Which x_0 to use in a Taylor series expansion?

[morenopo2012]

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?

 

[fresh_42]

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.

 

[Mark44]

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 $a_2 = 1$.

 

[mfb]

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.

 

[FactChecker]

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.

 

[Stephen Tashi]

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.