Timeline for Uniformly approximating a function and its derivative using polynomials
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
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| Oct 8, 2020 at 2:47 | history | became hot network question | |||
| Oct 7, 2020 at 23:37 | comment | added | mw19930312 | @IlyaBogdanov I got what you meant. I forgot we can approximate $f’$ if we assume span of $\{p_k’\}$ is also dense. Thanks! | |
| Oct 7, 2020 at 22:37 | comment | added | Ilya Bogdanov | Well, OK, you should have all functions which are constant on each segment in $K$. In this case, you may proceed as in @Algernon’s comment, approximating $f’$ by the derivatives, and then correcting the approximation of $f$ by those constants. | |
| Oct 7, 2020 at 22:29 | comment | added | mw19930312 | @IlyaBogdanov It’s fair to assume that $S$ consists of constants. What I wonder here is even we assume the span of $\{p_k’\}$ is dense, there is still no guarantee on existence of $g_k$, is there? Or am I missing anything straightforward? | |
| Oct 7, 2020 at 22:19 | comment | added | Ilya Bogdanov | Hm, right! It seems that this works only if constants are in the span (which I somehow implicitly assumed...). Otherwise that would be true only if we work in the space of functions vanishing at a certain point. | |
| Oct 7, 2020 at 22:11 | comment | added | mw19930312 | @IlyaBogdanov Thanks so much for the straightforward counter-example here. But I'm a bit slow on the reformulation here. How can we argue that if $\{\p_k'}$ is also dense, then we can find $g_k$ such that $g_k\to f$ and $g'_k\to f'$ at the same time? | |
| Oct 7, 2020 at 21:31 | answer | added | Pietro Majer | timeline score: 3 | |
| Oct 7, 2020 at 21:22 | answer | added | Ilya Bogdanov | timeline score: 9 | |
| Oct 7, 2020 at 21:17 | comment | added | Ilya Bogdanov | An equivalent reformulation: if the span of $\{p_k\}$ is dense in $C(K)$, is the span of $\{p_k’\}$ also dense? | |
| Oct 7, 2020 at 20:45 | answer | added | Algernon | timeline score: 3 | |
| Oct 7, 2020 at 19:57 | comment | added | Algernon | Still, what I said reduces the problem to the case where $f$ is itself a polynomial. | |
| Oct 7, 2020 at 19:38 | comment | added | Algernon | Oooh, I see. Sorry for my misunderstanding. | |
| Oct 7, 2020 at 19:37 | comment | added | mw19930312 | @Algernon Thanks for the answer here. I'm more interested in the case where $S$ is a subset of all polynomials. The tricky part in your comment is that $\int_a^x h(z)dz$ does not necessarily lie in $S$. | |
| Oct 7, 2020 at 19:31 | comment | added | Algernon | Let's start with the case in which $K$ is an interval $[a,b]$. If $h$ is a polynomial that $\varepsilon$-approximates $f'$, then $H(x):=f(a)+\int_a^x h(z)\,\mathrm{d}z$ is a polynomial that $\delta$-approximates $f$, where $\delta:=\varepsilon |b-a|$. | |
| Oct 7, 2020 at 19:07 | history | edited | mw19930312 | CC BY-SA 4.0 |
added 4 characters in body
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| Oct 7, 2020 at 19:07 | comment | added | mw19930312 | @LSpice Yes. It should be $C^1(K)$. I've changed the original post. It's actually an intermediate step in a theorem that I would like to prove in my research work. I tried to look up real-analysis references. But I didn't find anything useful... | |
| Oct 7, 2020 at 18:58 | comment | added | LSpice | Is $C(K)$ supposed to be $\mathrm C^1(K)$? Otherwise, it's not even clear what $f'$ means. (Also, where does this come from? It looks like a homework question.) | |
| Oct 7, 2020 at 18:46 | history | asked | mw19930312 | CC BY-SA 4.0 |