$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$ I've been going through some series notes from my lecture and got stuck at this equality: $$\left(\sum_{j=0}^\infty\frac{z^j}{j!}\right)\left(\sum_{k=0}^\infty\frac{w^k}{k!}\right)=\sum_{n=0}^\infty\sum_{j=0}^n\frac{z^jw^{n-j}}{j!(n-j)!}$$
Where $z,w\in\mathbb C$.

Rather than a proof, I'm looking more for a way of understanding this equality, so next time I see something similar I go "oh, right, I know that". Right now I only have vague ideas of why that would be true.

Thanks a bunch for any help!

This is an example of the change of order of summation (regrouping terms). Formally (without concern for convergence), the product
$$
\begin{align}
\left(\sum_{i=0}^\infty a_i\right)\left(\sum_{j=0}^\infty b_j\right)
&=\sum_{i=0}^\infty\sum_{j=0}^\infty a_ib_j\tag{1}\\
&=\sum_{k=0}^\infty\sum_{j=0}^k a_{k-j}b_j\tag{2}
\end{align}
$$
$(1)$ is the distributing multiplication over addition.

$(2)$ is a change of variables: $i+j=k$

Each product in $(1)$ appears once and once only in $(2)$.

Given the series can be written

$=e^z e^w =e^{z+w}=\sum_{n=0}^\infty =\sum \sum....$


sum of reciprocals of derivative of polynomial at its roots If $P(x)$ is a polynomial of degree $n > 1$ with only simple roots $a_1,\ldots,a_n$, is it true that $\frac 1{P'(a_1)} + \cdots + \frac 1{P'(a_n)} = 0$, and, if so, what is the proof?  I can see this directly for $n = 2,3,4$ with some brute force for $4$.

The partial fraction decomposition yields

$$\frac{1}{P(x)}= \sum_{i=1}^n \frac{c_i}{x-a_i}  $$

Multiplying by $x-a_j$ yields

$$\frac{x-a_j}{P(x)}= \sum_{i=1}^n \frac{c_i(x-a_j)}{x-a_i}  (*)$$

from where $c_j =\frac{1}{P'(a_j)}$.  [ED: $P(x)=(x-a_j)Q(x)$, sub it back up and note that $P'(a_j)=Q(a_j)$]

Hence

$$1=\sum_{i=1}^n \frac{1}{P'(a_i)} \frac{(x-a_1)\cdots(x-a_n)}{x-a_i}$$

The coefficient of $x^{n-1}$ is

$$0= \sum_{i=1}^n \frac{1}{P'(a_i)}$$

Here is a link to a youtube video which I have made proving this result using complex analysis and the residue theorem:

https://www.youtube.com/watch?v=m_LWdPl6npw


Prove that $\log_a(1/x)=-\log_a(x)$. I thought to write $$\log_a(1/x)=\log_a(x^{-1})=-\log_a(x).$$ But it has two problems: when x=0 and on the other problem it doesn't mention any condition. How should I solve it in each of them?

Exclusively by definition:

$$y=\log_a\frac1x\iff a^y=\frac1x=x^{-1}\iff a^{-y}=x\iff -y=\log_ax$$

log_x(x^-1)=-1=-log_x(x). log_a=log_x/log_x(a) is a multiple of log_x so behaves similarly.


How do I differentiate $f(x)=\log(x)$ in $\mathbb{Q}_2$? How do I differentiate these functions in $\mathbb{Q}_2$?

$f(x)=x^2$

$f(x)=1/x$

$f(x)=\log(x)$

I've been struggling with this for a couple of days now.

Please bear in mind I'm working without a teacher here so no laughing...

$\displaystyle f(x)=x^2: f'(x)=\lim_{\lvert a\rvert_2\to0}\frac{x^2-(x-a)^2}{a}=\lim_{\lvert a\rvert_2\to0}\frac{a^2+2ax}{a}=2x$

$f(x)=1/x:\displaystyle f'(x)=\lim_{\lvert a\rvert_2\to0}\frac{\frac{1}{x}-\frac{1}{x-a}}{a}=\lim_{\lvert a\rvert_2\to0}\frac{-1}{x(x-a)}=-x^{-2}$

But I think we would need to define $\displaystyle f'(x)=0$ for $x=0$?

As for $f(x)=\log(x)$ I have little idea where to start. Other than knowing that $\log$ is defined in $\mathbb{Q}_2$

Let's follow Wikipedia and define the 2-adic (Iwasawa) logarithm on $\Bbb{Q}_2^*$ as follows. Recall that the group $\Bbb{Q}_2^*$ is a direct product
$$
\Bbb{Q}_2^*=\langle 2\rangle\times(1+2\Bbb{Z}_2).
$$
The usual Taylor series of $\log(1+x)$ familiar from calculus
$$
x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n
$$
converges w.r.t. the 2-adic topology whenever $x\in2\Bbb{Z}_2$. This is because for all $n$ we have $\nu_2(x^n)=n\nu_2(x)\ge n$ but we easily see that $\nu_2(n)\le n/2$. Therefore the general term $x^n/n\to0$ which is sufficient for convergence in a complete non-archimedean case. We can thus use this series and define
$$
\log(1+x)=\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}n
$$
whenever $x\in 2\Bbb{Z}_2$. The familiar power series identities then show that the rule $$\log(xy)=\log x+\log y\qquad(*)$$ holds whenever $x,y\in1+2\Bbb{Z}_2$.
We can extend the definition of the logarithm to all of $\Bbb{Q}_2$ (actually even further, see WP) in such a way that $(*)$ holds by declaring that $\log(2)=0$, and, as a consequence of $(*)$ that
$$
\log(2^n(1+x))=\log(1+x)=x-\frac{x^2}2+\cdots
$$
for all integers $n$ and all $x\in 2\Bbb{Z}_2$.

The usual business of termwise differentiation of a power series carries over to the 2-adics. Therefore when $x,a\in1+2\Bbb{Z}_2$ we get
$$
\lim_{x\to a}\frac{\log x-\log a}{x-a}=1-(a-1)+(a-1)^2-(a-1)^3+\cdots=\frac1a
$$
because the geometric series converges as $a-1\in2\Bbb{Z}_2$.

As we took care to have $(*)$, the usual calculation of the real limit goes thru, and the result $\log'(x)=1/x$ holds everywhere.

Hint

$$f(x)=\log(x)\implies f'(x)=\lim_{ a\to 0}\frac{\log(a+x)-\log(x)}{a}=\lim_{ a\to 0}\frac{\log\left(1+\frac a x\right)}{a}$$ Now, using equivalents $$\log\left(1+\frac a x\right)\sim  \frac a x$$


The proof of $\sqrt{2}$ is not rational number via fundamental theorem of arithmetic. I assume that $\sqrt{2}$ is positive number satisfies $(\sqrt{2})^2=2$.

proof) Let $m$, $n$ as natural number,$\ $ $M$ is the number of prime factor of $m$,$\ $ $N$ is also the number of prime factor of $n$. For example, $m=12=2^2\cdot3$, $M$ is $3$.

Then, if $\sqrt{2}$ were rational number, it could be expressed as a fraction $\frac{m}{n}$ in lowest terms.

If $\sqrt{2}=\frac{m}{n}$, $m^2=2\cdot n^2$. Then, $m^2$ has $2M$ prime factor, $n^2$ has $2N$ prime factor. LHS has even prime factor, RHS has odd prime factor.  This is a contradiction to fundamental theorem of arithmetic.

Therefore the initial assumption—that $\sqrt{2}$ can be expressed as a fraction—must be false.

Is there any problems? If not, I think this proof is simple and easier than ordinary proof -contradiction to the property of lowest terms.

p.s. This proof may apply to any root of $n$-th power $\sqrt[n]{a}$,  $n\in\mathbb{N}$ iff $a$ is prime.

It looks OK, apart from one little typo:

$\sqrt{2}=\frac{n}{m}\implies 2m^2=n^2$, you got it the other way around.

The step Bill Dubuque refers to can be made more explicit with this argument:

If P(x) is the prime factorization of x then by definition P(x)P(x) is equal to xx. The combination of the sets P(x) and P(x) [allowing duplication of elements] is when multiplied out, by FTA, the prime factorization of only one number, which because it is P(x)P(x) can only be xx. Consequently if M(P(x)) is the size of P(x), M(P(xx)) = M(P(x)) + M(P(x)).


Discrete Probability: Four dice are thrown, what's the probability that... Four dice are thrown, what's the probability that:

a) None of them fall higher than three?

b) None of them fall higher than four?

c) That four is the highest number thrown?

So for a, I wanna think about the denominator first. There are 6 possible outcomes for each dice, and we have four dice. So we're technically picking r out of n objects, aka picking 4 possible values out of 6. So the denominator/sample space should be $6^4$ right?

Now this is where I'm getting tripped up. Order technically shouldn't count, because we only care about the quadruplets that don't have a value higher than 3. I might be wrong, maybe I am, but could someone explain a bit more? My professor stated that typically in the sample space/denominator, we want order to count.

So I'm trying to think of the numerator now, so we want to find the probability that no values appear higher than 3, so some events that can occur are: $(1, 1, 2, 3), (1, 2, 2, 3), (1, 1, 1, 1)$ etc. Again, I can't see why order should count here, because if we only are concerned about what appears rather than how they appear, then we can say $(1, 1, 2, 3)$ is equal to $(1, 2, 1, 3)$? So if order does not count, and we have replacement, then does this lead the case where we use $$\binom{n-1+r}{r}$$to find out the probability?

I'm pretty sure if I can do a), I could prob do b), but if someone could maybe lead me in the right direction for c) I would appreciate that too!

a) The probability that a die does not fall higher than $3$ is given by
$$P(X\le3)=\frac{3}{6}=\frac{1}{2}$$
So as each event is independent we can find the probability that no die falls higher than three by raising this result to the power of $4$
$$(P(X\le3))^4=\frac{1}{2^4}=\frac{1}{16}$$

b) Similarly the probability that a die does not fall higher than $4$ is given by
$$P(X\le4)=\frac{4}{6}=\frac{2}{3}$$
So the final probability is
$$(P(X\le4))^4=\frac{2^4}{3^4}=\frac{16}{81}$$

c) If $4$ is the highest rolled value then every die rolled has a value $\le 4$. But we also need at least one $4$ to be rolled - so we need to subtract the probability of rolling every dice $\le3$. The answer is then
$$(P(X\le4))^4-(P(X\le3))^4=\frac{16}{81}-\frac{1}{16}=\frac{175}{1296}$$
which is the difference of the answers from a) and b).

$a)$
The probability that one dice rolls a number equal to $3$ or lower is $\frac{3}{6}=\frac{1}{2}$.
Hence, the probability that all of them roll $3$ or lower is $(\frac{1}{2})^4=\frac{1}{16}$.

Can you do the same for $b)$?

$c)$
Firstly, There is a probability of $(\frac{4}{6})^4=(\frac{256}{1296})$ that all numbers rolled are $4$ or lower.

However, at least one of the numbers must equal $4$, so we subtract the probability that they are all 3 or lower. We already calculaties this in $a)$.

Hence, the probability that the highest number rolled is $4$, equals
$$\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$


Prove if $n^2$ is even, then $n$ is even. I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

It could use a little more explanation, but yes, it works. I’d expand it to point out explicitly why $n(n+1)$ is even and that $n=(n^2+n)-n^2$ is then the difference of two even numbers and as such is even (assuming that you already have this fact available to use).

An alternative approach is to show that if $n$ is odd, then $n^2$ is odd; the desired result is the contrapositive and therefore follows at once.

You have a true conclusion "if n$^2$ is even, then n is even" and thus you automatically have a valid argument, since every argument with a true conclusion is valid.

That said, it might come as interesting to look at something more general.  Let z indicate any positive natural number, and n any integer.  We'll prove that if n$^z$ is even, then n is even.

"If n$^z$ is even, then n is even" is logically equivalent to "if n is not even, then n$^z$ is not even."  All integers either are even (equal to a number of the form 2k) or odd (equal to a number of the form (2k+1)).  So, "if n is not even, then n$^z$ is not even." is logically equivalent to "if n is odd, then n$^z$" is odd."  It follows that, "If n$^z$ is even, then n is even" is logically equivalent to "if n is odd, then n$^z$" is odd."  Consequently, if we can prove that "if n is odd, then n$^z$" is odd.", it will follow that "if n$^z$ is even, then n is even."

If n is odd, then it can get written in the form (2k+1).  Since z is a positive natural number, we can use the binomial theorem on (2k+1)$^z$ to see if it's odd or even.  The last term of the expansion of (2k+1)$^z$ is 1$^z$ which equals 1.  So, is the sum of the other terms even?  The other terms of the sum go (2k)$^z$, (2k)$^z$$^-$$^1$a, ..., (2k)y.  Thus, for the sum of those terms, we can factor out 2 using the distributive property.  Consequently, the sum of those terms equals 2x where x is some integer, and thus that sum is even.  Since 1 is the last term, it follows that (2k+1)$^z$ equals (2x+1) where x is some integer.  But, that means that (2k+1)$^z$ is odd.  So, if n is odd, then n$^z$ is odd.

Since "if n is odd, then n$^z$ is odd." is logically equivalent to "if n$^z$ is even, then n is even." it logically follows that "if n$^z$ is even, then n is even" is true.

Since "if n$^z$ is even, then n is even" is true, by letting z=2, it follows that "if n$^2$ is even, then n is even." is true also.


Find the remainder in the following case Find the remainder when $19^{729}$ when divided by 107?

My approach

$\frac{19^{723}}{107}$=$\frac{19^{636}}{107}$*$\frac{19^{87}}{107}$

by Euler theorem, we know that remainder of $\frac{19^{636}}{107}$ =1

so the whole problem turned into the following

$\frac{19^{87}}{107}$

now

19 divided by 107 leaves remainder 19

19*19 divided by 107 leaves remainder 40

19*19*19 divided by 107 leaves remainder 11

19*19*19*19 divided by 107 leaves remainder 102

so how to approach this question??


Why is this derivative not undefined at a given point? I'm working on a problem from Keisler's Calculus (not homework, for my own amusement.) One of the problems is confusing me a bit. The first part goes like this:

Suppose $g(x)$ is differentiable at $x = c$ and $f(x)=|g(x)|$. Show that:
$$f'(c)=
\begin{cases}
g'(c) & \text{if }g(c)>0  \\
-g'(c) & \text{if }g(c)<0
\end{cases} $$

Easy enough:
$$\frac{d(f(x))}{dx}=\frac{d(|g(x)|)}{dx}$$
$$f'(x)=\frac{d(\sqrt{g(x)^2})}{dx}$$
$$f'(x)=\frac{g(x)g'(x)}{|g(x)|}$$
Plugging $c \text{ and } g(c)$ into the above yields the desired result. The second part of the problem is where I get confused. It goes like so:

Prove that $f'(c)=0 \text{ if } g(c)=0 \text{ and } g'(c)=0$.

Prove that $f'(c)$ does not exist if $g(c) = 0 \text{ and } g'(c) \ne 0$.

Shouldn't the derivative not exist in either case since division by zero occurs in both? What am I misunderstanding here?

If $g(c) = g'(c) = 0$, we have $\lim_{x \to c} { g(x) \over x-c} = 0$ and
$\lim_{x \to c} { -g(x) \over x-c} = 0$, so by continuity of the $\max$ function, we have $\lim_{x \to c} { \max({|g(x)| \over x-c},-{|g(x)| \over x-c} )} = \lim_{x \to c} { |g(x)| \over x-c} = 0$.

If $g(c) =0$, and $ g'(c) \neq 0$, first show that in some neighbourhood of $c$, we have $g(x) \neq 0$ for $x \neq c$, and that $g$ has one  sign in $(c-\epsilon,c)$ and the opposite sign in $(c, c+ \epsilon)$.
Then show that $\lim_{x \uparrow c} { f(x) \over x-c} = 0$ and
$\lim_{x \downarrow c} { f(x) \over x-c}$ are different.

for the first proof:

let g(c)=0 and g'(c)=0.

There are three cases:

case 1:
   0 is a local maximum for g(x) (g(x) lies below the x-axis)

case 2:
   0 is a local minimum for g(x) (g(x) lies above the x-axis)

case 3:
   0 is neither a local minimum nor a local maximum for g(x), which means (which is like the graph of y=x^3 near x=0)

Analyze case 1:

0 is a local maximum for g(x)
   -->for some g on the interval [a,b], g<=0
   -->f(x)=abs(g(x)) on the interval [a,b] is equal to -g(x) on that interval
   if the derivative g(x) is called g'(x) on [a,b], then the derivative of -g(x)=f(x) with respect to x is -g'(x) by chain rule.
   -->if g'(c)=0, then f'(c)=-g'(c)=0
QED for case 1

Case 2 is basically the same proof

Case 3:

 if 0 is neither a maximum nor a minimum on on g(x) at x=c,
 then we can conclude that there exists an interval

   [a,c] for which g<=0

   [c, b] for which g>=0

   (there are actually two cases. this one is just for g as an increasing function around x=c)

   on [a,c] where g<=0, f(x)=-g(x). f'(x)=-g'(x). you need to prove that the lim (x-->c-) f'(x)=0

   on [c,b] where g>=0, you need to prove that lim(x-->c+)f'(x)=0

   therefore, lim(x-->c)f'(x)=0

   I think you have to prove that f(x) is continuous at x=c. then it gives you that f'(c)=0

=======

Hope this helps. Good luck! :)


Show that recursive sequence is decreasing I'm required to show that the above series is decreasing. However, I encounter a problem when I realize that in the inductive step, I have  a term for a(n) in both the numerator and denominator, which makes it difficult to show that a(n+1) > a(n+2). Any Help would be appreciated.

You want to show that $a_{n+1}<a_n$ for all $n\geq1$. Writing this out yields
$$\frac{a_n}{2}+\frac{1}{a_n}<a_n,$$
and subtracting $\frac{a_n}{2}$ from both sides leaves us with
$$\frac{1}{a_n}<\frac{a_n}{2}.$$
This is equivalent to $a_n^2>2$. Can you prove this by induction?

Here’s a hint. Consider when the sequence would not decrease, namely where it would stay constant. To do this, solve:

$$x=\frac{x}{2}+\frac{1}{x}$$
$$x^2=2 \implies x=\{-\sqrt{2},\sqrt{2}\}$$

Since this sequence is always positive, the positive solution seems most relevant. As matter of fact, whenever $ a_n> \sqrt{2}$, $a_{n+1}$ decreases. I’ll leave it to you to rigorously prove the above claim, and show that $x>\sqrt{2} \implies \frac{x}{2}+\frac{1}{x}>\sqrt{2}$


How to evaluate $\sqrt{5+2\sqrt{6}}$ + $\sqrt{8-2\sqrt{15}}$? My exams are approaching fast and I found this question in one of the unsolved sample papers.
I tried squaring the whole term but couldn't work out the answer.

I am a ninth grader so please try to explain in simple terms.

HINT:

What is $\displaystyle (\sqrt{2} + \sqrt{3})^2 $ and $\displaystyle (\sqrt{5} - \sqrt{3})^2 $?

The basic 'trick', so to say, behind such questions is to identify that the surd can be expressed as a square. For example, consider your first surd $\sqrt{5+2\sqrt{6}}$. Here, there's a $2\cdot\sqrt{\text{something}}$. Now, if you see, the factors of that $\sqrt{\text{something}}$ are $\sqrt2$ and $\sqrt3$. A quick check shows that the sum of their squares give $5$. So, this is of the form $a^2 + b^2 + 2ab = (a+b)^2$

It's worth noting that $(\sqrt a \pm \sqrt b)^2 = (a+b) \pm 2\sqrt{ab}$.
So, if the answer to $\sqrt{M \pm 2\sqrt N}$ is going to be of the form $
\sqrt a \pm \sqrt b$ then we need $N=ab$ and $M = a+b$.
$\sqrt{5+2\sqrt{6}}$
Note that $6 = 2 \times 3$ and $5 = 2 + 3$.
Hence $$(\sqrt 3 + \sqrt 2)^2 = 5+2 \sqrt 6$$

$\sqrt{8-2\sqrt{15}}$
Note that $15 = 5 \times 3$ and $8 = 5 + 3$.
Hence $$(\sqrt 5 - \sqrt 3)^2 = 8 - 2 \sqrt{15}$$
\begin{align}
   \sqrt{5+2\sqrt{6}} + \sqrt{8-2\sqrt{15}}
   &= (\sqrt 3 + \sqrt 2) + (\sqrt 5 - \sqrt 3) \\
   &= \sqrt 5 + \sqrt 2
\end{align}

Added 1/18/2021
Another way of looking at $\sqrt{M + 2\sqrt N} = \sqrt a + \sqrt b$.
\begin{align}
   \sqrt{M + 2\sqrt N} &= \sqrt a + \sqrt b \\
   M + 2 \sqrt N &= (a+b) + 2\sqrt{ab} \\
\hline
   M &= a + b \\
   N &= ab \\
\hline
  x^2 - Mx + N &= (x-a)(x-b) \\
  \{a,b\} &= \dfrac{M \pm \sqrt{M^2-4N}}{2}
\end{align}
So if $\sqrt{5 + 2\sqrt 6} = \sqrt a + \sqrt b$
$$\{a,b\} = \left\{\dfrac{5 \pm \sqrt{5^2 - 4(6)}}{2}\right\} = \{3,2\}$$
$$x^2-5x+6 = (x-2)(x-3)$$
So
$$\sqrt{5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3 $$


what happens when in a combination, $n \choose k$, $n < k$? Is this 0? What happens when in a combination, $n \choose k$, $n < k$? Does this result in a zero, or is this undefined?

Not necessarily $0$. We have definition
$$\binom{n}{k} = \frac{n(n-1)...(n-k+1)}{k!}$$
So for example if we take $n = -3$ and $k = 2$, we have
$$\binom{-3}{2} = \frac{(-3)\cdot(-4)}{2!} = 6$$
If you don't believe, see https://www.wolframalpha.com/input/?i=C(-3,2). Selection of objects is just an analogy for natural number values of $n$ and $k$, I suggest you to use the definition instead, for a more general result.

We have $n \choose r$. (The common notation is $n$ and $r$)

\begin{align}
n \choose r & = \frac{n!}{(n-r)!r!}\\
\end{align}

$n \lt r \implies n-r < 0$. The factorial is not defined for negative numbers, so $n \choose r$ does not have an answer if $n < r$.


Show that $\binom{2n}{  n}$ is divisible by 2? Possible Duplicate:
  prove that $(2n)!/(n!)^2$ is even if $n$ is a positive integer

Show that $\binom{2n}{ n}$ is divisible by 2?

Any help would be appreciated..

There are several ways to show this. I gave this as a homework exercise once (after having given the theory for computing a binomial coefficient modulo two in terms of the binary expansions), and a student surprised me with
$$
{2n\choose n}={2n-1\choose n-1}+{2n-1\choose n}=2{2n-1\choose n-1}.
$$
IOW he used Pascal's triangle rule once and then the symmetry ${n\choose k}={n\choose n-k}$.

An algebraic approach:

$$ { 2n \choose n } = \frac{(2n)!}{n! \cdot n!} $$

$$ = \frac{ 1 \cdot 2 \cdot 3 \cdots n \cdot (n+1) \cdots (2n-1)\cdot (2n) }{n! \cdot n!}$$

$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [2 \cdot 4 \cdot 6 \cdots (2n)]}{n! \cdot n!}  $$

$$ = \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [(2.1) \cdot (2.2) \cdot (2.3) \cdots (2.n)]}{n! \cdot n!}  $$

$$= \frac{ [1 \cdot 3 \cdot 5 \cdots (2n-1)] \cdot [ 2^n \cdot (n)! ]}{n! \cdot n!}  $$

$$ = 2^n \frac{ 1 \cdot 3 \cdot 5 \cdots (2n-1)}{n!} $$


Just using $+$ ,$-$, $\times$ using any given n-1 integers, can we make a number divisible by n? (NO bracket allowed!!) Probably title is slightly ambiguous but I could not see any way of shortening the problem.
I am sure many of you have seen the problem like
$2$ $\square$ $2$ $\square$ $2$ $\square$ $2$ $\square$ $2$  $=$ $20$
something like this where between the numbers we can use some sort of operation depending on the question.
Well I think the question is motivated by this.
Let $n>2$. Given any set of $n-1$ numbers (we can swap the order) and we can only use addition, subtraction and multiplication, can we make a multiple of $n$ always?
Another way of putting it is:
Given $n-1$ elements of non-integral domain $\mathbb{Z}/n\mathbb{Z}$, can we make zero only using $+,-,\times$? Oh and by the way bracket is not allowed!!!!
I was asked this in an interview randomly and I could not figure it out and the interviewer did not know the answer either.

As people suggested I will post the 'proof' if bracket is allowed.
If bracket is allowed then this is easy. I will use the second form of the problem.
Firstly let $n>2$ and $a_1,...,a_{n-1} \in \mathbb{Z}/n\mathbb{Z}$
If any of the $a_i$ is equal to zero then we could just multiply all the $a_i$s so assume $a_i$ are all non-zero.
Also if bracket is allowed, if $a_i=a_j$, we could simply put $(a_i-a_j)$ and multiply all other elements together.
So we could assume $a_1,...,a_{n-1}$=1,...,n-1
In which case if bracket is allowed, $(1+(n-1))$ times all other elements would give the wanted answer.
But hey. NO Brackets allowed!
I have checked for small $n$s that this is possible and my intuition is that for big $n$ this should easily be possible. But I can't think of a way to show that.

Edit: If $n\gt 5$, it seems we can make everything equal $0$ just using $\{-,\times\}$. That is, it seems $+$ is not needed! (This was verified for $n=6,7,8,9,10,11,12$.)
But, I do not know how to prove this.

This is a long comment, as requested.

There are only $\binom{2n-3}{n-1}$ multisets $a_1,...,a_{n-1} \in \mathbb{Z}/n\mathbb{Z}$ to consider, for given $n$.
We have $a_1,...,a_{n-1}\ne 0$, because otherwise we can multiply everything with $0$.
I have written a simple brute force search in python. I observed:

It seems we only need to use $2$ operations at once ($+,\times$ or $-,\times$ or $+,-$).

It seems we can always assume $a_1\ge a_2\ge \dots \ge a_{n-1}$.

These two observations have been verified for all $n\le12$.

Below is my (disclaimer: inefficient) python code and its output.
For all possible (decreasing) mutisets of numbers:

It tries all $3\cdot 2^{n-2}$ ways to put two of three operations between numbers.

If solution isn't found, it tries it again but for all permutations of numbers.

If solution isn't found, it tries all $3^{n-2}$ ways to put three operations.

If solution isn't found, it tries it again but for all permutations of numbers.

If solution isn't found, it is a counter-example to the problem.

Code:
import itertools

def exprf(A,b,O):
    return ''.join([f'{A[_]}{O[b[_]]}' for _ in range(len(b))])+str(A[-1])

def solve(A,n,OP):
    k = len(OP[0])
    for o in range(k**(n-2)):
        b=[]
        while o!=0:
            o, r = divmod(o, k)
            b.append(r)
        while len(b)!=n-2:
            b.append(0)
        for op in OP:
            expr = exprf(A,b,op)
            if eval(expr)%n==0:
                return [expr, op]
    return False

for n in range(3,20):
    m = 0
    for a in itertools.combinations(range(1,2*(n-1)), n-1):
        m += 1
        A = [a[i]-i for i in range(n-2,-1,-1)]
        solved = solve(A, n, [['+','*'],['-','*'],['+','-']])
        if not solved:
            for P in itertools.permutations(A):
                solved = solve(P, n, [['+','*'],['-','*'],['+','-']])
                if solved:
                    break
            if not solved:
                solved = solve(A, n, [['*','+','-']])
                if not solved:
                    for P in itertools.permutations(A):
                        solved = solve(P, n, [['*','+','-']])
                        if solved:
                            break
                    if not solved:
                        print("COUNTER-EXAMPLE:", A)
                    else:
                      print(f"{n}: All operations and permutation needed:", solved[0])
                else:
                    print(f"{n}: All operations needed:", solved[0])
            else:
                print(f"{n}: Permutation needed: ", solved[0])
        else:
            pass #print(solved[0])
    print(f'{n}: {m} multisets solved.')

Output:
3: 3 multisets solved.
4: 10 multisets solved.
5: 35 multisets solved.
6: 126 multisets solved.
7: 462 multisets solved.
8: 1716 multisets solved.
9: 6435 multisets solved.
10: 24310 multisets solved.
11: 92378 multisets solved.
12: 352716 multisets solved.

Btw, if "$\ge$" is changed to "$\le$" in observation 2., then
10: Permutation needed:  1*1*1*1*1*4*8-1-1

is the only observed exception to the modified observation.

If $n=2k+1$, take
$$\sum_{i=1}^{n-1} i = \frac{n (n-1)}{2} = \frac{(2k+1) 2k}{2} = k (2k+1) = k n $$
which is divisible by $n$ .
If $n=2k$ and $n>4$, take
$$\prod_{i=1}^{n-1} i = 2k \prod_{i=1,i \neq 2,k}^{n-1} = n\prod_{i=1,i \neq 2,k}^{n-1} i $$
which is divisible by $n$ .
Finally, for $n=4$, take
$$ 3-2-1 = 0 $$
which is divisible by 4.


Original works of great mathematicians In almost every mathematical text there is a line as This was first proved by Gauss or This formula first appeared in a work of Riemann, but for me it's more like My friend told me once that...

For my Bachelor thesis and other papers I'm working on I would prefer to add a scan of the original paper rather that a quote from a book1 that quoted book2 that found it in a book3. I was already looking for some original papers from some great mathematicians (Riemann, Euler, Cantor, Hilbert etc...) on the internet, but I got quite disappointed by results. I expected some organization that collects scans from the old scientific works and makes them publicly available, but either I was looking elsewhere or it just doesn't exist...

So my question is:
Do you know about any place (website?) where scans of original work of great old mathematicians are collected?

  Look, the scan of the first page of Riemann's original work Über die Anzahl der Primzahlen unter einer gegebenen Grösse !! Such a mathematical and historical gold...

http://archive.org

Just search for "Bernhard Riemann", for example.

This is more of a science thing, but you might be interested in De Re Metallica for its Renaissance metallurgy (no alchemy here, the thing is a prelude to modern chemistry and geology and engineering).  Both the book itself and the translation  are amazing works of scholarship.  And the woodcut drawings are amazingly detailed, helpful for seeing mechanical relations, as well as just beautiful.

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


Expected value of $\inf\{n|X_{1}+...X_{n}\ge1\}$ for $\{X_n\}$ i.i.d. uniform on $(0,1)$ Let $\{X_{n}\}_{n}$ be a sequence of i.i.d. random variables of uniform distribution on $(0,1)$. Find $E\tau$ if $$\tau=\inf{\{n\mid X_{1}+\dots+X_{n}\ge1\}}$$

Should I compute probability of $\tau$ being equal to $n$ for all $n$'s?

Notemos primero que $\tau$ es una variable aleatoria discreta, que toma valores enteros no negativos. Podemos notar que $\tau\geq2$ pues las variables aleatorias $X_j$ toman valores en $(0,1)$.  Entonces estamos interesado en determinar

\begin{equation}
\mathbb{E}[\tau]=\sum_{n=2}^\infty n\mathbb{P}(\tau=n)
\end{equation}
Veamos que $\{\tau=n\}\Leftrightarrow \{S_n>1\}$ y $\{S_{n}<1\}$, donde $S_n=\sum_{j=1}^nX_j$. Por lo tanto

\begin{equation}
\mathbb{P}(\tau=n)=\mathbb{P}(S_n>1)-\mathbb{P}(S_{n-1}>1)
\end{equation}
Ahora debemos determinar la función de distribución de probabilidad de la variable aleatoria $S_n$, notemos que nos es conveniente determinar como se distribuye solo para valores en el intervalo $(0,1)$. Hagamos esto mediante inducción. Para $n\geq 1$ deseamos probar que

\begin{equation}
\mathbb{P}(S_n=x)=\frac{x^{n-1}}{(n-1)!}, \quad 0<x<1.
\end{equation}

Para $n=1$, es inmediato que $S_1=X_1\sim U(0,1)$, lo cual cumple la relación.
Para $n=2$, $S_2=X_1+X_2$, toma valores sobre $(0,2)$. Sin embargo veamos que para valores en (0,1)

\begin{equation}
f_{S_2}(x)=\int_0^xf_{X_1}(y)f_{X_2}(x-y)dy=\int_0^xdy=x
\end{equation}
lo cual cumple la hipótesis.
Asumiento válida la hipóteis para $n$, y $0<x<1$. Comprobemos ahora para $n+1$.

\begin{equation}
\begin{aligned}
f_{S_{n+1}}(x))&=\int_0^xf_{S_n}(y)f_{X_{n+1}}(x-y)dy=\int_0^x\frac{y^{n-1}}{(n-1)}dy=\frac{x^n}{n!}
\end{aligned}
\end{equation}
donde hemos asumido válida la relación para $S_n$ en $(0,1)$

Con todo lo anterior tenemos que

\begin{equation}
\begin{aligned}
\mathbb{P}(S_n>1)=1-P(S_n\leq 1)=\int_0^1f_{S_n}(y)dy=1-\frac{1}{n!}
\end{aligned}
\end{equation}
y asi

\begin{equation}
\mathbb{P}(\tau=n)=\left(1-\frac{1}{n!}\right)-\left(1-\frac{1}{(n-1)!}\right)=\frac{n-1}{n!}
\end{equation}
Por lo tanto

\begin{equation}
\mathbb{E}(\tau)=\sum_{n=2}^\infty n\frac{n-1}{n!}=\sum_{n=2}^\infty \frac{1}{(n-2)!}=\sum_{n=0}^\infty \frac{1}{n!}=e\end{equation}

You can easy compute $P(\{\tau = n\}$ if you consider following:

$$\tau = n \iff X_1 + \ldots X_n \ge 1 \text{ but }X_1 + \ldots X_{n-1} < 1$$

So: $$\begin{align*} \{\tau = n\} &= \{X_1 + \ldots X_n \ge 1, X_1 + \ldots X_{n-1} < 1\} \\ &= \{X_1 + \ldots X_{n-1} < 1\}\setminus\{X_1 + \ldots X_n < 1, X_1 + \ldots X_{n-1} < 1\} \\ &= \{X_1 + \ldots X_{n-1} < 1\}\setminus\{X_1 + \ldots X_n < 1\}\end{align*}$$

Where the last equality follows due to $X_n \ge 0$.
And with continuity of $S_n = \sum_\limits{k=1}^n X_k$ it follows:

$$P(\{\tau = n\}) = P(S_n \le 1) - P(S_{n-1} \le 1) = F_{S_n}(1) - F_{S_{n-1}}(1)$$

So all you have to do is calculate the distribution of $S_n = \sum_\limits{k=1}^n X_k$


A property of a separable space Suppose $(X,\tau)$ is a separable topological space. Let $C$ be a basis for the topology $\tau$. Is it true that any open set in $\tau$ can be written as a countable union of elements in $C$?

If not, is the above true in case of a separable metrizable space?

Thanks,
Phanindra

The answer to the first question is no. Let $\mathscr{D}$ be an uncountable family of almost disjoint subsets of $\Bbb N$. Let $X=\Bbb N\cup\mathscr{D}$. Each $n\in\Bbb N$ is isolated in $X$, and a basic open nbhd of $D\in\mathscr{D}$ is any set of the form $\{D\}\cup(D\setminus F)$ such that $F$ is a finite subset of $D$. It’s easy to check that this does define a completely regular topology on $X$. $\Bbb N$ is clearly a countable dense subset of $X$, so $X$ is separable. Let $$\mathscr{B}=\Big\{\{n\}:n\in\Bbb N\Big\}\cup\Big\{\{D\}\cup(D\setminus F):D\in\mathscr{D}\text{ and }F\subseteq D\text{ is finite}\Big\}\;;$$ then $\mathscr{B}$ is a base for $X$, but $X$ itself is not the union of any countable subset of $\mathscr{B}$.

The answer to the second question is yes. Every separable metric space is second countable, i.e., has a countable base. Second countability is clearly a hereditary property, and every second countable space is clearly Lindelöf, so every separable metric space is hereditarily Lindelöf. Now let $\mathscr{B}$ be any base for the separable metric space $X$, and let $U$ be any non-empty open set in $X$; there is a subset $\mathscr{U}\subseteq\mathscr{B}$ such that $U=\bigcup\mathscr{U}$. $\mathscr{U}$ is an open cover of $U$, and $U$ is Lindelöf (as a subspace of the hereditarily Lindelöf space $X$), so $\mathscr{U}$ has a countable subcover $\mathscr{U}_0$. But then $\mathscr{U}_0$ is a countable subfamily of $\mathscr{B}$ whose union is $U$.

Added: One way to get an uncountable almost disjoint family of subsets of $\Bbb N$ is as follows. For each irrational number $x$ let $\langle q_x(k):k\in\Bbb N\rangle$ be a sequence of rational numbers converging monotonically to $x$. It’s easy to see that if $x$ and $y$ are distinct irrationals, the sequences $\langle q_x(k):k\in\Bbb N\rangle$ and $\langle q_y(k):k\in\Bbb N\rangle$ can have only finitely many terms in common. Now let $\varphi:\Bbb Q\to\Bbb N$ be any bijection, for each irrational $x$ let $D(x)=\{\varphi(q_x(k)):k\in\Bbb N\}$, and let $\mathscr{D}=\{D(x):x\in\Bbb R\setminus\Bbb Q\}$; $\mathscr{D}$ is an uncountable family of almost disjoint subsets of $\Bbb N$.

Let $X=\mathbb{R}_{\ell}$ be the Sorgenfrey line, this is separable but not second countable. It is true if $X$ is metrizable because in metrizable spaces separable and second countable are equivalent.


A problem for the New Year What better to start the year than a dazzling integral?

$$\int_{0}^{\infty}\left[1+\left(\frac{2013}{x+2013}+\cdots +\frac{2}{x+2}+\frac{1}{x+1}-x\right)^{2014}\,\right]^{-1}\,dx$$

Happy New Year to the mathematical community!

(I am not too familiar with the posting policies on this site, hopefully this is not a major breach of rules)

The integral from $-\infty$ to $\infty$ is $$\frac{2\pi}{2014}\csc\left[\frac{\pi}{2014}\right]$$
See M.L. Glasser, A remarkable property of definite integrals, Math. Comp. 40, 261 (1981).

Enfim, depois de passar alguns dias a fio tentando resolver esse desafio, acredito que a resposta numérica final seja $0{,}5 \cdot e \approx 1{,}35914$, onde $e \approx 2{,}71828$ é o número de Euler.

Anyway, after spending a few days on end trying to solve this challenge, I believe that the final numerical answer is $0{,}5 \cdot e \approx 1{,}35914$, where $e \approx 2{,}71828$ is the Euler number.

This is a numerical approximation, not the exact result. The function being integrated relies only on the sum $n / (x + n)$, where $n$ is an integer and belongs to the interval $[1, 2013]$. The sum has a single root in the interval $[0, \infty)$. That is, $x ≈ 938{,}17268...$ the value of the sum is zero. Well, then the function being integrated is equal to $1$. Therefore, we can easily see that the function being integrated in the interval $[0; 938{,}17268...)$ is growing. And in the interval $(938{,}17268; \infty)$ is decreasing.

Using the method of trapezoids in the vicinity of $x \approx 938{,}17268$ we can determine an approximation for the numerical integration. As the function values are dwindling, there is no significant change in the first 5 digits houses. So I suggested that $0{,}5 \cdot e \approx 1{,}35914$ is an approximation to the result. Furthermore, it is possible to prove this result using Maple with interactive integration algorithm.

It is important to remember that the function is not continuous the full extent of the real numbers. How can there be integral from $-\infty$ to $\infty$, mentioned in the first reply?


Discontinuity of $\sin\ (1/x) $ at $x=0$? Why is $\sin\ (1/x)$ discontinous at $x=0$ ? Isn't the value of sine always between $1$ and $0$? Why doesnot $\sin\ (1/x)$ exist?

The function $f(x)=\sin(1/x)$ isn't discontinuous at $x=0$, it is undefined. That's a different thing.

However, if you remedy that by defining it (say we set the value to $f(0)=0$), then it will necessarily be discontinuous. This follows quite immediately from (the negation of) any reasonable definition of continuity, for instance the $\epsilon$-$\delta$ definition using $\epsilon=\frac13$.

When x = 0, for sin(1/x) you have the fraction 1/0. Division by 0 is undefined, therefore there is a discontinuity at x= 0 + pik


Find $\lim_{x\to 0}\frac{|\sin x|}{x}$ Can I say that $$\lim_{x\to 0}\frac{\sin x}{x}=1 \implies \lim_{x\to 0}\frac{|\sin x|}{x}=1$$?

Exercise: show that
$$
\lim_{x\to 0}\frac{|\sin x|}{x}
$$ does not exists.

I know three ways to attack this. The sandwich theorem, L'Hospital's Rule or Taylor series.
I will show you the way with the Taylor series. Let's look at the series expansion of $\sin{x}$ at the point $0$:
$$\sin{x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...
$$
Now you can see that
$$
\frac{\sin{x}}{x}=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n}=1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...
$$
Taking the limit now will be trivial if you belief in Taylor series. Maybe you can proof the series expansion of $\sin{x}$.


Is a ball a polyhedron? In the book Introduction to Linear Optimization by Bertsimas Dimitri, a polyhedron is defined as a set $ \lbrace x \in \mathbb{R^n} | Ax \geq b \rbrace $, where A is an m x n matrix and b is a vector in $\mathbb{R^m}$. What it means is that a polyhedron is the intersection of several halfspaces.

A ball can also be viewed as the intersection of infinitely many halfspaces. So I was wondering if a ball is also a polyhedron by that definition or by any other definition that you might use?

Thanks and regards!

No a ball is not a polyhedron, even by this definition. In your definition the matrix $A$ is of size $m\times n$, where $m\in\mathbb{N}$ thus the matrix is finite. The integer $m$ is an upper bound on the number of halfspaces which intersect to form the polyhedron.

The reason $m$ is an upper bound is because suppose $A$ has two rows identical. Then there are two hyperspaces which are parallel so at least one of them does not form any part of the polyhedron.

polyhedron is not ball cause its a solid figure bounded by plane polygons or faces


How many different paths from origin to $(4,4)$ if you can't cross $y = x$? Assuming you can only go right 1 step or up 1 step, and you're trying to get from the origin to $(4,4)$ or any $(n,n)$ or $(m,n)$, what is an algorithm to help you determine the number of paths possible without crossing $y=x$?
I can do it just by experimenting and drawing, but I was told it's better to use an inductive proof.
I'm not sure how I'd do that.
Intuitively, I know that it's the number of paths from the origin to $(n,n)$, which is $\binom{2n}{n}$ minus the number of paths that cross the $y=x$ line.
The base case is, for example, $(1,1)$. Theres two possible paths, and one crosses $y=x$. But what next?

In prinziple you always have two choices, going up or going to the right. But obviously you have two limitations.
Depending on your first step, you are either under or above the line. You can view the sqauare with the corner points (0,0) and (4,4). The line y=x divides it into two.
W.l.o.g you can assume, that your first step is one to the right, so you have to stay under the line. Whatever you do, u have to go 4 steps to the right and 4 steps to the top in total.
The question is, how many choices do you have? Assume you are still on heigth 0, so your y=0. Than you have 4 choices to go one step up. On (0,1) to (1,1), from (0,2) to (1,2), from (0,3) to (1,3) and from (0,4) to (1,4).
The same goes for y=1, there u have three possibilities to go one step up. For y=2 you have two choices and for y=3 only one.
In total you have 4+3+2+1 = 10 possiblities.
Since you can only go right and up, you have to chose one way up for each y. For y=0 you have to take one of your 4 choices. For y=1 one of your three choices and so on.
In total you have 4 * 3 * 2 * 1 = 4! choices to go up. You can mirror this explanation to the triangle above the line y=x.
So in total you have 2 * 4! = 48 possible ways to get from (0,0) to (4,4).
You can easy generalize this problem on (n,n) or even (n,m).


The joint density of $X$ and $Y$ is given by : $f(x,y)=\frac{1}{\sqrt{2 \pi}}e^{-y}e^{-\frac{(x-y)^2}{2}}$. The joint density of $X$ and $Y$ is given by
  $$f\left(x,y\right)=\frac{1}{\sqrt{2\pi}}e^{-y}e^{-\frac{\left(x-y
\right)^{2}}{2}}$$
  Here $0<y<\infty$ and $-\infty<x<\infty$. Compute the individual MGFs of $X$ and $Y$.

I could find the marginal distribution of $Y$ which is ${\rm Exp}\left(1\right)$, so I can find its MGF. But I cannot find the marginal distribution of $X$. Please help!

Assuming "MGF" means "moment generating function":

In this set-up, $Y$ is exponential and conditional on $Y$, the rv. $X$ is $N(Y,1)$.  That is, $X$ can be represented as $Y+Z$, where $Z\sim N(0,1)$ is independent of $Y$.  So the MGF of $X$ is $E\exp (uX) = E\exp(uY+uZ) = E\exp(uY)\exp(uZ)=\frac 1{1-u}\exp(u^2/2)$.  The MGF of $Y$ is $E\exp(vY)=1/(1-v)$.

You don't need to find the distribution of $X$ to derive its MGF.

As @kimchilover has already mentioned, from the joint pdf of $(X,Y)$ it is clear that $X$ conditioned on $Y=y$ has a $\mathcal{N}(y,1)$ distribution, where $Y$ itself is distributed as $\text{Exp}(1)$.

MGF of $X$ is

$$M_X(t)=E\,[e^{tX}]=E\,E\,[e^{tX}\mid Y]=E\,[e^{Yt+t^2/2}]=e^{t^2/2}E\,[e^{tY}]$$

The only fact used above is the MGF of a univariate Normal distribution.

Now MGF of $Y$ is $E\,[e^{tY}]=\frac{1}{1-t}$ for $t<1$.

Therefore the required MGF is $$M_X(t)=\frac{e^{t^2/2}}{1-t}\quad,\,t<1$$


Cardinality of the set $S$ Let $S=\{f \in C^1(\mathbb{R}): f(0)=0,f(2)=2 \; \mbox{and}\; |f'(x)|\leq\frac{3}{2} \forall x \in \mathbb{R}\}$,
find cardinality of $S$.

I think $S$ is an uncountable set. Am I right?

It should be possible to fit criteria for a parabola $f(x)=ax^2+bx+c$ for $x\in[0,2]$ in uncountably many ways:
$$0=f(0)=c\implies c=0$$
$$2=f(2)=4a+2b+c\implies 2a+b=1$$
so we will have $f(x)=ax^2+(1-2a)x$ and $f'(x)=2ax+1-2a$.
As $f'(x)$ is a linear function, it reaches maximum and minimum at the ends of the interval $[0,2]$, so we only need to make sure that $|f'(0)|\le 3/2$ and $|f'(2)|\le 3/2$. This means that the constraints are $|1-2a|\le 3/2, |1+2a|\le 3/2$. One can easy see that any $a$ in $[-1/4, 1/4]$ satisfies these inequalities, and there are uncountably many of those.
Now let's define the function $f$ on the whole $\mathbb R$:
$$f(x)=\begin{cases}(1-2a)x&x<0\\ax^2+(1-2a)x&0\le x\le 2\\(1+2a)x-4a&x>2\end{cases}$$
which matches the original function $f$ on $[0,2]$ and linearly extends it on $(-\infty, 0)$ and $(2,\infty)$.

There is $g\in C^\infty(\mathbb R )$ such that
i) $g=0$ on $(-\infty,0]\cup [2,\infty)$
ii) $g=3/2$ on $[1/3,5/3]$
iii) $0<g<3/2$ elsewhere.
Note that $\int_0^2 g > (4/3)\cdot(3/2) =2.$ We can thus choose $c\in (0,1)$ such that $\int_0^2 cg = 2.$
Now define $G(x) = \int_0^x cg(t)\,dt,$ $x\in \mathbb R.$ Then $G\in \mathbb C^\infty,$ $G(0)=0, G(2)=2.$ We also have $G'(x)=cg(x),$ which implies $|G'(x)|  < 3/2$ everywhere. Note that $G'(x)=0$ for $x\notin [0,2].$
To finish, define
$$f_a(x) = G(x)+G(x-a),\, a>3.$$
Then $\{f_a\}$ is an uncountable collection of $C^\infty$ functions that satisfy the criteria.


What do you call this kind of matrix product? Let’s say I am using multiple polls to evaluate a politician’s popularity. Each poll’s rating goes into the column vector $v$. Then, I want to take the weighted average of these entries, using weights that I store in the column vector $w$. Then the weighted average is $w^T v = p$.

Now, I want to track how her popularity changes over time. I collect the popularity vectors $v_i$ at times $i = 1$ through $n$. My poll rating system is constantly being refined based on the outcomes of other races, so for every popularity vector $v_i$, there is a corresponding weight vector $w_i$ that gives the weights used at time $i$.

Then we have
$$\begin{align}
W = \begin{bmatrix}
\mid & \mid & & \mid \\
w_1 & w_2 & \cdots & w_n \\
\mid & \mid & & \mid \\
\end{bmatrix}
& \qquad V = \begin{bmatrix}
\mid & \mid & & \mid \\
v_1 & v_2 & \cdots & v_n \\
\mid & \mid & & \mid \\
\end{bmatrix}\end{align}
$$

I want to multiply $W$ (or $W^T$) and $V$ together in such a way that

$$W \boxtimes V = \begin{bmatrix} w_1^T v_1  \\ w_2^T v_2  \\ \vdots \\ w_n^T v_n \end{bmatrix} = \vec p$$

and the resultant vector contains the weighted averages at each time.

I can see that $\vec p = \mathrm{diag}( W^T V )$, but this requires lots of unnecessary computations.

A better definition might be to sum the rows of $W \circ V$ and then transpose, where $\circ$ is the Hadamard product. This is how I do it in numpy:

np.random.seed(123)
w = np.random.rand(5, 7)
w = w / np.sum(w, axis=0)
v = np.random.rand(5,7)

np.sum(w*v, axis=0)

yields

array([0.42125858, 0.50186797, 0.82132014, 0.63440834, 0.48432466,
       0.52752764, 0.31076101])

(np.diag(w.T @ v) also works.)

It seems like this product should have a name, as it has many applications in statistics and optimization.

$${\rm W} \boxtimes {\rm V} := \mbox{diag} \left( {\rm w}_1^\top, {\rm w}_2^\top, \dots, {\rm w}_n^\top \right) \mbox{vec} \left( {\rm V} \right)$$


What does it mean when we say $\frac{dy}{dx} = f(x)$? If we have an equation that's true for all $(x,y)$ in a set $S$, for example:
$y = 2x^2 + 3x$

We say that its derivative is:
$\frac{dy}{dx} = 4x + 3$

However, if we use the definition of the derivative on the left half $f(x) = y$, we get:
$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$

This would mean that either:

Whenever $y = 2x^2 + 3x \rightarrow 0=4x+3$, which is false
$\frac{dy}{dx}$ doesn't always mean "the derivative of $f(x) = y$"

The same problem also happens with implicit differentiation:

$x^2 + y^2 = 1$
$2x + \frac{dy^2}{dx} = 0$
$2x + lim_{h\to0}\frac{y^2 - y^2}{h} = 0$
$2x = 0$  (not equivalent to $x^2+y^2=1$)

All I did to get that result was to use the limit definition of the derivative on $\frac{dy}{dx}$ instead of leaving it as it is and isolating it.

What am I doing wrong?

Also, by $x$ and $y$, I am referring to variables in a two-variable equation. I'm not using "y" to mean "f(x)".

The expression $\frac{dy}{dx}$ is read "the derivative of the variable $y$ with respect to the variable $x$". Actually, this refers to the derivative of the function that relates the variables.

If $y=2x^2+3x$ and $y=f(x)$, then $f(x)=2x^2+3x$ and $f(x+h)=2(x+h)^2+3(x+h) \neq y$. This is your mistake.

The definition is for the derivative of a function, not of a variable.
$y=f(x)$ and $y=f(x+h)$ are different ways of relating the variables, and that is what causes a zero result (if both hold for all small values of $h$, this would mean that $y$ is constant with respect to $x$, so the derivative is zero).

You have the right to say that $x$ and $y$ are variables and that $y$ is not only an expression for $f(x)$, but it must be clear what function you are trying to differentiate. And remember that $y=f(x)$ is an equation relating the variables as much as $x^2+y^2=1$ is

Your reasoning is wrong in here:

$\lim_{h\to0}\frac{f(x+h) - f(x)}{h} = \lim_{h\to 0} \frac{y - y}{h} = 0$

In fact, $\lim_{h\to0} f(x+h) = f(x) + h \cdot g(x)$ and it is exactly this $g(x)$ which is the value of interest and which is the derivative. In many cases, $g(x)$ can be evaluated without using calculus.


Find all complex numbers $z$ satisfying the equation I need some help on this question. How do I approach this question?

Find all complex numbers $z$ satisfying the equation

$$
(2z - 1)^4 = -16.
$$

Should I remove the power of $4$ of $(2z-1)$ and also do the same for $-16$?

HINT: How many solutions does $x^4+16=0$ have?

   $x_{1,2,3,4}= \sqrt{2}(\pm 1 \pm i)$, substitute $x$ with $2z-1$, solve for $z$ and you're done.

The answer to this problem lies in roots of a polynomial.
From an ocular inspection we know we will have complex roots and they always
Come in pairs! The power of our equation is 4 so we know will have a pair of
Complex conjugates. I believe it is called the Fundamental
Theorem.
Addendum
A key observation is how to represent -1 in its general term using Euler's identity and the concept of odd powers in EF.


Primes of the forms $4m+1$ and $4m+3$ I ask if it is possible to demonstrate that any prime number $p$ of the form $4m+1$ can be written as $$p=q_1+2q_2=(4n_1+3)+2(4n_2+3)$$ and, conversely, that any prime number $q$ of the form $4n+3$ can be written as $$q=p_1+2p_2=(4m_1+1)+2(4m_2+1)$$
For the first primes we have:
$$3=1+2\cdot1\qquad5=-1+2\cdot3$$
$$7=5+2\cdot1\qquad13=7+2\cdot3$$
$$11=1+2\cdot5\qquad17=3+2\cdot7$$
$$19=17+2\cdot1\qquad29=7+2\cdot11$$
$$23=13+2\cdot5\qquad37=23+2\cdot7$$
Thanks a lot for your attention.


The number which are divisible by $8$ is also divisible by $2$ and $4$.

The number which are divisible by $9$ is also divisible by $3$.

The number of which is divisible by $6$ is also divisible by $2$ and $3$.

The number which are divisible by $10$ is also divisible by $2$ and $5$.

Also also the number we expect is divisible by $11$ and $7$.

So the number is in the form $=P×2^{3i}×3^{2j}×5^k×7^m×11^n$,
Where $i$, $j$, $k$, $m$, & $n$ are positive any positive integer and $P$ is any positive integer integer.Using this condition we will produce a required ten digit number.


How do we solve $\sin(\sin(x))=a$? While I was solving a sum, I got stuck at how to go further when I incurred
$\sin(\sin(x^2))=1$ or $0$, where $x\in \mathbb{R}$.
Can someone help me to proceed this step and also a general way of approaching $\sin(\sin(x))$ type problems?

Hint:

You can invert a relation like $$v=\sin(u)$$ with $$u=\arcsin(v)+2k\pi\lor u=\pi-\arcsin(v)+2k\pi.$$

Use the trick once to get $\sin(x^2)$ and a second time to get $x^2$.

Notice that the value of the sine is limited to the range $[-1,1]$, and as it is monotonous there,

$$\sin(\sin(t))\in[-\sin(1),\sin(1)]$$ and there are no solutions for $a$ outside this range. ($\sin(1)\approx0.841471$)

draw graphs of sin(x) and arcsin(a).
the solution will be the intersection point of these two graphs.


I have a simple problem which gives two different solutions in two different calculators. I hope that this kind of questions doesn't break the site's rules.

I have this simple problem which gives two different solutions in two different calculators (Wolfram Alpha and Symbolab).

What am I doing wrong in the calculation of
$$
[(-7)^6]^\frac{1}{2} \:\:?
$$
In one of the solutions, the solution is positive, and negative in the other. I think that the problem might arise when trying to solve a power of a power with a negative base. Is there a restriction to this kind of problems?

Here are the two links to the solutions:

Wolfram Alpha

Symbolab

Thanks in advance for your help!

Kind regards!

Symbolab used the rule $(a^m)^n=a^{m×n}$, which does not always work when $a$ is negative and $m$ and $n$ are not integers; see this answer.

Both answers are correct, in that $343^2$ and $(-343)^2$ both equal $(-7)^6$.

There is no the square root of anything (except $0$) because $\sqrt x$ or $x^{\frac12}$ is a 2-valued function.

That said, you may feel that thinking “the square root of a sixth power is the same as the cube” is over-clever. But on the other hand you may feel that taking it as the cube is rescuing you from the loss of sign information caused by raising to the sixth power in the first place.


Subgroups of finite Abelian groups I am interested in finding all of the subgroups (up to isomorphism) of a finite Abelian group $A$.

I know the following:

-- A finite Abelian group $A$ can be represented as a direct product of cyclic groups, $A_1, A_2, A_3, ...$.

-- The direct product of subgroups of a set of groups is always a subgroup of the direct product of the groups.

-- Each of the cyclic groups $A_1, A_2, A_3, ...$ will have a unique subgroup for each divisor of its order.

To start finding subgroups of $A$ I can therefore:

-- Write down all the subgroups of each of $A_1, A_2, A_3, ...$

-- Form all possible direct products of these subgroups

I realise that this method will not necessarily find all subgroups of $A$. (For example, if $A = \mathbb{Z}_2 \times \mathbb{Z}_2$, then it has a subgroup $ \{(0, 0), (1, 1) \}$ which is not a direct product of subgroups of $\mathbb{Z}_2$.)

However my question is this: will the above method find representatives from all isomorphism classes of subgroups of $A$? If no, can you provide a counterexample? If yes, can you provide a proof?

The first thing I would do is to reduce to abelian $p$-groups: If $A = G_1 \times \cdots \times G_r$ where each $G_i$ is a $p_i$-Sylow subgroup of $A$, then every subgroup $B$ of $A$ is a product of subgroups $B_i$ of $G_i$'s. (in fact $B_i$ will be the $p_i$-Sylow subgroup of $B$.)

So assume $A$ is an abelian $p$-group, say $A=\prod_{i=1}^k (\mathbb{Z}/p^{e_i} \mathbb{Z})^{r_i}$. In this case, clearly every $i$ can contributes up to $r_i$ direct factors of order dividing $p^{e_i}$. I think it is straightforward, although tedious, to show that no other subgroups can occur. (I haven't wrote the details, so one have to check...)

So I think the answer to your question is: YES

Birkhoff, Garrett;
Subgroups of Abelian Groups.
Proc. London Math. Soc. S2-38 no. 1, 385.


Do we assume $f_n$'s map into $\Bbb{R}$ or $\Bbb{C}$ in Theorem 7.8 Rudin's *Principles of Mathematical Analysis*? Theorem 7.8 The sequence of functions $\{f_n\}$ defined on $E$ converges uniformly on $E$ if and only if for every $\epsilon > 0$ there exists an integer $N$ such that $m \geq N, n \geq N, x \in E$ implies
  \begin{equation}
|f_n(x)-f_m(x)| \leq \epsilon
\end{equation}

For the backwards direction, since the codomain of $f$ is not given, how can we use Theorem 3.11 (Cauchy sequence in a compact metric space (or $\mathbb{R}^k$) converges to some point in the metric space) to prove pointwise convergence of $f$?

For each $x \in E$, the sequence $(f_n(x))_{n \in \mathbb N}$ is a Cauchy-sequence in $ \mathbb R$ or ($\mathbb C$).

I cannot quickly scan the whole text to check all references to 7.8. Based on the contents of chapter Uniform convergence and continuity (pages 149-151), Theorem 7.8 is used to prove Theorem 7.15, which is essentially about complex-valued, bounded and continuous functions $\mathcal{C}(X)$.

Theorem 3.11 is about metric spaces that are either compact, $\mathcal{R}^n$ or $\mathcal{C}$.

So $E$ in Theorem 7.8 is a typo. Actually, it should be $\mathcal{C}$ as we are dealing with complex-valued functions.

Yes, this book had 3 editions, so many views, but no mention of this bit in errata.

Errata from George Bergman does not include this error https://math.berkeley.edu/~gbergman/ug.hndts/m104_Rudin_notes.pdf

Notes from Jiří Lebl point to complex-valued functions for Theorem 7.8, page 3 of 21 in PDF
https://math.okstate.edu/people/lebl/uw522-s12/lec1.pdf


How does (r-1) complement for subtraction work? My instructor gave an algorithm for doing subtraction with (r-1)'s complement. For subtracting M - N, it goes like the following.

1) Find the (r-1)'s complement of N by using formula r^n - r^m - N. n is the number of non-fractionated digits and the m represent the number of digits after the radix point. For instance, if N = 324.16 in decimal, the formula would 10^3 - 10^-2 - 324.16

2) Next, Add the M to N's complement. If there is an end around carry, add it to the least significant digit. Else, take the r-1's complement of the sum.

The algorithm works but I just don't understand how it works. Can someone explain ? Thanks !

subtraction using complement is something like comparing two values indirectly using some standard third value(r's or r-1's complement). suppose there are two straight lines on paper and we want to fined difference in length between them then we draw third line greater then both of previous lines and compare both of these lines with third one and then easily find difference between them without comparing them directly.


Is being a Cauchy sequence equivalent to $ \lim_{n\to+\infty}d(x_{n+k},x_n)=0$ for every $k$? Is this statement true?

  In a metric spase $(E,d)$, a sequence $(x_n)$ is Cauchy if and only if $  \forall k\in \mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$

I proved that $\Rightarrow$ is true,indeed if $(x_n)$ is Cauchy then for all $\varepsilon>0,\exists n_0\in\mathbb{N}, \forall p,q\geq n, d(x_p,x_q)<\varepsilon$ it still right for all $ n\geq n_0$ and $m=n+k>n\geq n_0$

But I don't know if $\Leftarrow$ is true ?

The equivalence is not correct.

To see this, consider any divergent series of real numbers $\sum a_n$ such that
$a_n\to 0$ as $n\to \infty$; for example $a_n=\frac1n$. Then define $x_n=a_1+\dots +a_n$. This is a non-Cauchy sequence (as any non-convergent sequence of real numbers), but $x_{n+k}-x_n=a_{n+1}+\dots +a_{n+k}\to 0$ as $n\to \infty$ for any fixed $k\in\mathbb N$, since this is the sum of $k$ terms tending to $0$; that is, $d(x_n,x_{n+k})\to 0$ as $n\to\infty$. So the implication $\Leftarrow$ does not hold.

If you prefer, you can take $x_n=\log n$ or $x_n=\sqrt{n}$, and check that $x_{n+k}-x_n\to 0$ as $n\to \infty$, for any $k\in\mathbb N$.

On the other hand, your proof of the other implication is correct.

Last remark : in fact the assumption "$\forall k\in\mathbb N\; d(x_{n+k},x_n)\to 0$" is equivalent to the seemingly weaker "$d(x_{n+1},x_n)\to 0$", as you can easily check.

With the way you edited it, you actually just wrote the definition of a cauchy sequence in a different manner.

$\lim_{n\to+\infty}d(x_{n+k},x_n)=0$ means that for every $\varepsilon>0$ there exists $N$ such that $d(x_{n+k},x_n)<\varepsilon\ \forall\ n>N$.

But since $k$ can be any arbitrary natural number, this means that $d(x_n,x_m)<\varepsilon\ \forall\ n>N,m>N$.

Thus, you didn't really change anything there.


Prove that if $X$ is Hausdorff, $\Delta$ is closed in $X\times X$ Prove that if $X$ is Hausdorff, $\Delta=\{(x, x)\mid x\in X\}$ is closed in $X\times X$ (with the product topology).

My attempt:

Let $x_1, x_2\in X$ s.t. $x_1\ne x_2$.

There exist neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ that are disjoint.

$U_1\times U_2$ is a basis element in the product topology on $X\times X$. So, $U_1\times U_2$ is open in $X\times X$.

Let $x\in X$.

$(x, x)\in U_1\times U_2\implies x\in U_1$ and $x\in U_2\implies x\in U_1\cap U_2$, which contradicts the fact that $U_1$ and $U_2$ are disjoint.

So, $(x, x)\notin U_1\times U_2$.

I feel that I'm on the right track but don't know how to proceed. Could someone please help me out?

You're basically there, you just need to interpret your result. You found that for any point $(x_1,x_2)\in X\times X-\Delta$, there exists a neighborhood of $(x_1,x_2)$ contained in $X\times X-\Delta$. That is, $X\times X-\Delta$ is open. Therefore $\Delta$ is...

Let $(x, y)\in X\times X-\Delta$

$\implies(x, y)\in X\times X$ and $(x, y)\notin\Delta$

$\implies x, y\in X\text{ and }x\ne y$

There exist neighborhoods $U_x$ and $U_y$ of $x$ and $y$ respectively that are disjoint.

$U_x\times U_y$ is a basis element in the product topology on $X\times X$.

$(x, y)\in U_x\times U_y$ --------------------------------------------- (1)

Let $(u, v)\in U_x\times U_y$. --------------------------------------------- (2)

$\implies u\in U_x$ and $v\in U_y$

Since $U_x\subset X$ and $U_y\subset X$,

$u, v\in X$

$\implies(u, v)\in X\times X$ --------------------------------------------- (3)

$u=v\implies u=v\in U_x\cap U_y$, a contradiction as $U_x$ and $U_y$ are disjoint.

So, $u\ne v$.

$\implies(u, v)\notin\Delta$ --------------------------------------------- (4)

From (3) and (4),

$(u, v)\in X\times X-\Delta$ --------------------------------------------- (5)

From (2) and (5),

$U_x\times U_y\subset X\times X-\Delta$ --------------------------------------------- (6)

Combining (1) and (6),

$(x, y)\in U_x\times U_y\subset X\times X-\Delta$

So, $X\times X-\Delta$ is open in $X\times X$.

So, $\Delta$ is closed in $X\times X$.

QED


Rational root coefficient I saw this question in my exam recently,

  If a, b, c are distinct rational roots of $x^3+ax^2+bx+c=0$, find the values of a, b, c.

Can someone give me a hint or answer? I tried factoring it and then expand it all out, but it doesn't seem very useful as they are all "rational". Can someone help me with it?

Apologies for the length of the argument.  I tried to format it to be more comprehensible.

Multiplying out $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so we have the equations:

$1)$ $a=-a-b-c$

$2)$ $b=ab+ac+bc$

$3)$ $c=-abc$

Notice that if $a = 0$ or $b = 0$, then by the $(3)$, $c=0$ contradicting the fact that they must be distinct.  Therefore $a,b\neq 0$.

Now if $c=0$, then by $(1)$ and $(2)$, $2a=-b$ and $b=ab$, which has the solution $a=1, b=-2, c=0$.  The goal now is to show no other solution exists.

Consider when $c \neq 0$.  Cancelling the $c$'s in $(3)$ gives us $1=-ab \implies a=-\frac{1}{b}$.  Then we can simplify $(1)$ into $b^2+cb-2=0$

$\implies b=\frac{-c \pm \sqrt{c^2+8}}{2}$

and $(2)$ simplifies to $(c-1)b^2-b-c=0$

$\implies b=\frac{1\pm\sqrt{4c^2-4c+1}}{2(c-1)}=\frac{1\pm(2c-1)}{2(c-1)}$

for $c\neq 1$

(if $c=1$, then $(c-1)b^2-b-c=0\implies b=-1 \implies a=1$, contradicting distinct values again).

Setting the equations for $b$ equal yields

$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\pm(2c-1)}{2(c-1)}$

or

$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\mp(2c-1)}{2(c-1)}$

Changing these two equations to the one

$c^2+8=\left(\frac{1\pm(2c-1)}{c-1}+c \right)^2$

makes it slightly easier.

Taking the $-$ gives us $c=-1 \implies b=-1$ contradicting distinct again, and taking the $+$ gives us the more ugly cubic equation

$c^3-2c^2+4c-2=0$

By the Rational Root Theorem, one need only test the values $\pm 1, \pm 2$, none of which yield a solution to this cubic, so this cubic has no rational solutions.

Therefore, the only solution was the first one we found; namely $\boxed{a=1, b=-2,c=0}$.

The idea provided by @JasonM is correct. You should be able to solve the problem as follows.

Notice that the cubic polynomial is monic, meaning that the coefficient of $x^3$ is $1.$ This is nice because now we can write the polynomial as $(x - a)(x - b)(x - c)$ in factored form. Now, we can equate this to the given form:
$$(x - a)(x - b)(x - c) = x^3 + ax^2 + bx + c$$
$$x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc = x^3 + ax^2 + bx + c.$$

We can see from the equation that $-abc = c,$ meaning that $ab = -1.$ We also have that $a + b + c = a,$ meaning that $b + c = 0,$ or $b = -c.$ With $ab + bc + ca = b,$ substituting values in, we get $-1 - c^2 + 1 = - c,$ which can be rearranged to $c^2 - c = c(c - 1) = 0.$ The solution $c = 0$ cannot be accepted because in that case, $b = c = 0$ and the roots are not distinct. But if $c = 1,$ then $b = -1$ and $a = 1,$ in which case $a = c.$ So there is $\boxed{\text{No solution}}$ to the requested problem.


in any triangle abc prove that bisectors of the interior angle A and the exterior angles at B and C are concurrent i assumed the exterior angles meet at point D, then joined A with D and tried proving AD is angle bisector of angle A.
But i cant get to the proof. any idea will be helpful

It is enough to consider that a bisector (internal or external) of the angle made by two distinct lines $\ell_1,\ell_2$ meeting at some point $P$ is (part of) the locus of points $Q$ for which $d(Q,\ell_1)=d(Q,\ell_2)$.

If some point $Q$ belongs to the external angle bisector through $B$ and to the external angle bisector through $C$, it fulfills
$$ d(Q,BA) = d(Q,BC) = d(Q,CA) $$
so by $d(Q,BA)=d(Q,CA)$, it belongs to the internal angle bisector through $A$, too.

we construct a parallel line to BC that passes through A and a parallel line to BC that passes through C, we obtain

A parallelogram ABCD.

Consider triangles ABC and ADC

ÐADC º ÐABC.  So, the angle bisectors are the same. We know that the angle bisectors of a triangle are concurrent.  So, the bisectors of ÐABC, ÐDAC, and ÐDCA are concurrent.


Variance of $X_{i}+X_{j}$ if $X_{1} + \ldots + X_{N}=1$ I have random variables $X_{1}, X_{2}, \ldots, X_{N}$, where $X_{i} \in \{0,1\}$ and
$$X_{1} + \ldots + X_{N}=1$$
I.e. exactly one of the $X_{i}$'s are $1$, and the rest are $0$.

In addition, denote $P(X_{i} = 1) = p_{i}$, where $\sum_{i} p_{i}=1$. I do not want to assume that the $X_{i}$'s are identically distributed, meaning we may have $p_{i} \neq p_{j}$ for some $i\neq j$.

I now want to calculate the variance of $X_{i}+X_{j}$ for any $i \neq j$, but am having some difficulty. Can anyone provide a formula and an explanation of how to do so for this example? Much appreciated!

$X_i + X_j$ takes values in $\{0,1\}$, so is a Bernoulli random variable.

This is because at most one of the $X_k$ is equal to $1$.

$\mathbb{P}(X_i + X_j = 1) = \mathbb{P}(X_i =1 \text{ or } X_j = 1) = \mathbb{P}(X_i = 1) + \mathbb{P}(X_j = 1) = p_i + p_j$

The second equality holds because the events $\{ X_i = 1\}, \{ X_j = 1\}$ are disjoint.

Thus, $X_i + X_j \sim \text{Ber}(p_i + p_j)$
If $Y \sim \text{Ber}(q)$, then $\text{Var}(Y)=q(1-q)$
Thus, the variance of $(X_i + X_j)$ is $(p_i + p_j) \cdot (1 - p_i - p_j)$

Let $S_N := \sum\limits_{k=1}^N X_k$

If the $N$ variables are independent and identically distributed (are they?) then when given that exactly one of them has the value of $1$, there is equal probability that it can be any one of them.

$$\mathsf P(X_i=1\mid S_N=1) = \tfrac 1N$$

Now similarly evaluate the following conditional joint probabilities, for any $i,j\in\{1,..,n\}$ such that $i\neq j$

$$p_{0,0}~=~\mathsf P(X_i=0, X_j=0\mid S_N=1)\\ p_{0,1}~=~\mathsf P(X_i=0, X_j=1\mid  S_N=1)\\ p_{1,0}~=~\mathsf P(X_i=1, X_j=0\mid S_N=1)\\ p_{1,1}~=~\mathsf P(X_i=1, X_j=1\mid S_N=1)$$

Then use the definition of variance:

$$\mathsf {Var}(X_i+X_j\mid S_N=1)~=~\sum_{x=0}^1\sum_{y=0}^1 (x+y)^2p_{x,y} -\left(\sum_{x=0}^1\sum_{y=0}^1 (x+y)p_{x,y}\right)^2$$

Or else similarly find $\mathsf {Var}(X_i\mid S_N=1)$ and $\mathsf {Cov}(X_i,X_j\mid S_N=1)$ and then use:

$$\mathsf {Var}(X_i+X_j\mid S_N=1)~=~\mathsf {Var}(X_i\mid S_N=1)+\mathsf{Var}(X_j\mid S_N=1)+2\mathsf{Cov}(X_i,X_j\mid S_N=1)$$


Does $\lim \sup x_{n+1}-x_n=+\infty \implies \lim \dfrac{n}{x_n}=0$ If $(x_n)$ is a real sequence such that $\lim \sup x_{n+1}-x_n=+\infty$ , then must we have $\lim \dfrac{n}{x_n}=0$ ?

Some non-monotone examples were given in comments. But even if $(x_n)$ is additionally assumed to be increasing, the answer remains negative. For $n =1,2,3,\dots $ define
$$x_{n^3}=n^2, \qquad x_{n^3+1}=n^2+n $$
and extend to the rest of indices by linear interpolation.

If you assume $\liminf (x_{n+1}-x_n)=+\infty$, then the conclusion is true (and easy to prove).

Let it be: $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$ and $\lim_{x\to \infty } \, |f(x)|=+\infty$ then:

$\lim_{x\to \infty } \, |f'(x)| \neq +\infty$ , if it is $+\infty$ then, by L'Hôpital's rule $\lim_{x\to \infty } \, \frac{x}{f(x)}=\lim_{x\to \infty } \, \frac{1}{f'(x)}=0$ which is in conflict with assumption that: $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$, so: $$\lim_{X\to \infty } \, \int_X^{X+1}f'(x)dx=\lim_{X\to \infty } \, f(X+1)-f(X)\neq\infty$$ $$(\lim_{x\to \infty } \, |f'(x)| \neq +\infty)$$

If $\lim_{x\to \infty } \, |f(x)|\neq \infty$ then we can't use L'Hôpital's rule,but situation is obvious and $\lim_{X\to \infty } \, f(X+1)-f(X)$ is also not $\infty$.

So, I've proved that if $\lim_{x\to \infty } \, \frac{x}{f(x)}\neq 0$ then $\lim_{X\to \infty } \, f(X+1)-f(X)\neq\infty$ which means if $\lim_{X\to \infty } \, |f(X+1)-f(X)|=+\infty$ then  $\lim_{x\to \infty } \, \frac{x}{f(x)}=0$.

Note:
There must exist $x_0$, that for any $x$ greater than $x_0$ , $f(x)$ has real values and derivative.

If $\lim_{x\to \infty } \, f'(x)$ exists and is equal to k then $k=\lim_{x\to \infty } \, \frac{f(x)}{x}=\lim_{x\to \infty } \, f(x+1)-f(x)$, which is useful and related with $O(n)$ functions.

Summary:

Theory that: $$\lim\sup x_{n+1}-x_n=+\infty\ \rightarrow \lim \frac{n}{x_n}=0$$ is false if we consider all sequences with real values:

Sample:

Let's consider: $$x(n) = (-1)^n\cdot\sqrt{n}$$ $\lim \sup x(n+1)-x(n) = +\infty$ , but $\lim \frac{n}{x(n)}$ is not zero.
But theory is true for large family of sequences $f(n)$ : $f(x)$ has real values and derivative for any real $x$ greater than some real value $x_0$ related with considered function.


Why is the minimal uncountable well-ordered set $S_{\Omega}$ unique? In section 10 of Topology by Munkres, the minimal uncountable well-ordered set $S_{\Omega}$ is introduced. Furthermore, it is remarked that,

  Note that $S_{\Omega}$ is an uncountable well-ordered set every section of which is countable. Its order type is in fact uniquely determined by this condition.

However, how to justify its uniqueness?

This stems from the fact that given two well-ordered sets $X, Y$ exactly one of the following holds (see this previous question):

$X$ is order-isomorphic to $Y$;
There is a (unique) $a \in X$ such that $\{ x \in X : x < a \}$ is order-isomorphic to $Y$; or
There is a (unique) $b \in B$ such that $\{ y \in Y : y < b \}$ is order-isomorphic to $X$.

So suppose $X$ is another well-ordered set which is not order-isomorphic to $S_\Omega$.  Then either

there is a (unique) $b \in S_\Omega$ such that $\{ y \in S_\Omega : y < b \}$ is order-isomorphic to $X$, and since  $\{ y \in S_\Omega : y < b \}$ is countable, it follows that $X$ is also countable.
there is a (unique) $a \in X$ such that  $\{ x \in X : x < a \}$ is order-isomorphic to $S_\Omega$.  But then $\{ x \in X : x <_X a \}$ is itself an initial section of $X$ which is uncountable.

So any well-ordered set which is not order-isomorphic to $S_\Omega$ is either countable, or has an uncountable proper initial section.

Suppose that $\alpha$ and $\beta$ are two uncountable ordinals.  Suppose further that for every $\gamma<\alpha$, $\gamma$ is countable.  Likewise, suppose that for very $\eta<\beta$, $\eta$ is countable.  You want to show that $\alpha=\beta$.

Suppose that $\alpha<\beta$.  Then $\eta=\alpha$ is uncountable, and satisfies $\eta<\beta$.  This is a contradiction.  Thus $\alpha\not<\beta$.  Likewise $\beta\not<\alpha$.


Matlab function with array input problem My question is simple but I just don't understand. I just wanted to plot Scorer function which is $$Hi(x)=\int_0^\infty{\exp(xt-\frac{t^3}{3})}dt.$$

First I tried to write, for example, x=0:0.1:1.
Then I write

y =  1/(pi)*integral(@(s) exp(x*s -(s^3)/3),0,inf).

It shows error using * and inner matrix dimensions must agree.

I also tried .* and .^ but the error then becomes only, matrix dimensions must agree.

I know my question seems stupid but I am just starting to learn matlab. What is the problem and what should be the correct steps of plotting this function? Thank you!

You need to loop through the different x inputs.For instance, the following will do what you are looking for:

$\texttt{x=0:0.1:1;}$

$\texttt{y=zeros(size(x));}$

$\texttt{for k=1:length(x)}$

$\quad \texttt{y(k)=1/pi*integral(@(s) exp(x(k)*s-s.^3/3),0,Inf);}$

$\texttt{end}$


Infinite solutions of Pell's equation $x^{2} - dy^{2} = 1$ Let $d > 1$ be a squarefree integer. Prove that the equation $x^{2} - dy^{2} = 1$ has infinitely many solutions in $\mathbb{Z} \times \mathbb{Z}$.

What I have done: let $ \ \mathbb{K} = \mathbb{Q}[\sqrt{d}]$. If $ d \not\equiv 1$ mod $4$, then $O_{\mathbb{K}} = \mathbb{Z}[\sqrt{d}]$ and the statement follows by the Dirichlet Unit Theorem. What about the case $ d \equiv 1 $ mod $4$ ?

This is known as Pell's equation. As is explained in the Wikipedia article, one of the convergents of $\sqrt{d}$ (one of the rational approximations of the irrational number $\sqrt{d}$ found by truncating the continued fraction expansion) is a solution to the equation. From this solution all other solutions can be found - the way in which these solutions are generated shows that there are indeed infinitely many of them.

If x solves $x^2-4 = dy^2$ then so does $x^2-2$

Put $(x^2-2)^2 = x^2 - 4x + 4$ , which is $d(xy)^2 + 4$

Dividing $x$, $y$ by $2$ will give this equation in units.

Now, if $x'$ and $x''$ are consecutive solutions, then $x x'' - x'$ is also a solution.  Put

Since we have $2$ and $x$ gives $x^2-2$, it then follows that $x_{n+1} = x x_n - x_{n-1}$ os also a solution, by

$x_{n+1} + x_{n-1} = x x_n$

One shows that by this rather exotic arrangement, that if two values solve this equation, then the third one does also, and thus that this series is recursive over powers of $x$.

Suppose $a, b, c, d$ form a series such that $xb = a+c$.  Then the measure $b^2 - ac$ is a constant, since $d = xc-b$.

Now, $c^2 - bd = c^2 - xcb + b^2  = (c-xb)c + b^2 = b^2-ac$

Thus, if eg $2, x, x-2, x^3-3x$ form such a series based on this solution


Proving the length of a circle's arc is proportional to the size of the angle How can I prove that:

  The length of the arc is proportional to the size of the angle.

Every book use this fact in explaining radians and the fundamental arc length equation $s = r\theta$. However no book proofs this fact.

Is this fact some axiom, some natural law like $\pi$ and the triangle side proportions?

Can I proof the above? Or is it something that you should just accept?

I believe this is a matter of definition. (One) Radian measure is defined as the angle subtended at the centre by an arc of length equal to the radius of a circle. So for any arc of length $l$, the angle (in radians) is $\theta = \dfrac l r$. So $ l = r\cdot\theta $

Let o = theta(angle subtended by the arc)
Ur equation can be written as s/r=o

We surely have two radii of circle subtending that angle o and forming arc s.
In the space between them we can fit more such radii.

The number of that radii will be decided by theta and they will form arc s.
Now divide arc s with the radius to get theta.


$\frac{1}{2 \pi i} \int_{a-i \infty}^{a+ i \infty} e^{st} \hat{f}(s) ds = \frac{2 e^{at}}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut du$ The claim is from this paper: https://www.researchgate.net/publication/220668741_Numerical_Inversion_of_Laplace_Transforms_of_Probability_Distributions. The whole thing goes as:

And where $\hat{f}$ is the Laplace transfrom of $f$.

But I dont get the last line. Where does that come from? Function $f$ is assumed to be real, so I understand how imaginary parts cancel each other away, but the last line seem to assume $\Re(\hat{f}(a+iu))\cos ut = -\Im(\hat{f}(a+iu))\sin ut$ (!?), which kind of result I have never heard of and could not find from the references.

Thanks for help.

If $f$ is  real valued and $a,u$ are real then $\hat{f}(a+iu)  = \overline{\hat{f}(a-iu)}$ so that (assuming the Fourier inversion theorem)

$$e^{-at} f(t) = \frac{e^{-at}}{2 \pi i} \int_{a-i \infty}^{a+ i \infty} e^{st} \hat{f}(s) ds = \frac{1}{2\pi}\Re( \int_{-\infty}^\infty  \hat{f}(a+iu))e^{iut}dt)\\= \frac{ 1}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut \ dt- \frac{ 1}{\pi} \int_0^{\infty} \Im(\hat{f}(a+iu)) \sin ut \ du$$
one term is even and the other one is odd so it must be
$$ =  \frac{e^{-at} f(t)+e^{at} f(-t)}{2} \quad -\quad\frac{e^{at} f(-t)-e^{-at} f(t)}{2}$$
whence for $t > 0$, if also $f$ vanishes on $t < 0$  it is
$$f(t) = 2  \frac{e^{-at} f(t)+e^{at} f(-t)}{2}  = \frac{ 2}{\pi} \int_0^{\infty} \Re(\hat{f}(a+iu)) \cos ut \ dt$$

The result follows from monotone class theorem for functions Wikipedia. Let $\mathcal{H}$ be a set of functions $f$ which admits the condition
$$
\int_{-\infty}^{\infty} \Re \hat{f}(a+ui) \cos(ut) du = -\int_{-\infty}^{\infty} \Im \hat{f}(a+ui) \sin(ut) du,
$$
or equivalently
$$
\int_{0}^{\infty} \int_{0}^{\infty}e^{-a \xi} \sin (u \xi) \cos(ut)f(\xi) d \xi  du = \int_{0}^{\infty} \int_{0}^{\infty} e^{-a \xi} \sin(u \xi)  \sin(ut)f(\xi) d\xi du.
$$
Assuming $a>0$ one can verify the following cases:

(1) If $A \subset \mathbb{R}_+ \cup \{0\},$ then by evaluating the integrals one can conclude that $ f(x) = \mathbb{1}(x \in A) \in \mathcal{H}.$

(2) If $g,f \in \mathcal{H},$ then by linearity $g+f, cf \in \mathcal{H}$ for $ c \in \mathbb{R}.$

(3) If $f_n \in \mathcal{H}$ is a sequence of non-negative functions that increase to a bounded function $f,$ then by monotone convergence theorem $f \in \mathcal{H}.$

Consequently all bounded integrable functions on $\mathbb{R}_+ \cup \{0\}$ belongs to $\mathcal{H}.$


What does Arrow's theorem say about Kaldor-Hicks social welfare functions with von Neumann-Morgenstern utility? Let $A$ be the set of all possible states of the world, let $G(A)$ be the set of all "lotteries" or "gambles", i.e. the set of all probability distributions over $A$.  Now consider an individual with a preference ordering of the various lotteries in $G(A)$.  Then the von Neumann-Morgenstern theorem states that, assuming the individual's preferences obeys certain rationality conbditions, there exists a function $u: A \rightarrow \mathbb{R}$, such that the individual's preference ordering maximizes the expected value of $u$.  Moreover, the function $u$ is unique up to linear transformations, i.e. maximizing the expected value of $u$ and maximizing the expected value of $a + bu$ yield equivalent results.

Now consider a society with N individuals, where each individual's preferences obey the von Neumann Morgenstern axioms.  Then we can define a social welfare function $W = a_1u_1 + a_2u_2 + ... + a_Nu_N$, where $u_i$ is the von Neumann-Morgenstern utility function for the $i^{\textrm{th}}$ individual, and $a_i$ is the reciprocal of the marginal utility of money for the $i^{\textrm{th}}$ individual.  As shown in this thread, $W$ is well-defined, because it's invariant under linear transformations of the $u_i$'s.  More importantly for our purposes, it is my understanding that maximizing $W$ will achieve a Kaldor-Hicks optimal result.  (Can someone back me up on this, and preferably tell me where I can find a proof?)

My question is, how does Arrow's impossibility theorem apply to a social preference ordering based on Kaldor-Hicks efficiency?  Specifically, given two outcomes in $A$, what would happen if we let the social ordering prefer the outcome that has a greater value of W?  Arrow's theorem, as usually stated, is about rules that are maps from $L(A)^N$ to $L(A)$, i.e. rules that take each individual's preference ordering on A, and then spit out a social preference ordering on A.  ($L(A)$ is the set of linear orders on the set $A$.)

But the rule I'm describing is not just based on each individual's preference ordering on $A$ (their preferences for certain outcomes), but on their von Neumannn-Morgenstern utility function $u$, i.e. on their preference ordering on $G(A)$ as well (their preferences under uncertainty).  So are there generalizations of Arrow's theorem that deal with maps from $L(G(A))^N$ to either $L(G(A))$ or failing that, maps from $L(G(A))^N$ to $L(A)$, as is the case with the rule I'm describing?  If an extension of Arrow's theorem does apply, what does it say about this rule?  What conditions does the rule obey or not obey?

Any help would be greatly appreciated.

Thank You in Advance.

Short answer

Arrow's theorem holds for social welfare function from the set of orderings profiles into the set of orderings, no matter whether the set of alternatives is a set of lotteries, or any other set of alternatives (see edit at the end of the post).
No, Arrow's impossibility theorem does not apply to social welfare function which take sets of utility profiles as domains, as Arrow acknowledged himself:

  From Wikipedia's article on Arrow's impossibility theorem:

  Voting systems that use cardinal utility (which conveys more information than rank orders; see the subsection discussing the cardinal utility approach to overcoming the negative conclusion) are not covered by the theorem.2

  Interview with Dr. Kenneth Arrow: "CES: Now, you mention that your theorem applies to preferential systems or ranking systems. Dr. Arrow: Yes CES: But the system that you're just referring to, Approval Voting, falls within a class called cardinal systems. So not within ranking systems. Dr. Arrow: And as I said, that in effect implies more information. Archived January 14, 2013, at the Wayback Machine.

Yes, there exists extensions of Arrow's impossibility theorem when the domain of the social welfare function is the set of vNM utility profiles, see corollary 4.1 of Social Choice with Interpersonal Utility Comparisons : A Diagrammatic Introduction, by Blackorby, Donaldson and Weymark

Long answer

The fact that -- as you mention -- the domain of your social welfare $W$ function is the space of vNM utilities implies that $W$ does not satisfy Arrow's "Independence of Irrelevant Alternative" axiom. Relaxing this axiom allows for the existence of social welfare functions satisfying the other Arrovian axioms.

To be more precise, as shown by Fleurbaey and Mongin's "The news of the death of welfare economics is greatly exaggerated" and others, Independence of Irrelevant Alternatives is equivalent to the combination of two weaker axioms:

Ordinal Noncomparability : the ranking of alternatives depends only on the individual orderings of the allocation. In utility terms, the social ranking is invariant to any increasing transformation of the individual utility levels.
Binary Independence : the ranking of two alternatives depends only on people's utilities at these two alternatives (and not on the relative ranking of these two alternatives with respect to some third, fourth, ... alternative)

Your social welfare function $W$ satisfies Binary Independence but violates Ordinal Noncomparability which is the reason why it does not satisfy the conditions of the Arrow's impossibility theorem. Obviously, as $W$ does not satisfy the conditions of the theorem, the theorem does not apply to it (hopefully this answers your first question?).

Now the whole question is "what do you replace Ordinal Noncomparability with ?".

(If your want to read more about what forgoing Ordinal Non-comparability implies -- and why it is a bad idea, you may want to read "Inequality, income, and well-being" by  Koen Decancq, Marc Fleurbaey and Erik Schokkaert.)

If you just discard the axiom without replacing it by a weaker constraint on the way your social welfare function should react to transformations of the utility profiles, then you allow for the existence of a plethora of social welfare functions satisfying efficiency, binary independence and non-dictatorship on a universal domains of preferences.

However, if you do add weaker restrictions, you might run into impossibility results again. Because you speak of vNM utility functions, it is interesting to consider the case of affine transformations. An alternative to  Ordinal Noncomparability which is relevant with these kinds of preferences is

Cardinal Noncomparability : the ranking of alternatives depends only on the individual orderings of lotteries. In utility terms, the social ranking is invariant to any affine transformation of the individual utility levels.

Then you recover an impossibility result if you slightly strengthen Arrow's efficiency condition, as shown in Corrolary 4.1 of Social Choice with Interpersonal Utility Comparisons : A Diagrammatic Introduction, by Blackorby, Donaldson and Weymark

  Corrolary 4.1  (roughly): If a social-welfare function satisfies
  Unrestricted domain, Binary Independence, Strong Pareto (i.e. if some
  are made better of and no-one is made worse-off, we have a social improvement,
  even if some individuals are not strictly better-off) and Cardinal
  Noncomparability, it must be a dictatorship

EDIT following OP question

The theorems I mentioned are valid for any set of alternatives. Whether the alternatives are lotteries or non-stochastic outcomes (or really whatever else) does not alter the validity of the theorems.

I think limiting the domain of vNM functions without weakening IIA would not be enough to prevent us from running into impossibilities. Here is a rather informal argument. vNM only restricts the ranking of lotteries, not the ranking of the "degenerated" lotteries through the function $u(.)$. So if $A = \{a,b,c,\dots\}$ is the set of degenerated lotteries, $u : A \rightarrow \mathbb{R}$ is not constrained.

So assume that Arrow's theorem does not hold on a vNM domain. This means that there exists a social welfare function $F$ satisfying the axioms of the theorem (except for unrestricted domain) on $G(A)$. Thus there exists a subrelation of $F$ over the set of degenerated lotteries $A$, say $\tilde{F}$, which also satisfies the axioms on $\tilde{F}$. But this implies that  $\tilde{F}$ satisfies the axioms of Arrow's theorem on a set of alternatives $A$ for an unrestricted domain of preferences over $A$, a contradiction.

Hope this helps.

Using the arrows imposibility theorem show that the “idea of social welfare function to determine a unique Point of maximum social welfare is not only utopian, but in principle impossible”


What exactly is a tensor product? This is a beginner's question on what exactly is a tensor product, in laymen's term, for a beginner who has just learned basic group theory and basic ring theory.

I do understand from wikipedia that in some cases, the tensor product is an outer product, which takes two vectors, say $\textbf{u}$ and $\textbf{v}$, and outputs a matrix $\textbf{uv}^T$. ($\textbf{u}$ being a $m\times 1$ column vector and $\textbf{v}$ being a $n\times 1$ column vector)

How about more general cases of tensor products, e.g. in the context of quantum groups?

Sincere thanks.

If you want to study a mathematical object, whether it is a set, manifold, group, vector space, whatever, it is often fruitful to look at natural collections of functions on that space.

Roughly, the purpose of the tensor product, $\otimes$, is to make the following statement true:
$$\text{functions}(X \times Y) = \text{functions}(X)\otimes \text{functions}(Y)$$

The specific details about which spaces of functions to choose depend on the type of mathematical object you are interested in.

Here's a pdf that explains it better than I can,
http://www.math.harvard.edu/archive/25b_spring_05/tensor.pdf

See whether this makes sense

http://stemandmusic.in/maths/mvt-algebra/matrixTP.php

In case it does not, please let me know in comments


$n^k \equiv n \pmod 5 $ for all $ n\in \mathbb Z $ iff $k \equiv 1\pmod 4 $ Let $k \in \mathbb N $ .Prove that $n^k \equiv n $ ( mod $5$ )  for all $ n\in \mathbb Z $ iff $k \equiv 1$ ( mod $4$ )

I was trying to solve this using the corollary of Fermat's little theorem , i.e $n^p \equiv n$ ( mod $p$)  for all $n \in \mathbb Z $  ,  but I got stuck.
Need some help.

Hint: if $n$ is a multiple of $5$, so is $n^k$. If not, try showing that $n^{k-1}\equiv 1$ mod $5$ if $k\equiv 1$ mod $4$.

Edit: missed the "iff". To show that it doesn't work if $k\not\equiv 1$ mod $4$, use the if part and $n=2$. E.g. if $k\equiv 3$ mod $4$ then $2^{k-2}\equiv 2$ so $2^k\equiv 8\not\equiv 2$.

First, let $k \equiv 1 (mod 4)$ then, $k=4m + 1$.
So, $n^k = n^{4m+1} = n^{4m} n$.
Now, we know, $n^4 \equiv 1 (mod 5)$ because, $1^4=1, 2^4=16, 3^4=81, 4^4=256, 6^4= 1296 ... 9^4=6561 $ which are all $1 (mod 5)$[EXCEPT FOR 5, and its multiples obviously].
So, $n^4 \equiv 1 (mod 5)$ and thus $n^{4m} \equiv 1(mod 5)$.
Thus, $n^k \equiv n^{4m+1} \equiv n^{4m} n \equiv 1.n \equiv n (mod 5)$.

Similarly, you can try it out the other direction to complete the $iff$ condition.


Knowing that $\prod_{i = 1}^na_i = 1$, prove that $\prod_{i = 1}^n(a_i + 1)^{i + 1} > (n + 1)^{n + 1}$. Given natural $n$ $(n \ge 3)$ and positives $a_1, a_2, \cdots, a_{n - 1}, a_n$ such that $\displaystyle \prod_{i = 1}^na_i = 1$, prove that $$\large \prod_{i = 1}^n(a_i + 1)^{i + 1} > (n + 1)^{n + 1}$$

We have that $$\prod_{i = 1}^n(a_i + 1)^{i + 1} \ge \prod_{i = 1}^n(2\sqrt{a_i}) \cdot \left(\sqrt[m]{\prod_{i = 1}^na_i^i} + 1\right)^m$$

where $\displaystyle p = \sum_{i = 1}^ni = \dfrac{n(n + 1)}{2}$, then I don't know what to do next.

I suspect that $\displaystyle \min\left(\prod_{i = 1}^n(a_i + 1)^{i + 1}\right) = 2^q$, occuring when $a_1 = a_2 = \cdots = a_{n - 1} = a_n = 1$, where $q = \dfrac{(n + 3)n}{2}$, although I'm not sure that $2^q > (n + 1)^{n + 1}, \forall n \in \mathbb Z^+, n \ge 2$.

(I've just realised this is just a redraft of problem 2, IMO 2012.)

For the numbers $2^{\frac{n}{2}}$ is greater than n+1 obviously satisfies, for n>5 can be proven by induction (hint: x$\sqrt{2}$ - x - 1>0 for x> $\sqrt{2}$ + 1)and for other n<6 check manually

According to the AM-GM inequality and telescopic products, we have that $$\prod_{i = 1}^n(a_i + 1)^{i + 1} = \prod_{i = 1}^n\left(a_i + i \cdot \frac{1}{i}\right)^{i + 1} \ge \prod_{i = 1}^n\left[a_i \cdot \frac{(i + 1)^{i + 1}}{i^i}\right] = (i + 1)^{i + 1}$$

The equality sign occurs when $\displaystyle a_i = \dfrac{1}{i},i = \overline{1, n} \implies \prod_{i = 1}^na_i = \prod_{i = 1}^n\dfrac{1}{i} \implies \prod_{i = 1}^ni = 1$, which is false, $\forall n \in \mathbb N, n \ge 2$.

$$\implies \prod_{i = 1}^n(a_i + 1)^{i + 1} > (i + 1)^{i + 1}$$


Continuous map from the ring on the unit circle Is there a surjective continuous map from the ring $r<x^2+y^2<1\,(0<r<1)$ on the unit circle $x^2+y^2<1$ ?

It seems NO, but how can it be done ?

  Edit: what if we add the injectivity to the map ?

Yes. Let me denote $R = \sqrt{r}$, so the ring is defined by $R < |p| < 1$ where $p=(x,y)$. Divide the ring into an inner subring
$$R < |p| \le (R+1)/2
$$
and an outer subring
$$(R+1)/2 \le |p| < 1
$$
Map the entire inner subring to the origin. Map the outer subring by pulling its inner boundary inward to the origin: for each angle $\theta$, the radial segment of the outer ring of angle $\theta$ connecting the circle of radius $(R+1)/2$ and the circle of radius $1$ is stretched by a constant factor over the radial segment of angle $\theta$ connecting the origin to the circle of radius $1$.

I find visual building of a map from an open ring to an open circle.

In the three dimension space imagine an open ring as a lateral surface of the cylinder. Now let pull the central line of the cylinder to the point. We have two cones with a common vertex and open boundaries. Finally, project (two-fold covering) these two cones to an open circle (two boundaries of cones map to the boundary of the circle).


How prove this diophantine equation $(x^2-y)(y^2-x)=(x+y)^2$ have only three integer solution? HAPPY NEW YEAR To Everyone! (Now Beijing time  00:00 (2015))

Let $x,y$ are integer numbers,and such $xy\neq 0$,
Find this diophantine equation all solution
$$(x^2-y)(y^2-x)=(x+y)^2$$

I use Wolf found this equation only have two  nonzero integer solution $(x,y)=(-1,1),(-1,-1)$,see wolf
$$\Longleftrightarrow x^3+y^3+x^2+y^2=xy(xy-1)$$
But How prove it?
and I found sometimes,and my problem almost similar with 2012 IMO shortlist:2012 IMO shortlist

(Apologies for the rather long solution.)

Write $x = ka$, $y = kb$, with $a, b$ relatively prime.
The equation $(x^2 - y)(y^2 - x) = (x + y)^2$ becomes
$$
\newcommand{\divs}{\; \mid \;}
(ka^2 - b)(kb^2 - a) = (a + b)^2 \tag{1}
$$
Or, rearranged,
$$
(ka + 1)(k b + 1) ab = (a + b)(kb^2 + b + ka^2 + a) \tag{2}
$$
Taking (2) modulo $a$, we get $kb^3 + b^2 \equiv 0$, and
dividing by $b^2$ (which is a unit mod $a$) we get $kb + 1 \equiv 0$.
So $a \divs kb + 1$ and likewise $b \divs ka + 1$.
Let $c = \text{gcd}(ka + 1, kb + 1)$,
$ka + 1 = bcA$, $kb + 1 = acB$ for relatively prime integers $A, B$.
Rewrite (2) as
$$
(bcA)(acB) ab = (a + b)(abcB + abcA)
$$
i.e.
$$
abc A B = (a + b)(A + B). \tag{3}
$$
Since $a$ and $b$ are relatively prime to $a + b$,
$$
ab \divs A + B \tag{4}
$$
Since $A$ and $B$ are relatively prime to $A + B$,
$$
AB \divs a + b \tag{5}
$$
Recall the following integer properties:

If $m \divs n$, then $|m| \le |n|$ OR $n = 0$.
$|m + n| \le |mn| + 1$ for all $m,n \ne 0$.
(Follows from triangle inequality and $(|m| - 1)(|n| - 1) \ge 0$.)

From these and (4), (5) we get that
$$
|a + b| \le |ab| + 1 \le |A + B| + 1 \le |AB| + 2 \le |a + b| + 2
$$
unless $A + B = 0$ or $a + b = 0$ or one of $A, B, a, b$ is $0$.

Case 1: $\boldsymbol{A + B = 0}$

Since $A$ and $B$ are relatively prime, we conclude WLOG $A = 1$, $B = -1$.
From (3), $abc = 0$.
But $a, b \ne 0$, so $c = 0$.
So $ka + 1 = kb + 1 = 0$.
So $x = ka = -1$ and $y = kb = -1$,
and in this case we get the solution $(x,y) = \boxed{(-1, -1)}$.

Case 2: $\boldsymbol{a + b = 0}$

Since $a$ and $b$ are relatively prime, we conclude WLOG $a = 1$, $b = -1$.
From (1),
$(k - 1)(k + 1) = 0$,
so either $k = 1$ or $k = -1$,
giving the solutions $(x,y) = (ka, kb) = (k, -k) = \boxed{(1, -1), (-1, 1)}$.

Case 3: $\boldsymbol{abAB = 0}$

From (3) this case implies $0 = (a + b)(A + B)$,
so this case is encompassed in the first two.

Case 4: $\boldsymbol{|a + b| \le |ab| + 1 \le |A + B| + 1 \le |AB| + 2 \le |a + b| + 2}$

From the inequality, we conclude that
$$
|a + b| \le |a| + |b| \le |ab| + 1 \le |a + b| + 2 \le |a| + |b| + 2
$$
Therefore,
$$
|ab| + 1 = |a| + |b| + r
$$
where $r \in \{0, 1, 2\}$.
In particular,
$$
(|a| - 1)(|b| - 1) \in \{0, 1, 2\}
$$
We know $a, b \ne 0$.
The first subcase is if $|a| = 1$ or $|b| = 1$ (WLOG $|a| = 1$).
Otherwise, $(|a|, |b|) = (2, 2), (2, 3), \text{ or } (3, 2)$,
Since $a$ and $b$ are relatively prime,
the first of these is impossible.
Otherwise we may assume $|a| = 2$.
By flipping the sign of $k$, we may assume $a$ is positive.
So we are left with three subcases: $a = 2$ and $b = 3$,
or $a = 2$ and $b = -3$, or $a = 1$.

(Case 4a) $a = 2$, $b = 3$

(1) gives
$(4k - 3)(9k - 2) = 25$,
which has no rational solutions.

(Case 4b) $a = 2$, $b = -3$

(1) gives
$(4k + 4)(9k - 2) = 1$,
which has no rational solutions.

(Case 4c) $a = 1$

From (4), $b \divs A + B$.
By the inequality we are still assuming,
$|A + B| + 1 =$ $|a + b|$, $|a + b| + 1$, or $|a + b| + 2$.
This implies $A + B = \pm(a + b)$
or $A + B \in \pm(a + b) \pm 1$.
With $a = 1$, we have
$A + B = b + 1, -b - 1, b + 2, b, - b, -b - 2$.

(Case 4c-i)
If $A + B = b + 1$ or $-b - 1$, then from (4) $b \divs A + B$ so $b = \pm 1$.
$a = 1$ and $b = -1$ implies $a + b = 0$, which we have already covered.
So $a = b = 1$.
Then (2) gives $(k + 1)^2 = 2(2k + 2)$,
so $k = -1$ or $k = 3$.
If $k = -1$ then $(x,y) = (-1, -1)$ which we already found.
If $k = 3$ then we have the solution $(x,y) = \boxed{(3,3)}$.

(Case 4c-ii)
If $A + B = b + 2$ or $-b - 2$, then from (4) $b \divs A + B$ so $b \divs 2$.
$b = \pm 1$ is covered in case 4c-i.
So assume $b = \pm 2$.
Then $A + B = 0$ or $\pm 4$.
If $A + B = 0$ this is covered by Case 1.
So assume $|A + B| = 4$.
Rewriting the inequality assumed in Case 4,
$$
|b + 1| \le |b| + 1 \le |A + B| + 1 \le |AB| + 2 \le |b + 1| + 2
$$
The middle term is $|A + B| + 1 = 5$, so $5 \le |b + 1| + 2$,
so $b = 2$.
Returning to (1), we get
$$
(k - 2)(4k - 1) = 9
$$
so $4k^2 - 9k - 7 = 0$,
which has no integer solutions.

(Case 4c-iii)
If $A + B = b$ or $-b$,
then write
$A + B = \frac{ka+1}{bc} + \frac{kb+1}{ac}$
so that $ka^2 + a + kb^2 + b = abc(A + B) = \pm b^2 c$.
Then from (2) we get
$$
(k + 1)(bk + 1) = \pm (b + 1) b^2 c
$$
Since no factors of $b$ are shared by $bk + 1$, $b^2 \divs k+1$.
But we also have $k + 1 + kb^2 + b = \pm b^2 c$,
implying $b^2 \divs  b$.
Since $b \ne 0$, $b \in \{-1, 1\}$.
So $(a,b)$ is one of $(1, -1)$ or $(1, 1)$.
But $(a,b) = (1, -1)$ satisfies $a + b = 0$, so was covered by Case 2.
And $(a,b) = (1,1)$ was covered in Case 4c-i.

Note that your equation is basically

$$-(x^2-y)^2 = (x+y)^2$$


Find the coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$ The coefficient of $x^9$ in the expansion of $(1+x)(1+x^2)(1+x^3)...(1+x^{100})$.

I tried the following concept, how to sum 9 using 1-9 only without repetition.

1:9, 2:1+8,3:7+2, 4: 6+3, 5: 5+4,

6:1+2+6 ,7:1+3+5, 8:2+3+4

The answer is 8.

How will it be solved using Binomial theorem.

Yes we can count directly the cases that is

$x^9$
$x\cdot x^8$
$x\cdot x^2\cdot x^6$
$x\cdot x^3\cdot x^5$
$x^2\cdot x^7$
$x^2\cdot x^3\cdot x^4$
$x^3\cdot x^6$
$x^4\cdot x^5$

to obtain $8$.


Calculate distance, knowing actual and perceived size What's equation would I use to calculate distance to an object, if I know it's actual and perceived size?

Say there is a line, and I know it's actual length is 65 (units), I perceive it as 62 (units).

I found various angular size equations on wikipedia and calculators like http://www.1728.org/angsize.htm however I don't know angle.

Suppose objects $X$ and $Y$ appear to be the same size. $X$ is actually of size $s_X$ and at distance $d_X$, and $Y$ is actually of size $s_Y$ and at distance $d_Y$.  Then:

$${s_X\over s_Y} = {d_X \over d_Y}$$

or equivalently:

$${d_X\over s_X} = {d_Y\over s_Y}$$

Any three of these will determine the fourth.

For example, a U.S. quarter dollar coin is about 2.5 cm in diameter. Held at a distance of 275 cm, it appears to just cover the full moon.  We can conclude that the  distance to the moon, divided by the diameter of the moon, is equal to about 275 cm / 2.5 cm = 110.

If we somehow know that the distance to the moon is around 385,000 km, then this tells us that the diameter of the moon is around 3500 km, or vice versa, but we cannot get either one without the other.

Similarly, we know from the existence of total solar eclipses that

$${d_☉\over s_☉} = {d_☾\over s_☾} $$

where ☉ is the sun and ☾ is the moon. So we know that ${d_☉\over s_☉}\approx 110$ also. Again, if we know that the distance to the sun is about $150·10^6$ km, we can conclude that the sun's diameter is about $1.36·10^6$ km, and if we know the diameter of the sun instead we can estimate its distance. But without one we can't get the other.

I looked at a 4 inch object which measured 800 inches away. At 12 inches from my eyes it measured 1/16th of an inch. For the perception cslculation factor (12 ÷ by .06 = 200). Then (800 ÷ by 200 = 4 inches). The formula matches good.