text:
# Help on contraction

1. Dec 11, 2006

### AiRAVATA

Hello guys. Here I am, bothering all of you again...

I am having troubles proving the following:

For $v \in V$ and $\omega \in \Lambda^k(V)$, show that if $v_1,v_2,...,v_n$ is a basis of $V$ with dual basis $\phi_1,\phi_2,...,\phi_n$ then

$$i_{v_j}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})=\left\{ \begin{array}{ll} 0 & j\neq \hbox{ any } i_\alpha \\ (-1)^{\alpha-1} \phi_{i_1}\wedge ... \wedge \hat{\phi}_{i_\alpha}\wedge...\wedge \phi_{i_k} & \hbox{if }j=i_\alpha \right.$$

where $i_v \omega (v_1,...,v_{k-1})=\omega(v,v_1,...,v_{k-1})$ and $\hat{\phi}_{i_\alpha}$ means we are omitting that term.

Use this result to show that for $\omega_1 \in \Lambda^k(V)$ and $\omega_2\in \Lambda^l(V)$ we have
$$i_{v}(\omega_1 \wedge \omega_2)=(i_v\omega_1)\wedge \omega_2+(-1)^k\omega_1\wedge(i_v\omega_2).$$

Here is what I don't understand. In order to prove the first result, I apply the right side to $(v_1,...,v_{j-1},v_{j+1},...,v_k)$, so
$$i_{v_j}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})(v_1,...,v_{j-1},v_{j+1},...,v_k)=(-1)^{j-1}(\phi_{i_1}\wedge ... \wedge \phi_{i_k})(v_1,...,v_j,...,v_k)=\sum_{\sigma \in S} (-1)^\sigma \sigma (\phi_{i_1}\otimes ...\otimes\phi_{i_k}) (v_1,...,v_j,...,v_k)$$

I'm stuck at this point. I know I need to evaluate the tensor product, but I'm going nowhere. Please someone enlighten me...

Last edited: Dec 11, 2006
2. Dec 12, 2006

### AKG

I don't think you mean to define iv on (v1, ..., vk-1) only (since that is just a list of the first k-1 basis vectors of V), but for any k-1 vectors in V, so you probably want to call them w1, ..., wk-1, i.e. you want to define:

$$i_v \omega (w_1, \dots ,w_{k-1}) = \omega (v,w_1,\dots ,w_{k-1})$$

To prove the first result, go back to the definitions.

Case 1, $j \neq i_{\alpha }$ for any $\alpha$. Let's denote vj by w0. We want to prove:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} ) = 0$$

This is true iff for all (k-1)-tuples of vectors in V, (w1, ..., wk-1), the following equation holds:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1}) = 0(w_1, \dots , w_{k-1})$$

The right side is equal to 0, obviously. The left side is:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1})$$

$$= (\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$= k! Alt(\phi _{i_1} \otimes \dots \otimes \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$= \frac{k!}{k!} \sum _{\sigma \in S_{\{ 0, 1, \dots , k-1\} }} sgn(\sigma )(\phi _{i_1} \otimes \dots \otimes \phi _{i_k} )(w_{\sigma (0)}, w_{\sigma (1)}, \dots , w_{\sigma (k-1)})$$

Let A denote S{0, 1, ..., k-1}. Then the above equals:

$$\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

Now for each $\sigma$, there is some n such that $\sigma (n) = 0$. For this n, we get:

$$\phi _{i_{n+1}}(w_{\sigma (n)}) = \phi _{i_{n+1}}(w_0) = \phi _{i_{n+1}}(v_j) = \delta _{i_{n+1}j} = 0$$

with the second last equality holding by definition of the action of an element of a dual basis on an element of the original basis (that $\delta$ being the Kronecker $\delta$), and the last equality holding because $j \neq i_{\alpha }$ for all $\alpha$, hence in particular, $j \neq i_{n+1}$.

So for each $\sigma$, one of the factors in the expression

$$\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

will be 0, hence each term of the sum will be 0, hence the sum is 0, as desired.

Case 2, $j = i_{\alpha }$: your turn.

3. Dec 12, 2006

### AiRAVATA

Oh... I see.

In the case $j=i_{n+1}$, for each $\sigma$ there wil be some n such that

$$= (\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_0, w_1, \dots , w_{k-1})$$

$$=\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi_{i_n}}(w_{\sigma(n-1)}) \cdot 1 \cdot \phi_{i_{n+2}}(w_{\sigma(n+1)})\dots\phi _{i_k}(w_{\sigma (k-1)})$$

and then, taking the change

$$i(j)=\left\{\begin{array}{ll} j+1 & \hbox{if } j<n \\ j & \hbox{if } j>n\end{array}\right.$$

the equality becomes

$$=\sum _{\sigma \in A'} sgn (i \circ \sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi_{i_n}}(w_{\sigma(n)}) \phi_{i_{n+2}}(w_{\sigma(n+1)})\dots\phi _{i_k}(w_{\sigma (k-1)})$$

$$=(-1)^{n} \frac{k!}{k!} \sum_{\sigma \in S_{\{1,2,...,k-1\}}} sgn(\sigma)(\phi_{i_1}\otimes\dots\otimes \phi_{i_{n}}\otimes\phi_{i_{n+2}}\otimes \dots \otimes \phi_{i_{k-1}})(w_{\sigma(1)},...,w_{\sigma(k-1)})$$

$$=(-1)^{n}(\phi_{i_1}\wedge \dots \wedge \hat{\phi}_{i_{n+1}}\wedge \dots \wedge \phi_{i_{k-1}})$$

What do you think?

I am guessing that in order to apply this result to the next identity I must use linearity.

Last edited: Dec 12, 2006
4. Dec 12, 2006

### AKG

Your proof ends up being okay, but the way it's written makes it hard to see how you've justified some of the lines. In the second line, it's technically okay to sum over all $\sigma \in A$, but it looks strange, because most of those terms are zero. Also, it kind of makes it look as though for all $\sigma$, $\phi _{i_n+1}(w_{\sigma (n)}) = 1$. Your permutation i doesn't even have a value at n (although the only remaining choice is i(n) = 0, which ends up being the correct choice). Even though $sgn (i\circ \sigma ) = sgn (\sigma \circ i)$, you really should write $sgn (\sigma \circ i)$ because that's the actual permutation you're applying to the subscripts of the w's. Your second last line goes from summing over A to summing over Sk-1. It's not immediately clear why this is justified. Essentially, it is saying that you can sum over this smaller set because for permtuations $\sigma \in A$ such that $\sigma (0) \neq 0$, the term

$$(\phi_{i_1}\otimes\dots\otimes \phi_{i_{n}}\otimes\phi_{i_{n+2}}\otimes \dots \otimes \phi_{i_{k-1}})(w_{\sigma(1)},...,w_{\sigma(k-1)})$$

is zero, and Sk-1 consists precisely of those elements in A which fix 0. But why is the above term 0 if $\sigma$ doesn't fix 0? The way you've written the whole proof, I can't clearly explain why this would be the case.

I would do it as follows:

Case 2, $j = i_{\alpha }$:

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} )(w_1, \dots , w_{k-1})$$

$$= \sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (0)}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$ (as computed in post #2)

Define $\tau = (0\, 1\, 2\, \dots \, a-2\, a-1)$. Then the above is:

$$= \sum _{\sigma \in A} sgn (\sigma \circ \tau )\phi_{i_1}(w_{\sigma \circ \tau (0)}) \dots \phi _{i_k}(w_{\sigma \circ \tau (k-1)})$$

$$= \sum _{\sigma \in A} sgn (\sigma )sgn(\tau )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha }}(w_{\sigma (0)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in A} sgn (\sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha }}(w_{\sigma (0)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in S_{k-1}} sgn (\sigma )\phi_{i_1}(w_{\sigma (1)}) \dots \phi _{i_{\alpha -1}}(w_{\sigma (\alpha - 1)})\phi _{i_{\alpha + 1}}(w_{\sigma (\alpha )}) \dots \phi _{i_k}(w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\sum _{\sigma \in S_{k-1}} sgn (\sigma )\phi_{i_1} \otimes \dots \otimes \hat{\phi _{i_{\alpha }}} \otimes \dots \otimes \phi _{i_k}(w_{\sigma (1)},\, \dots ,\, w_{\sigma (k-1)})$$

$$= (-1)^{\alpha - 1}\phi_{i_1} \wedge \dots \wedge \hat{\phi _{i_{\alpha }}} \wedge \dots \wedge \phi _{i_k}(w_1,\, \dots ,\, w_{k-1})$$

so in this case 2,

$$i_{v_j}(\phi _{i_1} \wedge \dots \wedge \phi _{i_k} ) = (-1)^{\alpha - 1}\phi_{i_1} \wedge \dots \wedge \hat{\phi _{i_{\alpha }}} \wedge \dots \wedge \phi _{i_k}$$

as desired.

Last edited: Dec 12, 2006
5. Dec 12, 2006

### AiRAVATA

Thanks a lot for your help.


text:
# The molarity of con.HCl is $1.15\;M$. What volume of con. HCl is required to make 1.00 of 0.1M HCl
$\begin{array}{1 1} (a) 56.9\;ml\\ (b) 89\;ml\\ (c) 86.9\;ml\\ (d) 80\;ml \end{array}$

format: plain


text:
# How to get microtype to work with mathematical minus sign and package lmodern?

Using the package microtype by including the line

\usepackage{microtype}

in the preamble subtly alters the appearance of a typeset document by, amongst other things, letting optically less dense characters, like dashes or punctuations, protrude slightly in the margins. When I use a mathematical minus sign in a document written in the class article, I write it like this:

$-$

If I include \usepackage{lmodern} in the preamble, the minus sign does not get moved out a little bit if it happens to become the first character of a line, but stays at the same horizontal position as for example an "m" letter. Microtype does work for the plus sign and dashes, including when the plus sign is written in math mode:

$+$

Does anyone know how to enable microtype also for the mathematical minus sign when using lmodern?

When not using lmodern, the problem can be fixed as suggested by Mico by using \usepackage[protrusion=allmath]{microtype}.

edit: MWE, for which the fix to use \usepackage[protrusion=allmath]{microtype} does not work.

Here is some example code that generates the incorrectly placed mathematical minus sign:

\documentclass{article}
\usepackage{lmodern}
\usepackage[protrusion=allmath]{microtype}
\begin{document}
\section{Test Section}
\label{Test Label}
Test text to reproduce the microtype mathematical minus sign problem that I experienced. I am putting in a few more words to get some text to demonstrate the effect. This is just some text with mathematical minusaaaaaaaaaaaa signs $-$inserted (that was a mathematical minus sign in front of inserted'')aaaaa. A +plus sign looks like that. In a longer text that I had, the plus sign looked fine, but the minus sign clearly looked like it was to far in. I am now putting in some more text, just to get one more line to show the look of the left margin.
\end{document}

Compiled result (first an enlargement to show the left margin, and then the entire page):

• Have you tried providing the protrusion=allmath option to microtype? – Mico Jul 16 '12 at 23:27
• I was not aware of that. I tried it (just now), but unfortunately it did not help. It only rendered several warning about microtype not finding protrusion list for font ... – hjb981 Jul 17 '12 at 0:06
• Aaah, now we're starting to get somewhere: Which (math) fonts do you use? According to the microtype manual, protrusion should be enabled for all of the "main" math font packages (Computer Modern (obviously), mathptmx, mathpazo, eulervm, and probably a few more). Please consider editing your question to include a standalone MWE (minimum working example) that generates the problem behavior you're trying to correct. – Mico Jul 17 '12 at 0:09
• Ah, protrusion=allmath does solve the problem for your MWE. It still doesn't make sense that + and - are treated differently without the allmath option. (Ping, @Mico.) – Hendrik Vogt Jul 17 '12 at 13:51
• @hjb981: Please try and remove anything from the preamble that does not contribute to the problem. You could do that by removing things piece by piece (and recompiling of course), or by removing 1. the first half of the preamble (+recompile), 2. the 2nd half of the preamble (+recompile), hoping that with either 1. or 2. the problem persists. Then repeat until you get some MWE. – Hendrik Vogt Jul 17 '12 at 14:36

\DeclareMicrotypeAlias{lmsy}{cmsy}

This seems to be an oversight in the microtype package since there is already a line

\DeclareMicrotypeAlias{lmr} {cmr}

so maybe report this to microtype author...

• That worked. I still get several errors of the kind: I cannot not find a protrusion list for font 'OML/lmm/m/it/10.95' Switching of protrusion for this font on line ... The numbers can also be exchanged for 6, 7, 8, 10, 12 and a few more---I guess that it refers to different sizes. Is there a way to enable microtype for those fonts also? – hjb981 Jul 17 '12 at 15:18
• I guess \DeclareMicrotypeAlias{lmm}{cmm} should do it. – Lev Bishop Jul 17 '12 at 15:36
• Yes, it worked. What does that line mean? Does one always write {l abc}{c abc}, where abc is the name of the font? – hjb981 Jul 17 '12 at 15:53
• @hjb981 it means "use the settings from this font for that one". Since lmodern mostly the same as computer modern, and the fonts are (mostly?) named the same just with interchange of c and l, it should work ok. However computer modern and lmodern are not identical. Microtype already knows to treat the \textquotedblleft et al slightly differently for example, so this quick fix may not be good enough (perhaps this is why it is currently missing from microtype). – Lev Bishop Jul 17 '12 at 23:17
• indeed, these two alias declarations were missing in microtype.cfg. I'll add them for the next version (coming "soon"). Thanks for bringing this to my attention. – Robert Jul 18 '12 at 1:59


text:
# Derivative of an integral

1. Dec 6, 2007

### Mothrog

I have a function of the form

$$\int^{r}_{0} tf(t)dt$$

I'm supposed to take the derivative with respect to r of this integral. By the fundamental theorem of calculus, is the derivative not

$$rf(r)$$

The problem being I need, for the problem to work out correctly, to have a df/dr term. So, am I mistaken in my solution?

2. Dec 6, 2007

### sutupidmath

well, it depends on what properties does the function under the integral sign has, tf(t).
THere are some conditions which this function has to fulfill in order for the answer to be what u said. THe first one is that f(t)>0, which implies directly that $$\int^{r}_{0} f(t)dt$$ is monotono increasing.

3. Dec 6, 2007

### HallsofIvy

No, there is no need that f(t)> 0. By the fundamental theorem of calculus, as long as g(x) is continuous if $$G(x)= \int_a^x g(t)dt$$, then $$\frac{dG}{dx}= g(x)$$. In this case, g(x)= xf(x) so, as long as f is continuous, if $$F(r)= \int_0^r tf(t)dt$$, then $$\frac{dF}{dr}= rf(r)$$. I do not understand why you would "need to have a df/dr term. Perhaps if you were to post the actual problem, it would make more sense.

4. Dec 6, 2007

### Mothrog

It is desired to show that if, for constant k,

$$\frac{d}{dr} g(r) = \frac{d}{dr}k\frac{r^2}{\int^{r}_{0} tf(t)dt} < 0$$

Then

$$\frac{d}{dr}f(r) > 0$$

When I take the derivative of g, I get

$$\frac{dg}{dr} = k\frac{2r\int^{r}_{0}tf(t)dt - r^2(rf(r))}{(\int^{r}_{0} tf(t)dt)^2}$$

If that is correct, I don't see how it is possible to show the desired relation. The definition of g given is the only information known about g. Am I missing something?

Last edited: Dec 6, 2007
5. Dec 7, 2007

Bump.

6. Dec 7, 2007

### Dick

Well, here's something. Let's throw k away. If it's positive it doesn't add anything to the problem and if it can be either sign, then it just plain isn't true. Now let

$$F(r)=\int^{r}_{0} tf(t)dt$$ and $$G(r)=\frac{r^2}{2} f(r)$$.

Now look at the numerator of dg/dr. If it's negative then G(r)>F(r). But G(0)=F(0)=0. So if G'(r)>F'(r) then I could conclude G(r)>F(r). G'(r)=rf(r)+(r^2/2)*f'(r). F'(r)=rf(r). So if f'(r)>0 then dg/dr<0 and vice versa. Trouble is, I can't see how the converse would hold, which seems to be what you are actually trying to prove.

7. Dec 7, 2007

### Mothrog

I'm not sure you can assume $$g(0) = F(0) = 0$$. Can you walk me through your logic on that step?

Last edited: Dec 7, 2007
8. Dec 7, 2007

### Dick

Not the little f and g. The big F and G. It's not an assumption, look at their definitions.

9. Dec 7, 2007

### Mothrog

But G(r) depends only on little f(r), so how can you conclude that if F(0) = 0, that G(r) = 0?

10. Dec 7, 2007

### Dick

G(0)=f(0)*0^2/2=0. F(0) is the integral from 0 to 0 of t*f(t) which=0.

11. Dec 7, 2007

### Mothrog

Ah, OK. That makes sense.

12. Dec 7, 2007

### Mothrog

Actually, looking back at the problem again, I needed to show that f'(r) > 0 implies g'(r) < 0, so I don't need to prove the converse. So, problem solved. Thanks for your help. I was tearing my hair out over that one.

13. Dec 7, 2007

### Dick

No problem. Good thing you don't have to prove the converse, It's not true.


text:
# Writing numbers on the Web

By Murray Bourne, 17 Apr 2007

Jakob Nielsen's Alertbox article today, Show Numbers as Numerals, is interesting for those of us who need to communicate mathematics via the Web.

After conducting eye-tracking research on Web users, Nielsen points out the differences between numeral writing for print ("Don't begin sentences with a numeral, use words instead") and the Web (use a numeral).

A key thing here is that most users scan Web pages and rarely read everything. (Actually, isn't that true of print these days, as well?)

Nielsen raises the following interesting point:

Instead, [Web users] scan the text and pick out headlines, highlighted words, bulleted lists, and links. [...] There are a few exceptions to this rule: lower-literacy users can't scan, while higher-literacy users read the entire page if they're really interested or in desperate need of the information.

I've noticed this - many students have trouble quickly scanning a page and gleaning the main facts.

Back to numbers and numerals vs words. What came to mind was my page on Scientific Notation, where I extol the advantages of a more convenient way to write large numbers.

I'm glad Nielsen mentions the difference between US "billions" and other-country "billions".

There's also an internationalization issue here: "billion" represents a thousand millions in American English, but a million millions in many European languages.

We really should standardise (oops, I mean standardize) the usage of "billion", "trillion" and the like.

In summary, I wonder why we bother writing complete sentences - does anyone read everything...?

Be the first to comment below.

### Comment Preview

HTML: You can use simple tags like <b>, <a href="...">, etc.

To enter math, you can can either:

1. Use simple calculator-like input in the following format (surround your math in backticks, or qq on tablet or phone):
a^2 = sqrt(b^2 + c^2)
(See more on ASCIIMath syntax); or
2. Use simple LaTeX in the following format. Surround your math with $$ and $$.
$$\int g dx = \sqrt{\frac{a}{b}}$$
(This is standard simple LaTeX.)

NOTE: You can't mix both types of math entry in your comment.


text:
LLVM  15.0.0git

#include "llvm/DebugInfo/PDB/Native/RawTypes.h"

[legend]

## Public Attributes

support::ulittle32_t Version

support::ulittle32_t Size

uint64_t FileTime

support::ulittle32_t Age

## Detailed Description

Definition at line 321 of file RawTypes.h.

## ◆ Age

Definition at line 325 of file RawTypes.h.

## ◆ FileTime

Definition at line 324 of file RawTypes.h.

Definition at line 326 of file RawTypes.h.

## ◆ Size

Definition at line 323 of file RawTypes.h.

## ◆ Version

Definition at line 322 of file RawTypes.h.

The documentation for this struct was generated from the following file:


text:
# The SSM Procedure

### Example 27.2 Panel Data: Random-Effects and Autoregressive Models

This example shows how you can use the SSM procedure to specify and fit the two-way random-effects model and the autoregressive model to analyze a panel of time series. The fitting of dynamic panel model for such data is illustrated in Example 27.11. These (and a few other) model types can also be fitted by the PANEL procedure, a SAS/ETS procedure that is specially designed to efficiently handle the cross-sectional time series data. However, because of the differences in their model fitting algorithms, generally the parameter estimates and other fit statistics produced by the SSM and PANEL procedures do not match. The SSM procedure always uses the (restricted) maximum likelihood for parameter estimation. The estimation method used by the PANEL procedure depends on the model type and the particular estimation options.

The cross-sectional data, Cigar, that are used in the section Getting Started: SSM Procedure are reused in this example. The output shown here is less extensive than the output shown in that section. The main emphasis of this example is how you can specify the two-way random effects model and the autoregressive model in the SSM procedure.

According to the two-way random effects model, the cigarette sales, lsales, can be described by the following equation:

This model represents lsales in region i and in year t as a sum of an overall intercept , the regression effects due to lprice, lndi, and lpimin, a zero-mean, random effect associated with region i, a zero-mean, random effect associated with year t, and the observation noise . The region-specific random effects and the year-specific random effects are assumed to be independent, Gaussian sequences with variances and , respectively. In addition, they are assumed to be independent of the observation noise, which is also assumed to be a sequence of independent, zero-mean, Gaussian variables with variance .

You can specify and fit this model by using the following statements:

 proc ssm data=Cigar;
id year interval=year;
parms s2g/ lower=(1.e-6);
array RegionArray{46} region1-region46;
do i=1 to 46;
RegionArray[i] = (region=i);
end;
/* region-specific random effects */
state zeta(46) T(I) cov1(I)=(s2g);
component regionEffect = zeta * (RegionArray);
/* year-specific random effect */
state eta(1) type=wn cov(D);
component timeEffect = eta[1];
irregular wn;
intercept = 1.0;
model lsales = intercept lprice lndi lpimin
timeEffect regionEffect wn;
run;

The PARMS statement defines s2g, a parameter that is restricted to be positive and is used later as the variance parameter for the region effect. Similarly the 46-dimensional array, RegionArray, of region-specific dummy variables is defined to be used later. The state subsection zeta corresponds to , which is the 46-dimensional vector of region-specific, zero-mean, random effects. The component regionEffect extracts the proper element of by using the array RegionArray. A constant column, intercept, is defined to be used later as an intercept term. The component timeEffect corresponds to , and wn specifies the observation noise . Finally the MODEL statement defines the model. Some of the tables that are produced by running these statements are shown in Output 27.2.1 through Output 27.2.5.

The model summary, shown in Output 27.2.1, shows that the model is defined by one MODEL statement, the dimension of the underlying state vector is 47 (because is 46-dimensional and is one-dimensional), the diffuse dimension is 4 (because of the four predictors in the model), and there are three parameters to be estimated.

Output 27.2.1: Two-Way Random-Effects Model: Model Summary

The SSM Procedure

Model Summary
Model Property Value
Number of Model Equations 1
State Dimension 47
Dimension of the Diffuse Initial Condition 4
Number of Parameters 3

Output 27.2.2 provides the likelihood information about the fitted model.

Output 27.2.2: Two-Way Random-Effects Model: Likelihood Summary

Likelihood Computation Summary
Statistic Value
Nonmissing Response Values Used 1380
Estimated Parameters 3
Initialized Diffuse State Elements 4
Normalized Residual Sum of Squares 1376.0001
Diffuse Log Likelihood 1459.0277
Profile Log Likelihood 1470.8628

Output 27.2.3 shows the regression estimates.

Output 27.2.3: Two-Way Random-Effects Model: Regression Estimates

Regression Parameter Estimates
Response Variable Regression Variable Estimate Standard
Error
t Value Pr > |t|
lsales intercept 2.798 0.1136 24.62 <.0001
lsales lprice -0.903 0.0365 -24.73 <.0001
lsales lndi 0.592 0.0246 24.08 <.0001
lsales lpimin 0.127 0.0398 3.18 0.0015

The ML estimate of s2g, a parameter specified in the PARMS statement, is shown in Output 27.2.4. It corresponds to , the variance of the region effect.

Output 27.2.4: Two-Way Random-Effects Model: Estimate of

Estimates of Named Parameters
Parameter Estimate Standard
Error
s2g 0.0241 0.00512

Output 27.2.5: Variance Estimates of and

Model Parameter Estimates
Component Type Parameter Estimate Standard
Error
eta Disturbance Covariance Cov[1, 1] 0.000681 0.000264
wn Irregular Variance 0.005698 0.000224

The estimates of the other unknown parameters in the model are shown in Output 27.2.5. It shows the estimate of the variance of the irregular component wn and the estimate of the variance of the time effect .

The remainder of this example describes how you can specify and fit the following first-order vector autoregessive model to the cigarette data:

This model represents lsales in region i and in year t as a sum of an overall intercept , the regression effects due to lprice, lndi, and lpimin, and the ith element of a vector error term . The multidimensional error sequence is assumed to follow a first-order autoregression with a diagonal autoregressive coefficient matrix and with a multivariate, white noise sequence as its disturbance sequence. The covariance matrix of , , is assumed to be dense. Note that the dimension of the vectors is the same as the number of cross-sections in the study (the number of regions in this example). Therefore, even for a relatively modest panel study, the total number of parameters to be estimated can get quite large. Therefore, in this example only the first three regions are considered in the analysis. The following statements specify and fit this model to the Cigar data set:

 proc ssm data=Cigar;
where region <= 3;
id year interval=year;
array RegionArray{3} region1-region3;
do i=1 to 3;
RegionArray[i] = (region=i);
end;
state zeta(3) type=varma(p(d)=1) cov(g) print=(ar cov);
component eta = zeta*(RegionArray);
intercept = 1.0;
model lsales = intercept lprice lndi lpimin eta;
run;

The vectors are specified in the STATE statement. The TYPE= specification signifies that the three-dimensional state subsection, zeta, follows a vector AR(1) model with a diagonal transition matrix and a disturbance covariance of a general form. The PRINT=(AR COV) option causes the SSM procedure to print the estimated AR coefficient matrix, , and the disturbance error covariance , respectively. The COMPONENT statement defines the appropriate error contribution (named eta), . Output 27.2.6 shows the estimated regression coefficients, Output 27.2.7 shows the estimate of , and Output 27.2.8 shows the estimate of :

Output 27.2.6: Autoregressive Model: Regression Estimates

The SSM Procedure

Regression Parameter Estimates
Response Variable Regression Variable Estimate Standard
Error
t Value Pr > |t|
lsales intercept 3.576 0.3621 9.88 <.0001
lsales lprice -0.230 0.0832 -2.77 0.0056
lsales lndi 0.218 0.0753 2.89 0.0039
lsales lpimin 0.068 0.0996 0.68 0.4952

Output 27.2.7: Estimate of the AR Coefficient

AR Coefficient Matrix for zeta
Col1 Col2 Col3
Row1 0.943078 0 0
Row2 0 0.989301 0
Row3 0 0 0.888549

Output 27.2.8: Estimate of the Disturbance Covariance

Disturbance Covariance for zeta
Col1 Col2 Col3
Row1 0.000941 0.000445 0.000346
Row2 0.000445 0.00226 0.000242
Row3 0.000346 0.000242 0.000882


text:
User pranay - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T16:17:48Z http://mathoverflow.net/feeds/user/12922 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109509/the-inverse-of-the-euler-totient-function The Inverse of the Euler Totient Function pranay 2012-10-13T03:18:05Z 2012-10-13T06:13:34Z
How can we calculate the cardinality of the inverse of Totient function of any positive integer n ? I tried going through this paper, but I couldn't understand the procedure.

Thanks
 http://mathoverflow.net/questions/59404/calculating-modular-exponentiation-if-the-exponent-is-in-base-3 calculating modular exponentiation if the exponent is in base 3 pranay 2011-03-24T08:35:39Z 2011-03-24T12:44:50Z
Hi , i am trying to solve such a problem as follows: say the base is 2(base 10) and exponent is 10(base 3) modulus is 10(base 10), then

there was a bug . Thanks .
 http://mathoverflow.net/questions/58702/number-of-divisors-of-an-integer-of-form-4n1-and-4n3 Number of divisors of an integer of form 4n+1 and 4n+3 pranay 2011-03-17T01:10:53Z 2011-03-18T16:29:32Z
Suppose $n$ is a large odd integer. Let $D_1(n)$ be the number of divisors of $n$ of the form $4k+1$ and let $D_3(n)$ be the number of divisors of the form $4k+3$. I would like to compute $(D_1(n),D_3(n))$.

As Joe Silverman points out, the number of representations of $n$ as a sum of two squares of integers is $4(D_1(n)-D_3(n))$. For example, $D_1(225)=6$ and $D_3(225)=3$, so there are $4(6-3)=12$ lattice points on the circle of radius $\sqrt {225}$ centered at the origin including $(0,15)$ and $(-9,-12)$.

Is there a faster way to find $(D_1(n),D_3(n))$ than factoring $n$?

Original:

Hi, one way to do so is to list all the divisors of the integer and check each if it is of the form 4n+1 or 4n+3. Is there any faster method to it, especially for large n?
 http://mathoverflow.net/questions/55179/partitioning-a-number-into-two-sets-based-on-sum-of-digits partitioning a number into two sets based on sum of digits pranay 2011-02-12T03:27:07Z 2011-02-13T18:05:18Z
hi, how can one determine whether a number belongs to such a set of numbers whose every element can be divided into two sets such that the sum of digits of each set of the number is same. E.g 23450 does belong to such a set , as it can be split into two sets : (3,4,0) and (2,5) such that sum of digits in each set is the same(7). Similary 91125 belongs to such a set: (9) and (1,1,2,5). but 567 , or 34523 do not belong to such a set of numbers .

Thanks.
 http://mathoverflow.net/questions/109509/the-inverse-of-the-euler-totient-function/109511#109511 Comment by pranay pranay 2012-10-13T08:04:10Z 2012-10-13T08:04:10Z thanks Eric, but for calculations what's o(1) ? http://mathoverflow.net/questions/99773/finding-the-value-of-x-for-an-equation Comment by pranay pranay 2012-06-16T10:14:51Z 2012-06-16T10:14:51Z thanks Chandrasekhar for the guidance http://mathoverflow.net/questions/59736/generating-a-series Comment by pranay pranay 2011-03-27T17:02:52Z 2011-03-27T17:02:52Z This is a sequence i need as a part of a question where i need to calculate the nth term. so in that case if 18th term is asked, then 18 in base 10 = 102 in base 4 = 325 after mapping = 61 in base 10 but answer is 73 http://mathoverflow.net/questions/59736/generating-a-series Comment by pranay pranay 2011-03-27T16:27:40Z 2011-03-27T16:27:40Z any C programme? http://mathoverflow.net/questions/59404/calculating-modular-exponentiation-if-the-exponent-is-in-base-3 Comment by pranay pranay 2011-03-24T09:05:47Z 2011-03-24T09:05:47Z sorry, corrected. http://mathoverflow.net/questions/58702/number-of-divisors-of-an-integer-of-form-4n1-and-4n3/58795#58795 Comment by pranay pranay 2011-03-22T11:56:03Z 2011-03-22T11:56:03Z Thanks a lot :) , i too found the same solution in one of the books . http://mathoverflow.net/questions/58702/number-of-divisors-of-an-integer-of-form-4n1-and-4n3/58709#58709 Comment by pranay pranay 2011-03-17T02:02:55Z 2011-03-17T02:02:55Z Thanks, but i was thinking if there exists any method not involving factorisation.. http://mathoverflow.net/questions/55179/partitioning-a-number-into-two-sets-based-on-sum-of-digits Comment by pranay pranay 2011-02-12T17:53:01Z 2011-02-12T17:53:01Z By simple brute-force method it would take long time to find such numbers among a range. So is there any better solution to it? Maybe by using bit operations?


text:
# What is the slope intercept form of the line passing through (15,3)  with a slope of 1/2 ?

Jun 18, 2018

See a solution process below:

#### Explanation:

The slope-intercept form of a linear equation is: $y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$

Where $\textcolor{red}{m}$ is the slope and $\textcolor{b l u e}{b}$ is the y-intercept value.

First, we can substitute the slope in the problem for $\textcolor{red}{m}$ and substitute the values from the point in the problem for $x$ and $y$ and solve for $\textcolor{b l u e}{b}$:

$3 = \left(\textcolor{red}{\frac{1}{2}} \times 15\right) + \textcolor{b l u e}{b}$

$3 = \textcolor{red}{\frac{15}{2}} + \textcolor{b l u e}{b}$

$3 - \frac{15}{2} = \textcolor{red}{\frac{15}{2}} - \frac{15}{2} + \textcolor{b l u e}{b}$

$\left(\frac{2}{2} \times 3\right) - \frac{15}{2} = 0 + \textcolor{b l u e}{b}$

$\frac{6}{2} - \frac{15}{2} = \textcolor{b l u e}{b}$

$\frac{6 - 15}{2} = \textcolor{b l u e}{b}$

$- \frac{9}{2} = \textcolor{b l u e}{b}$

$\textcolor{b l u e}{b} = - \frac{9}{2}$

We can now substitute $- \frac{9}{2}$ for $\textcolor{b l u e}{b}$ and the slope from the problem for $\textcolor{red}{m}$ in the original formula to write the equation:

$y = \textcolor{red}{\frac{1}{2}} x + \textcolor{b l u e}{- \frac{9}{2}}$

$y = \textcolor{red}{\frac{1}{2}} x - \textcolor{b l u e}{\frac{9}{2}}$


text:
# tilt.boot

0th

Percentile

##### Non-parametric Tilted Bootstrap

This function will run an initial bootstrap with equal resampling probabilities (if required) and will use the output of the initial run to find resampling probabilities which put the value of the statistic at required values. It then runs an importance resampling bootstrap using the calculated probabilities as the resampling distribution.

Keywords
nonparametric
##### Usage
tilt.boot(data, statistic, R, sim = "ordinary", stype = "i",
strata = rep(1, n), L = NULL, theta = NULL,
alpha = c(0.025, 0.975), tilt = TRUE, width = 0.5,
index = 1, …)
##### Arguments
data

The data as a vector, matrix or data frame. If it is a matrix or data frame then each row is considered as one (multivariate) observation.

statistic

A function which when applied to data returns a vector containing the statistic(s) of interest. It must take at least two arguments. The first argument will always be data and the second should be a vector of indices, weights or frequencies describing the bootstrap sample. Any other arguments must be supplied to tilt.boot and will be passed unchanged to statistic each time it is called.

R

The number of bootstrap replicates required. This will generally be a vector, the first value stating how many uniform bootstrap simulations are to be performed at the initial stage. The remaining values of R are the number of simulations to be performed resampling from each reweighted distribution. The first value of R must always be present, a value of 0 implying that no uniform resampling is to be carried out. Thus length(R) should always equal 1+length(theta).

sim

This is a character string indicating the type of bootstrap simulation required. There are only two possible values that this can take: "ordinary" and "balanced". If other simulation types are required for the initial un-weighted bootstrap then it will be necessary to run boot, calculate the weights appropriately, and run boot again using the calculated weights.

stype

A character string indicating the type of second argument expected by statistic. The possible values that stype can take are "i" (indices), "w" (weights) and "f" (frequencies).

strata

An integer vector or factor representing the strata for multi-sample problems.

L

The empirical influence values for the statistic of interest. They are used only for exponential tilting when tilt is TRUE. If tilt is TRUE and they are not supplied then tilt.boot uses empinf to calculate them.

theta

The required parameter value(s) for the tilted distribution(s). There should be one value of theta for each of the non-uniform distributions. If R[1] is 0 theta is a required argument. Otherwise theta values can be estimated from the initial uniform bootstrap and the values in alpha.

alpha

The alpha level to which tilting is required. This parameter is ignored if R[1] is 0 or if theta is supplied, otherwise it is used to find the values of theta as quantiles of the initial uniform bootstrap. In this case R[1] should be large enough that min(c(alpha, 1-alpha))*R[1] > 5, if this is not the case then a warning is generated to the effect that the theta are extreme values and so the tilted output may be unreliable.

tilt

A logical variable which if TRUE (the default) indicates that exponential tilting should be used, otherwise local frequency smoothing (smooth.f) is used. If tilt is FALSE then R[1] must be positive. In fact in this case the value of R[1] should be fairly large (in the region of 500 or more).

width

This argument is used only if tilt is FALSE, in which case it is passed unchanged to smooth.f as the standardized bandwidth for the smoothing operation. The value should generally be in the range (0.2, 1). See smooth.f for for more details.

index

The index of the statistic of interest in the output from statistic. By default the first element of the output of statistic is used.

Any additional arguments required by statistic. These are passed unchanged to statistic each time it is called.

##### Value

An object of class "boot" with the following components

t0

The observed value of the statistic on the original data.

t

The values of the bootstrap replicates of the statistic. There will be sum(R) of these, the first R[1] corresponding to the uniform bootstrap and the remainder to the tilted bootstrap(s).

R

The input vector of the number of bootstrap replicates.

data

The original data as supplied.

statistic

The statistic function as supplied.

sim

The simulation type used in the bootstrap(s), it can either be "ordinary" or "balanced".

stype

The type of statistic supplied, it is the same as the input value stype.

call

A copy of the original call to tilt.boot.

strata

The strata as supplied.

weights

The matrix of weights used. If R[1] is greater than 0 then the first row will be the uniform weights and each subsequent row the tilted weights. If R[1] equals 0 then the uniform weights are omitted and only the tilted weights are output.

theta

The values of theta used for the tilted distributions. These are either the input values or the values derived from the uniform bootstrap and alpha.

##### References

Booth, J.G., Hall, P. and Wood, A.T.A. (1993) Balanced importance resampling for the bootstrap. Annals of Statistics, 21, 286--298.

Davison, A.C. and Hinkley, D.V. (1997) Bootstrap Methods and Their Application. Cambridge University Press.

Hinkley, D.V. and Shi, S. (1989) Importance sampling and the nested bootstrap. Biometrika, 76, 435--446.

• tilt.boot
##### Examples
# NOT RUN {
# Note that these examples can take a while to run.

# Example 9.9 of Davison and Hinkley (1997).
grav1 <- gravity[as.numeric(gravity[,2]) >= 7, ]
grav.fun <- function(dat, w, orig) {
strata <- tapply(dat[, 2], as.numeric(dat[, 2]))
d <- dat[, 1]
ns <- tabulate(strata)
w <- w/tapply(w, strata, sum)[strata]
mns <- as.vector(tapply(d * w, strata, sum)) # drop names
mn2 <- tapply(d * d * w, strata, sum)
s2hat <- sum((mn2 - mns^2)/ns)
c(mns[2]-mns[1],s2hat,(mns[2]-mns[1]-orig)/sqrt(s2hat))
}
grav.z0 <- grav.fun(grav1, rep(1, 26), 0)
tilt.boot(grav1, grav.fun, R = c(249, 375, 375), stype = "w",
strata = grav1[,2], tilt = TRUE, index = 3, orig = grav.z0[1])

#  Example 9.10 of Davison and Hinkley (1997) requires a balanced
#  importance resampling bootstrap to be run.  In this example we
#  show how this might be run.
acme.fun <- function(data, i, bhat) {
d <- data[i,]
n <- nrow(d)
d.lm <- glm(d$acme~d$market)
beta.b <- coef(d.lm)[2]
d.diag <- boot::glm.diag(d.lm)
SSx <- (n-1)*var(d$market) tmp <- (d$market-mean(d$market))*d.diag$res*d.diag\$sd
sr <- sqrt(sum(tmp^2))/SSx
c(beta.b, sr, (beta.b-bhat)/sr)
}
acme.b <- acme.fun(acme, 1:nrow(acme), 0)
acme.boot1 <- tilt.boot(acme, acme.fun, R = c(499, 250, 250),
stype = "i", sim = "balanced", alpha = c(0.05, 0.95),
tilt = TRUE, index = 3, bhat = acme.b[1])
# }
Documentation reproduced from package boot, version 1.3-24, License: Unlimited

### Community examples

Looks like there are no examples yet.


text:
# Reflexive relation explained

In mathematics, a homogeneous binary relation R on a set X is reflexive if it relates every element of X to itself.[1] [2]

An example of a reflexive relation is the relation "is equal to" on the set of real numbers, since every real number is equal to itself. A reflexive relation is said to have the reflexive property or is said to possess reflexivity. Along with symmetry and transitivity, reflexivity is one of three properties defining equivalence relations.

## Definitions

Let

R

be a binary relation on a set

X

, which by definition is just a subset of

X x X.

For any

x,y\inX,

the notation

xRy

means that

(x,y)\inR

while "not

xRy

" means that

(x,y)\not\inR.

The relation

R

is called if

xRx

for every

x\inX

or equivalently, if

\operatorname{I}X\subseteqR

where

\operatorname{I}X:=\{(x,x)~:~x\inX\}

denotes the identity relation on

X.

The of

R

is the union

R\cup\operatorname{I}X,

which can equivalently be defined as the smallest (with respect to

\subseteq

) reflexive relation on

X

that is a superset of

R.

A relation

R

is reflexive if and only if it is equal to its reflexive closure.

The or of

R

is the smallest (with respect to

\subseteq

) relation on

X

that has the same reflexive closure as

R.

It is equal to

R\setminus\operatorname{I}X=\{(x,y)\inR~:~xy\}.

The irreflexive kernel of

R

can, in a sense, be seen as a construction that is the "opposite" of the reflexive closure of

R.

For example, the reflexive closure of the canonical strict inequality

<

on the reals

R

is the usual non-strict inequality

\leq

whereas the reflexive reduction of

\leq

is

<.

### Related definitions

There are several definitions related to the reflexive property. The relation

R

is called:
or : If it does not relate any element to itself; that is, if not

xRx

for every

x\inX.

A relation is irreflexive if and only if its complement in

X x X

is reflexive. An asymmetric relation is necessarily irreflexive. A transitive and irreflexive relation is necessarily asymmetric.
• If whenever

x,y\inX

are such that

xRy,

then necessarily

xRx.

[3]
• If whenever

x,y\inX

are such that

xRy,

then necessarily

yRy.

• If every element that is related to some element is also related to itself. Explicitly, this means that whenever

x,y\inX

are such that

xRy,

then necessarily

xRx

yRy.

Equivalently, a binary relation is quasi-reflexive if and only if it is both left quasi-reflexive and right quasi-reflexive. A relation

R

is quasi-reflexive if and only if its symmetric closure

R\cupR\operatorname{T

} is left (or right) quasi-reflexive.
Antisymmetric
• If whenever

x,y\inX

are such that

xRyandyRx,

then necessarily

x=y.

• If whenever

x,y\inX

are such that

xRy,

then necessarily

x=y.

[4] A relation

R

is coreflexive if and only if its symmetric closure is anti-symmetric.

A reflexive relation on a nonempty set

X

can neither be irreflexive, nor asymmetric (

R

is called if

xRy

implies not

yRx

), nor antitransitive (

R

is if

xRyandyRz

implies not

xRz

).

## Examples

Examples of reflexive relations include:

• "is equal to" (equality)
• "is a subset of" (set inclusion)
• "divides" (divisibility)
• "is greater than or equal to"
• "is less than or equal to"

Examples of irreflexive relations include:

• "is not equal to"
• "is coprime to" (for the integers

>1,

since 1 is coprime to itself)
• "is a proper subset of"
• "is greater than"
• "is less than"

An example of an irreflexive relation, which means that it does not relate any element to itself, is the "greater than" relation (

x>y

) on the real numbers. Not every relation which is not reflexive is irreflexive; it is possible to define relations where some elements are related to themselves but others are not (that is, neither all nor none are). For example, the binary relation "the product of

x

and

y

is even" is reflexive on the set of even numbers, irreflexive on the set of odd numbers, and neither reflexive nor irreflexive on the set of natural numbers.

An example of a quasi-reflexive relation

R

is "has the same limit as" on the set of sequences of real numbers: not every sequence has a limit, and thus the relation is not reflexive, but if a sequence has the same limit as some sequence, then it has the same limit as itself. An example of a left quasi-reflexive relation is a left Euclidean relation, which is always left quasi-reflexive but not necessarily right quasi-reflexive, and thus not necessarily quasi-reflexive.

An example of a coreflexive relation is the relation on integers in which each odd number is related to itself and there are no other relations. The equality relation is the only example of a both reflexive and coreflexive relation, and any coreflexive relation is a subset of the identity relation. The union of a coreflexive relation and a transitive relation on the same set is always transitive.

## Number of reflexive relations

The number of reflexive relations on an

n

-element set is
 n2-n 2

.

[5]

## Philosophical logic

Authors in philosophical logic often use different terminology.Reflexive relations in the mathematical sense are called totally reflexive in philosophical logic, and quasi-reflexive relations are called reflexive.[6] [7]

## References

• Levy, A. (1979) Basic Set Theory, Perspectives in Mathematical Logic, Springer-Verlag. Reprinted 2002, Dover.
• Lidl, R. and Pilz, G. (1998). Applied abstract algebra, Undergraduate Texts in Mathematics, Springer-Verlag.
• Quine, W. V. (1951). Mathematical Logic, Revised Edition. Reprinted 2003, Harvard University Press.
• Gunther Schmidt, 2010. Relational Mathematics. Cambridge University Press, .

## Notes and References

1. Levy 1979:74
2. Relational Mathematics, 2010
3. The Encyclopedia Britannica calls this property quasi-reflexivity.
4. Fonseca de Oliveira, J. N., & Pereira Cunha Rodrigues, C. D. J. (2004). Transposing Relations: From Maybe Functions to Hash Tables. In Mathematics of Program Construction (p. 337).
5. On-Line Encyclopedia of Integer Sequences A053763
6. Book: Alan Hausman. Howard Kahane. Paul Tidman. Logic and Philosophy — A Modern Introduction. 2013. Wadsworth. 1-133-05000-X. Here: p.327-328
7. Book: D.S. Clarke. Richard Behling. Deductive Logic — An Introduction to Evaluation Techniques and Logical Theory. 1998. University Press of America. 0-7618-0922-8. Here: p.187


text:
Remembering the golden ratio

I always forget how one can deduce the golden Ratio and its property. I hope somebody can explain me the chain of thoughts of its introduction in the book.

By property I mean that the reciprocal of the golden Ratio $$+1$$ is again the golden ratio. In the book the golden ratio is introduced by the reciprocal:

I.e. we know that the reciprocal $$h$$ of the golden ratio $$g$$ has the properties that $$1:h=h:(1-h)$$. That is the only thing I remember. It is something charachteristic what I don't see however is why the inverse of this number (the inverse of $$h$$ so I mean $$g$$) is so interesting? With other words can the golden ratio also be described in a way its reciprocal is described? It seems odd to me that this number has a name altough its key charachteritic actually lays in its inverse.

The book continues without further explaination by saying that from the above stated property one can say:

$$h^2=1-h,\quad g^2=1+g,\quad g=1+h$$

The order of the listed formulas give the Impression that we deduce them inductively. However I have two Problems, it took me some time to prove those formulas and more importantly I don't see how the second formula implies the last one. Maybe my chain of reasoning is not the most efficient or obvious one.

The first formula we get from $$h:1=1:(1-h)$$.It is something obvious the next step is not so obvious for me, because as I have said why do we look at the inverse?

We have defined $$g$$ as the inverse of $$h$$. So $$g^2=\frac{1}{h^2}$$. This step is also not intuitive for me.

So we have

$$g^2=\frac{1}{1-h}=\frac{1-h+h}{1-h}=\frac{1-h}{1-h}+\frac{h}{1-h}=1+\frac{1}{h}=1+g$$

The last formula I actually deduced by Looking at $$g$$ itself.

Prior to that I have tried to deduce the last formula from the second but I didn't get anything out from it:

$$g^2=1+g\iff g^2-1=g\iff \frac{1}{1-h}-1=\frac{1-(1-h)}{1-h}=\frac{h}{1-h}=\frac{1}{h}=g$$ There is nothing new.

By looking at $$g$$ itself however I could proof the third formula. So I don't see why we need the second formula because I didn't need it:

$$g=\frac{1}{h}=\frac{1-h+h}{h}=\frac{1-h}{h}+\frac{h}{h}=1+\frac{h^2}{h}=1+h$$

And because from the first formula one can say $$(h+\frac{1}{2})^2=1+\frac{1}{4}$$. And because $$1^2+(\frac{1}{2})^2=\frac{5}{4}$$. We can construct a right triangle with $$a=1$$ $$b=\frac{1}{2}$$. The hypothenuse must be the square root of $$\frac{5}{4}$$. Subtracting $$\frac{1}{2}$$ from it we get $$h$$. So we can construct $$h$$ with a circle and a ruler. And by adding 1 to $$h$$ we would get $$g$$ by the last formula.

The whole concept except for the fact $$1:h=h:(1-h)$$ just seems so random to me and I hope somebody can help me to connect the dots. I am pretty sure that I missed to unterstand the authors intention.

The golden ratio is usually denoted not by $$g$$, but by the Greek letter phi, often but not always written in its variant form $$\varphi$$. You've stated its determining characteristic correctly in words, and the proportionality, symbolically, is

$$1 : \varphi :: \varphi-1 : 1$$

It may be easier to express this in terms of fractions:

$$\frac1\varphi = \frac{\varphi-1}{1} = \varphi-1$$

If we multiply both sides by $$\varphi$$, we get

$$1 = \varphi^2-\varphi$$

which we can write as

$$\varphi^2 = 1+\varphi$$

which you wrote as $$g^2 = 1+g$$, or to put it in standard quadratic form,

$$\varphi^2-\varphi-1 = 0$$

At this point, we can use the quadratic formula (if you remember that), to obtain $$\varphi$$ itself:

$$\varphi = \frac{1 \pm \sqrt{5}}{2}$$

This expression gives two values for $$\varphi$$, one positive and one negative. By convention, the golden ratio is taken to be the positive value, since it has an easier geometrical interpretation. The negative value turns out simply to be the negative reciprocal of the positive value, anyway.

Does that help?

ETA: Suppose we denote the reciprocal of $$\varphi$$ by $$\Phi$$; that is, $$\Phi = 1/\varphi$$. Remember that

$$\varphi^2-\varphi-1 = 0$$

If we divide both sides by $$\varphi^2$$, we get

$$1-\frac1\varphi-\frac{1}{\varphi^2} = 0$$

which we can rewrite as

$$1-\Phi-\Phi^2 = 0$$

or, equivalently,

$$\Phi^2 = 1-\Phi$$

which you wrote as $$h^2 = 1-h$$.

Finally, multiply both sides of that last equation by $$\varphi$$, and you get

$$\varphi\Phi^2 = \varphi-\varphi\Phi$$

But $$\varphi\Phi = 1$$ (they're reciprocals), so this is really just

$$\Phi = \varphi - 1$$ $$\varphi = 1+\Phi$$

which you wrote as $$g = 1+h$$.


text:
## Polymorphic Bacteria

Published on Saturday, 20th April 2019, 01:00 pm; Solved by 201;
Difficulty rating: 45%

### Problem 666

Members of a species of bacteria occur in two different types: $\alpha$ and $\beta$. Individual bacteria are capable of multiplying and mutating between the types according to the following rules:
• Every minute, each individual will simultaneously undergo some kind of transformation.
• Each individual $A$ of type $\alpha$ will, independently, do one of the following (at random with equal probability):
• clone itself, resulting in a new bacterium of type $\alpha$ (alongside $A$ who remains)
• split into 3 new bacteria of type $\beta$ (replacing $A$)
• Each individual $B$ of type $\beta$ will, independently, do one of the following (at random with equal probability):
• spawn a new bacterium of type $\alpha$ (alongside $B$ who remains)
• die

If a population starts with a single bacterium of type $\alpha$, then it can be shown that there is a 0.07243802 probability that the population will eventually die out, and a 0.92756198 probability that the population will last forever. These probabilities are given rounded to 8 decimal places.

Now consider another species of bacteria, $S_{k,m}$ (where $k$ and $m$ are positive integers), which occurs in $k$ different types $\alpha_i$ for $0\le i< k$. The rules governing this species' lifecycle involve the sequence $r_n$ defined by:

• $r_0 = 306$
• $r_{n+1} = r_n^2 \bmod 10\,007$

Every minute, for each $i$, each bacterium $A$ of type $\alpha_i$ will independently choose an integer $j$ uniformly at random in the range $0\le j<m$. What it then does depends on $q = r_{im+j} \bmod 5$:

• If $q=0$, $A$ dies.
• If $q=1$, $A$ clones itself, resulting in a new bacterium of type $\alpha_i$ (alongside $A$ who remains).
• If $q=2$, $A$ mutates, changing into type $\alpha_{(2i) \bmod k}$.
• If $q=3$, $A$ splits into 3 new bacteria of type $\alpha_{(i^2+1) \bmod k}$ (replacing $A$).
• If $q=4$, $A$ spawns a new bacterium of type $\alpha_{(i+1) \bmod k}$ (alongside $A$ who remains).

In fact, our original species was none other than $S_{2,2}$, with $\alpha=\alpha_0$ and $\beta=\alpha_1$.

Let $P_{k,m}$ be the probability that a population of species $S_{k,m}$, starting with a single bacterium of type $\alpha_0$, will eventually die out. So $P_{2,2} = 0.07243802$. You are also given that $P_{4,3} = 0.18554021$ and $P_{10,5} = 0.53466253$, all rounded to 8 decimal places.

Find $P_{500,10}$, and give your answer rounded to 8 decimal places.


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## Friday, March 12, 2010

### Tamino vs random walk

Grant Foster has found a new inconvenient enemy, a random walk:
Not a Random Walk
He starts by saying that Euler has "proven" God by the following assertion:
Sir, (+ bn) / z = x, hence God exists. Reply!
In other words, Euler emitted some mathematical nonsense and considered it a proof of God. (See the slow or fast comments for details about this story.)

Off-topic: Local Theory of Relativity. Hat tip: Olda K. ;-)

Tamino's article continues as expected. Tamino emits some mathematical nonsense and considers it a proof of man-made global warming. His multi-kilobyte verbal emissions make no sense, except for making one point.

He's trying to "debunk" a comment by his reader who argued that the measured temperature series may be represented as a random walk - which is a much better model to describe the measured data than a linear trend combined with a white noise.

Of course, Tamino doesn't like it because it undermines the power of the AGW religious cult he faithfully believes. So he tries to produce some counter-arguments. The only counter-argument he produces is that random walks produce unbounded functions.

That's true except that it's irrelevant. To create a random-walk-based model that produces bounded functions, it's enough to replace it by an autoregressive model. Nonlinearities prevent the model from going "too far away" from the normal. However, during any reasonable timeframes, e.g. those shorter than 10,000 years, the model will be indistinguishable from a proper random walk.

Colors of noise

When we say that a function, "f(x)", resembles "white noise", it means that its values at different values of "x" are random and independent from each other. Such functions are inevitably completely discontinuous. If we use them as a model of temperatures, the temperature in the next year has nothing to do with the temperature of the previous year. It can suddenly jump to the temperatures seen in 1650.

Needless to say, this is not how temperatures behave. The temperatures are much more continuous than that. It's physically clear why it has to be so. One needs heat - energy - for the temperature of a large object to change. And heat - energy - can't flow infinitely quickly. We know this to be the case for theoretical reasons - and we also observe that quantities don't change infinitely quickly.

Is there a better, more continuous model of a random function (such as temperature series) than white noise? You bet. It's pink noise, an interpolation between white noise and red noise. So I have to explain you the red noise.

Red noise: random walk

A random walk is a function such that the increment, "f(x) - f(x-1)", is a random variable distributed symmetrically around zero. And for different values of "x", these increments are independent of each other. In other words, you may visualize a random walk as an integral of a white noise.

The character of the functions obtained as random walks is also called red noise - because it favors low-energy (red) components more than the white (color-neutral) noise does - or Brownian noise because the Brownian motion of pylon particles driven by the chaotic motion of water molecules is the oldest example of a random walk in Nature (observed by biologist Robert Brown, explained by Albert Einstein in 1905 and Marian Smoluchowski in 1906).

Random walks have one property that is so important that I will prove it for you. How far will you get after time "t" of random walk? What I want to prove is that the distance scales like "sqrt(t)", the square root of time. Why is it so?

First, a random walk may drive the pylon to the left or to the right: "f(t)" may be both positive and negative. We only want to know the absolute value. It's much more natural to talk about the squared absolute value, i.e. the squared value of "f(t)" (the sign squares to a "plus" in both cases). Generate many random walks with the same parameter and ask: what is the average value of "f(t)^2" computed out of all these random walks? Imagine that "t" is the number of steps.

We have
f(t) = f(0) + step(0 → 1) +
+ step(1 → 2) + ... + step(t-1 → t)
Let's choose the initial position "f(0)=0" and the steps of the random walk are random numbers equally likely to be positive and negative. The average value of the square is
[f(t)2] =
= [(step(0 → 1) + ... + step(t-1 → t))2] =
= ...
Are you following me? The "[...]" brackets represent the average value over many random walks. And inside these brackets, I have simply written the square of "f(t)". The latter is just the sum of "t" different steps because it's a random walk.

You should recall the formula for the square of a sum. When you expand this formula, there will be many "mixed terms" such as
[ step(0 → 1) . step(5 → 6) ]
However, the signs of the step from time 0 to time 1, and the step from time 5 to time 6, are independent of each other, by the definition of the random walk. So the average values of the type above are just zero! The positive contributions cancel against the negative ones. Consequently, only the squared individual terms contribute to the average value. Returning to the previous displayed equation, we learn that
[f(t)2] =
= [(step(0 → 1) + ... + step(t-1 → t))2] =
= [step(0 → 1)2 + ... + step(t-1 → t)2] =
= t . [step(0 → 1)2]
The average or "expectation value" we were computing is simply given by "t" (because there were "t" equal terms) times a universal parameter describing the typical step. It's more natural to take the square root of the result above. The typical
typical value f(t) = sqrt ( [f(t)2] ) =
= sqrt(t) x const
That's great. After time "t", the random walk drives you to a distance that is proportional to the square root of "t". A key formula for the random walk.

Matching temperature jumps with the random walk

You can actually see that this is a very good model for the most typical temperature changes (e.g. of the global mean temperature) after time "t" as long as "t" is shorter than 10,000 years or so. Just use this formula:
Typical temperature jump after time "t"
= sqrt(t / year) * 0.08 °C
You express the time in years, take the square root of it, and multiply the result by 0.08 °C. Look what it gives you for various values of "t":
1 year: 0.08 °C
10 years: 0.25 °C
100 years: 0.8 °C
1000 years: 2.5 °C
10,000 years: 8 °C
This table by itself looks excellent. The final figure, 8 °C, is the typical difference between the temperature during ice ages and interglacials. The middle figure, 0.8 °C, is approximately the estimated warming in the last 100 years. It just seems to work beautifully.

For timescales longer than 10,000 years, the observed temperature changes stabilized. They were never much bigger than those 8 °C, even during the millions of years. At the level of our theoretical model, it's easy to add nonlinear terms that discourage too big deviations from some idealized mean - deviations by more than 10 °C. Such terms can also be given a natural physical explanation. They're negative feedbacks of various types. They "complexify" the simple random walk model into an autoregressive model.

Now, if you study the detailed numbers more accurately, you will find out that the actual color of the noise describing various temperatures is not quite red: it's somewhere in between the white noise and the red noise, even when you compare e.g. the variations after centuries with those after decades.

We need to understand some basic power laws in a more coherent and unified way. The typical temperature jump is proportional to a power of the time separation. The exponent depends on the color of the noise:
white noise: jump in T goes like t^0
red noise: jump in T goes like t^(0.5)
pink noise: jump in T goes like t^(0.25)
The white noise had the exponent equal to zero. Anything to the zeroth power equals one: so the zeroth power means that the temperature jump was time-independent for the white noise. No matter how much time you wait, you will deviate from the original temperature by the same step.

Similarly, the 0.5-th power is the square root that I derived for the random walk - or red noise - previously.

Pink noise is a technical name for a random function whose typical jump as a function of time scales as a power law in "t" - a power law with an exponent exactly in between the white noise and the red noise. You can create "pink noise functions" e.g. by producing "white noise" for the Fourier components, multiplying them with a proper exponent of the frequency (corresponding to the desired power law), and by Fourier-transforming them back to the time variable.

Various temperature series are pretty close to the pink noise but you must be ready that the detailed exponent depends on the context, the local or global character of your measurements, and it also depends on the timescale "t" itself.

Also, there are big irregularities if "t" is just a few years because the El Nino / La Nina dynamics is not quite random. It is "more periodic" than a typical random function - although El Ninos and La Ninas surely don't alternate "exactly periodically".

At any rate, it is straightforward to construct statistical models based on pink noise, perhaps with some autoregression included, that describe all the statistical features of the measured data very well. Recall that I "predicted" the UAH temperature until 2100:

Note how natural, continuous this extrapolation beyond 2010 looks like. Whenever someone tells you that he has a nice theory of climate change - of the variations of the temperature - you should make some basic tests. You should check how large the "typical jump" of the temperature is, and how it depends on the time separation "t". In intervals comparable to a few years up to a century, the dependence will be closely matching a power law. The exponent is interesting and you should compare it between the models and the observations, too.

There are lots of deep physical data that can be obtained by a proper statistical analysis of the time series. Most of the data is thrown away by the IPCC types. They're not interested in any of that. They're not interested in the exponents that determine the color of the noise, the correlation lengths, the correlation times, and the processes that are responsible for them. They're interested in a "trend" which has really nothing to do with the variability because a description in terms of a "trend" assumes that the deviation of the temperature from a linear function (that they try to identify with the climate even though no such a function plays any important role in the climate, and no one knows how this function should look like, anyway) is white noise - but it surely is not.

The real deviation of the temperature from a "normal long-term temperature", whatever you choose it to be, is close to a pink noise and the precise parameters of this pink noise, especially the amplitude and the exponent, are extremely important. They're what the climate variability is all about. They're the most important parameters whose predictions by the theories and models should be tested against the observations.

I challenge Grant Foster or anyone else to find the simplest "pinkish noise" model of whatever temperature series they like. Fine-tune the amplitude and the exponent. Assume that the temperatures are "pinkish noise" given purely by these two simple parameters. And try to falsify this theory by a statistical test.

To be honest, I actually know the ways how to solve this task because I have learned not only the amazing depth included in this simple model but also many of its limitations. The real climate system is more complex and similar power laws, while very useful in many intervals, break down behind their range of validity, much like any phenomenological theory you can invent. In various regimes, you may also see that there are probably "two different kinds of noise" with two different exponents being added.

But these pinkish noise models are still extremely good, and infinitely more accurate descriptions of the real observations than a "linear trend plus white noise" model that the AGW types would like the people to buy. The white noise is perhaps easier for the laymen to understand than the red noise, and certainly more so than the pink noise, which is why people often buy this naive theory of the climate (trend plus white noise). That's a pity because a priori, it's a very ill-motivated physical model. And a posteriori, it's a model that doesn't agree with the observations well.

More generally, I want to emphasize that a working predictive theory of the climate can never "neglect" the "natural variability". Even if there were an important man-made trend, it's clear that the climate variability is important, too. It's damn important to know how large it is and how it depends on the timescale (and distances), so that we also know the typical timescale where the "man-made linear trend" could start to beat the natural variability. We surely know that the timescale from the previous sentence is longer than 15 years because it's pretty easy for the natural factors to beat the man-made effects for 15 years because there's been no statistically significant warming since 1995. But it may be much longer.

So even if the man-made trend existed and were large, it's completely self-evident that most of the research concerning "climate change" would have to focus on the variations which are obviously of natural origin and they have always existed, pretty much with the same amplitudes and exponents. The climate has been around for billions of years and it's been changing for billions of years.

You can learn exactly nothing if you deny all of this and you only focus on some hypothetical, politically motivated term that only existed for 100-150 years. The Earth's long history doesn't deserve to be denied in this way. While the Young Earth people at least admit that the Earth is older than 5,000 years, the AGW cultists want to deny any history of the climate before the year 1850 or so.

People like Grant Foster are just looking at the climate in a completely wrong way. It is a way based on pre-existing dogmas - and a way that prevents one from ever learning anything besides these dogmas. It's a method that discourages the people from studying anything that looks "random": all "random" things are just inconvenient and should be thrown away. They only want to see "non-random" signals which can never be isolated in the real climate because most of it is random and the details how it is random mater.

In other words, in the real climate, these random phenomena are the essence. And the color, character, and consequences of their "random" behavior hides the most important keys about the climate change, the most important collection of material that our theories should agree with. And it may potentially give us all the predictions of a future climate, too.

And that's the memo.

1. The anecdote concerning Euler's "proof of God" as I recall has Euler at a party where the hostess warned him that another rather obnoxious guest was present. This other guest, whom I think was also famous though his name escapes me, was an outspoken but mathematically challenged atheist. Euler approached him, uttered his irrelevant mathematical expression, and demanded that the other man refute it.

2. The anecdote about Euler, repeated by Eric Temple Bell in "Men of Mathematics," has been shown by historians to be apocryphal and likely dismissed as absurd by the intended audience even if it was true.

Lubos is correct about pink noise modeling and I think the following is true - for the exponent a on time t for o<a<1, the corresponding Fourier series representation of the function is absolutely convergent everywhere (for a=0 there are functions with divergent Fourier series everywhere)

3. Lubos,

The increase in global avg temp since 1880 is not merely random: A random increase would cause a negative energy imbalance or the extra energy would have to come from another segment of the climate system (eg the ocean, cryosphere, etc). Neither is the case: There is a positive energy imbalance and the other reservoirs are also accumulating energy.

I think it is fair to conclude that the observed increase therefore has not been random.

Moreover, there is a known positive radiative forcing; measurements indicate an enhanced greenhouse effect (more IR being emitted to space and being more IR being reflected back to the surface); changes in other segments of the earth system also indicate global warming.

A different question would be, if these data, without any physical constraints on it, could mathematically be described as "VS" does (purely stochastic; random walk). Perhaps it could; I refrain from an opinion on that. I do note though, that in light of the physical system that these data are a part of, this is a purely academic mathematics question. The physics of it all tells me that it hasn’t in fact been random, since it is inconsistent with other observations.

I provide a longer reasoning at my blog: http://ourchangingclimate.wordpress.com/2010/03/08/is-the-increase-in-global-average-temperature-just-a-random-walk/

Bart Verheggen

4. Dear Bart,

your "argument" that the changes are not random is completely nonsensical. You say that there's a positive energy imbalance. Well, I am not really sure.

But even if it is, the probability for the imbalance to be positive given a random evolution - e.g. for a Markov process - is 50%. So that's not shocking that it's positive. It can only be positive or negative. And by the way, if it were negative, you would conclude the (almost) same thing - but the "worry" would be global cooling.

After all, this is not just a speculation because this is what some people were worried about 35 years ago.

So while I know methods how to exclude various excessively simple random models, you haven't offered any valid argument - not even a glimpse of an argument - that would suggest that the temperature series were not random (similar to pink noise).

Cheers
LM

5. I have just posted a series of statistical tests on this issue:

Everybody is welcome to take a look and inspect my method / arguments, here:

Summarizing: Using these different statistical tests, we find:

ADF: Clear presence of a unit root

KPSS: Stationarity (no unit root) rejected at 5% and 10% sig, not at 1% sig.

PP: No presence of unit root, but only when using 'intercept and trend' in the test equation specification

DF-GLS: Clear presence of a unit root

In my view, this is enough statistical evidence to conclude that the GISS combined surface and land record is in fact integrated of the first order, or I(1).

VS

6. An illuminating post. In other words "climate science" is riddled with confirmation bias.


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# Finding point distribution by eigen vectors

First of all I want to tell that my mathematics is poor, so I can’t use correct terms. Sorry for that. I have a point data set. This data represents some cylindrical objects surfaces (not exactly cylindrical due to occlusion part of the surfaces, represent by point data. ). In addition to cylindrical objects there are noisy data. What I want is to extract points represent cylindrical surfaces out of the other points. My idea is to use eigen vector analysis for that. So I already computed relevant eigen values (x0, x1, x2) and their eigen vectors (v0, v1, v2) for each point. For this, I used neighbor points, i.e., within given distance (for example 0.2m), of each point. I roughly aware of the vector (v2), relevant to eigen value two (x2), represents the direction of point distribution along the surface.

My question is how can i compare each 3d vector (v2) with other neighbor vectors in order to extract points which have similar point distribution direction (having nearly equal vectors). I’m poor in vector analysis. I am not sure the dot product between any two adjacent vectors is enough for this.

Thank you in advance.

-
You might want to use least-squares for this. A cylinder is given by $x^2+y^2=r^2$, where the $xy$-plane is in the base of the cylinder, and the $z$-axis is along the center. Using that constraint, you could transform your data to get a cylinder, and then remove the $z$-coordinate. From there, I would think the cylinder constraint above would allow least-squares to work. This would get you an "average" for the cylinder, and then you could determine the distances of points from that average, and choose points where the distance is small enough to your liking. Is this like what you want? –  Eric Stucky Jun 3 '12 at 20:50
@Eric Stucky: thanks for your ideas but it is not what I wanted. Actually, it doesn’t have exact cylindrical shape (it likes small arc in 2D, so can not fit a correct cylinder), so that, I want to get only points which represent similar principal direction. I hope that we can assume small regions of these points as planar surfaces. So, I selected small regions based on their neighbors and calculated their eigen values and eigen vectors. Then, I want to find points which have very similar distribution direction. –  sana Jun 4 '12 at 11:01
@Eric Stucky: Actually, for this I check angle in between 2 vectors of near points to find whether it is too small or not (based on my given threshold). But I found angle difference is always very large for example more than 85.. 114...degrees., so I am confusing with this. Also, this object has an almost vertical shape in 3D. So I also check vectors (eigen vector related to the largest eigen value) of each points with vertical vector (0, 0, 1) for check their behavior and interesting I got very large angle difference too.So, really, I can not understand the behavior of eigen vectors. –  sana Jun 4 '12 at 11:03
@Eric Stucky: Cloud you have any idea about this? –  sana Jun 4 '12 at 11:04


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# Math Help - antiderivative problem

1. ## antiderivative problem

i have no idea where to start with this one:
1/x(1+lnx)

2. Sub $u = \ln{(x)}$

3. it always seems so obvious once it's pointed out to me.
thank you.


text:
# All Questions

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Question. To find the integral of- $$\int{1/(x^4+1)^{1/4}} \, \mathrm{d}x$$ I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?
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### what is the ordered triple of postive Integers, ABC = 2104000 [on hold]

What is the ordered triple of positive integers (a,b,c) satisfy abc = 2104000 Sorry but I do not know anything about this problem.
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### Are cyclotomic polynomials irreducible modulo a prime?

I am actually interested on cyclotomic polynomials of the form $x^n+1$ (i.e. $n$ is a power of 2, for large $n$). Are these polynomials irreducible modulo a prime $q$? I already saw that $X^2+1$ is ...
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### How many five-digit numbers are there that have number 4 as at least one digit?

How many five-digit numbers are there that have number 4 as at least one digit? How to do this? I don't know how to start.
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### Why is the complex plane shaped like it is?

It's always taken for granted that the real number line is perpendicular to multiples of $i$, but why is that? Why isn't $i$ just at some non-90 degree angle to the real number line? Could someome ...
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### Dependence of algebraic elements in a finite field

Lets work over the finite field $\mathbb{F}_p$ for a prime $p$. Consider a monic irreducible polynomial $f(X)=X^3+aX^2+bX+c$ in $\mathbb{F}_p[X]$. Let $x$ be a root of $f(x)=0$ (say, in the closure of ...
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### Regarding pedigrees

Here's a errrhmm.... a modelling problem. Applied math. Or something. We're given pair of sets $X,Y$ (for notational simplicity, assumed disjoint). Let $T_{X,Y}$ denote the following disjoint union:...
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### Nonlinear first-order differential equation with periodic bounded solution

Let $f:\mathbb R\times \mathbb R\longrightarrow\mathbb R$ be a $C^1$ function, periodic in the first variable, and such that $f(t,1)\leq > 0\leq f(t,0)$ for all $t$. Consider the ...
3 views

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### Example of a set and monotone class where monotone class is not a $\sigma$-algebra?

What is an example of a set $X$ and a monotone class $\mathcal{M}$ consisting of subsets of $X$ such that $\emptyset \in \mathcal{M}$, $X \in \mathcal{M}$, but $\mathcal{M}$ is not a $\sigma$-algebra?
37 views

### Construction of Lebesgue measure in Rudin's RCA book

This theorem from Rudin's RCA book. Here's one moment from it's proof which seems to me very weird. Rudin states that equality $\lambda(E)=m(E)$ holds for all Borel sets. But I think that it's ...
56 views

### Is there a concept of |x|<0?

I'm just curious, did any of the famous mathematicians consider |x|<0? Would the zero have to be then not the additive identity and what else would it be then? (assuming you could build a field ...
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### Is $\pi^1:C_c(G)\rightarrow \operatorname{End}(H)$ a homomorphism of the convolution algebra when $G$ is not unimodular?

Let $G$ be a Hausdorff locally compact group and $H$ a Banach space. Let $\pi:G\rightarrow \operatorname{GL}(H)$ be a representation and define $$\pi^1(\phi)v = \int_G\phi(x)\pi(x)vdx$$ for $v\in H$ ...
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### What is the derivative of $x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}$

What is the derivative of $$x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}$$ My effort: Let $$g(x)=x!^{x!^{x!^{x!^{x!^{x!^{x!^{.{^{.^{.}}}}}}}}}}\implies g(x)=x!^{g(x)}$$ Taking natrual log on ...


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# What kind of functions cannot be described by the Taylor series? Why is this?

It's true that I'm not familiar with too many exotic functions, but I don't understand why there exist functions that cannot be described by a Taylor series? What makes it okay to describe any particular functions with such a series? Is there any difference for different number sets? In the case of complex numbers maybe? Could somebody provide an example?

• A Taylor series exists if and only if the function is infinitely differentiable at some a. – Doug M Feb 2 '17 at 3:46
• And even then, the Taylor series doesn't always converge to the function in some neighbor. @DougM – Thomas Andrews Feb 2 '17 at 3:48
• @ThomasAndrews Indeed, I thought about going there, but then decided to keep it to the point. Radius of convergence is then a whole 'nuther thing. – Doug M Feb 2 '17 at 3:50
• – Winther Feb 2 '17 at 3:52
• The Taylor series represent the function when the Taylor remainder tends to zero, otherwise the Taylor series doesnt represent the function. – Masacroso Feb 2 '17 at 4:03

We have the somewhat famous function:

$$f(x)=\begin{cases}e^{-1/x^2}&x\neq 0\\ 0&x=0 \end{cases}$$

is infinitely differentiable at $0$ with $f^{(n)}(0)=0$ for all $n$, so, even though the function is infinitely differentiable, the Taylor series around $0$ does not converge to the value of the function for any $x>0$.

Technically, any function that is infinitely differentiable at $a$ has a Taylor series at $a$. Whether you find that Taylor series useful depends on what you want the series to do.

For example, if given a $g$ infinitely differentiable at $0$, the we know that there exists $C,\epsilon>0$ such that:

$$\left|g(x)-\sum_{k=0}^{n} \frac{g^{(k)}(0)}{k!}x^k\right|<Cx^{n+1}$$

for all $|x|<\epsilon$.

So the finite terms of the Taylor series are in some sense always the "best" polynomial for agreeing with function.

So what happens to our function $f$ above is that $f(x)$ converges to $0$ faster than any function $x^n$.

What we don't always get, for real functions, is a Taylor series that converges to the function in the interval.

In complex numbers, things become intriguing. It turns out, if you define differentiation on complex functions in a relatively simple way, then any function which is differentiable at a point is infinitely differentiable at that point, and the Taylor series converges in some "ball" centered on that point.

• I've never understood why so many find this surprising; I guess it is because many are taught (incorrectly) that Taylor Series must converge to the function. I was taught from the beginning that a Taylor Series simply provides an approximation in terms of derivatives around a point and might happen to converge. In the neighborhood of $0$, the function $e^{-1/x^2}$ is approximated quite well by $0$ – Brevan Ellefsen Feb 2 '17 at 4:02
• @BrevanEllefsen: That's not really why. It's because you would intuitively expect a smooth function to be predictable, and it seems weird that the derivatives of a function do not contain enough information to predict it at a nearby point. – Mehrdad Feb 2 '17 at 7:59
• Yes, but that function does have a Taylor series, it just doesn't agree with the function anywhere but at 0. – Klangen Feb 2 '17 at 14:21
• @BrevanEllefsen But any infinitely differentiable function in complex numbers does have a converging Taylor series, so the question for reals is interesting. I wouldn't say it is surprising that there are counter-examples in the reals, but it is interesting. – Thomas Andrews Feb 2 '17 at 15:24
• One of my favorite "proofs" that complex numbers "exist" is that the Taylor series at the real number $a$ for the function $\frac{1}{1+x^2}$ has radius of convergence $\sqrt{1+a^2}$. This is entirely a statement about real numbers, but it suggests there is some root of $1+x^2$ that is $\sqrt{1+a^2}$ distance away from each real number $a$. (If $p(x)$ has all real roots, then the radius of convergence at $a$ of $\frac{1}{p(x)}$ is always the distance to the nearest root to $a$.) @BrevanEllefsen – Thomas Andrews Feb 2 '17 at 15:43

If the limit of the Lagrange Error term does not tend to zero (as $n \to \infty$), then the function will not be equal to its Taylor Series.

You can also read more on this in Appendix $1$ in Introduction to Calculus and Analysis $1$ by Courant and John. Hope it helps.

I think the intuition you want is the fact that functions that are not complex-differentiable* (also known as holomorphic) are not described by a Taylor series.

And to give another example that is perhaps even more unexpected than the one given by Andrew:

$$f(z) = \begin{cases} e^{-\frac{1}{z}} && \text{if } z > 0 \\ 0 && \text{otherwise}\end{cases}$$

This function is smooth and zero over an infinitely long interval, and yet nonzero, because it is not holomorphic.

*If you're not familiar with complex differentiation, it's like real differentiation, with $h$ complex:

$$f'(z) = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h}$$

For details, see here.

The existence of functions that cannot be described by Taylor series is actually completely intuitive; take the indicator function of the rational numbers viewed as a subset of the reals, for example. Try to keep in mind that functions can be really... arbitrary.

Much more subtle is the existence of smooth functions that aren't analytic; Thomas Andrews gives the standard example of such a beast. Fwiw, my understanding of why this is possible is that okay, there's functions that change behaviour suddenly at a point, BUT the change in behaviour at that point is so gradual, so gentle, so smooth, that none of the function's derivatives can see the change happening; therefore, the Taylor series can't, either.

In addition to all the comments here, I would like to add the curious Weierstrass function, which is known for its quality of being nowhere differentiable despite the fact that it is continuous everywhere:

$$W(x) = \sum_{n=0}^\infty a^n\cos(b^n\pi x)$$

Consequently, it does not have a Taylor series.

You can find a visualization of $W$ here.

• This is a nice fractal! Are all fractals no differentiable? – dmtri Aug 23 '18 at 10:55
• @dmtri while I'm unsure of a formal definition of a fractal, differentiable functions must look "linear" if you "zoom in enough". Fractals self similarity on "zooming in" seems to rule this out. – Mark Aug 23 '18 at 16:02


text:
# A derivation in Jaynes' paper, linking stat-mech and Shannon's entropy

I have been going through E. T. Jaynes' 1957 paper, Information Theory and Statistical Mechanics. There is a step in his derivations, which has been giving me headaches for the past day; would appreciate some pointers on how to complete it!

A bit of notation: Lagrange multipliers are represented by $$\lambda$$'s, and $$\lambda_0$$ = ln Z where Z is the partition function. The probability of each state $$j$$ is

$$$$p_j = \frac{\exp \left[ -(\lambda_0 + \lambda_1 f_1(x_j) + \cdots \right]}{Z}.$$$$

The expectation values of the functions $$f's$$ are fixed/known, and provide the constraints for maximizing Shannon's entropy; for simplicity let's consider only $$f_1$$. Assume that the probabilistic states $$i$$'s are discrete.

Now onto the question. Jaynes says near Equation (5.1) of the paper that he's going to perturb $$f_1$$ such that $$f_1(x_j) \rightarrow f_1(x_j) + \delta f_1(x_j)$$, for all $$j$$. At the same time, the expectation value of $$f_1$$ is independently altered: $$\left \rightarrow \left + \delta\left$$.

How does the entropy, the derived probability distribution, and the Lagrange multiplier change? Equation (5.1) states that

$$$$\delta\lambda_0 = \delta \mathtt{ln} Z = -\left( \delta\lambda_1 \left + \lambda_1 \left<\delta f_1\right> \right).$$$$

But how? Here's my current approach:

$$\begin{eqnarray} \delta\lambda_0 &= ln Z' - ln Z \\ &= ln (Z' / Z) \\ &= ln (\frac{\sum_j \exp[(\lambda_1 + \delta\lambda_1)\times (f_1(x_j) + \delta f_1(x_j))]}{Z}). \end{eqnarray}$$

But then I am pretty lost on how to "get rid of" the logarithmic function...Thanks for reading, and looking forward to see what you think!

• I do not think the tag "research-level" is appropriate for this question. – lattitude Jun 9 at 7:21
• That's fair --- even though I am reading the paper for grad work, the math is admittedly "basic" relative to the curriculum. Tag removed. – Eve L Jun 10 at 20:35

Remember that $$\delta\lambda_1$$ and $$\delta f_1$$ are small so expanding the exponential and then the log to first order: \eqalign{ \delta\lambda_0&\simeq\ln\Big[{1\over{\cal Z}}\sum_je^{-\lambda_1 f_1(x_j)-\delta\lambda_1 f_1(x_j)-\lambda_1\delta f_1(x_j)}\Big]\cr &=\ln\Big[{1\over{\cal Z}}\sum_je^{-\delta\lambda_1 f_1(x_j)-\lambda_1\delta f_1(x_j)}e^{-\lambda_1 f_1(x_j)}\Big]\cr &\simeq \ln\Big[{1\over{\cal Z}}\sum_j\Big(1-\delta\lambda_1 f_1(x_j)-\lambda_1\delta f_1(x_j)\Big)e^{-\lambda_1 f_1(x_j)}\Big]\cr &=\ln\Big(1-\delta\lambda_1\langle f_1\rangle-\lambda_1\langle \delta f_1\rangle\Big)\cr &\simeq -\delta\lambda_1\langle f_1\rangle-\lambda_1\langle \delta f_1\rangle } Alternatively, a faster calculation is \eqalign{ \delta \ln{\cal Z}&={1\over{\cal Z}}\delta{\cal Z}\cr &={1\over{\cal Z}}\sum_j\Big(-\delta\lambda_1 f_1(x_j)-\lambda_1\delta f_1(x_j)\Big)e^{-\lambda_1 f_1(x_j)}\cr &=-\delta\lambda_1\langle f_1\rangle-\lambda_1\langle \delta f_1\rangle }

• Thank you thank you! My main problem was that I couldn't get the right expansion; it is great to see the steps worked out :) – Eve L Jun 10 at 20:41

The equation 5.1 in the Jaynes' paper can be understood from basic knowledge in differential equations. $$\delta lnZ$$ can be written as

$$\delta \lambda_0 = \delta lnZ = \frac{\delta lnZ}{\delta \lambda_1}\delta \lambda_1 + \sum_{i}\frac{\delta lnZ}{\delta f_1(x_i)}\delta f_i(x_i)$$

Here, we know that the first factor in the first term is the expectation value of the variable corresponding to that Lagrange multiplier. Similarly, the first factor in the summation turns out to be $$p_i$$, the probability, times $$\lambda_1$$. Thus,

$$$$\begin{split} \delta \lambda_0 & = -(\delta \lambda_1\langle f_1\rangle + \lambda_1\sum_i p_i \delta f_i(x_i))\\ & =-(\delta \lambda_1\langle f_1\rangle + \lambda_1\langle \delta f_1\rangle) \end{split}$$$$

You can derive this expression using your way of taking a difference between the new and old $$\mathcal{Z}$$, but the same method needs to be used to split the equation and simplify it.

• Thanks for providing a framework for solving the problem! I have to accept 1 of 2 answers, but it doesn't mean I like yours less :P – Eve L Jun 10 at 20:48


text:
# Issue with OLS Regression for Nelson Siegel Svensson parameters

I have been working on getting input parameters to the Non-Linear Optimization which gives the Nelson Siegel Svensson model parameters and am carrying out the OLS regression as described in this answer. However, the input parameters obtained from the OLS are too far off the actual parameters, which I checked against some parameters I actually do have. I am using the equations shown in 'Figure 5' on Page 12 of this paper, and obtain the yield data, by choosing Par Bonds and using their coupons as Par Yields to bootstrap from to get the Spot Rates, which appears to be an okay method based on Page 3 of this paper. The code that I use is below, where I've just implemented the formula in the previous link and have carried out the regression in Python. My query is if there is an issue with the way I set matrix_of_params or if it could be to do with the data in df itself.

I run the function above for different values of tau_1 and tau_2. I then have a function to get the params associated with the lowest residuals, which I am positive is correct.

#df is a Dataframe containing all the data about the Bonds
def obtainingparams(self, df, tau_1, tau_2, residuals):
values = []
face_values = df['FACE_VALUE'].values #Writing face values to an array
yields = (df['coupon'].values) #COUPON = YTM for Par Bonds
spot_rate = np.zeros((yields.shape[0]))

#Calculating Spot Rates
for x, value in np.ndenumerate(yields):
index = x[0]
if index == 0:
spot_rate[index] = (yields[index]/face_values[index]) * 100

else:
if index < spot_rate.shape[0]:
for i in range (0, index, 1):
spot_rate[index] = (2 * ((np.power(((((face_values[index] + ((value*face_values[index]/200)))/term_1))),1/(index+1)))-1))*100

matrix_of_params = np.empty(shape=[1, 4])
months_to_maturity_matrix = df.months_to_maturity.values #Writing months to maturity to an array

#Populating the Matrix of Parameter Coefficients
count = 0
for x, value in np.ndenumerate(months_to_maturity_matrix):
if count < months_to_maturity_matrix.shape[0]:
months_to_maturity = months_to_maturity_matrix[count]
years_to_maturity = months_to_maturity/12.0

#Applying the equation in the link
newrow = [1, ((1-np.exp(-years_to_maturity/tau_1))/(years_to_maturity/tau_1)), ((1-np.exp(-years_to_maturity/tau_1))/(years_to_maturity/tau_1))-(np.exp(-years_to_maturity/tau_1)), ((((1-np.exp(-years_to_maturity/tau_2))/(years_to_maturity/tau_2))))-(np.exp(-years_to_maturity/tau_2))]
count = count + 1

#Just adding the new row to the matrix of parameter coefficients
matrix_of_param_coefficients = np.vstack([matrix_of_params, newrow])

#Carrying out OLS Regression
params = np.linalg.lstsq(matrix_of_params,spot_rate)[0]
residuals = np.sqrt(((spot_rate - matrix_of_params.dot(params))**2).sum())

#To keep track of which params are associated with which residuals
values.append((tau_1, tau_2, residuals, params))
return values

Thank You

• @haginile With reference to your answer here would I need to divide newrow in the code above by 100? Thank You. – Jojo Aug 5 '15 at 21:56


text:
## Calculus (3rd Edition)

Published by W. H. Freeman

# Chapter 3 - Differentiation - 3.3 Product and Quotient Rules - Exercises - Page 122: 23

-80

#### Work Step by Step

We use the quotient rule to find the derivative of $\frac{x^{4}-4}{x^{2}-5}$ at $x=2$. $\frac{d}{dx}(\frac{x^{4}-4}{x^{2}-5})=\frac{(x^{2}-5)\times\frac{d}{dx}(x^{4}-4)-(x^{4}-4)\times\frac{d}{dx}(x^{2}-5)}{(x^{2}-5)^{2}}$ $=\frac{(x^{2}-5)4x^{3}-(x^{4}-4)2x}{(x^{2}-5)^{2}}=\frac{2x^{5}-20x^{3}+8x}{(x^{2}-5)^{2}}$ $\frac{dy}{dx}|_{x=2}=\frac{2(2)^{5}-20(2)^{3}+8(2)}{((2)^{2}-5)^{2}}=-80$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
## Nov 15th 2008How to have the next slide in a second screen in PDF presentations

I do my presentations in latex – beamer, and something that I miss from other software applications is to see the upcoming slide in the notes screen, so I can prepare my speech and make the transitions between slides smoother.

Although I usually use pstricks, and typeset everything with Tex + Ghostscript, for this presentation I’m going to use pdftex so I can include pdf as graphics.

Here is what I did, and you might find it useful too, tweak it as needed:

Compile your presentation without notes first, and include an extra slide (empty if you like).

Save it with another name, e.g.: reference.pdf

\usepackage[absolute,showboxes]{textpos}
\usepackage[]{eso-pic}
\TPGrid{210}{297}
\newcounter{pdfPage}
\newcommand{\insertslideintonotesA}{
\begin{textblock}{100}(125,177)
\includegraphics[page=\value{pdfPage} , scale=.4]{reference.pdf}
\end{textblock}
}

And these lines after your \begin{document}

\AtBeginNote{
\scriptsize
}\AtEndNote{
\insertslideintonotesA
}

Compile again your presentation WITH notes, and delete the extra slide you included in the first step.

Your next slide will appear in the bottom right of the notes screen.

You can use tdclock.sty (see more here) if you want to have a clock too, although in Mac, it doesn’t work with SplitShow.

You will need to add this line to the preamble:

\usepackage[timeinterval=30]{tdclock}

this line to your first slide:

\initclock

and this line to your \AtBeginNote{} section

\tdtime

Posted by julovi / Salt and pepper

## Apr 28th 2008Using Apple remote control with PDF presentations

I do all my presentations in Latex/Beamer and I wanted to control my presentation with the remote control of my MacBook. Apple support already Keynote, and there are applications for Microsoft PowerPoint. So, frustrated by that, I did it myself and here it is for you to use it freely:

1. Get and install iRed Lite from http://www.filewell.com/iRedLite/

This is the button / action pairs:

• Play/Stop: Enter in Full Screen Mode
• |<< : Backward one slide
• >>|: Forward one slide
• double |<<: First slide
• double >>|: Last slide
• +: Last Link/ Field/ Comment

This setup works for me, however, if you don’t like it, you can easily edit it using the iRed Lite – Editor. The truth is that it’s very simple.

Let me know if you find this helpful.

NOTE:

Acrobat Reader layer for iRed Lite

This program is free software: you can redistribute it and/or modify
the Free Software Foundation, either version 3 of the License, or
any later version.

This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
GNU General Public License for more details.

You should have received a copy of the GNU General Public License
along with this program.  If not, see <http://www.gnu.org/licenses/>

Posted by julovi / Salt and pepper


text:
# Finding the basic and free variables of this matrix

Assuming that A is already in row reduced echelon form

$$A= \begin{pmatrix} 1 & -1 & 0 & 0 & -5/3 & 6\\ 0 & 0 & 1 & 0 & -1 & -3 \\0 & 0 & 0 &1 & -2/3 & -2 \end{pmatrix}$$

(this is a row reduced echelon form matrix that i obtained from a system of linear equations, through gaussian elimination, that used the variables x1, x2, x3, x4,x5)

from my understanding, the basic variables are x1, x3 ,and x4. the free variables is just x2. But then we have a missing variable x5 that does not correspond to any pivot in the matrix, does this mean that x5 is NEITHER a free variable nor a basic variable

• Everything except $x_1, x_3$ and $x_4$ is a free variable. – user8277998 Dec 15 '17 at 20:30
• ah i understand now. So we have to use the logic that if a variable is not basic, then it must be free. thank you – Soon_to_be_code_master Dec 15 '17 at 20:32
• Is that last column sixth column of the matrix ? Or it is the augmented column ? – user8277998 Dec 15 '17 at 20:40
• it is an augmented column, sorry i forgot to mention that. – Soon_to_be_code_master Dec 15 '17 at 21:16