# Show that $|x-\ln(1+x)| \leq cx^2$

Use Taylor's Theorem to show that there is some fixed constant $c>0$ such that $$|x-\ln(1+x)| \leq cx^2$$ for all $|x|<\frac{2}{3}$.

My attempt: Let $f(x)=\ln(1+x)$. Then by calculating the first degree Taylor polynomial centered at $x=0$, we get $\ln(1+x)=x+R_1(x)$ where $R_1(x)=\dfrac{f^{(2)}(\xi) x^2}{2}$, where $\xi \in (0,x)$.

Notice that $$|R_1(x)|=\left|\frac{f^{(2)}(\xi) x^2}{2}\right|=\frac{x^2}{2(1+ \xi)^2}=cx^2.$$

Then we have $$|x- \ln(1+x)|=|R_1(x)| \leq cx^2.$$

In this question, I don't know how to show that the inequality holds for all $|x| < \frac{2}{3}$. Can anyone explain?

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As you show, if $|x|<1$ then $x-\ln(1+x)=x^2/2(1+\xi)^2$ for some $|\xi|\le |x|$. When $|x|<2/3$, we have $x^2/2(1+\xi)^2\le x^2/2(1-2/3)^2$. We are done.
The real $\xi\in(0,x)$ if $x>0$ and $\xi\in (x,0)$ if $-1<x<0$ so $0<|\xi|<|x|<\frac{2}{3}$ by assumption.
Now we have $$\frac{1}{3}=1-\frac{2}{3}<1+\xi<1+\frac{2}{3}=\frac{5}{3}$$ so $$|R_1(x)|=|\frac{f^{(2)}(\xi) x^2}{2}|=\frac{x^2}{2(1+ \xi)^2}\leq \frac{9}{2}x^2$$