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Apr
21
comment Solving the non-homogeneous recurrence relation: $g_{n} = 12g_{n-2}-16g_{n-3}+6\cdot 2^n+25n$
You're right about it being pretty nasty. The generating function is useful to get asymptotics (you only need to look at the singularities), but actually carrying out the partial fraction decomposition seems to require more persistence than is justifiable. :-)
Apr
21
comment Probability of at least one card matching when flipping through two separate decks.
+1; numerically, the probability of at least one match is $923/924 \approx 0.9989$.
Apr
21
comment Recurrence relation with generating function problem
I was going to post an answer suggesting this "simpler way", but I feel the OP's way actually turns out to be simpler here.
Apr
21
comment Combinatorial explanation for why $n^2 = {n \choose 2} + {n+1 \choose 2}$
Yes, seems more satisfying now. :-) By the way, your geometric picture along with the proof that $1 + 2 + \dots + n = \binom{n + 1}{2}$ might make for an elegant version!
Apr
21
answered Why do exponents not distribute over addition?
Apr
20
comment Combinatorial explanation for why $n^2 = {n \choose 2} + {n+1 \choose 2}$
What is the combinatorial argument that the latter number is $\binom{n+1}{2}$ ?
Apr
20
comment How do I prove that an order of a cycle is its length?
Might be worth adding a clarifying note that "order" here means its order in the permutation group, i.e. the smallest $n$ such that $\sigma^n$ is the identity permutation. (Someone had proposed an edit to the question with wording like "order of the cycle generator", which I find even more confusing.)
Apr
20
reviewed Reject suggested edit on How do I prove that an order of a cycle is its length?
Apr
20
comment Prove that there are two frogs in one square.
There is no probability in the quoted question; "prove that it is possible..." just means that there exists some way of arranging things.
Apr
20
comment Hint finding exact value of half-angle when $\tan (\theta) = {3}$
In the fraction under the square root (in my comment above), the numerator $(10 + \sqrt{10})(10 + \sqrt{10})$ is of course $(10 + \sqrt{10})^2$ (you can leave it like that, because you're going to take its square root next). The denominator $(10 - \sqrt{10})(10 + \sqrt{10})$ is $10^2 - (\sqrt{10})^2 = 100 - 10 = 90$. You've made some mistake somewhere in your calculation.
Apr
20
comment Hint finding exact value of half-angle when $\tan (\theta) = {3}$
You are doing right so far: your $$\sqrt{{10+\sqrt{10}}\over{10-\sqrt{10}}}$$ is equal to (by your idea of "multiplying the top and bottom by the conjugate") $$\sqrt{\frac{10 + \sqrt{10}}{10 - \sqrt{10}}\frac{10+\sqrt{10}}{10 + \sqrt{10}}}$$ If you get stuck again, note that $\sqrt{90} = 3\sqrt{10}$.
Apr
20
comment Hint finding exact value of half-angle when $\tan (\theta) = {3}$
@JasonS.: the formula gives $\tan(2y)$ in terms of $\tan y$. What you've done is find $\tan(2\theta)$, instead of finding $\tan(\theta/2)$ as the question asks. Instead, try letting $y = \theta/2$, and apply the formula and see what turns up.
Apr
20
comment combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
@user4140: The second one also works only for $x$ being positive integers, not arbitrary $x$. In fact, it holds for only countably many $x$ rather than the uncountably many in $[0, 1]$ :-)
Apr
20
revised combinatorial proof $\sum_{i=0}^m x^i=\frac{x^{m+1}-1}{{x-1}}$
added 47 characters in body
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!