Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am doing an exercise to see the error when solving this polynomial for $x = 1.00001$ using nested multiplication.

I believe the correct way to achieve this simplification (based on a lecture) is to multiply the polynomial by $\frac{1+x}{1+x}$; however, my algebra skills are not really up to par so I am failing to see the purpose of this. I assume it is to cancel out most of the terms in the polynomial - but which ones are being canceled?

share|cite|improve this question
up vote 8 down vote accepted

Here's the beginning:

$$\begin{align} &\ \ \ 1-x+x^2-x^3+\cdots+ x^{98}-x^{99}\\ &= \left(1-x+x^2-x^3+\cdots+ x^{98}-x^{99}\right)\frac{1+x}{1+x}\\ &= \frac{1(1+x)-x(1+x)+x^2(1+x)+\cdots +x^{98}(1+x)-x^{99}(1+x)}{1+x}\\ &=\frac{1+x-x-x^2+x^2+x^3-x^3\cdots +x^{98}+ x^{99}-x^{99} -x^{100}}{1+x}\\ \end{align}$$

Can you see where this goes?

share|cite|improve this answer
So you are essentially just left with $1 - x^{100}/(1 + x)$? That is very convenient :) – Logan Serman Feb 2 '12 at 8:26

this is a geometric sequence i.e.$(-x)^0+...+(-x)^{99}$,using the sum or geometric seuqence,i.e.$1+r+...+r^n=\frac{r^{n+1}-1}{r-1}$,so in this case put r=-x n=99 and you get the expression

share|cite|improve this answer
I see, thank you. It has been a long time since I have worked with geometric series. – Logan Serman Feb 2 '12 at 8:28

You could try to substitute x by another variable, for instance let x = -y in the formula, and see what happens...

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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