How do you use the binomial series to expand $1/(1-x^2)^(1/2)$ ?

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Answer 1 · 2016-01-05T22:05:32

$1 + 1/2 x^2 + 3/8 x^4 +15/48 x^6$

Rewrite $1/(1 - x^2 )^(1/2)$ as

$(1 - x^2)^(-1/2)$

since there is a negative index the only formula that can be used is

$1 + na +( n(n - 1 ))/(2!) a^2 +( n(n- 1)(n - 2 ))/(3! )a^3 + ...$

here $n = -1/2 , a = - x^2$

≣ 1 + $(-1/2 )(- x^2 ) +( (- 1/2)(- 3/2))/(2!) (- x^2 )^2 + ((-1/2)(-3/2)(-5/2)) /(3!) (-x^2)^3$

be very careful when expanding.

$≣ 1 + 1/2 x^2 + 3/8 x^4 + 15/48 x^6$

How do you use the binomial series to expand 1/(1-x^2)^(1/2)1/(1-x^2)^(1/2)?