Prove $\sin x + \arcsin x > 2x$ using Maclaurin series

Question

My teacher asked us to solve this problem using the Maclaurin series, but I could not figure out how to approach..

Prove that the inequality sin x + arcsin x > 2x holds for all values of x such that 0 < x ≤ 1.

I know that the Maclaurin series of sin(x) = x - $\frac{x^3}{3!}$ + $\frac{x^5}{5!}$ - $\frac{x^7}{7!}$ + ...

arcsin(x) = x + $\frac{1}{2}\cdot\frac{x^3}{3}$ + ($\frac{1}{2}\cdot\frac{3}{4}$)$\cdot\frac{x^5}{5}$ + ...

However, I do not know how to prove this using there series...Could anyone have some ideas?

Thank you!

Answer 1

You can try to prove that all coefficients of the sum are positive.

Notice that $\sin(x) + \arcsin(x) = 2x+\dfrac{x^5}{12}+ \dfrac{2x^7}{45}+\dfrac{5513x^9}{181440} \quad ... $

And for the domain where your equality holds, note that the Maclaurin series of $\arcsin$ only holds for $x$ in $]-1,1[$.

Answer 2

Hint: A nicer representation of the arcsin series is $$\arcsin(x) = \sum_{n = 0}^\infty \frac{1}{4^n} {2n \choose n} \cdot \frac{x^{2n + 1}}{2n + 1}$$ Since both $\sin(x)$ and $\arcsin(x)$ are both odd, we can look at the coefficients on all the odd-degree terms. Can you prove that for all $n \geq 1$, $$\frac{1}{4^n} {2n \choose n} \cdot \frac{1}{2n + 1} \geq \frac{1}{(2n + 1)!} \iff \frac{1}{4^n} {2n \choose n} \geq \frac{1}{(2n)!}$$ This would prove all coefficients of terms with degree greater than $1$ in the expansion of $\arcsin(x) + \sin(x)$ are positive, and thus $\sin(x) + \arcsin(x) = 2x + C$, where $C$ is strictly positive.

Answer 3

Hint:

As all coefficients of the development of $\arcsin x$ are positive, it is obvious that $\arcsin x>x$.

This cannot be said of the series for the sine, which alternates.

Now you can infer that the sum will work, if the respective coefficients of the arc sine compensate the negative coefficients of the sine.

Answer 4

Consider the infinite expansion $$\sin (x)+\sin ^{-1}(x)=2x+\sum_{n=1}^\infty \left(\frac{4^{-n} (2 n)!}{(2 n+1) (n!)^2}+\frac{(-1)^n}{(2 n+1)!} \right)x^{2n+1}$$ If $n$ is even, the coefficient in brackets is positive. So, for the case where $n$ is odd, we need to prove that the coefficient is positive $\forall n$ that is to say that $$\frac{4^{-n} (2 n)!}{(2 n+1) (n!)^2}>\frac1{(2 n+1)!}\implies b_n=\frac{4^{-n} (2 n)! (2 n+1)!}{(2 n+1) (n!)^2}>1 $$ Simplifying $$b_n=\frac{4^{-n} (2 n)! (2 n)!}{ (n!)^2}$$ Take logarithms, use Stirling approximation, go back to exponentials to show that $$b_n \sim 2\left(\frac {2n} e\right)^{2n}\exp\left(-\frac 1{12n} \right)$$ So, all coefficients are positive.

Answer 5

For $0 < x \le 1$ we have $$ \sin(x) + \arcsin(x) = 2 x + \sum_{n=2}^\infty \left(\frac{(-1)^n}{(2n+1)!} + \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} \right) x^{2n+1} $$ because the $x^3$ terms cancel. Therefore it suffices to show that $$ \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \frac{1}{2n+1} > \frac{1}{(2n+1)!} $$ for $n \ge 2$. This is equivalent to $$ \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots (2n)} \cdot (2n)! > 1 $$ or $$ \bigl( 1 \cdot 3 \cdots (2n-1) \bigr)^2 > 1 \, , $$ which is obvious.

Answer 6

For $0\le x\le1$ we have

$$\sin x\ge x-{1\over6}x^3\ge0\quad\text{and}\quad\arcsin x\ge x+{1\over6}x^3+{3\over40}x^5\ge0$$

which imply

$$\sin x\arcsin x\ge x^2+\left({3\over40}-{1\over36} \right)x^6-{1\over80}x^8=x^2+\left(34-9x^2\over720\right)x^6\ge x^2$$

By AGM we have

$${\sin x+\arcsin x\over2}\ge\sqrt{\sin x\arcsin x}\ge x$$

Remark: As Martin R astutely observes in comments, as soon as you have $\sin x\ge x-{1\over6}x^3$ and $\arcsin x\ge x+{1\over6}x^3$, you have $\sin x+\arcsin x\ge2x$, so tacking another (nonnegative) term onto the arcsine series, taking the product and using AGM is wholly unnecessary. I failed to notice this because I was approaching things backwards: I had decided to see if AGM could be used and then worked out how much of the two series were needed to arrive at the desired inequality.

The inequality $\sin x\ge x-{1\over6}x^3$ for $0\le x\le1$ can be seen from the fact that the series for $\sin x$ is an alternating series of decreasing terms.