2

To put things into perspective I've read some posts with the method of undetermined coefficients in it for differential equations. What I'm talking about is the method to determine a closed formula for a sequence. For those of you who are unfamiliar with it I'll try to explain it.

E.g.:

$$0, 2, 5, 9, 14,\ldots$$ $$2, 3, 4, 5,\ldots$$ $$1, 1, 1,\ldots$$

The $1^{st}$ sequence (it also can be a different sequence of course, this is just an example) is always given in an exemplary exercise (in my case I found it myself, just by the brute-force method). The $2^{nd}$ one is acquired by subtracting the number in the 1^{st}sequence by the following number of that sequence. And one is required to stop this algorithm after it get a constant sequence like in the $3^{rd}$ one.

So now one has performed the algorithm two times. Thus concluding we need a polynomial of degree 2. Now let's take a random polynomial of which we need to determine the coefficients:

$$P(x)= ax{^2} + bx + c$$

We know that $P(0)= 0,P(1)= 2$and $P(2)= 5$ .

So $\begin{cases} c= 0 \\ a + b =2 \\ 4a + 2b =5\\ \end{cases}$ which gives us that $P(x)= \frac 12 x^2 + \frac 32 x $

To rephrase my question from early on:

Why does the method of coefficients work for every the simpler sequences as stated above? Does there exist a proof of it? Please explain that proof to me?

PS.:

1) I know that it doesn't work for sequences whose solutions aren't reals or if there are multiple solutions of such a P(x). Then you need to work with generating functions.

2) I've chosen this sequence on purpose because it's the sequence to determine number of diagonals of a convex polygon. Just for those who wondered about it.

Anonymous196
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    The phrase "method of undetermined coefficients" usually connotes a topic in differential equations. I am more used to seeing this type of method called the "method of finite differences" or the "calculus of finite differences". These are classical and understood, and in fact Newton used the calculus of finite differences extensively in his development of differential calculus. If you search for "finite differences", you'll get lots of leads. – davidlowryduda Sep 21 '17 at 14:54
  • Ah ok. I live in Belgium as you maybe noticed and if you translate this method litteraly from Dutch to English you'll get that. Thanks for your clarification. – Anonymous196 Sep 21 '17 at 14:56
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    It might be worth looking into Newton's series for finite differences. These aren't always unique, but Carlson's theorem provides the necessary and sufficient conditions, though this relies on complex analysis, which might not be as elementary as you'd like (though nor is the fundamental theorem of algebra, or generalized binomial expansion). https://en.m.wikipedia.org/wiki/Carlson%27s_theorem – adfriedman Sep 21 '17 at 15:00
  • As finite difference equations and and differential equations are analogous, this seems like a completely reasonable name for the method. I was a bit confused initially as well, but they work the same, e.g., your first equation is just $$\Delta^2 P(n) = 1, ; P(0)=0,;\Delta P(0)=2$$ where $\Delta P(n) = P(n+1) - P(n).$ – adfriedman Sep 21 '17 at 15:11
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    @adfriedman That was exactly what I was looking for, regarding the link. – Anonymous196 Sep 21 '17 at 15:12
  • You are implicitly assuming that the solution is a polynomial in $n$.The first half bounds its degree,then the second half computes it via interpolation. What more do you seek to know? – Bill Dubuque Sep 21 '17 at 16:32
  • Whenever we know that the solution has some special form, we can exploit properties of that special form (here computing a polynomial by interpolation or undetermined coefficients). See the discussion of Sylvester's identity in this answer for a truly spectacular
    exploitation of polynomial form (see also this answer).     – Bill Dubuque Sep 21 '17 at 17:13
  • @BillDubuque On one I must say that your answers are always referring to your past answers. It's like an endless labyrinth. On the other hand your answers are always usefull. Thanks for the interesting proofs and explanations. – Anonymous196 Sep 21 '17 at 17:47
  • @AnonymousI Mathematics is a wondrously connected web of beautiful ideas. The endless surprising interconnections is part of what motivates many to study mathematics. I strive to emphasize these connections inasmuch as possible - hoping that they will motivate others to explore the beautiful web of mathematics. – Bill Dubuque Sep 21 '17 at 18:09
  • Yeah me too. It's just new to me that a user refers to his own questions which is quite smart to promote own questions to get more upvotes. I know you just want to explain maths to more and more people. But it's just clever to do it that way – Anonymous196 Sep 21 '17 at 18:13

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