Approximation by Taylor polynomial

Let $f(x) = (1 − x)^{-1}$ and $x_{0} = 0.$

(a) Find the nth Taylor polynomial $P_{n}(x)$ for $f(x)$ about $x_{0}$.

(b) Find the smallest value of $n$ necessary for $P_{n}(x)$ to approximate $f(x)$ to within $10^{-6}$ on $[0; 0.5]$.

For part (a) I get $P_{n}(x)=1+x^{2}+x^{3}+...+x^{n}.$ I don't know how to get started with part (b), is it finding $n$ such that $|f(x)-P_{n}(x)| \leq 10^{-6}$? Thanks!

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We could try to use various general expressions for the remainder. You may have learned already about the Lagrange form of the remainder.

These however often do not give good upper bounds for the error. We want a good upper bound because we were asked to find the smallest $n$.

Instead of attempting to use a general estimate, we work with our particular function. Note that the error we make by truncating just after the $x^n$ term is exactly equal to the missing tail, which is $$x^{n+1}+x^{n+2}+x^{n+3}+\cdots.$$ The sum of this geometric series is equal to $$\frac{x^{n+1}}{1-x}.$$ In our interval, this is biggest at $x=0.5$, when it is equal to $(0.5)^n$.

So we are trying to find the smallest integer $n$ such that $(0.5)^n\lt 10^{-6}$.

You can solve $(0.5)^t=10^{-6}$ using logarithms. But in this case we can do the problem in our heads, since $1000$ is very close to $2^{10}$.

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Thank you! I was having the exact same problem and this was a huge help! –  SSumner Mar 3 '13 at 5:13

a is correct. For part b, you are right you want $|f(x)-P_n(x)|\le 10^{-6}$ over the given interval. To do that, you need the error term in the Taylor series. It has the $n+1^{\text{st}}$ derivative in it, you find the maximum of that derivative and plug it in.

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So the problem is now reduced to finding smallest $n$ such that $\left| {\frac{{(k + 1)!{{(1 - \xi )}^{ - (k + 2)}}}}{{(n + 1)!}}{x^{n + 1}}} \right| \le {10^{ - 6}}$ where $x \in [0,0.5]$ and $\zeta$ is between $x$ and $0$. How can it be done? –  drawar Jan 26 '13 at 17:16