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Power series expansion

  • #1
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Homework Statement


Show that ##\displaystyle \frac{1}{1+x^2} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \frac{1}{x^8} + \cdots##

Homework Equations




The Attempt at a Solution


I know that the power series expansion of ##\displaystyle \frac{1}{1+x^2}## about ##x=0## is ##1-x^2 + x^4 - x^6 + \cdots##, but I don't see how this can translate into the expressions as above.
 

Answers and Replies

  • #2
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8,946
Substitute ##x=\frac{1}{y}## and write everything in terms of ##\frac{1}{y}##. At the end, re-substitute ##\frac{1}{y}=x##.
 
  • #3
1,456
44
Substitute ##x=\frac{1}{y}## and write everything in terms of ##\frac{1}{y}##. At the end, re-substitute ##\frac{1}{y}=x##.
I don't follow. If I do that then ##\frac{1}{1+1/y^2} = 1-1/y^2 + 1/y^4 - 1/y^6 + \cdots##. I don't see where that gets me. Maybe I am missing what you mean by "write everything in terms of 1/y".
 
  • #4
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8,946
I hope, I didn't make a mistake this time:
$$
\frac{1}{1+\frac{1}{y^2}}=\frac{y^2}{1+y^2}=y^2 \cdot (1-y^2+y^4-+ \ldots) = y^2 - y^4 +y^6-+\ldots = \frac{1}{\frac{1}{y^2}}-\frac{1}{\frac{1}{y^4}}+\frac{1}{\frac{1}{y^6}}-+\ldots
$$
 
  • #5
33,176
4,859

Homework Statement


Show that ##\displaystyle \frac{1}{1+x^2} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \frac{1}{x^8} + \cdots##

Homework Equations




The Attempt at a Solution


I know that the power series expansion of ##\displaystyle \frac{1}{1+x^2}## about ##x=0## is ##1-x^2 + x^4 - x^6 + \cdots##, but I don't see how this can translate into the expressions as above.
If you use long division to divide 1 by ##x^2 + 1## you'll get the expansion you show just above. Instead, if you divide 1 by ##1 + x^2## you'll get the expansion at the top of this post.
 
  • #6
vela
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement


Show that ##\displaystyle \frac{1}{1+x^2} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \frac{1}{x^8} + \cdots##

Homework Equations




The Attempt at a Solution


I know that the power series expansion of ##\displaystyle \frac{1}{1+x^2}## about ##x=0## is ##1-x^2 + x^4 - x^6 + \cdots##, but I don't see how this can translate into the expressions as above.
That expansion only converges for ##x^2<1##. You need a different expansion when ##x^2>1##.

The way I'd do it is to factor ##x^2## out of the denominator:
$$\frac{1}{1+x^2} = \frac{1}{x^2[(1/x^2)+1]} = \frac{1}{x^2}\left[\frac{1}{1+(1/x^2)}\right]$$
Then expand the term in the square brackets in powers of ##(1/x^2)## the usual way. When ##x^2>1##, you know that ##(1/x^2)<1## so the series will converge.
 
  • #7
pasmith
Homework Helper
1,735
410
That expansion only converges for ##x^2<1##. You need a different expansion when ##x^2>1##.
And the implication of a series in [itex]x^{-2}[/itex] is that you are not looking near [itex]x = 0[/itex].

The way I'd do it is to factor ##x^2## out of the denominator:
$$\frac{1}{1+x^2} = \frac{1}{x^2[(1/x^2)+1]} = \frac{1}{x^2}\left[\frac{1}{1+(1/x^2)}\right]$$
Then expand the term in the square brackets in powers of ##(1/x^2)## the usual way. When ##x^2>1##, you know that ##(1/x^2)<1## so the series will converge.
 

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