4

Trying to figure out how to calculate the 90th derivative of $\cos(x^5)$ evaluated at 0. This is what I tried, but I guess I must have done something wrong or am not understanding something fundamental:

$\cos(x) = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} {(x)}^{2n}$

$\cos(x^5) = \displaystyle\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} {(x^5)}^{2n}=$

$1-\dfrac{{x^{5\cdot2}}}{2!}+\dfrac{{x^{5\cdot4}}}{4!}-\dfrac{{x^{5\cdot6}}}{6!}+...-\dfrac{{x^{5\cdot18}}}{18!}+...-\dfrac{{x^{5\cdot90}}}{90!}+...+\dfrac{{x^{5\cdot190}}}{180!}+...$

$f(x) = \displaystyle\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!} \cdot {x^{n}} $

$\dfrac{f^{90}(0)}{90!}\cdot {{x^{90}}} = -\dfrac{{x^{5\cdot18}}}{18!}$

${f^{90}(0)} = -\dfrac{90!}{18!}$

Wolfram has it at some really large negative number.

JimmyK4542
  • 54,331
user17753
  • 414

2 Answers2

2

Taylor series is the right way to go, however, you need to be careful in how you obtain the Taylor series for $\cos x^5$.

Since $\cos \theta = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^n\theta^{2n}}{(2n)!}$, we have $f(x) = \cos x^5 = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^n(x^5)^{2n}}{(2n)!} = \displaystyle\sum_{n = 0}^{\infty}\dfrac{(-1)^nx^{10n}}{(2n)!}$

The $x^{90}$-coefficient is found only in the $n = 9$ term, so we have $\dfrac{f^{(90)}(0)}{90!}x^{90} = \dfrac{(-1)^9x^{90}}{18!}$.

Now, just solve for $f^{(90)}(0)$.

JimmyK4542
  • 54,331
  • Thanks this is essentially what I was trying to do, but made several simple mistakes, I completely understand it now thanks to you guys. – user17753 Dec 17 '15 at 20:46
1

$$\cos(x^5) = \sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} {(x^5)}^{2n}=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!} x^{10n}$$

To get the coefficient of $x^{90}$ in this series, just put $n=9$ to get $\dfrac{(-1)^9}{18!}$ (I think this is where you went wrong). Now if you differentiate the series $90$ times, and set $x=0$, the only non-zero term is $$\dfrac{(-1)^990!}{18!}$$

TonyK
  • 64,559