| bio | website | |
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| visits | member for | 1 year, 3 months |
| seen | 56 mins ago | |
| stats | profile views | 457 |
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Jun 5 |
comment |
Generating function for Banach's matchbox problem Using this relation between $p_{k,m,n}$ and $p_{m,n}$, We don't need generating functions. I want to study generating functions more closely, so I posted this question. |
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Jun 5 |
answered | Integral of $\int \frac{5x+1}{(2x+1)(x-1)}$ |
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Jun 5 |
revised |
Generating function for Banach's matchbox problem added 206 characters in body |
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Jun 5 |
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Generating function for Banach's matchbox problem Alright. I'm looking for anybody else interpreting the equation (2) directly, other wise I'll take your answer. |
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Jun 4 |
revised |
Generating function for Banach's matchbox problem added 1 characters in body |
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Jun 4 |
asked | Generating function for Banach's matchbox problem |
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Jun 4 |
suggested | suggested edit on How to solve $x_j y_j = \sum_{i=1}^N x_i$ |
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Jun 4 |
awarded | Citizen Patrol |
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Jun 4 |
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Explicit formula for recurrence relation $a_{n+1} = 2a_n + 1$ Duplicated: math.stackexchange.com/q/106036/23875 |
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Jun 4 |
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The Fermat prime 257 and binomial sum $\sum_{n=0}^\infty \frac{(-1)^n}{\binom {8n}{4n}}$? Could you figure out the reference for such formulas? |
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Jun 4 |
awarded | Benefactor |
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Jun 4 |
accepted | How to solve this recurrence |
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Jun 4 |
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How to solve this recurrence The recurrence is related to Lemma 2 in Knuth's first analysis. Am I right? |
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Jun 4 |
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How to solve this recurrence Thanks. I will first read some of these articles. |
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Jun 4 |
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How to solve this recurrence Yeah, it's really linear probing. I don't know it's an open problem, but what about my recurrence? It seems that the recurrence can be solved. |
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Jun 3 |
revised |
How to solve this recurrence added 93 characters in body |
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Jun 3 |
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When can we plug an arbitrary number into an equation on formal power series The case of rational functions are not too difficult. I'm afraid I'll manipulate a more complex one, for example, on $e^z$ or something else. |
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Jun 3 |
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When can we plug an arbitrary number into an equation on formal power series I'm still confused. I might know that $A(z)$, or your notation $A_m$, is really formal objects, for example, apples, where multiplying means a magic merge of two apples. The formula is true whenever the number of each kind of apple is same. My question is on substituting the apple with concrete number, e.g. $\omega$. $\omega$ is so lucky that some of elements in neighborhood converges. What about plugging $2\omega$ into $A_0 = A_1 + \cdots + A_m$? |
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Jun 1 |
revised |
How to solve this recurrence added 1 characters in body |
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Jun 1 |
comment |
How to solve this recurrence @leonbloy The bug of ambiguity between horse and house is fixed. |