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Consider $$ F(\mathbf{r})=F(x,y,z) = \frac{2z^2 - x^2 - y^2}{(x^2+y^2+z^2)^{5/2}} $$ where $x,y,$ and $z, $ are all $n^{\text{th}}$ order polynomial functions of a parameter $t$ with arbitrary coefficients. I want to approximate $F$ to $n^{\text{th}}$ order accuracy in $t$ with a polynomial function of $t$ on the interval $t\in[0,h]$. Basically, I need to know the best way to do this, I'm thinking it's probably best to find an $n^{\text{th}}$ order approximation as a function of $x, y$ and $z$ and then just sub in whatever polynomials they are, but I don't know. Also, I want to avoid dealing with the derivatives of $F$ if that's at all possible. This might be a bit of a stretch, but I'd also really like the result to be a function of the coefficients of $x, y$ and $z$ and not just true for one specific case, but I'll take what I can get.

Mr. G
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  • Have a look at this http://math.stackexchange.com/questions/415458/how-to-best-approximate-higher-degree-polynomial-in-space-of-lower-degree-polyno/415811#415811 – AnilB Jul 09 '13 at 16:36
  • @OccupyGezi The minimax idea looks like it could work, but what should the guess for $P^{}(x, y, z)$ be? Could you just do, say, $P^{}(x, y, z) = a + bx + cy + dz + ex^2 + fy^2 + gz^2 + \cdots$? – Mr. G Jul 09 '13 at 17:16
  • Actually it depends on you... Your formulation is correct for generic case. – AnilB Jul 09 '13 at 18:35
  • @OccupyGezi Okay, but the issue is that I would have to solve $\nabla(F - P^{*})=\mathbf{0}$, which seems basically impossible ... – Mr. G Jul 09 '13 at 21:57
  • I agree with you; you may use least squares more easily. Do you have the explicit formulations of$x,y,z$ for $t$. – AnilB Jul 10 '13 at 07:46
  • @OccupyGezi They are all fourth order polynomials with arbitrary coefficients, ie. $x(t) = x_{0} + x_{1}t + x_{2}t^2 + x_{3}t^3 + x_{4}t^4$ – Mr. G Jul 10 '13 at 13:53

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Use the trinomial expansion to get the Taylor of $(x^2+y^2+z^2)^{-5/2}$ and multiply by $2z^2-x^2-y^2$.

OR.
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  • Right, but the Taylor series of $F$ at $\mathbf{r}{\circ}$ is only accurate to $n^{\text{th}}$ order at $\mathbf{r}{\circ}$, not necessarily over the whole interval. – Mr. G Jul 09 '13 at 01:43
  • What do you mean by accurate to order $n^{th}$? Order is a local property as far as I know. You are not going to get a polynomial accurate to positive order at every point on an interval since, in particular, that means the value at every point is the same as the function's; and the function is not a polynomial. – OR. Jul 09 '13 at 01:49
  • For example: $p(x):=f(0)+f'(0)x$ is accurate to order $1$ to $f(x)$. In particular $p(0)=f(0)$. If we had that the same $p$ gives you order $1$ approximation in an interval then $p(a)=f(a)$ on that interval. Therefore $f$ would be a polynomial in that interval. – OR. Jul 09 '13 at 01:52
  • Say you want to approximate a function $f(x)$ with a third-order polynomial $p(x)$. The only methods I know of are just expanding the function in a Taylor series to third-order, or minimizing the coefficients $\int_{a}^{b}(f(x)-p(x))^2,dx$. I was under the impression that, since the Taylor series is really good close to the expansion point, but bad away from it, the second method was better for approximations on a defined interval. – Mr. G Jul 09 '13 at 02:03
  • Oh! You want 'least square approximation' by a degree $n$ polynomial. The thing is the 'approximation to order $n$' has some specific meaning which is what I understood. Yes, if you want some global approximation you need something like least squares. – OR. Jul 09 '13 at 02:09