Solving Taylor Series: (1-x^2)^(-0.5) Help

In summary: However, multiplying the original series by the series for \sqrt{1+ x} would give a series for 1/\sqrt{1- x^2}.In summary, the conversation discusses finding the Taylor series for (1-x)^(-0.5) and using it to find the Taylor series for (1-x^2)^(-0.5). The person was able to easily find the first series but was unsure of how to find the second one. They considered multiplying the two expansions but realized it would involve tedious algebra. They then discovered that substituting x^2 into the first expansion would give the desired result.
  • #1
dan38
59
0

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
 
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  • #2
dan38 said:

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
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
dan38 said:
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
dan38 said:
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
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].
 

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, each of which is a multiple of a power of the variable x.

2. Why is the Taylor series important?

The Taylor series allows us to approximate complicated functions with simpler ones, making it easier to analyze and solve mathematical problems.

3. How do you solve a Taylor series?

To solve a Taylor series, you need to find the coefficients of the terms in the series. This can be done using the formula for the nth derivative of a function, evaluated at x = 0.

4. What is the Taylor series for (1-x^2)^(-0.5)?

The Taylor series for (1-x^2)^(-0.5) is 1 + x^2 + (3/4)x^4 + (5/8)x^6 + ...

5. How do you use the Taylor series for (1-x^2)^(-0.5) to approximate values?

You can use the Taylor series to approximate values by plugging in the desired value for x into the series and adding up the terms until you reach the desired level of accuracy. The more terms you use, the more accurate the approximation will be.

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