Series representation of a radical function without using taylor's expansion

Calculus Level 3

We can use taylor series expansion to find an infinite series representation of radical functions. However can you find an infinite series representation of 1 1 − x 2 \displaystyle \frac{1}{\sqrt{1-x^{2}}} without using taylor series? If this series can be represented as ∑ n = 0 ∞ ( − 1 ) n x A n ( 1 − x 2 ) B n + C D \displaystyle \sum_{n=0}^{\infty} \displaystyle\frac{\displaystyle\left(-1\right)^{n} x^{An}}{\left(1-x^{2}\right)^{ Bn +\frac{C}{D}}}

and C , D C, D are relatively prime find A + B + C + D A+B+C+D


The answer is 8.

Amal Hari by Amal Hari , 22, Canada

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1 solution

Amal Hari
Dec 4, 2019

We know ∫ 1 1 − x 2 d x = arcsin ⁡ x \displaystyle \int \frac{1}{\sqrt{1-x^{2}}} dx =\arcsin x

and , arctan ⁡ x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 2 n + 1 \arctan x =\displaystyle \sum_{n=0}^{\infty} \displaystyle\frac{\displaystyle\left(-1\right)^{n} x^{2n+1}}{2n+1}

Now arcsin ⁡ x = arctan ⁡ ( x 1 − x 2 ) \arcsin x=\arctan \left(\displaystyle \frac{x}{\sqrt{1-x^{2}}}\right)

substitute this to arctan ⁡ \arctan series expansion

arcsin ⁡ x = arctan ⁡ ( x 1 − x 2 ) = ∑ n = 0 ∞ ( − 1 ) n ( x 1 − x 2 ) 2 n + 1 2 n + 1 \arcsin x=\arctan \left(\displaystyle \frac{x}{\sqrt{1-x^{2}}}\right) =\displaystyle \sum_{n=0}^{\infty} \displaystyle\frac{\displaystyle\left(-1\right)^{n} \left(\displaystyle \frac{x}{\displaystyle \sqrt{1-x^{2}}}\right) ^{2n+1}}{2n+1}

arcsin ⁡ x = ∑ n = 0 ∞ ( − 1 ) n x 2 n + 1 ( 1 − x 2 ) n + 1 2 ( 2 n + 1 ) \arcsin x=\displaystyle \sum_{n=0}^{\infty} \displaystyle\frac{\displaystyle\left(-1\right)^{n} x^{2n+1}}{\left(1-x^{2}\right)^{n +\frac{1}{2}}\left(2n+1\right)}

we differentiate this to obtain 1 1 − x 2 \frac{1}{\sqrt{1-x^{2}}} on left side and on right side we will get ∑ n = 0 ∞ ( − 1 ) n x 2 n ( 1 − x 2 ) n + 3 2 \displaystyle \sum_{n=0}^{\infty} \displaystyle\frac{\displaystyle\left(-1\right)^{n} x^{2n}}{\left(1-x^{2}\right)^{n+\frac{3}{2}}} after simplification

0 Helpful 0 Interesting 1 Brilliant 0 Confused

This is not a series representation of the given function in the essence of the term. A series representation of a function f ( x ) f(x) gives a series formed by its argument x x . At the most it can be called an identity.

A Former Brilliant Member - 1 year, 6 months ago

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A series is a sum of a sequence, a function may be expressed as series if they are equal. I don't think that it has to be derived from the function itself as long as they are equivalent.

Amal Hari - 1 year, 6 months ago

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