Very very basic taylor series problem

PTTB
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Homework Statement



Consider f(x) = 1 + x + 2x^2+3x^3.
Using Taylor series approxomation, approximate f(x) arround x=x0 and x=0 by a linear function

Homework Equations





The Attempt at a Solution



This is the first time that I have seen Taylor series and I am totally lost on how to do it, I have looked all around the internet for some help and I just don't have a clue on where to start. Any Help would be appreciated
 
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Well start from the definition of a taylor series:
The taylor series of a function centered around a is given by
\sum_{n=0}^{\infty} f^n (0) \frac{(x-a)^n}{n!} where f^n is the nth derivative of f, not the nth power. The 0th derivative is just the original function.

So it wants you to find 2 series, one with center 0, or a=0, and the other with centre a=x_0. Use the formula to get the taylor series centered around those points, and then since it only wants a linear function, only use the first 2 terms, ie the constant and the x terms.
 
thanks a ton, i think I am starting to get this stuff, you were a huge help

i got 1+x and 1+x-2x_o^2-6x_0^3+(4x_0)x+(9x^2_0)x I am pretty sure that's rite, thanks a ton
 
PTTB said:
thanks a ton, i think I am starting to get this stuff, you were a huge help

i got 1+x and 1+x-2x_o^2-6x_0^3+(4x_0)x+(9x^2_0)x I am pretty sure that's rite, thanks a ton
Actually, it makes little sense to say "Taylor series approximation". The Taylor series is exact. It is the "Taylor polynomial" that is approximate and it is my guess that this is what is intended. Of course, the Taylor Polynomial that gives a linear approximation is just the 1st order approximation: f(x0)+ f '(x0)(x- x0)- and that's just the tangent line approximation.
Since f(x_0)= 1+ x_0+ 2x_0^2+ 3x_0^3 and f '(x_0)= 1+ 4x_0+ 9x_0^2, the tangent line approximation at x_0 is 1+ x_0+ 2x_0^2+ 3x_0^3+ (1+ 4x_0+ 9x_0^3)(x- x_0). After multiplying out the last term, that gives exactly what you have.
 
HallsofIvy said:
The Taylor series is exact.

*Grumble*
*Mutterings about analytic functions and radius of convergence*
*Grumble*
 
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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