Inequality of $\vert \text{tanh}(x)-x+\frac{x^3}{3}\vert\leq\frac{1}{8}$ for $x\in[-1/4,1/4]$

Question

I have an assignment, that I cannot seem to find a starting point for. I am not looking for the specific answer, but only where to begin. Any pointers are appreciated.

I have to prove the inequality $\vert \tanh(x)-x+\frac{x^3}{3}\vert\leq\frac{1}{8}$ for $x\in[-1/4,1/4]$ where $x-\frac{x^3}{3}$ is the 3rd order Taylor expansion.

I've tried with the triangle inequalities, Lagrange remainders, and a lot of other stuff, but I am no where near a correct argumentation. Does anyone have any suggestions?

Answer 1

Taylor's theorem with Lagrange remainder gives $$ \tanh(x)-x+\frac{x^3}{3} = \frac{\tanh^{(4)}(\xi)}{4!} x^4 $$ for some $\xi$ between $0$ and $x$.

For an estimate of $\tanh^{(4)}(\xi)$ use that

• $\tanh'(x) = 1 - \tanh(x)^2 $, this allows to express $\tanh^{(4)}(x)$ as a polynomial in $\tanh(x)$ with constant coefficients.
• $|\tanh(x)| < 1$ for $x \in \Bbb R$.

This gives $$ \tanh^{(4)}(x) = 16 \tanh(x) - 40 \tanh^3(x) + 24 \tanh^5(x) $$ and therefore $$ \left| \tanh(x)-x+\frac{x^3}{3} \right| \le \frac{80}{4!} \left( \frac 1 4 \right)^4 = \frac{5}{384} $$ for $|x| \le 1/4$.