Question

What is the arc length of `f(x)=x^2/(4-x^2)` on `x in [-1,1]`?

Answer 1

`L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))` units.

Explanation:

`f(x)=x^2/(4-x^2)=1-4/(4-x^2)`

`f'(x)=(8x)/(4-x^2)^2`

Arc length is given by:

`L=int_-1^1sqrt(1+(64x^2)/(4-x^2)^4)dx`

For `x in [-1,1]`, `(64x^2)/(4-x^2)^4<1`. Take the series expansion of the square root:

`L=int_-1^1sum_(n=0)^oo((1/2),(n))((64x^2)/(4-x^2)^4)^ndx`

Isolate the `n=0` term and simplify:

`L=int_-1^1dx+sum_(n=1)^oo((1/2),(n))4^(3n)int_-1^1x^(2n)/(4-x^2)^(4n)dx`

Rearrange:

`L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n)(1-1/4x^2)^(-4n)dx`

For `x in [-1,1]`, `1/4x^2<=1`. Take another series expansion:

`L=2+sum_(n=1)^oo((1/2),(n))1/4^nint_-1^1x^(2n){sum_(m=0)^oo((-4n),(m))(-1/4x^2)^m}dx`

Simplify:

`L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)int_-1^1x^(2n+2m)dx`

Integrate directly:

`L=2+sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/4^(n+m)[x^(2n+2m+1)]_-1^1/(2n+2m+1)`

Insert the limits of integration and simplify:

`L=2+2sum_(n=1)^oosum_(m=0)^oo((1/2),(n))((-4n),(m))(-1)^m/((2n+2m+1)*4^(n+m))`