Tag: prove

elementary number theory – prove that $ frac{sigma(1)}{1}+frac{sigma(2)}{2}+dots+frac{sigma(n)}{n} leq 2 n $

(HMMT 2004) For every positive integer $n$, prove that
$
frac{sigma(1)}{1}+frac{sigma(2)}{2}+dots+frac{sigma(n)}{n} leq 2 n
$

If $d$ is a divisor of $i,$ then so is $frac{i}{d},$ and $frac{i / d}{i}=frac{1}{d} .$

Summing over all divisors $d$ of $i$ (which is $sigma(i)$ ), we see that $frac{sigma(i)}{i}$ is the sum of all the reciprocals of the divisors of $i ;$

that is,
$
frac{sigma(i)}{i}=sum_{d | i} frac{1}{d}
$ …..

But how they concluded that $
frac{sigma(i)}{i}=sum_{d | i} frac{1}{d}
$ ??? ***

i am getting trouble in understanding this question in recent few days, what i till now understand is that

if
$d$ is divisor of $i$ then $i/d$ is also divisor of $i$ , so

$frac{i / d}{i}=frac{1}{d} .$ but they are not all divisors of i so how we get *** ,i also tried to take some examples but still can’t get it,

can anyone explain this ..

thankyou

general topology – I don’t know how to prove the statement.

The set of all matrices in $M_2(mathbb{R})$ such that none of the eigenvalue is real is open.

I tried to prove this by considering the characteristic polynomial to be a continuous function and then the inverse image of ${ mathbb{C} backslash mathbb{R} }$ is open. I don’t know if my argument is correct. Any help is highly appreciated.

real analysis – Gradient of convex $f$ is $L$-Lipschitz, then how to prove this $f(x^{k+1}) leq f(x) + (x^{k+1}-x)^Tnabla f(x^{k}) + (L/2)|x^{k+1} – x^k|$?

I have been struggling to prove the following inequality given in this paper–please see eq. (34)

$$f(x^{k+1}) leq f(x) + (x^{k+1}-x)^Tnabla f(x^{k}) + frac{L}{2}|x^{k+1} – x^k|$$ for an iterative scheme — ADMM — where $f$ is a convex function whose gradient is $L$-Lipschitz and $k$ is an iteration number.

Although they show some proof but I can barely follow. Please help to clarify this proof. Below I have pasted for your convenience.
begin{align}
f(x) – f(x^{k+1}) – left( x – x^{k+1} right)^T nabla f(x^{k}) &geq ? tag 1\
f(x^{k}) + left( x – x^{k} right)^T nabla f(x^{k})
– f(x^{k+1}) – left( x – x^{k+1} right)^T nabla f(x^{k}) &= ? tag 2\
f(x^{k}) – f(x^{k+1}) – left( x – x^{k+1} right)^T nabla f(x^{k}) &geq frac{-L}{2} | x^{k+1} – x^{k} |^2 tag 3
end{align}

How did they achieve $(1)$ and also $frac{-L}{2} | x^{k+1} – x^{k} |^2 $ ?

From the Lipschitz condition, the bound that can be used as a starting point is
begin{align}
f(y) leq f(x) + left( y – x right)^T nabla f(x) + frac{L}{2} | y – x |_2^2.
end{align}

If I set $y = x^{k+1}$ and $x = x^{k}$, then we have
begin{align}
f(x^{k+1}) leq f(x^{k}) + left( x^{k+1} – x^{k} right)^T nabla f(x^{k}) + frac{L}{2} | x^{k+1} – x^{k} |_2^2.
end{align}

Then, what to do? Can someone please enlighten me and show the steps in between? I would be so grateful to you.

programming languages – How to prove that replacing a character in a string in both C and JavaScript is equivalent?

I would like to try some different proofs, specifically in proving equivalence of the implementation of some feature in two different programming languages (C and JS in this question).

This is about proving that replacing one character in a string is equivalent in C and JS.

In JS, for example, strings are immutable, while in C, strings are mutable. So in JS you might have to do this:

functioni replaceAt(string, index, replacement) {
  return string.substr(0, index) + replacement + string.substr(index + replacement.length)
}

While in C you might just do something like this:

#include<stdio.h>

int
main() {
  char string(11) = "hello world";
  string(1) = 'i';
  printf("%s", string);
  return 0;
}

Basically, I am trying to come up with an example where, the perceived effect or desired outcome is for all intents and purposes the same. That is, the end result is that the character was replaced at a specific position (the same position in each language). Even though in one language the string was mutable, while in the other, it was immutable. What needs to be taken into account here to make a proof saying these are equivalent? How do we capture the notion of “partial equivalence” or “perceived equivalence”? (By that I mean, the outcome is roughly the same, so we want to make a proof statement that these are the same with regard to some spec).

elementary set theory – Prove that if $x in A$ and $x notin (B cap C^c)$, then $(x in A$ and $x notin B)$ or $(x in A$ and $x notin C^c)$

This is part of a larger proof where I am attempting to show that
$$A setminus (B setminus C) = (A setminus B) cup (A setminus C^c).$$

I have already shown that $(A setminus B) cup (A setminus C^c) subseteq A setminus (B setminus C)$.

I also know that $A setminus (B setminus C) = A setminus (B cap C^c)$. My proof so far only consists of

“If $x in A setminus (B cap C^c)$, then $x in A$ and $x notin B cap C^c$. If $x notin B cap C^c$ then $x in B$ and $x in C^c$. Thus if $x in A setminus (B cap C^c)$, then $(x in A$ and $x notin B)$ or $(x in A$ and $x notin B).$”

The problem I am having is that I can’t use the distributive properties of conjunctions and disjunctions to prove this (as we have not done them so far). I’m not quite sure how to go about showing the above without using propositional logic. Any help would be appreciated.

Prove you can weigh any number between 1 and $frac{3^{n+1} -1}{2}$ using $n+1$ weights – Discrete

You have $n+1$ weights, with each weighing $1,3,9, dots, 3^n$

Prove that you can weigh with a traditional scale ( The one with two bowls) each integer weight between $$1 ~~~~~text{And }~~~~~frac{3^{n+1} -1}{2}$$

My go:
This was very confusing to me because I did not understand well how you can possibly weigh, let’s say $4$ (Maybe it’s just putting $1+3$ ?)

I tried proving using induction:

  1. If $n=0$ then we have $1$ weight weighing 1, and so we can weigh $frac{3^1 -1}{2} ~~~ text{And} ~~~ 1 = 1$ which is obvious.

  2. Assume you have $n+1$ weights and you can weigh each integer between $1$ and $frac{3^{k+1} -1}{2}$

  3. Now we prove for $n = k+1$ so: $1$ to $frac{3^{k+2}-1}{2}$ using the fact we can weigh $frac{3^{k+1} -1}{2}$
    However I am stuck from here, I am clueless on how to use the fact we now have a weight of $1,3,9,dots,3^k,3^{k+1}$

This seems like a known question, but I could not find anything on the web!
Thank you!

Will it be the most humiliating day of every libs life if studies prove that hydroxicholoroquine works as a preventative measure?

That is not the way that sensible people handle things. 

. The question is not whether it works as a preventative measure, as much as whether it is a relatively safe chemical/drug to take..If it was suggested that swallowing battery acid might work, it is a bad idea even if it did kill the virus or the common cold or anything..I think you are just hoping for a little more drama  

group theory – Given H subgroup of G, prove the function f(gH)=Hg^-1 is a bijection [without relying on a specific g]

The question explicitly states that we must assume the function is well defined and works without relying on a specific g. I assume this is because even if g1!=g2 g1H==g2H is possible, and the function actually is defined from a group of left cosets to a group of right cosets.

I’ve tried proving the groups are of equal size but that doesn’t help in the infinite case, and without a specific group I don’t know how to prove that there are as many right cosets as there are left cosets, and I don’t know what to do with the inverse.

Prove that {a^*b^*c^* – {a^n b^n c^n | n >=0}} is not a regular language, by using the pumping lemma

Suppose that $L$ is regular. Take the complement of $L$ and intersect it with $a^*b^*c^*$. You are left with $M = {a^n b^n c^n mid n ge 0}$. Due to the closure properties of regular languages, $M$ is also regular.

Let $n_0$ be the pumping length of $M$. By the pumping lemma there is some $x in {1, dots, n_0}$ such that all words $a^{(n_0-x)+ix} b^{n_0} c^{n_0}$ for $i ge 0$ belong to $M$.

Pick $i=0$ to obtain $a^{n_0 – x} b^{n_0} c^{n_0} in M$, a contradiction.

nt.number theory – Prove that a ring homomorfism $f: mathbb{Z}_m rightarrow mathbb{Z}_n $ is injective iff n|m and $gcd(n/m,m) = 1$

Prove that a ring homomorfism $f: mathbb{Z}_m rightarrow mathbb{Z}_n $ is injective iff n|m and $gcd(n/m,m) = 1$

I already did the proof of f being injective under the hypotesis, but I have no idea about the first implication; in particular I dont know how to show that $mcd(n/m,m)=1$.