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The Maclaurin series for a function $f$ is $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(0)x^n}{n!}$$ Suppose that instead of the $x^n$ we picked up a function $g_n$? We can write

$$f(x)=a_0+a_1g_1(x)+a_2g_2(x)+\ldots\\f'(x)=b_1a_1+b_2a_2g_2'(x)+...$$ hence $$f'(0)=b_1a_1$$ but i'm not sure whether this is right. What would it then look like? $$f(x)=\sum_{n=0}^\infty\Big(\underline{\color{white}{what}}{}\Big)g_n(x)$$

Have there been any work on that in the literature?

user153330
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  • Please explain how did you obtain $f'(x)=b_1a_1+b_2a_2g_2(x)+...?$ – NoChance Feb 03 '16 at 22:01
  • In addition, if you replace $x^n$ with $g(x)$, the L.H.S must be $f(g(x))$ not just $f(x)$. – NoChance Feb 03 '16 at 22:09
  • @NoChance i choosed a specific function $g$ that behaves a bit like $x^{n}$, so to get the same thing (if this is unclear to you please state that) i made a mistake in formula will correct it – user153330 Feb 03 '16 at 22:21
  • why the downvote? – user153330 Feb 03 '16 at 22:26
  • @NoChance basically i will try to deal with the same thing using $x^n$ (you should be thinking of $g_n$ by analogy) we have $$f(x)=a_0+a_1x^1+a_2x^2+...$$ right? we differentiate (i will write the constants that arise as a sequence $(b_i)$) $$f'(x)=a_1+b_1a_2x+...$$ setting $x=0$ yields $f'(0)=a_1$ (that's where concept of $g_n(0)=0$ comes from, for convenience I choose $0$, i could be more general and generalize that for any $c$), now if we repeat the same we will eventually get the taylor series, i hope you see the analogy – user153330 Feb 03 '16 at 22:29
  • Thanks for the explanation, I will try to get it! By the way, I did not down vote. – NoChance Feb 03 '16 at 22:36
  • @NoChance so do you have any idea now? – user153330 Feb 11 '16 at 13:14
  • Unfortunately, I still don't get your point. – NoChance Feb 11 '16 at 13:39
  • @NoChance okay, i will make a major edit with more clear explanations, will ping you when done – user153330 Feb 11 '16 at 19:50

1 Answers1

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But of course.

First of all, note that your question may be rewritten in this way: given a set of functions $\{ g_1, \dots, g_n \dots \}$, what conditions do I need to have any other (continuous?) function approximated by a sum of the $g_i$s?

The Stone-Weierstrass theorem gives an answer to this question: https://en.wikipedia.org/wiki/Stone%E2%80%93Weierstrass_theorem

I also suggest you take a look at the Fourier series of a function, in which we use $\sin(nx)$ and $\cos(nx)$ instead of $x^n$.

AnalysisStudent0414
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    There are two quite distinct questions:
    1. Can the function be arbitrarily well approximated on some domain by finite linear combinations of $g_1, g_2, \ldots$?
    2. Is there a series $\sum_j c_j g_j$ that converges to the function on some domain?

    Even in the case $g_j = x^j$ , domain $[-1,1]$ the answers are quite different: For any continuous function on the interval, (1) is true, but for (2) you need the function to have an extension that is analytic in the disk $|z| < 1$.

     – Robert Israel Dec 09 '15 at 23:17
  • Thanks for answering, need more answers for more information – user153330 Dec 20 '15 at 14:51