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So I am asked to find the Taylor polynomial of degree 2n-1 for $f(x)=\sinh(x)$ centered at 0.

I have the general formula for a Taylor series but I'm not really sure how to proceed. I can't really take the 2n-1'st derivative or anything.

Can someone suggest how I start this problem?

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  • $\begingroup$ Taking the $2n-1$ first derivative is not the good way. Use the classical expansion $sinh(x)=x+x^3/3!+x^5/5!+\cdots+x^{2n-1}/(2n-1)!$. The proof of it being what @Claude Leibovici indicates. $\endgroup$
    – Jean Marie
    Commented Nov 6, 2016 at 8:03
  • $\begingroup$ Isn't it 2n+1? Not 2n-1? $\endgroup$ Commented Nov 6, 2016 at 8:17
  • $\begingroup$ You are asked a degree $2n-1$ polynomial... $\endgroup$
    – Jean Marie
    Commented Nov 6, 2016 at 8:31
  • $\begingroup$ Right. I forgot the exponent is the degree of the polynomial. Thanks. That means I can't get a closed form expression, I just have to use a $ ... x^{2n-1}/(2n-1)!$ $\endgroup$ Commented Nov 6, 2016 at 8:33
  • $\begingroup$ Thanks. I got it now. Just had to wrap my head around the fact the degree wasn't an explicit number which threw me off. $\endgroup$ Commented Nov 6, 2016 at 8:35

1 Answer 1

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Hint

Why not to start from $$\sinh(x)=\frac{e^x-e^{-x}}2$$

Edit

I suppose that your professor would not enjoy the following lazy way : starting from the trigonometric identity $$\sin(ix)=i \sinh(x)\implies \sinh(x)=-i \sin(ix)$$ then, using Taylor expansion of the sine,

$$\sinh(x) =- i\sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} (ix)^{2n+1}=- \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}i^{2n+2} x^{2n+1}= \sum_{n=0}^\infty \frac{ x^{2n+1}}{(2n+1)!}$$

If he/she has a sense of humour, it could be interesting to ask him/her what would have been his/her reaction facing such an approach.

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  • $\begingroup$ Yeah I know I can do that but how does the 2n-1 come into play? I feel like I am missing something very trivial. $\endgroup$ Commented Nov 6, 2016 at 8:14
  • $\begingroup$ In any case, here is the derivation: $$\sinh x = \frac{1}{2} ( e^{x} - e^ {-x}) = \frac{1}{2}(\sum_{n=0}^\infty \frac{x^n}{n!} - \sum_{n=0}^\infty \frac{(-x)^n}{n!}) = \frac{1}{2}\sum_{n=0}^\infty \frac{(1-(-1)^n ) x^n}{n!} = \sum_{n=0}^\infty \frac{ x^{2n+1}}{(2n+1)!}$$ Where do I do from there? $\endgroup$ Commented Nov 6, 2016 at 8:21
  • $\begingroup$ Stop the summation at $2n-1$ and use the big $O$ notation. $\endgroup$ Commented Nov 6, 2016 at 8:25
  • $\begingroup$ Ohh I see what you're saying. There is no possible way to get a closed form polynomial is there? I just have to use the O notation. $\endgroup$ Commented Nov 6, 2016 at 8:30
  • $\begingroup$ @SubhashisChakraborty. This is correct. Cheers. $\endgroup$ Commented Nov 6, 2016 at 8:44

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