1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Series Sum

Tags:
  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove that,
    $$\sum _{n=1,3,5...} \frac{1}{n} e^{-nx} \sin{ny} = \frac{1}{2}\tan^{-1} (\frac{\sin{y}}{\sinh{x}})$$

    2. Relevant equations

    $$\tan^{-1}{x} = x - \frac{x^3}{3} +\frac{x^5}{5} - ... $$

    3. The attempt at a solution

    $$\sum _{n=1,3,5...} \frac{1}{n} e^{-nx} \sin{ny}$$
    $$= \sum _{n=1,3,5...} \frac{1}{n} e^{-nx} Im(e^{i ny})$$
    $$= \sum _{n=1,3,5...} \frac{ Im(e^{(-x+iy)n})}{n} $$
    $$=Im( \sum _{n=1,3,5...} \frac{z^n}{n}) $$
    where ##z=e^{-x+iy}##.
    I am stuck here. Any help will be appreciated.
     
  2. jcsd
  3. Sep 15, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Looking at ##\sum_n \frac {(iz)^n}{n}## looks promising. Note the sum over all n >=1.
     
  4. Sep 17, 2016 #3
    • Member warned that giving full solutions is not allowed at this site
    Preliminaries... I'm going to use 2 identities.
    Identity 1: $$\displaystyle \sum_{n\,\textrm{odd}}\frac{x^n}{n}=\frac 1 2 \ln \left(\frac {1+x} {1-x} \right)$$
    Proof: From the Maclaurin expansion of ##\ln (1+x)## and ##\ln(1-x)##, we get,
    $$\ln (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots$$
    $$\ln (1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots$$
    Subtracting the 2nd expression from the 1st, we get,
    $$\ln (1+x)-\ln (1-x)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\ldots\right)=2\sum_{n\,\textrm{odd}}\frac{x^n}{n}$$
    $$ \therefore \sum_{n\,\textrm{odd}}{\frac{x^n}{n}} = \frac{1}{2} \ln \left( \frac{1+x}{1-x} \right) $$


    Identity 2: $$\displaystyle \tan^{-1}x=\frac{i}{2}\ln\left(\frac{i+z}{i-z}\right)$$
    Proof: Let, ## \tan^{-1}x=w ##. So,
    $$ x=\tan w=\frac{\sin w}{\cos w}= \frac{\frac{e^{iw}-e^{-iw}}{2i}}{\frac{e^{iw}+e^{-iw}}{2}}= \frac{1}{i}\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} $$
    \begin{align}
    & \therefore x=\frac{1}{i}\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \nonumber \\
    & \Rightarrow x=-i\cdot\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{-iw}} \nonumber\\
    & \Rightarrow x(e^{iw}+e^{-iw})=-i(e^{iw}-e^{-iw}) \nonumber \\
    & \Rightarrow e^{iw}(i+x)=e^{-iw}(i-x) \nonumber \\
    & \Rightarrow e^{i2w}=\frac{i-x}{i+x} \nonumber \\
    & \Rightarrow i2w=\ln\left(\frac{i-x}{i+x}\right) \nonumber \\
    & \Rightarrow w=\frac{1}{2i}\cdot\ln\left(\frac{i-x}{i+x}\right)=-\frac{i}{2}\ln\left(\frac{i-x}{i+x}\right)=\frac{i}{2}\ln\left(\frac{i+x}{i-x}\right) \nonumber \\
    & \therefore \tan^{-1}x=\frac{i}{2}\ln\left(\frac{i+z}{i-z}\right) \nonumber \\
    \end{align}

    Now, for the proof of the problem...
    \begin{align}
    \sum_{n\, \textrm{odd}} \frac{1}{n} e^{-nx} \sin{ny} & = \sum_{n\, \textrm{odd}} \frac{1}{n} e^{-nx}\cdot \frac{e^{iny}-e^{-iny}}{2i} \nonumber\\
    & = \sum_{n\, \textrm{odd}} \frac 1 n\cdot \frac{e^{-nx+iny}-e^{-nx-iny}}{2i} \nonumber\\
    &= \frac{1}{2i} \left[\sum_{n\, \textrm{odd}} \frac{(e^{-x+iy})^n}{n} -\sum_{n\, \textrm{odd}} \frac{(e^{-x-iy})^n}{n}\right] \nonumber\\
    &= \frac{1}{2i} \left[\sum_{n\, \textrm{odd}} \frac{z^n}{n} - \sum_{n\, \textrm{odd}} \frac{{\bar z}^n}{n}\right]\qquad (z=e^{-x+iy},\; \bar z = e^{-x-iy}) \nonumber\\
    &= \frac{1}{2i}\left[\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)-\frac{1}{2}\ln \left(\frac{1+\bar z}{1-\bar z}\right)\right]\quad \textrm{(Identity 1)} \nonumber\\
    &= \frac 1 2 \left[\frac{1}{2i}\ln \left(\frac{1+z}{1-z}\right)-\frac{1}{2i}\ln \left(\frac{1+\bar z}{1-\bar z}\right)\right] \nonumber\\
    &= \frac 1 2 \left[-\frac{i}{2}\ln \left(\frac{i+iz}{i-iz}\right)+\frac{i}{2}\ln \left(\frac{i+i\bar z}{i-i\bar z}\right)\right] \nonumber\\
    &= \frac 1 2 \left[-\tan^{-1}iz+\tan^{-1}i\bar z\right] \qquad \textrm{(Identity 2)} \nonumber\\
    &= \frac 1 2 \left[\tan^{-1}i\bar z-\tan^{-1}iz\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{i\bar z-iz}{1+i^2z\bar z}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x-iy}-ie^{-x+iy}}{1-e^{-x-iy}\cdot e^{-x+iy}}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x}(e^{-iy}-e^{iy})}{1-e^{-2x}}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{ie^{-x}(-2i\sin y)}{1-e^{-2x}}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{2\sin y}{e^x-e^{-x}}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left[\frac{\sin y}{\frac{e^x-e^{-x}}{2}}\right] \nonumber\\
    &= \frac 1 2 \tan^{-1}\left(\frac{\sin y}{\sinh y}\right) \nonumber\\
    \end{align}

    And... we're done. :D :D :D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Series Sum
  1. Sum series (Replies: 1)

  2. Sum of Series (Replies: 10)

  3. Sum of a series (Replies: 11)

  4. Sum of a series (Replies: 15)

  5. The sum of a series (Replies: 4)

Loading...