1

If $f$ and $g$ have the Taylor expansions given by

$$f(x) = f_{0} + f_1x + f_2x^2 + \mathcal{O}(x^3), $$

$$g(x) = g_0 + g_1x + g_2x^2 + \mathcal{O}(x^3), $$

is it possible to find the Taylor expansion of the product $f(x)g(x)$ up to $\mathcal{O}(x^3)$?

I was thinking of multiplying termwise, like this:

$$f(x)g(x) = f_0g_0 + f_1g_1x + f_2g_2x^2 + \mathcal{O}(x^3), $$

but I'm not so sure if that's allowed.

  • It doesn't work like that, you have to distribute according to the powers of x – Pink Panther Mar 14 '19 at 12:05
  • Potentially helpful (look at the formula for $c_n$ there in the top answer, equation $(*)$): https://math.stackexchange.com/questions/106649/multiplying-taylor-series-and-composition. Basically, you just multiply like you would polynomials. The formula you proposed unfortunately cannot be right, because it fails for polynomials for instance. – Minus One-Twelfth Mar 14 '19 at 12:08

1 Answers1

2

This is just a simple calculation, i.e. $$f(x)g(x)=f_0g_0 + (f_0g_1+f_1g_0)x+(f_0g_2+f_1g_1+f_2g_0)x^2+O(x^3)$$

Pink Panther
  • 841
  • How'd you get the $x^2$ term? –  Mar 14 '19 at 12:30
  • Shouldn't it be $2f_{1}g_{1}$? –  Mar 14 '19 at 12:35
  • Why would it be $2f_1g_1$? You just have to look at $f$ and $g$ and distribute with standard multiplication rules (it is a field axiom, and i assume your functions have coefficients in $\mathbb R$, or $\mathbb C$) – Pink Panther Mar 14 '19 at 15:03