#2**+5 **

A Taylor series expansion of f(x) about point x=a can be written as:

$$f(x)=f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+...$$

If we set a = 0 here:

f(x) = (2-x)^{1/2} f(0) = 2^{1/2}

f'(x) = -(1/2)(2-x)^{-1/2} f'(0) = -(1/2)2^{-1/2}

f''(x) = -(1/4)(2-x)^{-3/2} f''(0) = -(1/4)2^{-3/2}

... etc.

so

f(x) = √2 - x/(2√2) - x^{2}/(8*2^{3/2}) ...

(this could be tidied up by shifting all the √2s in the denominators to the numerators - by multiplying by √2/√2)

Similarly for the other one, which is best written as f(x) = (1 - x^{2})^{-1/2} to start with.

.

Alan
Nov 10, 2014

#2**+5 **

Best Answer

A Taylor series expansion of f(x) about point x=a can be written as:

$$f(x)=f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+...$$

If we set a = 0 here:

f(x) = (2-x)^{1/2} f(0) = 2^{1/2}

f'(x) = -(1/2)(2-x)^{-1/2} f'(0) = -(1/2)2^{-1/2}

f''(x) = -(1/4)(2-x)^{-3/2} f''(0) = -(1/4)2^{-3/2}

... etc.

so

f(x) = √2 - x/(2√2) - x^{2}/(8*2^{3/2}) ...

(this could be tidied up by shifting all the √2s in the denominators to the numerators - by multiplying by √2/√2)

Similarly for the other one, which is best written as f(x) = (1 - x^{2})^{-1/2} to start with.

.

Alan
Nov 10, 2014