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1

Here is a similar approach using basic complex variables. Suppose we seek to evaluate $$\sum_{k\ge 0} {2k\choose k} {n+k\choose m+2k} \frac{(-1)^k}{k+1}$$ where $n\ge m.$ Introduce the integral representation $${n+k\choose m+2k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{m+2k+1}} (1+z)^{n+k} \; dz.$$ This yields the following integral for the ...


1

The moment generating function for the binomial distribution $B_{n,p}$, whose discrete density is $\binom{n}{k}p^k(1-p)^{n-k}$, is defined as $$ \begin{align} M_{B_{n,p}}(t) &=\mathrm{E}(e^{tk})\\ &=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}e^{tk}\\ &=\sum_{k=0}^n\binom{n}{k}\left(pe^t\right)^k(1-p)^{n-k}\\ &=\left(pe^t+(1-p)\right)^n \end{align} ...


0

The Moment Generating Function of the Binomial Distribution Consider the binomial function (1) b(x;n,p) = (n!/x!(n−x)!)p^x q^(n−x) with q = 1 − p. Then the moment generating function is given by (2) Mx(t) = nX ext n! x=0 x!(n − x)!pxqn−x nX (pet)x n! x=0 x!(n − x)!qn−x = (q + pet)n, where the final equality is understood by recognising that it represents ...


1

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} ...


1

If we define $$ f_n(t)=\sum_{k=0}^\infty\binom{2k+n}{k}t^k\tag{1} $$ then we have $$ \begin{align} f_n'(t) &=\sum_{k=0}^\infty\binom{2k+n}{k}kt^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n-1}{k-1}(2k+n)t^{k-1}\\ &=\sum_{k=0}^\infty\binom{2k+n+1}{k}(2k+n+2)t^k\\[6pt] &=(n+2)f_{n+1}(t)+2tf_{n+1}'(t)\tag{2} \end{align} $$ If we define $$ ...


3

The following method could be explained to any student that understands equations for lines and knows how to sum an arithmetic sequence. However, you may find it at least as tedious as the method you already have! The Idea Think of points $(x,y)$ in the plane as representing "$x$ 2-cent coins and $y$ 3-cent coins".   Now, the grid points in the first ...


1

It is possible to shorten a bit the argument by noticing that $$\operatorname{Res}\left(\frac{1}{(1-z)(1-z^2)(1-z^3)},z=1\right)=-\frac{17}{72},$$ hence the contribute given by the simple poles $-1,\omega,\omega^2$ lying on the unit circle cannot exceed $\frac{17}{72}$, and is exactly equal to $\frac{17}{72}$ when $6\mid n$. The pole in $z=1$ is a triple ...


3

I'll only show how to obtain a linear recurrent relation from the problem, it has been explained at various places how to solve these. Let: $a(n)$ be solutions (ways how to change $n$) using only $1$c, $b(n)$ solutions using only $1,2$c and at least one $2$c, $c(n)$ solutions using all three coins and at least one $3$c. Clearly each solution fits ...


1

I am getting: $\left(x+x^2+x^3+x^4+...+x^{15}\right)\left(x^{15}+x^{20}+x^{25}+...+x^{85}\right)$ for the first kid. $(\left(x^3+x^4+x^5+...+x^{15}\right) \left(x^{15}+x^{20}+x^{25}+...+x^{45}\right))^3$ for the next 3 kids. Then multiply both of them and check the coefficient of $x^{100}$


1

Using Maple and the information given at http://en.wikipedia.org/wiki/Hermite_distribution it is possible to obtain The first ten probabilities are These results can be obtained as follows. The generating function for the Hermite polynomials is $${{\rm e}^{2\,xt-{t}^{2}}}=\sum _{n=0}^{\infty }{\frac {H_{{n}} \left( x \right) {t}^{n}}{n!}}$$ Then ...


0

$G$, using your notation, is known in probability as the probability generating function. See here for more details. It is seen on this link that $$p(k) = \dfrac{\left.G^{(k)}(s)\right|_{s=0}}{k!}$$ where $G^{(k)}$ denotes the $k$th derivative.


0

Here is a proof using complex variables. We seek to show that $$\sum_{k=0}^n {n\choose k} {n+k\choose k} F_{k+1} =\sum_{k=0}^n {n\choose k} {n+k\choose k} (-1)^{n-k} F_{2k+1}.$$ Start from $${n+k\choose k} = \frac{1}{2\pi i} \int_{|z|=1} \frac{1}{z^{k+1}} (1+z)^{n+k} \; dz.$$ This yields the following expression for the sum on the LHS $$\frac{1}{2\pi i} ...


0

Suppose we seek to evaluate $$\sum_{k=0}^{n+1} {n+1+k\choose 2k} (-1)^{n-k} 4^k = (-1)^n \sum_{k=0}^{n+1} {n+1+k\choose n+1-k} (-4)^k.$$ Start from $${n+1+k\choose n+1-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+2-k}} (1+z)^{n+1+k} \; dz.$$ This yields the following expression for the sum $$\frac{(-1)^n}{2\pi i} \int_{|z|=\epsilon} ...


0

Multiplying formal power series in multiple indeterminates is not really much more complicated than doing so for univariate ones. The main difference is that each term has not a single degree, but separate degrees for each of the $k$ indeterminates, so the coefficient is attached to a point of $\def\N{\Bbb N}\N^k$ rather than of$~\N$. With an infinite ...


1

When branching processes are concerned, conditioning on the first generation often uncovers the recursive structure of the whole object, hence its properties. In the present case, call $N$ the total size of the random tree $T$ and $L$ the progeny of the ancestor, assuming the branching process starts from an initial population of $1$. If $L=0$, then ...


8

My question is how to find the easiest way to find the number of non-negative integer solutions to $$x+2y+4z=400$$ I think the following way is easy (I'm not sure if it's the easiest, though). Since $x+2y+4z=400$, $x$ has to be even. So, setting $x=2m$ gives you $$2m+2y+4z=400\Rightarrow m+y+2z=200.$$ Since $m+y$ has to be even, setting $m+y=2k$ gives ...


1

You were on the right track, but instead of truncating the factors, just consider the coefficient of $x^{400}$ in: $$(1+x+x^2+x^3+\ldots)(1+x^2+x^4+x^6+\ldots)(1+x^4+x^8+x^{12}+\ldots)=\frac{1}{(1-x)(1-x^2)(1-x^4)},\tag{1}$$ then write the RHS of $1$ as a sum of terms like $\frac{A}{(1-\xi x)^k}$, with $\xi\in\{1,-1,i,-i\}$, and exploit the identities: ...


0

Using other method, I got a different answer; I'm sure about my answer $$E=\text{Coefficient of } \large \alpha^s\text{ in }(\alpha^0+\alpha^1+\alpha^{-1})^N\\ =\text{Coefficient of } \large \alpha^s\text{ in }\alpha^{-N}(1+\alpha^1+\alpha^2)^N$$ Let power of $\alpha$ from bracket be k,then since $s=-N+k$,$k=s+N$ $$ E=\text{Coefficient of } \large ...


0

What is the $x^r$ coefficient in $(1+x)^n$? What is the $x^r$ coefficient in $(1+x)^{n-1}$? What is the $x^r$ coefficient in $x(1+x)^{n-1}$? After answering these questions, take another look at the identity $(1+x)^n = (1+x)^{n-1}+x(1-x)^{n-1}$ and equate the $x^r$ coefficients on both sides.


0

I'd write it as $$ a_{n+1}=\frac{3}{4}a_n+4e $$ so we can start counting from $0$. We also have $$ a_{n+2}=\frac{3}{4}a_{n+1}+4e $$ so $$ a_{n+2}=\frac{3}{4}a_{n+1}+a_{n+1}-\frac{3}{4}a_n $$ or $$ 4a_{n+2}-7a_{n+1}+3a_n=0 $$ which has, as characteristic polynomial $4X^2-7X+3$. The roots are $1$ and $3/4$, so the general solution is $$ a_{n}=\alpha ...


0

Generating functions are a good method, but for this straightforward problem there's another one, since you can get rid of the constant: $$ a_ n = \frac{3}{4} a_{n-1} + 4e\\ a_{n-1} = \frac{3}{4}a_{n-2} +4e\\ a_n - a_{n-1} = \frac{3}{4}(a_{n-1} -a_{n-2}) $$ Now set $\Delta a_n = a_n - a_{n-1}$ $$ \Delta a_n = \frac{3}{4} \Delta a_{n-1} = ...


2

As Did notes in comments above, getting a generating function is a bit of overkill here. That said, I figured I'd show how it works in this case for both exponential and ordinary generating functions. Exponential GF We introduce $A_e (x)=\sum\limits_{n=0}^\infty \dfrac{a_n}{n!}x^n$ as the exponential generating function of $\{a_n\}$. Then \begin{align} ...


0

I can't provide a nice formula as in your referenced question, but I can show you at least half of the way. Let's start with your Example: There are two dice with 4 and 6 sides. How many ways are there, so that rolling both dice once result in 8 pips? We encode the dies with polynomials and use the exponents to label the pips \begin{align*} ...


1

It may interest the reader that the generating function $P(z, y)$ of this sum can be evaluated using the technique of annihilated coefficient extractors (ACE). Start by recalling the bivariate generating function of the Stirling numbers of the second kind, which is $$G(w, u) = \exp(u(\exp(w) -1)).$$ Now the sum is $$\sum_{j=0}^n {n\choose ...


1

Fibonacci is defined as $$F_1 = 1, F_2 = 1\\F_{n+1} = F_n + F_{n-1}$$ and never hits zero since it is obviously increasing. If you would take $$G_1=0,G_2=0\\G_{n+1} = G_n + G_{n-1},$$ then of course the sequence will be constantly zero. EDIT: According to your link, a sequence is causal if $x_n=0$ for $n<0$, however, this means that the definition ...



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