Is There a Power Series for X^2?

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Discussion Overview

The discussion centers around the existence and formulation of power series for the function x^2, exploring different points of expansion, including around zero and an arbitrary point a. Participants examine definitions, provide examples, and discuss the implications of these series in various contexts.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that x^2 can be represented as a power series where all coefficients except a_2 are zero when expanded about zero.
  • Others explain that a power series about an arbitrary point x_0 can be derived by expanding (x - x_0 + x_0)^2 in terms of (x - x_0).
  • A participant provides the Maclaurin series for x^2, stating it is simply x^2 itself, and discusses the Taylor series expansion around a point a, leading to the expression (x - a)^2 = x^2 - 2ax + a^2.
  • Another participant mentions that any polynomial has its own unique power series around zero.
  • Some participants clarify that the expression provided for the power series around a should be rearranged to fit the standard form of a power series, emphasizing the need for a constant plus a linear term in (x - a).
  • One participant introduces an alternative approach using the exponential function and logarithms, suggesting that e^(2lnx) could also converge to x^2, although this would not be a power series in x.
  • Another participant reiterates the previous point about using e^(2lnx) as a series, noting that the original question was not specific enough to exclude this method.

Areas of Agreement / Disagreement

Participants express differing views on the formulation of the power series for x^2, particularly regarding the correct representation around different points. There is no consensus on a single method or expression, and multiple approaches are presented.

Contextual Notes

Some discussions involve assumptions about the definitions of power series and the conditions under which they are derived, which may not be explicitly stated. The dependence on the choice of expansion point also introduces complexity that remains unresolved.

Who May Find This Useful

This discussion may be of interest to those studying mathematical series, particularly in the context of calculus and polynomial functions, as well as individuals exploring different methods of series expansion.

Mabbott608
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title is pretty much the jist of it.
 
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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.
 
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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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).
 
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).
 
You could also do e^x and then e^(2lnx) as a series, which would converge to x^2.
 
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.
 

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