it's simple until you get to the integral expression, from that moment on, I think he overcomplicated it. I would just look it up on an integral table, none of those integrals is particularly unusual.
I evaluated this before watching the video, and I immediately noticed that this series is the dilogarithm function evaluated at 1/2. What I did is derive the reflection formula for the dilogarithm function using the integral form of the function and evaluate it at 1/2.
Given that we packaged up the series into an ln(1-x) only to later unpackage it back into an infinite series, is there any way to "cut out the middle man", so to speak, and do that whole section just with the series? It seems like the key step was disassembling the integral into two halves, changing variables on one half, and then reassembling it into a single integral across an interval of twice the length; is there a reason we couldn't have done that earlier and saved the trouble of at least some series manipulation?
Let S = -int_0^{1/2} ln(1-t)/t dt integration by parts gives = -ln(t)ln(1-t)|_0^{1/2} - int_0^{1/2} ln(t)/(1-t) dt = -ln(2)^2 - int_0^{1/2} ln(t)/(1-t) dt Swapping t with 1-t gives = -ln(2)^2 - int_{1/2}^1 ln(1-t)/t dt Adding S to both sides 2S = -ln(2)^2 - int_0^1 ln(1-t)/t dt Then using the fact that sum x^n/n^2 = - int_0^x ln(1-t)/t dt the result follows.
Well, we had to pull out the -1/2 times log(2)^2 at some point. Noticing that the sum we want to calculate is equal to a part of the product of the other two known sums given, it should be possible to assail this problem by taking this product of two infinite sums and subtracting all the parts that don't belong to the infinite sum we're interested in, but that seems more laborious than the method in the video (and is probably not a method of a nature different to the method used here as integrals are just infinite sums, too).
Let f(x) = Σ x^n/n^2. Then f'(x) = -1/x ln(x). If A is the integral of f from 0 to 1/2, and B is the integral from 1/2 to 1, you get B - A = ln(2)^2 by integrating by parts and changing variables, and B + A = f(1) = pi^2 / 6. Therefore f(1/2) = A = 1/2 (pi^2 / 6 - ln(2)^2).
@@Technopolo I think there is a typo here - as previously noted by @Spaghetti (and later more laboriously by me) f'(x) = -1/x ln(1-x) which is essential to the solution you outline involving IBP and the change of variable.
yes, there tons of ways to solve integrals, Michael seems to like this particular trick with infinite series. Or you can just look it up in a table of integrals.
The opening observations do give us a benchmark we can use to sanity check our final answer, as this sum must be less than either of those two (being less than or equal to them on a term by term basis).
You can even improve that benchmark by realizing that 1/2 is the largest term of the (1/2)^n series. Then we already know that the result has to be less than (pi^2)/12
i used to tinker with this stuff back in high school. around 2002. Back then we used graphing calculators. I stumbled across the Polylogarithm, hypergeometric series, and the Zeta function just from this kind of tinkering. of course it was VERY beyond me at the time.
I wonder if there's a shorter solution taking advantage of Fourier series and convolutions? Since the terms of this series is the term wise product of two sequences
As usual the translation from sums to integrals is always a cool thing to see. It did go a bit over my head at times, but that's because I need to get better at my analysis. Sorry for looking at this video a day late and thank you for posting it.
∑[1,∞] (x^n/n) = -ln(1-x) is used in the final steps of the proof. So, ∑[1,∞] (x^(n-1)/n), near the beginning, is just -ln(1-x)/x, eliminating the second integral (in y) and making the original sum S = -∫[0,1/2] ln(1-x)/x dx. Integrating by parts gives -ln²(1/2) - ∫[0,1/2] ln(x)/(1-x) dx. This latter integral = ∫[1/2,1] ln(1-y)/y dy by substituting y = 1-x and interchanging the limits. So, S + S = 2S = -ln²(1/2) - ∫[1/2,1] ln(1-y)/y dy - ∫[0,1/2] ln(1-x)/x dx = -ln²(1/2) - ∫[0,1] ln(1-y)/y dy. The rest of the proof is as above.
Excellent video with a great topic. Students find hard time when it comes to series. Your video provides a great motivation for those who are struggling with the topic. Keep us the good work. Thank you.
11:15 shouldn’t that second term be subtracted based on the previous line? It seems like it should be added based on the final result, just not sure where the minus sign went
Notice that he's splitting the integral in two, leaving the integral of ln(1/2)/1-y and then, the integral of -ln(y)/1-y. So ln(1/2) in the first one can simply be factored out, leaving 1/x with x = 1-y.
That's the second order polylogarithm evaluated at 1/a. I think the only "nice" value is at a=2 though (other than a=1, which is zeta(2), or the Basel problem), based on some testing with WolframAlpha.
The steps in this solution are to a certain extent standard methods of solving analysis problems, but to realise what method is needed at a particular step would take time no matter what level of mathematician you are. It just takes lots of practice. So to answer your question, I would expect he came up with this solution himself as he is a professional mathematician, but it would have taken him some time to figure out the steps of the solution
@@gearoidmccarthy8408 That is something I considered, but I'd like to hear from Michael himself. Some of these problems seem comparable in complexity to the Basel Problem. They ought to take on the order of days or weeks, but he is able to post most every day.
Just a quick criticism, You kept changing the upper boundary of integration from ½ to 1 in random times without explanation at the beggining and multiplied the entire thing by -1 at the end without even mentioning it. Are you sure the result isn't the negative of what you got in the end?
First glance try: change 2 with X, differentiate once, multiply by X, differentiate again, and you have the equation xy"+y'=1/(1-x). Then use variation of parameters to obtain that the particular solution is Li2(x) Then realise you were stupid and the original expression is the definition for Li2(1/2).
@Noam .Menashe [Edit - Hmm , just realised this is similar to @Spaghetti earlier comment , anyway ...] Actually, assuming that by y you mean the f(x) defined by Michael to generalise the infinite sum, where f(1/2) is the quantity to be found then, ignoring Li2(x) etc, I think this differentiation approach is nicer than Michael's nested integrals. It also loses the infinite sums sooner. Recall we are given that f(1)=π^2/6, and obviously f(0)=0. As you derived, (I just differentiated the series twice and used the geometric series sum) xf''+f'=1/(1-x) but LHS is (x f')' so integrating twice gives f(x)=Integral[- ln(1-z)/z,{z,0,x}] - an integral Michael also produces. Recall though that we already know that f(1) is the Basel sum, which he had to demonstrate. This integral is not trivial but it has an interesting integration by parts: particularly useful for x=1/2. Integral[ ln(1-z)/z] = ln(1-z)ln(z)+Integral[ln(z)/(1-z)]. So, f(1/2)=-[ln(2)]^2-Integral[ln(z)/(1-z),{z,0,1/2}]. Now time for Michael's change of variable (r=1-z) in that last integral, switching to Integral[- ln(1-r)/r,{r,1/2,1}], but this is just f(1)-f(1/2). So 2 f(1/2)= -[ln(2)]^2+π^2/6 and result follows. Apologies for the clunky rendering of the equations.
One could mention that this is a special case of the inversion relation for the polylogarithm of weight 2 (for example see www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0844-20.pdf and set z=2 in the formula on the second page).
I haven't taken a calculus class since the 80s, so I wasn't following most of what you did, but at one point, you defined a new x for the right half, and then proceeded as if it was all the same x for the whole thing. That seems kind of shaky to me. Assuming that you know what you are doing, and given that I don't know what you are doing, I probably just didn't understand what you did there.
What he did is ok, but I know it can be confusing. Just don't care about the variable; think of one integral as evaluating an area below a function from 0 to 1/2, and the other integral the area from 1/2 to 1. How you call the variable that parts the integral doesn't matter here. So the sum of the 2 integrals is like "ok, let's get the total area" which is the sum of the other 2 areas.
@@Ligatmarping I was just thinking of an integral as the reverse of a derivative. I forgot that it is also the area beneath a curve. Thanks! That helped.
My approach was to construct the goal sum starting from the geometric series. Consider S = Sum(n>=0, x^n) = 1/(1-x). Integrate both sides, reindex, and use x=0 to get C=0. Sum(n>0, x^n/n) = -ln(1-x) Then, Sum(n>0, x^(n-1)/n) = -ln(1-x)/x Integrate from 0 to 1/2 and from 1 to 1/2. Adding those, the LHS is 2S - pi^2/6. On the right-hand side, we manipulate the 1 to 1/2 integral to cancel the 0 to 1/2 integral. First, use IBP to turn 1/x into lnx and -ln(1-x) into 1/(1-x). The uv term converges to 0 at the endpoint x=1 using L'Hopital's rule 3 times. This leaves -ln^2(2) - Integral(1, 1/2, lnx/(1-x)). Next, using u=1-x, that integral becomes the 0 to 1/2 integral, and they cancel. Thus, 2S - pi^2/6 = -ln^2(2). A more general problem is Sum(n>0, x^n/n^2). Both my approach and Michael's approach rely on x = 1/2. The generalized sum is the standard definition of the polylogarithm of order 2 (n^2). Other special cases occur when x is 2 or various powers of +- phi (golden ratio). That might make for a good video topic depending on how difficult it is to derive the necessary functional relations.
It seems like you’re playing fast and loose with the integration limits. In one step the upper limit is 1/2, then in the next step it’s 1, the in the next step it’s back to 1/2. Could you please explain what’s ging on?
Partial fractions separates polynomials, 2^n is exponential. If you try to use partial fractions, it won't work because the assumption that 1/(2^n * n^2) = A/2^n + B/n^2 is wrong. You end up getting 1 = A*n^2 + B*2^n. There are no A and B you can choose to make that statement true for all n. You can change the numerators to something else, but it still won't work as exponentials will never cancel polynomials.
Yet again you finished by telling us the answer without telling us the answer. The answer is c. 0.58224, but it would have been more interesting if you had told us.
A "tool" which is usually useful is to try to solve something changing those limits carelessly and, if it turns out to go to the solution, verify then that the change is ok, which happens in a lot of cases.
@@Ligatmarping your quote symbols enveloping tool is noted and appreciated. I love this channel and learn a lot, but sooo much of the stuff is not in the spirit of discovery, BUT more how to use unmotivated tricks after one already knows the solution.
@@MyOneFiftiethOfADollar first sorry if my english isn't accurate... I hope to express the following properly. I'm a math teacher at the university of Buenos Aires and I get that questioning a lot when showing solutions. "Where does that come from?". It is hard to draw the line of what is a trick and what is not. I agree that just seeing the solution does not motivate the trick, but a lot of the techniques to solve thing when investigating a topic, come from a "seemingly useless trick" which you saw in the solution of a particular case. What I do prefer, is to let someone "find" the trick or at least think about the problem some time, so that you afterwards kinda appreciate what's the idea on the trick and which step it helps to solve. I usually speak normally in english but this is something a little complicated to express, so sorry if I didn't manage to show my thoughts about it.
Thank you to the few commenters who are always so quick to point out shortcuts involving functions with no closed form, derived from the lowest levels of effort in Mathematica.
I’m confused in the substitution y=1-x. I understand how we’ll make that substitution but I can’t understand why we can combine the new integral with the one on the left.
They are integrals of the same function, and share a limit of integration (1/2). Thinking of the integral as the area under a curve, it's like saying two areas under the curve that border each other. So, we can add those areas together and descirbe them as one big area, all in one piece (aka as just one big integral).
For any integrable f(x), the integral from a to b of f(x) + the integral from b to c of f(x) is equal to the integral from a to c of f(x). This follows immediately from the linearity of integration, or you can get it from FTOC. From FTOC: since f(x) is integrable, there exists an antiderivative F(x) such that F'(x) = f(x). The first integral is defined as F(b) - F(a) by FTOC; similarly, the second integral is defined as F(c) - F(b) by FTOC. Adding the two together gives F(b) - F(a) + F(c) - F(b). Canceling the + and - terms of F(b) yields F(c) - F(a). Again, by FTOC, this is nothing more than the integral from a to c of F'(x), and since F'(x) = f(x), this is the integral from a to c of f(x), as was to be shown.
Taking the time to input series into Wolfram Alpha and dutifully report. We all gained immense insight into the solution process due to your contribution.
Hello again Mr Penn, In an attempt to raise my chances of you seeing this message I will state my ask once more . Could you perhaps try to solve lim x -> infinity of x/(tan((pi/2)-pi/x)) in a future video. I think the result might surprise, though I wouldn't know how to solve this using classical calculus techniques
First, let u=pi/x. The limit is then lim(u -> 0, pi / (u * tan(pi/2 - u))). Recall that tan((pi/2)-x) = cot(x). So, we have lim(u -> 0, pi * tan(u)/u) = pi * lim(u -> 0, sin(u)/u) / lim(u -> 0, cos(u)) = pi * 1 * 1 = pi What gives it away is knowing trig identities and the sinx/x or tanx/x limit.
"...but as we'll see it's a little bit more complicated", that's a new definition of "little bit" 😄
yeah; ...really
the first thing he did made no sense to me...
@@seanmcghee2373 lol it makes sense to me but how he figured that out is beyond me. It would have taken a lot longer for me to get there
@@lukealadeen7836 It's probably a classic trick.
it's simple until you get to the integral expression, from that moment on, I think he overcomplicated it. I would just look it up on an integral table, none of those integrals is particularly unusual.
What I learned from the video: the UA-cam suggestion algorithm utterly overestimates my math skills :-o
I evaluated this before watching the video, and I immediately noticed that this series is the dilogarithm function evaluated at 1/2. What I did is derive the reflection formula for the dilogarithm function using the integral form of the function and evaluate it at 1/2.
What is the reflection formula?? Is thst a commonly known thing?
@@leif1075
It's a relation between f(a-x) and f(x). e.g. Euler's reflection formula Γ(1-z)Γ(z)=π/sin(πz) or even/odd functions for a=0.
@@leif1075in this case, the identity is
Li₂(x) + Li₂(1-x) = π²/6 - ln(x)ln(1-x)
Given that we packaged up the series into an ln(1-x) only to later unpackage it back into an infinite series, is there any way to "cut out the middle man", so to speak, and do that whole section just with the series? It seems like the key step was disassembling the integral into two halves, changing variables on one half, and then reassembling it into a single integral across an interval of twice the length; is there a reason we couldn't have done that earlier and saved the trouble of at least some series manipulation?
Let S = -int_0^{1/2} ln(1-t)/t dt
integration by parts gives
= -ln(t)ln(1-t)|_0^{1/2} - int_0^{1/2} ln(t)/(1-t) dt
= -ln(2)^2 - int_0^{1/2} ln(t)/(1-t) dt
Swapping t with 1-t gives
= -ln(2)^2 - int_{1/2}^1 ln(1-t)/t dt
Adding S to both sides
2S = -ln(2)^2 - int_0^1 ln(1-t)/t dt
Then using the fact that sum x^n/n^2 = - int_0^x ln(1-t)/t dt the result follows.
Well, we had to pull out the -1/2 times log(2)^2 at some point.
Noticing that the sum we want to calculate is equal to a part of the product of the other two known sums given, it should be possible to assail this problem by taking this product of two infinite sums and subtracting all the parts that don't belong to the infinite sum we're interested in, but that seems more laborious than the method in the video (and is probably not a method of a nature different to the method used here as integrals are just infinite sums, too).
Let f(x) = Σ x^n/n^2. Then f'(x) = -1/x ln(x). If A is the integral of f from 0 to 1/2, and B is the integral from 1/2 to 1, you get B - A = ln(2)^2 by integrating by parts and changing variables, and B + A = f(1) = pi^2 / 6. Therefore f(1/2) = A = 1/2 (pi^2 / 6 - ln(2)^2).
@@Technopolo I think there is a typo here - as previously noted by @Spaghetti (and later more laboriously by me) f'(x) = -1/x ln(1-x) which is essential to the solution you outline involving IBP and the change of variable.
yes, there tons of ways to solve integrals, Michael seems to like this particular trick with infinite series. Or you can just look it up in a table of integrals.
Small mistake from 8.45 to 8.52
That is y=0➡️x=1, y=1/2➡️x=1/2
And not x=0➡️y=1,x=1/2➡️y=1/2
The opening observations do give us a benchmark we can use to sanity check our final answer, as this sum must be less than either of those two (being less than or equal to them on a term by term basis).
You can even improve that benchmark by realizing that 1/2 is the largest term of the (1/2)^n series. Then we already know that the result has to be less than (pi^2)/12
i used to tinker with this stuff back in high school. around 2002. Back then we used graphing calculators. I stumbled across the Polylogarithm, hypergeometric series, and the Zeta function just from this kind of tinkering. of course it was VERY beyond me at the time.
3:06 Why does the upperbound keep switching between 1/2 and 1? Is that a mistake or am I missing something?
Just a mistake
@@feliperodriguezbarrera I was wondering the same thing. Usually Prof. Penn clears mistakes up either with a wrist-erase or post-production.
I was wondering the same. I assumed it had to be 1/2 though, but it was slightly confusing.
Just classic Michael Penn mistake
@@davidcroft95 i probably make more mistakes than all of UA-cam combined rofl 🤣 i really should brush off my math rusty-nessz
This can be easily evaluated at z= 1/2 in the identitiy L(z) + L(1-z) = zeta(2) where L(z)= dialog(z) + (1/2) log(z) log(1-z).
It would be good to see this approach using properties of the dilogarithm. The playlist for the dilogarithm is a short one.
I followed everything perfectly, what a beautiful process
This math is a big rabbit hole, behind which a large labyrinth with many branches and cross-connections opens up.
I wonder if there's a shorter solution taking advantage of Fourier series and convolutions? Since the terms of this series is the term wise product of two sequences
@ 2:17 I'm having some trouble seeing how it went from upper limit of 1 to upper limit of 1/2. And then it switches again... Can anyone explain?
As usual the translation from sums to integrals is always a cool thing to see. It did go a bit over my head at times, but that's because I need to get better at my analysis.
Sorry for looking at this video a day late and thank you for posting it.
13:12
It blows my mind how fluent this man is in Maths.
So young to have mastered his craft.
Passion is a powerful force 🙂
where does this upper integration limit of 1/2 come from at 2:15? the final result is correct, but this step seems just wrong to me
From the fact that u have to plug in X=1/2 in order to get back to the original expression
There are two places where the Prof mistakenly wrote the integral from 0 to 1.
I think it's correct at 1:14 but the next step should be from 0 to 1/2 and at 2:15 it's right again
@@mavenfromheaven1190 ah yes, i just assumed that 0 to 1 was correct in the first place without checking xD
your sum is dilog(1/2) with |x[
∑[1,∞] (x^n/n) = -ln(1-x) is used in the final steps of the proof. So, ∑[1,∞] (x^(n-1)/n), near the beginning, is just -ln(1-x)/x, eliminating the second integral (in y) and making the original sum
S = -∫[0,1/2] ln(1-x)/x dx. Integrating by parts gives -ln²(1/2) - ∫[0,1/2] ln(x)/(1-x) dx. This latter integral = ∫[1/2,1] ln(1-y)/y dy by substituting y = 1-x and interchanging the limits.
So, S + S = 2S = -ln²(1/2) - ∫[1/2,1] ln(1-y)/y dy - ∫[0,1/2] ln(1-x)/x dx = -ln²(1/2) - ∫[0,1] ln(1-y)/y dy. The rest of the proof is as above.
You know what they say about magicians... ain't no magic trick without misdirection
Excellent video with a great topic. Students find hard time when it comes to series. Your video provides a great motivation for those who are struggling with the topic. Keep us the good work. Thank you.
11:15 shouldn’t that second term be subtracted based on the previous line? It seems like it should be added based on the final result, just not sure where the minus sign went
power series of ln(1-x) is -x^n/n so - * - = +
I lost track of how many minus signs I lost track of...
@@krishdewani4949 thx
Damn good work. I’m here not even in school and watched this video while it was halftime
Thank you, professor.
Please mention the theorems used in the proof. Thank you.
9:00 When extracting the (ln(1/2)) constant from the sum under the integral sign, how can it be extracted as a factor?
Notice that he's splitting the integral in two, leaving the integral of ln(1/2)/1-y and then, the integral of -ln(y)/1-y.
So ln(1/2) in the first one can simply be factored out, leaving 1/x with x = 1-y.
What if instead of 2^n at the denominator, we had a more general a^n (with a>1)?
That's the second order polylogarithm evaluated at 1/a. I think the only "nice" value is at a=2 though (other than a=1, which is zeta(2), or the Basel problem), based on some testing with WolframAlpha.
@@1s3k3b5 actually theres more
Could someone explain the zeroth integral trick? I’ve seen him use it before but I don’t completely understand
The idea is to think of the term as F(b)-F(a) (antiderivative) then write this as an integral of f(x) from a to b.
Michael, question -- are these solutions on your channel original, or do you get the seemingly arbitrary steps from other sources?
The steps in this solution are to a certain extent standard methods of solving analysis problems, but to realise what method is needed at a particular step would take time no matter what level of mathematician you are. It just takes lots of practice.
So to answer your question, I would expect he came up with this solution himself as he is a professional mathematician, but it would have taken him some time to figure out the steps of the solution
@@gearoidmccarthy8408 That is something I considered, but I'd like to hear from Michael himself. Some of these problems seem comparable in complexity to the Basel Problem. They ought to take on the order of days or weeks, but he is able to post most every day.
Excellent
=Li2(1/2)
Just a quick criticism, You kept changing the upper boundary of integration from ½ to 1 in random times without explanation at the beggining and multiplied the entire thing by -1 at the end without even mentioning it. Are you sure the result isn't the negative of what you got in the end?
First glance try: change 2 with X, differentiate once, multiply by X, differentiate again, and you have the equation xy"+y'=1/(1-x). Then use variation of parameters to obtain that the particular solution is Li2(x)
Then realise you were stupid and the original expression is the definition for Li2(1/2).
@Noam .Menashe [Edit - Hmm , just realised this is similar to @Spaghetti earlier comment , anyway ...] Actually, assuming that by y you mean the f(x) defined by Michael to generalise the infinite sum, where f(1/2) is the quantity to be found then, ignoring Li2(x) etc, I think this differentiation approach is nicer than Michael's nested integrals. It also loses the infinite sums sooner. Recall we are given that f(1)=π^2/6, and obviously f(0)=0. As you derived, (I just differentiated the series twice and used the geometric series sum) xf''+f'=1/(1-x) but LHS is (x f')' so integrating twice gives f(x)=Integral[- ln(1-z)/z,{z,0,x}] - an integral Michael also produces. Recall though that we already know that f(1) is the Basel sum, which he had to demonstrate. This integral is not trivial but it has an interesting integration by parts: particularly useful for x=1/2. Integral[ ln(1-z)/z] = ln(1-z)ln(z)+Integral[ln(z)/(1-z)]. So, f(1/2)=-[ln(2)]^2-Integral[ln(z)/(1-z),{z,0,1/2}]. Now time for Michael's change of variable (r=1-z) in that last integral, switching to Integral[- ln(1-r)/r,{r,1/2,1}], but this is just f(1)-f(1/2).
So 2 f(1/2)= -[ln(2)]^2+π^2/6 and result follows. Apologies for the clunky rendering of the equations.
is there a way to calculate the sum, without the application of integral calculus?
You keep on writing 1 instead of 1/2 for the upper limit of the integral.
Interesting illustration that sum(ab) =\= sum(a)sum(b)
It sounds like you're making stuff up. I feel like I'm watching Time Travel Understander.
approx 0.582241, where the first term contributes 0.500
Just as I was looking for a maths video! Great timing!! First
Polylog[2](1/2)=pi^2/12-(log(2))^2/2
I don’t understand the vertical line thing.
It’s integral evaluation short-hand.
@@NotoriousSRG but then why is the integral sign (S) used together with |?
I really liked this one.
What about it did you like?
One could mention that this is a special case of the inversion relation for the polylogarithm of weight 2 (for example see www.kurims.kyoto-u.ac.jp/~kyodo/kokyuroku/contents/pdf/0844-20.pdf and set z=2 in the formula on the second page).
do you mean the third centered formula on the second page?
It's a case of Li2(X)+Li2(1-x).
Is there a way to calculate X when X^X=Y ?
Using the Lambert function W, you have x=e^(W(ln y)). Maybe there are better ways.
I haven't taken a calculus class since the 80s, so I wasn't following most of what you did, but at one point, you defined a new x for the right half, and then proceeded as if it was all the same x for the whole thing. That seems kind of shaky to me. Assuming that you know what you are doing, and given that I don't know what you are doing, I probably just didn't understand what you did there.
What he did is ok, but I know it can be confusing. Just don't care about the variable; think of one integral as evaluating an area below a function from 0 to 1/2, and the other integral the area from 1/2 to 1. How you call the variable that parts the integral doesn't matter here. So the sum of the 2 integrals is like "ok, let's get the total area" which is the sum of the other 2 areas.
@@Ligatmarping I was just thinking of an integral as the reverse of a derivative. I forgot that it is also the area beneath a curve. Thanks! That helped.
dominance??
11:29 deja vu from 2:22
My approach was to construct the goal sum starting from the geometric series.
Consider
S = Sum(n>=0, x^n) = 1/(1-x).
Integrate both sides, reindex, and use x=0 to get C=0.
Sum(n>0, x^n/n) = -ln(1-x)
Then,
Sum(n>0, x^(n-1)/n) = -ln(1-x)/x
Integrate from 0 to 1/2 and from 1 to 1/2. Adding those, the LHS is 2S - pi^2/6.
On the right-hand side, we manipulate the 1 to 1/2 integral to cancel the 0 to 1/2 integral.
First, use IBP to turn 1/x into lnx and -ln(1-x) into 1/(1-x). The uv term converges to 0 at the endpoint x=1 using L'Hopital's rule 3 times. This leaves
-ln^2(2) - Integral(1, 1/2, lnx/(1-x)).
Next, using u=1-x, that integral becomes the 0 to 1/2 integral, and they cancel. Thus,
2S - pi^2/6 = -ln^2(2).
A more general problem is Sum(n>0, x^n/n^2). Both my approach and Michael's approach rely on x = 1/2. The generalized sum is the standard definition of the polylogarithm of order 2 (n^2).
Other special cases occur when x is 2 or various powers of +- phi (golden ratio). That might make for a good video topic depending on how difficult it is to derive the necessary functional relations.
It seems like you’re playing fast and loose with the integration limits. In one step the upper limit is 1/2, then in the next step it’s 1, the in the next step it’s back to 1/2. Could you please explain what’s ging on?
Just 2 mistakes. All of them should be 1/2
Hi Dr. Penn!
I can not follow what the upper limits of all of these integrals are, he flips from 1/2 to 1, so what?
There must be a 1/2 at 2:00
Would it not be nicer to just use partial fractions from the beginning?
Partial fractions separates polynomials, 2^n is exponential. If you try to use partial fractions, it won't work because the assumption that 1/(2^n * n^2) = A/2^n + B/n^2 is wrong.
You end up getting 1 = A*n^2 + B*2^n. There are no A and B you can choose to make that statement true for all n. You can change the numerators to something else, but it still won't work as exponentials will never cancel polynomials.
Oh yes, a failed to realise this. Thanks
Yet again you finished by telling us the answer without telling us the answer. The answer is c. 0.58224, but it would have been more interesting if you had told us.
No.
An exact solution is better than an approximation
Dominating convergence theorem ignorance is why I would have never found the very first step in your proof.
A "tool" which is usually useful is to try to solve something changing those limits carelessly and, if it turns out to go to the solution, verify then that the change is ok, which happens in a lot of cases.
@@Ligatmarping your quote symbols enveloping tool is noted and appreciated. I love this channel and learn a lot, but sooo much of the stuff is not in the spirit of discovery, BUT more how to use unmotivated tricks after one already knows the solution.
@@MyOneFiftiethOfADollar first sorry if my english isn't accurate... I hope to express the following properly. I'm a math teacher at the university of Buenos Aires and I get that questioning a lot when showing solutions. "Where does that come from?". It is hard to draw the line of what is a trick and what is not. I agree that just seeing the solution does not motivate the trick, but a lot of the techniques to solve thing when investigating a topic, come from a "seemingly useless trick" which you saw in the solution of a particular case. What I do prefer, is to let someone "find" the trick or at least think about the problem some time, so that you afterwards kinda appreciate what's the idea on the trick and which step it helps to solve. I usually speak normally in english but this is something a little complicated to express, so sorry if I didn't manage to show my thoughts about it.
Is there an error in the solution?
#include
#include
int main(void)
{
int n=1;
int j;
float a,b,c,d,e,f;
b=0;
for(j=0;j
true
I don't see any problems other than a lot of unused variables and redundant code
Thank you to the few commenters who are always so quick to point out shortcuts involving functions with no closed form, derived from the lowest levels of effort in Mathematica.
there is a mistake 1:59 to 2:20 or so. got me hang up for awhile.
good problem... i thought he went to use taylor series... but... it was much easier the way he done
:45 Just multiply them together, problem solved! ;)
I’m confused in the substitution y=1-x. I understand how we’ll make that substitution but I can’t understand why we can combine the new integral with the one on the left.
They are integrals of the same function, and share a limit of integration (1/2). Thinking of the integral as the area under a curve, it's like saying two areas under the curve that border each other. So, we can add those areas together and descirbe them as one big area, all in one piece (aka as just one big integral).
For any integrable f(x), the integral from a to b of f(x) + the integral from b to c of f(x) is equal to the integral from a to c of f(x). This follows immediately from the linearity of integration, or you can get it from FTOC. From FTOC: since f(x) is integrable, there exists an antiderivative F(x) such that F'(x) = f(x). The first integral is defined as F(b) - F(a) by FTOC; similarly, the second integral is defined as F(c) - F(b) by FTOC. Adding the two together gives F(b) - F(a) + F(c) - F(b). Canceling the + and - terms of F(b) yields F(c) - F(a). Again, by FTOC, this is nothing more than the integral from a to c of F'(x), and since F'(x) = f(x), this is the integral from a to c of f(x), as was to be shown.
Ah, thanks for the replies. I understand it now. It has been quite some time since I did any integration 😅
Liked that one better.
The result is Li2(1/2); where Li(x) is the polylogarithm function.
Wolfram notation Li(2, 1/2)
Taking the time to input series into Wolfram Alpha and dutifully report. We all gained immense insight into the solution process due to your contribution.
@@MyOneFiftiethOfADollar
I can sum your cynicism as your dollars diminish from x= 1/50 to zero.
Советские школьники такое в уме решали
way cool
Hello again Mr Penn,
In an attempt to raise my chances of you seeing this message I will state my ask once more . Could you perhaps try to solve lim x -> infinity of x/(tan((pi/2)-pi/x)) in a future video. I think the result might surprise, though I wouldn't know how to solve this using classical calculus techniques
First, let u=pi/x. The limit is then
lim(u -> 0, pi / (u * tan(pi/2 - u))).
Recall that tan((pi/2)-x) = cot(x). So, we have
lim(u -> 0, pi * tan(u)/u)
= pi * lim(u -> 0, sin(u)/u) / lim(u -> 0, cos(u))
= pi * 1 * 1
= pi
What gives it away is knowing trig identities and the sinx/x or tanx/x limit.
There’s a link in the description to a form to suggest problems. Have you tried that?
i was the first
wooooowwww