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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?

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3 Answers 3

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?

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

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

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