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How do I compute the taylor series for $\cos(x)^{\sin(x)}$ ? I tried using the $e^x$ rule but I still am not getting to the result:

$$\cos(x)^{\sin(x)}=1-\frac{x^3}{2}+\frac{x^6}{8}+o(x^6).$$

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Very informally:

Noting $\bigl(\cos x \bigr)^{\sin x } =\exp\bigl ( \sin (x) \ln (\cos x) \bigr)$.

Start with $\cos x = 1-{x^2\over 2!}+{x^4\over 4!}-\cdots$.

Then use the Taylor series $$\ln(1+z)=z-{z^2\over2}+{z^3\over3}-\cdots$$ with $z=-{x^2\over 2!}+{x^4\over 4!}-\cdots$ to obtain the first few terms of the expansion of $\ln(\cos x)$.

Multiply this by the first few terms of the Taylor series for $\sin x$.

This will give you some polynomial expression $P(x)$; which you would then substitute into the Taylor series for $e^x$.

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    Would it be simpler to calculate a few derivatives of $exp(\sin(x) \ln(\cos x))$ and use the formula for a Taylor series? I guess you would need to calculate 6 derivatives and that is going to get pretty complicated.2012-01-13
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    Once you know what to do, as explained by @David, the evaluation is taylor-made for any computation package that knows how to substitute one power series for the variable in another. So you get $1 - (1/2)x^3 + (1/8)x^6 - (1/80)x^7 - (37/1512)x^9 + (1/160)x^{10} - (71/57600)x^{11} + (107/24192)x^{12} - \cdots$2012-01-14
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    thanks david, this was very helpful2012-01-22
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Your formula ($\cos(x)^{\sin(x)}=1-\frac{x^3}{2}+\frac{x^6}{8}+o(x^6).$) has been achieved from the definition of The Taylor Series: $$f(x) = \sum_{i=0}^{\infty}\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$ Where $f^{(n)}(x)$ is $n$th derivative of $f(x)$ with respect to $x$.

(Notice that $f\in c^{\infty}$)

put $x_0=0$ and calculate the coefficients of $x^0$, $x^1$, ... $x^6$.