How do I find the Taylor series of $\cos^{20}{(x)}$, for $x_0 = 0$?

Question

How do I find the Taylor series of $\cos^{20}{(x)}$ for $x_0 = 0$, knowing the Taylor series of $\cos{(x)}$?

Answer 1

If you are looking for a limited number of terms, just compose $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320 }-\frac{x^{10}}{3628800}+O\left(x^{12}\right)$$

Now, square a few times $$\cos^2(x)=1-x^2+\frac{x^4}{3}-\frac{2 x^6}{45}+\frac{x^8}{315}-\frac{2 x^{10}}{14175}+O\left(x^{12}\right)$$ $$\cos^4(x)=1-2 x^2+\frac{5 x^4}{3}-\frac{34 x^6}{45}+\frac{13 x^8}{63}-\frac{514 x^{10}}{14175}+O\left(x^{12}\right)$$ $$\cos^8(x)=1-4 x^2+\frac{22 x^4}{3}-\frac{368 x^6}{45}+\frac{1957 x^8}{315}-\frac{48428 x^{10}}{14175}+O\left(x^{12}\right)$$ $$\cos^{16}(x)=1-8 x^2+\frac{92 x^4}{3}-\frac{3376 x^6}{45}+\frac{41462 x^8}{315}-\frac{2501536 x^{10}}{14175}+O\left(x^{12}\right)$$

$$\cos^{20}(x)=\cos^{4}(x) \cos^{16}(x)$$

Now, look at the patterns

Edit

A bit faster is to compute $$\log (\cos (x))=-\frac{x^2}{2}-\frac{x^4}{12}-\frac{x^6}{45}-\frac{17 x^8}{2520}-\frac{31 x^{10}}{14175}+O\left(x^{12}\right)$$ $$\log (\cos^{16} (x))=-8 x^2-\frac{4 x^4}{3}-\frac{16 x^6}{45}-\frac{34 x^8}{315}-\frac{496 x^{10}}{14175}+O\left(x^{12}\right)$$ $$\cos^{16} (x)=e^{\log (\cos^{16} (x))}=$$ $$1-8 x^2+\frac{92 x^4}{3}-\frac{3376 x^6}{45}+\frac{41462 x^8}{315}-\frac{2501536 x^{10}}{14175}+O\left(x^{12}\right)$$

Answer 2

We can use the even symmetry of the cosine function to determine that $ \ (\cos x)^{20} \ $ is likewise an even function, thus the Maclaurin series for this function must contain only terms with even powers of $ \ x \ \ . \ $ This means that we will only be interested in higher derivatives of the function of even order.

We can also exploit properties of the derivatives of the cosine function. If we take $ \ u \ = \ \cos x \ \ , \ $ then $$ u^{(2n)} \ \ = \ \ (-1)^n·u \ \ \Rightarrow \ \ u^{(2n)}(0) \ = \ (-1)^n \ \ \ , \ \ \ u^{(2n+1)} \ \ = \ \ (-1)^n·(-\sin x) \ \ \Rightarrow \ \ u^{(2n+1)}(0) \ = \ 0 \ \ . \ $$ It will also be helpful to apply $ \ (u')^2 \ = \ (-\sin x)^2 \ = \ 1 - u^2 \ \ . $

Starting with the second derivative, we obtain

$$ (u^{20})'' \ \ = \ \ ( \ 20·u^{19}·u' \ )' \ \ = \ \ (20·19·u^{18}·u')·u' \ + \ 20·u^{19}·u'' $$ $$ = \ \ ( 20·19·u^{18})·(u')^2 \ + \ 20·u^{19}·(-u) \ \ = \ \ ( 20·19·u^{18})·(u')^2 \ - \ 20·u^{20} $$ $$ \Rightarrow \ \ (u^{20})''(0) \ \ = \ \ 20·19·1^{18}·0^2 \ - \ 20·1^{20} \ \ = \ \ -20 \ \ . $$ [Of course, this term could be extracted easily enough from $ \ (\cos x)^{20} \ \approx \ \left(1 - \frac12x^2 \right)^{20} \ \approx \ 1 - 10x^2 \ \ . \ ] $

In proceeding to the fourth derivative, we observe that we have already done some of the work: $$ (u^{20})^{(4)} \ \ = \ \ [ \ ( 20·19·u^{18})·(u')^2 \ - \ 20·u^{20} \ ]'' \ \ = \ \ [ \ ( 20·19·u^{18})·(1 - u^2) \ - \ 20·u^{20} \ ]'' $$ $$ = \ \ [ \ 20·19·u^{18} \ - \ (20 \ + \ 20·19) ·u^{20} \ ]'' $$ $$ = \ \ ( \ 20·19·18·u^{17}·u' \ )' \ - \ (20 \ + \ 20·19)·[ \ 20·19·u^{18} \ - \ (20 \ + \ 20·19) ·u^{20} \ ] $$ $$ = \ \ ( \ 20·19·18·17·u^{16}·u'·u' \ + \ 20·19·18·u^{17}·u'' \ ) $$ $$ - \ (20 \ + \ 20·19)·[ \ 20·19·u^{18} \ - \ (20 \ + \ 20·19) ·u^{20} \ ] $$

$$ \Rightarrow \ \ (u^{20})^{(4)}(0) \ \ = \ \ 20·19·18·17·1^{16}·0^2 \ + \ 20·19·18·1^{17}·(-1) $$ $$ - \ (20^2·19 \ + \ 20^2·19^2)·1^{18} \ + \ (20 \ + \ 20·19)^2 ·1^{20} $$ $$ = \ \ -6840 \ - \ 7600 \ - \ 144,400 \ + \ 160,000 \ \ = \ \ 1160 \ \ . $$

Although it may seem daunting, we will work to obtain the sixth derivative, which we may hope will be far enough to see a pattern. (We note along the way that we have found that the first and third derivatives of $ \ u^{20} \ $ only have terms with factors that are odd derivatives of $ \ u \ \ , \ $ making odd-power terms of the Taylor series equal to zero, as expected.)

$$ (u^{20})^{(6)} \ \ = \ \ [ \ 20·19·18·17·u^{16}·(1 - u^2) \ + \ 20·19·18·u^{17}·(-u) $$ $$ - \ (20^2·19 \ + \ 20^2·19^2)·u^{18} \ + \ (20 \ + \ 20·19)^2 ·u^{20} \ ]'' $$

$$ = \ \ [ \ 20·19·18·17·u^{16} \ - \ (\overbrace{20·19·18·17 \ + \ 20·19·18 \ + \ 20^2·19 \ + \ 20^2·19^2}^{\mathcal{C}})·u^{18} $$ $$ + \ (20 \ + \ 20·19)^2 ·u^{20} \ ]'' $$

$$ = \ \ 20·\ldots·15·u^{14}·(u')^2 \ + \ 20·\ldots·16·u^{15}·(u'') \ - \ \mathcal{C}·18·17·u^{16}·(u')^2 \ - \ \mathcal{C}·18·u^{17}·(u'') $$ $$ + \ (20 \ + \ 20·19)^2 · [ \ 20·19·u^{18} \ - \ (20 \ + \ 20·19) ·u^{20} \ ] $$

$$ \Rightarrow \ \ (u^{20})^{(6)}(0) \ \ = \ \ 20·19·18·17·16·1^{15}·(-1) $$ $$ - \ (20·19·18·17 \ + \ 20·19·18 \ + \ 20^2·19 \ + \ 20^2·19^2)·18·1^{17}·(-1) $$ $$ + \ (20 \ + \ 20·19)^2 · 20·19·1^{18} \ - \ (20 \ + \ 20·19)^3 ·1^{20} $$ $$ = \ \ -1,860,480 \ + \ 4,952,160 \ + \ 60,800,000 \ - \ 64,000,000 \ \ = \ \ -108,320 \ \ . $$

Thus we have produced by direct computation the first four terms of the Taylor series $$ (\cos x)^{20} \ \ = \ \ 1 \ + \ \frac{-20}{2!}·x^2 \ + \ \frac{1160}{4!}·x^4 \ + \ \frac{-108320}{6!}·x^6 \ + \ \ldots $$ $$ = \ \ 1 \ - \ 10·x^2 \ + \ \frac{145}{3}·x^4 \ - \ \frac{1354}{9}·x^6 \ + \ \ldots \ \ , $$ in agreement with the result from WolframAlpha (which, interestingly, declines to generate any further terms, at least at "Pro" level).

I have not found a concise way to express the general term of this series (in the time I've had available), but the calculation of higher derivative values at $ \ x \ = \ 0 \ $ is not difficult, though a little tedious.