Find $f^{(k)}(0)$ of $f(x)=\arctan(x^n), n\in \Bbb N$

Question

Let $n\in \Bbb N$, and $f:\Bbb R \rightarrow \Bbb R$, $f(x)=\arctan(x^n)$. For any $k\in \Bbb N$, find $f^{(k)}(0)$.

Now, I found the first derivative:$$f'(x)= \frac{nx^{n-1}}{1+x^2}$$

I was thinking of maybe using Leibniz' rule but that doesn't seem like the best idea. What else can I do?

Answer 1

Note that the geometric series formula gives $$ f'(x) = \frac{nx^{n-1}}{1+x^{2n}} = nx^{n-1} \sum_{j=0}^\infty (-x^{2n})^j= \sum_{j=0}^\infty (-1)^{j}nx^{2{jn}+n-1} = \sum_{\tilde \jmath =n-1}^\infty a_\tilde \jmath x^\tilde \jmath, $$ where $$ a_\tilde \jmath = \begin{cases} 0 & \tilde \jmath<n-1 , \text{ or } \tilde \jmath+1-n \neq 0 \mod {2n} \\ n(-1)^{\frac{\tilde \jmath+1-n}{2n}} & \text{otherwise} \end{cases}$$ By uniqueness of Taylor series, we know that $$ f'(x) = \sum_{j=0}^\infty \frac{f^{(j+1)}(0)}{j!}x^j = \sum_{\tilde \jmath =n-1}^\infty a_\tilde \jmath x^\tilde \jmath $$ so one can read off the values of $f^{(j+1)}(0)$ as coefficients of $x^{j}/j!$ in the above expansion. The answer is $$ f^{(k)}(0) = \begin{cases} 0 & k<n , \text{ or } k-n \ne 0 \mod 2n \\ (k-1)!n(-1)^{\frac{k-n}{2n}} & \text{otherwise} \end{cases}$$

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The previous version of this answer did not coincide with Bernard's answer. This is because I thought you computed the first derivative correctly, but you didn't. Specifically, you applied 'chain rule' in the form $$ (fg)'(x) = f'(x) g'(x)$$ when the correct chain rule is $$ (fg)'(x) = f'(g(x)) g'(x)$$ so the correct first derivative is $nx^{n-1}/(1+x^{2n})$, and not $nx^{n-1}/(1+x^{2})$.

Answer 2

Hint Substitute $y = x^n$ in the Maclaruin series $$y - \frac{1}{3} y^3 + \frac{1}{5} y^5 - \frac{1}{7} y^7 + \cdots$$ for $\arctan y$.

Answer 3

Hint:

Use the substitution $x^n=u$: $$\arctan u=\sum_{i\ge 0}\frac{(-1)^i}{2i+1}\,u^{2i+1}=\sum_{i\ge 0}\frac{(-1)^i}{2i+1}\,x^{(2i+1)n},$$ and remember the coefficient of $x^k$ is $\;\dfrac{f^{(k)}(0)}{k!}.$