Are there power series for X^2

1. May 1, 2010

Mabbott608

title is pretty much the jist of it.

2. May 1, 2010

Mute

x^2 is pretty much the power series for x^2. A power series (about zero) is defined by

$$f(x) = \sum_{n=0}^\infty a_n x^n$$

x^2 is just a power series where all of the a_n except a_2 are zero.

Of course, that's not the whole story. That's for a power series about the point x = 0. About some arbitrary point x_0 a power series is defined by

$$f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n$$

In this case then we can write a slightly non-trivial power series for x^2. Noting that x = x - x_0 + x_0, and expanding (x-x_0+x_0)^2 in terms of (x-x_0) will get you the power series.

3. May 2, 2010

HallsofIvy

The power series for $x^2$ about x= 0 (its MacLaurin series) is just $0+ 0x+ 1x^2+ 0x^2+ \cdot\cdot\cdot= x^2$ itself. To find its power series about x= a (the general Taylor's series), let u= x- a. Then the power series for x, about x= a, is the power series for u about u= a- a= 0, $u^2$. And since u= x- a, the power series for $x^2$ about x= a is $(x- a)^2= x^2- 2ax+ a^2$

Last edited by a moderator: May 2, 2010
4. May 2, 2010

JSuarez

Any polynomial in the form:

$$a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x + a_0$$

Its own unique power series (around 0, in this case).

5. May 2, 2010

Mute

Careful, technically what you've given is the power series of $(x-a)^2$ about the point x = 0. It is not the power series of x^2 about x = a. That should look like const + A*(x-a) + B*(x-a)^2.

To the OP, to get the power series of x^2 about the point x = a, take the form HallsofIvy gives and rearrange it to give

$$x^2 = -a^2 + 2ax + (x-a)^2.$$

Now, fiddle with the first to terms of the right hand side of the equation to make them look a constant plus another constant*(x-a).

6. May 2, 2010

Anonymous217

You could also do e^x and then e^(2lnx) as a series, which would converge to x^2.

7. May 2, 2010

g_edgar

So that would be a power series in log x, and not a power series in x.
I guess the OP was not specific, so this, too, answers the question.