We can easily generalize this by putting a>1 instead of 2. We'll get the result of ln(a/(a-1)). The fact that a>1 has been used many times, from the inequality pushing the floor to n+1, to the convergence of the series
Just wanted to post basically the same thing, for every a>1 the sum_[1...] 1/(n * a^n) converges to log(a/(a-1)), and everything else in the original proof can be reused.
what this makes no sense..how does that t equal the log and I dont see why anyine would ever think of that--and besides if it goes to infinty then the denominator would just approach zero..surely there is another way to solve this that ppl would actually think of??
Hahaha Step 1: Let's make this easier to understand by eliminating the integral altogether. Step 2: Let's make this easier to understand by reintroducing an integral.
@@Happy_Abe the floor function isn't part of the substitution, just everything inside the floor function, so all that's left inside the floor is a solitary u. Then since u goes from 1 to inf, it's easy to split into intervals on which floor(u) is constant.
@@giron716 ah yes I see that, but what about the dx term? You’d have to differentiate 1-log_2(x) so it’s not just as simple as integrating 1/floor(u) right?
@@Happy_Abe Right, so you have to use a similar strategy to the one in the video. After substituting and swapping the bounds of integration, the bounds will be from 1 to positive infinity. Now you separate this integral into intervals from n to n + 1. On each of these intervals, floor (u) will be equal to n, so you can substitute that to get rid of the floor function. At this point you have to integrate 2^(1-u), which is pretty trivial. After solving this integral and simplifying you should arrive at the same infinite sum Michael did at the end.
For the sum at the end, I reindexed the sum and wrote 1/2 as -1*-1/2. Then, I brought another -1 Inside the sum by multiplying by -1 outside the sum. This gave me -(sum from n=1->inf of (-1)^(n+1)*(-1/2)^n / n. This is the series expansion of -ln(1+x) at x = 1/2. This gives -ln(1-1/2) = -ln(1/2) = ln(2).
how did you replace dx with dy though--since its nota natural log base so we dont really know how to take the derivative of log base 2 x--did you rewrite in terms of ln x/ln 2 I presume first and then solve?
@@The1RandomFool oh meaning you replaced dx with that expression for dy in theintegral--well then my approach isslightly different fromyours but equally valid right? just rewrite log2x as lnx*ln2 or whatever and then subsittue y=ln x and then you can solve or with integration by parts..
I just used the substitution x=2^-u, from which it follows dx=-ln(2)2^-u du as well as the integral bounds infty and 0, so after switching the bounds, you get I=ln(2) int_0^infty 2^-u 1/floor(1+u) du. I think this is a lot easier to split into a sum, because you can just split it into the integrals from n to n+1: I=sum_{n=0}^infty ln(2) int_n^{n+1} 2^-u 1/floor(1+u) du Now n < u < n+1, so floor(1+u)=1+n. That gives you: I=sum_{n=0}^infty ln(2)/(1+n) int_n^{n+1} 2^-u du And the integral int_n^{n+1} 2^-u du is easily calculated as int_n^{n+1} 2^-u du = -1/ln(2) [2^-u]_n^{n+1} = -1/ln(2) (2^-(n+1) - 2^-n) = 1/ln(2) 2^-(n+1) so in total you get I=sum_{n=0}^infty ln(2)/(1+n) 1/ln(2) 2^-(n+1) =sum_{n=0}^infty 1/(n+1) (1/2)^(n+1) And this is the exact same result as what was achieved at 7:30. Personally I found that this approach felt a lot less tricky than choosing the correct integral bounds from the start and showing that the floor function behaves correctly for those. Basically, simplify the contents of the floor() function as far as possible, and that way it'll give you the summation more directly.
Rewriting that 1/2 as an "X" is a tremendously clever move .. I wonder if some of the kids didn't get that. -- also dominated convergence, that uh .. seems like something most of them would simply have never heard of. At least, traditionally a lot of the best students at finding integrals are not math majors, right?
We can easily generalize this by putting a>1 instead of 2. We'll get the result of ln(a/(a-1)). The fact that a>1 has been used many times, from the inequality pushing the floor to n+1, to the convergence of the series
I was going to say the same thing.
The actual result is (a-1)*ln[a/(a-1)] because the step where michael wrote (1/2)^n-(1/2)^(n-1)=(1/2)^(n-1) only works for the case of 1/2.
Just wanted to post basically the same thing, for every a>1 the sum_[1...] 1/(n * a^n) converges to log(a/(a-1)), and everything else in the original proof can be reused.
what this makes no sense..how does that t equal the log and I dont see why anyine would ever think of that--and besides if it goes to infinty then the denominator would just approach zero..surely there is another way to solve this that ppl would actually think of??
Hahaha
Step 1: Let's make this easier to understand by eliminating the integral altogether.
Step 2: Let's make this easier to understand by reintroducing an integral.
10:40
you are a hero
Very interesting! I just used the initial substitution u = 1 - log_2(x) and got the same result.
How’d you calculate the du term with the flor function there?
@@Happy_Abe the floor function isn't part of the substitution, just everything inside the floor function, so all that's left inside the floor is a solitary u. Then since u goes from 1 to inf, it's easy to split into intervals on which floor(u) is constant.
@@giron716 ah yes I see that, but what about the dx term?
You’d have to differentiate 1-log_2(x) so it’s not just as simple as integrating 1/floor(u) right?
@@Happy_Abe Right, so you have to use a similar strategy to the one in the video. After substituting and swapping the bounds of integration, the bounds will be from 1 to positive infinity. Now you separate this integral into intervals from n to n + 1. On each of these intervals, floor (u) will be equal to n, so you can substitute that to get rid of the floor function. At this point you have to integrate 2^(1-u), which is pretty trivial. After solving this integral and simplifying you should arrive at the same infinite sum Michael did at the end.
@@ultimatedude5686 thank you!
This evaluation invokes so many different concepts I for sure would not have been able to solve it myself in an appreciable amount of time.
For the sum at the end, I reindexed the sum and wrote 1/2 as -1*-1/2. Then, I brought another -1 Inside the sum by multiplying by -1 outside the sum. This gave me -(sum from n=1->inf of (-1)^(n+1)*(-1/2)^n / n. This is the series expansion of -ln(1+x) at x = 1/2. This gives -ln(1-1/2) = -ln(1/2) = ln(2).
What I ended up doing is a substitution, 1-log_2 x = y, to simplify the floor function to make the series easier to work with.
how did you replace dx with dy though--since its nota natural log base so we dont really know how to take the derivative of log base 2 x--did you rewrite in terms of ln x/ln 2 I presume first and then solve?
@@leif1075 If 1-log_2(x) = y, then you solve for x and take the derivative. So log_2(x) = 1-y, x = 2^(1-y), and dx = -ln(2)*2^(1-y) dy.
@@The1RandomFool oh meaning you replaced dx with that expression for dy in theintegral--well then my approach isslightly different fromyours but equally valid right? just rewrite log2x as lnx*ln2 or whatever and then subsittue y=ln x and then you can solve or with integration by parts..
That was great.
Thank you, professor!
A sketch of the graph of the integrand at the start might have more clearly motivated the intro of -ve powers of 2. Ingenious soln as always
I just used the substitution x=2^-u, from which it follows dx=-ln(2)2^-u du as well as the integral bounds infty and 0, so after switching the bounds, you get I=ln(2) int_0^infty 2^-u 1/floor(1+u) du.
I think this is a lot easier to split into a sum, because you can just split it into the integrals from n to n+1:
I=sum_{n=0}^infty ln(2) int_n^{n+1} 2^-u 1/floor(1+u) du
Now n < u < n+1, so floor(1+u)=1+n. That gives you:
I=sum_{n=0}^infty ln(2)/(1+n) int_n^{n+1} 2^-u du
And the integral int_n^{n+1} 2^-u du is easily calculated as
int_n^{n+1} 2^-u du = -1/ln(2) [2^-u]_n^{n+1} = -1/ln(2) (2^-(n+1) - 2^-n) = 1/ln(2) 2^-(n+1)
so in total you get
I=sum_{n=0}^infty ln(2)/(1+n) 1/ln(2) 2^-(n+1)
=sum_{n=0}^infty 1/(n+1) (1/2)^(n+1)
And this is the exact same result as what was achieved at 7:30. Personally I found that this approach felt a lot less tricky than choosing the correct integral bounds from the start and showing that the floor function behaves correctly for those. Basically, simplify the contents of the floor() function as far as possible, and that way it'll give you the summation more directly.
wow, after watching so many your video, what a surprise I can do it on my first trial with the exact approach!
¡Impresionante!
Very cool video and the best part of it is that I managed to do the integral myself :)
Hi micheal how are you ,bravoo it's a beatiful solution
Amazing!
lb(x) can be used for log of x in base 2.
This was beautiful
I came up with something that might be a cool integral. It's the integral from 1 to infinity of dx/[1+2+3+4+...+floor(x)]
Rewriting that 1/2 as an "X" is a tremendously clever move .. I wonder if some of the kids didn't get that. -- also dominated convergence, that uh .. seems like something most of them would simply have never heard of. At least, traditionally a lot of the best students at finding integrals are not math majors, right?
I set y=2/x to make it int_2^{\infty}. No need for limits.
An integral with an infinite upper bound also implicitly uses limits, e.g. \int_a^\infty f(x) dx is the same as lim_{b->/infty} /int_a^b f(x) dx
Can we generalize this!!?
What happened to the logró base 2 during the video?
Hard ❌
Interesting ✅
I tried it by newton lebnitz rule bt still difficult to sove for me plz need its solution to
لذيذ
how am I comment #13
high-caliber math olympiad contenders: 🦾
me: 🥀