Let $f(x) = (1-x)^{-1}$, and i want to find the degree $n$ of a Taylor polynomial centered at $0$, in the interval $[0,0.5]$ such that the error is no greater than $10^{-8}$ when approximating $f(x)$. This has been solved before, expressing the remainder as a geometric series , however i want to solve this using the remainder formula: $$\frac{\lvert f^{(n+1)}(c)\lvert}{(n+1)!} \cdot \lvert(x-\bar{x})^{(n+1)}\lvert\le 10^{-t}$$ The $(n+1)$ derivative of $f(x)$ goes like this: $$(n+1)!(1-x)^{-(n+2)}$$ So now, plugging in the derivative and the values in the remainder formula: $$(1-c)^{-(n+2)} \cdot (0.5)^{n+1}\le10^{-8}$$ I'm stuck here, i tried to plug in the max value on the interval (that is, $0.5$), and try with different values of $n$ until i find the answer, but both $n$ get cancelled. How can i proceed now?, The only way to solve this problem is using the geometric series?



using the fact that $$0 <c <0.5$$ and

$$0.5 <1-c <1$$

We will look for $n $ such that $$\frac {1}{2^{n+1}}<10^{-8} $$

You can finish by logarithm.


The error for a given $x$ and a given $n$ is exactly $\;\dfrac{x^{n+1}}{1-x}$.

Now, if $0\le x\le0.5$, $1-x\ge 0.5$ and $x^{n+1}\le \dfrac1{2^{n+1}}$, hence the error is $\le \dfrac1{2^n}$, so we have to solve $$\frac1{2^n}\le 10^{-8}\iff n\ge\frac{8}{\log 2}\approx 26.6$$ and we'll take $\; n=27$.

  • $\begingroup$ Oh! yes. Fixed.Thanks for pointing it! $\endgroup$ – Bernard Aug 26 '17 at 1:02
  • $\begingroup$ The degree $n$ Taylor polynomial actually has remainder $x^{n+1}/(1-x)$. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 1:02
  • $\begingroup$ Clearly, it's getting late here… $\endgroup$ – Bernard Aug 26 '17 at 1:05
  • $\begingroup$ xD Okay man, you can head to bed. Tiny things are of no great importance compared to sleep. $\endgroup$ – Simply Beautiful Art Aug 26 '17 at 1:05

Here's a nice algebraic approach: (likely what Bernard was aiming for)


For $x\in[0,0.5]$, we have

$$0\le x^{n+1}\le0.5^{n+1}$$



In particular, you want to solve



Recall that





In general, to get the error under $10^{-k}$, we have

$$2^n\ge10^k\\2^n=(2^{10})^{n/10}>10^{3n/10}\ge10^k\\\frac{3n}{10}\ge k\\n\ge\frac{10k}3$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.