How to find the value of the $20^\text{th}$ derivative of the function in concrete point?

Question

We have function $\arcsin(x)$. How to find it's $20^\text{th}$ derivative in $x = 0$? Actually i don't have any idea except to get that derivatives manually one by one. Also, i've tried to get it with computer help, the function i get is something terrible. Also it can be solved with Leibniz formula, but it is too hard (to my mind).

Answer 1

Hint: Since we have $$(\arcsin(x))'=\frac{1}{\sqrt{1-x^2}}$$ we can write

$$(1-x^2)^{-1/2}=(1-x)^{-1/2}(1+x)^{-1/2}$$ $$((1-x)^{-1/2})'=-\frac{1}{2}(1-x)^{-3/2}(-1)=1/2(1-x)^{-3/2}$$ This is $$f^{(20)}(x)=1856156927625\,{x \left( 65536\,{x}^{18}+5603328\,{x}^{16}+95256576\,{ x}^{14}+555663360\,{x}^{12}+1354429440\,{x}^{10}+1489872384\,{x}^{8}+ 744936192\,{x}^{6}+159629184\,{x}^{4}+12471030\,{x}^{2}+230945 \right) \left( - \left( x-1 \right) \left( x+1 \right) \right) ^{- {\frac{39}{2}}}} $$

So $$f^{(20)}(0)=0$$

Answer 2

$$y'=\dfrac{dy}{dx}=\dfrac1{\sqrt{1-x^2}}$$

$$\implies(1-x^2)(y')^2=1$$

Differentiate both sides wrt $x$

$$(-2x)(y')^2+2(1-x^2)y'' y'=0$$

As $y'\ne0,$ $$xy' - (1-x^2)y''=0\ \ \ \ (1)$$

By General Leibniz rule $n(\ge2)$th derivative $$xy_{n+1}(x)-\binom n1y_n(x)-(1-x^2)y_{n+2}(x)-\binom n1(-2x)y_{n+1}(x)-\binom n2(-2)y_n(x)=0$$

At $x=0,$ $$-\binom n1y_n-y_{n+2}(0)+2\binom n2y_n(0)=0$$

$$\iff y_{n+2}(0)=n(n-2)y_n(0)$$

Now by $(1)$ at $x=0, y''=-xy'=0$

Answer 3

The answer is zero.

Note that in the Taylor series of $$\sin ^{-1} x$$ the coefficients of the even powers of $x$ are $0$ because we are dealing with an odd function.

Well, the twentieth derivative appears in the coefficient of $x^{20}$ which is $$\frac {f^{(20)}(x)}{20!}$$

Thus the $$ f^{(20}(x)=0$$

Answer 4

The differentiation will act on a series decomposition $$ f(x) = \sum a_n x^n $$ which for an odd function only has terms with odd powers. As $$ (x^n)' = n x^{n-1} $$ every differentiation will flip an odd function to an even one, and every second differentiation will result in an odd function again.

$20=10\cdot 2$ so we end up with an odd function.

Answer 5

Suppose you have an odd function $f(x)$. Then we can prove that the derivative of this function is an even one. Differentiating that again gives as back an odd function. Continuing like that we can have the following rule: $$f^{(2n-1)}(x) \text{ is even and }f^{(2n)}(x) \text{ is odd for } n\in\mathbb{N} $$ In our case setting $f(x) = \arcsin(x)$ we can conclude that f is odd since $\arcsin(x)$ by definition is odd. By taking into account the above rule we can say that $f^{(20)}(x)$ is odd. But we know that for any odd function $g(x)$ we have that $g(0)=0$ (I assume that you are aware of it). So $f^{(20)}(0)=0$.

I wish I helped!!!

Answer 6

By the generalized binomial formula,

$$(1-t)^{-1/2}=1+\frac12t+\frac12\frac32\frac{t^2}2+\frac12\frac32\frac52\frac{t^3}{3!}+\cdots=1+\sum_{n = 1}^{\infty}\frac{(2n - 1)!!t^n}{(2n)!}.$$

Then if we substitute $t=x^2$ and integrate,

$$\arcsin x=x+\sum_{n = 1}^{\infty}\frac{(2n - 1)!!x^{2n+1}}{(2n)!(2n+1)}.$$

This gives you a closed form formula for all derivatives at $0$.

(Anyway, the parity argument is must faster.)