Very very basic taylor series problem

In summary, the student attempted to find a solution to a homework equation but was lost due to not knowing how to do Taylor series. The equation was solved using the tangent line approximation.
  • #1
PTTB
2
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
HallsofIvy said:
The Taylor series is exact.

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

FAQ: Very very basic taylor series problem

What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, each with a higher order derivative of the function evaluated at a specific value.

What is the purpose of a Taylor series?

The purpose of a Taylor series is to approximate a function, allowing for easier calculation and analysis of the function's properties.

How is a Taylor series calculated?

A Taylor series is calculated using the Taylor series formula, which involves taking the derivatives of the function at a specific point and plugging them into the formula.

What are some common applications of Taylor series?

Taylor series are commonly used in physics, engineering, and other fields to approximate and analyze various natural phenomena and functions.

Are there any limitations to using a Taylor series?

Yes, Taylor series are only accurate for functions with continuous derivatives and may only provide an approximation of the function within a certain interval. They may also become increasingly complex and difficult to calculate as the order of the series increases.

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