- #1
yungman
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In page 11 of http://math.arizona.edu/~zakharov/BesselFunctions.pdf, I am trying to follow the derivation using binomial theorem to get this step:
[tex](e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^k[/tex]If you read the paragraph right above this equation, it said only the terms that has the ##e^{jn\theta}## alone will not be integrated to zero. So the equation becomes:[tex](e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^k[/tex]
As it's the ##(k+1)^{th}## term.
Is it because when substitute into (53)
[tex]I_{k,n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(e^{j\theta}-e^{-j\theta})^{n+2k}e^{-jn\theta}d\theta[/tex]
Only the term that has ##e^{jn\theta}## alone can cancel out the ##e^{-jn\theta}## term in (53). Any ##e^{jn\theta}## type exponential term will result in zero after integrate from ##-\pi## to ##\pi##. Am I correct.
Thanks
[tex](e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^k[/tex]If you read the paragraph right above this equation, it said only the terms that has the ##e^{jn\theta}## alone will not be integrated to zero. So the equation becomes:[tex](e^{j\theta}-e^{-j\theta})^{n+2k}≈\frac{(n+2K)!}{k!(n+k)!}(e^{j\theta})^{n+k}(-e^{-j\theta})^k[/tex]
As it's the ##(k+1)^{th}## term.
Is it because when substitute into (53)
[tex]I_{k,n}=\frac{1}{2\pi}\int_{-\pi}^{\pi}(e^{j\theta}-e^{-j\theta})^{n+2k}e^{-jn\theta}d\theta[/tex]
Only the term that has ##e^{jn\theta}## alone can cancel out the ##e^{-jn\theta}## term in (53). Any ##e^{jn\theta}## type exponential term will result in zero after integrate from ##-\pi## to ##\pi##. Am I correct.
Thanks
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