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!
Best Answer
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 $]-1,1[$.