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How can I find the Maclaurin series for $e^{\cos 2x-1}$ ?

I already know Maclaurin series for $$\cos2x= 1-\frac{4x^2}{2!}+\frac{16x^4}{4!} - ...$$

and

$$\cos 2x -1 = -\frac{4x^2}{2!}+\frac{16x^4}{4!} - ...$$

However, I couldn't figure out how to find it when it's power of $e$.

ancient mathematician
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saelassa
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    This may help: https://math.stackexchange.com/questions/25182/composition-of-power-series – Gary Jan 13 '24 at 12:36
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    You know a similar series for the exponential - try, now substitute the cosine series into the exponential series and get the general formula. – zkutch Jan 13 '24 at 12:47
  • Another way is finding $\left.\frac{d^n}{dx^n}e^{\cos(2x)}\right|_0$ and expand $e^{\cos(2x)}$ or $e^y$ as a series – Тyma Gaidash Jan 13 '24 at 15:08
  • WLOG we may consider $e^{\cos x}.$ The $2n$ derivative is of the form $p_n(\cos^2x) e^{\cos x},$ where $p_n$ is a polynomial of $n$th degree. Perhaps it is possible to find a recurrence relation for $p_n$. – Ryszard Szwarc Jan 13 '24 at 18:50

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As days pass, there are no new answers, and old ones are deleted, then let me show how the little $o$ technique can be used to get an answer:

I'll use well know formulas $$ \begin{align*} &y=\cos 2x-1 = - \frac{2^2}{2!}x^2+\frac{2^4}{4!}x^4-\cdots+(-1)^n\frac{2^{2n}}{(2n)!}x^{2n}+o(x^{2n+1}) \\ &e^y=1+y+\frac{y^2}{2!}+\cdots+\frac{y^n}{n!}+o(y^{n})\end{align*}$$ Now, if substitute $y= \cos 2x - 1$ first line in second and grouping members with same power up to, for example, $6$ we get $$e^{\cos 2x - 1}=1- \frac{2^2x^2}{2!}+\left(\frac{2^4}{4!}+\frac{1}{2!}\left(-\frac{2^2}{2!}\right)^2\right)x^4+\\+\left(\frac{-2^6}{6!}+ \frac{1}{2!}\cdot 2\cdot (-2^2)\cdot \frac{2^4}{4!} - \frac{2^3}{3!}\right)x^6+o(x^{7})=\\ =1-2x^2+\frac{8}{3}x^4-\frac{124}{45}x^6+o(x^{7})$$ where I use, that in zeros neighbourhood small little $o$ power "eats" big one. For clarity let me explain, for example, braces for $x^6$: first summund comes from $y$ - there is only one member with $x^6$, second from $y^2$ - there are two members with $x^6$ and third from $y^3$.

zkutch
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