Equality of perturbed Ramanujan's sum and -1/12
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Consider the series $sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)$. If one lets $varepsilon$ be small enough (for example 0.0001) and $n$ large enough (for example 1.000.000), one sees by numerical computations that this sum will converge to $-frac{1}{12}$. See also https://en.wikiversity.org/wiki/MATLAB/Divergent_series_investigations or test it by yourself using any programming language.
Besides this rather interesting fact, my question is now how to solve $varepsilon$ in terms of $n$ in the following equation $$sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)=-frac{1}{12}.$$ If one solves this, one can for example obtain sufficient conditions in order to apply zeta function regularization in areas where is made a lot of use of perturbed series (such as physics).
I tried to integrate both sides of the expression (w.r.t. $varepsilon$) but arrived at no interesting relationships so far. Maybe one does have heard of this expression, knows more about it or even know how to solve it for $varepsilon$?
See also my slightly related question Closed form expression for (periodic) generalized harmonic numbers? which might provide clearer insights.
calculus sequences-and-series complex-analysis problem-solving divergent-series
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
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add a comment |
$begingroup$
Consider the series $sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)$. If one lets $varepsilon$ be small enough (for example 0.0001) and $n$ large enough (for example 1.000.000), one sees by numerical computations that this sum will converge to $-frac{1}{12}$. See also https://en.wikiversity.org/wiki/MATLAB/Divergent_series_investigations or test it by yourself using any programming language.
Besides this rather interesting fact, my question is now how to solve $varepsilon$ in terms of $n$ in the following equation $$sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)=-frac{1}{12}.$$ If one solves this, one can for example obtain sufficient conditions in order to apply zeta function regularization in areas where is made a lot of use of perturbed series (such as physics).
I tried to integrate both sides of the expression (w.r.t. $varepsilon$) but arrived at no interesting relationships so far. Maybe one does have heard of this expression, knows more about it or even know how to solve it for $varepsilon$?
See also my slightly related question Closed form expression for (periodic) generalized harmonic numbers? which might provide clearer insights.
calculus sequences-and-series complex-analysis problem-solving divergent-series
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
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1
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Today's Ramanujan's bday btw ;)
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– Ankit Kumar
Dec 22 '18 at 16:58
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Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
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$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
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– reuns
Dec 22 '18 at 23:21
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03
add a comment |
$begingroup$
Consider the series $sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)$. If one lets $varepsilon$ be small enough (for example 0.0001) and $n$ large enough (for example 1.000.000), one sees by numerical computations that this sum will converge to $-frac{1}{12}$. See also https://en.wikiversity.org/wiki/MATLAB/Divergent_series_investigations or test it by yourself using any programming language.
Besides this rather interesting fact, my question is now how to solve $varepsilon$ in terms of $n$ in the following equation $$sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)=-frac{1}{12}.$$ If one solves this, one can for example obtain sufficient conditions in order to apply zeta function regularization in areas where is made a lot of use of perturbed series (such as physics).
I tried to integrate both sides of the expression (w.r.t. $varepsilon$) but arrived at no interesting relationships so far. Maybe one does have heard of this expression, knows more about it or even know how to solve it for $varepsilon$?
See also my slightly related question Closed form expression for (periodic) generalized harmonic numbers? which might provide clearer insights.
calculus sequences-and-series complex-analysis problem-solving divergent-series
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
$endgroup$
Consider the series $sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)$. If one lets $varepsilon$ be small enough (for example 0.0001) and $n$ large enough (for example 1.000.000), one sees by numerical computations that this sum will converge to $-frac{1}{12}$. See also https://en.wikiversity.org/wiki/MATLAB/Divergent_series_investigations or test it by yourself using any programming language.
Besides this rather interesting fact, my question is now how to solve $varepsilon$ in terms of $n$ in the following equation $$sum_{k=1}^n k e^{-varepsilon k}cos(varepsilon k)=-frac{1}{12}.$$ If one solves this, one can for example obtain sufficient conditions in order to apply zeta function regularization in areas where is made a lot of use of perturbed series (such as physics).
I tried to integrate both sides of the expression (w.r.t. $varepsilon$) but arrived at no interesting relationships so far. Maybe one does have heard of this expression, knows more about it or even know how to solve it for $varepsilon$?
See also my slightly related question Closed form expression for (periodic) generalized harmonic numbers? which might provide clearer insights.
calculus sequences-and-series complex-analysis problem-solving divergent-series
calculus sequences-and-series complex-analysis problem-solving divergent-series
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
edited Dec 22 '18 at 16:50
Frozenharp
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
asked Dec 22 '18 at 16:42
FrozenharpFrozenharp
244
244
1
$begingroup$
Today's Ramanujan's bday btw ;)
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– Ankit Kumar
Dec 22 '18 at 16:58
$begingroup$
Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
$begingroup$
$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
$endgroup$
– reuns
Dec 22 '18 at 23:21
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03
add a comment |
1
$begingroup$
Today's Ramanujan's bday btw ;)
$endgroup$
– Ankit Kumar
Dec 22 '18 at 16:58
$begingroup$
Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
$begingroup$
$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
$endgroup$
– reuns
Dec 22 '18 at 23:21
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03
1
1
$begingroup$
Today's Ramanujan's bday btw ;)
$endgroup$
– Ankit Kumar
Dec 22 '18 at 16:58
$begingroup$
Today's Ramanujan's bday btw ;)
$endgroup$
– Ankit Kumar
Dec 22 '18 at 16:58
$begingroup$
Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
$begingroup$
Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
$begingroup$
$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
$endgroup$
– reuns
Dec 22 '18 at 23:21
$begingroup$
$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
$endgroup$
– reuns
Dec 22 '18 at 23:21
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03
add a comment |
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1
$begingroup$
Today's Ramanujan's bday btw ;)
$endgroup$
– Ankit Kumar
Dec 22 '18 at 16:58
$begingroup$
Hey yeah, I see! Nice coincidence :D!
$endgroup$
– Frozenharp
Dec 22 '18 at 17:46
$begingroup$
$sum_{k=0}^n k e^{-x k}2cos(x k) =frac{d}{dx} sum_{k=0}^n frac{1}{1+i}e^{-x (1+i) k}+frac{1}{1-i}e^{-x (1-i) k}) = frac{d}{dx}( frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}}) = ...$
$endgroup$
– reuns
Dec 22 '18 at 23:21
$begingroup$
I see that actually $frac{1}{6}=-sum_{k=0}^n 2ke^{-xk}cos(xk) = frac{d}{dx}(frac{1}{1+i}frac{1-e^{-x(1+i)(n+1)}}{1-e^{-x(1+i)}}+frac{1}{1-i}frac{1-e^{-x(1-i)(n+1)}}{1-e^{-x(1-i)}})=-(frac{(n+1)e^{x(-i-1)(n+1)}(1-e^{x(-i-1)})-e^{x(-i-1)}(1-e^{x(-i-1)(n+1)})}{(1-e^{x(-i-1)})^2}+frac{(n+1)e^{x(i-1)(n+1)}(1-e^{x(i-1)})-e^{x(i-1)}(1-e^{x(i-1)(n+1)})}{(1-e^{x(i-1)})^2})$. I don't know how to reduce it to an expression in which we can solve it for $x$ however...
$endgroup$
– Frozenharp
Dec 23 '18 at 15:03