Find exponential generating function for following sequence:

$s_{n} = \sin{nt}$

the answer should be in terms of trigonometric functions. The exponential generating function is defined as:

$S(x)= \sum_{n\geq 0} s_{n} \frac{x^n}{n!} $


You can determine a function that simultaneously "generates" both $sin(nt)$ and $cos(nt)$. If you assume $x$ to be real, then you're looking for the imaginary part of the coefficient of $\frac{x^n}{n!}$ in the following expression.

$$ \sum_{n=0}^\infty \left(cos(nt) + i.sin(nt)\right)\frac{x^n}{n!} \\ = \sum_{n=0}^\infty e^{int}\frac{x^n}{n!} \\ = \sum_{n=0}^\infty \frac{e^{n(ln(x) + it)}}{n!} = e^{e^{ln(x) + it}} \\ = e^{x.e^{it}} $$ Is this what you are looking for?

  • $\begingroup$ Thank you for replying. It looks make sense, but I cannot understand why you says "if you assume $x$ to be real, the you're looking for the imaginary part of the coefficient...". And how can I get the generating function of $S_{n} = \sin{nt}$ from this expression? $\endgroup$ – cinvro Feb 17 '14 at 5:47
  • $\begingroup$ Another thing I doesn't understand is why $\sum_{n=0}^\infty \frac{e^{n(ln(x) + it)}}{n!} = e^{e^{ln(x) + it}}$ ? $\endgroup$ – cinvro Feb 17 '14 at 6:32
  • $\begingroup$ If $x$ and $t$ are real, then the generating function for $cos(nt)$ and $sin(nt)$ correspond to the real and imaginary parts of the final answer. To answer your second question, $$ \sum_{n=0}^\infty \frac{e^{n(ln(x) + it)}}{n!} \\ = \sum_{n=0}^\infty \frac{\left[e^{ln(x)+it}\right]^n}{n!} \\ = e^\left[e^{ln(x)+it}\right] $$ $\endgroup$ – sid Feb 17 '14 at 18:54
  • $\begingroup$ For the generating function you want, write out the final answer in the form $a + ib$, then $b$ will be the one you're looking for. $$ e^{x.e^{it}} = e^{x\left[cos(t) + i.sin(t)\right]} \\ e^{x.cos(t)}\left[cos(x.sin(t)) + i.sin(x.sin(t))\right] $$ So your generating function would be $e^{x.cos(t)}sin(x.sin(t))$. $\endgroup$ – sid Feb 17 '14 at 19:14
  • $\begingroup$ Great answer! Thank you very much! $\endgroup$ – cinvro Feb 18 '14 at 0:02

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