Find $f^{(200)}(1)$ for $f(x)=(x-1)\arctan(x-1)$

Question

Find $f^{(200)}(1)$ for $f(x)=(x-1)\arctan(x-1)$.

I know that:

$f'(x)=\arctan(x-1)+\frac{x-1}{x^2-2x+2}$

$f''(x)=...=\frac{2}{(x^2-2x+2)^2}$

Now, how do I find the $n$th derivative? Do I derive it once more and then use Leibniz' rule or something like that?

Answer 1

Equivalently we may look at $g^{(200)}(0)$ for

$$g(x)=x\arctan(x) = \sum_{m\geq 0}\frac{(-1)^m x^{2m+2}}{2m+1}.$$ By considering $m=99$ we have $$ [x^{200}]g(x) = -\frac{1}{2\cdot 99+1} = \frac{g^{(200)}(0)}{200!} $$ hence $$ g^{(200)}(0) = -\frac{200!}{199} = -200\cdot 198!.$$

Answer 2

Hint: Use the Taylor series: $$\arctan(x-1) = \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1}{(x-1)}^{2n+1}.$$

Answer 3

HINT: It is the same as to find $g^{(200)}(0)$ for $g(x)=x\arctan x$. But $$ \arctan x=\int_0^x \frac1{1+t^2}\,dt. $$ We know the series of $1/(1+t^2)$, so we compute the series for $\arctan$.

Answer 4

let $y = (x-1)$ and $g(x) = x\arctan x$

Then $f(x) = g(y)$ and $f(1) = g(0)$ and $f^{(n)}(1) = g^{(n)}(0)$

The Taylor series of $g(x)$ is easier to derive.

Answer 5

Hint: For a function $g$, expanded in its disk of convergence centred at $0$, the coefficient of $x^{2n}$ is equal to $$\dfrac{g^{(2n)}(0)}{(2n)!}.$$

So you have to make the substitution $u=x-1$ and expand $g(u)=u\arctan u$ up to order $200$.