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Find power series expansion for :

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

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

Guest Nov 9, 2014

Best Answer 

 #2
avatar+26745 
+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) = 21/2

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) - x2/(8*23/2) ...  

(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 - x2)-1/2 to start with.

.

Alan  Nov 10, 2014
 #1
avatar+92775 
0

Alan, if you are in the mood I thing that this one is for you. :)

Melody  Nov 10, 2014
 #2
avatar+26745 
+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) = 21/2

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) - x2/(8*23/2) ...  

(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 - x2)-1/2 to start with.

.

Alan  Nov 10, 2014
 #3
avatar+92775 
0

Thank you Alan:)

Melody  Nov 10, 2014

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