monoid β] (s : finset α) (f : α → β) : β := (s.1.map f).prod

#### Mario Carneiro (Sep 13 2018 at 18:56):

what is this about nat? I don't see it in the question

#### Reid Barton (Sep 13 2018 at 18:56):

nat is the codomain of finset.card (I assume)

#### Mario Carneiro (Sep 13 2018 at 18:57):

sure but that doesn't really matter; we are doing the choosing on finset A which is not decidable (unless you assume so)

#### Mario Carneiro (Sep 13 2018 at 18:58):

all that decidability of nat gives you is decidability of the relation ≼ given by s ≼ t ↔ card s ≤ card t

#### Kenny Lau (Sep 13 2018 at 18:58):

I guess Kevin ran away from all this decidable nonsense :P

#### Reid Barton (Sep 13 2018 at 18:59):

I'm a bit confused.
All I am saying is that the original thing should be a direct conclusion of some lemma about max of a multiset nat, applied to F.1.map finset.card.

#### Reid Barton (Sep 13 2018 at 19:00):

That lemma being the analogue for multisets of finset.max_of_ne_empty and finset.le_max_of_mem

#### Reid Barton (Sep 13 2018 at 19:03):

But I guess maybe decidable_linear_order is stronger than decidable_eq anyways, so it doesn't really matter.

#### Mario Carneiro (Sep 13 2018 at 19:07):

the point is this isn't a question about decidable linear orders, but decidable preorders. Lack of uniqueness of the maximum makes it hard to define a function like max directly

#### Reid Barton (Sep 13 2018 at 19:08):

You mean something like a "pre-total order"?

#### Mario Carneiro (Sep 13 2018 at 19:13):

is_total_preorder exists in core btw

#### Mario Carneiro (Sep 13 2018 at 19:13):

but totality is not needed

#### Reid Barton (Sep 13 2018 at 19:14):

Finding the maximum value of a function on a finite set is a pretty standard thing to do. Of course there might not be a unique element of the set which maximizes the function, but the maximum value of the function is still well-defined. And often you only need some statement about the existence of a maximizing element.
It's not very obvious how you're supposed to get this from only the ability to take the maximum element of a finite set (in a totally ordered type).

#### Mario Carneiro (Sep 13 2018 at 19:15):

like kenny and chris have shown, you can take the image of the finset and get the maximum

#### Mario Carneiro (Sep 13 2018 at 19:16):

unless you care about constructivity and then you have to prove it from scratch

#### Mario Carneiro (Sep 13 2018 at 19:17):

that works in any total preorder

#### Mario Carneiro (Sep 13 2018 at 19:17):

in a partial order, I don't see any easy way to get it except zorn's lemma

#### Kenny Lau (Sep 13 2018 at 19:22):

like kenny and chris have shown, you can take the image of the finset and get the maximum

I don't think I did anything

#### Bryan Gin-ge Chen (Sep 13 2018 at 19:23):

I care a little bit about constructivity, but maybe I'll see how far I can get with Chris's solution first.

#### Mario Carneiro (Sep 13 2018 at 19:56):

oops, I didn't see that you just quoted chris's solution

Last updated: May 11 2021 at 23:11 UTC


text:
## Calculus 8th Edition

The cosine function is continuous and decreasing when $x\in[0,\pi]$. This means that $h(x)=1+\cos x$ is also continuous and decreasing on $[0,\pi]$. So, for every $x_1,x_2\in[0,\pi]$ (which is the domain) $x_1h(x_2)$ and if $x_2$ is greater then $h(x_2)>h(x_1)$.


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00-4   Berichtsreihe des Mathematischen Seminars der Universität Kiel

## Ramsey and Freeness Properties of Polish Planes

Suppose that $X$ is a Polish space which is not $\sigma$-compact. We prove that for every Borel colouring of $X^2$ by countably many colours, there exists a monochromatic rectangle with both sides closed and not $\sigma$-compact. Moreover, every Borel colouring of $[X]^2$ by finitely many colours has a homogeneous set which is closed and not $\sigma$-compact. We also show that every Borel measurable function $f:X^2 \rightarrow X$ has a free set which is closed and not $\sigma$-compact. As corollaries of the proofs we obtain that, firstly, the product forcing of two copies of superperfect tree forcing does not add a Cohen real, and, secondly, that it is consistent with ZFC to have a closed subset of the Baire space which is not $\sigma$-compact and has the property that for any three of its elements, none of them is constructible from the other two. A similar proof shows that it is consistent to have a Laver tree such that none of its branches is constructible from any other branch. The last four results answer questions of Goldstern and Brendle (see [2, Questions 6.7, 6.8, 6.10]).

Mathematics Subject Classification (1991): 03E15, 26B99, 54H05

Bibliographical note: Proceedings of the London Mathematical Society (3) 82 (2001), 31-63.

Keywords: Polish space, Baire space, Borel function, homogeneous set, free set, Cohen forcing, Miller forcing, Laver forcing, constructibility


text:
## Learn to Use ITensor

main / itensor_conventions

# Tensor and Index Conventions

## Arrows

• IQIndex Arrows (represented by the Arrow enum) have the values In or Out and should not be thought of as pointing in an absolute direction such as up/down/left/right.

• Conceptually Arrows represent the flow or flux of quantum numbers out of a tensor. Equivalently, they specify whether a particular index transforms as a ket (a vector, or raised index in classic tensor notation) or a bra (a covector, dual vector, or lowered index).

• An Out Arrow means the index transforms as a ket (for example, a site of an MPS), an In Arrow means an index transforms as a bra.

• Conjugating a tensor (applying dag(T)) reverses Arrow directions in addition to taking the complex conjugate.

## Index Types

• The two primary index types (represented by the IndexType enum) are Site and Link.

• Site indices are physical, or real, indices such as the sites of a lattice model.

• Link indices (sometimes called virtual indices) label internal, or gauge, degrees of freedom of a tensor network. For example, the bonds of a matrix product state are Link indices.

• There is a special IndexType All which can be used to specify that a certain function or transformation should be applied to all indices regardless of type.

## Matrix Product States

• An orthogonality center of a matrix product state or MPS is any tensor whose indices all label orthonormal states. For example, every site tensor of a canonical MPS is an orthogonality center. During a conventional DMRG calculation, where an MPS is in the so-called mixed canonical gauge, only one tensor (the left or right of the two "center" sites depending on sweep direction) is an ortho center.

• All Arrows (including both Link and Site indices) flow Out from the orthogonality center of an MPS. For a normalized MPS in a well-defined gauge, therefore, a tensor is an ortho. center if and only if all its indices point Out.

## Operators

• Single-site operator tensors have one unprimed Site index S and one primed Site Index S'. For IQTensor operators, IQIndex S has an In Arrow and S' an Out Arrow.

• Each tensor of a matrix product operator (MPO) follows the same convention as a single-site operator above.

• Unless an MPO is in a definite gauge (typically not the case), Link arrows should flow out from the first site tensor.

Back to Main


text:
Charge 1 is positive and is located at the origin and charge 2 is negative. The...

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show work

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An experiment is conducted to compare the maximum load capacity tons (the maximum weight that can be tolerated without breaking_ for Iwo alloys = A and B is known that the two standard deviations load capacity are equal at tons each: The experiment is conducted on 50 specimens of each alloy (A and B...
The Dubs division of Fast Company (the parent company) produces wheels for off-road sport vehicles. One-half...
The Dubs division of Fast Company (the parent company) produces wheels for off-road sport vehicles. One-half of Dub's output is sold to the Hoon division of Fast; the remainder is sold to outside customers. Dub's estimated operating profit for the year is shown in the table. Internal Sa...
Please don't copy another solution. Make it original! If the significance level of a hypothesis test...
Please don't copy another solution. Make it original! If the significance level of a hypothesis test is increased from ɑ = 0.05 to ɑ = 0.10, a. Will the probability of a Type I error increase, decrease, or stay the same? b. Why?...
QUESTION 19 For an independent-measures research study, what is measured by Cohen's dori?? a. Whether or...
QUESTION 19 For an independent-measures research study, what is measured by Cohen's dori?? a. Whether or not treatment means are significantly different b. The size of the difference between the two treatements OC. The risk of committing a Type I error d. Whether the difference between the two t...
From the perspective of those who are planning for their retirement, The SECURE Act of 2019...
From the perspective of those who are planning for their retirement, The SECURE Act of 2019 includes many provisions that individuals who are planning for their retirement will find helpful. However, one of the provisions of the SECURE Act will not be helpful to many of these individuals. This provi...
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A canteen is covered with wet canvas. If 15 g of waterevaporates from the canvas and if 50% of the thermal energy used toevaporate the water is supplied by the 400g of water in thecanteen, what is the temperature change of the water in thecanteen? Please explain in details/pictures or diagrams
A canteen is covered with wet canvas. If 15 g of water evaporates from the canvas and if 50% of the thermal energy used to evaporate the water is supplied by the 400g of water in the canteen, what is the temperature change of the water in the canteen? Please explain in details/pictures or diagrams...
Find an equation for the ellipse. Graph the equationcenter at ( - 2,2); vertex at ( - 2,10); focus at ( - 2,7)Type the left side of the equation of the ellipseWhich graph shown below is the graph of the ellipse?
Find an equation for the ellipse. Graph the equation center at ( - 2,2); vertex at ( - 2,10); focus at ( - 2,7) Type the left side of the equation of the ellipse Which graph shown below is the graph of the ellipse?...
EXE 23, For Wy P7o, Shww that 4 e-udx 5 {S Conwerdent _ Qnd olculate its Volue 7 6 Sok: Foy each Azo, lo A e-Fdx 12 [-e-"IA Diy!) 1  = - e-m) 15 I6 Hence 17 fe" dx=k Rerik-ky Huce") I7 At A+ p ie. edk Owbgps , Od te Vdr t5
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Question 5 View Policies Show Attempt History Current Attempt in Progress port * Your answer is incorrect. In its first month of operations, Waterway Industries made three purchases of merchandise in the following sequence: (1) 370 units at 54 (2) 470 units at $6, and (3) 570 units at$7. Assuming t...
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Art II (Open Book, Open notes) 42pts Do any two Use the following data to estimate the Ulattice of RbF(s) in kJlmol F(g) +328.0 kJlmol Firg ionizationgpotential of Rb 403 kJlmol First -549.3 kJlmol AH' for RbF(s) 159.0 kJlmol Bond Enthalpy for F_ F (g) 81.0 kJlmol AHsub Rb(s) >Rb(g) BeH(4) where the number Consider the linear moletela into two different fra II. Divide the molecule species: to belong to Dzh: the center orbitals assumed outer orbitals for the outer at the other the normali
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# Why are there no key_frames created with Python code?

Thanks to the answer on my question Python convert curve to object, I can now extrude SVG curves and move and rotate the objects in Python.

Now I want to start with animation in Python, and I was hoping to insert some keyframes in a similar fashion as done in the example below which I found here

import bpy

positions = (0,0,2),(0,1,2),(3,2,1),(3,4,1),(1,2,1)
start_pos = (0,0,0)

ob = bpy.context.active_object

frame_num = 0

for position in positions:
# bpy.context.scene.frame_set(frame_num)
ob.location = position
ob.keyframe_insert(data_path="location", index=-1, frame=frame_num)
frame_num += 10

This works and lets a sphere move around. For my project with movement and rotation I came up with the following code:

import bpy
from math import *

context = bpy.context
scene = context.scene
bpy.ops.object.select_all(action='SELECT')
bpy.ops.import_curve.svg(filepath = "~/tmp/barkruk.svg")
newcurves = [c for c in context.scene.objects if not c.select]
# deselect all
bpy.ops.object.select_all(action='DESELECT')
# create an empty make it the active  object
empty = bpy.data.objects.new("Holder", None) # use filename maybe
scene.objects.active = empty
empty.select = True
# loop thru the curve objects
for c in newcurves:
print("Curve: %s", c.name)
# make it selected
c.select = True
# make the parent the empty if not already set
if not c.parent:
c.parent = empty
# change some curve details
curve = c.data
curve.dimensions = '2D'
curve.extrude = 0.003
# sane object names
c.name = curve.name
# location
#c.location.x += 0.332

empty.scale *= 2

bpy.ops.object.origin_set(type='GEOMETRY_ORIGIN', center='BOUNDS')
scene.objects[1].location.x=-1
scene.objects[3].location.x=1

bpy.context.scene.frame_set(0)
scene.objects[3].keyframe_insert(data_path="rotation_euler", frame=10)
scene.objects[1].keyframe_insert(data_path="rotation_euler", frame=20)
scene.objects[2].location.z=0.12
scene.objects[2].keyframe_insert(data_path="location", frame=30)
scene.objects[1].location.x=0
scene.objects[1].keyframe_insert(data_path="location", frame=40)
scene.objects[3].location.x=0
scene.objects[3].keyframe_insert(data_path="location", frame=50)

But this doesn't do anything more than the answer in my previously asked question. At least I don't see the keyframes appearing on the timeline. Where do I go wrong and how can I generate keyframes?

Each curve object is still in the newcurves collection, hence newcurves[index] can be used rather than scene.objects[index]

The behaviour in the example sphere script you posted appears to rely on obj.location = (x,y,z) being set as opposed to obj.location.x = x etc.

My advice would be call scene.frame_set(f) update your props, insert keyframe, repeat. The frame argument of keyframe_insert defaults to scene.frame_current. Also to keyframe in only the x location pass index=0.

# set initial locs / rots here
newcurves[0].location.x=-1
newcurves[2].location.x=1

bpy.context.scene.frame_set(0)
# keyframe in all at frame 0
for curve in newcurves:
curve.keyframe_insert(data_path="rotation_euler")
curve.keyframe_insert(data_path="location")

# call frame set (updates the scene as well)
scene.frame_set(10)
# update the properties
# insert the keyframes
newcurves[2].keyframe_insert(data_path="rotation_euler")

scene.frame_set(20)
newcurves[0].keyframe_insert(data_path="rotation_euler", frame=20)

scene.frame_set(30)
newcurves[1].location.z=0.12
newcurves[1].keyframe_insert(data_path="location", index=2)

scene.frame_set(40)
newcurves[0].location.x=0
newcurves[0].keyframe_insert(data_path="location", index=0)

scene.frame_set(50)
newcurves[2].location.x=0
newcurves[2].keyframe_insert(data_path="location", index=0)


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# M5-S5: Introduction to Circular Motion

• ### conduct investigations to explain and evaluate, for objects executing uniform circular motion, the relationships that exist between:

- centripetal
- mass
- speed

• Uniform circular motion is defined as when an object travels in a circular motion at constant speed. Hence, it does not experience acceleration in the direction of its rectilinear motion. However, it does experience acceleration in another direction due to centripetal force.

F_c=(mv^2)/r

Fc is the centripetal force of the circular motion, m is the mass (kg) of the object undergoing the motion, v (m/s) is the linear or tangential velocity of the object and r (m) is the radius of this circular motion.

Centripetal force provides the curvature to an object’s circular motion. Its direction is orthogonal (right-angled) to the direction of the object’s motion (velocity). In other words, it is always towards the centre of the circular motion. Centripetal acceleration (which can be calculated by Newton’s second law) is in the same direction as centripetal force.

### Centripetal force is a non-real force

This means that centripetal force always caused by a real force.

For example:

 ·       When an object with mass is attached to a rope and swung in circles, it undergoes circular motion. If the motion has constant velocity, the circulation motion is uniform. The centripetal force in this example is provided by the tension in the rope, which has a direction towards the centre of the circular motion. ·       In the case of a car travelling around a bend, it experiences centripetal force caused by kinetic friction between the car’s tyres and the ground. ·       When a spacecraft or satellite is orbiting the Earth, gravitational force provides the centripetal force for circular motion.

### Relationship between centripetal force, mass, speed and radius

From the formula, we can deduce the following:

• Centripetal force is directly proportional to:
• Mass of the object undergoing circular motion
• Tangential velocity of the object squared
• Centripetal force is inversely proportional to:

Effect on centripetal force

• A stronger centripetal force is required to maintain an object’s uniform circular motion if:
• Its mass increases
• Its tangential velocity increases

 Change Centripetal force required to maintain circular motion Increase in mass Increase in centripetal force Increase in velocity Increase in centripetal force

Changes in centripetal force

• Conversely, changes in centripetal force can impact velocity and radius. Mass does not change.
• If centripetal force increases, velocity needs to increase in order to maintain the radius of circular motion, otherwise the object spirals inwards to reach a new circular path with smaller radius.
• If centripetal force decreases, velocity needs to decrease to maintain the radius of circular motion, otherwise the object spirals outwards to reach a new circular path with larger radius.

 Change Condition Effect Increase in centripetal force Velocity remains constant Radius decreases Increase in centripetal force Velocity increases Radius remains constant Decrease in centripetal force Velocity remains constant Radius increases Decrease in centripetal force Velocity decreases Radius remains constant

## Concept Question 1

A car rounds a bend on a road that follows the arc of a circle with radius r. The car has mass m and is travelling at a velocity v. Explain the following situations:

(a) Why are drivers advised to slow down during wet weather, specifically when they are making a bend.

(b) Assuming the friction between the tyres and the road does not change, describe the path of a car with mass 2m when it rounds the bend at velocity v?

(c) A motor cyclist rounds the same bend at velocity 2v. If the mass of the motorcycle is 0.25m, what would be different about the centripetal force acting on the motorcycle compared to that on the car?

## Concept Question 2

HSC Q30 2013

The diagram shows a futuristic space station designed to simulate gravity in a weightless environment.

(a) Explain how the rotating space station simulates gravity.

(b) Describe the effect of the space station’s rotation speed on the magnitude of centripetal force.

(c) To maintain ‘gravity’, does the space station’s rotation speed need to change if its radius changes? Explain your answer.

(d) Does the mass of the space station influence the rotation speed required to simulate gravity? Justify your answer.

## Concept Questions Solutions

Question 1

(a) During wet weather, the kinetic friction between a car's tyres and the ground is reduced. This means the centripetal force acting on the car during its bend is reduced. As a result, velocity needs to decrease to maintain radius of the curvature.

(b) Since centripetal force remains constant, the radius of curvature is doubled. This means for a car with mass 2m travelling at the same speed v, it requires a greater distance to complete the bend.

(c)

F_c=(0.25m(2v)^2)/r

F_c=(0.25m*4v^2)/r

F_c=(mv^2)/r

As seen above, substituting 0.25m and 2v into the equation F_c=(mv^2)/r will yield a magnitude of centripetal force identical to one with mass m and velocity v. This means the centripetal force acting on the motorcycle remains unchanged and so does its radius.

Question 2

(a) The rotating motion of the spacecraft exerts normal force (centripetal force) on the astronaut. Due to Newton's third law, the astronaut exerts reaction force on the outer perimeter of the spacecraft. The acceleration resulted from this reaction force simulates gravity.

(b) As v increases, the magnitude of centripetal force increases (since radius remains constant). Changes in centripetal force are always proportional to the square of change in velocity.

(c) Yes, to maintain gravity, the magnitude of centripetal force cannot be changed. A decrease in radius needs to be compensated by a reduction in velocity v. This means the rotational speed to simulate gravity is lower for smaller space stations.

(d) No, the mass of the space station does not affect centripetal force nor acceleration. In addition, the mass of the astronaut does not influence the rotational speed required to achieve 1g of gravitational acceleration. This is because simulated gravity is independent of mass: a_c=v^2/r.


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Abstract

In 2007, the government of St. Vincent and the Grenadines enacted legislation Act #25 0f 2006 to introduce the Value Added Tax (VAT). The Value Added Tax is set at a rate of 15 percent and replaced the consumption tax originally placed on imported goods at the Customs and Excise Department. A minimum thresh hold level of EC$60,000/annum was set for an establishment to be vatable, which was later raised to EC$120,000.00. Within the framework of the Value Added Tax are exemptions and zero rated items or commodities which constitute basic and essential food and non-food items, deliberately set as policies of the government in support of the poor. Since the introduction of the VAT, local meat and meat products, eggs and vegetable prices which are vatable items have increased at the supermarket. The percentages increase in price for these vatable items have been substantially higher than similar imported vatable items and the zero rated food products. Moreover, trade statistics have shown increases in the importation of similar products. It is therefore hypothesized, that the application of the VAT on local produce is increasing food prices at the supermarkets causing consumers to purchase cheaper imports at the supermarkets. There is no VAT charged on local produce outside of the supermarkets. Despite this scenario, most consumers have developed preferences for convenient shopping – the one stop shop at the supermarket. Research has shown that over 60 percent of consumers in St. Vincent and the Grenadines prefer to shop at the supermarket compared to the open vegetable market since it is more convenient, quality is usually better and less time consuming. The paper will show that while the VAT is a revenue measure, the revenue collected is infinitesimal, compared to the value of imports. Thus the paper will also show that a burden is being placed on domestic consumers through the reduction of their purchasing power at the supermarkets, and that there are also negative impacts on production, whereby farmers are experiencing difficulties disposing of their produce, since they take a longer time to go through the marketing channels. The paper will argue that the government can remove VAT from local produce sold in the supermarkets and its revenue collection will not be affected negatively. However the removal of the VAT will reduce the prices of the local produce to consumers, hence, making such products more competitive with imported goods. The concerns of preferential treatment of local produce is considered a non-issue since over 80 percent of the imports are from third country where domestic support to agriculture exceeds 40 percent while farmers in St. Vincent and the Grenadines are not provided with any subsidy.


text:
# Is $\left(\frac{a^2}{5}\right)=1$ for all $a$ not divisible by 5?

Note $$\left(\frac{a^2}{5}\right)$$ is the Legendre symbol.

I used wolfram alpha to see if, $$\left(\frac{a^2}{5}\right) = 1$$ and this is true for integers from $$1$$ to $$10$$ and it is except $$5$$ and $$10$$, which are divisible by $$5$$. So is this true in general for any $$a$$ not divisible by 5?

• @zaira That is simply the definition of the symbol. May 1, 2022 at 17:22
• If you base it off the definition $\Big(\frac{a}{p}\Big) = a^{\frac{p-1}{2}}$ mod $p$, then the result follows immediately from Fermat's Little Theorem. May 1, 2022 at 17:23
• The legendre symbol is multiplicative in the upper entry, so yes! May 1, 2022 at 17:29

The expression $$\left(\frac{a^2}{5}\right)$$ asks if there exists a natural number $$x$$ such that $$x^2\equiv a^2 \bmod 5,$$ from the definition of the Legendre Symbol. You simply let $$x=a$$ to satisfy the equation, so such an $$x$$ does exist, that is why Wolfram-Alpha returned a $$1$$ for this Legendre Symbol.
When $$5 \mid x$$, it is clear that $$5 \mid a^2$$, so the Legendre gives $$0$$ by definition.
Euler's criterion tells us that the Legendre symbol is $$\big(\dfrac {a^2}5\big)=(a^2)^\frac{5-1}2\equiv (a^2)^2\equiv a^4\equiv 1\pmod 5,$$ by Fermat's little theorem.
This is true whenever $$(a,5)=1.$$
Alternatively $$\big(\dfrac {a^2}5\big) =\big(\dfrac a5\big) ^2=1.$$


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# Nodal Analysis - Circuit with Dependent Voltage Source

Determine the power of each source after solving the circuit by the nodal analysis.

Answers: $P_{I_x}=0.497W, P_{1A}=-1.806W, P_{2A}=4.254W, P_{3V}=-3.87W,$ and $P_{5V}=-3.552W$

Solution

I. Identify all nodes in the circuit.
The circuit has 6 nodes as highlighted below.
Continue reading

# Nodal Analysis - Supernode

Solve the circuit with nodal analysis and find $I_x$ and $V_y$ .

Solution
1) Identify all nodes in the circuit. Call the number of nodes $N$ .
There are four nodes in the circuit:
Continue reading

# Nodal Analysis Problem with Dependent Voltage and Current Sources

Solve the circuit with the nodal analysis and determine $i_x$ and $V_y$ .

Solution
1) Identify all nodes in the circuit. Call the number of nodes $N$ .
The circuit has 5 nodes. Therefore, $N=5$ .
Continue reading


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Über dieses Buch

In 1821, Augustin-Louis Cauchy (1789-1857) published a textbook, the Cours d’analyse, to accompany his course in analysis at the Ecole Polytechnique. It is one of the most influential mathematics books ever written. Not only did Cauchy provide a workable definition of limits and a means to make them the basis of a rigorous theory of calculus, but he also revitalized the idea that all mathematics could be set on such rigorous foundations. Today, the quality of a work of mathematics is judged in part on the quality of its rigor, and this standard is largely due to the transformation brought about by Cauchy and the Cours d’analyse.

For this translation, the authors have also added commentary, notes, references, and an index.

Inhaltsverzeichnis

Chapter 1. On real functions.

[31] When variable quantities are related to each other such that the value of one of the variables being given one can find the values of all the other variables, we normally consider these various quantities to be expressed by means of the one among them, which therefore takes the name the

independent variable

. The other quantities expressed by means of the independent variable are called

functions

of that variable.

Robert E. Bradley, C. Edward Sandifer

Chapter 2. On infinitely small and infinitely large quantities, and on the continuity of functions. Singular values of functions in various particular cases.

[37] We say that a variable quantity becomes

infinitely small

when its numerical value decreases indefinitely in such a way as to converge towards the limit zero. It is worth remarking on this point that one ought not confuse a constant decrease with an indefinite decrease. The area of a regular polygon circumscribed about a given circle decreases constantly as the number of sides increases, but not indefinitely, because it has as its limit the area of the circle.

Robert E. Bradley, C. Edward Sandifer

Chapter 3. On symmetric functions and alternating functions. The use of these functions for the solution of equations of the first degree in any number of unknowns. On homogeneous functions.

[71] A

symmetric

function of several quantities is one which conserves the same value and the same sign after any exchange made among its quantities.

Robert E. Bradley, C. Edward Sandifer

Chapter 4. Determination of integer functions, when a certain number of particular values are known. Applications.

[83] To determine a function when a certain number of particular values are taken to be known is what we call to

interpolate

. When it is a matter of a function of one or two variables, this function can be considered as the ordinates of a curve or of a surface, and the problem of

interpolation

consists of fixing the general value of this ordinate given a certain number of particular values, that is to say, to make the curve or the surface pass through a certain number of points. This question can be solved in an infinity of ways, and in general the problem of interpolation is indeterminate. However, the indeterminacy will cease if, to the knowledge of the particular values of the desired function, we add the expressed condition that this function be integer, and of a degree such that the number of its terms becomes precisely equal to the number of particular values given.

Robert E. Bradley, C. Edward Sandifer

Chapter 5. Determination of continuous functions of a single variable that satisfy certain conditions.

[98] When, instead of integer functions we imagine any functions, so that we leave the form entirely arbitrary, we can no longer successfully determine them given a certain number of particular values, however large that number might be, but we can sometimes do so in the case where we assume certain general properties of these functions.

Robert E. Bradley, C. Edward Sandifer

Chapter 6. On convergent and divergent series. Rules for the convergence of series. The summation of several convergent series.

[114]We call a series an indefinite sequence of quantities, u0, u1, u2, u3, …, which follow from one to another according to a determined law. These quantities themselves are the various terms of the series under consideration. Let be the sum of the first

n

terms, where

n

denotes any integer number. If, for ever increasing values of

n

, the sum

s

n

indefinitely approaches a certain limit

s

, the series is said to be

convergent

, and the limit in question is called the sum of the series. On the contrary, if the sum

s

n

does not approach any fixed limit as n increases indefinitely, the series is

divergent

, and does not have a sum. In either case, the term which corresponds to the index

n

, that is

u

n

, is what we call the

general term

. For the series to be completely determined, it is enough that we give this general term as a function of the index

n

.

Robert E. Bradley, C. Edward Sandifer

Chapter 7. On imaginary expressions and their moduli.

[153] In analysis, we call a

symbolic expression

or

symbol

any combination of algebraic signs that do not mean anything by themselves or to which we attribute a value different from that which they ought naturally to have. Likewise, we call

symbolic equations

all those that, taking the letters and the interpretations according to the generally established conventions, are inexact or do not make sense, but from which we can deduce exact results by modifying and altering either the equations themselves or the symbols which comprise them, according to fixed rules. The use of symbolic expressions or equations is often a means of simplifying calculations and of writing in a short form results that appear quite complicated. We have already seen this in the second section of the third chapter where formula (9) gives a very simple symbolic value to the unknown x satisfying equations (4).

1

Among those symbolic expressions or equations which are of some importance in analysis, we should distinguish above all those which we call

imaginary

. We are going to show how we can put them to good use.

Robert E. Bradley, C. Edward Sandifer

Chapter 8. On imaginary functions and variables.

When the constants or variables contained in a given function, having been considered real are later supposed to be imaginary, the notation that was used to express the function cannot be retained in the calculation except by virtue of new conventions able to determine the sense of this notation under the new hypotheses. Thus, for example, by virtue of the conventions established in the preceding Chapter, the values of the notations.

Robert E. Bradley, C. Edward Sandifer

Chapter 9. On convergent and divergent imaginary series. Summation of some convergent imaginary series. Notations used to represent imaginary functions that we find by evaluating the sum of such series.

Be the sum of the first

n

terms of this series. Depending on whether or not

s

n

converges towards a fixed limit∈dex{limit} for increasing values of

n

, we say that series (3) is

convergent

and that it has this limit as its sum, or else that it is

divergent

and it does not have a sum. The first case evidently occurs if the two sums.

Robert E. Bradley, C. Edward Sandifer

Chapter 10. On real or imaginary roots of algebraic equations for which the left-hand side is a rational and integer function of one variable. The solution of equations of this kind by algebra or trigonometry.

where

n

represents the degree of this equation and

a

0

,

a

1

,

a

2

, …,

a

n−1

, an, are constant coefficients, real or imaginary. A root of this equation is any expression,real or imaginary, that when substituted in place of the unknown value

x

, makes the left-hand side equal to zero.

Robert E. Bradley, C. Edward Sandifer

Chapter 11. Decomposition of rational fractions.

Imagine now that we separate the function F(

x

) into two factors where the first, instead of being linear, corresponds to several roots of the equation F(

x

) = 0. For example, take for the first factor the factor of second degree.

Robert E. Bradley, C. Edward Sandifer

Chapter 12. On recurrent series.

[321]A series ordered according to the ascending integer powers of the variable

x

, is called

recurrent

when in this series, starting after a given term, the coefficient of any power of the variable is expressed as a linear function of a fixed number of the coefficients of lesser powers, and consequently it suffices to

run back

1

to the values of these last coefficients to deduce the one we are seeking.

Robert E. Bradley, C. Edward Sandifer

Backmatter

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Für produzierende Unternehmen hat sich Product Lifecycle Management in den letzten Jahrzehnten in wachsendem Maße zu einem strategisch wichtigen Ansatz entwickelt. Forciert durch steigende Effektivitäts- und Effizienzanforderungen stellen viele Unternehmen ihre Product Lifecycle Management-Prozesse und -Informationssysteme auf den Prüfstand. Der vorliegende Beitrag beschreibt entlang eines etablierten Analyseframeworks Herausforderungen und Lösungsansätze im Product Lifecycle Management im Konzernumfeld.


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# Which empire's accomplishments include the ishtar gate, the hanging gardens of babylon, and a knowledge of astronomy and medicine?

###### Question:

Which empire's accomplishments include the ishtar gate, the hanging gardens of babylon, and a knowledge of astronomy and medicine?

### The area of a rectangle is 30 square feet. The perimeter is 22 inches. The length is 1 inche longer than the width. What are the length and width of the rectangle

The area of a rectangle is 30 square feet. The perimeter is 22 inches. The length is 1 inche longer than the width. What are the length and width of the rectangle...

### B. Kay does not want to lose money on her business. Her mother told her she needs to sell enough ornaments to at least cover her expenses (costs for materials and table rental). Kay figures that if she sells 8 ornaments, she covers her expenses and does not lose any money. Do you agree? Explain and show work to support your answer.

b. Kay does not want to lose money on her business. Her mother told her she needs to sell enough ornaments to at least cover her expenses (costs for materials and table rental). Kay figures that if she sells 8 ornaments, she covers her expenses and does not lose any money. Do you agree? Explain and ...

### Plz Help!! With 5,6,7,8

Plz Help!! With 5,6,7,8...

### How does an ion differ from an electrically-neutral atom? A) There are a different number of protons in an ion compared to a neutral atom. B) There are a different number of electrons in an ion compared to a neutral atom. C) There a different number of neutrons in an ion compared to a neutral atom. D) There are no differences in subatomic particles.

How does an ion differ from an electrically-neutral atom? A) There are a different number of protons in an ion compared to a neutral atom. B) There are a different number of electrons in an ion compared to a neutral atom. C) There a different number of neutrons in an ion compared to a neutral...

### What did the No Child Left Behind Act mandate?

What did the No Child Left Behind Act mandate?...

### What was the Suffolk Resolves?

What was the Suffolk Resolves?...

### They are the most animals that eats other animals

they are the most animals that eats other animals...

### To Kill a Mocking BirdIn a few sentences, explain how you feel about the people of Maycomb county based on Scout's description in chapter 16. Include the name Dolphus Raymond in your response.

To Kill a Mocking BirdIn a few sentences, explain how you feel about the people of Maycomb county based on Scout's description in chapter 16. Include the name Dolphus Raymond in your response....

### An animal cell placed in a hypotonic solution will: •take on water •loose water •divide •die

An animal cell placed in a hypotonic solution will: •take on water •loose water •divide •die...

### Tom wants to order tickets online so that he and three of his friends can go together to a water park. The cost of the tickets is $16.00 per person. There is also a$2.50 one-time service fee for ordering tickets online. Write an expression in terms of n that represents the cost for ordering n tickets online.

Tom wants to order tickets online so that he and three of his friends can go together to a water park. The cost of the tickets is $16.00 per person. There is also a$2.50 one-time service fee for ordering tickets online. Write an expression in terms of n that represents the cost for ordering n ticke...

### "I am dying of shame!" This sentence is an example of... Question 1 options: Hyperbole Personification Simile Metaphor

"I am dying of shame!" This sentence is an example of... Question 1 options: Hyperbole Personification Simile Metaphor...

### Bonjour est-ce que quelqu’un peut m’aider svp? Lisez le premier paragraphe du Père Goriot de Balzac (jusqu'à « dans son cœur peut-être ») et expliquez, en une petite synthèse de 20 lignes, en quoi cet incipit (= première page d'un roman) est réaliste, et trouvez un tableau d'esthétique réaliste qui pourrait l'illustrer.

Bonjour est-ce que quelqu’un peut m’aider svp? Lisez le premier paragraphe du Père Goriot de Balzac (jusqu'à « dans son cœur peut-être ») et expliquez, en une petite synthèse de 20 lignes, en quoi cet incipit (= première page d'un roman) est réaliste, et trouvez un tableau d'esthétiqu...

### When oxygen is available following glycolysis, pyruvic acid continues along the energy-producing pathway of cellular respiration called A) Aerobic respiration B) Anaerobic respiration C) Fermentation D) Photosynthesis

When oxygen is available following glycolysis, pyruvic acid continues along the energy-producing pathway of cellular respiration called A) Aerobic respiration B) Anaerobic respiration C) Fermentation D) Photosynthesis...

### The daily flag salute by school children in the u.s. is an attempt to use _____ to build _____.

The daily flag salute by school children in the u.s. is an attempt to use _____ to build _____....

### A single year on Saturn is equal to 29.4 years on earth. About how many Earth-years are equal to 3 years on Saturn

A single year on Saturn is equal to 29.4 years on earth. About how many Earth-years are equal to 3 years on Saturn...

### In what ways are prokaryotic and eukaryotic cells similar and different? What are the two eukaryotic cells and What are the three features that distinguish one from the other and why do they exist?

In what ways are prokaryotic and eukaryotic cells similar and different? What are the two eukaryotic cells and What are the three features that distinguish one from the other and why do they exist?...

### Solve: 4x - 3 = 2x + 7 Solve: 2x + 6 = 7x - 14

Solve: 4x - 3 = 2x + 7 Solve: 2x + 6 = 7x - 14...

### The thirteenth fourteenth and fifthteenth amendments were all important additions to the construction because

The thirteenth fourteenth and fifthteenth amendments were all important additions to the construction because...

### Need help with this please no spam

Need help with this please no spam...

### Find the indicated side of the right triangle. 45° y 6 45 Х y = [?] /

Find the indicated side of the right triangle. 45° y 6 45 Х y = [?] /...

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