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I have this: $$ \frac{x^2}{x^2+1} $$ Wolfram Alpha suggests that I should do long division to get this: $$ 1- \frac{1}{x^2+1} $$ But I don't understand how it can be that, please explain.

rablentain
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    $x^2=x^2+1-1$.${}$ – David Mitra Feb 28 '14 at 16:21
  • http://www.purplemath.com/modules/polydiv2.htm – Ian Coley Feb 28 '14 at 16:23
  • This question bewilders me, @theva: can't you do the fractions' substraction $;1-\frac1{x^2+1};$ and check it indeed equals the original expression?? Because if you can't then David's hint won't likely work... – DonAntonio Feb 28 '14 at 16:25
  • If you don't know how to perform polynomial long division by hand, you should definitely learn this. It is extremely easy and will give this answer immediately. – MPW Feb 28 '14 at 16:47

2 Answers2

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While this can be done "by inspection" it is instructive to recall the key idea of the polynomial division algorithm: $\rm\color{#c00}{scale}$ the divisor so that it has the same degree and leading coef as the dividend, then subtract it from the dividend, thereby canceling the leading term of the dividend; then recursively apply this process to the smaller degree remaining part of the dividend, viz.

$$ (a x^{k+n} + f) - \color{#c00}{a x^k} (x^n + g)\ =\ f-ax^kg$$

$$\ \Rightarrow\ \dfrac{a x^{k+n}+f}{x^n+g}\, =\ \color{#c00}{a x^k} +\!\!\! \underbrace{\dfrac{f-ax^k g}{x^n + g}}_{\large\rm recurse\ on\ this}$$

where the second equation arises from the first by dividing through by $\,x^n + g.$

In your example the dividend and divisor already have the same degree and same lead coef, therefore we cancel the leading term of the dividend by simply subtracting them

$$ x^2 -\color{#c00}{1}\cdot(x^2 +1)\ =\ \color{#0a0}{{-1}} $$ $$\Rightarrow\ \ \dfrac{x^2}{x^2+1}\ =\ \color{#c00}1+\dfrac{\color{#0a0}{-1}}{x^2+1}$$

Bill Dubuque
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This is the calculation taking place:

$$\frac{x^2}{x^2 + 1} = \overbrace{\frac{x^2 + 1 -1}{x^2 + 1}}^{Adding\,+1-1} = \underbrace{\frac{x^2 + 1}{x^2 + 1}}_{=1} - \frac{1}{x^2 + 1} = 1 - \frac{1}{x^2 + 1}$$

naslundx
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