Yes it is possible, applying Riemann-Stieltjes's integral definition, we come up with the integral from 0 to 1 of ln(1+x)d(x^2)=integral from 0 to 1 of 2x ln(1+x)dx=1/2, after applying exponential we have our solution: e^(1/2)
Nice use of the squeeze theorem :D. The brute force approach is Ln the whole expression then its the sum of ln(1+i/n^2) for i =1 to n then bring in the power series of ln(1+x)... So you get (1/n^2 + 2/n^2 +... n/n^2) - (1/n^4 + 4/n^4 + 9/n^4 + n^2/n^4) + (1/n^6 +8/n^6 +27/n^6 +... n^3/n^6) +... The first bracket is (n(n+1)/2) (1/n^2) the second bracket is (n(n+1)(2n+1)/6)(1/n^4) the third bracket is (n(n+1)/2)^2 (1/n^6) etc the first bracket simplifies to 1/2 +1/2n the second simplifies to 2/n +3/n^2 +1/n^3 the third 1/4n^2 + 1/2n^3 +1/4n^4 etc we notice that all the terms are devided by some power of n except for the 1/2 from the first bracket thus taking the limit as n tends to infinity we get 1/2. Since Ln of the original tends to 1/2 the original itself tends to e^1/2
In german we call the squeez theorem the police theorem for some reason. It's much more funny to just say police this and you have your answer :D My professor always cracked the joke: Have you seen any Policeman converging? I guess it's funny in lectures not in rl. Nice solution btw keep it up!
Because lim(a×b) = lim(a) × lim(b), we can take the middle term out of the limit and deal with it separately while still pairing the remaining terms like before. But the limit of the middle term by itself will just be 1, so we don't have to worry about it!
Hey mu prime math, I could do it applying the Riemann Stieltjes's integral definition, you can notice the next fact, when we apply logarithm we are going to have \lim_{n \to infty} \sum_{k=1}^{n}ln(1+k/n^2) as you can notice here we have our d(alpha) = d(x^2) due to the grade of the denominator, then we will build our integral, we have the definition of the integral like int_{a}^{b}f(x)d(alpha(x))=lim_{n \to infty}\sum_{k=1}^{n}f(delta(x) +a)(alpha(a+delta(x)k)-alpha(a+delta(x)(k-1)), then we have int_{0}^{1}ln(1+x)d(x^2)=1/2 , then after applying exp we come up with e^(1/2)
Interesting! You would have to enumerate more clearly what your function α would be, and what algebra gets that (α_k - α_{k-1}) term multiplied on the outside without changing the answer, but it seems like that would be doable.
Mr. prof. Mu Prime Math.I like your proof,maby a little bit overwrite, and like Bleckpenredpen ,you write with a pen with INK finished. Thanks for understanding.
If you take more care of your spelling and some things do not explain them on the air, then your videos will look more professional and easier to understand. Thanks for your time.
You’re welcome!! You earned the shirt! And thanks for the shout out!
I am still thinking this problem. I wonder if we can somehow turn that into a Riemann sum and integral then solve.
My guess us that it's not possible in this case, since you would have to take the ln and then there's nowhere to get a 1/n from.
YES, WE CAN.
What about
\log(L) = \sum_{k=1}^n \log(1+(k/n)/n) ~ \int_0^1\log(1+x dx) = \int_0^1 (x dx + O((dx)^2)) = 1/2, as dx is infinitessimal.
Yes it is possible, applying Riemann-Stieltjes's integral definition, we come up with the integral from 0 to 1 of ln(1+x)d(x^2)=integral from 0 to 1 of 2x ln(1+x)dx=1/2, after applying exponential we have our solution: e^(1/2)
@@joanmartinsuarezloaiza2767 Can you elaborate on how precisely you come up with int_0^1 ln(1+x) d(x^2) ?
I really like the style of your explanations: clear, concise, and well presented. You are up there in the global maximum of UA-cam math channels. :)
Nice use of the squeeze theorem :D. The brute force approach is Ln the whole expression then its the sum of ln(1+i/n^2) for i =1 to n then bring in the power series of ln(1+x)... So you get (1/n^2 + 2/n^2 +... n/n^2) - (1/n^4 + 4/n^4 + 9/n^4 + n^2/n^4) + (1/n^6 +8/n^6 +27/n^6 +... n^3/n^6) +... The first bracket is (n(n+1)/2) (1/n^2) the second bracket is (n(n+1)(2n+1)/6)(1/n^4) the third bracket is (n(n+1)/2)^2 (1/n^6) etc the first bracket simplifies to 1/2 +1/2n the second simplifies to 2/n +3/n^2 +1/n^3 the third 1/4n^2 + 1/2n^3 +1/4n^4 etc we notice that all the terms are devided by some power of n except for the 1/2 from the first bracket thus taking the limit as n tends to infinity we get 1/2. Since Ln of the original tends to 1/2 the original itself tends to e^1/2
There's a nice path... Try ln(limit), then Taylor series for ln(1+x), then sum and you're done...
Mathematictians when they see an infinite product:
E
Hey, you can also go forward directly: \log(L)= \sum_{k=1}^n \log(1+k/n^2) and then since x-x^2/2
Nice!
very nicely explained, there is another way to solve this, let x be a real number:
x-(x^2/2)
In german we call the squeez theorem the police theorem for some reason. It's much more funny to just say police this and you have your answer :D My professor always cracked the joke: Have you seen any Policeman converging? I guess it's funny in lectures not in rl. Nice solution btw keep it up!
This was just incredible, the explanation was very clear, and the solution was very clever.
Awesome, love seeing clever ways of evaluating limits.
I got a request from a viewer but I could not solve. Fematika and Mu prime, are you interested?
blackpenredpen I’m interested!
Very nice and clever solution
Very nice solution. I wouldn't have thought of that.
Brilliant solution with brilliant t-shirt
Hi mak Vinci today I have made a integral battle 😀
@@chirayu_jain Yes i watched it and liked, keep it up! 😁
underrated channel
Nice limits by the way... Are you reading Demidovich?
I learned calculus using Stewart!
I come from bprp, I was the one who wondered if you were able to lick your elbow :v, Now seriously nice video, better person :)
But what if n is odd you can't multiply the middle term by any other
Because lim(a×b) = lim(a) × lim(b), we can take the middle term out of the limit and deal with it separately while still pairing the remaining terms like before. But the limit of the middle term by itself will just be 1, so we don't have to worry about it!
@@MuPrimeMath lim(a×b) = lim(a) × lim(b)
are there any restrictions to it? I don't think it works in every case; right?
The primary condition is that all 3 limits in the equation must exist
n is tending to infinity, so saying n is odd doesn't really make sense.
Hey mu prime math, I could do it applying the Riemann Stieltjes's integral definition, you can notice the next fact, when we apply logarithm we are going to have \lim_{n \to infty} \sum_{k=1}^{n}ln(1+k/n^2) as you can notice here we have our d(alpha) = d(x^2) due to the grade of the denominator, then we will build our integral, we have the definition of the integral like int_{a}^{b}f(x)d(alpha(x))=lim_{n \to infty}\sum_{k=1}^{n}f(delta(x) +a)(alpha(a+delta(x)k)-alpha(a+delta(x)(k-1)), then we have int_{0}^{1}ln(1+x)d(x^2)=1/2 , then after applying exp we come up with e^(1/2)
Interesting! You would have to enumerate more clearly what your function α would be, and what algebra gets that (α_k - α_{k-1}) term multiplied on the outside without changing the answer, but it seems like that would be doable.
Which university are you student?
I am in high school!
Wow! İts great you are very skilled.
@@MuPrimeMath which is your country???
USA
@@MuPrimeMath oo
Mr. prof. Mu Prime Math.I like your proof,maby a little bit overwrite, and like Bleckpenredpen ,you write with a pen with INK finished. Thanks for understanding.
Nice solutions .
Another way to solve
Tasty sandwich!
If you take more care of your spelling and some things do not explain them on the air, then your videos will look more professional and easier to understand. Thanks for your time.