Can you make a power series for $y=x^2$?

Question

I tried to make a power series for $y=x^2$ by starting with $f^{-1}(x)=\sqrt{x}$ and applying Lagrange Inversion theorem with $a=1$, but it didn't converge. In fact, the best you could observe from the graph was that $1^2=1$, everywhere else it diverged.

Trying to apply Taylor series on $f(x)=x^2$ results in $f(x)=x^2$, which isn't really anything new.

Are there anyways I can make a power series for a simple polynomial such as $x^2$?

You have it right. $x^2$ is the Taylor series (at 0) for $x^2$. You can move the point your expanding around but you'll still get a polynomial. — Derek Elkins left SE, Jan 29 '16 at 13:28
@DerekElkins So that's it? I can't make a power series for $x^2$ other than $x^2$? :/ — Simply Beautiful Art, Jan 29 '16 at 13:29
@SimpleArt: As Derek says, you can make a power series around a different point than $x=0$, but the only power series around $0$ the function has is $x^2$. (A function cannot have two different power series expansions around the same point). — hmakholm left over Monica, Jan 29 '16 at 13:33
(The power series around $x=a$ will be $a^2(x-a)^0+2a(x-a)+(x-a)^2+0(x-a)^3+\cdots$, of course). — hmakholm left over Monica, Jan 29 '16 at 13:36
I know it's not what you really want, but you can make a Fourier series for $y=x^2$ by taking the domain $-1 \le x \le 1$ and using that to build a periodic sequence. Your series will only be valid for $-1 \le x \le 1$ :-) — tomi, Jan 29 '16 at 15:31
@HenningMakholm Huh. You think that'll converge for more than $x\le a$? — Simply Beautiful Art, Jan 29 '16 at 23:19
@SimpleArt: It converges for all $x$, since all terms with degree above 2 are zero. — hmakholm left over Monica, Jan 30 '16 at 00:21

Can you make a power series for $y=x^2$?

0 Answers0