Closed form for nth term - generating functions

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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.

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Alex

Updated on August 09, 2020

• Alex over 3 years

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."

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

• $(1+x)^{10}$

• Jack D'Aurizio over 8 years
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.
• Alex over 8 years
For binomial expansion I got ∑C(10,k)1^(10-k)x^k
• Alex over 8 years
So I would take the coefficient, which is C(10,k) , right?
• Alex over 8 years
So first I must find the series representation and then, its coefficient will equal the closed form?
• Brian M. Scott over 8 years
@Alex: Yes: find the series representation $\sum_{n\ge 0}a_nx^n$, and then find an expression for $a_n$ in terms of $n$.
• Alex over 8 years
and when I am adding some terms together, I multiply their functions by each other?
• Brian M. Scott over 8 years
@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.
• Alex over 8 years
six being one greater than the number of terms
• Alex over 8 years
am I correct that -x^2 will be represented by 2/(1+x)
• Brian M. Scott over 8 years
@Alex: No, it’s six because six turn out to be needed. I suggest that you begin by converting the fraction to a series; that will give you two of the cases.
• Brian M. Scott over 8 years
@Alex: No, $-x^2$ is simply one term of the infinite series. Since the coefficient of $x^2$ in the series expansion of $\frac1{1-x^3}$ is $0$, the coefficient of $x^2$ in the whole generating function is $-1$, and we now know that $a_2=-1$.
• Alex over 8 years
Okay thank you, I will work on it