Taylor Series help?

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Homework Statement


Have to find the Taylor series for (1-x)^(-0.5)
Then use this to find the Taylor series for (1-x^2)^(-0.5)

Homework Equations





The Attempt at a Solution


Was able to do the expansion for the first one quite easily, but not sure how to do the second one. My initial thought was to multiply the first expansion with the expansion for (1+x)^(-0.5)
since (1-x^2) = (1-x)(1+x)
But this is a one mark question, so this seems like too much working out to be the right way...
Is there a trick to this question? :S
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Have to find the Taylor series for (1-x)^(-0.5)
Then use this to find the Taylor series for (1-x^2)^(-0.5)

Homework Equations



The Attempt at a Solution


Was able to do the expansion for the first one quite easily, but not sure how to do the second one. My initial thought was to multiply the first expansion with the expansion for (1+x)^(-0.5)
since (1-x^2) = (1-x)(1+x)
But this is a one mark question, so this seems like too much working out to be the right way...
Is there a trick to this question? :S
If [itex]\displaystyle f(x)=\frac{1}{\sqrt{1-x}}\,,[/itex] then what is [itex]f(x^2)\ ?[/itex]
 
  • #3
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ah yes that's it, so I would just sub in x^2 to the expansion
thanks!
Would multiplying the two expansions have also worked?
 
  • #4
SammyS
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ah yes that's it, so I would just sub in x^2 to the expansion
thanks!
Would multiplying the two expansions have also worked?
Multiplying which two expansions ?
 
  • #5
Curious3141
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ah yes that's it, so I would just sub in x^2 to the expansion
thanks!
Would multiplying the two expansions have also worked?

Much, much more troublesome. You have to figure out which terms contribute to the coefficient of each term in the final expansion. For example, the [itex]x^3[/itex] term in the product is determined by the sum of the [itex]x^3[/itex] terms of both expressions, as well as the sum of the products of the [itex]x[/itex] term of one with the [itex]x^2[/itex] term of the other (and vice versa). The general term would involve lots of tedious algebra.

It wouldn't be too bad if they just asked you to find the expression correct to the first few terms, though.
 
  • #6
HallsofIvy
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No, squaring the original series would give [itex]\left(1/\sqrt{1- x}\right)^2= 1/\sqrt{1- x}[/itex], not [itex]1/(1- x^2)[/itex].
 

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