• Jul 29th 2006, 03:08 PM
given

f(x) = 36/(6+x)^2 and g(x) = 36/(1+x)^2

By writing f(x) = 1/(1+1/6x)^2

I found the Taylor series about 0 for f up to term in x^3,

= 1 - 1/3x + 1/12x^2-1/54X^3 .........(a)

Valid for -6<x<6

Then g(x) can be written as 1/(1+(x-5)/6)^2

Replace x in solution 1 by x-5 ???????

Solution

1 - 1/3(x-5) + 1/12(x-5)^2-1/54(X-5)^3 ......(b)...?????

Valid for -1<x<11

Not sure how to check the first four terms in the Taylor seris found in part b by finding the first, second and third derivatives of g and using these to find the cubic Taylor polynomial about 5 for g.

Thanx a lot.
• Jul 29th 2006, 06:24 PM
ThePerfectHacker
Quote:

given

f(x) = 36/(6+x)^2 and g(x) = 36/(1+x)^2

By writing f(x) = 1/(1+1/6x)^2

I found the Taylor series about 0 for f up to term in x^3,

= 1 - 1/3x + 1/12x^2-1/54X^3 .........(a)

Valid for -6<x<6

Then g(x) can be written as 1/(1+(x-5)/6)^2

Replace x in solution 1 by x-5 ???????

Solution

1 - 1/3(x-5) + 1/12(x-5)^2-1/54(X-5)^3 ......(b)...?????

Valid for -1<x<11

Not sure how to check the first four terms in the Taylor seris found in part b by finding the first, second and third derivatives of g and using these to find the cubic Taylor polynomial about 5 for g.

Thanx a lot.

It looks like a right substitution to me. Though the second series is a Taylor's series for 'g' centered at x=5. Do you want to "uncenter" it?
• Aug 4th 2006, 06:15 AM
kendon
Is the range of validity in the first post correct for the second series. Should it not be -11>x>1? Sorry if thats what you are hinting at in the previous post.
Just keen to learn!