The Integral Suggester Strikes Back

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7 сен 2024

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Комментарии : 77

@johnginos6520 2 года назад

Can you do a video where you solve a problem from scratch? Like something you know is solvable or had a closed for solution but that you haven’t seen. This would allow us to see how a mathematician deals with a new problem, makes adjustments and checks their own work.

@tomatrix7525 2 года назад

Yep. These videos are great and always fine with his ‘hint’ but it’s coming up with the hint that we would be problematic. I can’t see how you’d know to expand e^x without having seen the result before or being very familiar with fourier series structures to notice the 1/n+1 would be generated

@BiscuitZombies 2 года назад

This but unironically

@The1RandomFool 2 года назад

I attempted this prior to watching the video, and used the Maclaurin series expansion of sine instead. I noticed that the integral can be stated in terms of the Dirichlet eta function and the gamma function. I turned that into a double sum, changed the order of summation, and was able to use the geometric series to simplify it to the point of the tool. I forgot how to derive the tool with Fourier series, so I ended up directly evaluating the series with complex analysis.

@MathSolvingChannel 2 года назад

Contour integral solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-jXEH0O5P3Ws.html

@SuperSilver316 2 года назад

Hey I used the same contour!!

@MathSolvingChannel 2 года назад

@@SuperSilver316 😆😆

@The1RandomFool 2 года назад

@@MathSolvingChannel I considered a contour integral at the start, but held off to evaluate the series I got at the end instead. I find it interesting that I took a completely different series approach, but still ended with the exact same series as the video did.

@MathSolvingChannel 2 года назад

@@The1RandomFool Great! Yes, it is more fun to switch to a very different approach.

@md2perpe 2 года назад

It should be noted that the expansion @0:45 is valid on (-pi, pi), not on the whole real line.

@juanixzx 2 года назад

Yes, or course, and is valid as well if función repeated periodically every 2*pi.

@dmitrybogdanovich3767 2 года назад

The way of finding the integral of exp(x)cos(nx) is kind of hard. You can do it without "inventing" these coefs. You can assume your integral value as I. Then you do integration by parts two times and you are left with I = c1+ c2 * I. You solve linear equation with respect to I and this is your integral value. This technique is very common in Russia when you learn calculus

@DeanCalhoun 2 года назад

this is the method I always use for these types of integrals as well. I also use tabular integration by parts to expedite the process

@mohan153doshi Год назад

Solving the integral e^x * cos(nx)) was just awesome, multiplying the integrand with a "Michael Constant" which is the very unique 1. I wonder how Michael Penn comes up with such awesome values for unity. If calculus is taught the way Michael Penn teaches it, then it would be the subject everyone would just adore. I envy his students, who have probably the best teacher dealing with the subject.

@goodplacetostop2973 2 года назад

14:56 Homework 16:00 Good Place To Stop

@MathSolvingChannel 2 года назад

Contour integral solved 😉 Solution: ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-jXEH0O5P3Ws.html

@chandranichaki9580 2 года назад

ru-vid.com/video/%D0%B2%D0%B8%D0%B4%D0%B5%D0%BE-S8-97elM3ho.html 😊😊

@nullplan01 2 года назад

OK, yes, you can evaluate the integral of exp(x)cos(nx) like that. Or you can use the tabular method and be done a million times faster. Or you can pre-calculate the integral of exp((a+bi)x) once (and put the result into Cartesian form), and have all integrals of the form exp(ax)cos(bx) and exp(ax)sin(bx) solved immediately.

@tomatrix7525 2 года назад

Fourier series is only valid for periodic functions or bounded so really your start is valid of e^x if x is in the -pi, pi interval. However, obviously taking x as 0 is all good and the result is perfectly valid. Important to point out though I think.

@mathunt1130 2 года назад

Computing the coefficient a_n was like pulling teeth. The normal way it's done in integration by parts twice and you're done. You could have treated it like a complex exponential to get the same result. The method used was too obscure to be a general method.

@BikeArea Год назад

12:35 👌👌 More of such transitions, please! 😊😊😊

@rublade1 2 года назад

Using complex numbers : sin x = im( e^ix) , this is not too hard, I think.

@mictecacihuat665 2 года назад

I would've done the e^x cos(nx) integral using Euler's formula on cos(nx) and I believe you should get the same result after some simplification. Good video tho! This reminds me of Nahin's book which I started reading once out of boredom but then got really busy very quickly and I kinda forgot about it.

@jorgen_persson 2 года назад

I read the goal and that's the point I told myself - This is a good place to stop

@oremilak 2 года назад

What if we do this variable change: u = ix? Sin becomes cosh and denominator becomes inverse of sin. We get same integrals but without the series complications and then take the real part.

@emanuellandeholm5657 2 года назад

Before I watch the video: I would do complex integration. Rewrite sin(x) in terms of e^jx, and then hope for some kind of partial integration or substitution to make it into a geometric sum.

@robertveith6383 2 года назад

That would be in terms of e^(jx) instead.

@emanuellandeholm5657 2 года назад

@@robertveith6383 obviously. Who sees e^jx and thinks e^j times x?

@SuperSilver316 2 года назад

Would love to see you replace the sine with cosine, and tackle that integral, cause it’s a doozy!

@calcubite9298 2 года назад

What's a foe-yay series and where can I learn more about them. I have a basic familiarity with Taylor polynomials, but I do not know how to spell Foyet??

@pooydragon5398 2 года назад

The sigma should have been inside the integral sign first. And if you are going to switch it, proving/stating uniform convergence would be necessary. I just completed my real analysis so I am glad to spot this!!

@skylardeslypere9909 2 года назад

Most of the videos are about calculus, not real analysis. The computation is the focus here. But technically you're right.

@skylardeslypere9909 2 года назад

Also, I don't think uniform convergence is the term you're looking for, but dominated convergence?

@etto487 2 года назад

@@skylardeslypere9909 for the Reimann integral you need uniform convergence, for the Lebesgue integral you have the dominated covergence theorem

@andreasavraam6898 2 года назад

@@etto487 actually uniform convergence is not enough in this case because you are integrating over a set of infinite measure,so you need the dominated convergence theorem here,also the uniform convergence is also a theorem in lebesgue integration and if the theorem doesn't work for the lebesgue integral,it obviously doesn't work for the riemann integral either

@andreasavraam6898 2 года назад

but basically to switch the sum with the integral u need the sum to converge absolutely to an integrable function

@voyageur8001 2 года назад

I= Integral (0 to 1) (lg(base 10) (1+x))/(1+x^2)

@henryginn7490 2 года назад

I haven't tried it, but this looks like contour integration would be good

@artempalkin4070 2 года назад

You totally leave out why you can bring the sum out of the integral. All other things are more or less trivial for those who took calculus course. Can you perhaps dedicate a video to explaining that bit? I'm pretty sure for most it's much more challenging to prove that than actually count all that was counted. And I think that is why this explanation is always missing even with math professors :) But without that explanation everything else is invalid unfortunately.

@General12th 2 года назад

I think you could have made the integral of e^x * cos(nx) its own video because, as you showed, it has a really clever solution that takes a bit of nuance to figure out. Then it could be an "appetizer" for this video.

@user-qc5qn7yp2z 4 месяца назад

12:11 wrong, the pi should be out of the circle

@AlanAlan2001 2 года назад

9:27 ❤️

@user-wu8yq1rb9t 2 года назад

I just enjoyed. That's it.

@alainbarnier1995 2 года назад

Nice ! I would be interested by some knowledge about this Fourier stuff expansion that I don't know at all... Thanks a lot. The Chess Player ;-)

@tgx3529 2 года назад

There Is alternative solution of this problem. This integral we can write as integral (sinx/e^x)* (1/(1+e^-x)dx= integral[ (sinx/e^x)* suma(-e^(-x))^n]dx=suma[(-1)^n *integral e^(-x(n+1))dx .After the substitution x(n+1)=y we have integral e^(-y)*sin(y/(n+1)) dy,i used for primitive function sin(y/(n+1))= imaginar part exp(iy/(n+1)),integral( e^-y)sin(y/(n+1))dy where y in(0; Infinity)=(n+1)/[(n+1)^2+1]*(1/1), we have then suma(-1)^n/[1+(n+1)^2] n=0..... The series (-1)^n (sinx)e^(-x(n+1)) convergents uniformly, (-1)^n * sinx has bounded uniformly S_n; e^(-x(n+1)) this function convergents uniformly to 0, it is decreasing( Dirichtet's test for uniform convergention).

@jbtechcon7434 2 года назад

Your Fourier expansion was confusing. e^x is not a periodic function,. But when you replaced the (n*pi*x/L) we usually see in a Fourier expansion, you were implicitly treated e^x as if it were periodic on some interval of length 2pi. Please explain!

@jerrysstories711 2 года назад

Yeah, this whole think is confusing because of that.

@skylardeslypere9909 2 года назад

The Fourier series only holds for x in the interval (-pi,pi). If f is 2π-periodic it holds for all x, but as you said, e^x is not. You could fix this by cutting off its graph between -π and π and "copy-pasting" it, but then you wouldn't really have e^x anymore

@Dymodeus1 2 года назад

@@skylardeslypere9909 it works cause he took x=0 later which is in the interval

@skylardeslypere9909 2 года назад

@@Dymodeus1 yep, that's right!

@tahirimathscienceonlinetea4273 2 года назад

No it's ok you can choose any continue function works just make sure cos(nx),sin(nx) those are periodic functions

@tahirimathscienceonlinetea4273 2 года назад

Thanks Michael for a good solution I may suggest to use euler formula of sin (x) product combining with Taylor expansion check this works??

@Reliquancy 2 года назад

I was kind of wondering what a calculus class that just assumed you could do any integral that came up with computer software would be like? Lwarnjng how to integrate all these functions seems like a means to an end but it’s kind of an endless process with all the different methods. So what if you just start like you’ve already done that?

@cernejr 2 года назад

About 0.364, about 53% of what it could be without the sin(x).

@khaledjebari1874 2 года назад

Hey Micheal I have suggested a Nice integral I hope it will be the topic of one of your videos Thank you in advance

@robertapsimon3171 2 года назад

I was a little perplexed about using the Fourier series of e^x as that’s non-periodic, did I miss something with that?

@shouligatv 2 года назад

It's implied that he uses the Fourier series of the function you get when you consider exp on (-pi,pi) then extend it everywhere else by periodicity. It's a common trick that you can use to find the value of the sum of 1/n^2

@henk7747 2 года назад

Hyperbolic cosecant 😳

@becomepostal 2 года назад

Exponential is not a periodic function.

@hoseinshooryabi3318 2 года назад

Very nice

@Happy_Abe 2 года назад

Where does the pi in front of the csch come from?

@walkerbill2081 2 года назад

(e^π - e^-π)/π should be 2sinh(π)/π (Michael missed the π denominator). Then, the reciprocal of 2sinh(π)/π is π/(2sinh(π)) which is same as πcsch(π)/2

@Happy_Abe 2 года назад

@@walkerbill2081 thank you It missing in the denominator threw me off