Recent content by Also sprach Zarathustra

Hello, I need a help with the following: 1. Let $A$ be a transitive set, prove that $A\cup \{A \}$ is also transitive. 2. Show that for every natural $n$ there is a transitive set with $n$ elements.

Hello! I've a problem understanding the following lines(Arzela lemma, and first two sentences of a proof) from Fichtengoltz's book. I know, that some(2 members?) of you know Russian, help me please translate these line into English, with a short explanation on bold lines. Thank you!

What do you want?

Return to your algebra books from high-school!

Let us define the following infinite sum: $$\sum_{n=1}^{\infty} \frac{(\ln n)^n}{[\ln (n+1)]^{n+1}} $$ That sum is converges, use Cauchy root test, hence, $$\lim_{n\to\infty}\frac{(\ln n)^n}{[\ln (n+1)]^{n+1}}=0$$

I'm little late to work, so just try to answer my question: What is $$\sum\limits_{i=2}^n \frac{i}{x^i}$$ ? $$\sum\limits_{i=2}^n \frac{i}{x^i}=\frac{2}{x^2}+\frac{3}{x^3}+\frac{4}{x^4}+...+\frac{n}{x^n}$$ Yes? So, why then the above equals to what I wrote in my previous post? Good-luck! :)

$$\sum\limits_{i=2}^n (n - (n-i))x^{n-i}=\sum\limits_{i=2}^n ix^{n-i}=x^n\sum\limits_{i=2}^n \frac{i}{x^i}$$$$\sum\limits_{i=2}^n \frac{i}{x^i}=2(\frac{1}{x^2}+\frac{1}{x^3}+\frac{1}{x^4}+\frac{1}{x^5}+...+\frac{1}{x^n})$$...

Step 2*: True(induction hypothesis) : $z_{n+1}>z_{n}$ Need to prove: $$ z_{n+2}>z_{n+1}$$$$z_{n+2}=\sqrt{a+z_{n+1}}>\sqrt{a+z_n}=z_{n+1}$$

After substitution $x=\frac{1}{t}$, you will get: $$ \int_{0}^{\frac{\pi}{2}} \frac{\ln\cos (x)}{x^2}dx $$ Which can be written as:$$ \int_{0}^{\frac{\pi}{4}} \frac{\ln\cos (x)}{x^2}dx+ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\ln\cos (x)}{x^2}dx $$ Could you proceed? (Hint: this integral is...

Funny integral. $$ \int_0^{1}f(x) \ \ XD $$

Good observation!

In real analysis, above integral equals $\arctan(2\pi)$ (the area most exist, can't be $0$)

It's ok. :)There is few mistakes in my post, I will try to fix them, but this is one way to the right answer...

Let $f(x)$ be generating function of $T_n$: $$ f(x)=T_0+T_1x+T_2x^2+...$$Hence, we have: $$ f(x)=\sum_{i=0}^{\infty}T_ix^i=1+ \sum_{i=2}^{\infty}T_ix^i=1+\sum_{i=2}^{\infty}(aT_{i - 1} + bn)x^i $$ $$=1+\sum_{i=2}^{\infty}aT_{i - 1}x^i+ \sum_{i=2}^{\infty}bix^i= 1+a\sum_{i=2}^{\infty}T_{i -...

Say that, for some n \geq k, the followinf true: $$ \frac{a_{n+1}}{a_n}\le\frac{b_{n+1}}{b_n} $$ We'll write: $$ \frac{a_{k+1}}{a_k}\le\frac{b_{k+1}}{b_k} $$ $$ \frac{a_{k+2}}{a_{k+1}}\le\frac{b_{k+2}}{b_{k+1}} $$ $$ \frac{a_{k+3}}{a_{k+2}}\le\frac{b_{k+3}}{b_{k+2}} $$ . . . $$...