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How to derive the taylor series of $e^{\sin x}$, up to $x^5$?

i just don't know how to get the answer $$f(x) = 1 + x + \frac{x^2}{2} - \frac{x^4}{8} -\frac{x^5}{15}$$

really need some help. Thanks

Michael Hardy
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user139519
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  • If you lack a better idea, why not just compute the first 5 derivatives of $f(x) = e^{\sin x}$, and use that the taylor series around $c$ (by definition!) is $\sum_{n=0}^\infty (x-c)^n\frac{f^{(n)}(c)}{n!}$. Note that $f^{(n)}$ is the $n$-th derivative of $f$ here. – fgp Apr 01 '14 at 15:24

3 Answers3

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$$\sin x \sim x - \frac{x^3}{6} + \frac{x^5}{120}$$

$$e^x \sim 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \frac{x^5}{120}$$

so $$e^{\sin x} \sim 1 + \sin x + \frac{\sin ^2 x}{2} + \frac{\sin ^3 x}{6} + \frac{\sin ^ 4 x}{24} + \frac{\sin ^5x}{120} $$

Now substitute the expansion of $\sin x$, and you should get to the result

(remember to eliminate all those terms that have a degree higher than 5! :-) )

Ant
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Let $f(x) = e^{\sin x}$.

\begin{equation*} \begin{split} f'(x) &= e^{\sin x} \cos x = f(x) \cos x\\ f''(x) &= f'(x) \cos x - f(x) \sin x \\ f'''(x) &= f''(x) \cos x - f'(x)\sin x - (f(x)\cos x + f'(x) \sin x)\\ &= f''(x) \cos x - 2f'(x) \sin x - f'(x)\\ f^4 (x) &= f'''(x) \cos x - f''(x) \sin x - 2(f'(x)\cos x + f''(x)\sin x) - f''(x)\\ &= f'''(x) \cos x - 3f''(x) \sin x - 2f'(x) \cos x - f''(x) \end{split} \end{equation*}

Calculating higher order derivatives at $x=0$,

\begin{equation*} \begin{split} f(0) &= 1\\ f'(0) &= 1 \times 1 = 1\\ f''(0) &= 1 - 0 = 1\\ f'''(0) &= 1-0-1 = 0 \\ f^4 (0) &= 0 - 0 -2 -1 = -3\\ f^5 (0) &= f^4 (0) \cos 0 - 3 f''(0) \cos 0 - 2f''(0) \cos 0- f'''(0)\\ &= - 3 -3-2-0 =-8 \end{split} \end{equation*}

Using Maclaurin's expansion for infinite series,

\begin{equation*} \begin{split} f(x) &= f(0) + xf'(0) + \frac{x^2}{2!}f''(0) + \frac{x^3}{3!}f'''(0) + \\ & \frac{x^4}{4!}f^4 (0) + \frac{x^5}{5!}f^5 (0) + \frac{x^6}{6!}f^6 (0) + \ldots\\ e^{\sin x} &= 1 + x + \frac{x^2}{2!}\times 1 + 0 + \frac{x^4}{4!}\times (-3) + \frac{x^5}{5!} \times (-8) \ldots\\ &= 1 + x + \frac{x^2}{2} - \frac{x^4}{8} - \frac{x^5}{15} \ldots\\ \end{split} \end{equation*}

Suman
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If $$ f(x) = a_1x + \frac{a_2}{2}x^2 + \frac{a_3}{6} x^3 + \frac{a_4}{24} x^4 + \cdots + \frac{a_n}{n!} x^n + \cdots $$ then $$ e^{f(x)} = 1 + a_1 x + \frac{a_2+a_1^2}{2!} + \frac{a_3 + 3a_2a_1+a_1^3}{6} x^3 + \frac{a_4+4a_3a_1 + 3a_2^2 + 6a_2a_1 + a_1^4}{24} x^4 + \cdots $$

The pattern is this: Consider the last case shown above, where $n=4$. There are several ways to partition the number $4$: $$ \begin{array}{r|c} \text{integer partition} & \text{number of set partitions} \\ \hline 4 & 1 \\ 3+1 & 4 \\ 2+2 & 3 \\ 2+1+1 & 6 \\ 1+1+1+1 & 1 \end{array} $$ The numbers of set partitions are the coefficients. See this page: http://en.wikipedia.org/wiki/Exponential_formula

Now try this with $$ a_1 = 1,\quad a_2=0,\quad a_3=-1,\quad a_4=0,\quad a_5=1,\quad\ldots\ldots $$

$$ e^{\sin x} = 1 + x + \frac 1 2 x^2 + \frac{(-1)+0+1^3}{3} x^3 + \cdots $$

Michael Hardy
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