Using $\ln(e+x)$ to approximate $\ln(e+x^2)$

Question

I made a MacLaurin series which was $\displaystyle1+\frac{x}{e}-\frac{x^2}{2e^2}+\frac{x^3}{3e^3}\dots$

I had to use the first three terms of this series to approximate $\displaystyle\int_0^1 \ln(e+x^2)\,dx$.

So I just went with my gut instinct and plugged $x^2$ for $x$ in $\displaystyle\int_0^1 \left (1+\frac{x^2}{e} - \frac{x^4}{2e^2} \right)\,dx$

I got $\displaystyle1+\frac{3}{e}-\frac{1}{10e^2}$.

I checked in the answer manual and it was right. However, why is it valid to substitute $x^2$ for $x$? Is it because $x^2$ and $x$ are close on the interval $(0,1)$?

Answer 1

There is no coincidence here. If you have a function $f$ given by a power series $$f(x)=a_0+a_1x+a_2x^2+\cdots$$then the function $f\circ g$ is given by $$ f(g(x))=a_0+a_1g(x)+a_2(g(x))^2+\cdots $$In your case, $f(x)=\ln(e+x)$ and $g(x)=x^2$.

Of course, if the series for $f$ has a finite radius of convergence, then the above series for $f\circ g$ only works whenever $g(x)$ is within that radius. And if you instead insert a series for $g$ and expand using the binomial theorem, then you have to take into account the radius of convergence of that series too.

Answer 2

Hint: If $|x|<1$, then $|x^2|\leq |x|<1$. Think about the radius of convergence.