$B_{i}^{n}(x)={{n}\choose{i}}x^i(1-x)^{n-i}$, prove that $B_{i}^{n}(cu)=\sum\limits_{j=0}^{n}B_{i}^{j}(c)B_{j}^{n}(u)$ I tried to to solve it from the right side: $\sum\limits_{j=0}^{n}B_{i}^{j}(c)B_{j}^{n}(u)=\sum\limits_{j=0}^{n}{{j}\choose{i}}c^i(1-c)^{j-i}{{n}\choose{j}}u^j(1-u)^{n-j}$. Since ${{j}\choose{i}}$ is $0$ for $j<i$, then the sum is $\sum\limits_{j=i}^{n}B_{i}^{j}(c)B_{j}^{n}(u)=\sum\limits_{j=0}^{n-i}B_{i}^{i+j}(c)B_{i+j}^{n}(u)=\sum\limits_{j=0}^{n-i}{{i+j}\choose{i}}c^{i}(1-c)^{j}{{n}\choose{i+j}}u^{j+i}(1-u)^{n-j-i}$. But I’m stuck…

Numerically calculating the sum of the squares of the $m$th row of Pascal’s triangle, I found that for at least the first $10$ or so cases $$\sum_{i=0}^m \binom{m}{i}^2=\frac{(4m-2)!!!!}{m!}$$ Where $(4k-2)!!!!=(4k-2)(4k-6)(4k-10)\cdots 6\cdot2=2^m(2m-1)!!$. Of course the problem reduces to, using the well-known formula, $$\frac{(2m)!}{(m!)^2}=\frac{2^m(2m-1)!!}{m!}$$ $$(2m-1)!!=\frac{(2m)!}{m!2^m}$$ Is there a direct proof for the above equation, or (second-best) is […]

I need to write a program to input a number and output it’s factorial in the form: $4!=(2^3)(3^1)$ $5!=(2^3)(3^1)(5^1)$ I’m now having trouble trying to figure out how could I take a number and get the factorial in this format without actually calculating the factorial. Say given 5 need to get result of $(2^3)(3^1)(5^1)$ without […]

I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$. Why?

Bernstein polynomials are defined like this $B_i^n={{n}\choose{i}}x^i(1-x)^{n-i}$.I need to prove that $r$’th derivative of it is equal to: $(B_i^n)^{(r)}=\dfrac{n!}{(n-r)!}\sum\limits_{k=0}^{r}(-1)^k{{r}\choose{k}}B_{i-r+k}^{n-r}=\dfrac{n!}{(n-r)!}\sum\limits_{k=0}^{r}(-1)^k{{r}\choose{k}}{{n-r}\choose{i-r+k}}x^{i-r+k}(1-x)^{n-i-k}$. Here’s my partial solution end eventual getting stuck: $(B_i^n)^{(r)}=({{n}\choose{i}}x^i(1-x)^{n-i})^{(r)} = {{n}\choose{i}}(x^i(1-x)^{n-i})^{(r)} \stackrel{Leibniz}{=} {{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}(x^i)^{(k)}((1-x)^{n-i})^{(r-k)}={{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}\dfrac{i!}{(i-k)!}x^{i-k}\dfrac{(n-i)!}{(n-i-r+k)!}(1-x)^{n-i-r+k}(-1)^k $ What am I doing wrong?

How do you prove the following without induction: 1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$ 2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$ 3)$\prod\limits_{k=1}^n2k-1<n^n$ I think AM-GM-HM inequality is the way, but am unable to proceed. Any ideas. Thanks beforehand.

Expression I have somehow stumbled upon this expression (I believe I have proved it, but that is not important right now), which I have tried to simplify by writing it like something like this (I have been playing with powers of natural numbers and something similar to Pascal’s triangle) : n! = (t-1)^n , t^a=(a+1)^n […]

I need help proving the following identity: $$\frac{(6n)!}{(3n)!} = 1728^n \left(\frac{1}{6}\right)_n \left(\frac{1}{2}\right)_n \left(\frac{5}{6}\right)_n.$$ Here, $$(a)_n = a(a + 1)(a + 2) \cdots (a + n – 1), \quad n > 1, \quad (a)_0 = 1,$$ is the Pochhammer symbol. I do not really know how one converts expressions involving factorials to products of the Pochhammer […]

I know this has something to do with factorials, and combinations and permutations. I’ve been puzzling over this for a little while, and I can’t come up with an answer. My question is, How would one figure out how many ways there are to use exactly 10 numbers from 1 to 12 to sum 70. […]

I was playing around with factorials on Wolfram|Alpha, when I got this amazing result : $$\sum \limits_{n=0}^{\infty} \dfrac{n!}{(n+1)!+(n+2)!} = \dfrac{3}{4}.$$ Evaluating the first few partial sums makes it obvious that the sum converges to $\approx 0.7$. But I am not able to prove this result algebraically. I tried manipulating the terms and introducing Gamma Function, […]

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