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Oct
12 |
awarded | Yearling |
Oct
12 |
awarded | Nice Question |
Apr
21 |
awarded | Scholar |
Apr
21 |
accepted | Generating-functions: is there a relationship between a generating function and the corresponding squared generating function |
Apr
21 |
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Generating-functions: is there a relationship between a generating function and the corresponding squared generating function
Know this is quite late, but thank you haha. |
Oct
10 |
awarded | Autobiographer |
May
28 |
awarded | Popular Question |
May
20 |
awarded | Nice Answer |
Nov
28 |
awarded | Commentator |
Nov
28 |
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Singular value decomposition over finite fields?
You can still use an unordered field, you just need that the inner-product maps to non-negative reals. |
Nov
28 |
answered | Thorough Introduction to Singular Value Decomposition |
Nov
28 |
answered | A random walk matrix has eigenvalue 1 with multiplicty 1 - why? |
Nov
20 |
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Computing the maximum salary
This wouldn't work for 2 people would it? |
Nov
20 |
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Computing the maximum salary
I like this question, hehe, simple but interesting. |
Nov
20 |
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Infinite Ramsey theorem with infinitely many colours
Kristal, I may be missing something, but when you say countably finite isn't that redundant, I mean all finite sets are countable by definition no? |
Nov
20 |
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How to compute the rank of a matrix?
haha I think its fine to have a misleading title as long as its all fun and games. |
Nov
19 |
answered | Ways to Synthesize Topics in Linear Algebra |
Nov
11 |
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Math History Question about the exponential function
"entities must not be multiplied beyond necessity", thats all. |
Nov
9 |
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Math History Question about the exponential function
thanks for the answer, I suspect your reasoning is correct. Correct me if I am wrong but f continuous on isn't a sufficient condition. For example, take $f(x) = 1 \text{ for } x \in \mathbb{Q}$ and $f(x)=0 \text{ for } x \in \mathbb{R} \ setminus \mathbb{Q} $ then that method would imply f(x) is 1 everywhere, since f(x) is continuous on the rationals |
Nov
9 |
awarded | Supporter |