Is There a Power Series for X^2?

In summary: A power series (about zero) is defined byf(x) = \sum_{n=0}^\infty a_n x^nx^2 is just a power series where all of the a_n except a_2 are zero. However, if you want the power series of x^2 about the point x = a, you need to take the form given by HallsofIvy and rearrange it to givex^2 = -a^2 + 2ax + (x-a)^2.
  • #1
Mabbott608
6
0
title is pretty much the jist of it.
 
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  • #2
x^2 is pretty much the power series for x^2. A power series (about zero) is defined by

[tex]f(x) = \sum_{n=0}^\infty a_n x^n[/tex]

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

[tex]f(x) = \sum_{n=0}^\infty a_n (x-x_0)^n[/tex]

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
The power series for [itex]x^2[/itex] about x= 0 (its MacLaurin series) is just [itex]0+ 0x+ 1x^2+ 0x^2+ \cdot\cdot\cdot= x^2[/itex] 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, [itex]u^2[/itex]. And since u= x- a, the power series for [itex]x^2[/itex] about x= a is [itex](x- a)^2= x^2- 2ax+ a^2[/itex]
 
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  • #4
Any polynomial in the form:

[tex]a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots +a_{1}x + a_0[/tex]

Its own unique power series (around 0, in this case).
 
  • #5
HallsofIvy said:
And since u= x- a, the power series for [itex]x^2[/itex] about x= a is [itex](x- a)^2= x^2- 2ax+ a^2[/itex]

Careful, technically what you've given is the power series of [itex](x-a)^2[/itex] 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

[tex]x^2 = -a^2 + 2ax + (x-a)^2.[/tex]

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
You could also do e^x and then e^(2lnx) as a series, which would converge to x^2.
 
  • #7
Anonymous217 said:
You could also do e^x and then e^(2lnx) as a series, which would converge to x^2.

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.
 

FAQ: Is There a Power Series for X^2?

What is a power series?

A power series is an infinite series of the form ∑n=0∞ an(x-a)n, where a is a constant and an is a sequence of coefficients. It is a useful mathematical tool for representing functions as a sum of polynomial terms.

How do you determine if a function has a power series representation?

A function has a power series representation if it can be expressed as an infinite sum of polynomial terms. This can be determined by checking if the function is continuous and differentiable at a point and if its derivatives have a pattern that can be expressed using the coefficients of a power series.

Can any function be represented by a power series?

No, not all functions can be represented by a power series. The function must be continuous and differentiable at a point in order to have a power series representation.

How do you find the coefficients of a power series?

The coefficients of a power series can be found using the formula an = f(n)(a)/n!, where f(n)(a) represents the nth derivative of the function at the point a. Alternatively, the coefficients can also be found by using the Taylor series expansion of the function.

Are there power series for X^2?

Yes, the power series representation for X^2 is ∑n=0∞ (x-0)n/n! = ∑n=0∞ xn/n!. This power series converges for all values of x and represents the function x^2 for |x| < 1.

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