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Hello I have some problems concerning Taylor series.

Given the function $$f(x)=e^{\sin{x}} $$

I concluded that the Taylor series expansion would be

$$f(x) = \sum^\infty_{n=0}\frac{1}{n!}f^{(n)}(x)(x-x_0)^n$$ But Wolfram wrote a much simpler form

$$f(x)=\sum^\infty_{n=0}\frac{\sin^k{x}}{k!}$$

My question is: how come? Secondly, could somebody explain to me how to get the second sum's radius of convergence?

Thank you very much for your time!

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  • $\begingroup$ You might need to revise your conception of what a Taylor series is saying: the factor $f^n(x)$ should probably read $f^{(n)}(x_0)$ (thus, two mistakes). $\endgroup$
    – Did
    Mar 8, 2014 at 15:42
  • $\begingroup$ @Did of course you're right - I didnt write the parentheses and followed my previous statement mindlessly - edited $\endgroup$
    – Simon
    Mar 8, 2014 at 15:45
  • $\begingroup$ Worlfram's series, while equal to your function $f$ (by series composition), is not a power series, so it isn't the Taylor series of anything, doesn't have a radius, etc. $\endgroup$
    – Andrew D. Hwang
    Mar 8, 2014 at 15:49

2 Answers 2

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Take in to major account Did's comment. If you properly apply the definition and build your series at $x=0$, you will obtain $$1+x+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^5}{15}-\frac{x^6}{240}+\frac{x^7}{90}+\frac{3 1 x^8}{5760}+O\left(x^9\right)$$

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  • $\begingroup$ OK, got it, thanks - but calculating the derivatives of e^sinx just gives me longer and longer formulae - how can I simplify it and get the radius of convergence? $\endgroup$
    – Simon
    Mar 8, 2014 at 15:55
  • $\begingroup$ Did you just write in $\LaTeX$? Congrats! $\endgroup$
    – Git Gud
    Mar 9, 2014 at 15:32
  • $\begingroup$ @GitGud. Most of it is done by my wife ! The remaining is done when I can generate the TeX format ! Cheers. $\endgroup$
    – Claude Leibovici
    Mar 9, 2014 at 15:45
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For range of convergence, you may want to check out my working at Range of convergence for Taylor's series (about 0) for e^(sin x). Not sure if I did anything wrong though.

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