How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$?

Question

How to show that $f(x)=\frac{\arcsin x}x$ is increasing when $x\ge0$?

My Attempt:

$f'(x)=\frac{\frac x{\sqrt{1-x^2}}-\arcsin x}{x^2}$

Since $0\le x\le1\implies0\le\sqrt{1-x^2}\le1$

And $0\le\arcsin x\le1.57$

But not able to show that $f'(x)\ge0$

Answer 1

You can also find a series expansion of $\arcsin(x)$ as follows:

$$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\dots$$ Put $x=\frac{t^2}{1-t^2}$, and we have $$\frac{1}{\sqrt{1-t^2}}=1+\frac{t^2}{2}+\frac{3t^4}{8}+\frac{5t^6}{16}+\dots$$ Now integrating we have $$\arcsin(x)=\int_0^x\frac{1}{\sqrt{1-t^2}}dt=x+\frac{x^3}{6}+\frac{3x^5}{40}+\frac{5x^7}{112}+\dots$$ and so $$\frac{\arcsin(x)}{x}=1+\frac{x^2}{6}+\frac{3x^4}{40}+\frac{5x^6}{117}+\dots$$ which is clearly increasing for $x\geq 0$.

Answer 2

We want to show that $\frac {\arcsin x} x \leq \frac {\arcsin y} y$ if $0 \leq x\leq y$. This can be written as $y \arcsin x \leq x\arcsin y$. The derivative of $y \arcsin x - x\arcsin y$ w.r.t. $y$ is $\arcsin x -\frac x {\sqrt {1-y^{2}}}$. If we show that this is non-positive we can finish the proof by observing that $y \arcsin x - x\arcsin y$ vanishes when $y=0$. So we have to prove that $\arcsin x -\frac x {\sqrt {1-y^{2}}} \leq 0$. For this differentiate w.r.t. $x$. The derivative is $ \frac 1 {\sqrt {1-x^{2}}} -\frac 1 {\sqrt {1-y^{2}}}$ which is non-positive for $x \leq y$. Since $\arcsin x -\frac x {\sqrt {1-y^{2}}}=0$ when $x=0$ we are done.

Answer 3

After copper.hat's comment,

putting $x=\sin t$,

$f'(\sin t)=\frac{\tan t-t}{\sin^2t}$

When $t\ge0, \tan t\ge t\implies f'(\sin t)\ge0\implies f'([0,1])\ge0\implies f'(x)\ge0$

Thus, $f(x)$ is an increasing function.

Answer 4

Alternatively one can argue as follows.

Let $$f(x)=g(u)=\frac{u}{\sin u},$$ where for $x\in [0,1],$ $u=\arcsin x$ is monotonely increasing with $u\in [0,\pi/2]$.

Now by the Chain Rule, one has $$f’(x)=\frac{dg}{dx}=\frac{\sin u-u\cos u}{\sin^2 u}\cdot \frac 1{\sqrt{1-x^2}}>0,x\in (0,1),$$ since $\tan u>u$ for $u\in (0,\pi/2).$ Combining with the Mean Value Theorem, this actually shows strict increasing of $f$.