Hyperbolic “Fourier” type infinite series for $log(sinh x)$ and $log(cosh x)$ The 2019 Stack Overflow Developer Survey Results Are InOn the zero of $f(x)=int_0^x ln(sinh(z))dz$Making use of Fourier series to evaluate an infinite sumCompute the fourier coefficients, and series for $log(sin(x))$Computing the series of log and sineProving $pi coth pi a= frac1a+ sum_n=1^inftyfrac2an^2+a^2$ using the Fourier series for $cosh ax$Fourier series, infinite seriesFourier series of: $[log(sin x)]^2$Fourier series for $cosh(x)$fourier series of $cosh(ax)$Find Fourier series of $cosh(ax)$Finding Fourier series by evaluating sum of infinite series
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Hyperbolic “Fourier” type infinite series for $log(sinh x)$ and $log(cosh x)$
The 2019 Stack Overflow Developer Survey Results Are InOn the zero of $f(x)=int_0^x ln(sinh(z))dz$Making use of Fourier series to evaluate an infinite sumCompute the fourier coefficients, and series for $log(sin(x))$Computing the series of log and sineProving $pi coth pi a= frac1a+ sum_n=1^inftyfrac2an^2+a^2$ using the Fourier series for $cosh ax$Fourier series, infinite seriesFourier series of: $[log(sin x)]^2$Fourier series for $cosh(x)$fourier series of $cosh(ax)$Find Fourier series of $cosh(ax)$Finding Fourier series by evaluating sum of infinite series
$begingroup$
Hyperbolic Fourier type infinite series for $log(sinh x)$ and $log(cosh x)$ analogous to Fourier $cos$ series for $log(sin x)$ and $log(cos x)$
$$log(sinh(x))=-frac12 (i pi )-log (2)-sum _k=1^infty fraccosh (2 k x)k tag1$$
$$log(cosh(x))=-log (2)-sum _k=1^infty frac(-1)^k cosh (2 k x)k tag2$$
These analogous formula were found using Mathematica, but I am not sure about the proof. Do I convert $cosh(2kx)$ into its exponential form and proceed that way?
sequences-and-series fourier-series
$endgroup$
add a comment |
$begingroup$
Hyperbolic Fourier type infinite series for $log(sinh x)$ and $log(cosh x)$ analogous to Fourier $cos$ series for $log(sin x)$ and $log(cos x)$
$$log(sinh(x))=-frac12 (i pi )-log (2)-sum _k=1^infty fraccosh (2 k x)k tag1$$
$$log(cosh(x))=-log (2)-sum _k=1^infty frac(-1)^k cosh (2 k x)k tag2$$
These analogous formula were found using Mathematica, but I am not sure about the proof. Do I convert $cosh(2kx)$ into its exponential form and proceed that way?
sequences-and-series fourier-series
$endgroup$
add a comment |
$begingroup$
Hyperbolic Fourier type infinite series for $log(sinh x)$ and $log(cosh x)$ analogous to Fourier $cos$ series for $log(sin x)$ and $log(cos x)$
$$log(sinh(x))=-frac12 (i pi )-log (2)-sum _k=1^infty fraccosh (2 k x)k tag1$$
$$log(cosh(x))=-log (2)-sum _k=1^infty frac(-1)^k cosh (2 k x)k tag2$$
These analogous formula were found using Mathematica, but I am not sure about the proof. Do I convert $cosh(2kx)$ into its exponential form and proceed that way?
sequences-and-series fourier-series
$endgroup$
Hyperbolic Fourier type infinite series for $log(sinh x)$ and $log(cosh x)$ analogous to Fourier $cos$ series for $log(sin x)$ and $log(cos x)$
$$log(sinh(x))=-frac12 (i pi )-log (2)-sum _k=1^infty fraccosh (2 k x)k tag1$$
$$log(cosh(x))=-log (2)-sum _k=1^infty frac(-1)^k cosh (2 k x)k tag2$$
These analogous formula were found using Mathematica, but I am not sure about the proof. Do I convert $cosh(2kx)$ into its exponential form and proceed that way?
sequences-and-series fourier-series
sequences-and-series fourier-series
edited Jan 24 at 22:35
James Arathoon
asked Jan 24 at 22:21
James ArathoonJames Arathoon
1,608423
1,608423
add a comment |
add a comment |
1 Answer
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$begingroup$
Hint:
$$sin(ix)=isinh(x)$$
and $$cos(ix)=cosh(x)$$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Hint:
$$sin(ix)=isinh(x)$$
and $$cos(ix)=cosh(x)$$
$endgroup$
add a comment |
$begingroup$
Hint:
$$sin(ix)=isinh(x)$$
and $$cos(ix)=cosh(x)$$
$endgroup$
add a comment |
$begingroup$
Hint:
$$sin(ix)=isinh(x)$$
and $$cos(ix)=cosh(x)$$
$endgroup$
Hint:
$$sin(ix)=isinh(x)$$
and $$cos(ix)=cosh(x)$$
answered Mar 23 at 16:14
clathratusclathratus
5,1141439
5,1141439
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