Taylor Expansion Analysis

Question

I have been given a "help function" (direct translated) $g(t)=\ln(2+t)$, for $|t|< 1$. Show that: $$ \biggl|g(t)-\Bigl(\ln(2)+\frac{t}{2}\Bigr)\biggr| \leq \frac{1}{2}t^2$$ For all $|t|\leq 1$. I am a bit lost here as it does not look like an exercise we previously have solved. Normally I would evaluate the remainder and see that it is less than $\frac{M_n}{(n+1)!}|x-x_0|^{n+1}$. However, it does not seem possible in this assignment. We have also been given a function $f(x)=\ln(2+x^3) \ , \ |x|\leq1$ where I've found $T_3(x)=\ln(2)+1/2 x^3$. I cannot see how this would help me here though...

Answer 1

Hint:

Rewrite $g(t)=\ln2\bigl(1+\frac t2\bigr)=\ln 2+\ln\bigl(1+\frac t2\bigr)$ and observe that $$\ln\Bigl(1+\frac t2\Bigr)=\frac t2-\frac{t^2}8+\frac{t^3}{24}-\dotsm$$ is an alternating series for $0<t<1$, so you can apply Leibniz' estimate for the error.

If $-1<t<0$, it is still more obvious.