Marriage of Geometric and Harmonic Series: MIT Integration Bee (25)

[@ThAlEdison]

When you integrated f'(x) you needed a +c, and to determine the constant. Since you had f(x) term-by-term, you could just plug in f(0) to find the constant (which equals 0). You could have saved that step by noting it before differentiation, but if you did it was too subtly for me.

[@LetsSolveMathProblems]

You are absolutely right! It is a complete mistake on my part--I apologize for failing to mention the important fact. Yes, the constant of integration should be included when antidifferentiating 1/(1-x), and its value should be found by examining f(0). As you mentioned, the value of the constant is 0 in this case, so the final answer in the video is correct. Thank you so much for pointing out this missing step! I will strive to avoid similar mistakes in the future videos.

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@alkankondo89]

I DO NOT have the mathematical acumen to solve integrals as complex as this on my own! But, I love math and I learned a lot from this well-explained example. So, I subscribed. Thank you for taking time to explain every detail! This is a very curious example.

[@metrogman2409]

That was beautiful. Every twist was clever.

[@architjain811]

We can also use the taylor series of -ln(1+x) though...It yields the same result

[@sibasishpadhy9172]

yes

[@filipeoliveira7001]

he derived the taylor series bro

[@KevinS47]

MIC drop at the end there..! This was great, seeing how you can try and tackle problems like this is incredible, I don't think I would have thought of differentiating the series to go back and integrate it. That was incredile! Thanks.

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@unho126]

that was an adventure!

[@surodeepspace]

You deserve to be much more popular! Great work!

[@dhruvsharma8522]

Write 1 in start as log2(2) Take log func formed in numerator by change of base ( for base 2) formula or simply flip the thing in log2( f(x)) Then again base e formula Take constant out Integrate by parts While putting limit 1 You get problem for log(1-x/2) get limit when x tend to one by dl'hospital rule rest all no problem

[@thismadhulikano]

That's amazing...but wont the step function in the denominator affect the flow of steps mentioned, as it cannot be integrated directly right

[@josephmartos]

Dude this is pure genious... Btw why you stopped posting videos???

[@kamehamehaDdragon]

Great video. Integrals have never been so exposed! Love your videos, man!

[@krisbrandenberger544]

What I learned from this video is that for every positive integer n, if u is between 1/2^n and 1/2^(n-1), the value of f(u) is 1/n.

[@stivensilva8121]

This result is pi^2/12-log^2(2)/2.

[@visheshmangla2650]

i m getting 1/4 ln(2) by modifying it to log2(2/(1-x)) and then take inside part as substitution.

[@eleazaralmazan4089]

How long does it take you to solve these kind of problems?

[@LetsSolveMathProblems]

For me, the time it takes to discover a solution varies significantly from a problem to problem. Many integration bee problems only take me less than a few minutes (some even less than a minute) due to their more basic nature or due to their coverage of the topics/tactics that I am well-versed in. On the other hand, some problems require me 30 minutes or far more time to decipher. Of course, there are some problems that totally confound me; in such a case, I research appropriate topics to aid my problem-solving journey. After I solve the problem, I tend to spend more time in organizing my solution and researching more related topics to increase the quality of the respective videos--often, the time it takes in organizing the solution is far longer than the time it takes to discover the solution in the first place. =)

[@pawananubhav12]

LetsSolveMathProblems thanks a lot for such a helpful reply!

[@leif1075]

@@LetsSolveMathProblems Don't you find that process too depressing and frustrating and,boring , especially becaise its so time-consuming? Thanks and I hope you can answer..

[@leif1075]

Cant you still,differentiate any infinite,series,term,by term since the derivative,of a sum equals the sum of,the derivatives? Whether it converges or not?

[@mastershooter64]

@@leif1075 Not if you enjoy problem solving, and many people who like maths and physics love solving problems

[@rickybobby5584]

we have - ln(1-x)= sum(n=1->inf) x^n/n then sum(n=1->inf) (1/2)^n/n=-ln(1-1/2)=ln(2)

[@WillCummingsvideos]

This can be generalised for log_n to yield the answer (1-n)log(1-1/n)

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@sonamthisside]

Nice question...I solved it very easily in my first attempt. solving it as x=1/y makes it easier to understand. I don't understand why you used differentiation and series convergence, the series which we got were just expansion of -(log(1-x)) with covergency for

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@towhidurrahman4686]

Very beautiful problem and solution!!!

[@sibasishpadhy9172]

thank you for this..good revision

[@coolclips101]

What a beautiful problem

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@novelas3536]

Excellent video as always. :)

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@ShoaibAli-er4id]

Why you stopped making videos???

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@jeremybrettfan7742]

For f(x) = x + x^2 / 2 + x^3 / 3 + .... could we have used the formula that ln( 1 - x ) = - ( x + x^2 /2 + x^3 /3 +....) instead of the differentiation ?

[@thephysicistcuber175]

This was awesome :)

[@CoolCat6131]

bro i fucking love you so much

[@زينالعابدينماجد-خ1خ]

thanks for you

[@reinerwilhelms-tricarico344]

Now of course I wonder if you would get ln[b] if you use log_b instead of log_2 ... but I will most likely screw up something when I try this. ;-)

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@sansamman4619]

watching this after the series of f(x) was a big pain for me since I'm familiar with Taylor series ;-;

[@0x8055]

that was beautiful

[@humbertocueva3815]

Why you changed the bound?

[@ayushjangid]

What if there was no floor function and we have to calculate the integral I think I have the solution for that case also

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@VaradMahashabde]

10:47 the proof has been left to the viewer

[@dhruvsharma8522]

Seeing MIT bee problems I salivate I can solve all that you make videos of in less time

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@tasteful_cartoon]

If it had been log_3(1-x), would the result be ln(3)?

[@erazorheader]

For log_a (1 - x) you'd get (a - 1) ln(a/(a - 1)). The integral is obviously reduced to the series \sum_1^\infty (a - 1)/(n a^n). Next, you take 1/n = \int_0^\infty dx e^{-n x} and get (a - 1) \int_0^\infty dx \sum_1^\infty e^{-n(x + ln(a))} = (a - 1) \int_0^\infty dx/(a e^x - 1) = (y = a e^x) = (a - 1) \int_a^\infty dy/(y (y - 1)) = (a - 1) ln(1 - 1/y)_a^\infty = (a - 1) ln(a/(a - 1)). So, for the base 3 you'd get 2 ln(3/2).

[@sankalanghosh5503]

well that's anothor nice one,by finding the area

[@eshward8448]

Yes,I agree, but since ln(1+x)=x-x^2/2+x^3/3-x^4/4+......, and ln(1-x)=-x-x^2/2-x^3/-x^4/4.....,S can straightaway written as -ln(1-x), which would make the solution much simpler.

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@armandobuzzini9171]

I from Argentina. It's very genial

[@erazorheader]

I have done it in my head actually lol The part to get it to the series is trivial, then I just used that 1/n = int_0^\infty e^(-n x) dx, so the series is reduced to \int_0^\infty dx \sum_1^\infty exp(-n (x + ln(2))) = \int_0^\infty dx (1/(2 e^x - 1)) = (y = 2 e^x) = \int_2^\infty dy (1/(y(y-1))) = ln(1 - 1/y)_2^\infty = ln(2).

[@guillermoalfaro2633]

Do you have a platform where we can share math problems with you? Or an email?

[@GhostyOcean]

Guillermo Alfaro I'm pretty sure you can still PM people on KZbin. I typically just Google KZbin inbox to send people stuff

[@guillermoalfaro2633]

GhostyOcean thanks fellow I'll try

[@aryan24jan]

kzbin.info/www/bejne/r5C7qIShmM2qqKc

[@yashvardhan2093]

The title is justified

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@filip-kochan]

wow Wolfram Alpha wasn't able to solve this...

[@rijulb5676]

The answer to this question is 2 and not ln2

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@anmolempire1197]

U done wrong ...as When u. log_2u. [1-log_2u] 1 0. 0 1/2 -1. -2

[@kreglfromworld]

"ooh"

[@maliktrevormalul3ke607]

Lol Chief you are Key

[@padraiggluck2980]

👍

[@user-vm6qx2tu3j]

Pretty😍

[@paolo6219]

Easiest u sub ever

[@abhijeet_1102]

The answer is Binod

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@Schizophrenic176]

Great video but I don't understand English enough.

[@beoptimistic5853]

kzbin.info/www/bejne/joKsk6FobMmCoKc 💐👍

[@faizanqaiser4027]

inbeitveee...