For how many $\lambda\in\mathbb{C}, \det(\lambda E-A)=0$ Suppose the polynomial $p(\lambda)=\det(\lambda E-A)$ not identially $=0$,
 i.e $p(\lambda)\neq 0$, for fixed $E,A\in M_n(\mathbb{R})$ and $\lambda$ is a complex variable.

Here are my questions

$1.$ Is it always true that degree of $p$ is always $n$? If $E=I_n$, then we know that it is a characteristic polynomial of $A$ and since $A$ is of $n\times n$ so $p$ is of $n$ degree polynomial, right? is it true in general for any invertible $E$? what is $E$ is singular?

$2.$ If degree of $p$ is $n$ then there are $\lambda_1,\dots,\lambda_n$ such that $p(\lambda_i)=0$ for $i=1,2,\dots ,n$

does this depends on rank of $E$? I mean if $E$ is singular and $rank(E)=n_1<n$, then what can be said about the degree of $p$ and number of zeros of $p$

Thanks

It is a bit more complicated than that.  You may rewrite it in a form like:
$$ {\rm const} \times \det  \left( \begin{matrix} \lambda 1_{n_1} - A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix} \right) $$
where the upper left matrix has size $n_1={\rm rank}\ E$. Now the coefficient to $\lambda^{n_1}$ is $\det A_{22}$ so if this is zero there is no term with $\lambda^{n_1}$ and you will have to look at the next coefficient which also involves $A_{11}$, $A_{12}$ and $A_{21}$. I don't know if there is an easy general description of the degree.

The degree of $p(\lambda)$ is at most the rank of $E$.

To see this, note that there exist invertible matrices $U$ and $V$ such that $UEV$ is diagonal with $1$s and $0$s, and that
$$
\det(\lambda E - A)= \det(U^{-1})\det(\lambda UEV - UAV)\det(V^{-1})
$$
By the fundamental theorem of algebra, a degree $n$ polynomial has $n$ roots.


Convergence radius of a power series I came across the following question :

Find the convergence radius for the series $\sum_{n=1}^\infty n!x^{n!}$.

My initial intuition led me to believe it should converge for $x\in(-1,1)$.

I would really appreciate if anyone could review my proof and point out if I`m missing anything.

The proof goes as following :

Let $a_n$ be the sequence of coefficients of $x^n$. For every $n_m=k!$ for some $k\in \mathbb{N}$, $a_{n_m}=n_m$.

By Cauchy-Hadamard theorem, $\limsup (|a_n|^\frac 1 n)=\frac 1R$, and by D'Alembert criterion, if the limit $\frac {a_{n_m+1}} {a_{n_m}}$ exists , it equals to  $\lim (|a_{n_m}|^\frac 1 {n_m})$ , and indeed $\lim \frac {a_{n_m+1}} {a_{n_m}}=\frac {{n_m+1}} {{n_m}}=1$. Therefore we have $1$ as a partial limit of $|a_n|^{\frac 1 n}$. In addition, for every $n\neq n_m$, $a_n = 0$, meaning $1$ is the limit supremum. (The partial limits are either 1 or 0)

From here we can deduce R=1.

Divergence for $x=1$ is obvious, since $n!$ doesn't converge to $0$, and divergence for $x=-1$ for the same reason (since $n!$ are all even after $n=2$, we get the same series as $x=1$)

Is my proof correct?

Note that $$\sum_{n=1}^M n!x^{n!}\le \sum_{n=1}^{M!} n|x|^{n}$$

So what can we say about the limit case?

It certainly does NOT converge for $|x| \geq 1.$ The interval of convergence is some $(-R,R).$ It is bounded above (for $x \gt 0$) by $\sum mx^m.$


Averaging method (parameter variation) Use the averaging method (parameter variation) to determine a uniform first-order approximation for:
$$ y''+y+ \epsilon y'^5=0$$
for $\epsilon\ll1$.

First transform into polar coordinates, $y=r\sin(\phi)$, $y'=r\cos\phi$. Then compute the derivatives in polar coordinates
\begin{align}
rr'&=y'(y''+y)=-ϵy'^6=-ϵr^6\cos^6\phi
\\
r^2\phi'&=y'^2-yy''=r^2-y(y+y'')=r^2-ϵr^6\cos^5\phi\sin\phi
\end{align}
Now take the average as $\phi$ varies over $[0,2\pi]$
\begin{align}
\bar r'&=-ϵ\bar r^5·\frac{20}{64}
\\
\bar \phi'&=1
\end{align}
so that $\bar \phi(t)=t+\phi_0$ and
$$
\bar r(t)=\frac{r(0)}{\sqrt[4]{1+\frac54ϵr(0)^4t}}.
$$


Cauchy sequence convergence Suppose that $a_n$ and $b_n$ are Cauchy sequences, and that $a_n < b_n$ for all n. Prove that $\lim_{x \to \infty}a_n \le \lim_{x \to \infty}b_n$ for all n.

Is it sufficient to say that we know both Cauchy sequences must converge to the limit, and since $a_n$ is always less than $b_n$, the limits will follow the desired inequality?

Edit: since this is not true, what would be the appropriate strategy to prove?

Let $a_n$ and $b_n$ be your two sequences such that $a_n<b_n$ for all $n$. Consider the
sequence $\{b_n-a_n\}$ which is greater than $0$ for all $n$. This sequence can also be seen to be Cauchy. Of course, all Cauchy sequences in $\mathbb{R}$ converge. Call its limit $x$. Suppose its limit was strictly
less than $0$. Since $x$ is less than $0$, there is a $\varepsilon>0$ such that
$(x-\varepsilon,x+\varepsilon)\subseteq (-\infty,0)$. Now since $\lim_{n\rightarrow\infty} \{b_n-a_n\}=x$, there exists an $N$ such that $b_N-a_N\in(x-\varepsilon,x+\varepsilon)$; that is, $b_N-a_N<0$ which is a contradiction. We must have the limit is at least $0$. In the case
that both limits are finite, it's an easy lemma that $\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ (which is our case as both $a_n$ and $b_n$ are Cauchy sequences). Thus $0\leq\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ which implies $\lim_{n\rightarrow\infty} a_n \leq \lim_{n\rightarrow\infty} b_n$.

It is not true, take $b_n=1/n=-a_n$, the strict inequality is not respected at the limit.


Show that $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x>-1$ is pointwise convergent. Title says it all, really. I have to show, that:
 $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x>-1$
is pointwise convergent, but I have no idea where to start. I am thinking, that I have to show it by comparison, but I am lost on what to compare it with.

Anyone got any ideas?

Much appreciated.

Hint: When $x>0$ $$0<\ln\left(1 + \frac{x}{n}\right) < \frac{x}{n}$$ for all positive $n$, and when $-1\le x<0$ $$0>\ln\left(1 + \frac{x}{n}\right) > \frac{x}{n-1} $$ for $n>1$.


The sum of 2 numbers is 9 and the product of these same 2 numbers is 20. What is the sum of the squares of these 2 numbers? The sum of 2 numbers is 9 and the product of these same 2 numbers is 20. What is the sum of the squares of these 2 numbers?

We have $a+b=9,ab=20$

We need $a^2+b^2$ which equals to $(a+b)^2-2\cdot ab$

if you want to find the values of $a$ and$b$ and further solve.$a$ and $b$ are the roots of the quadratic equation $$x^2-(a+b)x+ab=0$$ie$$x^2-9x+20=0$$which are$5$ and$4$.   further $4^2+5^2=41$


(A iff B), then (not A iff not B)? Suppose $A\Leftrightarrow B$, i.e. statement A holds if and only if statement B holds.

Does then $\neg A\Leftrightarrow\neg B$?

I think this is true since
$A\implies B$ implies $\neg B\implies \neg A$

and

$B\implies A$ implies $\neg A\implies\neg B$.

Hence, if $A\Leftrightarrow B$, then
$$
\neg A\implies\neg B\implies\neg A
$$
which means that
$$
\neg A\Leftrightarrow\neg B.
$$

I think, we even have
$$
(A\Leftrightarrow B)~\Leftrightarrow~(\neg A\Leftrightarrow\neg B)
$$

The symbol $\iff$ means if inputs are both true or false. So if A and b = 0 or 1, then A $\iff$ B is true otherwise false.

The $\lnot$ means "Not", so if A = 0, $\lnot$A = 1, same thing as B and $\lnot$B. Which brings me to that A and B have to be 0 or 1, otherwise $\lnot A \iff \lnot B$ with be false.


Continuity of a function in two variables Function $f(x,y)$ is continuous in each variable separately. Prove that there exists a point where it is continuous in two variables.

I do not quite understand how to act here. I know the definition of continuity. But how to use them, I do not know. This is my homework on mathematical analysis.

For $\varepsilon>0$ denote
$$X_{\varepsilon} = \Big \{ (x,y) ~|~ \exists \eta >0, \forall (h,k) \in [-\eta,\eta]^2,~ |f(x+h,y+k)-f(x,y)|< \varepsilon  \Big\}.$$
Then $f$ is continuous on $\bigcap_{\varepsilon \in \mathbb{Q}_{>0}} X_{\varepsilon}$.

Lemma: For any $\varepsilon>0$, and open $O$ : $X_{\varepsilon} \cap O$ has non empty interior.

Proof: WLOG we can assume that $O=(-1,1) \times (1,1)$.
Since $x \mapsto f(x,0)$ is continuous at $0$, there exists $\alpha \in (0,1)$ such that $|f(x,0)-f(0,0)| \leq \frac{\varepsilon}{4}$ for all $|x| \leq \alpha$.
Denote
$$E_{\beta} = \Big \{ x \in (-\alpha,\alpha) ~|~ \forall |y| < \beta, ~ |f(x,y) - f(x,0)| \leq \frac{\varepsilon}{4} \Big\}.$$
Then : (1) $E_\beta$ is closed because it is an intersection of closed sets, (2) $\cup_\beta E_\beta = (-\alpha,\alpha)$ because $y \mapsto f(x,y)$ is continuous for any $x$.
Baire theorem implies that $E_{\beta_0}$ contains an open $I$ for some $\beta_0$.
Then for any $(x,y)$ and $(x',y') \in I \times (-\beta_0,\beta)$, we have
$$\begin{array}{rcl}
|f(x,y)-f(x',y')| &\leq& |f(x,y)-f(x,0)| + |f(x,0)-f(x',0)| + |f(x',y')-f(x',0)| \\
&\leq& \varepsilon \\
\end{array}$$
Hence $X_\varepsilon \cap O$ has non-empty interior. QED

Baire theorem implies that $\bigcap_{\varepsilon \in \mathbb{Q}_{>0}} X_{\varepsilon}$ is dense.

Let $(x_p, y_p)$ be a point and let ${(x_n , y_n)}_{n\in \mathbb{N}}$ be any sequence such that $(x_n , y_n) \rightarrow (x_p, y_p)$.

Then
$$
\lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_p , y_p)| = \lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_n , y_p) + f(x_n , y_p) - f(x_p , y_p) + f(x_p , y_p) - f(x_p , y_n) + f(x_p , y_n) - f(x_p , y_p)| \leq \lim_{n\rightarrow \infty} |f(x_n , y_n) - f(x_n , y_p)| + |f(x_n , y_p) - f(x_p , y_p)| + |f(x_p , y_p) - f(x_p , y_n)| + |f(x_p , y_n) - f(x_p , y_p)|
$$

Each of these terms goes to zero since $f$ is continuous in each variable.

Therefore $$\lim_{n\rightarrow \infty} f(x_n,y_n) = f(x_p,y_p)$$ By definition this means f is continuous at $(x_p, y_p)$.

(This proof is straight forward to generalize in arbitrary metric spaces, though I figured that's not what you were looking for.)


complex analysis: If $f$ is analytic and $\operatorname{Re}f(z) = \operatorname{Re}f(z+1)$ then $Im\;f(z) - Im\;f(z+1)$ is a constant I am having trouble deciphering the reason behind a line in a complex analysis textbook (Complex made Simple by Ullrich, page 360 5 lines down in Proof of Theorem B, for those who are interested).

Basically it says that since $f(z)$ is a holomorphic function with  $\operatorname{Re} f(z+1) = \operatorname{Re}f(z)$, that for all $z$:

$$\begin{align}
f(z+1) - f(z) &= \operatorname{Re}f(z+1) + i\cdot\operatorname{Im}f(z+1) - \operatorname{Re}f(z) - i\cdot\operatorname{Im}f(z)\\
&= i\cdot\operatorname{Im}f(z+1) - i\cdot\operatorname{Im}f(z)\\
&= i\cdot \text{constant}.
\end{align}$$

I do not understand why we must have that for all $z$ $\quad\operatorname{Im}f(z+1) - \operatorname{Im}f(z) = $ a constant.

Can anyone help?

Thanks

I will give it a try, feel free to correct me. Since $f(z)$ is analytic then $f(z+1)$ is analytic aswell. write $f(x)= u + iv$ and $f(z+1)=u^{\ast} + iv^{\ast}$ so for $f(z)$ we have by the Cauchy-riemann equations: $$\frac {du}{dx}=\frac {dv}{dy}$$ and for $f(z+1)$ we have: $$\frac {du^{\ast}}{dx} = \frac {dv^{\ast}}{dy}$$ Now since we are given the information $\Re(f(z)) = \Re(f(z+1))$ it follows that: $$\frac {du}{dx} = \frac {du^{\ast}}{dx} \implies \frac {dv}{dy} = \frac {dv^{\ast}}{dy}$$ Integrate the last expression above and you will obtain $v-v^{\ast} = C$ for some constant $C \in \mathbb{C}$, in other words, $\Im(f(z))-\Im(f(z+1))$ = C$

That is wrong. Consider $f(z) = \cos 2 \pi z$, there $f(z + 1) = f(z)$, but $\Im(f(z))$ isn't constant.


Can a quadratic equation not equal to zero Well I wanted to know whether or not $y = x^2 + x + 7$ is a quadratic equation since the general form is $ax^2 + bx + c = 0$ here the equation
$y=x^2+x+7$. Isn't equal to zero so I'm a bit confused

When you write $y=x^2 + x+7$, that is not generally considered a "quadratic equation" in the commonly used sense. Most of the time, that is taken to mean a functional relationship between two variables, namely $y$ and $x$. Because the right hand side takes the form of a quadratic polynomial, you are justified in calling it a "quadratic function" of $x$. When writing the relationship between two variables in this form, you're looking to answer questions like: what is the value of $y$ for a given value of $x$? What does a plot of $y$ against $x$ look like? And so forth.
A quadratic equation is (already, or can easily be rearranged into) something of the canonical form $ax^2 + bx +c =0$. The last term on the left hand side and is a constant term while the right hand side is zero.
So these are quadratic equations:

$x^2 + x +7 =0$ (already in the canonical form)

$x^2 + x + 7 = 2$ (can be immediately rearranged into the canonical form)

$x^2 + x + 7 = k$ (where $k$ is specified as a constant, even if it's not a known constant, allowing rearrangement into the proper form)

Note that the quadratic functional relationship $y = x^2 + x +7$ can be made into a quadratic equation if we ask and try to answer questions like:

What value(s) of $x$ makes $y = 10$? In this case $x^2 + x +7 =10$, which is a quadratic equation with two real roots, so you have your two possible $x$ values.

Does the curve $y = x^2 + x +7$ intersect the $x$ axis? The answer is 'no' because the quadratic equation $x^2 + x + 7 =0$ has no real roots, only complex ones.

I hope I've understood the essence of your question and answered clearly enough.

The general form is not $ax^2+bx+c=0$, but $y = ax^2+bx+c$. To have a function, you must be able to put one number in ($x$ in this case), and output one number out ($y$).
$ax^2+bx+c =0$ when $y = 0$, so when you write this, you are trying to find which $x$ values the function outputs $0$ (or the roots). This doesn't help you figure out if $y=x^2+x+7$ is a quadratic equation.


Domain of $\operatorname{arccos}(x-1)$ All we know that domain of $\operatorname{arccos}(a)$ is $[-1; 1]$. So, if $a = x - 1$, then $-1 \leq x - 1 \leq 1$. I substracted one from all parts and got $-2\leq x\leq0$. But the answer is $[0;2]$. Where is mistake?

You have two inequalities:
$$
-1\le x-1 \quad \land \quad x-1\le1
$$
solve separately and you find:
$$
-1+1 \le x \quad \land \quad x\le 1+1
$$

The range of $arccos(x-1)$ is the same as the range of $arccosx$ which is $[0,\pi]$ It seems like you are confused with the domain of $arccos(x-1)$ which can be found by solving the inequality stated by others...


what is the limit as $x$ approaches infinity of $x-\sqrt{x^2-x+2}$ $$
\lim_{x \to \infty} \left( x - \sqrt{x^2 - x +2 } \right)
$$

I've tried rationalizing the expression but after repeated applications of L'Hospital's rule, it doesn't feel like I'm getting anywhere.


Given a joint density function, what is the conditional expectation $E[Y|x]$? The random variables $X$ and $Y$ have the joint density $$f_{X,Y}(x,y)=
\left\{\begin{matrix}e^{-y}, \mbox{ } 0\leq x \leq y \le \infty
\\ 0,  \mbox{   otherwise}\end{matrix}\right.$$
Evaluate the conditional expectation $E[Y|x]$.

I'm not sure if I'm doing this correctly. Please let me know if something is off in my solution. Thanks for reading.

My solution:
(My expected value is negative, which I'm pretty sure can never be the case. Also in line 4 and 8, are the domains correct?)

$$E[Y|x] = \int_{-\infty}^{\infty}yf_{Y|X}(y|x)dy.$$

$$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_{X}(x)}.$$

$$f_{X}(x) = \int_{-\infty}^{\infty}f_{X,Y}(x,y)dy$$

$$= \int_{0}^{\infty}e^{-y}dy $$

$$=\left [ -e^{-y} \right]\Big|_0^\infty $$

$$ = -e^{-\infty}-(-e^{0})=1. $$
$$f_{X}(x) = 1, x \geq 0.$$
Thus,

$$E[Y|x] = \int_{0}^{\infty}-ye^{-y}dy$$

$$= -\int_{0}^{\infty}ye^{-y}dy$$
After integration by parts,
$$E[Y|x] = \left [ ye^{-y}+e^{-y} \right]\Big|_0^\infty$$
$$=0-[e^{0} + 0]=-1$$
Therefore, $E[Y|x] = -1.$

Thanks in advance for reading this long problem and responding!

The limits in your fourth line are wrong since you know the joint density is $0$ unless $0\leq x \leq y \le \infty$ so it should be  $$f_{X}(x) = \int_{x}^{\infty}e^{-y}dy , \mbox{ } 0\leq x  \le \infty $$ leading to $$f_{X}(x)=
\left\{\begin{matrix}e^{-x}, \mbox{ } 0\leq x  \le \infty
\\ 0,  \mbox{   otherwise}\end{matrix}\right.$$

That will then give you

$$f_{Y \mid X=x}(y)=
\left\{\begin{matrix}\dfrac{e^{-y}}{e^{-x}}, \mbox{ } 0\leq x  \leq y \le \infty
\\ 0,  \mbox{   otherwise}\end{matrix}\right.$$ and $$E[Y \mid X=x]=
\int_{x}^{\infty}y\frac{e^{-y}}{e^{-x}}dy = x+1$$

which is an example of the memorylessness property of the exponential distribution

Yes, that negative expected value is absurd and a sign you did something wrong.

  $$f_X(x) = \int_0^{\infty} e^{-y}\,dy$$

This line is the first mistake. The density function is zero unless $y\ge x$, so that should be $\int_x^{\infty}$ instead, and the probability you're dividing by will depend on $x$ - in a way that makes it a single-variable density. A constant density of $1$ on $[0,\infty)$ for $f_X(x)$ is also an absurdity you could have caught.

  $$E(Y|x) = \int_0^\infty -ye^{-y}\,dy$$

All right, this should be divided by what we just got for $f_X$ - but that's not the only problem. How, exactly, did a minus sign appear here out of nowhere? The integrand should be a probability density function times $y$, and densities are nonnegative.

Also, again, this will be an integral from $x$ to $\infty$, as the density is zero unless $y \ge x$.

Oh, and here's a neat interpretation of this distribution: Choose $x$ and $z$ independently from exponential distributions of parameter $1$, and then let $y=x+z$.


Proving determinant with variables I have a problem that asks:

Prove that det$\begin{pmatrix}
1 && 1 && 1 \\
a && b && c \\
a^2 && b^2 && c^2 \end{pmatrix} = (c-a)(c-b)(b-a)$

I was thinking of solving it like normal and see if I can end up with $(c-a)(c-b)(b-a)$

What I did:

det$\begin{pmatrix}
b && c \\
b^2 && c^2 \end{pmatrix}$ -
det$\begin{pmatrix}
a && c \\
a^2 && c^2 \end{pmatrix}$ +
det$\begin{pmatrix}
a && b \\
a^2 && b^2 \end{pmatrix}$

which gives

$(bc^2 - cb^2)-(ac^2-ca^2)+(ab^2-ba^2)$

which simplifies to

$(b-a)c^2 + (a-c)b^2 + (c-b)a^2$

and this is where I got stuck. Am I even approaching this problem correctly? Any help would be appreciated.

If a=b, b=c, c=a then determinant value is 0 so
  (a-b)(b-c)(c-a) are factor of the determinant now comparing the leading coefficient so the Value of the determinant is (a-b)(b-c)(c-a)


Is it inuitively correct to say that $\lim_{n \to \infty} \frac{x}{n} = \mathrm{d}x$? We know that from the Riemann sum that $\Delta x = \mathrm{d}x$ where $n \to \infty$ and $\Delta x = \dfrac{b - a}{n}$. If, however, $x$ represents some length of interval, can we also say that $\displaystyle\lim_{n \to \infty}\dfrac{x}{n} = \mathrm{d}x$?

In short: No. The limit of that sort of expression is, generally speaking, going to be a real expression (provided the limit exists and is finite). A "real expression" in this context means a mathematical expression in terms of $x$, where you can plug in real numbers as possible values of $x$ and get out real numbers as a result (like a function). On the other hand, $\mathrm{d}x$ is not a real expression; it is a different kind of object called a differential form, which is (to greatly oversimplify) "a thing that you can put under an integral sign."
The defining property of the real numbers is that you can take limits on them, so much so that a limit is usually assumed to be taken over the reals unless context implies some other set of numbers (e.g. the complex numbers, if an $i$ is present). But another important property is the archimedean property, which states that there are no infinities or infinitesimals. In other words, no real number is infinitely larger or smaller than any other (except for zero, which is infinitely smaller than all non-zero values). More formally, if you have two real numbers $x$ and $y$, you can add $x$ (or $-x$) to itself repeatedly, and the sum will eventually be greater than $y$ (or $-y$).
If you want to think of $\mathrm{d}x$ as an infinitesimal, then it immediately follows that $\mathrm{d}x$ cannot be a real number. However, in standard analysis, differentials are actually a much stranger sort of object, which arises out of abstract algebra (see link above). You don't need to know this algebra in order to do simple manipulations like the chain rule (cancelling differentials when multiplying two derivatives), but it is important to understand if you want to start talking about things like $\mathrm{d}x$ by itself (and not under an integral or part of a derivative).
Nonstandard analysis, which actually does use infinitesimals, most commonly uses the hyperreal numbers rather than the reals, and (usually) doesn't bother much with limits as such at all. In that context, it might make sense to write something like "Let $\mathrm{d}x = \frac{x}{\infty}$" (although we must take care not to confuse this $\mathrm{d}x$ with the previous paragraph's definition!). But this is a somewhat misleading thing to write, because the hyperreals do not just have one infinity - they have infinitely many. So you need to specify which infinity you want, or at least clarify that you're always using the same infinity. For simple cases, we might write $\omega$ to mean an arbitrary infinity, and thus have $\mathrm{d}x = \frac{x}{\omega}$. Then, if we write $\omega$ in another place, it will be clear that we mean the same infinity, and not $2\omega$ or something else.

I think this is quite a tricky topic to talk about because of how people interpret things vs how they are actually defined. I find it best to think of $dx$ as a separate value, rather than a function or scaling of the variable $x$. whilst we are talking about a small change in $x$ the key word here is change, so rather than $x/n\to0$ really we want the change in $x$, or the interval that we are looking at to be very small rather than the initial or end values.
One nice way to think of this is by looking at your average integral:
$$I_{a,b}[f]=\int_a^bf(x)dx$$
now think of the integral symbol as a stretched out $S$, which is the first letter of sum. Now inside the integral (the part we are summing) is $f(x)\times dx$. In other words we can think of an integral as a sum over our domain $[a.b]$ where the height of each one is $f(x)$ and the width is $dx$. However for our error to tend to zero we require the rectangles to fit the function $f(x)$ well, and actually tend to it exactly. The best way for us to do this is to reduce the width $dx$ to a very small value close to zero, but it cannot actually equal zero or be "infinitely small". Whilst this may seem strange it does not actually matter as we do not need a numerical method for $dx$, unless using some methods for approximating and in that case the statement is not strictly true. Hope this helps


Find a unit vector that is parallel to $\nabla f(\cos\theta,\sin\theta)$ Suppose $f(x,y)$ is differentiable for all $(x,y),f(x,y)=17$ on the unit circle $x^2+y^2=1$, and $\nabla f$ is never zero on the unit circle. For any real number $\theta$, I have to find a unit vector that is parallel to $\nabla f(\cos\theta,\sin\theta)$.

My thoughts: When I say this question I feel that it relates to level curves in some way. However not sure how it helps in solving this problem.

If $f$ is a given function of $\theta$ only, then we have the following:

$\nabla f(\cos \theta,\sin \theta)=\frac{df}{d\theta}\nabla \theta=\frac{df}{d\theta} \hat \theta$

Where $\hat \theta$ is a unit vector that points in the direction of increasing $\theta$.  Thus a unit vector parrallel to $\nabla f$ is indeed $\hat \theta$.  In Cartesian coordinates this unit vector is given by $(-\hat y x+\hat y x)/(x^2+y^2)=-\hat x y+\hat y x$ since we are on the unit circle.

However, it seems that $f$ is actually a general function of both $r$ and $\theta$, with $f(1,\theta)=17$.  Then, we note that the directional derivative $\frac{df}{ds}$, where $s$ is the arc length parameter along the unit circle, is zero since $f=17$ is constant on the unit circle.  We can write

$$\frac{df}{ds}=\nabla f\cdot \frac{d\vec r}{ds}=0.$$

But, $\frac{d\vec r}{ds}=\hat \theta$.  Thus, $\nabla f$ is perpendicular to $\hat \theta$ and parallel to $\hat r=\hat x \cos \theta+\hat y \sin \theta$.

i do not think there is sufficient info to find gradient of f for f(x,y) where x and y are on unit circle.

we only know f(x,y) = 17 on unit circle, but there is no info for f(x,y) infinitely close to the unit circle.


Finding $\lim\limits_{n \rightarrow \infty} n(e^{x/n}-1)$ Since $\lim\limits_{n \rightarrow \infty} (1+x/n)^n = e^x$

I should be able to find $\lim\limits_{n \rightarrow \infty} n(e^{x/n}-1)$ in some related way, but I have been struggling.

Can anyone provide any insight?

Another approach (if you are not familiar with Taylor expansions): setting $h=\frac{1}{n} \xrightarrow[n\to\infty]{} 0$, you can rewrite
$$
n(e^{\frac{x}{n}} -1) = \frac{e^{h\cdot x}-1}{h} = \frac{e^{h\cdot x}-e^{0\cdot x}}{h-0}.
$$
Does that expression look familiar?

Further hint:

   Consider the function $g(t) = e^{tx}$, which is differentiable at $0$.


Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule? I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..
Edited
COSINE RULE
$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)
After finding the angle, I then use trigonometry to find
$$Cos{A} = \frac{PS}{60}$$

Writing $h$ for $RS$ and $a$ for $PS$, we have $a^2 + h^2 = 60^2$ (angle $PSR$ is right because angle $QSR$ is) and $(126-a)^2 + h^2 = 102^2$.  If I subtract the second equation from the first I learn that $a^2 - (126-a)^2 = 60^2 - 102^2$.  If I do some algebra I see that the $a^2$ terms on the left hand side of this equation drop out, and I can solve directly for $a$. Knowing $a$ makes the rest easy.

Use Heron's formula to find the area, where the semi-perimeter $s$ is $\frac{60+102+126}{2} = 144$.
Then $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{144(144-60)(144-102)(144-126)} = 3024$, and $A = \frac{1}{2}bh \Rightarrow h = \frac{A}{(1/2)b} = \frac{3024}{1/2 \cdot 126} = 48$, hence $RS = 48$.
$PS$ and $QS$ can be found using Pythagoras: they are $\sqrt{60^2-48^2} = 36$ and $\sqrt{102^2-48^2} = 90$ respectively.

As a sanity check, $PS+QS = 36+90=126$ which is the length of $PQ$.


Is it possible to find a context in which $A\to B$ and $\neg A\to\neg B$? I know that "if $A$ then $B$" does not mean "if $\neg A$ then $\neg B$" in the logical sense but in real life can it be possible to find a scenario that satisfies both statements and still perseveres the logic? For example, I can say if the car engine works, then the car moves. However, similarly, I can still say if there is a failure in the car engine, then the car does not move. Both statements are plausible to comprehend but not so at the logical sense, is it?

if A then B and if B then A ( also called A if and only if B (in short hand, A iff B)) implies not A then not B

When you declare something like : "A => B" , you are not specifically stating on A or B , you are declaring that the relation "=>" is true, no matter the trueness of A and B

Yes, it occurs that  "A => B" and "not A => not B". You may find a lot of examples, particularly when "A <=> B".

Indeed , $(A <=> B)  <=> (! A <=> ! B)$ and then it is an opportunity to say that we have both $(A => B)$  and $(! A => ! B)$ true.


Proving multivariable limit using epsilon-delta definition Let $\displaystyle g(x, y) = \frac{\sin^2(x - y)}{|x| + |y|}$. Prove
  that $\lim_{(x, y)\to(0,0)} g(x) = 0$.

Any help on this would be appreciated. So far I think that the fact that $|x + y| \le |x| + |y|$ and $|\sin(x + y)| \le |x + y|$ has to be utilized but I am not sure how to do so in the epsilon delta proof.

Remember that $|\sin(z)| < z$ for $z \ne 0$.  So $0 \le \sin^2(z) < z^2$.  Set $z = x-y$.  Your denominator satisfies $|z| = |x-y| \le |x| + |y|$ by the triangle inequality.

$$0 \le \lim_{(x,y) \to (0,0)} \frac{\sin^2(x-y)}{|x|+|y|} \le \lim_{(x,y) \to (0,0)} \frac{z^2}{|z|} = \lim_{(x,y) \to (0,0)} |z| = 0$$

You can easily convert this proof, which uses the squeeze theorem, into an $\epsilon-\delta$ proof by noting the inequalities and setting $\delta = \epsilon$.  For $\Vert ( x,y ) \Vert < \epsilon$, we have $|z| < \epsilon$ since $|x-y| \le \sqrt{x^2 + y^2}$.

An idea:

$$|x-y|\le|x|+|y|\implies \frac{\sin^2(x-y)}{|x|+|y|}\le\frac{\sin^2(x-y)}{|x-y|}=|\sin(x-y)|\frac{|\sin(x-y)|}{|x-y|}$$

Now just remember that $\,\displaystyle{\lim_{t\to 0}\frac{\sin t}t=1}\;$ and stuff...


How to show this formula to get a square root of a number in "just few seconds" is true? I don't remember in which topic I found it but I know it was there. And I still have not find a proof of this nice approximation.

Let $x$ be a non perfect square number. If $y$ is the closer perfect square to $x$ such that $y < x$ then
$$\sqrt{x}\approx \sqrt{y}+\frac{x-y}{2\cdot \sqrt{y}}$$
And it gives at the maximum two correct decimals after the decimal point.

My first reaction to try to find from where this formula goes, was to expand it, I found $$2\sqrt{xy}\approx y +\sqrt{2y}\cdot(x-y)$$ and I tried to find a remarkable identity but I failed and I'm still stuck there. Also I don't know how should I prove the maximum of two correct digits after the comma.
If we look for example to the sqare root time of $1000$, $961$ is the closer perfect square which verifies the condition. Then we have $\sqrt{1000} \approx 31 + \frac{39}{2*31} = 31.62903...$ and with a calculator we have $\sqrt{1000} = 31.6227766017...$ which is quite good.
Any hints would be appreciate, thank you in advance.

To understand why this is a good approximation note that $(\sqrt{x})' = \frac{1}{2\sqrt{x}}$.

So the formula is of the form $f(x) = f(y) + f'(y)(x-y)$. See Taylor's theorem for more details.

To find the square root of $x$ start with a guess of $y$, then  evaluate an updated value of $y$ using $(y+(x/y)) /2$. Repeat as many times as desired to get a more accurate answer to more decimal places.
You may start with any value of $y$. Even 1 000 000 works.


Finding the total surface area of cuboid model using volume I am stuck with a math question about application of differential calculus from my university math class.

  You are to design a cuboid model with a square base that has a volume of $12m^3$. In order to have a minimum total surface area for the cuboid, what are the values for the height of the cuboid and the length of a side in the square base?

So far, this is what I have done:

$$\text{Volume = }L.B.H = B^2.H$$
$$12 = B^2.H$$
$$H = \frac{12}{B^2}$$
$$\text{S.A. = } 2.B^2 + 4.B.H$$
$$= 2.B^2 + 4.B.(\frac{12}{B^2})$$
$$= 2.B^2 + \frac{48}{B}$$

I am stuck after this step. How should I go about to solve this question? I feel that there is not enough information to complete this question.

In order to find the minimum value of an expression, we take its derivative and equate it to zero. So continuing your solution, and taking the derivative of surface area, we get,

$$\frac{dB}{dy}=4B-\frac{48}{B^2}$$

Equating it to zero, we get

$$B=12^{\frac{1}{3}}$$

Substituting this in:

$$H = \frac{12}{B^2}$$

we get,

$$H=12^{\frac{1}{3}}$$

And $H=L$, so,

$$L=12^{\frac{1}{3}}$$

Finally $L=12^{\frac{1}{3}}, B=12^{\frac{1}{3}}$ and $H=12^{\frac{1}{3}}$


Advantages of IMO students in Mathematical Research Everyone in this community I think would be familiar with International Mathematical Olympiad, which is an International Mathematics Competition held for high school students, with many countries participating from around the world.

What's interesting to note is that many of the IMO participants have gone to win the Fields Medal. Notable personalities include Terence Tao (2006), Ngo Bao Chau (2010), Grigori Perelman (2006), etc.

I would like to know: What advantages does an IMO student possess over a 'normal' student in terms of mathematical research? Does the IMO competition help the student in becoming a good research mathematician or doesn't it?

Training for competitions will help you solve competition problems - that's all. These are not the sort of problems that one typically struggles with later as a professional mathematician - for many different reasons. First, and foremost, the problems that one typically faces at research level are not problems carefully crafted so that they may be solved in certain time limits. Indeed, for problems encountered "in the wild", one often does not have any inkling whether or not they are true. So often one works simultaneously looking for counterexamples and proofs. Often solutions require discovering fundamentally new techniques - as opposed to competition problems - which typically may be solved by employing variations of methods from a standard toolbox of "tricks". Moreover, there is no artificial time limit constraint on solving problems in the wild. Some research level problems require years of work and immense persistence (e.g. Wiles proof of FLT). Those are typically not skills that can be measured by competitions. While competitions might be used to encourage students, they should never be used to discourage them.

There is a great diversity among mathematicians. Some are prolific problem solvers (e.g. Erdos) and others are grand theory builders (e.g. Grothendieck). Most are somewhere between these extremes. All can make significant, surprising contributions to mathematics. History is a good teacher here. One can learn from the masters not only from their mathematics, but also from the way that they learned their mathematics. You will find much interesting advice in the (auto-)biographies of eminent mathematicians. Time spent perusing such may prove much more rewarding later in your career than time spent learning yet another competition trick. Strive to aim for a proper balance of specialization and generalization in your studies.

IMO participants are way higher in mathematical problem solving capability than their peers.

The question to ask is

1) How many of the non IMO students, made it as a conscious decision?. Like, i dont want to participate in IMO because it is on technique. How many did go that route?. Honestly lot of the guys who are good in maths, would have tried IMO, and national level olympaids, would not have succeeded (nothing wrong it), or made a defensive decision that it can affect their school grades, or their parents made that decision for them. I think we can reasonably assume that the students who decided not to pursue IMO because, it didn't fit the principles will be extremely less.  This is the category where, they can qualify in the national olympiads if they wanted to , but didnt.

2) Some of the folks may have developed interest in maths in their undergrad years, and so would not have prepared or given IMO level exams.

It is outright silly to say that Maths is for the long run and to dismiss the skill levels of IMO folks. Looks to me like sour grapes.


How to integrate $\int\frac{1}{\sqrt{9-x^2}}dx$? I've tried to integrate $$\int\frac{1}{\sqrt{9-x^2}}dx$$ through substitution. I changed the equation to $$\int\frac{1}{\sqrt{9-x^2}}dx\equiv\int 1(9-x^2)^\frac{-1}2$$ and took $u=9-x^2$ whereby $du=-2xdx$ ($\frac{-du}{2x}=dx$) but then I got stuck because where do I go on from there? The answer is $\arcsin(\frac{1}{3}x)+c$, but how?

This is a form that we learn to recognize as related to inverse trigonometric functions. When you have an expression in your integrand that looks like $x^2\pm a^2$ or, as in this case, $a^2-x^2$ (for $a=3$), and you can't just use $u$-substitution, the strategy is to try trig substitution.

In this case, it's actually easier than that, because if you know that $\frac{d}{dx} \sin^{-1}x=\frac{1}{\sqrt{1-x^2}}$, then you realize that this integral can be transformed into that by juggling constants and doing a $u$-substitution of $u=\frac{x}{3}$.

It's just $\arcsin\frac{x}{3}+C$ because it's the table integral.

Also, $$\left(\arcsin\frac{x}{3}\right)'=\frac{1}{\sqrt{1-\frac{x^2}{9}}}\cdot\frac{1}{3}=\frac{1}{\sqrt{9-x^2}}.$$


Distance to circle inside triangle I apologise for the lack of a precise term or title, math isn't my strong suit.

I'm trying to calculate the length of L, given angle A and radius of circle D, so that lines b and c tangents with circle D.

Can anyone help me?

Hint: Using the definition of sine and the orthogonality of the tangent one has
$$\sin\left(\alpha/2\right)=\frac{\text{radius}}{\text{distance from center}}$$

L = R / Sin(A/2)

where R is the radius of the circle and A is the angle


Representing any number using only 0 I am afraid to even ask this question, but this is something that was asked over a breakfast table by a friend of mine.

  How do I represent a number, say 4, by using just zeros as a numeral in tandem with any function or mathematical operator(s)?

Now one way to do this is by using a factorial, since $0!=1$:
$$0! + 0! + 0! + 0!$$
Another way uses $\cos 0=1$:
$$\cos 0+\cos 0+\cos 0+\cos 0$$
Also, the topmost  $1$ on Pascal’s Triangle is the zero$^{th}$ row, zero$^{th}$ entry, i.e.,  $0C$$_0$ = $1$.

The Partition function , P(n) is , by convention , defined as $1$ for $n=0$, i.e. , $P(0) = 1$.

Is there any other way I could possibly achieve this?I would be more interested to know the functions which when applied on zero produce non-zero results , like the ones mentioned earlier in the question, rather than semantic manipulations or visual representations.

Peano says: with the successor function.

$$4=S(S(S(S(0))))$$

Changing my answer...
Per the question, use only zeroes $$\frac{\color{blue}{zero} + \color{blue}{zero} + \color{blue}{zero} + \color{blue}{zero}}{\color{blue}{zero}}$$
Note: $zero\ne0$ $\color{lightgray}{:-)}$


Priving that the degree of an irreducible nondegenerate subvariety of $\mathbb{P^n}$ satisfies $d\ge n-k+1$ I'm reading through Dolgachev's "Classical algebraic geometry" and theorem 8.1.1 states the following

The references are not very precise, do you happen to know where, precisely, I can find the proof in any source?

One source is On Varieties of Minimal Degree by Eisenbud and Harris. This is available here at MSRI's website, for instance. The bibliography lists previous proofs by Bertini (1907), Harris (1981), and Xambo (1981), if you're looking for other sources.


Universal Mapping Property of Free Abelian Groups Let S be a set and $F=F_S$ the free group on S. Let $F'$ be the commutator subgroup of $F$. Set $A=A_S = F/F'$, and call it the free Abelian group on $S$. Prove the universal mapping property of the free Abelian group: for any function $f:S \rightarrow G$, where G is an Abelian group, there exists a unique group homomorphism $\varphi:A \rightarrow G$ so that the diagram

$S \xrightarrow{f} G$,
$S \xrightarrow{a \mapsto [a]} A$,
$A \xrightarrow{\varphi} G$

commutes.

We did a proof of this for free groups in class but I am not sure how to apply it to this. I think part of my issue is that I am having a hard time figuring out what the elements of A look like. Will $A=\{[s](a^{-1}b^{-1}ab) \vert a,b,s \in S\}$? If so, then could I just send $\varphi(w)$ to $f(w)$, with $w \in F$?

There is nothing special about the freeness. Given a group $G$ and a morphism $f:G \to A$ with $A$ abelian, $f$ must factor through the abelianization of $G$, which is $G / [G,G]$.

To be precise: This means that there exists a unique $\tilde f$ with $\tilde f \circ \pi = f$ where $\pi$ is the quotient map.

First of all free abelian group (in your sense) is obviously commutative because u kill commutator. Now if we have any abelian group $G$ and function $f:S\rightarrow G$ then we have unique morphism $f_*:F\rightarrow G$. Now we know that $[F,F]\subset ker(f_*)$ so we have unique factorization $f_*=gh$ where $h:F\rightarrow A$ is canonical projection and $g:A\rightarrow G$ unique morphism u looking for.
Secondly one can easily proof that $A_S=\bigoplus_S\mathbb Z\subset \prod_S\mathbb Z$. This is subgroup of product generated by sequences witch are not $0$ only on one coordinate.


Diophantine equations and Hilbert's 10th Problem, how did MRDP do it? I'm having a bit of trouble understanding the Wiki explanation of MRDP's (Matiyasevich, Robinson, Davis, Putnam)'s Theorem, which explains that Hilbert's 10th problem is unsolvable.

  The MRDP theorem states that a set of integers is Diophantine if and only if it is computably enumerable. A set of integers $S$ is recursively enumerable if and only if there is an algorithm that, when given an integer, halts if that integer is a member of $S$ and runs forever otherwise.

Source: http://en.wikipedia.org/wiki/Diophantine_set

What exactly is the set $S$ in the case of Hilbert's 10th Problem? Is it the set of solutions $n$ for any ${x_1}^n+{x_2}^n+...+{x_{k-1}}^n={x_k}^n$ Diophantine equation?

Also are we assuming that the $x_1,...,x_k$, are unchanging? That is, we choose this list of constants $(x_1,x_2,...,x_k)$, and we look for all the $n$ that make the equation true?

MOST confusing to me: why does it matter if this set is c.e.?

Sorry for so many questions, I've been sifting through Wiki pages and haven't been able to find a clear explanation for this.

Matiyasevich showed that for any recursively enumerable set $S$ of natural numbers, there is a polynomial $P_S(y,x_1,x_2,\cdots,x_k)$ with integer coefficients such that for any non-negative integer $y$, we have
$$y\in S \quad\text{if and only if}\quad \exists x_1 \exists x_2\cdots\exists x_k\left(P_S(y,x_1,x_2,\dots, x_k)=0\right)   .\tag{1}$$
Here the variables $x_i$ range over the non-negative integers, but one can let them range over the integers.

The polynomial $P_S$ depends on $S$. However, one can show that there is a polynomial $P(e,y,x_1,\dots,x_k)$ such that for any recursively enumerable subset $S$ of the natural numbers, there is a natural number $e_S$ such that
$$y\in S \quad\text{if and only if}\quad \exists x_1 \exists x_2\cdots\exists x_k\left(P(e_S, y,x_1,x_2,\dots, x_k)=0\right)   .\tag{2}$$

Let $y\in S$. Typically, the $(x_1,x_2,\dots,x_k)$ (there may be many) such that $P_S(y,x_1,\dots,x_k)=0$ will depend on $y$. And there are infinitely many $P_S$ that do the job of "representing" $S$ in the sense of Formula (1).

The condition that $S$ be recursively e
numerable is necessary. This is because for any polynomial $P(y,x_1,x_2,\dots,x_k)$ with integer coefficients, the set
$$\{y: \exists x_1\cdots \exists x_k(P(y,x_1,\dots,x_k)=0)\}$$
is recursively enumerable.

Remarks: $1.$ Technically, the example $x_1^n+\cdots+x_{k-1}^n=x_k^n$ is not a Diophantine equation in the sense of the theorem, since a variable occurs as an exponent. It is an exponential Diophantine equation. The fact that every recursively enumerable predicate is exponential Diophantine was known before the work of Matiyasevich.

$2.$ It has been known since the time of Godel that there is a recursively enumerable predicate 9set) $S$ which is not recursive. Let $P_S(y,x_1,\dots,x_k)$ be a polynomial that represents this $S$ in the sense of Formula (1). Then there is no algorithm that will determine, given input $y$, whether the equation $P(y,x_1,\dots, x_k)=0$ has a solution in natural numbers. It follows that there is no algorithm for determining, given a Diophantine equation $Q(z_1,\dots,z_n)=0$, whether the equation has a solution in natural numbers. For if there were such an algorithm, it could be used in particular to determine, given $y$, whether $P_S(y,x_1,\dots,x_k)$ has a solution $(x_1,\dots,x_k)$.

Not to be confused fundamentally insoluble algebraic equations of higher degrees with insolubility in general.
It is certainly possible to treat everything the same and stop the equation to solve, but we can try to understand what the equations amenable to solution.
For an example, the solution in the general form of a quadratic form.

Formula generally look like this: Do not like these formulas. But this does not mean that we should not draw them.  To start this equation zayimemsya,  well then, and others.  $aX^2+bXY+cY^2=jZ^2$

Solutions can be written if even a single root.$\sqrt{j(a+b+c)}$ , $\sqrt{b^2 + 4a(j-c)}$ , $\sqrt{b^2+4c(j-a)}$

Then the solution can be written.

$X=(2j(b+2c)^2-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+2(b+2c)(\sqrt{j(a+b+c)}\mp{j})sp+(j\mp \sqrt{j(a+b+c)})p^2$

$Y=(2j(2j-b-2a)(b+2c)-(b^2+4c(j-a))(j\pm\sqrt{j(a+b+c)}))s^2+ +2((2j-2a-b)\sqrt{j(a+b+c)}\mp{j(b+2c)})sp+(j\mp\sqrt{j(a+b+c)})p^2$

$Z=(2j(b+2c)^2-(b^2+4c(j-a))(a+b+c\pm\sqrt{j(a+b+c)}))s^2+ +2(b+2c) ( \sqrt{j(a+b+c)} \mp{j})sp + ( a + b + c \mp \sqrt{j(a+b+c)})p^2 $

In the case when the root $\sqrt{b^2+4c(j-a)}$ whole.

Solutions have the form.

$X=((2j-b-2c)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2++2(4ac+b(2j-b)\pm{(2j-b-2c)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$

$Y=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(2j-b-2a\mp\sqrt{b^2+4c(j-a)}))s^2++2(4ac+b(2j-b)\pm{(b+2a)}\sqrt{b^2+4c(j-a)})sp+(2j-b-2a\pm\sqrt{b^2+4c(j-a)})p^2$

$Z=((b+2a)(8ac+2b(2j-b))-(b^2+4a(j-c))(b+2c\mp\sqrt{b^2+4c(j-a)}))s^2++2(4ac+b(2j-b)\pm {(b+2a)}\sqrt{b^2+4c(j-a)})sp+(b+2c\pm\sqrt{b^2+4c(j-a)})p^2$

In the case when the root $\sqrt{b^2+4a(j-c)}$ whole.

Solutions have the form.

$X=(2j^2(b+2a)-j(a+b+c)(2j-2c-b\pm\sqrt{b^2+4a(j-c)}))p^2+ +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(2j-2c-b\mp\sqrt{b^2+4a(j-c)})s^2$

$Y=(2j^2(b+2a)-j(a+b+c)(b+2a\pm\sqrt{b^2+4a(j-c)}))p^2+ +2j(\sqrt{b^2+4a(j-c)}\mp{(b+2a)})ps+(b+2a\mp\sqrt{b^2+4a(j-c)})s^2$

$Z=j(a+b+c)(b+2a\mp\sqrt{b^2+4a(j-c)})p^2+ +2((a+b+c)\sqrt{b^2+4a(j-c)}\mp{j(b+2a)})ps+ (b+2a\mp\sqrt{b^2+4a(j-c)})s^2$

Since these formulas are written in general terms, require a certain specificity calculations.If, after a permutation of the coefficients, no root is not an integer. You need to check whether there is an equivalent quadratic form in which, at least one root of a whole. Is usually sufficient to make the substitution $X\longrightarrow{X+kY}$  or more $Y\longrightarrow{Y+kX}$  In fact, this reduces to determining the existence of solutions in certain Pell's equation. Of course with such an idea can solve more complex equations. If I will not disturb anybody, slowly formula will draw.  number $p,s$ integers and set us.   I understand that these formulas do not like. And when they draw - or try to ignore or delete.

Formulas but there are no bad or good. They either are or they are not.

In equation $aX^2+bY^2+cZ^2=qXY+dXZ+tYZ$

$a,b,c,q,d,t$ integer coefficients which specify the conditions of the problem.

For a more compact notation, we introduce a replacement.

$k=(q+t)^2-4b(a+c-d)$

$j=(d+t)^2-4c(a+b-q)$

$n=t(2a-t-d-q)+(2b-q)(2c-d)$

Then the formula in the general form is:

$X=(2n(2c-d-t)+j(q+t-2b\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{n})ps++(2b-q-t\pm\sqrt{k})s^2$

$Y=(2n(2c-d-t)+j(2(a+c-d)-q-t\pm\sqrt{k}))p^2+2((d+t-2c)\sqrt{k}\mp{ n })ps++(q+t+2(d-a-c)\pm\sqrt{k})s^2$

$Z=(j(q+t-2b\pm\sqrt{k})-2n(2(a+b-q)-d-t))p^2+2((2(a+b-q)-d-t)\sqrt{k}\mp{n})ps+(2b-q-t\pm\sqrt{k})s^2$

And more.

$X=(2n(q+t-2b)+k(2c-d-t\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps++(d+t-2c\pm\sqrt{j})s^2$

$Y=(2n(2(a+c-d)-q-t)+k(2c-d-t\pm\sqrt{j}))p^2++2((q+t+2(d-a-c))\sqrt{j}\mp{n})ps+(d+t-2c\pm\sqrt{j})s^2$

$Z=(2n(q+t-2b)+k(d+t+2(q-a-b)\pm\sqrt{j}))p^2+2((2b-q-t)\sqrt{j}\mp{n})ps++(2(a+b-q)-d-t\pm\sqrt{j})s^2$

$p,s$ are integers and are given us. Since formulas are written in general terms, in the case where neither the root is not an integer, it is necessary to check whether there is such an equivalent quadratic form in which at least one root of a whole. If not, then the solution in integers of the equation have not.

I noticed that there do not like to write solutions of equations. Billing solutions read bad taste.  Although when it, but it should start doing.

There are equations that can be solved, and there are equations that are not amenable to solution. And I think the approach may be different.


$x^3 - x +1$ has a root in $\mathbb{Q}(a)$? "Decide if $x^3 - x +1$ has a root in $\mathbb{Q}(a)$, with $a \in \mathbb{C}$ such that $a^3 + a + 1 = 0$."

I need a hint!

Based on the idea given by Jyrki Lahtonen, I think the following is relatively elementary.

Let $K$ be the splitting field of $x^3+x+1=0$ over $\mathbb{Q}$. Then the Galois group $G(K,\mathbb{Q})$ cannot fix the root of $x^3-x+1=0$ and therefore both cubics split in $K$.

However, this would mean that $K$ contained $L=\mathbb{Q}(\sqrt{-31},\sqrt{-23})$ and we would have $4=[L,\mathbb{Q}]$ as a factor of $6=[K,\mathbb{Q}]$.

We can write an arbitrary element of $\Bbb Q(a)$ as $x=\alpha+\beta a+\gamma a^2$, with $\alpha,\beta,\gamma\in\Bbb Q$. Let's plug that element into our equation: $$x^3-x+1=(\alpha+\beta a+\gamma a^2)^3 - (\alpha+\beta a+\gamma a^2) + 1.$$

You can expand this, reducing powers of $a$ by using the formula $a^3=-a-1$. Finally, you obtain something of the form $A+Ba+Ca^2$, where, $A$, $B$ and $C$ are expressions in $\alpha, \beta,\gamma$. Set all three to zero. If there is a rational solution, then you've shown that $x\in\Bbb Q(a)$


Trigonometric analytical proof How would you go about showing that

$-4\cos^{3}(t) +3\cos(t) = -\cos(3t)$

It's part of a larger problem, and I know that to be true, but I just can't figure out how to get there (I might be too tired, though).
Thanks for your help!

Write $\mathrm{cos}(3x)$ as $\mathrm{cos}(2x+x)$, use the formula for the cosine of a sum, and insert $\mathrm{cos}(2x)=\mathrm{cos}^2(x)-\mathrm{sin}^2(x)$ and $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$.


Prove that a Lipschitz function is continuous Define $f:[a.b] \rightarrow \mathbb{R}$ as Lipschitz
By definition, then, $\exists M>0, M \in \mathbb{R}:|f(x)-f(y)\leq M|x-y|$  $ $ $\forall x,y\in[a,b]$
Rearranging this, we have $\frac{|f(x)-f(y)|}{x-y} \leq M$, positing that the slope must be finite on the interval $[a,b]$
I understand conceptually that Lipschitz functions must be continuous but I'm having trouble showing it.
Should I go about this by showing assuming the negation of Lipschitz and showing that for some $\epsilon,  \exists \delta>0: |f(x)-L|>\epsilon$ ?

It is not necessary to proceed by contradiction. Use the $\varepsilon-\delta$ definition of continuity and choose $\delta=\varepsilon/M$.

If $\delta=\frac{\epsilon}{M}$. The definition of continuity says that we can suppose that $$0<|x-x_0|<\delta$$ from here: $$M|x-x_0|<\epsilon$$ but we now that $$|f(x)-f(x_0)|<M|x-x_0|$$ because $f\in Lip(\mathbb{R})$, then $|f(x)-f(x_0)|<\epsilon$ and therefore $f$ is continuous.


confusion of proving f(A)∩f(B) a subset of f(A∩B)? f:A↦B and S⊆A,T⊆A, then f(S)∩f(T)⊆f(S∩T)

We know that it is not.By setting A:{1,-3} B:{1,3},f(x)=$x^2$.

But what if f is one-to-one function?

How can we prove that ∀y(y∈(f(S)∧f(T)))⇒∃x(x∈S∧x∈T))?

Since you are talking about $f(A) $ and $f(B) $ we can assume the function is onto also. Then you see the inverse exists and inverse images are well behaved with $\cup$ and $\cap$. And by bijectivity onto the image $f^{-1} (f(A)) =A$.
In general if you don't want to use the above argument you can say the function has left inverse and using that you can achieve the same result.

So $f^{-1} (f(A) \cap f(B)) = f^{-1}(f(A)) \cap f^{-1}(f(B)) =A\cap B $ applying $f$ both the sides you will get the result.


Two definitions of Euler characteristics In third chapter of "Differential Topology" book by Guillemin and Pollack ,Euler's characteristic is defined as I(∆,∆) and in the end of chapter, an intuitive proof is given to show how this is same as one calculated by using triangulations. They defined vector field on triangulated manifold which has source on vertex ,sink on face and saddle on edge .Is there any rigorous as well as generalized ( to n dim.)  proof of this theorem by using tools from book or other fairly basic tools ?

Yes, the "usual proof" involves some (actually, quite a bit of) hand-waving. As an alternative, take a look at a careful write-up of a proof given by Jonathan Libgober, written as a part of his REU:

Euler characteristc, Poincare-Hopf theorem and applications.

With all the details, the proof takes about 16 pages, too long to be reproduced in my answer. An outline is that one first develops Morse theory which connects gradient vector fields of Morse functions and cell decompositions of manifolds. Once the machinery is there, it is fairly easy to see that the index of such a gradient vector field on a manifold $M$ equals $\chi(M)$. Lastly, one argues that any two vector nondegenerate fields  have the same index (for this, Guillemin and Pollack suffice).

Lastly, Peter May (at University of Chicago) had done an amazing job with his undergraduate students writing useful REU projects, such as the one quoted above.

There is nothing non-rigorous about that proof. It works in arbitrary dimensions once you know that manifolds are triangulable. You need some proof like this because the usual definition of Euler char is in terms of triangulations...


Is this can be an integer $\sqrt[2z-1]{2mz+p_{n}}$? let the integers $z>2$, $m>1$ and a prime number $p_{n}\geq 3$.

is the following can be an integer for any value of $z,m$ and $p_{n}$ ?

$\sqrt[2z-1]{2mz+p_{n}}$

Well no, because lets take z = 3, m = 2 and p = 3. Then the expression would be

$(2*2*3 + 3)^(0.2) = 15^(0.2)$

Which is clearly not an integer.


Proof explanation: If $p$ is prime, then $x^2 \equiv 1 \:(\mathrm{mod}\:p) \Leftrightarrow x \equiv \pm1 \:(\mathrm{mod}\:p)$ I was asked to prove the following theorem:

  Let $p$ be prime. Then for any $x$, the following holds true: $x^2
\equiv 1 \:(\mathrm{mod}\:p) \Leftrightarrow x \equiv \pm 1
\:(\mathrm{mod}\:p)$

My proof is as follows:

$\begin{alignat}{2} && x^2 \equiv 1 \:(\mathrm{mod}\:p) \\ &\Leftrightarrow\quad &x^2 -1\equiv 0 \:(\mathrm{mod}\:p) \\ &\Leftrightarrow &p \mid (x^2 - 1)
\\ &\Leftrightarrow &p \mid (x - 1)(x + 1)
\\ &\Leftrightarrow &p \mid (x - 1) \vee p \mid(x + 1)
\\ &\Leftrightarrow &x \equiv 1 \:(\mathrm{mod}\:p)\vee x\equiv -1 \:(\mathrm{mod}\:p) \end{alignat}$

My problem: I didn't (actively) use the fact that $p$ is prime, which is suspicious to me, because it means that either my proof is wrong, or that I used it somewhere in an implicit way without realizing it.

Can somebody point out where I used the fact that $p$ is prime?

The step
$$p \mid (x-1)(x+1) \iff p\mid (x-1) \lor p\mid (x+1)\tag{$\ast$}$$
uses that $p$ is prime.

Actually not quite, because the equivalence $x^2 \equiv 1 \pmod{m} \iff x \equiv \pm 1 \pmod{m}$ also holds for some composite moduli (and it trivially holds for $\lvert m\rvert \leqslant 1$). It holds whenever the group of units modulo $m$ is cyclic, and for composite $m > 1$ that happens when $m = 4$, $m = p^k$, or $m = 2p^k$ for an odd prime $p$ and $k \geqslant 1$.

So the specific case $(\ast)$ of
$$p \mid ab \iff p\mid a \lor p \mid b$$
does not characterise primes (or units or $0$) like the general case does.

For $p\ge2$, We have either $p|x-1$ or $p|x+1$.
So $x+1 \equiv 0 \mod p$ or $x-1 \equiv 0 \mod p$.
Implying
$$x\equiv \pm 1 \mod p$$


Prove that if for all $aba=bab$ then $|G|=1$. Let $G$ be a group such that for all $a,b \in G$ we have $aba=bab$. Prove that $|G|=1$.

So I have to show that $G =\left\{e \right\} $.
Because for any $a,b \in G$ we have $aba=bab$, let $b=e$. Then $aea=eae$ so $a^2 = a$ hence $a = e$. Because $a$ is any we have $G=\left\{e \right\} $.

Does it work?

Your proof works, but it lacks mindless symbol pushing, so I offer two proofs by symbol pushing to restore balance to the world.

Proof I.
Write $abab$ in two ways: $babb=aaba$. Then: $$aaba=b(ab)b=abbab$$ And cancel to get $aba=bbab$, but $aba=bab$, so $b=e$. $\square$

Proof II. We start with the natural equation
$$baab=aabaa=ababa=babba$$ Now, cancel $ba$ on the left, so that $ab=bba$. But $$aba=bab\implies ab=baba^{-1}$$ Therefore, $bab=bbaa$ and so $ab=baa$ as well. This means $bba=baa$; cancelling on both left and right, $b=a$. Since $a,b$ are abitrary, we are done. $\square$

I make it more detail:

G is a group => e $\in$ G. i.e., |G| >= 1

for some x $\in$ G

1, if $x^{-1}$ not x: |G| >= 3 (at least x, $x^{-1}$, e)

$x^{-1}$x$x^{-1}$ = x$x^{-1}$x, so x=$x^{-1}$

This is to say, for all x $\in$ G, x=$x^{-1}$

2, for any x $\in$ G, since x=$x^{-1}$: x=exe=xex=xe$x^{-1}$=e. so x=e

So G={e}, |G|=1

QED


Sum of total scores odds will end even or odd Friend and i are betting on the ending total scores of basketball game, he says odds are greater it will end with an even sum, i say the odds are equal. Who is right?

In theory @gev is correct, but I don't think it's a reasonable assumption that it's equally likely for either team to score odd vs even.

I analyzed every NCAA game since 2010 http://www.sports-reference.com/cbb/play-index/tourney.cgi and got

Odd: 262
Even: 222

So 54% odd.

If I had to guess which is more probable I'd say odd. Suppose that team A scored X points.  Then team B can score anything BUT X, meaning there's one less possibility to get an even sum.

even = even + even

even = odd + odd

i.e.: If both are similar (both are even or odd) then the answer is even.

But if both are different then the answer is odd.

odd = even + odd

odd =  odd + even


What is the sum of
$\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$? What is the sum of the following expression:
$$\sum_{k=0}^\infty \frac{(-1)^{k+1}}{2k+1}$$

I know it is convergent but I cannot evaluate its sum.

Consider adding $x^{2k+1}$ to the summand:

$$f(x)=\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{2k+1}$$

Then...

$$f'(x)=\sum_{k=0}^\infty(-1)^kx^{2k}=\frac1{1+x^2}$$

Thus,

$$f(x)=\arctan(x)$$

And

  $$\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}=f(1)=\arctan(1)=\frac\pi4$$

This is Leibniz's formula for $\pi$.

hint
use the fact that
$$\frac {x^k}{2k+1}=\frac {1}{\sqrt {-x}}\frac {(\sqrt{-x})^{2k+1}}{2k+1} $$
for $x<0$


Why does convolution of delta function commute (test distribution perspective)? If I understand correctly, for test functions $f$ we define

$$ \langle\delta, f\rangle = f(0)$$

and convolution is defined as

$$ \langle g * T, f\rangle = \langle T, g^- * f\rangle,$$

where $f$ and $g$ are test functions.

From this it follows that

$$ \langle g * \delta, f\rangle = \langle\delta, g^- * f\rangle = (g^- * f)(0) = \int g^-(-y) f(y) dy = \int g(y) f(y) dy = \langle g, f\rangle$$

so that $g * \delta = g$.

I don't see how any of this would imply that $
\delta * g = g$ as well. In fact, I don't even see how this expression makes sense, since $\langle \delta * g, f\rangle = \langle g, \delta * f\rangle $. Since $\delta$ is only defined as a distribution, I don't think $\delta * f$ has any meaning, let alone is a test function. Can someone clear this up for me?

The canonical definition of convolution of two arbitrary distributions with well associated supports is the following.

Let $S$ and $T$ be two distributions, respectively, with supports $A$ and $B$, and let $\varphi$ is a test function with support $K$. You define
$$
\langle S*T,\varphi\rangle=\langle S_x\otimes T_y,\alpha(x)\varphi(x+y)\rangle,
$$
where $\alpha$ is any test function that is equal to $1$ in a neighborhood of $A\cap(K-B)$. Then $S*T=T*S$ for any distributions with well associated supports.

Recall that two closed sets $A$ and $B$ are said to be well associated if $A\cap (K-B)$ is compact for any compact set $K$.

To your question: Your definition of convolution is a map

$$ *:(\mbox{test functions})\times (\mbox{distributions})\to \mbox{distributions}. $$

So you're right, "$\delta * g$" does not make sense, as $\delta$ is not a test function.


Simple example of a continuous onto function mapping $[0,1]$ to $\mathbb R$ There should exist such a function, but I cannot think of any example.
Onto continuous functions mapping $(0,1)$ to $\mathbb R$ are easy to find.

Edit: Sorry - I mentioned Tietze extension theorem proved the existence of such a function - that was wrong. I mixed it up with a different question.

No such function exists. $[0,1]$ is compact, $\mathbb{R}$ is not compact, and the continuous image of a compact space is compact.

No such function exists from $\mathbb{R}$ to $\mathbb{R}^2$. If you omit some lines from $\mathbb{R}$, it's not connected anymore, but $\mathbb{R}^2$ stays connected.


A double integral for $\frac{\pi}{2} \ln 2$. Show that
  \begin{eqnarray*}
I=\int_0^1 \int_0^1 \frac{dx \,dy}{\sqrt{1-x^2y^2}}  = \frac{\pi}{2} \ln 2.
\end{eqnarray*}

My try ... from this question here we have
\begin{eqnarray*}
\int_0^1 \frac{ \sin^{-1}(x)}{x} \,dx = \frac{\pi}{2} \ln 2 .
\end{eqnarray*}
And from this question here we have
\begin{eqnarray*}
\int_0^1 \ln \left( \frac{1+ax}{1-ax} \right) \frac{dx}{x\sqrt{1-x^2}}=\pi\sin^{-1} a,\qquad |a|\leq 1.
\end{eqnarray*}
It is easy to show
\begin{eqnarray*}
\int_0^1  \frac{dy}{1+yz}=\frac{1}{z} \ln(1+z).
\end{eqnarray*}
So we have (with a little tad of algebra)
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 &=&  \int_0^1 \frac{ \sin^{-1}(x)}{x}\, dx \\
&=& \frac{1}{\pi} \int_0^1  \int_0^1 \ln \left( \frac{1+xt}{1-xt} \right) \frac{dt}{xt\sqrt{1-t^2}} x\, dx \\
&=& \frac{2}{\pi} \int_0^1  \int_0^1 \int_0^1 \frac{dx \,dy\, dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
\end{eqnarray*}
This suggests we should consider the integral (sub $t=\sin(\theta)$)
\begin{eqnarray*}
\frac{2}{\pi}  \int_0^1 \frac{ dt}{(1-x^2y^2t^2)\sqrt{1-t^2} } \\
= \int_0^{\pi/2} \frac{d \theta }{1- x^2  y^2 \sin^2(\theta)}.
\end{eqnarray*}
Now it is well known (Geometrically expand, integrate term by term & sum the familiar plum) that
\begin{eqnarray*}
\frac{2}{\pi} \int_0^{\pi/2}  \frac{d \theta }{1- \alpha \sin^2(\theta)}=\frac{2}{\pi} \frac{1}{\sqrt{1-\alpha}}
\end{eqnarray*}
and using this we have
\begin{eqnarray*}
\frac{\pi}{2} \ln 2 =\int_0^1 \int_0^1 \frac{dx\, dy}{\sqrt{1-x^2y^2}}  .
\end{eqnarray*}

The above double integral reminds of
\begin{eqnarray*}
\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2}=\int_0^1 \int_0^1 \frac{dx\, dy}{1-x^2y^2}  = \frac{\pi^2}{8}
\end{eqnarray*}
which can be evaluated using the substitution $x= \frac{\sin u}{\cos v}$, $y= \frac{\sin v}{\cos u}$.

My solution above used some pretty heavy machinery to establish the result. So my question is: is there an easier method ?

$$\int_0^1 \int_0^1 \frac{dxdy}{\sqrt{1-x^2y^2}}\overset{xy=t}=\int_0^1 \frac{1}{y}\int_0^y \frac{1}{\sqrt{1-t^2}}dtdy=\int_0^1 \frac{\arcsin y}{y}dy
$$
$$\overset{IBP}=-\int_0^1 \frac{
\ln y}{\sqrt{1-y^2}}dy\overset{y=\sin x}=-\int_0^\frac{\pi}{2}\ln(
\sin x)dx=\frac{\pi}{2}\ln 2$$
See here for the above integral.

Why not go head-on? It seems to work here. That is, write the integral as $$\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{\sqrt{1-x^2y^2}}=\int_0^1\int_0^1\frac {\mathrm d x \mathrm d y}{y \sqrt{\frac{1}{y^2}-x^2}}.$$ Setting $x=\frac1y\sin\phi$ as usual helps us evaluate the first integral, or you can simply note that it has the form of an $\arcsin.$ If you make this substitution, the first integral becomes $$\int_0^{\arcsin y}\frac1y\mathrm d \phi=\frac{\arcsin y}{y}.$$ Thus, the integral reduces to $$\int_0^1\frac{\arcsin y}{y}\mathrm d y,$$ which may be done by parts. Forgetting limits for now, this begins as $$\arcsin y\log y-\int\frac{\log y}{\sqrt{1-y^2}}\mathrm dy,$$ etc.


stimulating budding math kids? (8 yo) My 8 yo child loves maths, he considers it "his hobby". Without pressuring him, I'd like to offer him more challenge, wonder and marvel. The school can't really offer him more (tables of multiplication, yay), and my clumsy attempts aren't helping.
I'm looking for resources/links/references I could use. A youtube maths for kids channel, or books or even a "mothly loot crate" subscription for maths?
Thanks in advance!
Edit: the question got closed (not sure why), but I would like to thank every single person who commented and answered. You've really helped me here!

Here are some math books my son has enjoyed:

Bedtime Math

The number devil

The boy who loved math

Number Crunch

Go Figure

Zero the hero

Wumbers

Also, we homeschool him and use "Beast Academy", which has an online curriculum for 2nd - 5th that I really like. It has some challenging problems, and some material not typically taught in elementary school in the U.S. It also has puzzles that my son likes.

Not a really an answer to your question, but too long for a comment.
I would suggest that you try to widen the scope of his interest, especially more towards 'pure maths': the more abstract topics like set theory, algebra etc. I know - he's only 8 years old and all, and it may sound like absurdly complicated stuff, but I'm not talking about full-on, formal algebra or axiomatic set theory. There is a lot of very useful stuff to learn about the motivations for these things, which will help him when he later gets to the formal maths.
I remember around that age, I started having a real interest in knowing more about the background and playing around with the edges of the concepts I had learned in school, and I found it hugely frustrating that nobody else thought about these things.
For example, when we were doing endless calculations in school to learn long multiplication, I really wanted to know why it worked - there is a simple enough explanation for anyone who is motivated enough to learn; my guess is that your boy would like to explore that.
Or set theory - what is it actually for? This is actually one of the deepest and most fundamental subject in mathematics, and it is a very accessible subject for any child who is interested enough.


How do we solve for a $a = (5^4\ (\text{mod}\ 7))^3\ (\text{mod}\ 13)$

I know that I have to evaluate $5^4\ (\text{mod}\ 7)$ first, but how do you do that without using a calc?

then I have to evaluate  $x^3\ (\text{mod}\ 13)$. Any tip?

You can use the fact that multiplication distributes over addition.  So, (5)(5)(5)(5)=(25)(5)(5)=(20+5)(5)(5)=((5)20+(5)5)5=(100+25)(5)=(125)5=(100+20+5)5=(100(5)+20(5)+5(5))=(500+100+25)=625.  (700-70)=630, so 623 comes as the closest multiple of 7 to 625 which is not greater than 625.   So, 625 (mod 7)≡2.  Since two the third power equals eight, and 8 mod 13 is congruent to 8, we get 8.  Personally, I like Argon's method better, but I didn't know that

a(b) (mod x)≡(a (mod x))(b (mod x)).


What is the error in this fake proof which uses series to show that $1=0$? A common "trick" for obtaining a closed form of a geometric series is to define
$$ R := \sum_{k=0}^{\infty} r^k, $$
then manipulate the series as follows:
\begin{align}
R - rR
&= \sum_{k=0}^{\infty} r^{k} + \sum_{k=0}^{\infty} r^{k+1} \\
&= (1 + r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\
&= 1 + (r + r^2 + r^3 + \dotsb) - (r + r^2 + r^3 + \dotsb) \\
&= 1.
\end{align}
On the other hand, $R-rR = (1-r)R$.  Hence
$$ (1-r)R = 1
\implies R = \frac{1}{1-r}. $$
In this example, the formula is obtained by a sequence of relatively elementary algebraic manipulations.

By a similar kind of manipulation, suppose that
$$ S := 1 + 1 + 1 + 1 + \dotsb = \sum_{k=0}^{\infty} 1. $$
$S$ is unaffected by addition of $1$, and so $S = 1+S$.  Canceling $S$ from both sides gives $0 = 1$, which is clearly nonsense.

  Question: What went wrong with the second computation?  Why do these arguments work well for summing the geometric series, but not for the series of ones?

Your argument hinges upon the assumption that $S$ is a number you can do arithmetic with. It isn't a number, you can't do arithmetic with it, and this is what you have shown (by contradiction).

In fact, for $|r|<1$,
$$R=\lim_{n\to\infty}\sum_{k=0}^n r^k$$ and
$$(1-r)R=\lim_{n\to\infty}\left(\sum_{k=0}^n r^k-\sum_{k=0}^n r^{k+1}\right)=\lim_{n\to\infty}(1-r^{n+1})=1.$$ All steps are rigorous.

If we try a similar argument with
$$S=\lim_{n\to\infty}\sum_{k=0}^n1$$
we obtain
$$S+1=\lim_{n\to\infty}\sum_{k=0}^{n+1}1$$
and by substraction,
$$S+1-S=\lim_{n\to\infty}\left(\sum_{k=0}^{n+1}1-\sum_{k=0}^n1\right)=\lim_{n\to\infty} 1\ne0.$$


$X,Y \sim N(0,1)$, which is the distribution of $W = \frac X Y$? Problem: Let $X,Y \sim N(0,1)$ independent random variables.

Which is the distribution of $W = \frac X Y$?

Attempt:
Something wierd happens. In fact $$ \mathbb P\left(\frac X Y \leq c\right) = \mathbb P(X -cY \leq 0) = \mathbb P(N(0,c^2+1) \leq 0)= \frac 1 2 $$ which is quite impossible. What's wrong?

The region of the plane in which $X/Y\le c$ is that in which $X\le cY$ if $Y>0$ and $X\ge cY$ if $Y<0.$ In polar coordinates, half of that region is described by $\pi > \theta > \arctan (1/c),$ if $c>0,$ and the other half has the same probability assigned to it, so let's double the probability assigned to the case $Y>0.$

The probability measure is $\displaystyle \frac 1 {2\pi} e^{-(x^2+y^2)/2} \, dx \, dy.$ In polar coordinates that becomes $\displaystyle \frac 1 {2\pi} e^{-r^2/2} r\,dr\,d\theta.$ Therefore
$$
\Pr\left( \frac X Y \le c \right) = \frac 1 {2\pi} \int_{\arctan(1/c)}^\pi \int_0^\infty e^{-r^2/2} r\,dr\,d\theta = \frac 1 \pi \left(\pi - \arctan\frac 1 c \right).
$$
Differentiating with respect to $c,$ we get the density:
$$
\frac 1 {\pi(1+c^2)},
$$
i.e. this is a standard Cauchy distribution.

You know that X~N(0,1) so you know the density function of X (fx(x)). You know also that Y~N(0,1) so you know fy(y).
Furthemore you have that X,Y are independent so you know the density function of f_(X,Y)(x,y) (i.e. f_(X,Y)= fx(x)*fy(y) )
Now you need to calculate the distribution function of X/Y so what you re gonna do is: P( W<= c) = P(X/Y <= c) = P(X<= Yc) and now you need to calculate the double integral where: y will be in all R and x will be bounded by (-infinite, xc).


Question about basis step of strong induction proof Let $P(n)$ be any collection of $n$ coins that can be obtained using a combination of $3$ cent and $5$ cent coins. Use strong mathematical induction to prove that $P(n)$ is true for all integers $n \ge 14$.

Basis: $P(14), P(15), P(16)$ can be made up of $5$ cents and $3$ cents.

Inductive Step: $P(14)$ through $P(k)$ are true

$P(k + 1): k + 1 = (k + 1 – 3) + 3$ where if $k \ge 16$, then $(k + 1 – 3) \ge 14$ so $k + 1 \ge 17$, so $P(k + 1)$ is true by hypothesis.

If this makes sense, why do we need to show the basis steps explicitly? Aren't we assuming them to be true in the inductive step?

This type of induction works for any property $\,P(n)\,$ that is preserved under a shift, such as here where $\,P(n)\,$ true $\,\Rightarrow\,P(n\!+\!3)\,$ true. Writing $\,P\,$ for the subset of naturals where $\,P\,$ is true, the induction works as follows.

Theorem $\ $ Suppose $\,P\subseteq \Bbb N\,$ satisfies $\,n\in P\,\Rightarrow\, n\!+\!3\in P,\ $ for all $\,n\ge a.\ $ Then

$$\,a,a\!+\!1,a\!+\!\color{#c00}2\in P\,\Rightarrow\,n\in P{\rm\, \ for\ all\,\ } n\ge a$$

Proof $\ $ If not there is a least counterexample $\,\ell\not\in P.\,$ Note $\,\ell \ge a\!+\!\color{#c00}3\,$ so $\,\ell\!-\!3\ge a,\,$ Therefore, by our shift-closure hypothesis, $\,\ell = (\ell\! -\! 3)+ 3\in P,\,$ contradiction.

Remark $\ $ Clearly the proof generalizes from shift increment $\,k=3\,$ to arbitrary $\,k\ge 1,\,$ with $\,k\,$ consecutive integers $\,a,a\!+\!1,\ldots,a\!+\!k\!-\!1\,$ serving as the base cases, i.e. the foundation of the induction. Notice that the case $\,k=1\,$ is simply ordinary induction.

Yes, you are assuming them to be true in the later step. (But you can assume anything you want.)

Precisely to make the argument valid you need to establish (or to prove) the base cases somewhere.

How do you know that $P(17)$ is true? You know it because you can reduce it to the truth of $P(14)$ (this is what is done in the induction step), and the truth of this was established explicitly.

If you do not do this, then you can only say $P(17)$ is true if $P(14)$ is true. (This is the induction step.) But you need to show the latter somewhere.

Of course, and this is the charm of induction, you can do longer chains.

Why is $P(20)$ true? It is true if $P(17)$ is true. And $P(17)$ is true if $P(14)$ is true. And $P(14)$ is in fact true. So $P(20)$ is true.
Yet, if you do not know the truth of $P(14)$  then you cannot conclude anything.

To look at it differently. You could just as well do the argument for the induction step with $k \ge 3$ and assuming $P(1), P(2), P(3)$ true. But you cannot get $4$ with $3$ and $5$ coins! Why? because you cannot get $1$ either.  But under the (false) assumption you can get $1$, you could also get $4$.

So you need to check that your base assumption are not false. This is what the base step is for.


Stuck on matrix exponential problem I want to show that the property $e^{tA}e^{tB}=e^{t(A+B)}$ implies that $AB=BA$. Here $A,B$ denotes matrices and $t\in \mathbb{R}$.

Im stuck, have tried to expand both sides with their taylor series but not sure if this is the way to go.

Could anyone give me a hint?

Differentiate 2 times with respect to time $t$ the identity
$$e^{tA}e^{tB}=e^{t(A+B)}$$
and then set $t=0$.

This is not what has been asked but to show the identity is valid if $A$ and $B$ commutes, you can...

Define
$$
f(t) = e^{t(A+B)}e^{-tB}e^{-tA}
$$

Then,
$$
f'(t) = (A+B)e^{t(A+B)}e^{-tB}e^{-tA} - e^{t(A+B)}Be^{-tB}e^{-tA}
- e^{t(A+B)}e^{-tB}Ae^{-tA}
$$

if $A$ and $B$ commutes, we can factor out A and B from the last two.
$$
f'(t) = (A+B) f(t) - A f(t) - B f(t) = 0
$$

Therefore $f(t)$ is a constant, you can show that it's identity.


Proof that $P(A \mid C) = 1$ implies $P(A \mid B∩ C) = 1$ This isn't homework. I'm reading Probabilistic Graphical Models by Koller et al.$\space$, and an easy problem in chapter $3$ made me think of a more general problem (which I'm now stuck on). I have everything in place except for this:

Let $P(B \cap C) \neq 0$. Then $P(A \space|\space B ∩ C) = 1$ if $P(A \mid C) = 1$. It's intuitively obvious, but I haven't been able to formally prove it. I get

$$P(A \mid B∩C) = \dfrac{P(A∩B∩C)}{P(B∩C)} = \dfrac{P(C) P(A\mid C) P(B\mid A∩C)}{P(B∩C)} = \dfrac{P(B\mid A∩C)}{P(B|C)}$$

I don't see why that ratio on the right hand side is $1$. Is it? What am I not seeing?

Assume that $\mathbb P(\ \mid C)$ and $\mathbb P(\ \mid B\cap C)$ both exist, that is, that
$$\mathbb P(B\cap C)\ne0.
$$
Then note that $\mathbb P(A\mid C)=1$ if and only if $\mathbb P(A\cap C)=\mathbb P(C)$ if and only if $\mathbb P(C\setminus A)=0$. Likewise, $\mathbb P(A\mid B\cap C)=1$ if and only if $\mathbb P((B\cap C)\setminus A)=0$. But $(B\cap C)\setminus A\subseteq C\setminus A$, hence the latter implies the former, that is,
$$
\mathbb P(A\mid C)=1\implies \mathbb P(A\mid B\cap C)=1.
$$

$\mathbb{P}\{A|C\}=1$ implies $\mathbb{P}\{A,C\}=\mathbb{P}\{C\}$, which implies $A\subset C$. Therefore $\mathbb{P}\{A|B,C\}=1$.


Max(F(x),F(y)) is a joint CDF? If F(•) is a cumulative distribution function:

•) Is F(x,y) =max[F(x),F(y)] a joint cumulative distribution function?

I intuitively think that it is not a cumulative distribution function

:( I have no idea how to start this exercise, for your help, thank you very much.

Hint:

If $F(x,y)$ is a joint cdf, and $x_1<x_2$ and $y_1<y_2$, then you must have
$$
F(x_2,y_2)-F(x_1,y_2)-F(x_2,y_1)+F(x_1,y_1)\ge 0,
$$
since the LHS is the probability of an event (what event?). Let $y_1=x_1$ and $y_2=x_2$, and show the above is not satisfied in general for your joint cdf.

Perhaps even simpler; any joint cdf must satisfy
$$
\lim_{x\to-\infty} F(x,y)=0
$$
for any fixed $y$. But $\lim_{x\to-\infty}\max(F(x),F(y))=F(y)$.

We already know that
$$
\lim_{x\to \infty} F(x)=1.
$$
The thing to note is that to be a joint CDF, we must have
$$
\lim_{x\to\infty, y\to \infty}F(x,y)=1
$$

Here is the proof: we have
$$
\lim_{x\to\infty, y\to \infty}F(x,y)
=\lim_{x\to\infty, y\to \infty}\max(F(x),F(y))\geq\lim_{x\to \infty} F(x)=1,
$$
since $\max(F(x),F(y))\geq F(x)$.

But we also have $F(x)\leq 1$ for all $x$, so $\max(F(x),F(y))\leq 1$.

This means $\lim_{x\to\infty, y\to \infty}F(x,y)=1$.

There are other details to consider. $F(x,y)$ need to be increasing with $x$ and $y$. So, let's suppose that for some $x_1<x_2$, $F(x_1,y)>F(x_2,y)$. A moment's thought could lead to that we must have $F(x_1)\leq F(y)$ and $F(x_2)\geq F(y)$, a contradiction. (I have omitted the details) Actually, in general, the "max" of two increasing functions is increasing.

The conclusion is that $F(x,y)$ is a joint CDF. The same is, however, not true for PDF.


Evaluate the limit or show that is does not exist $$\lim_{x \to 0}\frac{1-\cos(3x)}{2x^2}$$

So far I have noted

$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)}$$

and then used the identity $$\sin^2x = 1-\cos^2x$$ to reduce this to

$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
Any tips on how to proceed?

Given: $L=\lim_{x \to 0} \frac{1-\cos(3x)}{2x^2}$

$\implies L=\lim_{x \to 0} \frac{1-\cos(3x)}{(3x)^2} \times \frac{(3x)^2}{2x^2}$

$\implies L=\lim_{x \to 0} \frac{1}{2} \times \frac{9x^2}{2x^2}$   [Using $\lim_{x \to 0} \frac{1-\cos(x)}{x^2}=\frac{1}{2}$]

$\implies \boxed{L= \frac{9}{4}}$

EDIT: I have changed my solution based on the changes in your question.

You would have  have it here as follows if your algebra were correct.
$$\frac{\sin^2(9x)}{2x^2+\cos(9x)}.$$
As $x\to 0$, $\sin^2(9x) \to 0$ and $2x^2 + \cos(9x)\to 1.$
The limit of the ratio is zero.

However, the algebra leading up to your expression is flawed. I made the mistake
of not checking your preceding calculations.

$$\frac{1-\cos(3x)}{2x^2} = \frac{1-\cos(3x)}{2x^2} \cdot \frac{1+\cos(3x)}{1+\cos(3x)} ={\sin^2(3x)\over 2x^2(1 + \cos(3x))} \sim {\sin^2(3x)\over 4x^2}\to {9\over 4} $$
as $x\to 0.$


Understanding the translation of a graph (horizontally) I have been having trouble understanding the translation of a graph. I understand the 'rule in the sense that the '$+$' shifts to the left and the '$-$' to the right when dealing with something like e.g. $f(x + 2) = f(x)$.

In the book I am using it has written that if $g(x) = f(x-c)$, where $c > 0$ then the value of $g$ at $x$ is the same as the value of $f$ at $x-c$.

So ...if $c = 2$ and my function for $g$ was $g(x) = x^2$ and I put in $x=3$ I'd get $9$. For $f(x-c)$ I don't quite understand this part. Am I wanting to put in the same $x$ (i.e. $x = 3$) which would give me $f(3-1) = f(1) = g(3) = 9$. When I do this it seems to work against how I thought it would go. To me this implies that in $g(x)$ when I input $3$ for $x$ I'd get a $y$ value of $9$. Then when I input $x=3$ in the function for '$f$' I'd end up with $f(1)$ provides me with $9$. It seems $g(1)$ is going back $2$ spaces to the left when compared to $f(3)$ in order to get the answer $9$ (based on $3-1 =2
$, from within the functions respective brackets)

Clearly I misunderstand something here.....

I personally think of it like this. Let's use the graphs of $f(x) = \sqrt{x}$ and $g(x) = \sqrt{x - 3}$.

In this picture, $f(x) = \sqrt{x}$ is in red. Now, when we look at $g(x) = \sqrt{x+3} = f(x+3)$, it's getting a head start, so it hits all the same $y$-values, but earlier than regular $\sqrt{x}$.

So, while $\sqrt{x} = 1$ when $x = 1$, the function $\sqrt{x+3} = 1$ when $x = -2$, which is $3$ units earlier than $f(x)$.

And this keeps happening! The function $f(x)$ is always $3$ units behind $g(x) = f(x + 3)$, if you think of the $x$-axis as time; the further to the left $g(x)$ is, the more of a head start it's got. So while $f(x) = 1$ when $x = 1$, the function $g(x)$ has already raced ahead, and $g(x) = 2$ when $x = 1$. It takes $f(x)$ an extra $3$ units to hit $f(x) = 2$, when $x = 4$.

That's what helps me understand the counter-intuitive horizontal shifts, anyway: it's going to take $f(x)$ longer than $f(x + 3)$ to reach the same spot in their journey. Being to the right on the $x$-axis means its taking longer (higher $x$-values) to get to the same spot.

Original function: $g(x)= F(x-h)$ -used for horizontal translations only

If a positive number were used or plugged in for "$h$", then essentially it would become negative thus "$+$" equals left, to the negatives side or left. Similarly, if a negative number was used, then the two negatives would become a positive, so the graph would move to the positive side or right. This is why horizontal translations are counteractive.


Rings with noncommutative addition I was wondering if "rings" with noncommutative addition are studied at all? Of course, if a ring $R$ has a $1$, then for all $a, b\in R$, $a+a+b+b=(1+1)a+(1+1)b=(1+1)(a+b)=(a+b)+(a+b)=a+b+a+b$, from which it follows from cancellation that $a+b=b+a$. Thus, rings with $1$ automatically must have commutative addition.

So, are there interesting, necessarily non-unital, "rings" with noncommutative addition? Of course, you can take any non-abelian group and make such a "ring" by having all multiplications give the additive identity. Are there more interesting examples?

Vectors on a sphere? Their addition is essentially multiplication of rotations, which is non-commutative.


total variation and monotonicy I am stuck with this problem : I have a finite variation function $f$ and I have proved that the total variation on the interval $[0,t]$, denoted with $S_t^f$ for $t\in[0,T]$, is increasing. How can I prove that the function $S_t^f+f(t)$ is increasing too ?

Thank you very much for any help you can provide :)

Well you are not getting much help here and perhaps the homework deadline is looming--so here is a push.

The function $f:[0,T]\to \mathbb{R}$   has bounded variation there. You already have seen, for any $0\leq a<b\leq T$, that $$V(f,[a,b]) + f(b)-f(a)\geq 0$$ is trivial.

What you need to show now is that $$V(f,[0,a]) + V(f,[a,b])= V(f,[0,b]).$$
As is often the case, to show equality you show inequality--two of them.  That is you have two inequalities to establish:
$$V(f,[0,a]) + V(f,[a,b]) \leq V(f,[0,b])$$ and
$$V(f,[0,a]) + V(f,[a,b]) \geq V(f,[0,b]).$$
One is easy and the other will require you to get your hands dirty with partitions and the like.  (Think of a partition of $[0,b]$ and what would happen if you add "$a$" to it.)

[Note that I am delaying the notation $S^f_t=V(f,[0,t])$ for later since it is not very intuitive and can easily mislead you.]

Alternatively, you can just search around for textbooks that give a proof of Jordan's theorem for BV functions, that characterizes them as differences of monotone functions (which is what you are doing here).  Some use exactly this method; Royden's Real Analysis uses the clever idea of defining two half-variations (a positive variation and a negative variation) that ends up in the same place.  My favorite presentation of this topic is in Zygmund and Wheeden, Measure and Integral.  The techniques needed to solve this problem are, while elementary, absolutely essential to an ability to understand and do real analysis.  It is not a waste of time to linger on this problem until it is all completely transparent to you.

For any fixed partition $a=x_0\leq\dots\leq x_n =b$ we have
$$
  S^f(b) -S^f(a) + f(b) - f(a) \geq \sum_{i = 0}^{n-1} |f(x_{i+1}) - f(x_i)| + \sum_{i = 0}^{n-1} (f(x_{i+1}) - f(x_i)) \geq 0
$$
since $y + |y| \geq 0$ for all $y$. Hence, when you take $\sup$ over partitions, the inequality still holds.


Question regarding an analytic function and a meromorphic one Is it possible to have an analytic function on the unit disk $\mathbb{D}$ that has infinitely many isolated zeros? What is a good example? I guess then that would make this analytic function nontrivial, correct?

Also, what is an example of a meromorphic function on the complex plane with simple poles and points log$n$, for $n \geq 0$? All I know right now is probably that the principal part of this function would be of the form $$\frac{1}{z- log n}$$ , but I'm not so sure about that. Any guidance would be appreciated.

Better example for one of them would have been an infinite Blaschke product
$$B(z)= e^{i \theta} \prod_{n=1}^{\infty} \frac{z - a_n}{1 - \bar{a}_{n}z},$$
where $a_1,a_2,…$ are points in $\mathbb{C}$ (?) and $0 \leq \theta \leq 2 \pi$.


Proving that: $800 + n \log n + 600\sqrt n\log n = \Theta(n\log n)$ I am trying to prove that

  $$800 + n \log n + 600\sqrt n\log n = \Theta(n\log n)$$

  (where $\log$ is base $2$)

Basically so far, I've reduced this expression into an inequality:

$$0 \le c_1 \le \frac{800}{n\log n} + 1 + \frac{600}
{\sqrt n} \le c_2$$

From what I've been told, we only need to find ANY set of constants where this holds (where n goes to infinity).

ANY set of constants? Okay in that case: $n_0=5, c_2=400$ and $c_1=1$ works.

But, have I completed the proof or am I missing something? Are my constants valid or do they have to be more specific?

Thanks in advance.

The set $n_0=5$, $c_2=400$ and $c_1=1$, and the steps leading to it in the question, are perfect.

After a while, in many situations, one can even omit computing some explicit $(n_0,c_1,c_2)$, for example every time the ratio actually converges. For example, in your case, to note that $$\frac{800}{n\log n} + 1 + \frac{600}{\sqrt{n}}$$ converges to some positive limit (here, the limit is $1$) is enough to ensure the $\Theta(n\log n)$ property.

That appears (at a glance...) to be correct. But to complete the proof (at least in a first course!) you need to show that for all $n > n0$, your function's values lie between $c_1$ and $c_2$. That involves writing a proof. For this one, a little calculus would probably do the trick, but so would many other approaches.

IF you were writing a paper for a journal of computer science, you could probably just write down $n_0, c_1$, and $c_2$ and folks would look and think "yeah, that looks right." More likely, you'd just say "and because THIS is O(THAT), ..." and everyone would nod. But to get to that point, you have to prove you know what you're doing. :(

The remainder of your proof might look like this:

Pick $n_0 = 5$, $c_1 = 1$, $c_2 = 401$. (You'll see why I picked 401 a little lower down).

Then observe that for $n > n_0 = 5$, we have $n \log n > 1$, so $\frac{1}{n \log n} < 1$.
Further, for $n > n_0$, $\sqrt{n} > 2$, so $\frac{1}{\sqrt{n}} < 2$.
And finally, it's evident that $800 < 100 n \log n$, since $n > 5$, and $\log n > 1.6$.

Thus
\begin{align}
0 &\le 800 + n \log n + 600\sqrt{n} \log(n) \\
&\le 800 + n \log n + 600 \frac{n}{\sqrt{n}} \log n\\
&\le 800 + n \log n + 600 \frac{n}{2} \log n\\
&\le 800 + n \log n + 300 n \log n\\
&\le 800 + 301 n \log n\\
&\le 100 n \log n + 301 n \log n\\
&\le 401 n \log n\\
\end{align}

Your version of this proof would have put more things in the denominator -- it might be worth writing it out! -- but I find that it's easy, when including a division, to lose the sign-change in an inequality, so I prefer to avoid divisions when possible. It's  a matter of taste, I suppose (and experience with screwing things up!).


Show $u$ satisfy poisson equation Let $f\in \mathcal S(\mathbb R^n),n\geq 3$. How can I show that $$u(x)=C_n\int_{\mathbb R^n}|x-y|^{2-n}f(y)dy$$
where $C_n$ is a suitable constant solve Poisson equation ? i.e. $$\Delta u=f.$$

The big problem is to permute $\frac{\partial }{\partial x_i}$ and the integral, but also if I can, I still have the fact that $|x-y|^{2-n}$ is not differentiable (and also not defined) when $y=x$.

This is proved in Evans's PDE book, where the necessity of $n>2$ is also well explained.

Proof (Evans; I adapt.): The right-hand side of your formula provides a solution to $$\Delta u=f .$$ Because $f$ is Schwartz and

$$\int|x-y|^{2-n} \longrightarrow 0$$

outside large enough balls (note that this holds if $n>2$), we can see that the right hand side is in fact a bounded solution to $\Delta u=f$.

By uniqueness of bounded solutions (Liouville's Theorem) you can fix the constant so that $u$ is of the right hand side form.

About how you prove that RHS "is" indeed a solution, you have to prove that despite the discontinuity at $x=y$ the integral is continuously differentiable twice, and then calculate the Laplacian. You will then prove that it equals f. Fix an $x$, write the very definition of partial derivative as $h\longrightarrow 0$ of the difference quotient. In all other calculations, again work with fixed $x$.

Hope this helps. I know I did not present in the most clear way, but... Feel free to ask.


greates number of baskets to put equal number of fruits I am grade 6 student and trying this for the first time, so kindly excuse if I am making any mistakes.
I have this problem in my math book which I get a head around after lot of thinking, seeking help here
A grocery store clerk has to arrange 16 oranges, 20 apples and 24 pears in such a way that he puts equal number of Apples, oranges and Pears
in each basket. What is the greatest number of baskets that can be made so that no fruits is left?

This problem is in GCF/LCM chapter, but I am not sure what method will apply to solve this problem, please help.
Thanks in advance.

Hint:

You will need to find the GCF of the three numbers. That is the greatest number of baskets that the clerk can have and still have the same number of each fruit in each basket.

-

Solution:
The GCF of 16, 20, and 24 is 4. Each of the 4 baskets will have 4 oranges, 5 apples, and 6 pears.

The answer is 4 baskets.Look for the highest number that all these given numbers (16, 20, and 24) can be divided by. They are all divisible by 4. In each of the 4 baskets, there will be 4 oranges, 5 apples, and 6 pears (these are the results of each division).


k-nearest Neigbors kNN: Weights, Tie-Breaks, Majority Vote alternatives? I am programming a kNN algorithm and would like to know the following:

Tie-Breaks:

What happens if there is no clear winner in the majority voting? E.g. all k nearest neighbors are from different classes, or for k=4 there are 2 neighbors from class A and 2 neighbors from class B?
What happens if it is not possible to determine exactly k nearest neighbors because there are more neighbors which have the same distance? E.g. for the list of distances (x1;2), (x2;3.5), (x3;4.8), (x4;4.8), (x5;4.8), (x6;9.2) it would not be possible to determine the k=3 or k=4 nearest neighbors, because the 3rd to 5th neighbors all have same distance.

Weights:

I read it is good to weight the k-nearest neighbors before selecting the winning class. How does that work? I.e. how are the neighbors weighted and how is then the class determined?

Majority Vote alternatives:

Are there other rules/strategies to determine the winning class other than majority vote?

Regarding weights and vote alternatives, you're probably considering weights being 'uniform' in your kNN-classifier. Recall that they don't have to be! You can use other weighting-functions to make closer points having greater influence.


Is $\{w:\lim_{n \rightarrow \infty}S_n/n \in A\}$ in tail $\sigma$-field? Suppose $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Set $S_n:=\sum_{i=1}^{n}X_i$.

Suppose $\frac{1}{n}S_n \rightarrow Y \;a.s.$ for some real-valued random variable $Y$. How to show that for any set $A \subset R$
$$E:=\{w:Y \in A\} \in \mathcal{T}?$$
Here $\mathcal{T}$ represents the tail $\sigma$-field.

The trouble I meet:

If for any $t \in N^{+}$,
$$\{\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \in A\} = \{\lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \in A\},$$
we actually finished the proof.

My puzzle about above equation:
It's clear that the left hand side implies the right hand side. But I think the right hand side does not necessarily implies the left hand side. What if $X_j(w) = \infty$ for some $j < t$?

I didn't use the fact that $Y$ is a real-valued random variable. I think it may be the key. But How to use it?

Case 1: For all $n\in \mathbb{N}^+$, $X_n$ are real-valued random variables.

In this case, for all $n\in \mathbb{N}^+$ and for all $w\in \Omega$, $X_n(w)\neq \pm\infty$.

So, we have that, for any $A\subseteq \mathbb{R}$ and for any $t\in \mathbb{N}^+$

$$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i(w)}{n} \in A\right\}$$
and we can easily conclude that
$$\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\in \mathcal{T} \:\:\:\:\:\:\:\: (*)$$
where $\mathcal{T}$ is the tail $\sigma$-field.

Remark for case 1: On the other hand, since $\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \rightarrow Y \;a.s.$, we have that
$$P\left(\left\{w : \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}\:\Delta\:\{w:Y(w) \in A\}\right)=0$$
But, this information, even combined with $(*)$, is NOT enough to prove that that
$\{w:Y(w) \in A\} \in \mathcal{T}$.

Case 2: The random variables $X_n$ may have infinity values ($\pm\infty$).

In this case, the result is false, as shown in the following counterexemple.

Let, for each $n\in \mathbb{N}^+$, let $\Omega_n=\{0,1\}$, $\mathcal{F}_n=2^{\Omega_n}$ and $P_n$ the probability defined on $\mathcal{F}_n$ by $P_n({0})=1$, $P_n({1})=0$.

Let $\Omega=\prod_{n=1}^{+\infty}\Omega_n$, $\mathcal{F}=\prod_{n=1}^{+\infty} \mathcal{F}_n$ and $P=\prod_{n=1}^{+\infty} P_n$. For each $n\in \mathbb{N}^+$, $\pi_n: \Omega \to \Omega_n$ is the projection on the $n$th factor.

Let us,  for each $n\in \mathbb{N}^+$, define the random variable $X_n$ by:
$$
X_n(w)= \left\{\begin{aligned} &0 &\textrm{ if } \pi_n(w)=0; \\ &+\infty  &\textrm{ if }  \pi_n(w)=1  \end{aligned} \right.
$$
It is easy to see that $\{X_n\}$ are independent random variables on $(\Omega, \mathcal{F},P)$. Let $Y$ be a random variable defined on $(\Omega, \mathcal{F},P)$ by
$$
Y(w)= \left\{\begin{aligned} &0 &\textrm{ if, for all } n\in \mathbb{N}^+, \pi_n(w)=0; \\ &1  &\textrm{ if there is }  n\in \mathbb{N}^+ \textrm{ s.t. }  \pi_n(w)=1 \end{aligned} \right.
$$
Now, let $$E= \left\{w : \textrm{ for all } n\in \mathbb{N}^+, \pi_n(w)=0 \right\}$$
It is easy to see that
$$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} = Y(w)\right\}=E$$
and
$$P(E)=1$$
So $\frac{\sum_{i=1}^{n}X_i}{n}$ converges $Y$ a.s.

Now take $A=\{0\}$, we have
$$\left\{w :\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i(w)}{n} \in A\right\}=\left\{w : Y(w) \in A \right\}=E$$

And, it is easy to see that $E\notin \mathcal{T}$.

If the random variables take values in $(\mathbb R, \mathscr B(\mathbb R))$ rather than $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$, then none of the random variables or $Y$ can ever be infinite. I ran into a similar problem.

If the random variables indeed take values in $(\overline{\mathbb R}, \mathscr B(\overline{\mathbb R}))$...

Say for example $t=2$ and $X_1(\omega) = \infty$, then...I still don't see what's the problem. Actually, we know that if indeed

$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} \ \text{a.s.}$$

then $\forall t \in \mathbb N$,

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \lim_{n \rightarrow \infty}\frac{\sum_{i=t}^{n}X_i}{n} \ \text{a.s.}$$

This is because the limit of a sequence of numbers or functions doesn't depend on any finite subset of terms. Hence, the equality in question follows from the aforementioned.

So yeah, assuming the limit indeed exists, I don't think it should matter w/c of the random variables are infinite. Are you thinking something like:

$$Y = \lim_{n \rightarrow \infty}\frac{\sum_{i=1}^{n}X_i}{n} = \infty \ \text{a.s.}$$

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} < \infty \ \text{a.s.}$$

?

That would contradict the above fact. If for example $Y = X_1 = \infty \ \text{a.s.}$, then it indeed follows that

$$\lim_{n \rightarrow \infty}\frac{\sum_{i=2}^{n}X_i}{n} = \infty \ \text{a.s.}$$

Another thing to note is almost surely. Perhaps the equality in question actually holds only almost surely.


Solving for $y_3$ in a complicated expression I need to solve this equation for $y_3$:

$$y_3=z_1+(z_1+z_2)\sin\left(\arctan\left(\frac{y_3-z_1}{\sqrt{(z_1+z_2)^2–(y_3-z_1)^2}}\right)\right)$$

Is there a solution to this equation in terms of $z_1$ and $z_2$?

Consider the function
$$
f(x)=z_1+(z_1+z_2)\sin\left(\arctan\frac{x-z_1}{\sqrt{(z_1+z_2)^2-(x-z_1)^2}}\right)
$$
which is defined for $x\in I=(z_1-|z_1+z_2|,z_1+|z_1+z_2|)$. Calculating the derivative we see immediately that $f'(x)=1$ for every $x\in I$. So $f(x)=x+c$ for some constant $c$. But
$f(z_1)=z_1$, thus $f(x)=x$ for every $x\in I$. Thus the interval
$$I=(z_1-|z_1+z_2|,z_1+|z_1+z_2|)$$
represents the set of solutions to this equation.$\qquad\square$

Let $\displaystyle\arcsin\frac bc=\theta$

$\displaystyle\implies\sin\theta=\frac bc$ and $\displaystyle-\frac\pi2\le\theta\le\frac\pi2$

$\displaystyle\implies\cos\theta\ge0\implies\cos\theta=+\frac{\sqrt{c^2-b^2}}c$

$\displaystyle\implies\tan\theta=+\frac b{\sqrt{c^2-b^2}}\implies\arctan \frac b{\sqrt{c^2-b^2}}=\theta=\arcsin\frac bc$

Here $b=y_3-z_1$ and $c=z_1+z_2$


Block Matrix Zero Determinant Implication? Recently I've been working with a number of square (order of 2n) matrices whose determinants are zero. That is,

$$\det\begin{bmatrix}A&B\\C &D \end{bmatrix} = 0$$
where each of A,B,C, and D are also matrices.

I couldn't help but notice that in all of my cases, det(A) = det(D) = 0 as well. This is fairly surprising to me, and I'm curious: what is the limit of validity of this special case? Alternatively, under what conditions is this true?

If it matters, I'm working with passive electrical transmission networks, which typically never have zeroes on the main diagonal.

Thanks!

If $A$ and $D$ are of the same size, and $\det A = \det B = \det C = \det D = 0$ then of course $\det \begin{bmatrix}A&B\\C &D \end{bmatrix} = 0$.

But the converse does not hold:  It is easy to produce square matrices of determinant zero none of whose blocks have determinant zero.  An easy example is
$$A= \begin{bmatrix}4I&2I\\2I &I \end{bmatrix}
$$
where $A$ is $2n \times 2n$ and $I$ is an $n \times n$ unit matrix.

For instance, look at
$$A= \begin{bmatrix}4&0&2&0\\0&4&0&2\\2&0&1&0\\0&2&0&1 \end{bmatrix}
$$

The determinant of your matrix (or any such matrix in block diagonal form) is equal to the product of the determinants of the diagonal blocks, i.e. $\operatorname{det}(A)\operatorname{det}(D)$. If the determinant of the block diagonal matrix is $0$, this means $\operatorname{det}(A)\operatorname{det}(D)=0$ so that either $\operatorname{det}(A)=0$ or $\operatorname{det}(D)=0$, but not necessarily both.