Very very basic taylor series problem

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Homework Help Overview

The problem involves approximating the function f(x) = 1 + x + 2x^2 + 3x^3 using Taylor series around the points x = x0 and x = 0, specifically seeking a linear function as the approximation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the definition of the Taylor series and its application to the given function, with one participant expressing confusion about starting the problem. Others suggest using the Taylor series formula and focusing on the first two terms for a linear approximation.

Discussion Status

Some participants have begun to articulate their understanding of the Taylor series and polynomial approximations, while others are clarifying terminology and the nature of the approximation. There is an acknowledgment of the exactness of the Taylor series versus the approximation aspect of the Taylor polynomial.

Contextual Notes

One participant notes the distinction between "Taylor series approximation" and "Taylor polynomial," suggesting that the latter is what is intended in the problem context. There is also mention of the function's derivatives and their role in forming the linear approximation.

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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
[tex]\sum_{n=0}^{\infty} f^n (0) \frac{(x-a)^n}{n!}[/tex] 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 [itex]f(x_0)= 1+ x_0+ 2x_0^2+ 3x_0^3[/itex] and [itex]f '(x_0)= 1+ 4x_0+ 9x_0^2[/itex], the tangent line approximation at [itex]x_0[/itex] is [itex]1+ x_0+ 2x_0^2+ 3x_0^3+ (1+ 4x_0+ 9x_0^3)(x- x_0)[/itex]. 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*
 

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