Question on Inductive Proof of Implicit Function Theorem I am struggling with an inductive proof of the implicit function theorem, concretely with the final part of construction of a function, up to this final point everything is perfectly clear to me. First the following is known to be true:

Theorem 1: Let $F : G \subseteq \mathbb R^{n+1} \to \mathbb R$, $G \ne \emptyset$, $F \in C^1(G)$, $G$ open. For $x^0 \in \mathbb R^n, y^0 \in \mathbb R$ let
$$
 F(x^0, y^0) = 0 \quad \mbox{ and } \quad
 F_y(x^0, y^0) \ne 0.
$$
Then there exists some neighborhood $U(x^0) \subseteq \mathbb R^n$ and a function $f : U(x^0) \to \mathbb R$ such that
$$
 y = f(x_1, \ldots, x_n), \qquad
 y_0 = f(x_1^0, \ldots, x_n^0)
$$
and
$$
 F(x, f(x)) = 0
$$
and $f \in C^1(U(x^0))$.

Now the Implicit Function Theorem reads as

Theorem 2 (Implicit Function Theorem): Let $F : G \subseteq \mathbb R^{n+m} \to \mathbb R^m$, $G\ne \emptyset$,
$F \in C^1(G)$, $G$ open. Also $(x^0, y^0) \in G$ with $x^0 \in \mathbb R^n, y^0 \in \mathbb R^m, F(x^0,y^0) = 0$ and
$$
 \det\left( \frac{\partial F}{\partial y} \right)_{y=y_0} \ne 0.
$$
Then there exists neighborhoods $U(x^0) \subseteq \mathbb R^n, V(y^0) \subseteq \mathbb R^m$ and a function $g : U(x^0) \to V(x^0)$ such that $F(x, g(x)) = 0$ on $x \in U(x^0)$.

Proof: The proof proceeds by induction on $m$, if $m = 1$ it is Theorem 1, so we assume $m > 1$ and suppose (by induction hypothesis) the statement holds for $m-1$. Let $F : G \subseteq \mathbb R^{n+m} \to \mathbb R^n$ be given, then because of
$$
 \det\left( \frac{\partial F}{\partial y} \right)_{y=y_0} \ne 0
 \quad \mbox{ or } \quad
\det \begin{pmatrix}
  \frac{\partial F_1}{\partial y_1} & \frac{\partial F_1}{\partial y_2} & \cdots & \frac{\partial F_1}{\partial y_m} \\
  \vdots \\
  \frac{\partial F_m}{\partial y_1} & \frac{\partial F_m}{\partial y_2} & \cdots & \frac{\partial F_m}{\partial y_m}
 \end{pmatrix} \ne 0.
$$
So we can suppose that every row contains an entry $\ne 0$, suppose w.l.o.g. that
$\frac{\partial F_m}{\partial y_m} \ne 0$. By Theorem 1 we can solve (locally) for $y_m$, that means there is some neighborhood $U$ of $(x, y_1, \ldots, y_{m-1})$ and a function
$$
 y_m = \varphi(x, y_1, \ldots, y_{m-1})
$$
and $\varphi$ is continously differentiable with ($x \in \mathbb R^n$)
$$
 F_m(x, y_1, \ldots, y_{m-1}, \varphi(x, y_1, \ldots, y_{m-1})) = 0
$$
for all $(x,y_1,\ldots,y_{m-1}) \in U$. Now set
$$
 \Phi_i(x, y_1, \ldots, y_{m-1}) := F_i(x, y_1, \ldots, y_{m-1}, \varphi(x, y_1, \ldots, y_{m-1}))
$$
for $i = 1, \ldots, m-1$. Then $\Phi : U \subseteq \mathbb R^{n+m-1} \to \mathbb R^{m-1}$
and $\Phi$ fulfills als prerequisites to apply the induction hypothesis (here the proof shows this, but I omit it because it is quite long and does not apply to my question).
Then there exists neighborhoods $W \subseteq \mathbb R^n, V \subseteq \mathbb R^{m-1}$ and a function $g : W \to V$ with
$$
 \Phi(x, g(x)) = 0
$$
for all $x \in W$. Now set $h(x) = (g(x), \varphi(x,g(x))$, then we have
$$
 F(x, h(x)) = 0
$$
and the proof is finished. $\square$

My question is on the last part. Namely the construction of the function $h(x)$,

1) Because $h(x) = (g(x), \varphi(x, g(x))$ it must be the case that $(x, g(x) \in U$,
because $\varphi : U \to \mathbb R$, but I do not see that this must be the case?

2) The same issue if I want to show that $F(x, h(x)) = 0$, if $i \ne m$ it is
$$
 F_i(x, h(x)) = \Phi_i(x, g(x))
$$
by definition, but for
$$
 F_m(x, h(x)) = F_m(x, g(x), \varphi(x,g(x)) = F_m(x, g_1(x), \ldots, g_{m-1}(x), \varphi(x, g_1(x), \ldots, g_{m-1}(x))
$$
and to conclude that this equals $0$ it also must be that $(x, g(x)) \in U$, but I do not see that this must be the case?

As a minor point, you forgot to write that the implicit function $g$ on Theorem 2 is of class $C^1$.

More important, the theorems 1 and 2 that you enunciated do not claim the (local) uniqueness of the implicit solution. This feature is important.

By reading your question I guess that you are trying to understand the inductive proof of the implicit function theorem given in ``The Implicit Function Theorem - History, Theory, and Applications'', pp.39-41, by S. G. Krantz and H. R. Parks.

Regarding your specific question, notice that
$$0=\Phi_1(x,g(x))=F_1\Big(x,g_1(x),\ldots,g_{m-1}(x),\varphi\big(x,g_1(x),\ldots,g_{m-1}(x)\big)\Big).$$
Analogously for $F_2,\ldots,F_{m-1}$ and for $F_m$. Thus,
$$h(x)=\Big(g_1(x),\ldots,g_{m-1}(x),\varphi\big(x,g_1(x),\ldots,g_{m-1}(x)\big)\Big).$$

One approach to understand an inductive proof is to reproduce it for n=1, n=2, and n=3. Since you already understood the case n=1, I suggest that you develop a complete proof for the case n=2 (this case should be easy). Then, try to develop a complete proof for the case n=3 (if you overcome this case then you are almost done). Finally, try the general case.

I also suggest that you take a look at the article ``The Implicit and the Inverse Function Theorems: Easy Proofs'', Real Analysis Exchange, Vol. 39(1), pp. 207-218 (2013/2014), by O. de Oliveira. You can find a preprint on the internet.

Best wishes,

Oswaldo

The proof has to proceed by induction* on n because then you are always working with neighborhoods of P0.

F1(x1,x2,y1,y2)=0
F2(x1,x2,y1,y2)=0
Have a solution in a neighborhood of P0.  (F1->F2)

F1(x1,x2,y1,y2)=0
F2(x1,x2,y1,y2,y3)=0
P01 is entirely different than P02. (F1->’F2)

See appendix of “Rutherford Aris;  Vectors, Tensors, and the Basic Equations of Fluid Mechanics” for proof by induction on n.

EDIT: Elaboration and Examples of Aris Induction proof:

Fundamental Assumption (FA):
If F(x1,x2,..xn) satisfies conditions of Implicit Function Theorem (IFT),  F(x1,x2,..,xn) has a solution xn=f(x1,x2,…xn-1) in a neighborhood of P0=x10,x20,..,xn0.

Given, under conditions of IFT
F1(x1,y1,y2), F1(P0)=0
F2(x1,y1,y2), F2(P0)=0

Then
F1(x1,y1,y2)=0 has a solution y2=f1(x1,y1) in a neighborhood of P0 by FA.
Then
F2(x1,y1,f1(x1,y1))= 0 has a solution by FA in a neighborhood of P0, and
y1=y1(x1)
y2=y2(x1)
In a neighborhood of P0.

Geometrically, F1(x1,y1,z1)=0 and F2(x1,y1,z1)-0 are surfaces through P0 in box neighborhoods of P0 and their common solution is the line of intersection through P0 in a box neighborhood of P0.

Another Example
Given, under conditions of IFT:
F1(x1,x2,y1,y2,y3)=0
F2(x1,x2,y1,y2,y3)=0
F3(x1,x2,y1,y2.y3)=0..

Assuming you have solved F1 and F2,
y1=f1(x1,x2,y3)
y2=f2(x1,x2,y3)

Substituting into F3 gives an Equation in y3 which can be solved by FA. Then
y1=y1(x1,x2)
y2=y2(x1,x2)
y3=y3(x1,x2)
in a neighborhood of P0.

Note: A neighborhood of P0 arises from the requirement that dF/dxn =’ 0 is the same sign throughout a neighborhood of P0, which is true if dF/dxn  is continuous.

At this point the general procedure should be obvious from above.

The Inverse Function Theorem follows as a special case. Ex:
x1=f1(y1,y2) -> F1(x1,x2,y1,y2)=0
x2=f2(y1,y2) -> F2(x1,x2,y1,y2)=0
Then:
y1=y1(x1,x2)
y2=y2(x1,x2)

The terminology "induction on n" is Aris'. Though true in a sense, it's a little confusing, which is the reason for this post. The crucial notion is a common neighborhood P0 as you increase the number of variables you are solving for by 1.


Hadamard product with rank 1 matrix Let $\odot$ represent Hadamard or pointwise multiplication. If $\pmb{Y}$ is a $\underline{given}$ positive definite square matrix, can i estimate a positive semi-definite $\pmb{X}$ matrix such that
\begin{equation}
\pmb{Y} = \pmb{X} \odot \pmb{z}\pmb{z}^H
\end{equation}
where $\pmb{z}$ is a complex vector with phases only (i.e. each element of $\pmb{z}$ is of magnitude $1$). Also $(.)^H$ stands for conjuate-transposition, or Hermitian (Thanks @igael). I really don't know where to start. Maybe, i have formulated a problem that does not make sense.

$\mathbf Y = diag(\mathbf z)\mathbf X diag(\mathbf z^H)$. Since $diag(\mathbf z)diag(\mathbf z^H)=\mathbf I$, $\mathbf X$ can be estimated as $\mathbf X = diag(\mathbf z^H)\mathbf Y diag(\mathbf z)$, where $\mathbf z$ is any vector with your constraint.


why $F(a^2) \subset F(a)$? I have some  doubt in this post
Let $E$ be an extension field of $F$. If $a \in E$ has a minimal polynomial of odd degree over $F$, show that $F(a)=F(a^2)$.
let $n$ be the degree of the minimal polynomial $p(x)$ of $a$ over $F$ and $k$ be the degree of the minimal polynomial $q(x)$ of $a^2$ over $F$
Since $a^2 \in F(a)$, We have $F(a^2) \subset F(a)$, then $k\le n$
I don't understand  why  $F(a^2) \subset F(a)$?
My thinking: Take $ a  \in \mathbb{R} $, $a \subset a^2 \implies F(a) \subset F(a^2)$

Credit : Atticus Stonestrom( comment section)
For  $a = \sqrt 2$, $ F(a)=\mathbb{Q}({\sqrt 2})$ contain basis  $\{1, \sqrt 2\}$ but  $ F(a^2)=\mathbb{Q}(2)$ contain basis $1 \implies F(a^2) \subset F(a)$


Differential equation (Brachistochrone problem) I'm really only supposed to solve the differential equation $(1+(y')^2)y=k^2.$

I haven't encountered any problem with $(y')^2$. How do you start with a problem like this, I did try googling it but all I got was basic differential equation, should I substitute for something?

Grateful for any help at this point!

$$\left(1+\left(\frac{dy}{dx}\right)^2\right)y(x)=k^2$$
$$\frac{dy}{dx}=\pm \sqrt{\frac{k^2}{y}-1}$$
ODE of the separable kind :
$$dx=\pm \sqrt{\frac{k^2-y^2}{y}}\:dy \quad\to\quad x=\pm\int \sqrt{\frac{k^2-y^2}{y}}\:dy $$
$$\pm x=k^2\tan^{-1}\left(\sqrt{\frac{y}{k^2-y}}\right)-\sqrt{(k^2-y)y}\:+c$$
The solution is on the form of $x$ as a function of $y$. There is no simple closed form for the inverse function $y(x)$.

The result can be presented on parametric form , with $\quad\tan(\theta)=\sqrt{\frac{y}{k^2-y}}$ :

$$\begin{cases}
x=k^2\left(\theta -\sin(\theta)\cos(\theta)\right)+c \\
y=k^2\sin^2(\theta)
\end{cases}$$
which is a parametric form of equation of cycloid.

There is an algebraic mistake in the above answer:  In the third line,
dx should be sqrt[ y/(k^2 -y^2) ] dy
(The argument of the sqrt has the numerator and denominator
interchanged).


How to calculate defined limit using l'Hospital's rule? I want to calculate the following limit:

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} $$

It is clear to see that by plugging the value $x=0$ the result is 1. I want to calculate the limit using L'Hospital's rule, by applying $a^x=e^{x\ln(a)}.$ I got :

$$e^{2\ln(1+2x)/(3x+1)}$$

Then I tried to calculate the power limit by using l'Hospital's rule, and eventually calculate e^(power limit answer). still could not find the correct answer that way. Can somebody help me? Thanks in advance.

L'Hospital's rule works for limits of the form :

$$\frac{0}{0}$$

$$\frac{\infty}{\infty}$$

etc, which are called undetermined forms.

You cannot apply it on other cases, the limit of the derivatives of the denominator and numerator will not be the same as the limit of the initial expression.

As you've mentioned, this is a straight-forward limit to calculate, so :

$$\lim_{x\to 0} (1+2x)^{2/(3x+1)} = 1$$

let $y$ = $(1+2x)^{2/(3x+1)}$

$ln$ $y$ = $\frac{2}{(3x+1)}$ $ln$ $(1+2x)$

taking the limit then take the Exponential $e$ of the answer

is that what you do ?!

because  its not $\frac{0}{0}$ or $\frac{\infty}{\infty}$ the answer will be vary So ,

$\lim_{x\to0}$ $\frac{4}{3(2x+1)}$ apply L'H again to get $\lim_{x\to0}$ $\frac{0}{6}$ = 0

$e^{0}$ = 1


Integration with fraction I'm having difficulty understanding how below integral is evaluated :

$$\int_0^1\frac{(1-y)^2}{2}+(1-y)y dy = \frac{-1(1-y)^3}{6}+\frac{y^2}{2}-\frac{y^3}{3}\bigg|^1_0$$

What are steps involved in this evaluation ?

For $\frac{(1-y)^2}{2}$ this appears to be evaluated as  $\frac{(1-y)^{2+1}}{(3)2}$ but I'm how unsure - 1 is inserted ?

Antiderivative if $(1-y)^{2}$ is not $(1-y)^{3} /3$. If you differentiate the latter you get $-(1-y)^{2}$. Hence the antiderivative is $-(1-y)^{3} /3$.

Remember that where $f(x)$ is linear we have that:

$$\int[f(x)]^n dx=\frac{[f(x)]^{n+1}}{(n+1)f'(x)}+C$$


Help solving 2 second order differential equation $y'' + (y')^2 = y$

and

$y*y''= (y')^2 - (y')^3$

I know that both of these equations can be solved by using
$p(y) = y'$, but Im getting stuck in the middle, and can't seem to solve the thing after I get an expression which looks like something: $p^2$

any tips please?

For the first equation, $y'' + (y')^2 = y$, you might try the substitution $w(y)=(y')^2$ instead.

$$\frac{d}{dy}w(y)=\frac{d}{dy}(y')^2\\
=\frac{dy'}{dy}\frac{d}{dy'}(y')^2\\
=\frac{dx}{dy}\frac{dy'}{dx}\frac{d}{dy'}(y')^2\\
=\frac{1}{y'}(y'')(2y')\\
=2y''.$$

Hence the equation becomes the 1st order linear inhomogeneous equation with constant coefficients, $w'(y)+2w(y)=2y$. The general solution to this equation is found to be:

$$w(y)=c_1e^{-2y}+y-\frac12.$$

The equation $(y')^2=c_1e^{-2y}+y-\frac12$ can then be used to find an implicit formula for $y(x)$ in the form of an integral, but for most values of $c_1$ an explicit equation for $y(x)$ is likely unobtainable.

Hello everyone I am currently training and I have to solve a nonlinear differential equation and I totally lost I have a BAC + 2 and I never saw her going after large search I are always happens not:

I would like to solve this equation

A *y''(t) - B *(y'(t))^2 + C * y(t)

AND I AM FRENCH SORRY FOR THE BAD ENGLISH


Pointwise convergent of the sequence of functions Prove that if $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is a sequence of continuously differentiable functions such that the sequence of derivatives $\{f'_n : \mathbb R \rightarrow \mathbb R\}$ is uniformly convergent and the sequence $\{f_n (0)\}$ is also convergent, then $\{f_n : \mathbb R \rightarrow \mathbb R\}$ is pointwise convergent. Is the assumption that the sequence $\{f_n (0)\}$ converges necessary?

My attempt: Suppose $f'_n \rightarrow g$ uniformly and $f_n (0) \rightarrow a$. Then since $f'_n$ are integrable, $\int_0^x f'_n \rightarrow \int_0^x g$, i.e. $\lim_{n \rightarrow \infty} \int_0^x f'_n = \lim_{n \rightarrow \infty} (f_n(x)-f_n(0))=\int_0^x g$. Is this the right approach, cause I'm stuck here, if so, how to proceed?

$\newcommand{\dif}{\,\text{d}}$
By the fundamental theorem of calculus we have $f_n(x) = f_n(0) + \int_0^x f'_n(t) \dif t$. You know $f_n(0)$ converges to something and you know $\int_0^x f'_n(t) \dif t$ converges to something. Putting this together shows $f_n(x)$ converges to something.


Use the definition of the derivative to show that the function is not differentiable at the given value: (x-1)^(2/5) at x=1 I've been trying to figure out this problem for a while with no success. My main hurdle is figuring out the derivative of the function y=(x-1)^(2/5) using the definition of the derivative. At first I tried multiplying by a conjugate with a power of 4/5 but that did not work. I get the feeling I must use the (a^5-b^5) expansion, but I am not sure where to proceed from there. Thank you in advance.

Define the function $f: \mathbb{R} \mapsto \mathbb{R}$ by

$f(x)=(x-1)^{2/5}, \, \, \, x\in \mathbb{R}$.

We say that a function $f$ is differentiable in a point $x_0$ if the limit

$\lim_{h\to 0} \frac{f(x_0+h)-f(x_0)}{h}$,

exists in $\mathbb{R}$. Now with $x_0=1$,

$\frac{f(x_0+h)-f(x_0)}{h} = \frac{(1-1+h)^{2/5}-(1-1)^{2/5}}{h}
=\frac{h^{2/5}}{h}=h^{-3/5}$.

As $h\to 0$ from above, this limit converges to $+\infty$. Thus $f$ is not differentiable in $x_0=1$.


How to find that the alternating sum converges There's an exercise in Knuth's book, The art of computer programming sec 1.2.3, exercises first set, ex 8 that says:

Find an example of infinite series in which eq. (the following) is false:

$$
\sum_{R(i)}\sum_{S(j)} a_{ij} = \sum_{S(j)}\sum_{R(i)} a_{ij}
$$

An example is to take $a_{(i+1)i}=1$, $a_{i(i+1)}=-1$, others $a_{ij}=0$.
So this is a "sign sum", the left side starts with $+1$ and the right with $-1$. How do you prove that the left side is $-1$, and the right is $+1$? (The sum is infinite, starts at $i=0$, $j=0$)

The idea of the proof is that
$$\sum_{j\in\mathbb{N}} a_{ij} = \left\{\begin{array}{cc} -1 & i=1\\
0 & i\not = 1 \end{array}\right.,
$$
and
$$\sum_{i\in \mathbb{N}} a_{ij} = \left\{ \begin{array}{cc} 1&j=1\\  0& j\not=1\end{array}\right. $$

It would help to write down the values of $a_{ij}$ for $(i,j)\in \{1,2,3,4\}\times \{1,2,3,4\}$, say.

look in to this article. that should be the right tool:
https://en.wikipedia.org/wiki/Mathematical_induction


Show that function distances are preserved If for all $x,y \in \mathbb{R}^n$ that satisfies $|x - y| = t$ also satisfies $|f(x) - f(y)|=t$ (for some constant $t \in \mathbb{R}^{+}$), show that $|f(x) - f(y)| = |x-y|$ for all values of $x,y \in \mathbb{R}^n$.
This seems very much intuitive to me and that $f(x) = x$ or that the range set is a copy of $\mathbb{R}^n$. My initial goal was to show if $r \in l(x,y)$, then $f(r) \in l(f(x),f(y))$ such that $|f(r)-f(z)| = |r - z|$ where $z \in \{x,y\}$. Showing this is enough to solve the problem. The most intuitive idea, however that seems to work is creating equilateral triangles in some way and make it work.
$\color{green}{\text{Edit :}}$ In $\mathbb{R}$, by taking $f$ as the greatest integer function with a period of $t$, we get a clear contradiction to the initial statement. The idea doesn't seem to directly work for higher dimensions, albeit the first answer claims so. Can this (or the claim in the original question) be proven for $n > 1$?

This is true in $\mathbb{R}^n$ for $n\ge2$ and it is wrong in $\mathbb{R}$.
First not that $f:\mathbb{R}^n\to\mathbb{R}^n$ preserves the distance $t$ if and only if $\widetilde{f}(x)\triangleq\frac{1}{t}f(tx)$ preserves the distance $1$. So, we may suppose that $t=1$ without loss of generality.

If $n=1$ then the function
$$f(x)=\cases{x+1&if $x\in\mathbb{Z}$,\\
x&if $x\notin\mathbb{Z}$.}$$
is a counter-example.
If $n>1$ then it is Beckman-Quarles theorem that date back to 1953. The article is freely accessible here.

I don't think this is true without further hypothesis (continuity maybe), because we can build counter examples with periodic stuff or similar.
In $\mathbb R$ take $f(x)=\lfloor x\rfloor$ and $t=1$. Then $f(\frac12)-f(\frac 13)=0\neq \frac 12-\frac 13$.
I think we can extend that to higher dimensions without too much issues.


How to prove that this function satisfies the Cauchy Riemann equations? Let$$
f(z) = \left\{
        \begin{array}{ll}
            g(z) & \quad z \neq 0 \\
            h(z) & \quad z = 0
        \end{array}
    \right.
$$

where $g(z)=U_1+iV_1$ and $h(z)=U_2+iV_2$.

When we are asked to prove that $f(z)$ satisfies Cauchy Riemann equations at $z=0$ does that mean that we need to prove that ${U_1}_x={V_1}_y$ and ${U_1}_y=-{V_1}_x$ or ${U_2}_x={V_2}_y$ and ${U_2}_y=-{V_2}_x$?

If $h(z)=0$ does that mean that $f(z)$ satisfies Cauchy Riemann equations at $z=0$ for any $g(z)$?

The function $h$ is just a constant: it has little to do with the value of $f$, except, of course, that $f$ has to be continuous. All you need to do is check the Cauchy-Riemann equations for $g$ at $0$ (take the limit at $0$: it should exist if it is going to be analytic).


How can you prove that a function has no closed form integral? In the past, I've come across statements  along the lines of "function $f(x)$ has no closed form integral", which I assume means that there is no combination of the operations:

addition/subtraction
multiplication/division
raising to powers and roots
trigonometric functions
exponential functions
logarithmic functions

which when differentiated gives the function $f(x)$. I've heard this said about the function $f(x) = x^x$, for example.
What sort of techniques are used to prove statements like this? What is this branch of mathematics called?

Merged with "How to prove that some functions don't have a primitive" by Ismael:
Sometimes we are told that some functions like $\dfrac{\sin(x)}{x}$ don't have an indefinite integral, or that it can't be expressed in term of other simple functions.
I wonder how we can prove that kind of assertion?

It is a theorem of Liouville, reproven later with purely algebraic methods, that for rational functions $f$ and $g$, $g$ non-constant, the antiderivative
$$f(x)\exp(g(x)) \, \mathrm dx$$
can be expressed in terms of elementary functions if and only if there exists some rational function $h$ such that it is a solution to the differential equation:
$$f = h' + hg$$
$e^{x^2}$ is another classic example of such a function with no elementary antiderivative.
I don't know how much math you've had, but some of this paper might be comprehensible in its broad strokes:
https://ksda.ccny.cuny.edu/PostedPapers/liouv06.pdf
Liouville's original paper:

Liouville, J. "Suite du Mémoire sur la classification des Transcendantes, et sur l'impossibilité d'exprimer les racines de certaines équations en fonction finie explicite des coefficients." J. Math. Pure Appl. 3, 523-546, 1838.

Michael Spivak's book on Calculus also has a section with a discussion of this.

The general pattern is always the same: (1). Study properties of derivatives of closed form functions. (2). If your function $f(x)$ fails one of those properties, then it has no closed form antiderivative.

Here is how to get started: First, consider derivatives of rational functions. Notice that their pole orders at finite points are never 1 (more generally: the residue is 0). If $f(x)$ fails this property, then you know that its antiderivative can not be a rational function. Sounds simple, but the surprising part is that this can be generalized quite easily to logarithmic extensions (handling exponential extensions in Risch algorithm is more involved, and handling algebraic extensions is more involved still).

Take for example a function $f(x) \in \mathbb{C}(x, \ln(x))$. For each irreducible factor $p \in \mathbb{C}(x)[ \ln(x) ]$ in the denominator of $f(x)$, you can define a corresponding residue (which will be algebraic over $\mathbb{C}(x)$). A necessary condition for $f(x)$ to have an elementary anti-derivative is for these residues to be constants  (in the more basic case, $f(x) \in \mathbb{C}(x)$, residues are always constants, and thus, rational functions always have an elementary antiderivative). See "Liouville's theorem (differential algebra)" on wikipedia for more.


The math behind Warren Buffett's famous rule – never lose money This is a question about a mathematical concept, but I think I will be able to ask the question better with a little bit of background first.

Warren Buffett famously provided 2 rules to investing:

  Rule No. 1: Never lose money. Rule No. 2: Never forget rule No. 1.

I initially took this quote as tongue-in-cheek.  Duh, of course you don't want to lose money. But after better educating myself in the world of investing I see this quote more as words from the wise sensei of investing.  It means more than just be careful, or be conservative.  Losing money can destroy a portfolio because there is a mathematical disadvantage.

Consider two hedge fund managers:  Mr. Turtle and Mr. Hare.

Mr. Turtle is steady, he doesn't have high returns, but he also doesn't lose money.  Mr. Hare is aggressive, getting huge returns, but occasionally losing money. Here are their returns over the past 5 years

Mr. Hare has a higher average rate of return.  Further, he has made (significantly) more money in 4 out of the 5 years.  Mr. Turtle, however, has made his clients more money overall in the same timeframe.

This seems counter-intuitive. I would think you would want to maximize your average rate of return at all costs, but it's not that simple.

Why?

Why does one negative input have such a significant impact on an exponential growth function?

Why doesn't the average rate of growth always lead to the largest possible result?

How does one explain this (hopefully in layman's terms)?

There are two things I should point out. One is that the arithmetic mean doesn't properly measure annual growth rate. The second is why.

The correct calculation for average annual growth is geometric mean.

Let $r_1,r_2,r_3,\ldots,r_n$ be the yearly growth of a particular investment/portfolio/whatever. Then if you invest $P$ into this investment, after $n$ years your final amount of money is $Pr_1r_2\cdots r_n$. The (yearly) average growth rate of this investment is the number $r$ such that if the investment grew at a constant rate of $r$ every year then after $n$ years we'd have the same amount as we actually ended up with. In other words it is $r$ such that $Pr_1r_2\cdots r_n=Pr^n$. Thus we have
$$r=\sqrt[n]{r_1r_2r_3\cdots r_n},$$
which is the geometric mean, not the arithmetic mean.

If we use the geometric mean, we see that Turtle's average yearly growth is $\sqrt[5]{1.39}\approx 1.07$, and Hare's average yearly growth rate is $\sqrt[5]{1.36}\approx 1.06$, which is more in line with our expectations.

Why doesn't the arithmetic mean behave as expected?

Well, let's look at something over two years. Say its arithmetic mean growth is 1. Then the growth rate for one year will be $1+x$ and the other year will be $1-x$. Multiplying these together, we see that total growth is $1-x^2$. In other words, actual growth is always less than or equal to that predicted by the arithmetic mean (this is true for $n$ years as well, see the AM-GM inequality). Note further that the actual growth is closer to that predicted by the arithmetic mean when the individual annual growth rates are closer together. Thus if you are more consistent (your annual growth rates are closer together) then your arithmetic mean growth rate will closely approximate your true average annual growth rate as in Turtle's case. On the other hand, if your annual growth rates are more spread out, then your true average annual growth rate will be much lower than the arithmetic average growth rate (as in Hare's case).

When the economy is doing well, invest in the Hare. Just before the economy fails, sell your Hare investments and move them to Turtle/gold. Wait 2 years. When the economy gets better, sell Turtle/gold and buy Hare again. Everybody selling Hare at once actually causes the Hare stock to drop and, indirectly, the economy to fail. The trick is to guess when the recession will hit and be the first to move your money. Sell just before the bubble bursts. and buy back just after the market hit rock bottom.


How to prove a function has no real solutions So I have the equation:
$$
0.7=0.04(10+x)+\frac{2100}{(10+x)^3}
$$

And I have tried solving for x and have not succeeded, I believe it only has complex solutions.

So my question is, how do I go about actually proving this equation has no solutions for $x \ \varepsilon\ R$

Thank you for the help!

You can define the function $f(x)=0.04(10+x)+\frac{2100}{(10+x)^3}-0.7$.
If the equation has a real solution, then this function intersects the x axis at some point. It's a pretty "nice" function (continuous and differentiable at almost any point) which means you can use its derivatives and limits at specific points in order to practically know anything about its graph. Mainly - whether or not it intersects the x axis.

Multiplying by $$100(10+x)^3$$ and simplifying we get

$$2\,{x}^{4}+45\,{x}^{3}+150\,{x}^{2}-2500\,x+90000=0$$
This equation has only complex Solutions.


Derivative of Quadratic Form For the Quadratic Form $X^TAX; X\in\mathbb{R}^n, A\in\mathbb{R}^{n \times n}$ (which simplifies to $\Sigma_{i=0}^n\Sigma_{j=0}^nA_{ij}x_ix_j$), I tried to take the derivative wrt. X ($\Delta_X X^TAX$) and ended up with the following:

The $k^{th}$ element of the derivative represented as

$\Delta_{X_k}X^TAX=[\Sigma_{i=1}^n(A_{ik}x_k+A_{ki})x_i] + A_{kk}x_k(1-x_k)$

Does this result look right? Is there an alternative form?

I'm trying to get to the $\mu_0$ of Gaussian Discriminant Analysis by maximizing the log likelihood and I need to take the derivative of a Quadratic form. Either the result I mentioned above is wrong (shouldn't be because I went over my arithmetic several times) or the form I arrived at above is not the terribly useful to my problem (because I'm unable to proceed).

I can give more details about the problem or the steps I put down to arrive at the above result, but I didn't want to clutter to start off. Please let me know if more details are necessary.

Any link to related material is also much appreciated.

Let $Q(x) = x^T A x$. Then expanding $Q(x+h)-Q(x)$ and dropping the higher order term, we get $DQ(x)(h) = x^TAh+h^TAx = x^TAh+x^TA^Th = x^T(A+A^T)h$, or more typically, $\frac{\partial Q(x)}{\partial x} = x^T(A+A^T)$.

Notice that the derivative with respect to a column vector is a row vector!

I just learned a new trick when your independent variable is in more than two places within your formula: introduce a new (fake) parameter which will then disappear:

$$\frac{\partial}{\partial x} y^TAx = \frac{\partial y}{\partial x}[Ax]^T+y^TA $$
The transpose was to make the vector a row vector. Nothing deep there!

Now, if $y=x$ then
$$ \frac{d}{dx} x^TAx = x^TA^T+x^TA = x^T(A+A^T) \ . $$


What books must every math undergraduate read? I'm still a student, but the same books keep getting named by my tutors (Rudin, Royden).

I've read Baby Rudin and begun Royden though I'm unsure if there are other books that I "should" be working on if I want to study beyond Masters. I'm not there yet as I'm on a four year course and had a gap year between Years 3 and 4.

Please recommend for Algebra, Linear Algebra and Categories - Analysis, Set Theory, Measure theory (an area I have seen too little books dedicated for).

E.g. Spivak is very good for self learning basic real analysis, but Rudin really cuts to the heart.

EDIT: I now think that this list is long enough that I shall be maintaining it over time--updating it whenever I use a new book/learn a new subject. While every suggestion below should be taken with a grain of salt--I will say that I spend a huge amount of time sifting through books to find the ones that conform best to my (and hopefully your!) learning style.

Here is my two cents (for whatever that's worth). I tried to include all the topics I could imagine you could want to know at this point. I hope I picked the right level of difficult. Feel absolutely free to ask my specific opinion about any book.

Basic Analysis: Rudin--Apostol

Measure Theory: Royden (only if you get the newest fourth edition)--Folland

General Algebra: D&F--Rotman--Lang--Grillet

Finite Group Theory: Isaacs-- Kurzweil

General Group Theory: Robinson--Rotman

Ring Theory: T.Y. Lam-- times two

Commutative Algebra: Eisenbud--A&M--Reid

Homological Algebra: Weibel--Rotman--Vermani

Category Theory: Mac Lane--Adamek et. al--Berrick et. al--Awodey--Mitchell

Linear Algebra: Roman--Hoffman and Kunze--Golan

Field Theory: Morandi--Roman

Complex Analysis: Ahlfors--Cartan--Freitag

Riemann Surfaces: Varolin(great first read, can be a little sloppy though)--Freitag(overall great book for a second course in complex analysis!)--Forster(a little more old school, and with a slightly more algebraic bend then a differential geometric one)--Donaldson

SCV: Gunning et. al--Ebeling

Point-set Topology: Munkres--Steen et. al--Kelley

Differential Topology: Pollack et. al--Milnor--Lee

Algebraic Topology: Bredon--May-- Bott and Tu (great, great book)--Rotman--Massey--Tom Dieck

Differential Geometry: Do Carmo--Spivak--Jost--Lee

Representation Theory of Finite Groups: Serre--Steinberg--Liebeck--Isaacs

General Representation Theory: Fulton and Harris--Humphreys--Hall

Representation Theory of Compact Groups: Tom Dieck et. al--Sepanski

(Linear) Algebraic Groups: Springer--Humphreys

"Elementary" Number Theory: Niven et. al--Ireland et. al

Algebraic Number Theory: Ash--Lorenzini--Neukirch--Marcus--Washington

Fourier Analysis--Katznelson

Modular Forms: Diamond and Shurman--Stein

Local Fields:

Lorenz and Levy--Read chapters 23,24,25. This is by far my favorite quick reference, as well as "learning text" for the basics of local fields one needs to break into other topics (e.g. class field theory).
Serre--This is the classic book. It is definitely low on the readability side, especially notationally. It also has a tendency to consider things in more generality than is needed at a first go. This isn't bad, but is not good if you're trying to "brush up" or quickly learn local fields for another subject.
Fesenko et. al--A balance between 1. and 2. Definitely more readable than 2., but more comprehensive than 1. If you are wondering whether or not so-and-so needs Henselian, this is the place I'd check.
Iwasawa--A great place to learn the bare-bones of what one might need to learn class field theory. I am referencing, in particular, the first three chapters. If you are dead-set on JUST learning what you need to, this is a pretty good reference, but if you're likely to wonder about why so-and-so theorem is true, or get a broader understanding of the basics of local fields, I recommend 1.

Class Field Theory:

Lorenz and Levy--Read chapters 28-32, second only to Iwasawa, but with a different flavor (cohomological vs. formal group laws)
Tate and Artin--The classic book. A little less readable then any of the alternatives here.
Childress--Focused mostly on the global theory opposed to the local. Actually deduces local at the end as a result of global. Thus, very old school.
Iwasawa (read the rest of it!)
Milne--Where I first started learning it. Very good, but definitely roughly hewn. A lot of details are left out, and he sometimes forgets to tell you where you are going.

Metric Groups: Markley

Algebraic Geometry: Reid--Shafarevich--Hartshorne--Griffiths and Harris--Mumford

I second the endorsement of Schaum's Outlines.

I actually differ from the view that you have to read iconic books to "hang".  If you read any book and do a good job on it (mastery learning) that will drive your comprehension and skills.  Perfection of Schaum's will trump mediocre work through a tougher text any day of the week.  In addition, there are always different tricks or approaches to find if you look at different books.


Non translation polar Limits of integration for volume of cylinder not centered at origin where $D:=(x-1)^2+y^2≤1$,

How do we find $$  \iint_D dxdy?$$

I'm not sure how to find the volume of a cylinder not centered at the origin using a double integral with polar coordinates. I'm not sure how to setup the limits of integration.

Obviously, the volume of cylinder is independent of where we center it but are the bounds of integration the same as if the cylinder were centered at the origin?

Is the radius of this cylinder not centered at the origin two or one?
What I tried:

Let $ (x-1) = cos(\theta), y= sin(\theta)$

Rewrite $(x-1)^2 + y^2 \leq 1$ as $x^2 + y^2 \leq 2x$, which gives $r^2 \leq 2r\cos\theta$, or $r \leq 2\cos\theta$ (for $\cos\theta$ positive)

Therefore, the region is $\theta$ from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ (where cosine is positive) and $r$ from $0$ to $2\cos\theta$

The given inequality represents the interior of a circle with a radius 1 and centred at (1,0).

The given integral represents the area of this circle.

Limits of x will be from 0 to 2. Limits of y will be from -1 to 1.


How to solve a limit through rationalizing but the radicand is a linear relation and the denominator is a quadratic? The question is :

$$\lim_{x\to 0}{\frac{\sqrt {3x+4}-6}{x^2}}$$

My friend and I are absolutely stumped and can only attempt to solve through substitution. Any ideas?

At the limit,
the numerator is non-zero and negative
and the denominator is zero.

What does that tell you?


Is zero a prime number? Is zero a prime number?

When talking about prime numbers, it seems like the examples given $(2,3,5,7,11,13,...)$ have the property that they have no factors less than themselves and greater than one. But $0$ also has this property, so is it prime? If not, why not?

  Is zero odd or even?

When talking about even numbers, it seems like the examples given $(2,4,6,8,...)$ have the property that, when dividing them by $2,$ have a non-zero quotient and a zero remainder; odd numbers $(1,3,5,7,...)$ have a non-zero quotient and a remainder of $1.$ So, is $0$ odd? even? neither odd nor even?

  Is zero a number?

I've heard that every number is either odd or even, but if $0$ is neither odd nor even, does that mean it isn't even a number?

If you are willing to accept the integers as numbers, then you should have no trouble considering $0$ a number. For one willing to define even numbers as "integer multiples of $2$" then it's similarly clear that $0$ should be considered even. I don't want to spend a lot of space here rehashing the evenness of $0$ since there are already questions dedicated to that problem, but fortunately that makes it easy to direct you to the answer: Is zero odd or even?

I've also found some more discussions on the "numberness of zero" that you might find useful:  What's the hard part of zero? , Why do some people state that 'Zero is not a number'?

The question as to whether or not it should be considered prime is more interesting.

What should primes be?

After you learn about divisibility and factorization, this idea arises about breaking numbers down into smaller parts (sort of like describing matter with smaller and smaller parts). Divisibility makes a partial order on the nonegative integers. This just means that since $12=3\cdot 4$, the "smaller parts" 3 and 4 dividing 12, we can record this as $3\prec 12$ and $4\prec 12$. Furthermore $2\prec 4$ because $2$ divides $4$, and so on. Since $1$ divides everything, we would say that $1\prec n$ for any nonegative integer $n$.

In physics, we are interested in the smallest things from which everything is built from (the "atoms"!). The idea of atoms has two parts:

they should all be "small"
they should build everything else

Well, we can't let $1$ be such a thing, because it would be the only smallest thing, and moreover you can't build anything from $1$ alone. So it is in a sense, too simple.

The next best candidates are those things just above $1$. What just above means becomes clearer if you draw a picture:

This is a sort of Hasse diagram for the nonnegative integers partially ordered by divisibility. Since the diagram is infinite it's not really a Hasse diagram, and the lines to zero don't really come from any numbers, but this is good for our purposes.

From the diagram you can easily see that the primes lie in the first row above $1$, and so they are "as small as possible" without being $1$, and moreover, everything above them (excepting zero) is built out of various combinations of the primes. The gradeschool definition of prime number basically amounts to the fact that nothing lies between $1$ and $p$ for each prime.

Zero, paradoxically, is really aloof and nowhere near the rest of the primes: he doesn't seem very small after all. Moreover he is pretty useless for building numbers since $0n=0$ for any $n$.

So for reasons like these, $0$ is not considered as a prime: he doesn't make a good "atom."

Zero is not a prime number as prime numbers are defined for integers greater than 1.
Zero is an even number. Definition of an even number with modular arithmetic:

$\forall x\in \mathbb{Z},\, x$ is even if and only if $x\equiv 0 \pmod 2$

As $0$ satisfies the definition, then it is an even number.

Of course $0$ is a number, because it is a member of some sets who contains only numbers (such as integers, real numbers, complex numbers etc.). If your question is "Is $0$ a natural number?", it's controversial. In some definitions $0$ is a natural number, but in some of them not. Mathematicians do not have an agreement on that, but I'm with the ones who do not accept it as a natural number, because some theorems which are satisfied by all natural numbers are not satisfied by $0$.


How can I calculate the uncertainty of two combined uncertainties? If uncertainty can be calculated as half the range, and percentage uncertainty is the uncertainty over a mean all multiplied by 100, how can I find what the uncertainty is of A^2 * B^3 when A = 21.3 and the uncertainty of A is +-0.14, and B = 4 and the uncertainty is +-0.01?

Am I right in thinking you just do +-0.14*2 added to +-0.01*3?

No, just adding the uncertainties is wrong; and it is certainly wrong to ever square or cube uncertainties and use the results as uncertainties themselves.  You can see the latter by imagining a trivial re-scaling of the variables; quantities like an uncertainty cubed scale improperly.

The way to combine uncertainties depends on certain assumptions about the quantities being combined.  The usual assumptions applicable to your case are:

A and B are independent of one another.
The statistical distributions of the measurements for A and B are sufficiently similar to gaussians that you can use the math for combining gaussian random variates.
The uncertainites themselves are "small", meaning that certain calculations can be done dropping higher order powers of the uncertainties.

In the case of your problem, first we get the uncertainty on $A^2$.  Obviously $A$ and $A$ are not uncorrelated with each other.  The uncertainty with $\sigma_a = 0.14<A>$ is calculated in this dirty but justifiable way:
$$
( <A> \pm 0.14 <A> )( <A> \pm 0.14 <A> ) = <A>^2 \pm 0.28 <A> <A> + O(0.14^2) =\approx
<A>^2 \left( 1 + \pm 0.28 \right)
$$
Similarly, when the fractional error in $B$ is $0.01$,
$$
<B>^3 (1 \pm 0.01)^3 \approx <B>^3 \left( 1 + \pm 0.03 \right)
$$
Then we use the independence of $A$ and $B$ and the Gaussian property of the distributions to add the uncertainties "in quadrature":
$$
<A>^2 \left( 1 + \pm 0.28 \right) <B>^3 \left( 1 + \pm 0.03 \right) \approx
<A>^2 <B>^3 \left( 1 \pm \sqrt{.28^2+.03^2}\right)
$$
which gives a fractional error on $A^2B^3$ of $0.2816$.

If we assume that $A_\text{min} \leq A \leq A_\text{max}$ and $B_\text{min} \leq B \leq B_\text{max}$, and wish to work out $\min{A^2B^3}$ and $\max{A^2B^3}$, the following is useful. (See Mark's answer for discussions on probability, spread, independence etc). This is meant more to be a "calculate rough bounds for school physics" level answer.

The best way to do these questions is to work out the extremes of values and plug them in - though this might not be how you have been taught.

The lowest value is taking: $A = A_\text{min} = 21.3 - 0.14 = 21.16$, $B = B_\text{min} = 4 - 0.01 = 3.99$, thus min answer $ = A^2 B^3 = 28441.33736$

Similarly, $A_\text{max} = 21.3 + 0.14 = 21.44$, $B_\text{max} = 4.01$, max answer $= 29640.3058$.

Actual calculated answer $ = 21.3^2 \ast 4^3 = 29036.16$

So we have:

\begin{align*}
A^2 B^3 = 29036.16 \begin{cases} + & 604.145796 \\ - & 594.82264 \end{cases}
\end{align*}
(Round as appropriate)

Of course...

You may well have been taught "if you multiply or divide, add the percentage errors". This gives approximately the right answer in most cases, in the above, we get A%Error = 0.6572769953...%, B%Error = 0.25...%, so total percentage error is: $2 \ast 0.6572769953...\% + 3 \ast 0.25\% = 2.064553991...\%$

This gives a rough absolute final error of $\pm 29036.16 * 2.064553991\% = \pm 599.472 \approx \pm 600$.

Note that this is conceptually more confusing, almost more work than our previous working, and we get a less accurate answer. (Here ends my rant about education of calculating errors).


Is the set of all step functions countable or uncountable I've got $f:\mathbb{N}\to \mathbb{N}$ a step function if $f (1)= 1$ and for every $n\geq 2$ we have either
$f(n)=f(n−1)$ or $f(n)=f(n−1)+1.$
I need to show the set of all step functions is countable or
uncountable?
I know that if its countable i need to form a surjection between the set and the set of natural numbers but dont quite know how to show this.

For a given function $f$ define $ e_n \in \{0,1\}$ by $f(n)=f(n-1)+e_n$. So a given function is characterised by the vector $(e_1,e_2,\cdots)$. Which looks pretty uncountable ($2^{ \aleph_0}$).
Indeed a standard diagonal arguement will show it is uncountable.

One thing to note is that showing asset is countable means there exists an injective function into $\mathbb{N}$, not a surjective function. To show that a set is uncountable, you must show that there cannot exist an injective function


Product of two exponentially distributed random variables I am trying to find the close form expression of probability distribution of $Z$ such as $Z=X_1X_2$ where $X_1$ and $X_2$ are two independent exponentially distributed variables with PDF

$P(x1)=λ_1e^{−λ_1x_1}$;
$P(x2)=λ_2e^{−λ_2x_2}$

I know the PDF of $Z$ is

$λ_1 λ_2 \int_0^\infty \exp({-λ_1x_1-λ_2z/x_1)}dx_1$

My question is, is there any easy way to solve the integration part or do we have any close form? Please help.

The PDF of $X_1\cdot X_2$ depends on the Bessel K function, since:
$$\int_{0}^{+\infty}\exp\left(Ax-Bz/x\right)\,dx = 2\sqrt{\frac{Bz}{A}}\cdot K_1\left(2\sqrt{ABz}\right).$$

Closed form integration of above integral can be found in $Eq. 3.324.1$

Given as
\begin{equation}
\int_{0}^{\infty}\exp\bigg(-\frac{\beta}{4x}-\gamma x\bigg)dx= \sqrt{\frac{\beta}{\gamma}}K_1\big(\sqrt{\beta \gamma}\big)
\end{equation}
from the book by
I. S. Gradshteyn and I. M. Ryzhik,Table of Integrals, Series, and Products, 8e


Torsion and curvature of the curve $X(t) = (at, bt^2, ct^3)$ Hey all I am looking for help on a problem. I will post it, and than I will add what I have tried and my ideas etc.  The question has been up now for a few days, I'm sure someone out there can help! I even put a bounty, I have spent a lot of time on this question!

I am interested in calculating the torsion ($\tau$) and curvature ($\kappa$) of the curve $$X(t)=(at,bt^2,ct^3), \quad t   \ge   0 $$ and $a$, $b$, and $c$ are all positive constants.

So here is what I am having problems with. It seems like there are so many different formulas for curvature, and there are also the Frenet–Serret formulas so I am having issues deciding how to do it. I was thinking maybe I could reparametrize with respect to arc length, which would give me it in terms of unit length so I could use some of Frenet–Serret formulas, but I am not confident in that.

What I did so far was I calculated $X'(t)=(a,2bt,3ct^2)$ and $|X'(t)|=\sqrt{a^2+4b^2 t^2 +9c^2 t^4}$.

Then I  calculated $$X''(t)=(0, 2b, 6ct)$$ and $$|X''(t)|=2\sqrt{b^{2}+9c^{2}t^{2}}.$$

I also know that the unit tangent, $$T(t)={X'(t)\over|X'(t)|}$$ and the unit normal is $$N(t)={T'(t) \over |T'(t)|}$$ and that the binormal is $B= T \times N$.

But I am really not sure how to take it further.

I know that in arc length parametrzation we would have $dT/ds= \kappa N$ and $dN/ds=-\kappa T + \tau B$ and $dB/ds= -\tau N$.

Should I keep it in the form it is and use the equation $$\kappa={|X'(t) \times X''(t)|\over |X'(t)|^3}?$$

A few of the other things I am thinking of is that maybe I could solve for torsion by the following:

I know that $T$, $B$ and $N$ form an orthonormal basis of $\mathbb R^{3}$ so can and when we write $N'=\alpha T + \tau B$ , is the coefficient, $\tau$  the definition of torsion?

Moreover, noting $B= T \times N$, we have $$B'= T'\times N + T \times N' = T \times N' =T \times (\alpha T+\tau B) = \tau T \times B= -\tau N $$  (because $N=B \times T$ so $T \times B=-N)$. I think this is how I understand the derivation for that equation.

Is it only in arc length parametrization that I can use the Frenet equations for example?

However, I think I should be able to do it just using the regular formulas, and not an arc length parametrization as the integral would be tough. I did conform this as well, so in terms of the basics of this question I'd like to do it without arc length parametrization.

I am also interested in seeing a intuitive derivation of the torsion formula (the one with the triple product).

I apologize if I didn't show enough work, this is all I could do but I am very happy to learn it. Thanks a lot to any help!

UPDATE:

I am wondering about the validity of what I now have.

In addition to above I computed $X'''(t)=(0,0,6c)$

$X' \times X''= (6bct^2,-6act,2ab)$

$(X' \times X'') \cdot X''' = (12abc)$

$|X' \times X''| = 2 \sqrt {9b^2c^2t^4+9a^2c^2t^2+a^2b^2}$.

Now can I just apply the formulas

$$\tau = \frac{ (X' \times X'') \cdot X'''}{|X' \times X''|^2}.$$

Those calculations with space curves tend to be tedious;
therefore using a computer algebra system is not quite a luxury. Here goes:
$$
\vec{r}(t) = \left[ \begin{matrix} a t \\ b t^2 \\ c t^3 \end{matrix} \right]
\quad \Longrightarrow \\
\vec{\iota} = \, \stackrel{.}{\vec{r}} \, = \frac{d\vec{r}}{ds} =
\frac{d\vec{r}}{dt}\frac{dt}{ds} =
\left[ \begin{matrix} a \\ 2 b t \\ 3 c t^2 \end{matrix} \right] /
\sqrt{a^2 + 4 b^2 t^2 + 9 c^2 t^4}
\quad \Longrightarrow \\
\stackrel{..}{\vec{r}} \, = \frac{d^2 \vec{r}}{ds^2} =
\, \stackrel{.}{\vec{\iota}} \, = \frac{d\vec{\iota}}{ds} =
\frac{d\vec{\iota}}{dt}\frac{dt}{ds} = \left[ \begin{matrix}
-2at(2b^2+9c^2t^2)\\2b(a^2-9c^2t^4)\\6ct(a^2+2b^2t^2)
\end{matrix} \right]/(a^2+4b^2t^2+9c^2t^4)^2
\quad \Longrightarrow \\
\stackrel{...}{\vec{r}} \, =
\, \stackrel{..}{\vec{\iota}} \, = \frac{d\stackrel{.}{\vec{\iota}}}{dt}
\frac{dt}{ds} = \left[ \begin{matrix}
2a(2a^2b^2-24t^2b^4-162t^4b^2c^2+27a^2c^2t^2-405t^6c^4) \\
-8bt(27a^2c^2t^2-81t^6c^4+4a^2b^2) \\
6c(a^4-6a^2b^2t^2-63a^2c^2t^4-8b^4t^4-90b^2t^6c^2)
\end{matrix} \right] / (a^2+4b^2t^2+9c^2t^4)^{7/2}
$$
With $\rho^2 = (\stackrel{.}{\vec{\iota}} \cdot \stackrel{.}{\vec{\iota}})$
- inner product - we find for the curvature $\rho$ :
$$
\rho(t) =
2\sqrt{\frac{a^2b^2+9a^2c^2t^2+9t^4b^2c^2}{(a^2+4b^2t^2+9c^2t^4)^3}}
$$
With the (hopefully) well-known formula
$\;\det\left(\stackrel{.}{\vec{r}},\stackrel{..}{\vec{r}},\stackrel{...}{\vec{r}}\right)
= \rho^2 \tau \;$
we find herefrom for the torsion $\tau$ :
$$
\tau(t) = \frac{3 a b c}{a^2b^2+9a^2c^2t^2+9t^4b^2c^2}
$$

Let curvature $\kappa$ be defined by $\displaystyle \kappa = \frac{\big \|\mathbf T'(t) \big \|}{\big \| \mathbf r'(t)\big \|} $.

Since $\mathbf r(t) = (at, bt^2, ct^3)$,

$\mathbf r'(t)= (a, 2bt, 2ct^2)$ ,

$\|\mathbf r'(t) \| = \sqrt{a^2+4b^2t^2+9c^2t^4}$,

so $\mathbf T(t)= \frac{(a, 2bt, 2ct^2)}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$

and $\mathbf T'(t)= \frac{(0, 2b, 4ct)}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$.

and $\| \mathbf T'(t) \|= \frac{\sqrt{4b^2+16c^2t^2}}{\sqrt{a^2+4b^2t^2+9c^2t^4}}$

Now $\displaystyle \kappa = \frac{\big \|\mathbf T'(t) \big \|}{\big \| \mathbf r'(t)\big \|} = \frac{\sqrt{4b^2+16c^2t^2}}{a^2+4b^2t^2+9c^2t^4}$

For torsion do a similar thing but use the formula:

$\tau = \frac{\big | \mathbf r'(t) \ \mathbf r''(t) \ \mathbf r'''(t) \big| }{\kappa^2} $

where the numerator is the scalar triple product.


Continuous nonnegative supermartingale vanishes after it hits zero Let $(X_t)$ be a continuous nonnegative supermartingale and $T = \inf\{t\geq 0 \colon X_t = 0 \}$ then $X_t = 0$ for every $t\geq T$.

Idea of solution:

Since $T$ is stopping time, by Doob theorem:
$$E(X_{T+q} 1_{T < \infty} | F_T) \leq X_T 1_{T < \infty} =0 $$
for every $q \in \mathbb{Q}$

Then taking expectation we have that $E(X_{T+q}1_{T < \infty}) \leq 0$
. Since $X_{T+q}1_{T < \infty} $ is positive $X_{T+q}1_{T < \infty} = 0$ a.e.
(on a set $\Omega_q$ with $\mathbb{P}(\Omega_q)=1$). Taking $\Omega'=\cap_{q \in \mathbb{Q}^+} \Omega_q$ we will have that $$X_{T+t}1_{T < \infty} = 0$$
on $\Omega'$ for every $t\geq 0$.

The only problem is we are not allow to use Doob theorem since $T$ is not bounded and $X$ is not U.I. I try to use $T \wedge k $ to make the stoping time bounded but I couldn't take limit properly.

We can prove this result just using conclusions in Karatzas&Shreve Brownian Motion and Stochastic Calculus.

By Problem 1.3.16, we have $X_{\infty}\left(\omega\right)={\displaystyle \lim_{t\rightarrow\infty}X_{t}\left(\omega\right)}$ exists for a.e. $\omega\in\Omega$, and $E\left(\left|X_{\infty}\right|\right)<\infty$ and $\left\{ X_{t},\mathscr{F}_{t}:0\le t\le\infty\right\}$  is a supermartingale. Then clearly we have $\left\{ -X_{t},\mathscr{F}_{t}:0\le t\le\infty\right\}$  is a right-continuous submartingale.

Firstly, using Problem 1.2.7, since $X_{t}$ is continuous and $\left\{ 0\right\} $ is closed, $T$ is a stopping time. Then using Lemma 1.2.8, for all $0\le t<\infty$, $T+t$ is also a stopping time. Note that $T$ is $\mathscr{F}_{T}$-measurable.

Then using Theorem 1.3.22, we have $E\left[-X_{T+t}\chi_{\left\{ T<\infty\right\} }|\mathscr{F}_{T}\right]=\chi_{\left\{ T<\infty\right\} }E\left[-X_{T+t}|\mathscr{F}_{T}\right]\geq-\chi_{\left\{ T<\infty\right\} }X_{T}=0$

hence $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }|\mathscr{F}_{T}\right]\le0 $for all $0\le t<\infty$. Then we have $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }\right]\le0$. Since $X_{T+t}\chi_{\left\{ T<\infty\right\} }$ is nonnegative, we have $E\left[X_{T+t}\chi_{\left\{ T<\infty\right\} }\right]=0$. Hence $X_{T+t}\chi_{\left\{ T<\infty\right\} }=0$ a.e on $\left\{ T<\infty\right\} $ for all $0\le t<\infty$.

Then $X_{T+q}\chi_{\left\{ T<\infty\right\} }=0$ for all $q\in\left[0,\infty\right)\cap\mathbb{Q} $ holds a.s. on $\left\{ T<\infty\right\} $, i.e. there exists $A\subset\Omega$ with $P\left(A\right)=1$ such that for all $\omega\in A$ and $q\in\left[0,\infty\right)\cap\mathbb{Q}$, $X_{T+q}\chi_{\left\{ T<\infty\right\} }\left(\omega\right)=0$. Since $X$ has continuous path, for all $\omega\in A$ and $t\in\left[0,\infty\right)$, $X_{T+t}\chi_{\left\{ T<\infty\right\} }\left(\omega\right)=0$, that is $X_{T+t}=0$; $0\le t<\infty$ holds a.s. on $\left\{ T<\infty\right\}$.


Distribution of the maximum minus the minimum Assume $A_i\sim\exp(a)$. Define $G=\max\{A_i\}_{i\in\{1,...,n\}}-\min\{A_i\}_{i\in\{1,\dots,n\}}$. Find $P(G\le x)$.

My step:
$P(G\le x)$=$P(\max A_i\le\min A_i+x)=(\text{by Markov property})P(\max\ A_i\le \min\ A_i|\min\ A_i>x)=P(A_1=A_2=\dots A_n|\min A_i>x)=P(A_1=\dots=A_n)$ and $\min A_i>x)/P(\min\ A_i>x)$
I don't know which claims is wrong but there has to be something wrong cause i cannot get the answer

Thanks.

The lack of memory of the exponential distribution can be used to produce conceptual proofs that, for every $n\geqslant2$, $G$ is distributed as the maximum of $(n-1)$ i.i.d. random variables each exponentially distributed with parameter $a$. Since, however, the OP failed to explain their background, here is a direct, hands-on, approach.

Consider $U=\min\{A_k\,;\,1\leqslant k\leqslant n\}$ and $V=\max\{ A_k\,;\,1\leqslant k\leqslant n\}$, then $U\lt V$ almost surely and, for every $u\lt v$,
$$
[u\lt U,V\lt v]=\bigcap_{k=1}^n[u\lt A_k\lt v],
$$
hence, by independence of the random variables $(A_k)$,
$$
P(u\lt U,V\lt v)=(\mathrm e^{-au}-\mathrm e^{-av})^n.
$$
Differentiating this identity twice yields the density $f$ of $(U,V)$ as
$$
f(u,v)=n(n-1)a^2\mathrm e^{-au-av}(\mathrm e^{-au}-\mathrm e^{-av})^{n-2}\mathbf 1_{0\lt u\lt v}.
$$
By definition, $G=V-U$ hence, for every $x\gt0$,
$P(G\leqslant x)=(\ast)$ with
$$
(\ast)=\int_0^\infty\!\!\!\int_u^{u+x}f(u,v)\mathrm dv\mathrm du=\int_0^\infty na\mathrm e^{-au}\left[(\mathrm e^{-au}-\mathrm e^{-av})^{n-1}\right]_{v=u}^{v=u+x}\mathrm du,
$$
that is,
$$
(\ast)=\int_0^\infty na\mathrm e^{-au}(\mathrm e^{-au}-\mathrm e^{-au-ax})^{n-1}\mathrm du=(1-\mathrm e^{-ax})^{n-1}\int_0^\infty na\mathrm e^{-nau}\mathrm du,
$$
and finally,
$$
P(G\leqslant x)=(1-\mathrm e^{-ax})^{n-1}.
$$
To the OP: In the question you assert that $P(V\leqslant U+x)=P(V\leqslant U\mid U\leqslant x)$ and you explain that this holds "by Markov property". This is obviously wrong but the trouble is that I cannot even see what you think you are doing there... Please explain.

There is a Markova property as $\min A_i\sim exp$ And it is correct with your step as$\max A \le min\implies\max=min$. I think the difficulty is how to get the distribution function from the wrok you have done


Evaluate limit of $a_n = \bigg( 1 + \frac{2}{n} \bigg)^n$ $a_n = \bigg( 1 + \frac{2}{n} \bigg)^n$

$$\lim_{n\to\infty} \bigg(1+\frac{2}{\infty}\bigg)^\infty$$
so we need to use L'Hospital rule, I want to take the derivative of $\bigg( 1 + \frac{2}{n} \bigg)^n$, which I thought should be $(1 + \frac{2}{n})$, but am I doing this part wrong? Then I take the limit of the derivative and get

$$\lim_{n\to\infty} 1 + \frac{2}{n} = 1$$

But the book says the answer is $e^2$. I am really lost where $e$ comes into play

Look up how use l'Hopital's rule.  To use it on an indeterminate form $1^\infty$ as we have here, we need to take the logarithm, get an indeterminate form either $0/0$ or $\infty/\infty$, apply l'Hopital's rule to get the limit of that, then exponentiate.

\begin{align}
f(n) &= \left(1+\frac{2}{n}\right)^n,\qquad\text{indeterminate form }{1^\infty}
\\
\log f(n) &= n\log\left(1+\frac{2}{n}\right),\qquad\text{indeterminate form }\infty \times 0
\\
\log f(n) &= \frac{\log\left(1+2/n\right)}{1/n},\qquad\text{indeterminate form }\frac{0}{0}
\\
\log f(n) &\sim \frac{\frac{d}{dn}[\log\left(1+2/n\right)]}{\frac{d}{dn}[1/n]}\\
&\text{where I wrote $\sim$ for: "has the same limit as,}
\\
&\qquad\text{provided the limit on the right exists"}
\\
\log f(n) &\sim
\frac{-2/n^2}{(-1/n^2)(1+2/n)} = \frac{2}{1+2/n}
\\
\lim\log f(n) &= 2
\\
\log \lim f(n) &= 2
\\
\lim f(n) &= e^2 .
\end{align}

Question: Why is $1^\infty$ indeterminate?

Answer. Because we can get different results for limits of that form.  Example

$$
\lim_{n\to\infty}\left(1+\frac{2}{n}\right)^n = e^2
\\
\lim_{n\to\infty} 1^n = 1
\\
\lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{n^2} = +\infty
$$
and so on.


Do all 'or' cases need to be checked in a direct proof? There's a problem from Richard Hammack's book of proof, which asks to prove the proposition:

Suppose $x,y \in \mathbb{Z}$. If $5 \nmid xy$, then $5 \nmid x$ and $5 \nmid y$

In this case, the intent is to show the use of the contrapositive proof. So a direct proof on $$5\mid x \lor 5\mid y \implies 5 \mid xy$$

The proof is straightforward, and in the book the solution shows that when $5\mid x$ is true, then $5 \mid xy$ is true, and similarly for $5\mid y$.

In this case, the solution doesn't show the case where both $5\mid x$ and $5\mid y$ are true.

In this particular example the proof of the third case is trivial, and so I understand that it can be left out entirely.

However, I am wondering in general, when using direct proof, and when the $P(x,y)$ statement is compound statement made up of two statements conjuncted with an or, then is it necessary to check all 3 cases?

In other words, to link it back to the example proof, is it strictly necessary to show that $5\mid x$ AND $5\mid y$ to complete the proof? Or is it that since we have shown the T/F and F/T case, the T/T case is also always true?

The case $5|x$ and $5|y$ is already covered by the other cases, so it is not necessary to consider it separately. The important thing is to make sure that the cases you consider are exhaustive.

Consider it this way: If you have separately proven that $5|x$ is true and $5|y$ is true, you can linguistically say that you have proven that the statements $5|x$ and $5|y$ are both true. This is logically the same as $5|x \wedge 5|y$ being true, so you do NOT need to separately check this, since you have already done so logically. Typically, the "and" statement is not considered a separate case, so a proof by cases comprises only of proving the separate statements.


$f: X \rightarrow Y$ is continuous between two topological spaces. Show $\forall A \subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$ $f: X \rightarrow Y$ is continuous between two topological spaces. Show $\forall A \subseteq X$ we have $f(\overline{A}) \subseteq \overline{f(A)}$

Since we know that the function is continuous, we know that the preimage of a closed set in $Y$ is a closed set in $X$.

However, I don't see how this helps.

Another idea I have is to show that if every open set $V \subseteq Y$ containing $x$ where $x \in f(\overline{A})$ we have: $V \cup f(A) \ne \emptyset$.
That would imply that $x \in \overline{f(A)}$

However, I'm not quite sure if this is the most intuitive apprach.

Where would I start to show this inclusion?

You're on the right path but it needs more details in this implication you made (assuming that there is a typo in your question, ie, you wanted to write $f(A) \cap V \neq \emptyset$). What you need to show is: Given $y \in f(\overline{A})$, for every open set $V \subset Y$ such that $y \in V$ you have $V \cap f(A) \neq \emptyset$. This is true since $y \in f(\overline{A})$ implies that $y = f(x)$ with $x \in \overline{A}$, as $f^{-1}(V)$ is an open set (f is a continuous function) and $x \in f^{-1}(V)$ there exists $x_{1} \in f^{-1}(V) \cap A$ which gives you that $f(x_{1}) \in V \cap f(A)$, ie, $V \cap f(A) \neq \emptyset$.

Let $f(B)=\overline{f(A)}$, then by continuity B is closed and $A\subseteq B$. As $\overline{A}$ is an intersection of all closed sets containing $A$, we get $\overline{A}\subseteq B$ and $f(\overline{A})\subseteq f(B)=\overline{f(A)}$


What is rank of $f(A)$, where $f$ is the minimal polynomial of $A$? If $f(x)$ is minimal polynomial of the $4\times 4$ matrix
$$A=\begin{pmatrix}
0 &0 &0& 1\\
1 &0 &0 &0\\
0 &1 &0 &0\\
0 &0 &1 &0
\end{pmatrix}$$

Then what is rank of $f(A)$? I think $f(A)$ will be a zero matrix so its rank is 0. Am I right?

Yes, that's exactly right...  $\phantom{ok I'll add a little random text here to avoid automatic conversion}{}{}$

From the definition of minimal polynomial $f(A)=0.$ Hence $f(A)$ is the zero transformation. Hence rank is $0.$


Find $\int (x^3+1)(\cos(2x)) dx$ What I've tried for this problem is expanding it to $x^3\cos(2x) + \cos(2x)$ and then evaluating the respective functions as separate integrals. The first one uses tabular and the second one is simple u substitution. Is my procedure correct?

what we want to do is eliminate x for the integral, thus do the integral by parts.

$$
\begin{align}
\int (x^3+1)\cos (2x))\mathrm dx
&=\frac{1}{2}(x^3+1)\sin2x - \int 3x^2\sin 2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{2}\int 3x\cos2x\mathrm dx\\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x +\frac{3}{4}\int\sin 2x \mathrm dx \\
&=\frac{1}{2}(x^3+1)\sin2x ＋\frac{1}{4}3x^2\cos2x-\frac{1}{4}3x\sin2x -\frac{3}{8}\cos 2x +constant
\end{align}
$$

The first integral must be integrated by parts 3 times to whittle down the $x^3$.


Why is $\sin(x) = \sin(180^{\circ}-x)$ I cannot seem to understand why this is true. Same for $\cos(x) = -\cos(180^{\circ}-x)$ and $\tan(x) = -\tan(180^{\circ}-x)$. Without the use of the compound angle formulas.

Thanks

Hint  Rebember that
$
\tan( 180^o \pm x  ) = \frac{\sin( 180^o \pm x)}{\cos(180^o\pm x)}.
$
Now draw the  trigonometric cycle and observes the measures of  sine and cosine.

Because, if you take $\cos(180°-x)$ you are effectively getting the opposite of $\cos(x)$. Due to $180°$ being halfway over the circle diameter. So if you do take the opposite of that $-\cos(180°-x)$ you get $x$ back, because you switch the sign.

Easy to see with this image:


prove that minor arc of a great circle is the shortest distance How to prove that the minor arc of a great circle is the shortest path connecting two points on the surface of a sphere?

The key to solving the problems is that we must take all curves connecting the two points , not just great-circle or parametrized differentiable curve, into account!

A similar problem is the straight line distance(shortest distance between two points): we must take all curves connecting the two points into account!

Here is a low-tech approach which does not use derivatives much. Given distinct points $a,b$ on the sphere, let $\gamma$ be the shorter arc of great circle between them. Let $V$ be the surface obtained by rotating $\gamma$ about the line $ab$. Note that  $V$ is convex (i.e., bounds a convex body) and lies inside the sphere. The nearest-point projection onto a convex set does not increase Euclidean distance and therefore does not increase length of curves (defined by means of sums of Euclidean distances over partitions). Therefore, it suffices to prove that a curve connecting $a$ to $b$ on surface $V$ cannot be shorter than $\gamma$.

Let $\Gamma$ be any such curve. Pick partition points $p_k$ on $\Gamma$ to approximate its length. For each $k$, draw the plane through $p_k$ that is orthogonal to line $ab$. This plane crosses $\gamma$ at a point $q_k$. It is easy to see that $|p_k-p_{k+1}|\ge |q_k-q_{k+1}|$ for each $k$. (Just write the Euclidean distance in cylindrical coordinates with $ab$ as the axis.) Since $\sum_k |q_k-q_{k+1}|$ is greater than or equal to the length of $\gamma$, the conclusion follows.

You know that such a length minimizing curve must be a geodesic. Then since every geodesic of the sphere is a great circle, such a length minimizing curve must be an arc of a great circle.


Complex differentiability vs differentiability in $\mathbb{R}^2$ I was wondering about the following for quite a while: The euclidean topology on $\mathbb{R}^2$ is homeomorphic with the topology induced by the modulus function on $\mathbb{C}$. Hence limit behavior is the same in both. So, doesn't that mean that the limit which defines the complex derivative converges if and only it converges in $\mathbb{R}^2$?

Clearly, I am missing something, since not every differentiable function $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is smooth, but over $\mathbb{C}$ this is true.

P.S.: I read the question: Is Complex Analysis equivalent Real Analysis with $f:\mathbb R^2 \to \mathbb R^2$?, it does however not clear up my confusion.

Derivatives are linear maps that approximate the function within $o$. In other words, if $f$ is differentiable at $x$, then $f(x+t)=f(x)+A(t)+o(t)$, where $A$ is a linear map and $\lim_{t\to 0}\frac{o(t)}{|t|}=0$. Here $|t|$ denotes the magnitude of $t$, which, as you pointed out, is the same whether we view $t$ as a complex number or a $2$-dimensional real vector.

In the 2-dimensional real case, $A$ can be any linear map, ie. any $2\times 2$ matrix with entries in $\mathbb{R}$.

But in the complex case, $A$ is linear map of 1-dimensional complex vector spaces, ie. a $1\times 1$ matrix with entries in $\mathbb{C}$. In other words, $A$ is multiplication by some complex number $z$. If we think of $\mathbb{C}$ as $\mathbb{R}^2$, then $A$ is a rotation (by the argument of $z$) and scaling (by the magnitude of $z$). Not every $2\times 2$ linear map is of this form.

In sum, you are right to note that the convergence is the same in each space. Real differentiability means that approximations converge to some linear map, while complex differentiability means that approximations converge to a more specific type of linear map, which is why complex differentiability is a stronger condition than differentiability in $\mathbb{R}^2$.

The limit that defines $\mathbb{R}^n$ differentiability is:
$$\lim_{\mathbf{h}\to \mathbf{0}}\frac{\mathbf{f}(\mathbf{a}+\mathbf{h})-\mathbf{f}(\mathbf{a})-T(\mathbf{h})}{\left\|\mathbf{h}\right\|}=\mathbf{0}$$
where $T$ is a linear transform. The limit that defines $\mathbb{C}$ differentiability is not
$$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{\left|h\right|}=0$$
but rather
$$\lim_{h\to 0}\frac{f(z+h)-f(z)-T(h)}{h}=0$$
In addition, in the complex case we also require $T$ to be a complex linear transform, not just a linear map, that must be "nice" towards complex multiplication.These two facts create the so called Cauchy Riemann Equations. You can read more here pg 111.
 In other words, multiplication and division of complex numbers make $\mathbb{C}$  differentiabilty much stronger than $\mathbb{R}^2$ differentiability.


What is the probability of someone else in a room having the same birthday as me and no other people sharing a birthday? What is the probability that at least one person in a room of $n$ people (excluding me) has the same birthday as I and that nobody else in the room shares a birthday on a different date? For practical purposes, pretend that leap years don't exist.

As a follow-up, what is the expected number of people that have the same birthday as I in a room of $n$ people (excluding me) if nobody else in the room shares a birthday on a different day?

The probability is:

[Explanantion:Spoiler]

   Choose n birthdays from 364 days(except the one which you have), all different dates then select one person whose birthday you want to change to yours and arrange the rest.Total ways of selecting birthdays will be $365^n$.  In case of 366 persons select one person to have same birthday as you and arrange the rest 364 persons with the remaining 364 dates.

$$\frac{\binom{364}{n}\binom{n}{1}(n-1)!}{365^n}:\quad n\le 364\\\frac{\binom{365}1364!}{365^n}:n=365\\0:n\ge366$$
Sorry I know combinatorics and haven't yet covered probability or expected value.


Series $ \sum\limits_{n=1}^{\infty}\frac{1}{n}e^{2 \pi i n x} $ Prove that the sum
  \begin{equation}
\sum\limits_{n=1}^{\infty}\frac{1}{n}e^{2 \pi i n x}
\end{equation}
  converges for any $ x \notin \mathbb{Z} $.

By definition, the initial series converges if and only if $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\cos{(2 \pi n x)} $ and $ \sum\limits_{n=1}^{\infty}\frac{1}{n}\sin{(2 \pi n x)} $ both converge. Neither Dirichlet's nor Abel's test works. Integral test can't be applied because $ \frac{1}{t}\cos{(2 \pi x t)} $ is not monotonic function of $ t $. This problem slightly reminds me of how Equidistributional theorem is proved, but most probably it has nothing to do with it.

Hint We have for $x\neq 2\pi m$; $$\sum_{k=1}^ne^{ikx}=\frac{\sin(nx/2)}{\sin(x/2)}e^{i(n+1)x/2}$$

Thus Dirichlet's criterion works all right. This says the following:

  Let $\langle a_n\rangle$ be a sequence of complex numbers with bounded partial sums, and let $b_n$ be a sequence of real numbers decreasing monotonically to zero. Then $$\sum a_nb_n$$ converges.

Let $a_n(x)=\sum_{k=1}^n {\rm e}^{{\rm i}kx}$; even though  $\lim_{x\rightarrow 0}|a_n(x)|=n$ (and $|a_n(\pi)|=|\sin(n\frac{\pi}{2})|$), for each $x$ in $(0, \pi]$,  $ a_n(x)$ is indeed bounded. In fact, for large fixed $r$, $|a_n(\frac{\pi}{r})|=|\frac{\sin(\frac{n\pi}{2r})}{\sin(\frac{\pi}{2r})}|$ $\approx|\frac{\sin(\frac{ n\pi}{2r})}{\frac{\pi}{2r}}|$ $={\frac{2r}{\pi}}|\sin(\frac{n\pi}{2r})|$ is bounded. Also, the local maxima of $|a_n(x)|$, for $x\in(0, \pi]$, decrease with $x$.


How to find a explicit field morphism I want to find a explicit morphism from $K_1$ to $K_2$, where

$$K_1=\mathbb F_2[x]/(x^3+x+1)\mbox{ and }K_2=\mathbb F_2[x]/(x^3+x^2+1).$$

I've found that it must exist because these polynomials are irreducible, hence these fields have $2^3$ and $2^3$ elements and $3\mid 3$.

But how could I find the explicit morphism?

Since the multiplicative group $\mathbb F_8^*$ is the cyclic group of order $7$, any nontrivial element generates the full group. For example, in $K_1$, the group is $\{1,x,x^2, \ldots x^6\}$.
Next, we know that the Frobenius map $x \mapsto x^2$ is a field automorphism, so that $x,x^2,x^4$ must have the same minimal polynomial, and likewise with $x^3,x^5,x^6$.
Since the minimal polynomial of $x$ is $X^3+X+1$, this means that $x^3,x^5,x^6$ are the three roots of $X^3+X^2+1$ in $K_1$. Thus sending the $x$ of $K_2$ to any of those will work.

For example the map $K_2 \to K_1$ given by $x \mapsto x^{-1} (= x^6)$ is an isomorphism (and now we notice that the second polynomial was the first polynomial with the coefficients reversed)

Hint: Both fields are isomorphic to $\mathbb{F}_{2^{3}}$ and explicitly
the elements in both of them are of the form $[a+bx+cx^{2}]$ (by
the Euclidean algorithm) (where $a,b,c\in\mathbb{F}_{2})$.

So any homomorphism $\mathbb{F}_{8}\to\mathbb{F}_{8}$ will be a good
answer - given that you have explicitly isomorphisms between $\mathbb{F}_{8}$
to those fields. such isomorphism can be found by finding the multiplication
table of both field and comparing it with the one of $\mathbb{F}_{8}$.


Existence and value of $\lim_{n\to\infty} (\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x})$ for $x>0$ Does the limit
$$W(x)=\lim_{n\to\infty} \left(\ln\frac{x}{n}+\sum_{k=1}^n \frac{1}{k+x} \right)$$
exist for all $x>0$? If so, what is the limit
$$\lim_{x\to\infty}W(x)?$$

We have
$$\sum_{k=1}^n \dfrac1{k+x} = \int_{1^-}^{n^+} \dfrac{d \lfloor t \rfloor}{t+x} = \left. \dfrac{\lfloor t \rfloor}{t+x} \right\vert_{t=1^-}^{t=n^+} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \dfrac{n}{n+x} + \int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt$$
Now
$$\int_{1^-}^{n^+} \dfrac{\lfloor t \rfloor}{(t+x)^2} dt = \int_1^{n^+} \dfrac{t}{(t+x)^2} dt - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt$$
$$\int_1^{n^+} \dfrac{t}{(t+x)^2} dt = \int_1^{n} \dfrac{dt}{t+x} - x \int_1^n \dfrac{dt}{(t+x)^2} = \log(n+x) - \log(1+x) -x \left(\dfrac1{1+x} - \dfrac1{n+x}\right)$$
Hence, we get that
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log\left(\dfrac{x}n \right) + \dfrac{n}{n+x} + \log \left(\dfrac{n+x}{1+x}\right) -\dfrac{x}{1+x} + \dfrac{x}{n+x} - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = - \dfrac{x}{1+x} + 1 + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt\\
& = \dfrac1{1+x} + \log\left(\dfrac{x}n \cdot \dfrac{n+x}{1+x}\right) - \int_1^{n^+} \dfrac{\{t\}}{(t+x)^2} dt
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \dfrac1{1+x} + \log \left(\dfrac{x}{1+x}\right) - \underbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}_{\text{Converges since }\{t\} \in [0,1)}$$
$$0 \leq \overbrace{\int_1^{\infty} \dfrac{\{t\}}{(t+x)^2} dt}^{g(x)} \leq \int_1^{\infty} \dfrac1{(t+x)^2} dt = \dfrac1{1+x}$$ There might be some name for $g(x)$ (Probably some of the number theorists on this website might be able to identity this). For instance, $g(0) = 1-\gamma$, where $\gamma \approx 0.57721$ is the Euler Mascheroni constant.
Now $$\lim_{x \to \infty} W(x) = 0 + \log(1) + 0 = 0$$

Another method is as follows. From here, we have
\begin{align}
\sum_{k=1}^n \left(\dfrac1k - \dfrac1{x+k} \right) & = \sum_{k=1}^n \int_0^1 (y^{k-1} - y^{x+k-1})dy\\
& = \int_0^1 (1-y^x) \sum_{k=1}^n y^{k-1} dy\\
& = \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Hence, we have
\begin{align}
\log(x/n) + \sum_{k=1}^n \dfrac1{k+x} & = \log(x/n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right) + \sum_{k=1}^n \dfrac1k\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) + \sum_{k=1}^n \left(\dfrac1{k+x} - \dfrac1k \right)\\
& = \log(x) + \sum_{k=1}^n \dfrac1k - \log(n) - \int_0^1 (1-y^x) \dfrac{1-y^n}{1-y} dy
\end{align}
Now letting $n \to \infty$, we get that
$$W(x) = \log(x) + \gamma - \int_0^1 \dfrac{1-y^x}{1-y} dy$$
Now as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$ Hence, we get that
$$\lim_{x \to \infty} W(x) = 0$$

Let us prove why, as $x \to \infty$, we have $$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$
The proof is the same as before. We have
$$\sum_{k=1}^n \dfrac1k = \int_0^1 \sum_{k=1}^n y^{k-1} dy = \int_0^1 \dfrac{1-y^n}{1-y} dy$$
But we know that $\displaystyle \sum_{k=1}^n \dfrac1k = \log(n) + \gamma + \mathcal{O}(1/n)$. Hence, we get that
$$\int_0^1 \dfrac{1-y^n}{1-y} dy = \log(n) + \gamma + \mathcal{O}(1/n)$$
 Replacing $n$ by $x$ and because the integral is a smooth function of $x$, we can conclude that
$$\int_0^1 \dfrac{1-y^x}{1-y} dy = \log(x) + \gamma + \mathcal{O}(1/x)$$

I think it does not always converge: in fact for x->0, W(x) tends to -infinity, it seems

First of all, look at the definition of the Euler-Mascheroni constant and observe that it is very similar to your problem

http://en.wikipedia.org/wiki/Euler%27s_constant

You must develop the part of with the logarith ln(x/n)= ln x - ln(n)

Then, in the limit you may drop the x from the sumatory and if you substract ln(n) it is exactly the Euler constant=0.577 aprox. But ln(x) tends to -infinity. Thus, it does not always converge.

Hope this helps,

David, Barcelona, Catalonia


Is Scrabble's method of determining turn order fair? At least the way my family plays, turn order is determined by drawing tiles and seeing who has the letter closest to A (blanks taking precedence). If I recall correctly, there are 100 unevenly distributed letters.

Intuitively, it seems like it would be unfair, though I can't come up with a way to prove it. Obviously the distribution matters: my thoughts are if there are more tiles in the first half of the alphabet (including blanks), then the starting player holds an advantage since they're more likely to get a tile earlier in the alphabet than the next (and vice versa if there are fewer tiles in the first half). That doesn't seem too right, though. I'm probably just forgetting some basic prob stats.

I imagine this is likely a duplicate, but I couldn't find anything relating to it (maybe since I've been searching for Scrabble). My apologies if it is.

Please feel free to edit in appropriate tags.

Because of symmetry, no player has an advantage.  That is, if $\mathscr A$ is the set of outcomes where player $A$ gets the letter closest to A and $\mathscr B$ is the set of outcomes where player $B$ gets the letter closest to A, there is a one-to-one correspondence between $\mathscr A$ and $\mathscr B$ obtained by interchanging the letters players A and B get.  Since all outcomes are equally likely, they have the same probability of getting to go first.

The drawing order matters because tiles are drawn without replacement. If drawn with replacement, the order wouldn't matter.

Drawing without replacement changes the tile frequency, relatively more if an unfrequent tile is drawn. Drawing without replacement also changes the pool size, with an impact on tile frequencies.

It appears to me that frequent tiles of high value favour at a small extent the last player to draw.


Probability question where an object is chosen randomly out of two objects I am stuck at the following probability problem. Will appreciate your help.

Two opera singers, Mario and Clarissa both perform on the same night, in separate recitals. The independent probabilities that two newspapers X and Y publish reviews of their recitals are given below:

Probability of review in newspaper X

Mario's recital - 1/2,
Clarissa's recital - 2/3

Probability of review in newspaper Y

Mario's recital - 1/4,
Clarissa's recital - 2/5

Mario buys one of the newspapers at random. What is the probability that it has reviewed "both" recitals?

I did this way: P(reviewed both recitals) = P(buys paper X)*P(X reviews Mario)*P(X reviews Clarissa) = (1/2)(1/2)(2/3) = 1/6

But 1/6 is not the correct answer. Let me know where I am wrong and why.

Thanks in advance.

Let $B_x$ be the event "Mario buyed paper $X$", and let $B_x^c$ denote the complementary event . Let $R$ be the event "the buyed paper reviewed both recitals"

Then $P(R)= P(R \cap B_x)+P(R \cap B_x^c)=P(R\mid B_x)P(B_x)+P(R\mid B_x^c)P(B_x^c)$

Now, $P(B_x)=P(B_x^c)=\frac{1}{2}$. And $P(R\mid B_x)= \frac{1}{3} \frac{2}{3}$ ... Surely you can go on from here.

Should not it be:

\begin{align*}
P(\text{reviewed both recitals})& = P(\text{buys paper $X$})\times P(\text{$X$ reviews both Mario & Clarissa}) \\
&\qquad+ P(\text{buys paper $Y$})\times P(\text{$Y$ reviews both Mario & Clarissa}) \\
&= \frac{1}{2}\cdot\frac{1}{2}\cdot\frac{2}{3} + \frac{1}{2}\cdot\frac{1}{4}\cdot\frac{2}{5} \\
&= \frac{1}{6} + \frac{1}{20} = \frac{13}{60}
\end{align*}
(assuming independence of the appearance of the reviews)


Is the logic implication a tautology? I had a discussion with a person who says that logic implication is a tautology.

Reviewing this book, specifically on page $12$ ($19$ of the document), I see that there is a difference between what is a conditional, $⇒$, and what is an implication.
https://www.karlin.mff.cuni.cz/~krajicek/mendelson.pdf

The conditional has the next truth table:

http://www.metafysica.nl/nature/truth_table_impl.gif

And the book says on page $16$ that $p$ implies $q$ if and only if $p ⇒ q$ is a tautology. But, does this mean that an implication is a tautology? Because if it does, then we cannot say something does not imply another thing, I think.

A conditional has a condition: if A is true, then B will also be true.

An implication is a direct statement of fact: A is true, therefore B is true.

That becomes tautological because it is trivially true in all cases. A is true so B is true, so A is true so B is true.

No, it doesn't.

A tautology consists of a certain type of statement form which satisfies the given formation rules and always holds true under all evaluations.  Using bold letters, instead of capital letters, some tautologies are:

(B=>B)

and

(A=>(B=>A))

and

(($\lnot$A=>$\lnot$B)=>(($\lnot$A=>B)=>A))

On the other hand, though an implication will hold true for all valuations of the letters, it need not satisfy the formation rules for statement forms.  As an example, consider the following:

A logically implies ((A=>B)=>B)

Or we might write |= as a short-hand for logically implies and write:

A |= ((A=>B)=>B)

That doesn't satisfy the formation rules for statement forms.  To indicate that the premisses having the form "A" and the form "(A=>B)" logically imply B we might write

A, (A=>B) |= B

or

{A, (A=>B)} |= B

But, the component parts are different from those of tautologies.  With the idea of having a set of premisses in mind we could rewrite the above as:

{(A=>B), A} |= B

or

{A, A, A, (A=>B), A, A, (A=>B), (A=>B), A} |= B

Those immediately above all represent a single logical implication, at least as a single concept.

But, a tautology like

((A=>(A=>B))=>(A=>B))

is distinct from the tautology:

(((A=>(A=>B))=>(A=>(A=>B)))=>(A=>B)).

Those consist of two different statement forms.


Argument at branch cut I try to use residue to calculate this integral $$\int_1^2 \frac{\sqrt {(x-1)(2-x)}} {x}\ dx$$
I let $$f(z)=\frac{\sqrt {(z-1)(2-z)}} {z}$$ and evaluate the integral $$\int_{(\Gamma)} f(z)dz$$ along the contour $\Gamma$ consisting of: $(1)$ circle$(1;\epsilon)$; $(2)$ circle$(2;\epsilon)$; $(3) $circle$(0;R)$; $(4)$ segments $[1+\epsilon,2-\epsilon]$ - upper and lower sides of branch cut $[1,2]$, and $(5)$ segments $[2+\epsilon,R]$

My problem is how to define the argument of $z-1$ and $2-z$ at upper and lower sides of branch cut
In a similar example: http://en.wikipedia.org/wiki/Methods_of_contour_integration#Example_.28VI.29_.E2.80.93_logarithms_and_the_residue_at_infinity
why $\arg(z)$ ranged from $-\pi$ to $\pi$ while $\arg(3-z)$ ranged from $0$ to $2\pi$

This answer refers to the following
MSE post
and duplicate material has been omitted.

With these  types of integrals usually  what is being asked  for is to
use two  branches of  the logarithm whose  cuts cancel outside  of the
integration interval.

Suppose we seek to compute
$$Q = \int_1^2 \frac{1}{x} \sqrt{(x-1)(2-x)} dx.$$

Re-write this as
$$\int_1^2 \frac{1}{z}
\exp(1/2\mathrm{LogA}(z-1))
\exp(1/2\mathrm{LogB}(2-z)) dz$$
and call the function $f(z).$

We  must choose  two  branches of  the  logarithm $\mathrm{LogA}$  and
$\mathrm{LogB}$ so that  the cut is on the real axis  from $1$ to $2.$
This is accomplished when $\mathrm{LogA}$  has the cut on the negative
real  axis and  $\mathrm{LogB}$ on  the  positive real  axis with  the
arguments being $(-\pi,\pi)$ and $(0,2\pi).$

We use  a dogbone contour  which is traversed  counterclockwise.  Then
$\mathrm{LogA}$  gives   the  real  value  just  above   the  cut  but
$\mathrm{LogB}$ contributes  a factor of $\exp(2\pi  i \times 1/2).$
Below the  cut $\mathrm{LogA}$  again produces the  real value  but so
does $\mathrm{LogB}.$ This implies that

$$Q (1 - \exp(2\pi i \times 1/2))
= - 2\pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$
or
$$Q = - \pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$

For the residue at zero we get
$$\exp(1/2\mathrm{LogA}(-1))
\exp(1/2\mathrm{LogB}(2))
\\ = \exp(1/2\times (-i\pi))\exp(1/2\log2)
= -i\sqrt{2}.$$

Now for the residue at infinity we use the formula
$$\mathrm{Res}_{z=\infty} h(z)
= \mathrm{Res}_{z=0}
\left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right].$$

In  the following we  need to  distinguish between  the upper  and the
lower half-plane. Assume $z=R e^{i\theta}$ with $0\le\theta\lt2\pi.$
Upper half-plane.
Here we have $$\mathrm{LogA}(1/z-1) =
\mathrm{LogA}(1-z) - \mathrm{LogA}(z)$$
and $$\mathrm{LogB}(2-1/z) =
\mathrm{LogB}(2z-1) - \mathrm{LogB}(z).$$

This gives for the function to evaluate the  residue the term
$$- \frac{z}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(-1/2\mathrm{LogA}(z))\\ \times
\exp(1/2\mathrm{LogB}(2z-1))
\exp(-1/2\mathrm{LogB}(z)).$$

But we have in the upper half plane
$$\exp(-1/2\mathrm{LogA}(z))\exp(-1/2\mathrm{LogB}(z))
= \frac{1}{z},$$
so this becomes
$$- \frac{1}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(1/2\mathrm{LogB}(2z-1)).$$

Lower half-plane.
Here we have $$\mathrm{LogA}(1/z-1) =
\mathrm{LogA}(1-z) - \mathrm{LogA}(z)$$
and $$\mathrm{LogB}(2-1/z) = 2\pi i +
\mathrm{LogB}(2z-1) - \mathrm{LogB}(z).$$

This gives for the function to evaluate the residue the term
$$- \frac{z}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(-1/2\mathrm{LogA}(z))\\ \times
\exp(1/2\times 2\pi i)
\exp(1/2\mathrm{LogB}(2z-1))
\exp(-1/2\mathrm{LogB}(z)).$$

But we have in the lower half plane
$$\exp(-1/2\mathrm{LogA}(z))\exp(-1/2\mathrm{LogB}(z))
= -\frac{1}{z},$$
so this becomes
$$- \frac{1}{z^2}
\exp(1/2\mathrm{LogA}(1-z))
\exp(1/2\mathrm{LogB}(2z-1)).$$

We have  established matching  terms for the  upper and  the lower
half  plane.

The cut  from the first term starts  at one and extends  to the right.
The cut  from the second term starts  at one half and  also extends to
the right  and we have cancelation  of the overlapping  segments for a
final cut being $[1/2, 1].$

Hence we certainly have  analyticity of the exponential term in
a disk  of radius one half round the  origin.

We now evaluate the residue for this branch.
Here we have first that
$$\mathrm{LogA}(1-z) = - \mathrm{LogA}\frac{1}{1-z}$$
and second
$$\mathrm{LogB}(2z-1) = \pi i - \mathrm{LogB}\frac{1}{1-2z}.$$
where we choose  $\pi i$ from the upper half  plane to get analyticity
at the origin.

This finally yields
$$-\frac{1}{z^2} \exp(\pi i/2)
\exp\left(-\frac{1}{2}\mathrm{LogA}\frac{1}{1-z}\right)
\exp\left(-\frac{1}{2}\mathrm{LogB}\frac{1}{1-2z}\right).$$

We can extract coefficients from  this either with the Newton binomial
or recognizing the mixed  generating function of the unsigned Stirling
numbers of the first kind. Using the latter we find that
$$[z^n] \exp\left(u\log\frac{1}{1-z}\right)
= \frac{1}{n!} \times u(u+1)\cdots(u+n-1).$$

We need the first two terms of each. These are for the term
in $\mathrm{LogA}$
$$1 - \frac{1}{2} z +\cdots$$
and for the term in $\mathrm{LogB}$
$$1 - \frac{1}{2} 2z +\cdots$$

We have determined the residue at infinity of the original function to
be $$- \exp(\pi i/2) \times -\frac{3}{2} =
\frac{3}{2} i.$$

Recall that we had
$$Q = - \pi i \times
(\mathrm{Res}_{z=0} f(z)
+ \mathrm{Res}_{z=\infty} f(z))$$
so substituting the computed values for the residues gives
$$-\pi i \left(-\sqrt{2}i + \frac{3}{2}i\right)
= \pi i \left(\sqrt{2}i - \frac{3}{2}i\right)
= \pi \left(-\sqrt{2} + \frac{3}{2}\right).$$

In order to be rigorous we also need to show continuity across the two
overlapping   cuts  on   $(-\infty,  1)$   as  shown   in   this  MSE
link.

Remark. It really helps to think of  the map from $z$ to $-z$ as a
$180$ degree  rotation when one  tries to visualize what  is happening
here.

the function is analytic in the given domain so integral is zero


Evaluate $\lim _{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$ I think that $\lim\limits_{x\to \infty }\frac{1}{x}\int _0^x\cos\left(t\right)dt\:$ is divergent, I can prove with taylor series?

It can't be, since $\lvert \cos{t} \rvert<1$, so the integral is bounded between $x$ and $-x$. It could oscillate, but not diverge. However,
$$ \frac{1}{x}\int_{0}^{x} \cos{x} \, dx = \frac{1}{x} (\sin{x}-\sin{0}) = \frac{\sin{x}}{x}, $$
which tends to $0$ since it is absolutely bounded by $1/x$, which tends to $0$.

Note that, $\int_0^x \cot (t)dt = \sin(x)$. Now, $\lim_{x \to \infty} \sin(x)/x = 0$. Therefore,   $\frac1x \int_0^x \cot (t)dt = 0.$

Note- Sorry for typing it in a hurry, I was a bit busy while I typed the solution so thats why I couldn't type in latex. I'll update it as soon as possible.


How many faces does the resulting polyhedron have? Take a regular tetrahedron of edge one.

Also take a square-based pyramid, whose edges are all one (therefore the side faces are equilateral triangles of same size as the faces of the tetrahedron).

Glue a face of the tetrahedron to a triangular face of the pyramid so that their edges match up.

Considering the volume taken up by the two pieces as a single polyhedron, how many faces does it have?

I visualized the solution Nick provided:

(I found it because it had used this arrangement to build boats models with a magnet game)

Here's a variation on your solution to the problem.
If you truncate the regular tetrahedron at the midpoints (instead of the 1/3 points) this gives a regular octahedron. The above problem looks at 1/2 of the octahedron (a square pyramid) and puts back one of the tetrahedra that were truncated and removed in the first step. The two faces were coplanar when they were together originally.
To see that the above truncation gives an octahedron notice that each face on the original tetrahedron loses corner triangles but keeps an "old" equilateral triangle. The removal of the corner tetrahedron gives a new equilateral triangular face. 4 old equilateral faces + 4 new equilateral faces yield an 8 faced solid which has 4 faces at each vertex--a regular octahedron.


Cauchy's Integral Formula implies Holomorphicity? Is the converse direction of the Cauchy Integral Formula true? Meaning,

if $f:\mathbb{C}\supseteq U\rightarrow\mathbb{C}$, and

$$\forall a \in U \space \space f(a) = \frac{1}{2\pi i}\int_{|z-a|=R}\frac{f(z)}{z-a}dz$$

does it mean that f is complex-differenatiable?

If you parametrize your integral, you see that your equation is equivalent to

$$ f(a) = \dfrac{1}{2\pi} \int_0^{2\pi} f(a + r e^{i\theta})\; d\theta$$

and this is true for harmonic functions, not just holomorphic ones.

Well...yes, because then you would have an expression for the nth derivative of f, making f infinitely differentiable/analytic in U and thus holomorphic.

https://en.wikipedia.org/wiki/Analytic_function


Question which has no answers Sorry, I don't really know how to improve this question, this quiz has been given me today and I think there is no a true answer, because in the different cases of $a > 0, a < 0, b > 0, b < 0$(and nested cases) you have that every answer can be false.

Taken $a, b \in \mathbb{R}$, given $a^2 - b^2 = 0$, which one is true?

$ab > 0$
$a + b = 1$
$ab < -1$
$a > b$
$a + b = 0$

These are answers I gave myself(all the examples respect the given condition of $a^2 - b^2 = 0$:

If $a = 1 \land b = -1$ then $ab = -1$
If $a = 1 \land b = -1$ then $a + b = 0$
If $a = 0 \land b = 0$ then $ab = 0$
If $a = -1 \land b = 1$ then $a < b$
If $a = 1 \land b = 1$ then $a + b = 2$

Any suggestions? Thanks!

All of the options are false, your couter-examples are okay. Some statements you can get from $a^2 - b^2 = 0, a,b \in \mathbb R$ are:
$$\begin{align*}a^2 & = b^2\\|a|^2 & = |b|^2\\|a| &= |b|\\a&=\pm b\\b&=\pm a\end{align*}$$

since a2 - b2 = 0
by factorization
(a+b)(a-b) = 0

a + b = 0 and a - b = 0

Thus , among all options last one a + b = 0 is correct


Finding the inverse of $f(x)=|x|-2$ How would I find the inverse of the function $f(x)=|x|-2$? I have swapped $x$ and $y$, and tried to isolate $y$, reaching up to $x+2=|y|$

Whenever I see absolute values, I always break the problem up into cases.

Case $1$: $y\ge 0$, $|y|=y$

$y=x+2$

Case $2$: $y<0$, $|y|=-y$

$-y=x+2$
$\implies y=-x-2$

It doesn't seem right that I get two answers... Am I supposed to combine the two into one piecewise function like:

$$f^{-1}(x)=\begin{cases}x+2, \ x\ge 0 \\ -x-2, \ x<0 \\ \end{cases}$$

Any hints on how to do this?

Also, just to clarify, my homework assignment asks me to "Graph the inverse of $f(x)=|x|-2$"

$y=|x|-2$ is not one-to-one and hence not invertible. You can easily see this by comparing the values of the function at $-1$ and $1$ or simply drawing its graph: it doesn't pass the horizontal line test.

Maybe your homework is asking to reflect the graph of $|x|-2$ in $y=x$. This won't be a function but it will do the job.

Can you guess which graph(s) you want?


Natural maps between sets of linear maps Let $\mathcal{L}(V,W)$ denote the set of linear maps $T: V \to W$. For finite-dimensional vectors spaces $U,V,W$, I am trying to find a natural map $\mathcal{L}(V,W) \to \mathcal{L}(V,U)$ and one from $\mathcal{L}(U,V) \to \mathcal{L}(W,V)$. I'm struggling to get started due to how abstract the notion of a 'map of maps' is. In particular, I can't do anything of the form $Tv \mapsto Tw$ or even any kind of inverse or composition because the dimensions don't line up.
Any guidance would be appreciated.

Given a fixed linear transformation $\varphi : W \rightarrow U$, you can define the transformation $T : \mathcal{L}(V,W) \rightarrow \mathcal{L}(V,U) $ defined for all $u \in \mathcal{L}(V,W)$ by
$$T(u) =\varphi \circ u$$
I guess this is the most natural thing you can do !

Because your vector spaces are all finite-dimensional, the linear maps can be realized as matrices over the underlying field of the vector spaces (which are assumed fixed over all $U,V,W$. Now note that we have the following canonical (respectively, left- and right-) group actions via multiplication by of $\mathcal{L}(W,U)$,
$$\alpha\colon \mathcal{L}(W,U) \times \mathcal{L}(U,V)\to \mathcal{L}(W,V)\,,~\,~\,~(M,N)\mapsto MN$$
$$\beta\colon \mathcal{L}(V,W)\times \mathcal{L}(W,U)\to \mathcal{L}(V,U)\,,~\,~\,~(N,M)\mapsto NM\,.$$
You can then restrict to a fixed element of $\mathcal{L} (W,U)$ to obtain the desired natural maps.


If two random variables have CDFs that have the same value for all x, can we assume the random variables are equal? My text has the following theorem:

Let $X$ have a CDF $F$ and let $Y$ have CDF $G$. If $F(x) = G(x)$ for all $x$, then $\mathbb{P}(X \in A) = \mathbb{P}(Y \in A)$ for all $A$.

I don't see a way that X and Y could assign a different probability to the same event but still have their CDFs be equal at every point. If they disagreed at point j, then F(j) will not equal G(j). Therefore they must be the same?

Elementary probability:

They don't teach this is in elementary probability, but random variables have an explicit representation known as the Skorokhod representation.

Basically, we never really know the formulas for a lot of the $X$'s. We know the $X$'s mainly from the $F_X(x)$'s. It's kinda like talking about $f(x)=x^2+c$'s through their common derivative $f'(x)=2x$: When is $f$ increasing? When $f' > 0$. We know that if $f$ is not unique given an $f'(x)$. We can do that through integration, or just construct an explicit example $f(x)=x^2+5$ and $f(x)=x^2+4$.

How we do similarly here in probability?

For example, consider $X \sim Be(p)$ where $P(X=0):=p$ and $P(X=1):=1-p$ (Usually, textbooks use $p$ for the $P(X=1)$).

If both of the following $X_i$'s satisfy $X \sim Be(p)$, then we've given explicit Bernoulli random variables that can never be the same, i.e. $X \sim Be(p)$ doesn't have a unique Skorokhod representation.

$$X_1(\omega) := 1_{(0,1-p)}(\omega) := 1_{A_1}(\omega)$$

$$X_2(\omega) := 1_{(p,1)}(\omega) := 1_{A_2}(\omega)$$

If $\omega=\frac{1-p}{2}$, then $X_1(\omega)=1$ while $X_2(\omega)=0$.

Let us try to compute the CDF of $X_i$:

$P(X_i(\omega) \le x)$ is 0 for $x<0$ and 1 for $x \ge 1$.

As for $0 \le x < 1$, define

$$P(X_i(\omega) \le x) = P(X_i(\omega) = 0) = P(1_{A_i}(\omega) = 0) = P(\omega \notin A_i) = 1 - P(\omega \in A_i)$$

We have our result if $P(\omega \in A_1) = P(\omega \in A_2) = 1-p$. Is it?

Okay so here, we need to need to make some kind of assumption to say that the interval $(p,1)$ is not only as probable as $(0,1-p)$ but also that probability of each interval is $1-p$. Clearly, the intervals have the same length, but does that mean they have the same probability? Furthermore, if they do, is it equal to  That depends on how we define probabilities here. One such assumption is:

A uniformly distributed random variable $U$ on $(0,1)$ has Skorokhod representation $U(\omega) = \omega \sim Unif(0,1)$.

Hopefully this isn't circular, otherwise this half of the answer is nonsense.

Then $P(\omega \in A_i) = \frac{(1-p)-(0)}{1-0}$ or $= \frac{(1)-(p)}{1-0}$

$$P(\omega \in A_i) = \frac{1-p}{1-0} = 1-p$$

Advanced probability:

It can be shown that $$Y(\omega) = \omega \sim Unif(0,1)$$ for $\omega$ in $((0,1),\mathscr B(0,1),\mu)$ where $\mu$ is Lebesgue measure.

Hence,

$$P(\omega \in A_i) = \mu(A_i) = l(A_i) = 1-p$$

where $l$ is length.

For elementary probability:

I think it's just saying that if the cdf's are the same then the pdf's are the same. PDF's don't always exist in advanced probability.

Do you seriously prove that? Anyway, it likely comes down to how 'all' $A$ (subsets of $\mathbb R$?) that you can come up with is somehow 'related' to $(-\infty, x]$. For example,

$$A = (a,x] = (-\infty,x] \cap (-\infty,a]^c$$
$$A = (-\infty, x) = (-\infty, x] \cap \{x\}^c$$
$$A = \{x\} = (-\infty, x] \cap (-\infty, x)^c$$
$$A = \{x\} \cup [a,b] = [(-\infty, x] \cap (-\infty, x)^c] \cup [(-\infty,b] \cap (-\infty,a)^c]$$

Thus, if you can compute $P((-\infty, x])$ for all $x$, then you can compute $P(X \in A)$ for 'all' A. The thing is not all subsets of $\mathbb R$ are 'related' to $(-\infty, x]$. But that's okay because even in advanced probability, we don't mind those  subsets. In advanced probability, there may be random variables where that doesn't hold if you try those subsets.

From comments

  They just state the theorem, they don't ask you to prove it. I understand strictly speaking they have only claimed what you said: equal CDFs imply equal PDFs, but does my reasoning additionally prove equal CDFs imply the two random variables are equal (they must agree on the output for every input)? – Joseph Garvin Apr 20 at 19:22

  @JosephGarvin oh right that. Um, no. Just imagine them as two excel cells where you have =RAND() --> the formula is the same but the number is different. what you're thinking of is A1=RAND() and then B1=A1. All that P(X in A) = P(Y in A) is saying is that X and Y have an equal chance of being in A. Example: Jack and Jill have the same chance of falling down, but it doesn't mean that Jack falls if and only if Jill falls. Another example: X is 1 if a fair coin flip is heads and 0 for tails. Y is 1 for tails instead. X and Y both have an equal chance of being equal to 1 but they're never equal. – BCLC Apr 20 at 20:46

For advanced probability:

Use the uniqueness lemma:

If $\mu_1(A) = \mu_2(A)$ for $A \in \mathscr I$ where $\mathscr I$ is a $\pi$-system, then $\mu_1(A) = \mu_2(A)$ for $A \in \sigma(\mathscr I)$ where $\mu_1$ and $\mu_2$ are measures on $(S,\Sigma)$ with $\mu_1(S) < \infty$. In your case

$$\mathscr I = \{(-\infty,x)\}_{\{\text{'for all x'}\}} \to \sigma(\mathscr I) = \mathscr B(\mathbb R)$$

Note: The '$\sigma(\mathscr I) = \mathscr B(\mathbb R)$' is the non-baby version of the elementary probability stuff above. Technically, not all subsets of $\mathbb R$ work: See non-Borel sets.