Use Taylor Polynomial to approximate a definite integral.

Question

i have this problem:

Get the Taylor Polynomial of grade 3 of $\frac{1}{1-x}$ at the point a=0 Then use the polynomial to approximate the definite integral:

$$\int_0^\frac{1}{10} \frac{1}{1-x^2}$$

Here is what I did:

Taylor polynomial degree 3, center a=0:

$1+x+x^2+x^3$

then I replace $x$ for $x^2$ and I have:

$1+x^2+x^4+x^6$

then integrating

$(0.1)+\frac{(0.1)^3}{3}+\frac{(0.1)^5}{5}+\frac{(0.1)^7}{7}$

and that is the result.

Is this correct? I mostly taught myself through books and have no way of checking if i understood the topic correctly.

Thanks

Answer 1

So, for any integer $n\ge0$, we define the polynomial $$p_n(x)=\sum_{k=0}^{n}x^k$$ which is the $n$-th order Taylor polynomial for $\frac1{1-x}$. So the polynomial for $\frac1{1-x^2}$ is of course $$P_n(x)=\sum_{k=0}^{n}x^{2k}$$ and thus $$Q_n=\sum_{k=0}^{n}\int_0^{1/10}x^{2k}dx=\sum_{k=0}^{n}\frac{1}{10^{2k+1}(2k+1)}$$ which confirms your results.