TITLE: How to prove that as one zooms in on a point of the graph of a differentiable function, it looks more and more like the tangent at the point?
QUESTION [1 upvotes]: How can one prove that as one zooms in on a point of the graph of a differentiable function of one variable, the graph looks more and more like the tangent line at the point?
One can see this phenomenon with the aid of computer graph plotters and their zooming capabilities, but how can one really prove this?
Also, it is intuitively clear, but what does it mean mathematically "to look more and more like the tangent line"? I guess we first need a proper mathematical definition of this intuitive notion before we proceed to a proof of the original question, but I have no idea how to obtain one (maybe it considers the "error" made by this tangent approximation in comparison to that of other straight lines at the same point? I really don't know).

REPLY [2 votes]: Why not mimic the computer graph plotter idea, namely that after rounding off coordinates to the pixels on a screen, they form a straight line, and otherwise we could compute the standard deviation (via the sum of squares) from the tangent pixels to the graph pixels:

Let us make it simple and define the following screen: $[-b,b]\times(-\infty,\infty)$ pixels and the graph of $f$ translated such that the point of interest $(a,f(a))$ becomes centered at $(0,0)$ on the screen. Then we define a $k$-zoom as meaning that one pixel represents $1/10^k$ units on the graph. For each value of $x:=a+n/10^k$ and corresponding $y=f(x)$ we plot the point $$\newcommand{\round}{\operatorname{round}}P:=\left(\round(10^k(x-a)),\round(10^k(y-f(a))\right)$$ on the screen and compare point by point to the same type of plot of the tangent $$t(x)=f'(a)(x-a)+f(a)$$

Then what we have to prove is that the sum of squares of $y$-differences between the pixel plot of $f$ and $t$ converge to zero (as $k$ tends to $\infty$). Note that those differences can be upper bounded by differences of the form $|y_f-y_t|+1$ so this suggests that solving the problem without rounding off to integer pixel values will essentially give us the same metric of how well they match.


TITLE: Optics, image at infinity
QUESTION [0 upvotes]: I've been trying to look at other questions here but I cannot find to late on. I'm trying to replicate Young's original experiment, trying to find a fraunhofer diffraction, this is basically diffraction in infinity.
To solve this problem I'm using a lens. There I will have my light (collimated), striking the object, then through the lens and into the camera. He says I need f-f setup, that is the distance f from the object to the lens, and f from the lens to the camera. But when I use the thin-lens approximation, all I can get is that the length from the object to the lens should be f, what am I missing here ?

REPLY [1 votes]: You're focusing (pun intended) on the "f" from the slits to the lens.  That'll make the positions of the slits into angles past the lens, casting them out to focus at infinity.
But the part of the experiment that's interesting isn't the slits; they just sit there. It's the bright and dark fringes that emerge from the slits at various angles.  So you want an optical setup that converts angle at the lens to position (an image) at some point that you can then photograph.  That's the "f" spacing behind the lens:  The lens focuses rays of different angles to different points at a distance "f" behind it.
Here’s a found image that might help visualize it (ignore the caption, please):


TITLE: Optimizing Independent Variables to Maximize Dependent Variable
QUESTION [2 upvotes]: I looked around online and couldn't find anything that was answering my question so I thought I would take to the stack!
I'm interested in knowing if there is a statistical or mathematical way of optimizing ones dependent variable in regression analysis by shifting the values of your independent variables. For instance in market mix modeling generally the regression equation looks something like:
$$\text{Sales = TV Spend + Print Spend + Online Spend + Radio Spend}$$
So once you have your model based on the data how do you optimize the spend levels to have the max sales without just randomly shifting spend from one category to another?

REPLY [3 votes]: Regression models are not created to do what you want. For example, if you had each of those 4 variables in the model like this: $$y=b0+b1*x1+b2*x2+b3*x3+b4*x4$$
then we can increase the predicted y to infinity by increasing any x variable that has a positive coefficient. Furthermore, we cannot fit the model in one range of observed x values
and then use x values way out of that range to make predictions. The indep variables cannot be viewed in isolation -- if you advertise on radio, the impact of additional TV ads will have diminished impact so you need an interaction model. And something to include a decreasing marginal impact of ads in any one area. e.g, A(1-exp(-c*x))
If we created this highly non=linear model that actually described what was going on, then you have a math optimization model (nonlinear programming) to solve: maximize profit from advertizing (must take into account the cost of ads) subject to money < some number and maybe restrictions  on min or max money on each medium.


TITLE: A circuit and a magnet in a relative motion
QUESTION [3 upvotes]: First:
You have a circuit and a magnet. Will there be any difference in the induced current that flows in the circuit If

(a) Circuit is at rest, magnet is moved towards the circuit.
(b) Circuit is moved towards the magnet, the magnet is at rest.

There is no difference because of the principle of Galileo's relativity. (or because of another reason, but a and b are very easy to understand, intuitively.)
Question:
What if we move the circuit and magnet in the same direction and the same speed, will there any induced emf in the circuit? Moreover, what if we we just move them at the same speed but not the same direction, will there be any difference (so does direction effect the result here?)
Is there any relationship between this question and the special relativity principle (like in the moving wire and moving particle.)

REPLY [1 votes]: What if we move the circuit and magnet in the same direction and the same speed, will there any induced emf in the circuit?

There is one case where the electric conductor and the magnet move together and an EMF is induced. This device is a homopolar generator, also called a Faraday disc

The amazing thing is that not only a driven ring induces a voltage, but

If the magnetic field is provided by a permanent magnet, the generator works regardless of whether the magnet is fixed to the stator or rotates with the disc. From Wikipedia

To be precise, in the above described case the conductor moves in a circle and by this the influenced by the magnetic field electrons are under the influence of an acceleration too. Moving the disc together with the field on a straight line, no emf will be observable. Moving such a device on a zigzag trajectory one induce an AC current.

Moreover, what if we we just move them at the same speed but not the same direction, will there be any difference (so does direction effect the result here?)

For cases without curved trajectories the only phenomenon you have to take in account are the interdependencies between electric current, magnetic field and movement of the electric conductor. One of this dependencies is the Lorentz force, where a current carrying wire in a non parallel to the current applied magnetic field will move the wire perpendicular to the plain, formed by the directions of the current and the magnetic fields:

For your case of the induction of a current the conditions are the same. As long as you move a wire in relation to the magnetic field (or vary the magnetic field of an electric coil) AND this relative change in parameters happens not in directions, parallel to each other, a current will be induced.
To see the direction of the induced parameter one has to use right hand rules, which illustrate how the induction process happens in nature. As said above, the induced parameter has an direction of action and this direction is always perpendicular to the plain, formed by the two incoming parameters.
Of course occurs the question, in which direction of the plain the induced parameter acts. Because as you could image, were are two possible direction (because a plain has two surfaces, well?). The direction of action was found out empirical, means during experiments and than proved many times. After it became a low. And it turns out, that involved electrons and anti-protons show a opposite behavior in relation to positrons and protons:

All images from Wikipedia.


TITLE: Calculate $\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\ \{n\sqrt{2}\} }$
QUESTION [21 upvotes]: $$\text{Calculate :}\lim_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} } . $$

Note:
Weyl's equidistributed criterion. The following are equivalent:
$$x_n\quad\text{is equidistributed modulo 1}$$
$$\forall~ \text{continuous & 1-peridic} f: \quad\frac{1}{N}\sum_{n=1}^Nf(x_n)\rightarrow\int_0^1f $$
$$\forall~ k\in \mathbb Z^*:\quad \frac{1}{N}\sum_{n=1}^Ne^{2πikx_n}\rightarrow 0 $$

Background:
Im trying to approach this problem by weyl's criterion, so my thoughts so far are:
\begin{align}
&\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }\\
&=\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big)^{1/n}=\\
&=\exp\left(\frac{1}{n}\log\big(\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\}\big) \right)=\\
&=\exp\left(\frac{1}{n}\sum_{k=1}^n \log\big(\{k\sqrt{2}\}\big)\right)\\
\end{align}
So, since $\sqrt{2}$ is irrational  then it's simple to prove that the sequence $x_n=\{ n\cdot \sqrt{2}\}$ is equidistributed $\text{mod}\ 1$.
Let us define the continuous & $1$-periodic function $f(x):=\log(x-[x])$
by the weyl's criterion we get:
$$\begin{align*}
\frac{1}{n}\sum_{k=1}^n \log(\{k\sqrt{2}\})&\longrightarrow\int_0^1\log(\color{black}{\underbrace{\{x\}}_{=x-[x]}})dx\\[5pt]
&=\int_0^1\log(x)dx\\[5pt]
&=\bigg[x\log(x)\bigg]_0^1-\int_0^1dx\\[5pt]
&=-1\\[5pt]
\end{align*}$$
Hence
$$\lim\limits_{n \to\infty}\sqrt[n]{\{\sqrt{2}\}\{2\sqrt{2}\}\{3\sqrt{2}\}\cdots\{n\sqrt{2}\} }=e^{-1}  $$
Is there something wrong? Also, can we find this limit with some other way?
Let me know, thank you.

REPLY [7 votes]: Since $\{ n\sqrt{2} \}$ is equidistributed modulo $2$, the limit could be rewritten as the limit of the expected value of the geometric average of $n$ uniform random variables. The integral for this would be $$\lim_{n \to \infty}\int_0^1 \int_0^1... \int_0^1 (x_1x_2...x_n)^{\frac{1}{n}} dx_n...dx_2dx_1$$
This can actually be rewritten as $$\lim_{n \to \infty}\left(\int_0^1 x^{\frac{1}{n}} dx \right)^n$$
since each $x_i$ is independent of the others. The inner integral is then equal to $\frac{n}{n+1} = 1-\frac{1}{n+1}$, so the limit is $$\lim_{n \to \infty} \left(1 - \frac{1}{n+1}\right)^{n}$$
which is clearly $e^{-1}$.


TITLE: What is $\int_a^{b}(f'(t))^2 dt$?
QUESTION [5 upvotes]: Let $f:[a,b] \to \mathbb R$ be  nice differentiable function.  By Fundamental Theorem of Calculus (FTC), we have $$\int_a^{b} f'(t) dt = f(b) -f(a).$$
By Cauchy-Schwartz inequality, we have, $\int_{a}^{b} |f(t)| dt \leq  (b-a)^{1/2}(\int_a^b |f(t)|^2 dt)^{1/2} .$  Using this, and FTC, we have $\int_{a}^{b} (f'(t))^2 dt \geq  (b-a)^{-1} (f(b)-f(a))^2.$

Question: (1) Can compute $\int_a^{b}(f'(t))^2 dt$ precisely?  (2) Can we improve inequality $\int_{a}^{b} (f'(t))^2 dt \geq (b-a)^{-1} (f(b)-f(a))^2$ ?

REPLY [1 votes]: In reverse order: If
$$
f(x) = f(a) + \underbrace{\frac{f(b) - f(a)}{b - a}}_{m}(x - a),\quad a \leq x \leq b,
$$
then $f'(t) \equiv m$, so
$$
\int_{a}^{b} f'(t)^{2}\, dt = m^{2} (b - a)
= \frac{\bigl(f(b) - f(a)\bigr)^{2}}{b - a}.
$$
In words, the stated inequality is saturated for affine functions (as would be expected from Cauchy-Schwarz), so without further constraints on $f$ there is no tighter lower bound.
As for evaluating the integral of $(f')^{2}$, presumably that means something like "in terms of $a$, $b$, $f(a)$, $f(b)$, $f'(a)$, $f'(b)$". The answer is (tentatively, pending your exact meaning, but pretty firmly) no: Letting $\phi$ be a function that vanishes to second order at $a$ and $b$ (e.g., $(x - a)^{2} (x - b)^{2}$ multiplied by a smooth function), and replacing $f$ by $f + \phi$ does not change the value of any functional of the endpoint values, but does ("usually") change the integral of the derivative squared.


TITLE: How to prove that if $G$ is a group with $|G|= p^k*q$ with $p,q$ distinct primes, that $G$ has a subgroup of order $p^k$
QUESTION [1 upvotes]: I have to prove with induction on $k$ that if $G$ is a group with $|G|= p^k*q$ with $p,q$ distinct primes, that $G$ has a subgroup of order $p^k$. I already know that if $p$ is not a divisor of the center of $G$, $Z(G)$, that there is a subgroup of order $p^k$. I also know that if $G$ is abelian, G has an element of order $p$ and order q. I haven't got any further than this.

REPLY [1 votes]: For the induction step: suppose $G$ has order $p^k q$. If $p$ does not divide the order of the center of $G$, then you're done. Else, by your second lemma, $Z(G)$ has an element of order $p$. Call the cyclic subgroup generated by this element $P$. Then $P$ is a normal (in fact, central) subgroup of $G$. The quotient $G/P$ is a group of order $p^{k-1} q$, so by induction it contains a subgroup of order $p^{k-1}$. Apply the correspondence theorem and you're done.
Make sure you understand that you can take $k=1$ in the above argument, so the base step can be chosen to be the trivial $k=0$.


text:
\DOC PROVE_HYP

\TYPE {PROVE_HYP : thm -> thm -> thm}

\SYNOPSIS
Eliminates a provable assumption from a theorem.

\KEYWORDS
rule, assumption.

\DESCRIBE
When applied to two theorems, {PROVE_HYP} gives a new theorem with the
conclusion of the second and the union of the assumption list minus the
conclusion of the first theorem.
{
     A1 |- t1     A2 |- t2
   -------------------------  PROVE_HYP
     A1 u (A2 - {{t1}}) |- t2
}
If {t1} does not occurr in {A2} then the function simply returns the second
theorem {A2 |- t2} unchanged without including the assumptions {A1}.

\FAILURE
Never fails.

\EXAMPLE
{
  # let th1 = CONJUNCT2(ASSUME `p /\ q /\ r`)
    and th2 = CONJUNCT2(ASSUME `q /\ r`);;
  val th1 : thm = p /\ q /\ r |- q /\ r
  val th2 : thm = q /\ r |- r

  # PROVE_HYP th1 th2;;
  val it : thm = p /\ q /\ r |- r
}

\COMMENTS
This is sometimes known as the Cut rule. Although it is not necessary for the
conclusion of the first theorem to be the same as an assumption of the second,
{PROVE_HYP} is otherwise of doubtful value.

\SEEALSO
DEDUCT_ANTISYM_RULE, DISCH, MP, UNDISCH.

\ENDDOC

subset_name: curated


TITLE: Arranging balls in boxes
QUESTION [2 upvotes]: We have $n_1$ number of balls each labelled by the number $1$, $n_2$ number of balls each labelled $2$,..., $n_k$ number of balls each labelled $k$. In how many ways can we permute them in $r$ number of boxes assuming each box can hold exactly $1$ ball.
Edit:
Finding permutations for $n_1$ + $n_2$ + ... + $n_k$ = $r$ would be the same as finding permutations with repetition. Moreover if we have $n_1$ + $n_2$ + ... + $n_k$ < $r$ we may introduce $r$ - ($n_1$ + $n_2$ + ... + $n_k$) balls of another type and find permutations with repetition for $r$ total balls. So finding permutations for $n_1$ + $n_2$ + ... + $n_k$ > $r$ would be enough to answer this question.

REPLY [0 votes]: We have $ \sum_{i=1}^{k} n_i \leq r$. There are $ \binom{r}{n_1}$ ways to place the $1$'s. Then there are $ \binom{r-n_1}{n_2}$ ways to place the $2$'s. ... multiplying these binomial coefficients together gives the multinomial coeffiecient
\begin{eqnarray*}
\binom{r}{n_1 , n_2, \cdots , n_k,r-\sum_{i=1}^{k} n_i}.
\end{eqnarray*}


TITLE: $A$ is a rectangular matrix, and $(v_i)_1^k$ forms a basis of the rowspace. Is $(Av_i)_1^k$ linearly independent?
QUESTION [0 upvotes]: $A$ is a rectangular matrix of size $m\times n$ with real entries, and $\{v_1,v_2,\ldots,v_k\}$ forms a basis of the row space of $A$. Is the set $\{Av_1,Av_2,\ldots,Av_k\}$ linearly independent?

So we have that $\{v_1,\ldots,v_k\}$ is a basis of row space of $A$. We've to check if $\{Av_1,\ldots,Av_k\}$ is a linearly independent set. I started with
$$\sum_{i=1}^k\alpha_i Av_i =0 \stackrel{??}{\implies} \alpha_i = 0$$
but couldn't conclude anything about $\alpha_i's$. I also haven't found a counterexample to the result yet, so it may be true.
When $A$ is square and invertible, I could prove the claim.
Any hints to point me in the right direction?

REPLY [1 votes]: you're working over reals so you have a nice symmetric bilinear form given by the dot product, which is positive definite.
since  $\{\mathbf v_1, \mathbf v_2,\ldots,\mathbf v_k\}$ forms a basis, it is a linearly independent set.
Suppose for contradiction
$\{A\mathbf v_1, A\mathbf v_2,\ldots,A\mathbf v_k\}$ is linearly dependent
$\implies \mathbf 0 = \sum_{r=1}^k \alpha_r A\mathbf v_r = A\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big)$
where at least some $\alpha_j\neq 0 \implies \mathbf w :=\big(\sum_{r=1}^k \alpha_r \mathbf v_r\big) \neq \mathbf 0$ since $\mathbf v_r$ are linearly independent.,
$\implies \mathbf w \in \ker A$  but there must exist some x such that $\mathbf x^T A=\mathbf w^T$ giving
$ 0 = \mathbf x^T\mathbf 0= \mathbf x^T \big(A\mathbf w\big) = \big(\mathbf x^T A\big)\mathbf w = \mathbf w^T \mathbf w \gt 0$
which is a contradiction.
(i.e. to further expand the logic $\mathbf w^T\in \text{row space }A \iff \mathbf w\in \text{image }A^T$ which it is since it is written as a linear combination of basis vectors for such space, which means there is at least one vector in pre-image, i.e. some $\mathbf x$ satisfying $A^T\mathbf x = \mathbf w$.)


TITLE: Use central limit theorem to find variance
QUESTION [0 upvotes]: I've been stuck on this problem for ages now and don't even know where to start.
The problem states that women are supposed to give birth to children until they give birth to a boy. The probability of either a boy or a girl is 0.5. We have 100 discrete stochastic variables for the amount of children a woman has given birth to (So $X_1$ could be 4 children, 3 girls, 1 boy)
Then according to the central limit theorem we can say:
$$
\frac{X_1+X_2+X_3 + \cdots + X_{100}}{100} = N(\mu,\sigma^2)
$$
Now what is $\sigma^2$ ?

REPLY [1 votes]: It seems to me that $X_i$ has a geometric districution, which has variance $\frac{1-p}{p^2}=2$. For independent $X_i$, the sum has variance $2\times 100$ and $\sum_i X_i/100$ has variance $\frac{200}{100^2}=0.02$.


TITLE: Converting derivative of log to one without log using chain rule
QUESTION [0 upvotes]: If I have this:
$$
\frac{dlnx}{dln\eta}
$$
How can I pull the log out of this derivative using the chain rule?

REPLY [3 votes]: $$\frac{d\log[x]}{d\log[\eta]} = \frac{\frac{d\log[x]}{dx}}{\frac{d\log[\eta]}{d x}} = \frac{\frac{1}{x}}{\frac{d\log[\eta]}{d\eta}\frac{d\eta}{dx}} = \frac{\eta}{x}\frac{dx}{d\eta}$$