Basic Expression Expansion

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How can I expand this expression $(1-x^2)^-5/2$ as a series in x?



I'm looking eventually for a series in $x_1$ and $x_2$ where $x$, $x_1$, and $x_2$ are magnitudes of $vecx$,$vecx_1$, and $vecx_2$ and $vecx$ = $vecx_1$ + $vecx_2$




algebra-precalculus




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  • 1




    Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
    – Ross Millikan
    Aug 6 at 13:56










  • Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
    – M. Kans
    Aug 6 at 14:11











  • Alpha agrees with you. You can click more terms to get as many as you want.
    – Ross Millikan
    Aug 6 at 14:30










  • thank you for ur help
    – M. Kans
    Aug 6 at 14:33














up vote
0
down vote

favorite












How can I expand this expression $(1-x^2)^-5/2$ as a series in x?



I'm looking eventually for a series in $x_1$ and $x_2$ where $x$, $x_1$, and $x_2$ are magnitudes of $vecx$,$vecx_1$, and $vecx_2$ and $vecx$ = $vecx_1$ + $vecx_2$




algebra-precalculus




share|cite|improve this question















  • 1




    Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
    – Ross Millikan
    Aug 6 at 13:56










  • Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
    – M. Kans
    Aug 6 at 14:11











  • Alpha agrees with you. You can click more terms to get as many as you want.
    – Ross Millikan
    Aug 6 at 14:30










  • thank you for ur help
    – M. Kans
    Aug 6 at 14:33












up vote
0
down vote

favorite









up vote
0
down vote

favorite











How can I expand this expression $(1-x^2)^-5/2$ as a series in x?



I'm looking eventually for a series in $x_1$ and $x_2$ where $x$, $x_1$, and $x_2$ are magnitudes of $vecx$,$vecx_1$, and $vecx_2$ and $vecx$ = $vecx_1$ + $vecx_2$




algebra-precalculus




share|cite|improve this question











How can I expand this expression $(1-x^2)^-5/2$ as a series in x?



I'm looking eventually for a series in $x_1$ and $x_2$ where $x$, $x_1$, and $x_2$ are magnitudes of $vecx$,$vecx_1$, and $vecx_2$ and $vecx$ = $vecx_1$ + $vecx_2$





algebra-precalculus





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 13:52









M. Kans

1




1







  • 1




    Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
    – Ross Millikan
    Aug 6 at 13:56










  • Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
    – M. Kans
    Aug 6 at 14:11











  • Alpha agrees with you. You can click more terms to get as many as you want.
    – Ross Millikan
    Aug 6 at 14:30










  • thank you for ur help
    – M. Kans
    Aug 6 at 14:33












  • 1




    Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
    – Ross Millikan
    Aug 6 at 13:56










  • Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
    – M. Kans
    Aug 6 at 14:11











  • Alpha agrees with you. You can click more terms to get as many as you want.
    – Ross Millikan
    Aug 6 at 14:30










  • thank you for ur help
    – M. Kans
    Aug 6 at 14:33







1




1




Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
– Ross Millikan
Aug 6 at 13:56




Do you know how to expand $(1-y)^-5/2?$. You can do that with the binomial theorem. Now set $y=x^2$
– Ross Millikan
Aug 6 at 13:56












Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
– M. Kans
Aug 6 at 14:11





Please correct me: $(1-x^2)^-5/2 = 1 +frac52x^2 + frac358x^4+frac10516x^6$
– M. Kans
Aug 6 at 14:11













Alpha agrees with you. You can click more terms to get as many as you want.
– Ross Millikan
Aug 6 at 14:30




Alpha agrees with you. You can click more terms to get as many as you want.
– Ross Millikan
Aug 6 at 14:30












thank you for ur help
– M. Kans
Aug 6 at 14:33




thank you for ur help
– M. Kans
Aug 6 at 14:33















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