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I need to find the generating function for $a_n$ where:

$$a_{2k}=\dfrac{(-1)^k}{(2k)!}$$

$$a_{2k+1}=\dfrac{1}{(2k+1)}$$

$$k \geq 0$$

The solution is:

$$f(x)=\cos(x)+\operatorname{atanh}(x)$$

So far I've solved problems where everything began with a recurrence relation that I had to multiply with something to get sums (and substitute with $f(x)$).

I'm not sure how to even express $a_n$ here. I'd appreciate any help.

Nash J.
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frostpad
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  • When writing out the Taylor series expansion of $f(x)$ these are exactly the coefficients you obtain. In even degree you get those for the cosine and in odd degree those for the hyperbolic function. – NDewolf Aug 29 '18 at 21:34
  • $f(x)$ shown above is the solution to the problem. I'm not sure how I could expand it if I didn't find it yet (my apologies if I misunderstood your comment). Perhaps I should know that and recognize it? – frostpad Aug 29 '18 at 22:07
  • I indeed think you are supposed to recognize those coefficients. The cosine is well-known, but i must admit that knowing the coefficients for the hyperbolic functions by heart is a bit far-fetched. – NDewolf Aug 30 '18 at 08:56

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