# I 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?

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#### fresh_42

Mentor
2018 Award
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

Mentor
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

Mentor
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

Gold Member
2018 Award
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.