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Apr
20
revised combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
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Apr
20
comment combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
@user4140: What is the combinatorial interpretation of $x$ in that proof?
Apr
20
revised combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
added 1026 characters in body
Apr
20
comment combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
@user4140: Do you know of a combinatorial proof of any identity involving an arbitrary $x$ ?
Apr
20
answered combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
Apr
18
comment Kind of basic combinatorical problems and (exponential) generating functions
Also, you need ordinary generating functions here, not exponential ones (at least, not in the way you wrote them). Although the dice are distinguishable, there is no ordering in the set of results.
Apr
17
comment Consider the number $n= 2^{10^{33}} +1$
"Explain your answer!" Well, why don't you first explain what you've tried!
Apr
16
comment Get closed form of a series from its generating function
What you have is already a closed form for the generating function. Did you mean a closed form for the coefficients? (Also, I think you meant to have the same variable on both sides.) (Also, what have you tried?)
Apr
16
comment Prove that there exists a constant C such that $[z^n]\exp(z/(1-z)) = O(\exp(C\sqrt{n}))$
No, I meant this question.
Apr
15
comment Using generation functions solve the following difference equation
The same question was posted an hour ago by the same user, who's since deleted it for unclear reasons.
Apr
15
comment How to count the number of types
@Lembik: Code.
Apr
14
comment How to count the number of types
@Lembik: Oops, actually $k=6$ is possible and has many solutions e.g. $(2, 3, 8, 9, 9, 69)$. In fact all values of $k$ with $4 \le k \le 20$ have solutions.
Apr
14
comment How to count the number of types
@Lembik: Also, as I said in the answer, the exact answer is that $k$ is possible if $(0)$ and $(1)$ together have a solution. The inequalities are only analytic (i.e. if the $n_i$s are allowed to be arbitrary real numbers $\ge 1$, not necessarily integers). So for example the fact that $k=3$ or $k=6$ are not possible, while $k=4$ and $k=5$ are ($(2, 4, 10, 16, 68)$), is not explained by these bounds.
Apr
14
comment How to count the number of types
@Lembik: For example, k=4 is possible with (2, 8, 24, 66) or (6, 12, 14, 68).
Apr
13
comment Why does $| a_n + \frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} | \ge | a_n | - |\frac {a_{n-1}} {z} + \ldots + \frac {a_{0}} {z^n} |$ hold?
As in the other answer by user2345215, I find the inequality $|x+y| \ge |x| - |y|$ clearer when written as $|x-y| + |y| \ge |x|$.
Apr
12
comment Book on combinatorial identities
It's a great book, but it's the very opposite of combinatorial proofs: it describes very general algorithms to prove combinatorial identities mechanically by computer.
Apr
11
comment Combinatorics-Generating function
@gar: Thanks, it's clear now -- before you edited your answer I'd started writing one, and arrived at the same answer independently. +1.
Apr
11
answered Combinatorics-Generating function
Apr
11
comment Combinatorics-Generating function
Can you explain how you got the final expression with the binomial coefficients?
Apr
11
comment Does the order I multiply the characteristic equation's factors in the homogeneous solution matter?
Let's use different letters, to avoid confusion: your solution is $c_0 5^n + c_1 2^n$. The book's solution is $d_0 2^n + d_1 5^n$. These two sets of solutions are the same, as you can see by putting $c_0 = d_1$ and $c_1 = d_0$.