computing the Taylor series for $e^{1-\cos x}$

Question

I was told to compute the Taylor series for $e^{1-\cos x}$ to the fourth degree.
I plugged the Taylor series for $1-\cos x$ into the Taylor series for $e$.
This got me: $$1-\frac{x^2}2+\frac{x^4}{24}+\frac{(\frac{-x^2}2)^2}{2}$$A bit of simplification results in: $1+\frac{x^2}2+\frac{x^4}6$. Is this correct? The answer key to the question gave a different answer - it put the denominator of the last term as 12 instead of 6. However, I do not see the flaw in my calculations. Did I make the error or did they?

Answer 1

My answer agrees with the answer key:

$\cos x=1-\dfrac{x^2}2+\dfrac{x^4}{24}+O(x^6),$ so $1-\cos x=\dfrac{x^2}2-\dfrac{x^4}{24}+O(x^6),$

so $\exp(1-\cos x)=1 + \left(\dfrac{x^2}2 - \dfrac{x^4}{24} \right)+ \dfrac12\left(\dfrac{x^2}2\right)^2+O(x^6)=1 + \dfrac{x^2}2 + \dfrac{x^4}{12}+O(x^6).$

Answer 2

For a check, it's not really difficult to compute the derivatives: \begin{align} f(x)&=e^{1-\cos x} & f(0)&=1 \\ f'(x)&=e^{1-\cos x}\sin x & f'(0)&=0 \\ f''(x)&=e^{1-\cos x}(\sin^2x+\cos x) & f''(0)&=1 \\ f'''(x)&=e^{1-\cos x}(\sin^3x+3\sin x\cos x-\sin x) & f'''(0)&=0 \\ f''''(x)&=e^{1-\cos x}(\sin^4x+3\sin^2x\cos x-\sin x\cos x+3\cos^2x-\cos x) & f''''(0)&=2 \end{align} Thus $$ e^{1-\cos x}=1+\frac{x^2}{2}+\frac{x^4}{12}+o(x^4) $$