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The author of my book writes:

$$x^2+1=x^2\left(1+\frac{1}{x^2}\right)$$

$$=\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]$$

I do not understand the last step. How did the author write the last step. Please help.

M.S.E
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    I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – AlexR Jan 20 '15 at 12:50
  • @AlexR thanks for letting me know – M.S.E Jan 20 '15 at 12:57
  • Are you sure that this equation holds? – Alex Silva Jan 20 '15 at 13:26
  • Take $x = 1/2$. $\text{LHS} = 5/4$ and $\text{RHS} \rightarrow \infty$. – Alex Silva Jan 20 '15 at 13:29

4 Answers4

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I don't think they are equal. As others have said $$ \frac{x^2}{1+x^2} = 1 - \frac{1}{x^2} + \frac{1}{x^4} \dots \text{ for } \left\lvert \frac{-1}{x^2} \right\rvert < 1 $$

so for $x = \sqrt{2}$
$$ \frac{1}{x^2} \left( 1 - \frac{1}{x^2} + \frac{1}{x^4} \dots \right) = \frac{1}{2} \left( \frac{2}{3} \right) = \frac{1}{3}. $$

However $x^2+1 = 3 \neq \frac{1}{3}$.

Ailbe
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  • (+1) Thank you so much for taking me out of the indescribable frustration I have been for the past hour. – M.S.E Jan 20 '15 at 13:32
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I believe your author intended to write $$ \frac{1}{x^2+1}=\frac{1}{x^2\left(1+\frac{1}{x^2}\right)} \\ =\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right] $$

GEdgar
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Formula is not correct. Take the sum of a geometric series:

$$\sum_{i=0}^{\infty} ar^i=\frac{a}{1-r}$$

Lets work backward from your last equation:

$$\frac{1}{x^2}\left[1-\frac{1}{x^2}+\frac{1}{x^4}-\frac{1}{x^6}+\cdots\right]=\frac{1}{x^2}\left[\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i-x^2\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i+x^2\right]=$$

The two sums in the brackets are simply geometric sums with $r=\frac{1}{x^4}$, thus:

$$\left[\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i-x^2\sum_{i=0}^{\infty} \left(\frac{1}{x^{4}}\right)^i+x^2\right]=\left[\frac{1}{1-\frac{1}{x^4}}-\frac{x^2}{1-\frac{1}{x^4}} +x^2\right]=\left[\frac{1-x^2+(1-\frac{1}{x^4})x^2}{1-\frac{1}{x^4}}\right]$$

The last equation simplifies to:

$$\left[\frac{1-\frac{1}{x^2}}{1-\frac{1}{x^4}}\right]=\frac{x^2}{1+x^2}\implies x^2+1=\frac{1}{x^2+1}???$$ This is not correct

Community
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You may write $$x^2 + 1 = x^2(1 + \frac{1}{x^2})$$

as $$\frac{1}{(1 + \frac{1}{x^2})} = \frac{x^2}{1 + x^2}$$

and $$\frac{1}{1 + \frac{1}{x^2}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{x^{2n}}$$

Edit: To avoid any more downvotes, the author has made a mistake, clearly the equation does not hold, this was just an attemptive to reinterpret whatever the author's intention was.

Aaron Maroja
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  • I still don't get it :/ My bracket is $(1+\frac{1}{x^2})$ and not $\frac{1}{1+\frac{1}{x^2}}$ – M.S.E Jan 20 '15 at 12:56
  • I saw the edit . Your formula works ,except that it's confusing that my fraction is the inverse of yours. Mine is $\frac{1+x^2}{x^2}$ while yours is $\frac{x^2}{1+x^2}$ – M.S.E Jan 20 '15 at 13:01
  • (+1) for the continued support. I'm sorry I don't get it, let me try to explain what's worrying me, So $\frac{x^2}{1+x^2}=\sum_{n=0}^{\infty} \frac{(-1)^n}{x^{2n}} \implies \frac{1+x^2}{x^2} = \frac{1}{\sum_{n=0}^{\infty} \frac{(-1)^n}{x^{2n}}}$ which means $1+x^2=\frac{x^2}{\sum_{n=0}^{\infty} \frac{(-1)^n}{x^{2n}}} \neq \frac{1}{x^2} [1-1/x^2+1/x^4+\cdots]$ – M.S.E Jan 20 '15 at 13:10
  • Your equation is wrong then. No need to downvote. – Aaron Maroja Jan 20 '15 at 13:26
  • That is what I'm surprised about too, apparently the author and you and others are displaying workings that don't make sense. – M.S.E Jan 20 '15 at 13:30
  • What we wrote holds. It must have been a typing or something from the author. – Aaron Maroja Jan 20 '15 at 13:32