Creating a power series using long division
- Thread starter burt
- Start date
divide 
[MATH]\dfrac{1}{1-x^2} = 1 \text{ with remainder $x^2$}\\ \dfrac{x^2}{1-x^2} = x^2 \text{ with remainder $x^4$}\\ \dfrac{x^4}{1-x^2} = x^4 \text{ with remainder $x^6$}\\ \text{etc}[/MATH]
looks like you'll end up with [MATH]\dfrac{1}{1- x^2} = \sum \limits_{k=1}^\infty x^{2k}[/MATH]
[MATH]\dfrac{1}{1-x^2} = 1 \text{ with remainder $x^2$}\\ \dfrac{x^2}{1-x^2} = x^2 \text{ with remainder $x^4$}\\ \dfrac{x^4}{1-x^2} = x^4 \text{ with remainder $x^6$}\\ \text{etc}[/MATH]
looks like you'll end up with [MATH]\dfrac{1}{1- x^2} = \sum \limits_{k=1}^\infty x^{2k}[/MATH]
So \(\frac{1}{1-x^2}=x^2+x^4+x^6...\)?divide
[MATH]\dfrac{1}{1-x^2} = 1 \text{ with remainder $x^2$}\\ \dfrac{x^2}{1-x^2} = x^2 \text{ with remainder $x^4$}\\ \dfrac{x^4}{1-x^2} = x^4 \text{ with remainder $x^6$}\\ \text{etc}[/MATH]
looks like you'll end up with [MATH]\dfrac{1}{1- x^2} = \sum \limits_{k=1}^\infty x^{2k}[/MATH]
D
Deleted member 4993
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Check your answer using Taylor's expansion.
D
Deleted member 4993
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NO!
You lost the leading 1 of the series!! I was hoping you would catch it by now.......
Also the expansion would be valid only when x < 1 (otherwise the summation will not converge and the series cannot be terminated).
I did
Thank you for correcting me though!! Once I add in that one, I should be fine by my calculations.