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169

For any holomorphic function $G$, we can define a corresponding matrix function $\bar{G}$ via (a formal version of) the Cauchy Integral Formula: We set $$\bar{G}(B) := \frac{1}{2 \pi i} \oint_C G(z) (z I - B)^{-1} dz ,$$ where $C$ is an (arbitrary) anticlockwise curve that encloses the eigenvalues of the (square) matrix $B$. Note that the condition on $C$ ...

32

The gamma function is analytic. Use the power series of it. EDIT: already done: Some properties of Gamma and Beta matrix functions (maybe paywalled).

12

I don't have enough reputation points to comment on Travis' answer, but his numerical result is incorrect. Using Julia I get A = [1 3;2 1] EVD = eigfact(A) V = EVD[:vectors] g = gamma(EVD[:values]) gammaA = V * diagm(g) * inv(V) factA = A * gammaA 3.62744 8.84231 5.89488 3.62744 As long as cond(V) isn't too terrible, I've found the above procedure ...

12

It is zero. Actually $n!$ is a multiple of $10$ for $n\ge 5$, and raising a multiple of $10$ to any natural power results in a multiple of $10$ again. To elaborate, $235!$ ends in $\lfloor\frac{235}5\rfloor+\lfloor\frac{235}{25}\rfloor+\lfloor\frac{235}{125}\rfloor+\ldots = 57$ zeroes, hence $235!^{69}$ ends in $57\cdot 69=3933$ zeroes. The last non-zero ...

10

There are 403 numbers between 1 and 2015 are divisible by 5. Not all of them contribute just one factor of 5. The multiples of 25 contribute two factors of 5 (there are 80 such). We already have 483 factors of 5. A further 16 numbers are divisible by 125 and they contribute yet another factor of 5. We're now up to 499 factors of 5. Finally, the multiples ...

8

I would start from the logical definition of the matrix factorial, without assuming that we want to cover all properties that we know from factorial in set of reals. We define standard factorial as $1 \cdot (1+1) \cdot (1+1+1) \cdot ... \cdot (1+1+...+1+1)$ So first let us define $[n]!$ using the same logic replacing 1 with identity matrix. The obvious way ...

8

You can use $\nu_{p}(n!)=\sum \limits_{k\ge 1}\left[\dfrac{n}{p^k}\right]$ and that $[a+b]\ge [a]+[b]$

8

Let $U_{n,x}=\{(x_1,x_2,\dots,x_n)\mid 0<x_1<x_2<\cdots <x_n\leq x\}$. It is not hard to show, by a symmetry argument, that the hypervolume of this region is $\frac{x^n}{n!}$, because there are $n!$ ways to permute the $x_i$ to get a different order, and this covers "almost all" of the $n$-dimension hypercube with side $x$. An interesting ...

7

Yes, and that's precisely why $n!$ appears in the denominator of the term of a Taylor series containing $x^n$ (for simplicity, I'll assume the series is centered at $x=0$). That term is $\frac{f^{(n)}(0)}{n!}x^n$. When you take $n$ derivatives and plug in $x=0$, you get just $f^{(n)}(0)$ as desired. That's why the Taylor series has the correct derivatives ...

7

If we use as a lemma that the product of $k$ consecutive integers is divisible by $k!$ (proven e.g. here: The product of n consecutive integers is divisible by n factorial), we see that $(a+b)!/a!$ factors into the $b$ consecutive integers $a+1,\ldots,a+b$, hence is divisible by $b!$. Then $a!b!|(a+b)!$.

7

We have the following series: $$\sum_{n=0}^{\infty} \frac{(n!)^{2}}{(2n)!}$$ To prove convergence, use the Ratio Test: $$\lim_{n\to\infty}\left|\frac{\left((n+1)!\right)^{2}}{(2(n+1))!}*\frac{(2n)!}{(n!)^{2}}\right|$$ $$\lim_{n\to\infty}\left|\frac{(n+1)!(n+1)!}{(2n+2)!}*\frac{(2n)!}{(n!)(n!)}\right|$$ ...

6

We know that there are $\lfloor \frac{90}{2}\rfloor$ integers below $90$ that have at least one factor $2$, $\lfloor \frac{90}{4}\rfloor$ numbers that have $2$ factors $2$, etc. Thus, the number of factors $2$ in $90!$ is $$\sum_{k=1}^{\infty}\left\lfloor \frac{90}{2^k}\right\rfloor=86$$ We don't actually have to do this sum up to $\infty$, but only until ...

6

Note: I have added a failed attempt to prove that $x$ is transcendental by showing that $x$ is a Liouville number (see https://en.wikipedia.org/wiki/Liouville_number). Here is a formal proof of Winther's comment: Let $x =\sum_{p \in P} \frac1{p!}$, where $P$ is a sequence of strictly increasing positive integers. If $x = \frac{a}{b}$, $\begin{array}\\ ... 5 If you apply the ratio test you should get the limit of a(n+1)/a(n) tends to 1/4 hence we get convergence. 5 In the denominator$(2n)!= (2n)(2n-1) \dots (n+1)\cdot n!$. Each of the factors from$(n+1)$to$(2n)$are larger than$n$; there are$n$of these factors. So you can show that this sequence is less than$1/n!$. 4 By Stirling: $$\frac{(n!)^2}{(2 n)!} \sim \frac{n^{2 n} e^{-2 n} 2 \pi n}{(2 n)^{2 n} e^{-2 n} \sqrt{2 \pi 2 n}} = \frac{\sqrt{2 \pi n}}{\sqrt{2} 2^{2 n}}$$ so the sum certainly converges by comparison with $$\sum_{n=0}^{\infty} \sqrt{n} \, 2^{-2 n}$$ 4 Start from $$(x^n)^{(k)}=\dfrac{n!}{(n-k)!}x^{n-k},\quad\text{which is easy to prove by induction.}$$ and apply it replacing$k$with$n-k: $$(x^n)^{(n-k)}=\dfrac{n!}{(n-(n-k))!}\,x^{n-(n-k)}=\dfrac{n!}{k!}\,x^{k}.$$ Also $$\frac{n!}{k!}=\frac{n(n-1)\cdots(k+1)k(k-1)\cdots1}{k(k-1)\cdots1}=n(n-1)\cdots(k+1).$$ 4 Completing the square gives, $$\frac{k!}{(k-2)!} = k(k-1) = k^2-k+\frac{1}{4} = 2!(45)+\frac{1}{4}.$$ This gives \begin{align*} \left(k-\frac{1}{2}\right)^2 &= \frac{361}{4}\\ k-\frac{1}{2}&=\sqrt{\frac{361}{4}}\\ k &= \frac{19}{2}+\frac{1}{2}\\ k&=10 \end{align*} 3 Let us suppose that you knowr$and want to know$k$such that$\binom{k}{r}=c$, with$c\in \mathbb{N}$. Let us name$(*)$this equation. For$r=0$,$\binom{k}{r}=1$, and you can easly find the solution of$(*)$if there exists. If$r=1,2,3,4$you can solve$(*)$relatively easy, cause$\binom{r}{2}=\frac{r(r-1)}{2}$,$\binom{r}{3}=\frac{r(r-1)(r-2)}{3!}$... 3 Short answer is no, there's nothing quite so concise involving factorials of integers. There are some other tools which allow you to write these expressions in shorthand - for example, your product would be$3^n(2/3)_n$in the Pochhammer notation. There are some combinations of the type of product that you mention which can be written in the factorial form. ... 3 Without loss of generality, we can assume$x\le y$. We can quickly rule out any solution with$x=1$or$2$, since the equations $$y!=1+y!+z!\quad\text{and}\quad 2y!=2+y!+z!$$ are easily seen to have no solutions in positive integers$y$and$z$. Now suppose we had a solution with$x\lt y$. Then $$x!={x!\over y!}+1+{z!\over y!}\implies{z!\over ... 3 As pointed out by user236182, Legendre's Formula, proven in (3) of this answer, says that for any prime p, the number of factors of p in n! is$$ \frac{n-\sigma_p(n)}{p-1} $$where \sigma_p(n) is the sum of the digits in the base-p representation of n. Since 2015_{\text{ten}}=31030_{\text{five}}, the sum of the base 5 digits is 7 so the ... 3 Related to factorials is the gamma (\Gamma) function. For positve integers it satisfies: \Gamma(n)=(n-1)!. For non-integers it is given by: \Gamma(t)=\int_0^\infty x^{t-1}e^{-x}dx. Further reading: Wikipedia 3 If subtracting 1 successively from a number x eventually gives 1, then x must be an integer, too, and so the definition$$x! := x (x - 1) \cdots (2) (1)$$is only defined for positive integers x. (We can extend this by defining 0! to be the value 1 of the empty product; this is a good choice in the sense that the identity x! = x (x - 1)! then ... 3 Making use of:$$r.\left(r+1\right)!=\left(r+2-2\right).\left(r+1\right)!=\left(r+2\right)!-2.\left(r+1\right)!$$we write the summation as:$$\left[\left(n+2\right)!-2.\left(n+1\right)!\right]+\left[\left(n+1\right)!-2.n!\right]+\cdots+\left[4!-2.3!\right]+\left[3!-2.2!\right]$$leading to: ... 3 Look at the general expression$$y_i = \frac{{\rm d}^i x^n}{{\rm d}x^i}$$This is the i-th derivative of x^n for i=1\ldots n. This can be directly evaluated as$$ \boxed{ \frac{{\rm d}^i x^n}{{\rm d}x^i} = \frac{n!}{(n-i)!} x^{n-i} } $$So the i-th derivative of a n-th order polynomial contains terms of n!, (n-1)!, (n-2)! etc 3 This answer is a slightly different variation of the theme which affirms the result of @MarkoRiedel. Here we use exponential generating functions to count the number of configurations of labelled objects and apply the coefficient of operator [x^n] to denote the coefficient of x^n in a generating series. If we are looking for the number of ... 3 The identity you're seeking is closely related to finite differences. The finite differences \Delta^1f, \Delta^2f, \Delta^3f,\ldots of a function f are defined inductively as$$(\Delta^1f)(x) := f(x+1) -f(x)$$and$$(\Delta^{n+1}f)(x) := \Delta^1(\Delta^nf)(x), $$i.e. the (n+1)st finite difference of f is the first finite difference of the nth ... 3 Probably the order of the five white balls does not matter, so you have to divide by 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 to account for the different orders in which these could be drawn. Your calculation would be correct if the order did matter, i.e. if 1,2,3,4,5,6 wins, but not 2,1,3,4,5,6. 2 When you calculates your answer of 35 billion you counted that there are 35 billion ways to choose a ticket where ORDER MATTERS. What this means is that we have counted the tickets$$1 2 3 4 5 r10$$and$$5 4 3 2 1 r10$\$ as different tickets, but since order doesn't matter with white balls, they aren't actually different. This we need to change how we ...

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