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Dec
16
comment Solving $\sinh(ax) = bx$
@deight $x=0$ is always a solution to this equation. Concavity tells us that if $a\geq b$ there are no more real solutions, and if $a<b$ there are two more solutions, negatives of each other.
Dec
16
answered How many integers from 1 to 100,000 contain the digit 6 at least once?
Dec
15
answered Find the set of complex numbers $z$ which satisfy: $\left\lvert\frac{z-3}{z+3}\right\rvert=2$
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
This argument is regardless of $n+1\lesseqgtr p$. It just points out how $n+1$ is not always going to divide into $(p^{n+1}-1)p^n$. So your induction approach is not going to work.
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
No, it's not true that $K_{n-1}$ divides $K_n$, which is the whole point of the people here telling you that induction you are trying won't work. Please follow my suggestion and with $p=5$, and compute $K_1,K_2,K_3,\ldots$ to see this first hand. I think up to $K_3$ or $K_4$ should be enough.
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
By strengthening the induction hypothesis, I mean, instead of just "$K_n$ is an integer", you would need "$K_n$ is an integer and it is divisible by <certain things that allow the induction implication to be valid>".
Dec
14
comment Why Is $y^{-1}$ = $\frac{1}{y^1}$?
@Nishant Use $/$ if you prefer, but the point is to have whatever the opposite of $\cdot$ is. In my experience, if I use $/$ in something like $1/y/y$, lots of people forget their order of operations and read it as $1/(y/y)$ instead of $(1/y)/y$.
Dec
14
revised Why Is $y^{-1}$ = $\frac{1}{y^1}$?
added 149 characters in body
Dec
14
answered Why Is $y^{-1}$ = $\frac{1}{y^1}$?
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
Your premise is that $K_{n+1}/K_n$ is an integer, and that is just wrong, even for low $n$ where you claim you have the induction dealt with. Set $p=5$ and start computing $K_1,K_2,K_3,\ldots$. The ratios are not always integers, so your efforts to prove that $\frac{(p^{n+1}-1)p^n}{n+1}$ is in integer is futile. My posted "answer" explains a bit better why induction like this is not going to work unless you strengthen induction hypotheses a lot somehow.
Dec
14
revised Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
added 107 characters in body
Dec
14
answered Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
Oh I see! The source of the error and my misunderstanding too is that you have $(p^{n+1}-1)p^n=p^{2n+1}-1$. So you are not actually trying to show $n+1$ divides $p^{2n+1}-1$, but rather that it divides $p^{2n+1}-p^n$.
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
@Maman If $n=p-1$, then $n+1$ is $p$. And there is no way $p$ can divide $p^K-1$. If this is too general, just look at $n=4,p=5$. You are then claiming that $5$ divides $1953124$, clearly false.
Dec
14
comment Estimating the mode
Before asking about accuracy, how are you estimating the frequency of the mode at all based on a sample drawn with replacement? For example if $n=100$, and I sampled $150$ times, and my sample's mode is say $17$ with frequency $8$, how do you use all of this to estimate the population's mode's frequency? I'd want to start by answering that question (and maybe you already have an answer for it.)
Dec
14
comment Estimating the mode
I note you are sampling with replacement more times than the size of $A$. In practice, if it's at all possible, sample without replacement instead and you can sample the whole population and know the frequency of the mode with certainty.
Dec
14
comment Prove that $\frac{(p^{n}-1)(p^{n}-p)…(p^{n}-p^{n-1})}{n!} \in \mathbb{N}$ with $p$ a prime number and $n \in \mathbb{N}$
"Then I want to prove that $n+1$ divides $p^{2n+1}-1$." This fails for $n=p-1$. At least some of the time, the $n+1$ in your denominator cancels at least in part with $K_n$.
Dec
14
answered Question on the definition of vector spaces.
Dec
12
awarded  Enlightened
Dec
12
awarded  Nice Answer