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Fourier series of $f(x)=intlimits_0^x lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} text dt$

Convergence of the Fourier series of $f(t)=(t-pi)chi_{left(0,2piright)}$Relationship between Fourier coefficients of $fleft(xright)$ and $f^{-1}left(xright)$What is the Fourier series of $frac1Tsum^{infty}_{m=-infty}delta(f-frac mT)$?Fourier series of $sqrt{1 - k^2 sin^2{t}}$How do you find the Fourier series of $max(0, sqrt{1 - cos{theta}})$?Ways to justify this interchange of summation and integrationShow that Fourier Series $sum_{k=-infty}^{infty}frac{-ie^{ikx}}{k}$ converges to original $f(x)$$tan(pi x)$ defined on $left[-frac{1}{2},frac{1}{2}right]$ extended of period $frac{1}{2}$ has Fourier series of $sin$?Why does this method fail for finding the Fourier series for $cosleft(frac{x}{2}right)$ on the interval $-pi lt x lt pi$?Fourier series of non-periodic function $f(x)=e^{-frac{ax}{L}}$

1

1

$begingroup$

Find the Fourier series of the function
$$f(x)=intlimits_0^x lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} text dt$$
or show that it does not exist.

The first thing I have done is to find the domain of the under-integral function (call it $g$):

$$D_g=left{ x in mathbb R | x=(2k+1)pi vee x=left(2k+5/3right)pi, k in mathbb Z right}$$

Then, I established that
$$lim_{x to (2k+1)pi} g(t) = -infty$$
$$lim_{x to (2k+5/3)pi} g(t) = +infty$$
and this could cause me trouble evaluating the integral.

An idea that comes to mind is to expand $g$ into a Maclaurin series but I don't know what to do about the $frac{1}{2}$ factor or the absolute value sign.
Also, I do not know how to test whether the function is periodic or not, but graphing the function shows that it is.

What is a good way to approach this problem?

calculus fourier-analysis fourier-series periodic-functions

share|cite|improve this question

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

$endgroup$

add a comment | 

1

1

$begingroup$

Find the Fourier series of the function
$$f(x)=intlimits_0^x lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} text dt$$
or show that it does not exist.

The first thing I have done is to find the domain of the under-integral function (call it $g$):

$$D_g=left{ x in mathbb R | x=(2k+1)pi vee x=left(2k+5/3right)pi, k in mathbb Z right}$$

Then, I established that
$$lim_{x to (2k+1)pi} g(t) = -infty$$
$$lim_{x to (2k+5/3)pi} g(t) = +infty$$
and this could cause me trouble evaluating the integral.

An idea that comes to mind is to expand $g$ into a Maclaurin series but I don't know what to do about the $frac{1}{2}$ factor or the absolute value sign.
Also, I do not know how to test whether the function is periodic or not, but graphing the function shows that it is.

What is a good way to approach this problem?

calculus fourier-analysis fourier-series periodic-functions

share|cite|improve this question

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

$endgroup$

add a comment | 

$begingroup$

Find the Fourier series of the function
$$f(x)=intlimits_0^x lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} text dt$$
or show that it does not exist.

The first thing I have done is to find the domain of the under-integral function (call it $g$):

$$D_g=left{ x in mathbb R | x=(2k+1)pi vee x=left(2k+5/3right)pi, k in mathbb Z right}$$

Then, I established that
$$lim_{x to (2k+1)pi} g(t) = -infty$$
$$lim_{x to (2k+5/3)pi} g(t) = +infty$$
and this could cause me trouble evaluating the integral.

An idea that comes to mind is to expand $g$ into a Maclaurin series but I don't know what to do about the $frac{1}{2}$ factor or the absolute value sign.
Also, I do not know how to test whether the function is periodic or not, but graphing the function shows that it is.

What is a good way to approach this problem?

calculus fourier-analysis fourier-series periodic-functions

share|cite|improve this question

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

$endgroup$

share|cite|improve this question

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

share|cite|improve this question

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

asked Mar 11 at 19:09

Haris GusicHaris Gusic

2,725423

1 Answer
1

active

oldest

votes

1

$begingroup$

$g(t)=lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} =frac{1}{2}(ln|cos(frac{t}{2}-frac{pi}{3})| - ln|cos(frac{t}{2})|)$

Since $ln|2cos(frac{t}{2})|=sum_1^{infty}(-1)^{n+1}frac{cos(nt)}{n}, -pi<t<pi$ and the function is integrable at $pi$ (though infinity there), we can integrate term by term and get a result valid on the full interval $[-pi, pi]$,

$intlimits_0^xln|2cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2}$, hence $intlimits_0^xln|cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2} -xln2$

Similarly $ln|2cos(frac{t}{2}-frac{pi}{3})| = sum_1^{infty}(-1)^{n+1}frac{cos(n(t+2frac{pi}{3}))}{n}, -pi leq t leq pi, tneq -frac{pi}{3}$ but the integral works term by term, so

$intlimits_0^xln|2cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2})$, hence $intlimits_0^xln|cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2}) - xln2$

You can put things together now and massage them expanding $sin(n(x+2frac{pi}{3})) = sin(nx)cos(2nfrac{pi}{3})+ cos(nx)sin(2nfrac{pi}{3})$ if you want a result in classical Fourier terms as the non-periodic terms cancel out.

share|cite|improve this answer

edited Mar 11 at 21:42

answered Mar 11 at 21:35

ConradConrad

93345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your help. Just one question though. How did you get the series expansion for $ln|cos(t/2)|$? Did you use $$ln|cos(t/2)| = text{Re} left(lnfrac{1+e^{it}}{2}right)$$ or did you use another method?
  $endgroup$
  – Haris Gusic
  Mar 11 at 22:05
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  the easy one is $ln|2sin(frac{t}{2})|$ which is obtained by formally integrating the Poisson conjugate kernel which is $sum{r^nsin(nx)}$ (converging a.e when $r$ goes to 1 to an integrable function, so although for $r=1$ the original expansion is true only in an Abel summability sense, integrating term by term makes it true in the usual sense) and then substituting $t$ to $t+pi$ to get the logarithm of the cosine; note the constant 2 in front though; you can do it also term by term using the well known (and not hard to prove) fact that $intlimits_0^{pi}ln(2sin(frac{t}{2}))dt=0$
  $endgroup$
  – Conrad
  Mar 11 at 22:33
  

  
  
  
  

add a comment | 

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1

$begingroup$

$g(t)=lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} =frac{1}{2}(ln|cos(frac{t}{2}-frac{pi}{3})| - ln|cos(frac{t}{2})|)$

Since $ln|2cos(frac{t}{2})|=sum_1^{infty}(-1)^{n+1}frac{cos(nt)}{n}, -pi<t<pi$ and the function is integrable at $pi$ (though infinity there), we can integrate term by term and get a result valid on the full interval $[-pi, pi]$,

$intlimits_0^xln|2cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2}$, hence $intlimits_0^xln|cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2} -xln2$

Similarly $ln|2cos(frac{t}{2}-frac{pi}{3})| = sum_1^{infty}(-1)^{n+1}frac{cos(n(t+2frac{pi}{3}))}{n}, -pi leq t leq pi, tneq -frac{pi}{3}$ but the integral works term by term, so

$intlimits_0^xln|2cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2})$, hence $intlimits_0^xln|cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2}) - xln2$

You can put things together now and massage them expanding $sin(n(x+2frac{pi}{3})) = sin(nx)cos(2nfrac{pi}{3})+ cos(nx)sin(2nfrac{pi}{3})$ if you want a result in classical Fourier terms as the non-periodic terms cancel out.

share|cite|improve this answer

edited Mar 11 at 21:42

answered Mar 11 at 21:35

ConradConrad

93345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your help. Just one question though. How did you get the series expansion for $ln|cos(t/2)|$? Did you use $$ln|cos(t/2)| = text{Re} left(lnfrac{1+e^{it}}{2}right)$$ or did you use another method?
  $endgroup$
  – Haris Gusic
  Mar 11 at 22:05
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  the easy one is $ln|2sin(frac{t}{2})|$ which is obtained by formally integrating the Poisson conjugate kernel which is $sum{r^nsin(nx)}$ (converging a.e when $r$ goes to 1 to an integrable function, so although for $r=1$ the original expansion is true only in an Abel summability sense, integrating term by term makes it true in the usual sense) and then substituting $t$ to $t+pi$ to get the logarithm of the cosine; note the constant 2 in front though; you can do it also term by term using the well known (and not hard to prove) fact that $intlimits_0^{pi}ln(2sin(frac{t}{2}))dt=0$
  $endgroup$
  – Conrad
  Mar 11 at 22:33
  

  
  
  
  

add a comment | 

1

$begingroup$

$g(t)=lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} =frac{1}{2}(ln|cos(frac{t}{2}-frac{pi}{3})| - ln|cos(frac{t}{2})|)$

Since $ln|2cos(frac{t}{2})|=sum_1^{infty}(-1)^{n+1}frac{cos(nt)}{n}, -pi<t<pi$ and the function is integrable at $pi$ (though infinity there), we can integrate term by term and get a result valid on the full interval $[-pi, pi]$,

$intlimits_0^xln|2cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2}$, hence $intlimits_0^xln|cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2} -xln2$

Similarly $ln|2cos(frac{t}{2}-frac{pi}{3})| = sum_1^{infty}(-1)^{n+1}frac{cos(n(t+2frac{pi}{3}))}{n}, -pi leq t leq pi, tneq -frac{pi}{3}$ but the integral works term by term, so

$intlimits_0^xln|2cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2})$, hence $intlimits_0^xln|cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2}) - xln2$

You can put things together now and massage them expanding $sin(n(x+2frac{pi}{3})) = sin(nx)cos(2nfrac{pi}{3})+ cos(nx)sin(2nfrac{pi}{3})$ if you want a result in classical Fourier terms as the non-periodic terms cancel out.

share|cite|improve this answer

edited Mar 11 at 21:42

answered Mar 11 at 21:35

ConradConrad

93345

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much for your help. Just one question though. How did you get the series expansion for $ln|cos(t/2)|$? Did you use $$ln|cos(t/2)| = text{Re} left(lnfrac{1+e^{it}}{2}right)$$ or did you use another method?
  $endgroup$
  – Haris Gusic
  Mar 11 at 22:05
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  the easy one is $ln|2sin(frac{t}{2})|$ which is obtained by formally integrating the Poisson conjugate kernel which is $sum{r^nsin(nx)}$ (converging a.e when $r$ goes to 1 to an integrable function, so although for $r=1$ the original expansion is true only in an Abel summability sense, integrating term by term makes it true in the usual sense) and then substituting $t$ to $t+pi$ to get the logarithm of the cosine; note the constant 2 in front though; you can do it also term by term using the well known (and not hard to prove) fact that $intlimits_0^{pi}ln(2sin(frac{t}{2}))dt=0$
  $endgroup$
  – Conrad
  Mar 11 at 22:33
  

  
  
  
  

add a comment | 

$begingroup$

$g(t)=lnsqrt{frac{1}{2}left| 1+sqrt3 tanfrac{t}{2} right|} =frac{1}{2}(ln|cos(frac{t}{2}-frac{pi}{3})| - ln|cos(frac{t}{2})|)$

Since $ln|2cos(frac{t}{2})|=sum_1^{infty}(-1)^{n+1}frac{cos(nt)}{n}, -pi<t<pi$ and the function is integrable at $pi$ (though infinity there), we can integrate term by term and get a result valid on the full interval $[-pi, pi]$,

$intlimits_0^xln|2cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2}$, hence $intlimits_0^xln|cos(frac{t}{2})|dt=sum_1^{infty}(-1)^{n+1}frac{sin(nx)}{n^2} -xln2$

Similarly $ln|2cos(frac{t}{2}-frac{pi}{3})| = sum_1^{infty}(-1)^{n+1}frac{cos(n(t+2frac{pi}{3}))}{n}, -pi leq t leq pi, tneq -frac{pi}{3}$ but the integral works term by term, so

$intlimits_0^xln|2cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2})$, hence $intlimits_0^xln|cos(frac{t}{2}-frac{pi}{3})|dt=sum_1^{infty}(-1)^{n+1}(frac{sin(n(x+2frac{pi}{3}))}{n^2} -frac{sin(2nfrac{pi}{3})}{n^2}) - xln2$

You can put things together now and massage them expanding $sin(n(x+2frac{pi}{3})) = sin(nx)cos(2nfrac{pi}{3})+ cos(nx)sin(2nfrac{pi}{3})$ if you want a result in classical Fourier terms as the non-periodic terms cancel out.

share|cite|improve this answer

edited Mar 11 at 21:42

answered Mar 11 at 21:35

ConradConrad

93345

$endgroup$

share|cite|improve this answer

edited Mar 11 at 21:42

answered Mar 11 at 21:35

ConradConrad

93345

answered Mar 11 at 21:35

ConradConrad

93345

answered Mar 11 at 21:35

ConradConrad

93345