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A Math Challenge: Proof that sum(sin(x),x,0,infinity)=(1/2)*cotan(1/2)
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Topic: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=(1/2)*cotan(1/2) (Read 9644 times)
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Nosferatu Arucard 1983
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A Math Challenge: Proof that sum(sin(x),x,0,infinity)=(1/2)*cotan(1/2)
«
on:
November 08, 2012, 10:00:18 pm »
Since the infinity sum of sin(x) from x=0 to infinity (sin(0) + sin(1)+ ...) don't converge, we can use the AbelPlana 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.
«
Last Edit: November 09, 2012, 01:23:26 pm by Nosferatu Arucard 1983
»
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aeTIos
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #1 on:
November 09, 2012, 04:00:31 am »
Welcome to omni
I won't be able to proof that though.
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #2 on:
November 09, 2012, 05:53:44 am »
Welcome to Omnimaga indeed
And I am a little confused by your notation. When you say Sum, do you mean integral?
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aeTIos
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #3 on:
November 09, 2012, 05:54:58 am »
no I think it means the iterative sum >sigma notation?
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #4 on:
November 09, 2012, 07:47:06 am »
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) :
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Nosferatu Arucard 1983
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=(1/2)*cotan(1/2)
«
Reply #5 on:
November 09, 2012, 07:56:40 am »
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 AbelPlana formula was derived in Theorem 7.3 of my book
But I will translated to you...
1st Step:
Just apply f(t)=sin(t) to the AbelPlana 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 AbelPlana by steps.
«
Last Edit: November 09, 2012, 08:22:06 am by Nosferatu Arucard 1983
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Xeda112358
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #6 on:
November 09, 2012, 08:30:47 am »
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 EulerMaclaurin 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 EulerMaclaurin formula to show that:
sum(sin(n),n,0,x)=sum(sin(n),n,1,x)=integral(sin(x))+sin(x)/2+cos(x)/12cos(x)/(30*4!)+cos(x)/(42*6!)cos(x)(30*8!)+...
=B
_{0}
cos(x)+B
_{1}
sin(x)+B
_{2}
cos(x)/2!+B
_{4}
cos(x)/4!+B
_{6}
cos(x)/6!+B
_{8}
cos(x)/8!+B
_{10}
cos(x)/10!...
(where B
_{n}
is the Bernoulli numbers)
Then we have that sum(sin(n),n,1,x)=B
_{1}
sin(x)+cos(x)(B
_{0}
+B
_{2}
/2!+B
_{4}
/4!+B
_{6}
/6!+B
_{8}
/8!+...)=B
_{1}
sin(x)cos(x)+cos(x)sum(B
_{2n}
/(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}
B
_{2n}
/(2n)!,n,0,x)
Note that this is a constant: sum((1)
^{2n}
B
_{2n}
/(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(a
^{n}
,n,0,x)=a
^{0}
+a
^{1}
+a
^{2}
+a
^{3}
+...+a
^{x}
. If we add a
^{x+1}
, we will have the following:
a
^{x+1}
+sum(a
^{n}
,n,0,x)=sum(a
^{n}
,n,0,
x+1
)
Reindexing the right side a few times:
a
^{x+1}
+sum(a
^{n}
,n,0,x)=a
^{0}
+sum(a
^{n}
,n,1,
x+1
)
a
^{x+1}
+sum(a
^{n}
,n,0,x)=1+sum(a
^{n+1}
,n,0,
x
)
Now the sums have the same index, so we can combine them:
a
^{x+1}
1=sum(a
^{n+1}
,n,0,x)sum(a
^{n}
,n,0,x)
a
^{x+1}
1=sum(a
^{n+1}
a
^{n}
,n,0,x)
a
^{x+1}
1=sum(a
^{n}
(a1),n,0,x)
a
^{x+1}
1=(a1)sum(a
^{n}
,n,0,x)
(a
^{x+1}
1)/(a1)=sum(a
^{n}
,n,0,x)
Now how is this useful? Recall that e
^{bi}
=cos(b)+isin(b) (if you want, I can give a clever little proof of that using the Maclaurin series of e
^{x}
). 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(e
^{ni}
,n,0,x))=Im((e
^{i(x+1)}
1)/(e
^{i}
1))
So reindexing, we get:
sum(sin(n),n,1,x)=Im(sum(e
^{ni}
,n,1,x))=Im((e
^{i(x+1)}
1)/(e
^{i}
1)1)
sum(sin(n),n,1,x)=Im(sum(e
^{ni}
,n,1,x))=Im((e
^{i(x+1)}
1)/(e
^{i}
1)) (this is because "1" is not imaginary, so it doesn't matter)
Now we look back
Im((e
^{i(x+1)}
1)/(e
^{i}
1))=sin(x)/2+cos(x)sum((1)
^{2n}
B
_{2n}
/(2n)!,n,0,x)
Im((e
^{i(x+1)}
1)/(cos(x)(e
^{i}
1)))=tan(x)/2+sum((1)
^{2n}
B
_{2n}
/(2n)!,n,0,x)
Im((e
^{i(x+1)}
1)/(cos(x)(e
^{i}
1)))=tan(x)/2+sum((1)
^{2n}
B
_{2n}
/(2n)!,n,0,x)
And all I can get from this is that the constant "sum((1)
^{2n}
B
_{2n}
/(2n)!,n,0,x)" changes values
This is why I love the Bernoulli numbers when manipulating them with alternating functions.
«
Last Edit: November 09, 2012, 08:35:06 am by Xeda112358
»
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #7 on:
November 09, 2012, 08:34:43 am »
That's pretty interesting, I should show this to my math teacher. Also we need a prettyprint bbcode, like what Wikipedia have
And you deserve one (1) internet for crashing WolframAlpha.
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #8 on:
November 09, 2012, 10:52:19 am »
xeda strikes again O.o
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Nosferatu Arucard 1983
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #9 on:
November 09, 2012, 12:06:46 pm »
I will analyse the Xeda resolution method since it use the MacLaurin 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!
Using AbelPlana 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
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.
sum(sin(x),x,0,infinity) = 1 + sum(2/(4*pi^2*n^21),n,0,infinity).
And I will take a break to continue the proof in the next post.
Edit: Oh! I found a little mistake...
«
Last Edit: November 09, 2012, 01:24:55 pm by Nosferatu Arucard 1983
»
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Goplat
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #10 on:
November 09, 2012, 12:29:33 pm »
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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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #11 on:
November 09, 2012, 12:51:30 pm »
The final clue to solve this problem just needs the Euler Formula of sine:
sin(pi*x) / (pi*x) = product(1x^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^2n^2),n,1,infinity)
(This equation is main key to solve the Euler Problem to establish the semirational 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^21),n,0,infinity) = 2 + sum(2/(4*pi^2*n^21),n,1,infinity)
So is just the cotangent series, when it will make clear when split the common terms:
2/(4*pi^2*n^21) = 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^21),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)
Edit: Little error detected!
«
Last Edit: November 09, 2012, 01:25:51 pm by Nosferatu Arucard 1983
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Xeda112358
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #12 on:
November 09, 2012, 12:59:24 pm »
That is pretty cool o.o
@Goplat: That is brilliant o.o
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=(1/2)*cotan(1/2)
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Reply #13 on:
November 09, 2012, 01:29:21 pm »
* pimathbrainiac is mindblown
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Re: A Math Challenge: Proof that sum(sin(x),x,0,infinity)=2+(1/2)*cotan(1/2)
«
Reply #14 on:
November 09, 2012, 01:31:20 pm »
After some review, the sum of sin(x) from x=0 to infinity is really (1/2)*cotan(1/2).
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