A sum two ways!

We calculate the closed form of a nice infinite series two different ways.
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[@derendohoda3891]

[dominated convergence] "I promise I'm going to do a video on that in the future" great I can't wait!

[@jceepf]

And that will definitely be a good place to stop! (I love this guy!)

[@crackindenpockets6211]

I wish he would have checked the uniform convergence in this video, because he would have then realized that the integration summation switch WASNT legitimate. The geometric series does NOT converge uniformly on neither one of the intervals [0,1] and (0,1)

[@EddieEntertainment]

@@crackindenpockets6211 what up

[@CM63_France]

Hi Michael, You have 3 more sentences available for your tee-shirts: "Let's go ahead and do that" "All the way up to..." and the famous : "and so on and so forth" Like your channel

[@MonzennCarloMallari]

Also "I'll let you do the rest"

[@amandeep9930]

Also " and thats a good place to stop"

[@synaestheziac]

“So we’ve got something like this”

[@synaestheziac]

“Let me just clean up the board, put this at the top, and then we’ll go ahead and finish this off”

[@CM63_France]

@@synaestheziac you are right, or the variant : "and finish it off".

[@bsuperbrain]

The first solution was much more elegant.

[@crackindenpockets6211]

its not correct though (I believe)

[@goodplacetostop2973]

31:12

[@panagiotisapostolidis6424]

Wikipedia says that An infinite series of any rational function of n can be reduced to a finite series of polygamma functions, by use of partial fraction decomposition.This fact can also be applied to finite series of rational functions, allowing the result to be computed in constant time even when the series contains a large number of terms. en.wikipedia.org/wiki/List_of_mathematical_series Can you do a video on this?

[@d314159]

Third method: Let S_N be the partial sum of your series up to N, then S_N = 2(H_2N - log(2N)) - 2(H_N-log(N)) - 1/(2N+1) + 2log(2) - 1, where H_N is the harmonic series (1/n) summed to N. Take the limit as N tends to infinity noting that H_N - log(N) converges.

[@Blabla0124]

1:46 It is slightly easier to choose n=0, n=1/2 and n=-1/2. You get the A, B and C directly.

[@VerSalieri]

Seriously man... thank you. After a long day of tutoring, it's nice to unwind to your videos.

[@nilsastrup8907]

I understand all the steps perfectly fine, but how the hell do you come up with those manipulations?

[@Saki630]

[Penn]: Here we have a classic infinite sum problem. [Me]: Checks video and realizes its 31minutes. I'm going to diverge.

[@ThAlEdison]

23:25 When some steps were skipped, it included the cancelation of (1/y)ln(1+yz), which diverges as y->0. It cancels with (-1/y)ln(1-yz), but I would've liked to see the cancelation.

[@jordancourtemanche3164]

Your final integral you can also group together to rewrite as \int_0^1 [(1+z)ln(1+z) + (1-z)ln(1-z)]dz, then make the two substitutions and combine very quickly to \int_0^2 wln(w)dw. You do still get a limit evaluation, but much simpler and on a single piece.

[@luccavelier9514]

A simpler way is to multiply all terms by 2, at the very beginning and check that 4039,4041 gcd is one

[@田村博志-z8y]

We have 1/( n( 4n^2 - 1 ) ) = ( 4n n - ( 4n^2 - 1 ) )/( n( 4n^2 - 1 ) ) = 4n/( 4n^2 - 1 ) - 1/n = ( ( 2n + 1 ) + ( 2n - 1 ) )/( ( 2n + 1 )( 2n - 1 ) ) - 1/n = 1/( 2n - 1 ) + 1/( 2n + 1 ) - 1/n Easy way to divide the expression into fractions.

[@tomatrix7525]

Have you ever had a look at Blackpenredpen’s partial fraction shortcut. I’m sure he isn’t the one who created it, I’ve just seen him use it alot. Might be cool to start implemting that shortcut into your work.

[@dclrk8331]

Love to see a video about change of summation and integration!

[@nathanisbored]

8:10 how do we know you can add and subtract 1 to the series like that? In fact, it seems like it shouldn’t be able to because it doesn’t converge absolutely... which means where you put the +1 should matter right? If you’d put the +1 in the middle of the series instead, by the rearrangement theorem wouldn’t that suggest the answer might change?

[@pikupal8996]

If you remove first few terms of a conditionally convergent series then it does not affect the closed form of the sum that is it converges.

[@MELONOLDER101]

What a position at the end its pretty nice to say it when it comes to a huge product involved sum I believe the 3rd power will be baaaaad Hhhhhh I mean n(8n^3 - 1)

[@manucitomx]

Great! (That’s all can say aside from: Thank you!)

[@nicolascamargo8339]

wow me imagino con más factores diferentes en denominador más integrales y es un proceso super extenso ahí se ve la funcionalidad de las fracciones parciales jajaja.

[@alihaydar728]

11:40 I am pretty sure that -1 cancels when you plut in x=0 at the end and the answer should be just 2*ln2 edit: nevermind , now i see

[@backyard282]

No, when you put x =0 you just get ln(1) which is zero.

[@snejpu2508]

Not really, because -1 wasn't after the integral sign, it didn't depend on x. So the evaluation from 0 to 1 only refers to 2ln(1+x). In order to cancel out, it would have to be (-1+2ln(1+x)) from 0 to 1, but it's not possible actually.

[@Wurfenkopf]

I wonder if there is a general formula for the sum where we replace the coefficient 2 with a generic coefficient k>2 . This has to converge since it's bounded by the previous sum. So what may it be? 🤔

[@MarkusDarkess]

Say a square is. O,1/1,0 /0,-1/-1,0 back to 0,1 the square half the size would be .5,-.5/.5,.5/-.5,.5/-.5,-.5 Back to .5,-.5 the hypotenuse is congruent/ parallel to line segments (0,1/1,1)/(0,-1/-1,-1) or line segments (1,1/0,-1)(0,1/-1,-1) X,y. X going left and right.. Y going. Up and down And it has the area half of the previous square. 2^40= 6N+/-1 3^29= 6N+/-1 4^23= 6N+/-1 5^18= 15N+/- (2 or 4) 6^1= 6N+/-1 7^14 = 6N+/-1 8^14= 6N+/-1 9^13= 6N+/-1 10^13= 15N+/-1 Mersenne primes at bigger numbers are 2^n-1 And 2^n is divisble by 6 at 2^40 So mersenne primes are at 6n-1 at that point. So are there points beyond 2^40. That have primes at +/- 2 since mersenne search only searches for primes at 2^p Are there primes we are missing? And with my solution to riemann hypothesis I said twin primes lie at 15x even number is +/- 1 and 15x odd number is (+2/+4) or (-2/-4) And dont get me started on a square with an area of 8... (mind breaking) The checking I did was Google calculator converginging at those points. And my calculator app 2^n/6 at 2^46 while Google was at 2^40. But then bing didn't So does 2^n converge to 6n+/-1 yes or no? And if does/doesn't... do calculators world wide need to be calibrated? Or do computers take short cuts? And that the confirmed mersenne primes might be wrong past a certain point?

[@MarkusDarkess]

And with the math I did... twins lie at 15x even numbers +/-1.. 15x odd numbers is at (+2/+4) or (-2,+4) Twins 21,011/2013 and 21,017/21,019 prove it.

[@giuseppemalaguti435]

Ma scusa, al minuto 10 circa, non potevi riconoscere ln2 subito?

[@infas0tka693]

While doing partial fraction decomposition(especially when all the degrees are 1) it is(in my opinion) faster to multiply both sides by one term at a time and find one of the coefficient. E.g.: you had 1/(n(2n+1)(2n-1)) = A/n + B/(2n+1) + C/(2n-1), where we can multiply both sides by "n" and get 1/(4n^2 - 1) = A + Bn/(2n+1) + Cn/(2n-1). Then we can set n = 0 and see that A = - 1. Same way multiplying by "2n+1" we get 1/(n(2n-1)) = A(2n+1)/n + B + C(2n+1)/(2n-1), setting n = - 1/2 we found out, that B = 1.

[@maahaanmahmit9962]

Last integration could solve very easy. (1+-z)ln(1+-z)both have a similar integral like zlnz...! int(zlnz)=z2/2lnz-z2/4 that’s it...!

[@anastasissfyrides2919]

General question: During the part of the partial fractions,(lets denote the coefficients a1,a2, a3 etc instead of A, B,C,...) i noticed always(?) every k-1 fractions after becoming polynomials(when there are k fractions) have a common root. Since we need the equality to hold for any n, we can deliberately plug in those roots so that we can find a1, a2, a3,..., ak quicker, by exterminating k-1 polynomial terms every time, to find the coefficient of the remaining term almost instantly. The values i found are correct, nevertheless i noticed afterwards that we couldnt plug them in, since the polynomial was created from a rational function and these exact roots nullified the denominator. What is going on exactly?

[@TAT-pg3kj]

On the first way ,how to prove the follow lim integral_0^1 x^n(x+1)^-1 dx as n->infinity =0 ?

[@zanti4132]

When planning a trip with a group, the first suggestion comes from someone who says they can take the interstate. Then Prof. Penn says, "That works, but there is a second way. We can take these back roads through the mountains, board the ferry to cross the river, and after changing routes 15 more times we'll get there a few days later."

[@karakrisundkramm]

I like watching your videos, but the amount of commercials during the video makes it impossible to follow your arguments without losing focus. I am not going to finish this one for this particular reason.

[@amaysharma1909]

Big lol,i did this in 1 min

[@shafrazameer631]

glade to have a chat with you sir . ........ , sir can you please help me to solve this intergal which is given below ...... x^xsin(e^x-tanx)

[@math-4-science32]

Claim: If a, b, c ∈ Z and have the same parity, then the quantity Δ = b² - 4ac cannot be a perfect square. Is this claim true? If yes, prove it.

[@pbj4184]

Now I'm hooked as well :) Please let us know the solution if you find it

[@samsonblack]

Hint: can you find a quadratic polynomial with integer coefficients, all of the same parity, with integer roots?

[@pbj4184]

@@samsonblack I think you mean rational roots

[@pbj4184]

I think the question intended a,b,c to be relatively co-prime. Otherwise, you can come up with trivial counterexamples like 2x^2-4x+2=0 which really is (x-1)^2=0 Edit: Now that I think of it, shouldn't a,b,c be odd _and_ coprime for the question to make sense?

[@math-4-science32]

Let's say a, b, c are all odd or if even we have d = gcf(a,b,c) so a/d, b/d, c/d are all odd

[@giuseppemalaguti435]

Più semplice il primo metodo

[@rogerlie4176]

A faster way to find the coefficients of the partial fractions is to multiply by a factor in the denominator and set that factor to 0. E.g. to get B in 1/(n(2n - 1)(2n + 1)) = A/n + B/(2n - 1) + C/(2n + 1) multiply the equation with (2n - 1) and set n = ½. This directly give us 1 = B.

[@shafrazameer631]

x^xsin(e^x-tanx)

[@shafrazameer631]

x^xsin(e^x-tanx)

[@shafrazameer631]

x^xsin(e^x-tanx)

[@shafrazameer631]

x^xsin(e^x-tanx)

[@shafrazameer631]

x^xsin(e^x-tanx)

[@shafrazameer631]

x^xsin(e^x-tanx)

[@l1mbo69]

In the second solution when we were summing the geometric series, how do we know that the common ratio (xy²z²) is

[@goblin5003]

I think this has to do with how intégration over a [0;1] gives you the same answer as ]0;1[ Another way of explaining it would be to say that integrating over ]0;1[^3 is the same as integrating over [0;1]^3 since the boundary of the cube has no volume (it’s measure =0)

[@Grundini91]

couldn't the answer be rewritten as ln(4/e)?

[@lavneetjanagal]

Just want to point out there could be an intermediate method. In the first method you resolved into partial fractions, in the second method you did not. You can partially resolve into partial fractions as S = 1/2n [1/(2n-1)-1/(2n+1)] . Sum over n implied. Now, you can introduce the generating function as S = \int_0^1\int_0^t [1/(1-x^2)-x/(1-x^2)]dx dt which becomes , \int_0^1 Log(1+t) dt. Which gives -t +(1+t)Log(1+t) from 0 to 1. You get the answer -1+Log(4)

[@pandas896]

Fuck , I thought no one knew this method

[@Arbmosal]

If we multiply the whole series by 1/2 we can write it as 1/(2n-1)2n(2n+1), i.e. in the denominator we have three consecutive numbers. Now if we consider the power series f(x)=x^3+x^5+x^7+... and take the third derivative we find 1/(1*2*3) + x^2/(3*4*5) + ... So if we were to evaluate this function at 1, we get our original series back. HOWEVER, the obvious thing would be to recognize the geometric series in f, i.e. for |x|

[@pikupal8996]

I think your power series should be f(x)= 1+x^2+x^4+x^6 where |x|

[@Arbmosal]

@@pikupal8996 Haha silly mistake :D Thanks

[@user-A168]

Good

[@djvalentedochp]

wow nice problem 👍👍👍

[@nmkjnmnjm]

I don’t quite get what is the interest of the second way, which is clearly much more complex, with significant risks of errors along the way

[@jimskea224]

There are only risks of errors if you're human!

[@tomatrix7525]

He’s just demonstrating different ways of doing it, he’s not saying it’s the ideal way of doing it.

[@natepolidoro4565]

Great video

[@henk7747]

henk

[@oooo-vd1ry]

31:19

[@АлексейЯ-д3э]

n = 1, 4n^2-1 = 0, 1/0 = ? Or Summa from n = 2?

[@demenion3521]

4*(1)²-1=3 by the way...

[@MarkusDarkess]

Pythagream theorem. Right triangle... 3^2+4^2=5^2 (0,0/0,3/4,0) to make a square. From that The points would be 0,0/0,7/7,7/7,0 back to 0,0 To make a rectangle. 0,0/0,6/8,6/8,0 Both are the pythagream theorem of 3^2+4^2=5^2 Why because (0,0/0,1/0-1) The pythagream theorem would become (0,0/0,2/-2,2/-2,0 and back to 0,0)

[@MarkusDarkess]

kzbin.info/www/bejne/qmmvqYeXqrujgLM

[@wojteksocha2002]

31:12