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0

Not clearly more convenient, but an alternative form can be built by first establishing that: $$\int_0^1\frac{x^n}{n!}e^{-x}\,dx=1-\frac{1}{e}\sum_{k=0}^n\frac{1}{k!}$$ [This is best done by induction - if we call the integral on the left $I_n$, integrating by parts quickly gives us $I_n=I_{n-1}-\frac{1}{e\cdot n!}$.] Noting that ...

1

If $n>4$, then $(n!)^k \equiv 0\mod{10}$ $\implies$ the last digit is a $0$. To see this, note since $n>4$, $n!=(n)(n-1)\cdot\cdot\cdot(5)(4)(3)(2)(1)$. We can rewrite this as: $$10\cdot[(n)(n-1)\cdot\cdot\cdot(4)(3)(1)]$$ Now our modulo definition states that since 10 is a factor of $n!$, taking its mod will yield a zero remainder ($\exists h\in ... 12 It is zero. Actually$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 ... 0 Another approach is to show that $$\left(1+\frac1n\right)^{2n}=\left(\frac{n+1}{n}\right)^{2n}>2.$$ The inequality follows directly from the binomial theorem. 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$n$th ... 2 so we would need to show that for all$j$such that$n\ge j\gt0$$$\sum_{k=0}^n \binom n k \binom n j (-1)^k (k)^{n-j} = 0$$ that is $$\sum_{k=0}^n \binom n k (-1)^k (k)^{n-j} = 0$$ this involves a Stirling number of second kind $$\sum_{k=0}^n \binom n k (-1)^k (k)^{n-j} =(-1)^n (n!){n-j\brace n}$$ which is$0$since$j\gt0$3 This answer is a slightly different variation of the theme which affirms the result of @MarkoRiedel. Here we use exponential generating functions to count the number of configurations of labelled objects and apply the coefficient of operator$[x^n]$to denote the coefficient of$x^n$in a generating series. If we are looking for the number of ... 1$\require{cancel}$$$\frac{n!}{k!} = \frac{n(n-1)(n-2)\dots}{k(k-1)(k-2)\dots} = \frac{n(n-1)(n-2)\dots(k+1)k(k-1)(k-2)\dots}{k(k-1)(k-2)\dots} = \frac{n(n-1)(n-2)\dots(k+1)\cancel{k(k-1)(k-2)\dots}}{\cancel{k(k-1)(k-2)\dots}} = n(n-1)(n-2)\dots(k+1)$$ 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 $$\frac{n!}{k!}=\frac{n(n-1)\cdots(k+1)k(k-1)\cdots1}{k(k-1)\cdots1}=n(n-1)\cdots(k+1).$$ 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 ... 2 Travis' answer is very nice. It would be good to mention that (almost) any matrix function can be made into a power-series expansion, which eventually involves the values of the function on the eigenvalues of the matrix multiplied by the eigenvectors. In other words the matrix function is completely characterised by the values it takes on the eigenvalues ... 0 We use Eulers Formula Γ(1-z)Γ(z)=$\frac\pi{\sin(\pi z)}$with$z=\frac12$. We then have$\Gamma(1/2)^2=\frac\pi{\sin( \frac\pi2)}$. Since$\sin( \frac\pi2)=1$we now have$\Gamma(1/2)^2=\pi$. Finally therefore$\Gamma(1/2)=\sqrt\pi$. Not$-\sqrt\pi$because we know$\Gamma( 1/2) > 0$. 2 With these kinds of problems it can be useful to employ Stirling numbers of the second kind which encapsulate inclusion-exclusion. The count then becomes quite simple. Suppose we first count$n$-digit numbers with one, two, three and four different digits where we include those that start with one or more zeroes. This is given by ... 1 There's a truly beautiful proof by Tim Gowers. And a similar question has been asked before. 0 Hint: Your intuition isn't that bad. There are indeed$45$even factors, hence$2^{45}$. But some factors still have other powers of$2$(like$4=2^2$or$48=2^4\cdot3$). Look at what happens if you erase the odd factors and divide all the even ones by$2$... 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 ... 1 With the settings \begin{align*} [k]_q:=\frac{1-q^k}{1-q}\qquad\text{and}\qquad [k]_q!:=\prod_{j=1}^{k}[j]_q=\prod_{j=1}^{k}\frac{1-q^j}{1-q} \end{align*} the q-binomial coefficient\begin{bmatrix}a+b\\a\end{bmatrix}_qis defined as \begin{align*} \begin{bmatrix}a+b\\a\end{bmatrix}_q:=\frac{[a+b]_q!}{[a]_q![b]_q!} \end{align*} Since the following is ... 2 Possibly the answer that you don't want: $$\frac{(a+0)!}{a!\cdot 0!}=1=\frac{(0+b)!}{0!\cdot b!}$$ and $$\frac{(a+b)!}{a!\cdot b!}=\frac{(a+b)(a+b-1)!}{a!\cdot b!}=\frac{(a-1+b)!}{(a-1)!\cdot b!}+\frac{(a+b-1)!}{a!\cdot (b-1)!}.$$ Then by induction, Pascal's triangle is made of integers. 8 If we use as a lemma that the product ofk$consecutive integers is divisible by$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)!$. 9 You can use$\nu_{p}(n!)=\sum \limits_{k\ge 1}\left[\dfrac{n}{p^k}\right]$and that$[a+b]\ge [a]+[b]$1 According to your approach regarding$N(x)$it seems you have a function$x!:\mathbb{R}\rightarrow\mathbb{R}\begin{align*} x!:=\prod_{k=0}^{n-1}(x-k) \end{align*} consisting ofn$factors$x-k$in mind. If so, we can write the function using the Pochhammer symbol $$(x)_n=x(x-1)(x-2)\cdots (x-n+1)$$ which can be written as polynomial in$x$... 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 ... 172 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). 1 Differentiate the natural logarithm of$(n+1)!=(n+1)~n!~$You'll get$f(n+1)-f(n)==\dfrac1{n+1},~$where$f(n)=\Big[\ln(n!)\Big]'=\dfrac{(n!)'}{n!}.~$At the same time, we know that$H_{n+1}-H_n=\dfrac1{n+1},~$where$H_n=\displaystyle\sum_{k=1}^n\dfrac1k~$is the n-th harmonic number. So$f(n)-f(0)=H_n.~$But$f(0)=(n!)'_{n=0},~$since$0!=1.~$... 0 You can try defining a function$F(x)=\prod_{k=0}^{\lfloor x-1\rfloor} (x-k)$, where$\lfloor\cdot\rfloor$means the floor function. For any integer$n$, the function$F(n)=n!$, but it is also defined for$x\in\mathbb{R}^+$. It is continuous, and differentiable for$x\notin\mathbb{N}$. It will be interesting to try to differentiate it in a neighborhood of a ... 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 ... 2 Let$r=2,\,k=1$. Then $$x^{k+r-1}=x^2$$ and $$\frac{\mathrm d^{r-1}}{\mathrm dx^{r-1}}\left(x^2\right)=\frac{\mathrm d}{\mathrm dx}\left(x^2\right)=2x.$$ This proves$\left(r-1\right)!$is not an$(r-1)$-th derivative of$x^{k+r-1}$by counterexample. 1 the correct answer is$\frac{d^{r-1}(x^{k+r-1})}{dx^{r-1}} = \frac{(k+r-1)!}{k!} x^k$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 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 ... 0 Use Stirling approximation for large$n$: $$(2n)! = \sqrt{4\pi n}\left(\frac{2n}{e}\right)^{2n}$$ So $$\lim_{n\to +\infty} \frac{n^n}{(2n)!} = \lim_{n\to +\infty}\frac{n^n\cdot e^{2n}}{2\sqrt{\pi n} (2n)^{2n}} = \frac{1}{2\sqrt{\pi}}\left(\frac{e^2}{4}\right)^{n}\frac{1}{\sqrt{n}n^n} \leq \frac{C}{\sqrt{n}n^n}$$ For some$C > 0$with$C < ...

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Basically, $$(2n)!=2n\,(2n-1)\,(2n-2)\cdots\,(n+1)\,n!\ \geq\ n^n\,n!$$ because each of the factor before $n!$ is g.e.q. than $n$, and there are $2n-n=n$ of these factors. So: $$n^n\frac{1}{(2n)!}\leq\ n^n\frac{1}{n^nn!}$$ and therefore your sequence is squeezed.

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$$\frac{\frac{(n+1)^{n+1}}{(2(n+1))!}}{\frac{n^n}{(2n)!}} = \frac{\left(\frac{n+1}n\right)^n}{2(2n+1)}<\frac3{4n}$$ Now, by induction, $$\frac{n^n}{(2n)!}\le K\frac{3^n}{4^n(n-1)!}$$ for some $K>0$.

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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 $1/n!$.

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You would be right if Joe and Josh were identical twins. One way to think about it is you have a block of $3$ boys who can be placed in any of the $10$ spaces between or to either side of the other $9$, so the answer is $3! \cdot 9! \cdot 10$.

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The $10!$ comes from treating the three boys as one and so we are now considering 10 boys and asking how to line them up. However, in the grouping of the three boys, they can be permuted among themselves in $3! = 3 \times 2$ ways.

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It is just for definition of ! function 0=1! And n!=(n-1)!n

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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 $x = \frac{a}{b}$, $\begin{array}\\ ... 2 Here is the inductive step: from$\;\dfrac{1\cdot 3\cdots(2n-1)}{2\cdot 4 \cdots(2n)}<\dfrac{1}{\sqrt{2n+1}}$, you deduce $$\dfrac{1\cdot 3\cdots(2n-1)(2n+1)}{2\cdot 4 \cdots(2n)(2n+2)}<\dfrac{1}{\sqrt{2n+1}}\frac{2n+1}{2n+2},$$ hence it is enough to prove$\;\dfrac{1}{\sqrt{2n+1}}\dfrac{2n+1}{2n+2}<\dfrac1{\sqrt{2n+3}}. This is equivalent to ... 2 First observe that $$\frac{2n-1}{2n}<\frac{2n}{2n+1}\qquad\text{since }(2n-1)(2n+1)=4n^2-1<4n^2=(2n)(2n)$$ Then \begin{align} \left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)^2&<\left(\prod_{k=1}^{n}\frac{2k-1}{2k}\right)\left(\prod_{k=1}^{n}\frac{2k}{2k+1}\right)\\ &=\prod_{k=1}^{n}\frac{2k-1}{2k+1}\qquad\text{this product is "telescopic"}\\ ... 0 Letp$be a positive integer. We answer a more general question. Is the sum $$S_p(n) = \sum\limits_{k=0}^{n-1} k \cdot (k+p)!$$ given as a hypergeometric term plus a constant. We will be using Gosper's algorithm . Denote$t_n := n \cdot (n+p)!$. Calculate the ratio of the terms in the sum: r_k = ... 1 If you just want it to work with 0, use $$0!=1\\ (n+1)!=(n+1)\cdot n!$$ The other options have been covered in other answers. 0 The$n^n$function is not an exponential as the basis varies. To prove$n!=o(n^n)$, let's evaluate the ratio: $$\frac{n!}{n^n}=\frac1n\cdot\frac2n\cdots\frac{n-1}n\cdot \frac nn <\frac 1n$$ since each factor$\dfrac kn$is less than$1$if$k<n$. This inequality implies$\;\displaystyle\lim_{n\to\infty}\frac{n!}{n^n}=0$. 2 When thinking about combinations, we can derive a formula for "the number of ways of choosing k things from a collection of n things." The formula to count out such problems is$n!/k!(n-k)!$. For example, the number of handshakes that occur when everybody in a group of$5$people shakes hands can be computed using$n = 5$(five people) and$k = 2$($2$... 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: ... 0 Observe that$x, y, z > 2$. Obviously,$y < z$. By symmetry, assume$x \leqslant y$. Then, we have$x \leqslant y < z$. Let$x = y$. Then, $$x! = 2 + z! / x!.$$ Let$k = z - x > 0$, so the last equality becomes $$x! = 2 + (x + 1) (x + 2) \cdots (x + k).$$ Since$x | x!$, we should have$x | k! + 2$. If$k \geqslant x$, then, we get$x | 2$, which ... 0 Hint: Since$m\ge n$, write$m=n+k$with$k\ge 0$and do induction over$k$. (So actually the induction is over$m-n$). For$k=0$it is obvious (but let's do it also for$k=1$). So, Base case:$k=1\begin{align} (n-1)!(n-1)+(n-2)!n+(n-1)(n-2)&\ge n(n-1) \iff\\ (n-1)!+\frac{(n-2)!n}{n-1}+n-2&\ge n\iff\\ (n-1)!+\frac{(n-1)!n}{(n-1)^2}&\ge ... 0 The following formula denotes the number of times that a prime factorp$appears in$n!$: $$C_{n}(p)=\sum\limits_{k=1}^{\infty}\left\lfloor\frac{n}{p^k}\right\rfloor$$ Since$\lfloor\log_{\color\green{2}}2015\rfloor=\color\red{10}\$: ...

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