Numerical analysis Taylor $1/(1-x)$

Question

How does accuracy depend on the degree of the Taylor Polynomial and the distance from the point its being expanded about (say $x=0$). So I'm considering the function $f(x) =\frac{1}{1-x}$ centered at $0$. I have found up to the 4th degree Taylor polynomial, as I created a program on matlab to represent the errors graphically. I think because of the remainder term is depended on $1/(n+1)!$ by definition that its obvious that the errors will be decreasing as $n$ becomes large. However, I may be missing something, or looking at it incorrectly.

Answer 1

That function has a particularly easy Taylor series, namely $$ 1+x+x^2+x^3+\ldots, $$ in other words all coefficients are$~1$. To see this, easiest is not to bother with derivatives at all, but to observe that as a formal power series, multiplication by $1-x$ of that series produces just a constant term$~1$, and everything else vanishes; moreover it is the only power series with that property, a property which the Taylor series of $\frac1{1-x}$ must certainly satisfy.

Now whether taking more and more terms of the power series improves the approximation depends critically on the absolute value $|x|$: if it is less than $~1$ then each successive term will be smaller, and it is easy to see that the geometric series so obtain converges to $\frac1{1-x}$. However if $|x|>1$ then each successive term has larger absolute value, so that taking more terms only gets you further and further away. Indeed the geometric series can now easily seen to diverge to $\infty$. The critical case $|x|=1$ has the property that each successive term has absolute value $1$, so the series cannot converge; however the partial sums diverge to $\infty$ only if $x=1$; for all other values on the unit circle the partial sums remain bounded but do not converge.