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Sep
30
answered Manipulating harmonic series
Sep
30
comment Manipulating harmonic series
I cleared up your notation. In particular it's not true that $\sum_{k=1}^n {1 \over k} = \ln n$ exactly; it's only approximately true.
Sep
30
revised Manipulating harmonic series
cleaned up TeX
Sep
28
comment Dependence of certain random variables
Thanks. This is what happens when I write quickly.
Sep
28
comment Why is $f(x) = x\phi(x)$ one-to-one?
$n\phi(n) = \phi(n^2)$, which might help.
Sep
28
comment Rolling dice such that they add up to 13 — is there a more elegant way to solve this type of problem?
If you want to learn more about generating functions, you should take a look at Herb Wilf's book "generatingfunctionology", which can be found free online. (This is legal; Wilf has posted the book on his web page.) This book is probably best read after you know some combinatorics and some probability.
Sep
27
answered How would I solve $\frac{(n - 10)(n - 9)(n - 8)\times\ldots\times(n - 2)(n - 1)n}{11!} = 12376$ for some $n$ without brute forcing it?
Sep
27
answered Dependence of certain random variables
Sep
27
answered Rolling dice such that they add up to 13 — is there a more elegant way to solve this type of problem?
Sep
23
comment Why does volume go to zero?
Came into this thread to give this proof, but I see someone beat me to it.
Sep
20
comment Can you answer my son's fourth-grade homework question: Which numbers are prime, have digits adding to ten and have a three in the tens place?
I agree. But my gut feeling (which I admit may be wrong!) is that that non-independence only shows up in the constant factors that I've suppressed.
Sep
20
awarded  Nice Answer
Sep
20
answered Can you answer my son's fourth-grade homework question: Which numbers are prime, have digits adding to ten and have a three in the tens place?
Sep
20
comment How does one compute the sign of a permutation?
I think you can find this in Knuth. (No, I don't know where. It just feels like it's in there somewhere.)
Sep
20
comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have?
However, the algorithm I gave in my previous comment does not always give the optimal solution. Consider writing $14!$ as a product of four factors. $14!^{1/4} = 543.378$; the smallest factor of $14!$ greater than $543$ is $550$; the smallest factor of $14!/550$ greater than $(14!/550)^{1/3}$ is $567$; the smallest factor of $14!/(550\times567)$ greater than its square root is $546$. We get $550 + 567 + 546 + 512 = 2175$. But $560+540+546+528 = 2174$, and both sets multiply to $14!$.
Sep
19
comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have?
Another potential greedy algorithm: if we want to write $n$ as a product of $d$ factors, start by taking the factor of $n$ which is closest to (or closest to but above, or closest to but below) $n^{1/d}$; then write $n$ divided by that factor as a product of $d-1$ factors in the same way.
Sep
19
comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have?
Also, a quick proof that the particular set you gave is the best possible: to minimize $x+y+z+w$ subject to the constraint $xyzw=C$, you take $x=y=z=w=C^{1/4}$. (Use Lagrange multipliers if you've seen them; in any case this should seem reasonable.) So you can't do better than a sum of $4(10!)^{1/4} \approx 174.6$; in fact $40+42+45+48 = 175$.
Sep
19
comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have?
$10!^{1/4} = 43.65$ so it seems natural to choose the numbers close to $43.65$ which don't have any prime factors larger than $10$. This gives your set right away. I don't see if this can be generalized but it at least might help transform "brute force" into "intelligent brute force".
Sep
19
comment Proof that the sum of two Gaussian variables is another Gaussian
I like this proof very much, because it explicitly uses the rotational symmetry, and therefore makes it clear why the Gaussian has this property but other distributions do not.
Sep
16
comment Sequence of numbers with prime factorization $pq^2$
Numerically, this means $O_N/E_N \sim 0.8089 \cdots$.