Omnimaga
General Discussion => Other Discussions => Math and Science => Topic started by: Nosferatu Arucard 1983 on November 08, 2012, 10:00:18 pm
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Since the infinity sum of sin(x) from x=0 to infinity (sin(0) + sin(1)+ ...) don't converge, we can use the Abel-Plana formula to calculate a renormalized sum of the former divergent sum:
sum(f(t),t=0 to infinity) = f(0)/2 + integrate(f(t),t,0,infinity) - i * integrate((f(i*t) - f(-i*t))/(exp(2*pi*t)-1) , t, 0 ,infinity)
But today I only show the problem, waiting to someone to try solve this ridle. ;)
Edit: A minor fix to the result. :P
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Welcome to omni :D
I won't be able to proof that though.
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Welcome to Omnimaga indeed :D
And I am a little confused by your notation. When you say Sum, do you mean integral?
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no I think it means the iterative sum >sigma notation?
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I'm gonna think about that, it looks inteesting :)
And btw, you sir deserve 1 internet for making WolframAlpha crazy (first time I actually see something like that on it) :
(http://i.imgur.com/zpEVi.png)
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When I mean sum is the usual sigma notation.
I will show gradually the proof of this renormalized sum, so be aware.
I'm also the main author of a portuguese written book of Special Functions that I publish in Amazon: "Introdução à Função Zeta e Gama de Riemann": Introduction to Riemann Zeta and Gama Function.
The proof of Abel-Plana formula was derived in Theorem 7.3 of my book ;D(http://www.omnimaga.org/Themes/default/images/gpbp_arrow_up.gif)
But I will translated to you... :D
1st Step:
Just apply f(t)=sin(t) to the Abel-Plana Equation, and it give:
sum of sin(t) from t=0 to infinity = sin(0)/2 + integral(sin(x),x,0,infinity) + i * integral(2*sinh(t)/(exp(2*pi*t)-1),t,0,infinity)
sum of sin(t) from t=0 to infinity = integral(sin(x),x,0,infinity) - integral(2*sinh(t)/(exp(2*pi*t)-1),t,0,infinity)
Since sin(i*t) = i* sinh(t) and sin(-i*t) = - i * sinh(t).
If you ask it, I can publish the proof of Abel-Plana by steps.
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Wow, this looks absolutely intriguing. I cannot yet arrive to this same conclusion, but I have some math that may be of interest. Please note that I figured this all out on my own (including the Euler-Maclaurin formula. I only just recently learned that it was already discovered). Anyways, this does not at all answer your question, but I think you will appreciate this math, especially if you understand the Riemann Zeta function.
Use the Euler-Maclaurin formula to show that:
sum(sin(n),n,0,x)=sum(sin(n),n,1,x)=integral(sin(x))+sin(x)/2+cos(x)/12--cos(x)/(30*4!)+cos(x)/(42*6!)--cos(x)(30*8!)+...
=-|B0|cos(x)+|B1|sin(x)+|B2|cos(x)/2!+|B4|cos(x)/4!+|B6|cos(x)/6!+|B8|cos(x)/8!+|B10|cos(x)/10!...
(where Bn is the Bernoulli numbers)
Then we have that sum(sin(n),n,1,x)=B1sin(x)+cos(x)(-B0+|B2|/2!+|B4|/4!+|B6|/6!+|B8|/8!+...)=B1sin(x)-cos(x)+cos(x)sum(|B2n|/(2n)!,n,1,x)
To be mathematically tricky, I will make that -cos(x) go into the sum, even though all the other terms are positive. How? Note that 2^0=1, but 2^n, where n is an integer greater than 0 is even. Then, (-1)2n=-1 for n=0 and 1 for all the rest. So:
sum(sin(n),n,1,x)=sin(x)/2+cos(x)sum((-1)2n|B2n|/(2n)!,n,0,x)
Note that this is a constant: sum((-1)2n|B2n|/(2n)!,n,0,x)
Now we step completely away from this. How do we write sum(sin(n),n,1,x) in an alternative way? First, look at sum(an,n,0,x)=a0+a1+a2+a3+...+ax. If we add ax+1, we will have the following:
ax+1+sum(an,n,0,x)=sum(an,n,0,x+1)
Reindexing the right side a few times:
ax+1+sum(an,n,0,x)=a0+sum(an,n,1,x+1)
ax+1+sum(an,n,0,x)=1+sum(an+1,n,0,x)
Now the sums have the same index, so we can combine them:
ax+1-1=sum(an+1,n,0,x)-sum(an,n,0,x)
ax+1-1=sum(an+1-an,n,0,x)
ax+1-1=sum(an(a-1),n,0,x)
ax+1-1=(a-1)sum(an,n,0,x)
(ax+1-1)/(a-1)=sum(an,n,0,x)
:D
Now how is this useful? Recall that ebi=cos(b)+isin(b) (if you want, I can give a clever little proof of that using the Maclaurin series of ex). Then, if we take only the imaginary part of that, we will have sin(b). This means, if we want sin(0)+sin(1)+sin(2)+...+sin(x), we can simply do this:
sum(sin(n),n,0,x)=Im(sum(eni,n,0,x))=Im((ei(x+1)-1)/(ei-1))
So reindexing, we get:
sum(sin(n),n,1,x)=Im(sum(eni,n,1,x))=Im((ei(x+1)-1)/(ei-1)-1)
sum(sin(n),n,1,x)=Im(sum(eni,n,1,x))=Im((ei(x+1)-1)/(ei-1)) (this is because "-1" is not imaginary, so it doesn't matter)
Now we look back :3
Im((ei(x+1)-1)/(ei-1))=sin(x)/2+cos(x)sum((-1)2n|B2n|/(2n)!,n,0,x)
Im((ei(x+1)-1)/(cos(x)(ei-1)))=tan(x)/2+sum((-1)2n|B2n|/(2n)!,n,0,x)
Im((ei(x+1)-1)/(cos(x)(ei-1)))=tan(x)/2+sum((-1)2n|B2n|/(2n)!,n,0,x)
And all I can get from this is that the constant "sum((-1)2n|B2n|/(2n)!,n,0,x)" changes values XD This is why I love the Bernoulli numbers when manipulating them with alternating functions.
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That's pretty interesting, I should show this to my math teacher. Also we need a prettyprint bbcode, like what Wikipedia have :P
And you deserve one (1) internet for crashing WolframAlpha.
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xeda strikes again O.o
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:)
I will analyse the Xeda resolution method since it use the Mac-Laurin Equation which is very usefull to calculate former divergent series numerically, just to fill a possible analytic continuation of the function.
OK! Let's show my full resolution! :D
Using Abel-Plana formula, the original serie is equall to the following integral:
sum(sin(x),x,0,infinity) = integral(sin(x),x,0,infinity) - integral(2*sinh(x)/(exp(2*pi*x)-1),x,0,infinity)
The first one won't make sense (is divergent) unless we apply a renormalization factor:
integral(sin(x),x,0,infinity) = lim t=0 of integral(exp(-t*x)*sin(x),x,0,infinity)
Apply integration by parts twice and solve the equation to the original integral as an unknow:
integral(exp(-t*x)*sin(x),x,0,infinity) = 1 / (1 + t^2)
Then: integral(sin(x),x,0,infinity) = lim t=0 of 1/(1+t^2) = 1 :P
The second integral should be transformed in a series respecting to the denominator, since it is a nicelly absolute convergent integral, this means:
integral(2*sinh(x)/(exp(2*pi*x)-1),x,0,infinity) = - 1* sum of integral(2*sinh(x)*exp(2*pi*x*n),x,0,infinity) from n=0 to infinity
Now the new integral is easylly solved by integration by parts twice, and just apply the same trick just as before, to get:
integral(2*sinh(x)/(exp(2*pi*x)-1),x,0,infinity) = sum of 2/(4*pi^2*n^2 - 1) from n=0 to infinity.
And all this means I coulf convert a divergent series into a new one that is cleary convergent. ;D(http://www.omnimaga.org/Themes/default/images/gpbp_arrow_up.gif)
sum(sin(x),x,0,infinity) = 1 + sum(2/(4*pi^2*n^2-1),n,0,infinity).
And I will take a break to continue the proof in the next post.
Edit: Oh! I found a little mistake...
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This sum doesn't converge, but the average value of a finite sum sum(sin(x),x,0,i) is approximately 0.91524, and 2+(1/2)*cotan(1/2) = 2.91524386... I think that 2+ doesn't really belong.
Also, solving the infinite series the easy way (multiply and cancel) gives cot(1/2)/2 without the 2+:
sin(x) = (e^ix - e^-ix)/2i
sum(sin(x),x,0,inf) = (... - e^-3i - e^-2i - e^-i + 0 + e^i + e^2i + e^3i + ...)/2i
sum(sin(x),x,0,inf)*e^(-i/2) = (... - e^(-5i/2) - e^(-3i/2) + 0 + e^(i/2) + e^(3i/2) + e^(5i/2) + ...)/2i
sum(sin(x),x,0,inf)*e^(i/2) = (... - e^(-5i/2) - e^(-3i/2) - e^(-i/2) + 0 + e^(3i/2) + e^(5i/2) + ...)/2i
sum(sin(x),x,0,inf)*(e^(-i/2) - e^(i/2)) = (e^(-i/2) + e^(i/2))/2i
sum(sin(x),x,0,inf)*-2i*sin(1/2) = cos(1/2)/i
sum(sin(x),x,0,inf) = cot(1/2)/2
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The final clue to solve this problem just needs the Euler Formula of sine:
sin(pi*x) / (pi*x) = product(1-x^2/n^2,n,1,infinity)
Apply the logarithmic derivative of the Euler Formula of sine and you get:
pi*cotan(pi*x) = 1/x + sum(2*x/(x^2-n^2),n,1,infinity)
(This equation is main key to solve the Euler Problem to establish the semi-rational values of Riemann Zeta Function in terms of Bernoulli Numbers ::) )
This means if I split the series (since n=0 term is absent, so just expand it):
sum(2/(4*pi^2*n^2-1),n,0,infinity) = -2 + sum(2/(4*pi^2*n^2-1),n,1,infinity)
So is just the cotangent series, when it will make clear when split the common terms:
2/(4*pi^2*n^2-1) = 2 / (4 * pi^2 (n^2 - 1/(4 * pi^2)), this means x= -1/2*pi
And so, will get:
-pi*cotan(-1/2) = -(2*pi) - sum (2/(2*pi*(1/(4*x^2)-n^2)
To fix the discripancy, then just divide all terms by (2*pi), and we proofed that:
sum(2/(4*pi^2*n^2-1),n,1,infinity) = 1 - (1/2)*cotan(1/2), since the cotangent is a even function.
And finally at least, we obtain the final formula:
sum(sin(x),x,0,infinity) = 1 +1 - 2 + (1/2)*cotan(1/2) = (1/2)*cotan(1/2)
;D(http://www.omnimaga.org/Themes/default/images/gpbp_arrow_up.gif)
Edit: Little error detected!
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That is pretty cool o.o
@Goplat: That is brilliant o.o
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/me is mindblown
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After some review, the sum of sin(x) from x=0 to infinity is really (1/2)*cotan(1/2).