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Consider two continuous real valued functions $f$, and $g$ over the interval $[a,b]$. For a given $\epsilon > 0$, for each function the we can use the Weierstrass approximation theorem to express the function as polynomials $f \approx \mathcal{P}_f$ and $g \approx \mathcal{P}_g$, such that $\lvert f - \mathcal{P}_f\lvert < \epsilon$ and $\lvert g - \mathcal{P}_g\lvert < \epsilon$, over the interval $[a,b]$. Is there anything we can say about the approximating the product of the two functions $f \cdot g$, i.e., $\mathcal{P}_{f\cdot g}$ where $\lvert f\cdot g - \mathcal{P}_{f\cdot g}\rvert < \epsilon$ on the interval $[a,b]$. In other words, is there any relation between $\mathcal{P}_{f\cdot g}$, $\mathcal{P}_f$, and $\mathcal{P}_g$?

In particular, I am interested if from $\deg(\mathcal{P}_f)$ and $\deg(\mathcal{P}_g)$ we can say anything about $\deg(\mathcal{P}_{f\cdot g})$. I can appreciate that $\deg(\mathcal{P}_f) + \deg(\mathcal{P}_g)$ may be greater than $\deg(\mathcal{P}_{f\cdot g})$; as the mutliplication may conspire to cancel out higher terms (e.g., $g \ll f$, such that $\mathcal{P}_f$ has ''overfit'' $f$ to achieve $\epsilon$ in the composition.) However, I cannot see if one can guarantee that this is the upper limit, i.e., $$\deg(\mathcal{P}_{f\cdot g}) \leq \deg(\mathcal{P}_f) + \deg(\mathcal{P}_g)?$$

An example that the component degree sum is larger can be seen from $f(x) = 1+x$, and $g(x) = (1+x)^{-1}$. For small enough $\epsilon$, both have approximating polynomials greater than degree zero, while the product's approximating polynomial is trivially degree zero.

Is there a counter-example to the suggestion inequality, i.e., where

$$\deg(\mathcal{P}_{f\cdot g}) > \deg(\mathcal{P}_f) + \deg(\mathcal{P}_g)?$$

Disclaimer: This is a reformulation of a question I mistakenly asked about function composition.

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Novice C
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    In general I don't think you can get that bound. The problem is simple enough: $|xy-x'y'|=|xy-x'y+x'y-x'y'| \leq |x-x'||y| + |y-y'| |x'| \leq |x-x'| |y| + |y-y'| (|x|+|x-x'|)$. This is "basically" $|x-x'| |y| + |y-y'| |x|$. Thus what you really have is more like $\operatorname{deg}(P(f \cdot g,\varepsilon)) \leq \operatorname{deg}(P(f,\varepsilon/(2M_g)))+\operatorname{deg}(P(g,\varepsilon/(2M_f)))$. Again this isn't exactly correct because I suppressed a $\varepsilon^2$ term but that's not very important. – Ian Jun 19 '21 at 19:55
  • Thanks, I agree that this looks like a valid bound. It makes sense that the error bound cannot be preserved if you take the polynomial of $f \cdot g$ to be $\mathcal{P}_{f\cdot g} = \mathcal{P}_f \cdot \mathcal{P}_g$ (e.g., $|6-(2.01\times 3.01)| \neq 0.01$). I was hoping what gets lost in fuzziness of the inequalities when you reduce to terms of $\lvert f - \mathcal{P}_f\rvert$, could be tighter (assuming I interpret $x$ and $x'$ correctly, moreover one couldn't make use of $x'y-y'x=0$). But perhaps that cannot be shown so easily, I certainly have no idea and perhaps its ill posed. – Novice C Jun 19 '21 at 21:26
  • You can sorta see what's going on in the other question; you end up with a rough estimate (though it isn't rigorous because I'm ignoring the logarithm and the $\varepsilon^2$ term) of $M_f^2 d_g + M_g^2 d_f$ in this setting. So in particular your bound does in fact work if you can assume $|f|,|g| \leq 1$. – Ian Jun 19 '21 at 21:31
  • But the point is that just like with constants, the possibility that the other factor is big means that the error in each factor has to be a lot smaller than the actual error tolerance. – Ian Jun 19 '21 at 21:36

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