Real Analysis 8 | Example Calculation

[@PunmasterSTP]

Example calculation? More like "Exemplary conversation!" Thanks for presenting the material in such an easy-to-understand and interesting way. I definitely plan to watch this series through to its conclusion!

[@cptiglo5632]

This just looks like my stochastics course Rly interesting. I dont have much time for youtube but when I am here I rly enjoy ur video

[@PunmasterSTP]

How did your stochastics course go?

[@cptiglo5632]

@@PunmasterSTP I passed it. Exam was horrible tho

[@PunmasterSTP]

@@cptiglo5632 I'm glad to hear you passed, and I'm sorry the exam was horrible! I definitely remember that feeling from a few classes in college. Btw, I don't want to trigger any bad memories, but I was curious what topics were covered in your stochastics course. I am assuming there were a lot of Markov chains; am I right?

[@cptiglo5632]

@@PunmasterSTP oh God. I am a german Student I will try to list all the things we did. Discrete random variables and distribution at the beginning Then some other distributions like zipf or poisson and so on Markov chains Random variables with density(it means dichte in German it sounds rly stupid when translating it) Limits and Parameter estimation Confidence intervals Monte Carlo methods Aaand In German we call it "hypothesentest" Maybe hypothesis testing? And some other tests

[@PunmasterSTP]

@@cptiglo5632 That is really interesting and thanks for sharing! As a side note, when you mentioned Zipf, I immediately thought of this Vsauce video! kzbin.info/www/bejne/nHTRaa2pbpZlhac

[@duckymomo7935]

It’s not Analysis without deriving e

[@tatawhillman3783]

Hello at 8:52 when you used the inequality

[@brightsideofmaths]

[@ichkaodko7020]

I don't get the first part of proof. In order to determine monotonicity, why we use the an+1/1n and 1 here?

[@brightsideofmaths]

That is one possibility to show a_{n+1} > a_n

[@ichkaodko7020]

@@brightsideofmaths Thank you.

[@OscarGarcia-yb5vl]

min 4:50 why or what is the use of the binomial coefficient?

[@brightsideofmaths]

The binomial coefficient counts the correct occurrences in the binomial.

[@Hanulwalker]

Why did we use a binomial function? What are we measuring that has a binomial output?

[@none_of_your_business]

Shouldn't the ratio test for monotonicity of a sequence be a strict inequality for

[@leonfrolje6553]

No, that would be strict monotonicity then. Monotonically increasing/decreasing sequences (or generally functions) don't necessarily have to increase/decrease all the time, they can also be constant over an interval. Note that a constant sequence is also monotonical according to this definition and thus monotonically decreasing and increasing at the same time.

[@wqltr1822]

8:20, interesting, my analysis book bounds 1/(k!) under a geometric series instead of a telescoping sum. But this method is also cool.

[@brightsideofmaths]

Yes, I also like it :)

[@qiaohuizhou6960]

3:15 Bernoulli’s inequality

[@synaestheziac]

I tried proving Bernoulli’s inequality by induction and got stuck. I’ve got (1+x)^(n+1) = (1+x)(1+x)^n = (1+x)^n + x(1+x)^n >/ 1+nx + x(1+x)^n = 1 + (n+1)x - x + x(1+x)^n = 1 + (n+1)x + x(-1 + (1+x)^n) But I don’t know what to do with the big last term. I guess I probably need to use the binomial theorem?

[@demr04]

proposition P(k) : (1+x)^k >= 1+k*x base case P(k=1): (1+x)^1 >= 1+1*x 1+x >= 1+x; so base case is correct. Suppose that P(k=n) is correct, show that P(k=n+1) is also correct. First, choose k=n+1 to see what we are dealing... P(k=n+1): (1+x)^(n+1) >= 1+(n+1)*x Second, given that we can transform the left side of P(k=n) into P(k=n+1) we could try to multiply both sides of the inequality by (1+x) in P(k=n) to see what happend... (1+x)^n * (1+x) >= (1+n*x)*(1+x) (1+x)^(n+1) >= 1+x+n*x+n*x^2 = 1+x*(n+1)+n*x^2 (1+x)^(n+1) >= 1+x*(n+1)+n*x^2 Third, note that for x >= 0, we have that 1+x*(n+1)+n*x^2 >= 1+x*(n+1) which is the right side of the proposition for P(k=n+1), hence... (1+x)^(n+1) >= 1+x*(n+1)+n*x^2 >= 1+(n+1)*x (1+x)^(n+1) >= 1+(n+1)*x so P(k=n+1) is true for all x>0 in the real numbers and k in the natural numbers Fourth, given the alternating nature of (-1)^n between positive and negative, we impose the need that (1+x)^n >= 0, so 1+x >= 0 and then x >= -1 There is the proof by induction.

[@中文中国-t4u]

when will be update on liner algebra english version please ??? if i purchase your subscr.. can i have access to liner algebra eng version??

[@brightsideofmaths]

Yes, I update it with new videos soon :)

[@dassama7637]

Hii.. Sir, content not visible due to subtitles. Please keep content above the subtitles.

[@brightsideofmaths]

You can move the subtitles.

[@kristiyansmilyanski3447]

For me it's really hard to understand 😔

[@brightsideofmaths]

That is totally normal!

[@gdash6925]

I think you made a mistake. (n-1)/n is strictly less than 1

[@brightsideofmaths]

≤ includes < :)

[@gdash6925]

@@brightsideofmaths Yea but I mean it shouldn't include "="

[@brightsideofmaths]

@@gdash6925 Why not?

[@gdash6925]

@@brightsideofmaths because for the case that (n-1)/n = 1 for a natural number n. It makes no sense. Because you can rewrite this as n-1 = n. And that can't happen! However it does make sense if it's strictly less than 1 because then n-1 < n which suffices

[@brightsideofmaths]

@@gdash6925 Please always remember that ≤ means "less or equal". This is a logical "or" and in this sense we use it in mathematics.