I think I am mostly confused about what the question is asking. I read that "closed form" means that it should not be represented as as infinite sum, so I am not sure what they are asking for. Would they like a summation symbol with the summation of all numbers through the nth term?

"For the generating functions below, give a closed form for the nth term of its associated sequence."

  • $ 3x^4 +7x^3−x^2 +10 + \frac{1}{1-x^3}$

  • $(1+x)^{10}$

  • $\begingroup$ For the first one, exploit the fact that $\frac{1}{1-x^3}$ is a geometric series, $\sum_{n\geq 0}x^{3n}$. For the second one, the binomial expansion is enough. $\endgroup$ – Jack D'Aurizio Apr 2 '15 at 12:18
  • $\begingroup$ For binomial expansion I got ∑C(10,k)1^(10-k)x^k $\endgroup$ – Alex Apr 2 '15 at 12:42
  • $\begingroup$ So I would take the coefficient, which is C(10,k) , right? $\endgroup$ – Alex Apr 2 '15 at 12:42

No: a summation is not a closed form. Roughly speaking, a closed form is a function of $n$ not involving a summation or product whose length depends on $n$.

Here’s a simple example. The generating function $\frac1{1-2x}$ represents the series $$\sum_{n\ge 0}(2x)^n=\sum_{n\ge 0}2^nx^n\;,$$ whose associated sequence of coefficients is $\langle a_0,a_1,a_2,\ldots\rangle=\langle 2^0,2^1,2^2,\ldots\rangle$. A closed form for $a_n$ is $a_n=2^n$.

Added: For the first one your closed form will have several cases – six if you write it as I would. I would begin by expanding $\frac1{1-x^3}$ into its power series. For the second one you’ll want to use the binomial theorem.

  • $\begingroup$ So first I must find the series representation and then, its coefficient will equal the closed form? $\endgroup$ – Alex Apr 2 '15 at 12:49
  • 1
    $\begingroup$ @Alex: Yes: find the series representation $\sum_{n\ge 0}a_nx^n$, and then find an expression for $a_n$ in terms of $n$. $\endgroup$ – Brian M. Scott Apr 2 '15 at 12:52
  • $\begingroup$ and when I am adding some terms together, I multiply their functions by each other? $\endgroup$ – Alex Apr 2 '15 at 12:53
  • $\begingroup$ @Alex: No. If you’re thinking about the first problem, your closed form is going to have several cases — six, if I’ve counted correctly. $\endgroup$ – Brian M. Scott Apr 2 '15 at 12:57
  • $\begingroup$ six being one greater than the number of terms $\endgroup$ – Alex Apr 2 '15 at 12:58

Roughly speaking, a "closed form" is a formula that requires less work than just doing the computation. This typically means not having summation signs or other "repeat for all..." structures. But in any case, you'd have to agree on some set of "elementary" functions, i.e., powers, factorials, binomial coefficients are normally considered such.


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