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14h
comment Prove a $\pi$ inequality: $\left(1+\frac1\pi\right)^{\pi+1}<\pi$
$355/113$ and $22/7$ are both larger than $\pi$. In fact you need a tight approximation to $\pi$ from below. Probably $333/106$ is the one you're looking for.
15h
revised Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?
added 1526 characters in body
18h
answered Which is bigger: $(\pi+1)^{\pi+1}$ or $\pi^{\pi+2}$?
May
19
comment Cauchy product: how to calculate?
Yes, you can just multiply. Also, check against the other method: $\sum_n (n+2)(n+1)q^n=(d^2/dq^2) \left(q^2\sum_n q^n\right) = (d^2/dq^2)\left(q^2/(1-q)\right)=(d/dq)\left(2q/(1-q) + q^2/(1-q)^2\right)=(d/dq)\left(-1 + (1-q)^{-2}\right)=2(1-q)^{-3}$.
May
18
answered Cauchy product: how to calculate?
May
18
comment Definite integration about $\sin(2x) \cos(3x) \sin(4x)$
Pretty trivial to do using $\sin z = \frac{1}{2i}(e^{iz} - e^{-iz})$ and $\cos z = \frac{1}{2}(e^{iz} + e^{-iz})$.
May
16
answered Counting problem for the integers
May
16
comment Covergence of series
Since $\log(1+x)\sim x - \frac{1}{2}x^2$ for small $x$, the terms are alternating in sign with magnitude $\sim n^{-(1+\alpha)} - \frac{1}{2}n^{-(2+\alpha)}$ for large $n$. So for $\alpha \ge -1$ the series is alternating-decreasing and hence convergent. For $\alpha < -1$, the terms eventually increase in magnitude, so the series diverges.
May
14
comment Not so obvious probability puzzle.
@AndréNicolas: It doesn't matter what the probability distribution is; the numbers $a$ and $b$ can be completely arbitrary, as long as they're different. With the strategy I gave, you win with probability $1/2 + P(a \le X \le b) > 1/2$.
May
14
comment Not so obvious probability puzzle.
Choose a random number $X$ before opening the box, from any continuous distribution over the positive reals that you like. If the revealed number is greater than $X$, guess that the other number is smaller, and vice versa. Remarkably, this gives you a greater than 50% chance of winning.
May
11
comment Is $\phi$ convex if it is continuous and $\phi \Big(\frac{x+y}{2}\Big)\le \frac{1}{2}\phi (x) +\frac{1}{2} \phi (y)$
You can prove that $\phi(\lambda x + (1-\lambda) y) \le \lambda \phi(x) + (1-\lambda)\phi(y)$ whenever $\lambda = p/2^n$ and $p$ is odd and $0 < p < 2^n$, by induction on $n$. (For $n=1$, this is exactly your assumption.) Consider the cases $p < 2^{n-1}$ and $p>2^{n-1}$ separately. Then use continuity to prove the result for all $\lambda\in[0,1]$.
May
11
comment Finding the value of an antiderivative at a point, yet indefinite integral cannot be found
Is this multiple choice?
May
11
comment Finding out this combination
Working modulo $p$ will work fine.
May
11
answered Finding out this combination
May
10
comment Wilson's theorem states that if n is a prime number, it will divide (n-1)! + 1, using this find the smallest divisor of 12!+6! +12!×6! + 1?
The method is this: $12! + 6! + 12!\times 6! + 1 = (12! + 1)\times (6! + 1)$, so Wilson's theorem states that $7$ and $13$ divide it. And it is equal to $1$ modulo $6!$, so it's not divisible by $2$, $3$, or $5$. Its smallest divisor is therefore $7$.
May
10
comment Wilson's theorem states that if n is a prime number, it will divide (n-1)! + 1, using this find the smallest divisor of 12!+6! +12!×6! + 1?
$91=7\times 13$ isn't prime. It's not the smallest divisor of anything.
May
8
comment Random Walk Without Repetitions
@Meelo: While I was editing the post to include the exact result $P^{(3)}=4368595/13405842$, you beat me to the punch, and added $P^{(4)}$ etc. in the bargain! Very nice. I summed the series exactly by noting that for $r=3$, we have $N_k = 2N_{k-1} + N_{k-4}$ for $k\ge 7$.
May
8
revised Random Walk Without Repetitions
added 958 characters in body
May
8
revised Random Walk Without Repetitions
added 958 characters in body
May
8
answered Counting the tiles in the game Tsuro