## probability theory – Prove there exists a continuous funtion \$f:mathbb Rrightarrow mathbb R\$

Suppose $$X_1,X_2cdots$$ are i.i.d. with $$mathbb E:X_i=0$$ and $$mathbb E:X_i^2=1$$. and let $$Z$$ follow standard normal distribution ($$N(0,1)$$). Futher, suppose that $$Y_1,Y_2,cdots$$ are i.i.d. and finite $$a.s.$$, but $$mathbb E|Y_n|=infty$$ for all $$n$$. Prove there exists a continuous funtion $$f:mathbb Rrightarrow mathbb R$$ such that

$$frac{1}{n}left(left(sum_{i=1}^nX_iright)^2-left(sum_{i=1}^nX_iright)left(sum_{i=1}^{4n}X_iright)+left(frac{sum_{i=1}^{4n}X_i}{2}right)^2+Y_nright)implies f(Z)$$

I was really confused how to approach this problem. It will be great help if anyone help me to figure out this.

## cv.complex variables – Prove that a logarithmic integral is convergent

Define, $$I=int_{-infty}^inftyfrac{log^-|zeta(frac{1}{2}+it)|}{frac{1}{4}+t^2}dt$$
where $$zeta$$ denotes the Riemann zeta function and $$log^-|f|=max(-log|f|,0)$$.

Question Show that $$I$$ is integrable or $$int_{-infty}^inftyfrac{log^-|zeta(frac{1}{2}+it)|}{frac{1}{4}+t^2}dt

Hint: Use Riemann zeta theory about zero spacings.

## real analysis – How to prove the elementary inequalities? \$||{a+b}|^alpha – |a|^alpha| leq |b|^alpha\$

For $$a,b inmathbb{R}$$, some $$C_alpha>0$$,
begin{align*} &big ||a+b|^alpha – |a|^alpha big |leq|b|^alpha,&quad text{for }alphaleq 1,\\ &big ||a+b|^alpha – |a|^alpha big |leq C_alphabig(|a|^{alpha-1}+|b|^{alpha-1} big)|b|, &quad text{for }alpha>1 end{align*}
I kindly ask for hints how to prove these inequalities hold true. I could just notice that when $$|a+b|^alphageq|a|^alpha$$, then $$|a+b|^alphaleq|a|^alpha+|b|^alpha$$, but unfortunately I have no idea of how to e.g. apply the knowledge about the polynomial/exponential function $$c^alpha=e^{alphalog c}$$ or something else.

Thank you.

## proof writing – If \$X\$ is compact then prove that \$X\$ is complete and totally bounded.

I tried to do it in a way different from my textbook:

Let $$(X,d)$$ be a compact metric space then it is totally bounded.(I have been Ble to prove this).

My doubt lies in the following part that I have tried to prove:

Let $$X$$ be a compact metric space then $$X$$ is totally bounded.We choose a cauchy sequence $${x_n}$$ in $$(X,d)$$ .

Let $$A$$ be a subset of $$(X,d)$$ such that $$A={x_n}$$.Then $$A$$ will be totally bounded too as $$A subset (X,d)$$.

Let for an $$epsilon > 0$$ there exist points $${x_1′,cdots x_n’}$$ such that $$B_d(x_k’,epsilon)$$ contains infinitely many points of $$A$$.Then as $${x_n}$$ is a cauchy sequence so we can conlcude that $${x_n}$$ converges to $$x_k’$$. Also as $$(X,d)$$ is a metric space so $${x_n}$$ can converge to only one points and we have shown that $$x_k’$$ is a limit point of $${x_n}$$.

## Prove Sine of sum equals Sum of Sines by Induction

Use mathematical induction to show that there exists real numbers $$a_1+a_2+a_3+cdots+a_n$$ such that $$|a_i|le 1$$ for $$i=1, 2, 3, …, n$$ and such that $$sin (x_1+x_2+x_3+cdots+x_n)= a_1sin x_1 + a_2sin x_2 + a_3sin x_3 + cdots+a_nsin x_n$$.

## general topology – Using Arzela-Ascoli to prove there’s a converging sub-sequence

I am trying to solve the following problem in preparation for an upcoming exam in introductory topology and I’m unable to complete it entirely. My problem is with (b), but I’ve stated the full problem since it is likely that (a) is required for (b):

Let $$X$$ be a compact metric space.

a. Prove that every continuous function $$f:Xrightarrow mathbb{R}$$ is uniformly continuous.

b. Let $$T,S : X rightarrow X$$ be isometries.
Let $$f_1 in C(X,mathbb{R})$$. Denote $$f_{n+1} = frac{1}{2}Big(f_n(T(x))+f_n(S(x))Big) quadforall xin X, nin mathbb{N}$$ .
Prove that $${f_n}_{nin mathbb{N}}$$ has a converging subsequence using the sup norm.

So I’ve tried the following two approaches:

1. Use Arzela-Ascoli by first proving $$F:= {f_n}$$ is closed, bounded and equicontinuous. It seems impossible to prove it is closed – maybe I’m missing something? (tried assuming by contradiction there is a $$g$$ in $$cl(F)$$ not in $$F$$ but this didn’t take me too far).

2. Proving the space is complete (since $$X$$ is compact $$C(X,mathbb{R})= B_c(X,mathbb{R})$$ and using the fact that $$X$$ is a topological space and $$mathbb{R}$$ is complete) and then trying to prove that $${f_n}_{nin mathbb{N}}$$ is Cauchy, therefore converges, therefore has a convergent subsequence. Also got stuck in this case, although it seems more promising due to the isometries, but not sure.

What am I missing? Any help would be much appreciated!

## time complexity – Prove a lower bound

Prove: $$n^{5}-3n^{4}+logleft(n^{10}right)∈ Ωleft(n^{5}right)$$.

I always get stuck in these types of questions, where there is a $$”-(xy^{z})”$$ in the expression.
Whenever I see the solutions for these type of questions, I can’t identify a single method that works every time and it’s frustrating. How do I approach these types of questions?

## what are the prerequisite to prove this. The hypotenuse of a PPT is always odd.

can anyone tell me what are topics I need to study to solve this question?

## complexity theory – How to prove this josephus problem variation is a np-complete or np-intermediate problem?

have a problem that is a Josephus problem variation. It is described below:

There are m cards with number from 1 to m，and each of them has a unique number. The cards are dispatched to n person who sit in a circle. Note that m >= n.

Then we choose the person “A” who sits at the position “p” to out of the circle, just like the Josephus problem does. Next step we skip “k” person at the right of p while k is the number of the card toked by the person “A”, and we do the same thing until only one person left in the circle.

Question is given n person and m cards, can we choose n cards and allocate them to the n person, to make that whether start at which position(exclude the first position), the person survival at the end is always the first person in the circle.

For example, m = n = 5, the only solution is (4, 1, 5, 3, 2).

I think this problem is a np-complete or np-intermediate problem, but I can’t prove it. Anybody has a good idea to find a polynomial time solution or prove it’s np-hard?

— example solutions —

`````` 2: ( 1,  2)
2: ( 2,  1)
3: ( 1,  3,  2)
3: ( 3,  1,  2)
4: ( 4,  1,  3,  2)
5: ( 4,  1,  5,  3,  2)
7: ( 5,  7,  3,  1,  6,  4,  2)
9: ( 2,  7,  3,  9,  1,  6,  8,  5,  4)
9: ( 3,  1,  2,  7,  6,  5,  9,  4,  8)
9: ( 3,  5,  1,  8,  9,  6,  7,  4,  2)
9: ( 3,  9,  2,  7,  6,  1,  5,  4,  8)
9: ( 6,  1,  8,  3,  7,  9,  4,  5,  2)
10: ( 3,  5,  6, 10,  1,  9,  8,  7,  4,  2)
10: ( 4,  5,  2,  8,  7, 10,  6,  1,  9,  3)
10: ( 5,  1,  9,  2, 10,  3,  7,  6,  8,  4)
10: ( 6,  3,  1, 10,  9,  8,  7,  4,  5,  2)
10: ( 8,  5,  9, 10,  1,  7,  2,  6,  4,  3)
10: (10,  5,  2,  1,  8,  7,  6,  9,  3,  4)
11: ( 2,  1, 10, 11,  9,  3,  7,  5,  6,  8,  4)
11: ( 3,  7, 11, 10,  9,  8,  1,  6,  5,  4,  2)
11: ( 3, 11, 10,  9,  8,  1,  7,  2,  4,  5,  6)
11: ( 4,  1, 10,  2,  9,  8,  7,  5, 11,  3,  6)
11: ( 4,  2,  7, 11,  5,  1, 10,  9,  6,  3,  8)
11: ( 4,  7,  2,  3,  1, 10,  9,  6, 11,  5,  8)
11: ( 4,  7,  3,  9, 11, 10,  1,  8,  6,  5,  2)
11: ( 4, 11,  7,  2,  1, 10,  9,  6,  5,  3,  8)
11: ( 5, 11,  3,  9,  8,  7,  6,  1, 10,  4,  2)
11: ( 6,  1, 10,  2,  9,  8,  7,  5, 11,  3,  4)
11: ( 6,  2,  7, 11,  5,  1, 10,  9,  4,  3,  8)
11: ( 6, 11,  1,  3, 10,  2,  7,  5,  4,  9,  8)
11: ( 9,  5,  3,  1, 10,  2,  8,  7, 11,  6,  4)
12: ( 1,  7, 11, 10,  4,  9,  2, 12,  6,  5,  8,  3)
12: ( 3,  7, 12,  2, 11, 10,  9,  1,  6,  5,  4,  8)
12: ( 3,  8, 11,  2, 12,  9,  1,  7,  5, 10,  4,  6)
12: ( 4,  2,  5,  1, 11, 10,  9,  8, 12,  7,  3,  6)
12: ( 4,  3,  7,  6,  1, 11, 10,  9,  8, 12,  5,  2)
12: ( 5,  1,  6, 11,  9,  2, 10,  7, 12,  8,  3,  4)
12: ( 5,  2,  3, 12,  9, 10,  7,  6,  1, 11,  4,  8)
12: ( 5,  7, 12,  2, 10,  9,  8, 11,  1,  4,  6,  3)
12: ( 7,  1,  2,  3,  5,  9, 10,  8, 11,  6, 12,  4)
12: ( 8,  7,  1, 11,  9,  3,  5, 10,  6,  4, 12,  2)
12: ( 8,  7, 11, 10, 12,  3,  1,  9,  6,  5,  4,  2)
12: (12,  3, 11,  5,  1, 10,  8,  7,  6,  4,  9,  2)
12: (12,  7, 11,  1,  9,  3,  2, 10,  6,  5,  4,  8)
13: ( 2,  1,  4,  7, 11,  6,  3, 10, 13,  5,  8, 12,  9)
13: ( 2,  5, 13, 12,  4, 11,  3,  1,  9,  7,  8,  6, 10)
13: ( 2, 13, 12, 11,  3,  1,  9,  4,  8,  7, 10,  5,  6)
13: ( 3,  5,  2,  1, 12,  9, 11, 10,  7,  6, 13,  4,  8)
13: ( 3,  5, 13,  1, 11,  2,  9,  8,  7, 12,  6,  4, 10)
13: ( 4, 13,  3,  1, 12, 11, 10,  9,  7,  2,  5,  6,  8)
13: ( 6,  4,  3,  1, 10, 11, 13,  5,  9, 12,  7,  8,  2)
13: ( 6,  4, 13,  7,  5,  1, 12, 11, 10,  9,  8,  3,  2)
13: ( 6,  7,  3, 13, 12, 11, 10,  2,  1,  9,  5,  4,  8)
13: ( 6,  7, 13, 11,  2, 10,  9,  1,  8, 12,  5,  3,  4)
13: ( 6, 11,  7, 13,  1, 10,  2, 12,  9,  8,  5,  4,  3)
13: ( 7,  3,  2,  1, 11, 10,  9,  8, 13,  5, 12,  4,  6)
13: ( 7,  5, 13,  3, 10, 11,  2,  9,  1,  6,  8,  4, 12)
13: ( 7,  5, 13,  3, 11,  2,  9,  8,  1,  6, 12,  4, 10)
13: ( 7,  5, 13,  3, 11, 12,  2,  1,  9,  8,  6,  4, 10)
13: ( 7,  9,  1, 11,  3, 13,  2, 10, 12,  6,  5,  4,  8)
13: ( 8,  3,  5, 11, 13,  9, 10,  7,  1,  6,  4, 12,  2)
13: ( 8,  3, 13,  1,  5, 11, 10,  9, 12,  7,  6,  4,  2)
13: ( 9,  3, 13,  2, 10,  4,  1,  7,  6,  5, 12, 11,  8)
13: ( 9,  4,  7,  5,  1, 11, 13, 10, 12,  8,  6,  3,  2)
13: ( 9,  5,  4, 13,  2, 11,  8, 10,  1,  7, 12,  3,  6)
13: ( 9,  5, 13,  4, 11,  1,  8,  3,  7, 12,  6, 10,  2)
13: (10,  4,  3,  5, 13,  1,  9, 11,  7,  6,  8, 12,  2)
13: (11,  2,  7,  3, 12,  1, 10,  9,  6,  5, 13,  4,  8)
13: (11, 13,  5,  2, 10,  9,  8,  7,  1,  6,  4,  3, 12)
13: (11, 13,  7,  1, 12,  9,  2,  3, 10,  5,  4,  6,  8)
13: (12,  1,  3,  5, 11, 13,  4, 10,  9,  8,  7,  6,  2)
13: (12,  7, 13,  3, 11,  1,  9,  8,  6,  5, 10,  4,  2)
13: (12, 13,  7, 11,  2,  5,  1,  9, 10,  6,  4,  3,  8)
13: (13,  3,  1, 12, 11,  2,  9, 10,  7,  6,  4,  5,  8)
13: (13,  3,  7,  1,  5, 12,  4, 10,  9,  8, 11,  6,  2)
14: ( 3,  5, 13, 14,  1, 12, 11, 10,  9,  8,  7,  6,  4,  2)
14: ( 3,  9,  1, 13, 11, 10,  2,  4,  7, 14,  6,  8,  5, 12)
14: ( 3, 14,  4, 12, 11,  1,  9,  8,  2, 13,  7,  5, 10,  6)
14: ( 4, 11,  1, 13,  7, 10, 12,  2, 14,  9,  8,  5,  6,  3)
14: ( 4, 14,  2,  5, 13,  1, 12, 11,  7,  6, 10,  9,  3,  8)
14: ( 5,  7,  1, 13, 12, 11, 10,  2,  9,  8, 14,  6,  4,  3)
14: ( 6,  3, 14,  5, 11, 13,  2, 12,  9,  1,  7,  4,  8, 10)
14: ( 6, 14,  1, 12,  5, 13,  2, 11,  9,  7,  8,  4,  3, 10)
14: ( 7,  5, 13, 12,  1, 11,  4, 10,  2, 14,  9,  8,  6,  3)
14: ( 7, 11,  5, 13,  1,  3,  2,  4, 10,  9, 14,  6,  8, 12)
14: ( 7, 14,  1, 13,  2,  5, 11, 12, 10,  9,  8,  4,  3,  6)
14: ( 8,  7,  5, 13,  2, 11,  3,  9, 10, 12,  1, 14,  4,  6)
14: (11,  2, 10,  5,  8,  7,  9,  1, 13, 14, 12,  4,  3,  6)
14: (11,  3, 14,  2, 13,  1, 10,  8,  9,  7,  5, 12,  4,  6)
14: (11,  5,  3, 14,  2,  1, 13, 10,  8,  7,  6, 12,  4,  9)
14: (11, 14,  5,  3, 13,  1, 10,  2,  9,  4,  7,  8, 12,  6)
14: (12,  1, 14,  3, 13,  4, 10,  9,  2,  7,  6,  5, 11,  8)
14: (12, 11,  7,  5, 13,  3,  2, 14,  1,  9,  8,  4,  6, 10)
14: (12, 14,  7, 13,  6,  5, 11,  1, 10,  9,  8,  4,  3,  2)
14: (13,  1,  7,  2, 11,  3,  9, 14,  8,  6,  5, 10,  4, 12)
14: (13, 11,  3,  1,  4,  2,  7, 10,  9,  6, 14, 12,  5,  8)
14: (14,  1, 13,  3, 11,  5, 10,  9,  2,  6,  8,  7,  4, 12)
14: (14, 5, 1, 13, 12, 2, 11, 3, 7, 9, 6, 8, 4, 10)
``````

## bilinear form – How to prove that the following set is a linear subspace?

I need some help with the following problem. I have no idea how to go about a) and in b) I know the conditions of a linear subspace but I don’t know how to prove it and show the dimension.