Nice formula. We can find the 2nd sum by using ODE. The function y=sum x^(3n+2)/(3n+2)! has to verify y+y'+y"=exp(x) with y(0)=0 y'(0)=0. The answer is y"(1).
Fantastic! Suggestion: Make use of complex conjugates, when they arise. E.g., when you have e^(i·stuff) + e^(-i·stuff), where "stuff" is real, these are conjugates of each other, so you know _a priori_ that the imaginary parts cancel, and you need only evaluate the real parts, which are equal, so they double. In fact, e^(i·stuff) + e^(-i·stuff) = 2 cos(stuff), by Euler's formulas for sin & cos, anyway. But the same goes when e gets replaced by any other real number - you'll have conjugates, the imaginary parts cancel, and you need only evaluate the real parts. Nice job! Thumbs up!! Fred
I used the method to calculate Sum[(-1)^n/(3n+1),{n,0,Inf}]. Associated series: Sum[(-1)^n/(n+1) x^n,{n,0,Inf}]. Generating function: Ln[1+x]/x. Final result: (Ln[2]+Pi/Sqrt[3])/3.
Nah, I think leaving like it is makes it more mind-blowing. One could express it as a piecewise function where the answer depends on n modulo 4, but this formula is much more tidy.
I would like to know more of your thought process in coming up with the tool. I love formulations that generalize to solve classes of other problems. This one rocks.
@@fmakofmako Thanks for that. Remarkable. I guess it sometimes comes down to exposure. The use with generating functions is elegant. Who came up with the roots of unity filter?
Here's a practical problem. Come up with closed form formulae that give the sun's azimuth and altitude for any date and time and any latitude and longitude. I did that several years ago. IIRC, it had about three arc cosine functions in it. I think I used maybe a cross product and some dot products. No solar tracking feedback loop for me!
One other thing that might be helpful---this comes from a generalization of finding even and odd components of a function. You have probably seen examples like plugging in -1 to a function, and adding to cancel a bunch of terms. This comes from the thought "can we generalize this?"
I spent a long time puzzling over the fact the first result is wrong for n=0 until I realized the way he evaluates the terms only holds if n is larger than 0 because if you evaluate (1-x)^n at x=n=1 you get an indeterminate form. if you evaluate at n=0 before setting x=1 the correct result is returned. Just putting this here for anyone in the future who sees this and is also confused.
Hey Michael. I found a generalized form in terms of a sum for the problem, say a variable r in place of 3 for odd/even cases. I couldn't figure out a way to simplify it to be a single expression, however.
I'm watching these videos with the hope that someday, I'll be as smart as you! Thank you for being an inspiration to me and many others. Much love from an aspiring maths teacher.
Michael Penn: I have a problem for you. Take the factor 1/(1-n^k) and consider a product over n, k in N, n > 1. From exploring this product in Mathematica, it seems to me that the nominator contains a factor of k times pi to the floor((k-1)/2) power. Some of the products are given as products of Gamma() in terms of translated unity roots and the others are products of (hyperbolic) trig functions, basically sin()*..*sinh(). This is just beyond me, maybe you could make some headway here?
As you stated it, the product does not converge. Isn't it 1/(1-n^-k)? If so, and n runs over the primes, you get Zeta(k), but with n running over the Naturals, I don't know...
@@sergiokorochinsky49 Darn. I messed up the copy-paste/transcription. I meant the factor (1 - 1/k^n), ie. (k^n - 1)/k^n. This converges for at least n in N, n > 1. Thanks for bringing this to my attention!
I don't think it simplifies. This one simplified a little bit, but in general it shouldn't be expected to look nice. The formula is essentially a direct application of this method.
I did see a general formula for the sum of 1/(kn+c)! once, on a maths discord server. It involved sums of sums of exponentials. It was extremely cumbersome, and wasn't nice at all. I'm not sure if I would be able find it anymore, since I left the server. But it does mean that a closed formula is indeed possible
the tool looks like an extremely powerful method for calculating power series and normal series
I knew since the first day I found your channel that it will get the recognition it deserves!! Thank you for everything!!
Naila Ben Ali i remember when he only had hundred views.
At 18:23 is it not supposed to be 1/sqrt(e) rather than sqrt(e) or am I losing my mind?
Yeah the answer should be (e*sqrt(e)+2cos(sqrt(3)/2))/(3sqrt(e)) or . (e^2 + 2sqrt(e)*cos(sqrt(3)/2))/(3e).
You are right
Yes, you're right. But you're still free to lose your mind if you like.
Fred
Nice formula.
We can find the 2nd sum by using ODE. The function y=sum x^(3n+2)/(3n+2)! has to verify y+y'+y"=exp(x) with y(0)=0 y'(0)=0. The answer is y"(1).
excellent.
Fantastic!
Suggestion: Make use of complex conjugates, when they arise. E.g., when you have e^(i·stuff) + e^(-i·stuff), where "stuff" is real, these are conjugates of each other, so you know _a priori_ that the imaginary parts cancel, and you need only evaluate the real parts, which are equal, so they double.
In fact, e^(i·stuff) + e^(-i·stuff) = 2 cos(stuff), by Euler's formulas for sin & cos, anyway.
But the same goes when e gets replaced by any other real number - you'll have conjugates, the imaginary parts cancel, and you need only evaluate the real parts.
Nice job! Thumbs up!!
Fred
I used the method to calculate Sum[(-1)^n/(3n+1),{n,0,Inf}].
Associated series: Sum[(-1)^n/(n+1) x^n,{n,0,Inf}].
Generating function: Ln[1+x]/x.
Final result: (Ln[2]+Pi/Sqrt[3])/3.
Very powerful! I wonder it that can help in calculating any harmonic numbers or values of the Dirichlet' Beta function
Thank you for this very powerful tool and for the usage examples.
The first result should be arranged in a form that makes it easier to see that it is actually an integer.
Nah, I think leaving like it is makes it more mind-blowing. One could express it as a piecewise function where the answer depends on n modulo 4, but this formula is much more tidy.
J M Seconded, it makes you think for a bit and then you go aha!
Hi Michael, I’m from France and your tutorials are a great help for my studies !
I would like to know more of your thought process in coming up with the tool. I love formulations that generalize to solve classes of other problems. This one rocks.
You can look up a more detailed explanation under "roots of unity filter".
@@fmakofmako Thanks for that. Remarkable. I guess it sometimes comes down to exposure. The use with generating functions is elegant. Who came up with the roots of unity filter?
Here's a practical problem. Come up with closed form formulae that give the sun's azimuth and altitude for any date and time and any latitude and longitude. I did that several years ago. IIRC, it had about three arc cosine functions in it. I think I used maybe a cross product and some dot products. No solar tracking feedback loop for me!
One other thing that might be helpful---this comes from a generalization of finding even and odd components of a function. You have probably seen examples like plugging in -1 to a function, and adding to cancel a bunch of terms. This comes from the thought "can we generalize this?"
You can just use the complex definition of cosine to simplify the real and imaginary parts much quicker than expanding them.
I cannot unsee that Michael draws capital sigmas completely differently on the bottom third of the board than on the top two thirds.
I spent a long time puzzling over the fact the first result is wrong for n=0 until I realized the way he evaluates the terms only holds if n is larger than 0 because if you evaluate (1-x)^n at x=n=1 you get an indeterminate form. if you evaluate at n=0 before setting x=1 the correct result is returned.
Just putting this here for anyone in the future who sees this and is also confused.
absolutely amazing
this is great !
I just LOVE these kinds of identity sums😃
Another high quality video by Mr.Michael 😄😄😙👍👍👌👌
Great explanation
fantastic video,
It would be great if you did some Non-Euclidean geometry videos
Please do 2006 IMO problem 5 and thx for the great work
Where do you learn this sum manipulations? Can you suggest a book perhaps?
Generatingfunctionology. You can find it online.
I understand the maths, but how do you make things appear and disappear ??
magic.
Just like some functions, Michael's videos are not absolutely continuous!
@@Hiltok nah, that would be pushing your limits.
The first result does not hold for n=0 but it does for other (nonnegative) integers
Hey Michael. I found a generalized form in terms of a sum for the problem, say a variable r in place of 3 for odd/even cases. I couldn't figure out a way to simplify it to be a single expression, however.
does the first sum diverge?
I mentioned (very briefly) that n is a natural number and thus the first sum is finite. Its just simpler to write infinity as the upper bound.
Is this the idea behind the Fast Fourier Transform? Seems very familiar.
Very nice ... keep going
There is an error in writing the answer for the second part.It should be 1/sqr(e) .
Lets solve math problems (youtube channel) used this too. And he calls it as roots of unity filter
I'm watching these videos with the hope that someday, I'll be as smart as you! Thank you for being an inspiration to me and many others. Much love from an aspiring maths teacher.
I solved it using Laplace transform and it gives the perfect result
How?
Manu N with Laplace transform
But the video is not 4k. That's a contradiction!
nice
shouldn't it be relatively prime instead of indivisible by m?
Nice
5.34.how spread sum??
Good
This is brilliant !!!
I love your videos but sometimes I have students who have trouble following. What level of student are your videos geared towards?
Michael Penn: I have a problem for you. Take the factor 1/(1-n^k) and consider a product over n, k in N, n > 1. From exploring this product in Mathematica, it seems to me that the nominator contains a factor of k times pi to the floor((k-1)/2) power. Some of the products are given as products of Gamma() in terms of translated unity roots and the others are products of (hyperbolic) trig functions, basically sin()*..*sinh(). This is just beyond me, maybe you could make some headway here?
As you stated it, the product does not converge. Isn't it 1/(1-n^-k)?
If so, and n runs over the primes, you get Zeta(k), but with n running over the Naturals, I don't know...
@@sergiokorochinsky49 Darn. I messed up the copy-paste/transcription. I meant the factor (1 - 1/k^n), ie. (k^n - 1)/k^n. This converges for at least n in N, n > 1. Thanks for bringing this to my attention!
Nice video as always!
I have a problem for you, it is in your last video's comments
Is there a general formula for a for the sum of 1/(ak)! from k=0 to infinity?
we should give that a try
I don't think it simplifies. This one simplified a little bit, but in general it shouldn't be expected to look nice. The formula is essentially a direct application of this method.
I did see a general formula for the sum of 1/(kn+c)! once, on a maths discord server. It involved sums of sums of exponentials. It was extremely cumbersome, and wasn't nice at all. I'm not sure if I would be able find it anymore, since I left the server. But it does mean that a closed formula is indeed possible
Could you, please, solve this problem: find the minimum value of f(x,y) = sqrt(x^2 + 2x + y^2 - 2y + 2) + sqrt(x^2 - 6x + y^2 + 9) + sqrt(x^2 - 8x + y^2 + 16) + sqrt(x^2 - 8x + y^2 - 6y + 25)
Michael @11:48 you say the summation of (1-1)^n =0. But isn't the first term 0^0=1? It seems in this case 0^0=0.
No.
1 - x evaluated at 1:
1 - 1 = 0
This man looks quite like 3b1b Grant Sanderson !!!!!!!!!
Sir using cube roots of unity could have made things simpler and taking exponential series e^x
Is that not what he did?
Yes, that is exactly what he did.
OMG
9:52 Shouldn't it be there (1+x)^n=sum as k=0 to n of n choose k times x^k actually?
it comes back to the same because when k> n, the binomial coefficients are zero by definition