Taylor Polynomials: Find a0, a1, a2, a3, and a4

lmannoia
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Homework Statement


Let f(x)=x2 +3x -5, and let the summation (from n=0 to infinity) an (x-4)n be the Taylor series of f about 4. Find the values of a0, a1, a2, a3, and a4.

Homework Equations





The Attempt at a Solution


What am I supposed to do with the summation? And what does it mean to find a Taylor polynomial 'about 4'? I'm confused on how to relate both of those things to the f(x) that they gave me.
 
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lmannoia said:

Homework Statement


Let f(x)=x2 +3x -5, and let the summation (from n=0 to infinity) an (x-4)n be the Taylor series of f about 4. Find the values of a0, a1, a2, a3, and a4.

Homework Equations





The Attempt at a Solution


What am I supposed to do with the summation? And what does it mean to find a Taylor polynomial 'about 4'? I'm confused on how to relate both of those things to the f(x) that they gave me.
You are supposed to write f(x) = x2 + 3x - 5 as a power series in powers of (x - 4) instead of in powers of x. Since the function is of degree two, your Taylor's series will also be of degree two. IOW, it will be f(x) = a0 + a1(x - 4) + a2(x - 4)2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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