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?Originally Posted by fair_lady0072002
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 ???????
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.