5,292 reputation
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bio website quantdec.com
location Northeastern US
age 14
visits member for 4 years, 5 months
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Consultant (environmental and spatial stats a specialty), expert witness, and teacher. I can be reached through (outdated but still valid) links posted on my web site.

Twitter: @WilliamAHuber // ASA-P website: http://amstatphilly.org/


Why waste time learning, when ignorance is instantaneous?

--T(iger) Hobbes.

For any complex problem there is a simple solution. And it's always wrong.

--[Mis?]attributed to H.L. Mencken by Dava Sobel, Longitude.


Dec
29
comment What are some interesting cases of $\pi$ appearing in situations that are not / do not seem geometric?
This is David H's answer in disguise: the ARE is algebraically related to (a) $e^0=1$ and (b) $\int_\mathbb{R}e^{-x^2}dx=\Gamma(1/2)=\sqrt{\pi}$. The appearance of $\pi$ is due to the duplication formula for the Gamma function, $\Gamma(z)\Gamma(1-z)=\pi\csc(\pi z)$, exhibiting $\Gamma$ as a kind of "square root" of a trig function. This is because $\Gamma(z)=\Gamma(z+1)/z$ implies $\Gamma$ has (simple) poles at $0,-1,-2,\ldots$, whence $\Gamma(z)\Gamma(1-z)$ has poles at $\mathbb Z$, strongly suggesting periodicity--which is why $\pi$ should show up!
Dec
28
revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Appended some more discussion.
Dec
28
revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
[Edit removed during grace period]
Dec
28
comment Prove that $\lim_{x \to 0}\frac{x^2+x^3}{\sin^3x}$ doesn't exist.
Another way to appreciate @Chinny84's idea is to consider the limit of $1/x - (x^2+x^3)/\sin^3(x)$. This does exist and you can show it using any of the three techniques in the question. Now suppose, to assume the contrary of what you want to prove, that the limit of $(x^2+x^3)/\sin^3(x)$ exists. Then--adding the two results--you could conclude that the limit of $1/x$ exists, which it obviously does not (using any of the three methods).
Dec
28
revised Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
deleted 4 characters in body
Dec
28
comment Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
@Thank you for noticing that, Thomas. That is easily fixed by replacing one of the $c$'s by $1$; I will make the change.
Dec
24
comment Probability that there is sub-sequence of exact length
There is a lot of literature on this subject. The asymptotic distribution is known. I don't believe exact formulas for the case of finite $N$ are available.
Dec
24
revised Urn problem with balls
added 12 characters in body
Dec
24
comment Urn problem with balls
@Brian Thank you for that observation. It's very much in the spirit of this answer, which is an attempt to find as simple a solution as possible.
Dec
24
revised Urn problem with balls
added 2 characters in body
Dec
24
comment Using triangle inequality for upper limit of integral.
Why not just compute the integral? It's dead easy--the exponential part, being analytic throughout the triangle, integrates to zero; the remaining contribution from $\bar z$ will give $-2i$ times the area of the triangle.
Dec
24
comment The point of contact of between two circles and common tangent at this point.
+1 for the simplified elementary approach. But note that the solution is incomplete: it has not (yet) addressed whether the circles are internally or externally tangent.
Dec
24
comment Urn problem with balls
It might be even simpler just to count the ways of choosing one ball in one urn and the other in the other: $(n+1)^2$. Divided by $\binom{2n+2}{2}$, this gives the answer directly.
Dec
24
answered Urn problem with balls
Dec
24
answered Game to maintain distinct number of balls in glasses
Dec
24
comment Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
@Brian You're right; there are many inductions lurking here. The one you point to comes down to asserting that the integers are closed under addition and multiplication. I'll accept that as immediately obvious. The crux of the matter, though, is that scalars commute with matrices, whence $(2\mathbb P)^n = 2^n \mathbb P^n$: that shows where the powers of $2$ come from.
Dec
24
answered Prove without induction $2^n \mid (b+\sqrt{b^2-4c})^n + (b-\sqrt{b^2-4c})^n $
Dec
18
reviewed Approve Proof of the description of a set
Dec
15
comment Show that $\lim_{n\rightarrow\infty}{\displaystyle\sum_{i=1}^{n}{\frac{F_n}{2^n}}}=2$
The limit is $0$, not $2$. Use $(1 +\sqrt{5})/2$ instead of $2$ in the denominator to obtain a finite limit.
Dec
15
awarded  Caucus