Which x_0 to use in a Taylor series expansion?

In summary: For example when f(x) = 1/x, the Taylor series is different depending on where you want to expand it. In summary, when using Taylor series, the coefficients and thus the resulting series will change when different x0 points are used. This is because the coefficients are related to the derivatives of increasing order of the function at the point x0. For polynomials, the power series will always be the polynomial if you include enough terms. However, for non-polynomial functions, the Taylor series will be different depending on the chosen x0 point.
  • #1
morenopo2012
8
0
I already learn to use Taylor series as:

f(x) = ∑ fn(x0) / n! (x-x0)n

But i don´t see why the serie change when we use differents x0 points.

Por example:

f(x) = x2

to express Taylor series in x0 = 0

f(x) = f(0) + f(0) (x-0) + ... = 0 due to f(0) = (0)2

to x0=1 the series are totally differents

f(x) = 1 + 2(x-1) + (x-1)2 this is the parabola

Why happens this?

and, when we use the Taylor series, which situations we said that we can despreciate some terms, like cuadratic terms?
 
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  • #2
If you calculate both Taylor expansions, you get ##f(x)=x^2## in both cases. I cannot see, whether you did this right as you skipped the term ##\frac{1}{2!}f''(0)\cdot (x-0)^2 =x^2## in the first case. Differentiation is a local property, and a local approximation. Thus it makes sense to study local behavior as in general - other than your example - functions might behave differently at different locations. This becomes especially interesting, if the radius of convergence isn't infinite.
 
  • #3
A Taylor series is a series in powers of x - a. If you change a, the coefficients will change as well.
If we look at your example of f(x) = x2, the Taylor series in powers of x - 0 (the Maclaurin series), is
##x^2 = 0 + 0x + 1x^2##
Here the coefficients are ##a_0 = 0, a_1 = 0##, and ##a_2 = 1##.

If we look at the Taylor series in powers of x - 2, we have ##x^2 = 4 + 4(x - 2) + 1(x - 2)^2##. The coefficients of this Taylor series are ##a_0 = 4, a_1 = 4##, and ##a_2 = 1##.

If the Taylor series is expanded around some other number, you'll get different values for ##a_0## and ##a_1##, but you'll always get ##a_2 = 1##.
 
  • #4
For polynomials, the power series will always be the polynomial if you include enough terms (n+1 if the polynomial has degree n). There are other functions where this is not true, e.g. f(x)=1/x.
 
  • #5
The coefficients of the Taylor series are related to the derivatives of increasing order of the function at the point x0. When you change x0, all the derivative values can change and all the coefficients change.
 
  • #6
morenopo2012 said:
Por example:

f(x) = x2

to express Taylor series in x0 = 0

f(x) = f(0) + f(0) (x-0) + ... = 0 due to f(0) = (0)2

Instead of that, you should have:

##f(x) = x^2|_{x=0} + (2x)|_{x=0} (x-0) + (2)|_{x=x0} (x-0)^2/2 ##
## = 0 + 0 + x^2 ##
to x0=1 the series are totally differents

f(x) = 1 + 2(x-1) + (x-1)2 this is the parabola

Instead of that, you should have
##f(x) = 1 + 2(x-1) + 2(x-1)^2/2 ##
##= 1 + 2x - 2 + (x^2 - 2x + 1)##
## = x^2 ##

To illustrate you question, you need to pick a function f(x) that is not a polynomial.
 

1. What is x_0 in a Taylor series expansion?

X_0, also known as the center of the Taylor series, is the point around which the series is being expanded. It is usually the value for which the function has known values and is used as a starting point for the series.

2. Why is x_0 important in a Taylor series expansion?

X_0 plays a crucial role in a Taylor series expansion as it determines the accuracy and convergence of the series. Choosing the right value for x_0 can result in a more accurate approximation of the original function.

3. How do I choose the value for x_0 in a Taylor series expansion?

The choice of x_0 depends on the function being expanded and the desired level of accuracy. In some cases, it may be the midpoint of the interval, while in others it may be a specific value that makes the calculations easier. It is important to choose a value that is close to the point at which the function is being evaluated.

4. Can I use any value for x_0 in a Taylor series expansion?

No, the value for x_0 should be within the interval of convergence of the Taylor series. If x_0 is outside this interval, the series will not accurately approximate the function. It is also important to avoid choosing a value that would result in division by zero or other mathematical errors.

5. How does changing the value of x_0 affect the Taylor series expansion?

Changing the value of x_0 will change the coefficients of the Taylor series and therefore, the resulting approximation of the function. Choosing a value that is closer to the point of interest can result in a more accurate approximation, while choosing a value further away may result in a less accurate approximation or even a divergent series.

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