Taylor polynomial

Quote:

Originally Posted by roshanhero

Use a taylor polynomial to approximate $f(x)=\frac{1}{1+x}$ at x=0.1 with error less than 0.001.

The n-th derivative of f is $f^{(n)}(x)=\frac{(-1)^n n!}{(1+x)^{n+1}}$ . If you develop f around $x_0=0$ you get

$f(x) = 1-x+x^2-x^3+\cdots +\frac{(-1)^n x^n}{(1+\theta x)^{n+1}}$
Where $\theta \in (0,1)$ . To make the error term for $x=0.1$ smaller than 0.001, we solve the following (worst case) inequality for n

$\left|\frac{(-1)^n 0.1^n}{(1+0\cdot 0.1)^{n+1}}\right|< 0.001$ which gives, if I'm not mistaken, that $n>2.958$ . Thus we take $n=3$ and get:

$f(0.1) \approx 1-0.1+0.1^2=0.91$ . Compare this with the exact value of $f(0.1)=\frac{1}{1+0.1}=0.\overline{90}$ : the resulting error $0.91-0.\overline{90}=0.000\overline{90}$ is less than 0.001, as required.