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Problem. Find the $ n $-th derivative of $ {\cos^{n}}(x) $.

What I’m doing is substituting $ t = \cos(x) $ and then finding the $ n $-th derivative of the new function, but I’ve a feeling that this is wrong. Could anyone please point out the correct method? Thank you!

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  • $\begingroup$ @JackyChong do you mean to say I've to repeatedly apply Leibnitz's theorem of $ \frac{d^n}{dx^n}uv $ where u and v are two functions ? $\endgroup$ – sayan Sep 17 '16 at 9:28
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$$ cos(x) = \frac{e^{ix}+e^{-ix}}{2} $$

$$ cos^n(x) = \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n = \frac1{2^n}\sum_{k=0}^n \dbinom{n}{k} e^{ikx}e^{-i(n-k)x}$$

$$ cos^n(x) = \frac1{2^n}\sum_{k=0}^n \dbinom{n}{k} e^{i(2k-n)x}$$

$$ \frac{d^n}{dx^n}cos^n(x) = \frac{i^n}{2^n}\sum_{k=0}^n \dbinom{n}{k} (2k-n)^n e^{i(2k-n)x}$$

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  • $\begingroup$ Thanks, this approach is wonderful. This will also help in other problems involving hyperbolic functions where i can just use $ cos(x) = \frac{e^{x}+e^{-x}}{2} $ $\endgroup$ – sayan Sep 17 '16 at 9:30
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Doing it directly is likely to lead to a messy computation.

From trigonometry, we know that \begin{align*} 2^{n-1}\cos^n x = \cos nx +\binom{n}{1} \cos(n-2)x + \binom{n}{2}\cos(n-4)x + \cdots \end{align*}
Now, we calculate $\dfrac{d^n}{dx^n}(\cos kx)$ for an arbitrary $k$ as follows: \begin{align*} \frac{d}{dx}(\cos kx) &= -k\sin kx = k \cos\left(\frac{\pi}{2}+kx\right) \\ \frac{d^2}{dx^2}(\cos kx) &= \frac{d}{dx}\left( k \cos\left(\frac{\pi}{2}+kx\right)\right) \\ &= -k^2 \sin\left(\frac{\pi}{2}+kx\right) \\ &= k^2 \cos\left(2\cdot\frac{\pi}{2}+kx\right) \end{align*} Hence, more generally, \begin{align*} \frac{d^n}{dx^n}(\cos kx) &=k^n \cos\left(n\cdot\frac{\pi}{2}+kx\right) \end{align*} Thus, \begin{align*} 2^{n-1}\frac{d^n}{dx^n}(\cos^n x) &= n^n \cos\left(n \cdot \frac{\pi}{2} + nx\right)+\binom{n}{1}(n-2)^n \cos\left(n \cdot \frac{\pi}{2} +(n-2)x\right)+\binom{n}{2}(n-4)^n \cos\left(n \cdot \frac{\pi}{2} +(n-4)x\right)\cdots \end{align*}

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  • $\begingroup$ Hey, could you please explain where $ 2^{n-1}\cos^n x = \cos nx +\binom{n}{1} \cos(n-2)x + \binom{n}{2}\cos(n-4)x + \cdots $ comes from. I can prove it by induction but is there any other way? And thanks for answering! $\endgroup$ – sayan Sep 17 '16 at 9:36
  • $\begingroup$ Let $x = \cos x + i \sin x$. Then $2 \cos x = x +\frac{1}{x}$, and hence $(2\cos x)^n = \left(x +\frac{1}{x}\right)^n = \left(x^n + \frac{1}{x^n}\right)+\binom{n}{1}\left(x^{n-2} +\frac{1}{x^{n-2}}\right) + \cdots$ and note that $x^k + \frac{1}{x^k} = 2\cos kx$ $\endgroup$ – user348749 Sep 17 '16 at 9:48
  • $\begingroup$ Okay, now I get it. Thanks a lot! $\endgroup$ – sayan Sep 17 '16 at 15:10

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