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39

Well, it seems that you have just discovered a beautiful theory of (semi)group generators by yourself. To give some basics of it, let us consider a collection of "nice" functions on real values - e.g. bounded and having continuous derivatives. The action of operators $L^h$ on this space has a semigroup structure: $$ L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = ...


14

Let $X = \mathbb{R}^{\mathbb{Z}}$ be the space of real valued sequences defined over $\mathbb{Z}$. Let $R : X \to X$ be the operator on $X$ replacing the terms of a sequence by those on their right. More precisely, $$X \ni (\ell_n)_{n\in\mathbb{Z}} \quad\mapsto\quad ( (R\ell)_n = \ell_{n+1} )_{n\in\mathbb{Z}} \in X$$ The identities you have can be ...


13

$$ \begin{align} e^{d/dx} x^n & = \left(1+\frac{d}{dx} + \frac{(d/dx)^2}{2}+ \frac{(d/dx)^3}{6}+\frac{(d/dx)^4}{24}+\cdots+\frac{(d/dx)^k}{k!}+\cdots\right) x^n \\[10pt] & = x^n + nx^{n-1} + \frac{n(n-1)}{2}x^{n-2} + \frac{n(n-1)(n-2)}{6}x^{n-3}+\cdots \\[10pt] & \phantom{{}={}}\cdots+\frac{n(n-1)\cdots(n-k+1)}{k!}x^{n-k}+\cdots+0+0+0+\cdots ...


9

There is: for every integer $n\geqslant0$, $$ f(x+n)=\sum_{k=0}^n{n\choose k}\Delta^k[f](x). $$


9

For the first part let $k=n-i$: $$\sum_{i=0}^n(-1)^i\binom{n}iy(i)=\sum_{i=0}^n(-1)^i\binom{n}{n-i}y(i)=\sum_{k=0}^n(-1)^{n-k}\binom{n}ky(-k)\;.$$ Now note that $(-1)^{n-k}=(-1)^{n+k}$, and you have $$\sum_{i=0}^n(-1)^i\binom{n}iy(i)=(-1)^n\sum_{k=0}^n(-1)^k\binom{n}ky(0+n-k)=(-1)^n\Delta^ny(0)\;.$$ Improved version: For the second part, note that ...


8

Note that since $R$ and $1$ commute, $$ R^{2^k}-1=\left[\sum\limits_{j=0}^{2^k-1}R^j\right](R-1)\tag{1} $$ Therefore, $$ \begin{align} \prod_{k=0}^{n-1}\left(R^{2^k}-1\right)x^n &=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right](R-1)^nx^n\tag{2a}\\ &=\left[\prod_{k=0}^{n-1}\sum_{j=0}^{2^k-1}R^j\right]n!\tag{2b}\\ ...


7

Provided the values of $g$ lie in the domain of $f$ and $\Delta g(n)$ is an integer, you have the obvious rule $$ \Delta(f\circ g)(n)=\sum_{d=0}^{\Delta g(n)-1}\Delta f\bigl(g(n)+d\bigr), $$ where the summation must be interpreted as a sum of negated terms in case $\Delta g(n)<0$, similarly to integrals whose upper limit is lower than their lower limit. ...


6

UPDATE : Let's start by showing a solution of the difference equation : $$\Delta w+w-w^2-1=0$$ at least if this means $\ (w_{n+1}-w_n)+w_n=w_n^2+1$ because : $$w_{n+1}=w_n^2+1$$ admits the solution (for the specific case $w_0=1$) : $$w_n=\lfloor c^{2^n}\rfloor,\\\text{with}\quad c=\exp\left|\sum_{j=0}^\infty 2^{-j-1}\ln(1+w_j^{-2})\right|,\\c\approx ...


6

Actually, for Mathematica 7 and later versions, you have the functions Identity[], DiscreteShift[], and DifferenceDelta[]: Identity[f[x]] f[x] DiscreteShift[f[x], x] f[1 + x] DifferenceDelta[f[x], x] -f[x] + f[1 + x] The backward difference needs a bit more work: DifferenceDelta[DiscreteShift[f[x], {x, 1, -1}], x] -f[-1 + x] + f[x] Otherwise: ...


6

By linearity, it suffices to prove this for the polynomials $x(x - 1)\cdots(x - (n-1))$. This is just $n! {x \choose n}$. A basic property of the forward difference operator is that $\Delta {x \choose n} = {x \choose n-1}$, from which it follows that $$\Delta^k x(x - 1)\cdots(x - (n-1)) = n! {x \choose n-k} = k! {n \choose k} x(x - 1) \cdots(x - (n-k-1))$$ ...


6

The fact that it is second-order refers to the fact that the largest difference in indices is $2$. For example, $$ R_{n+4}=3R_{n+1}^2+R_n $$ is a fourth-order difference equation and $$ R_{n+3}=2R_{n+2}\cdot R_{n+1} $$ is a second order difference equation. If you're familiar with ODEs, the terminology is analogous.


5

For the general sum, Mathematica gives the closed-form expression $$\sum_{i=a}^b \sum_{j=c}^d (-1)^{N-i+j} = \frac{(-1)^{N-a-b}}{4} \left( (-1)^a + (-1)^b \right) \left( (-1)^c + (-1)^d \right).$$ Or, if you prefer a simpler answer but in piecewise form, write $$\sum_{i=a}^b \sum_{j=c}^d (-1)^{N-i+j} = (-1)^N \left(\sum_{i=a}^b (-1)^i\right) ...


5

how does one make sense of exponentiating or taking the logarithm of an operator? The operator is linear, and therefore so are its positive integer powers, hence any power series in that operator has a chance of making sense. At least the series is a limit of linear operators, and the series makes perfect sense without any limiting process when applied ...


5

As kindly suggested by Patrick Da Silva, I'm turning my comment into an answer. Let $x_0,x_1,\dots$ be distinct real numbers, let $f$ be a polynomial function on $\mathbb R$, and define $f[x_0,\dots,x_j]$ for $j=0,1,\dots$ recursively by $$ f[x_0]:=f(x_0), $$ $$ f[x_0,\dots,x_j]:=\frac{f[x_1,\dots,x_j]-f[x_0,\dots,x_{j-1}]}{x_j-x_0}\quad,\quad j\ge1. ...


5

The gamma function naturally generalizes the factorial to complex values. It satisfies the functional equation $x\Gamma(x)=\Gamma(x+1)$ for any $x$ (when both sides exist anyway). Hence $$\Gamma\big(x-(n-1)\big)\prod_{k=0}^{n-1}(x-k)=\Gamma(x+1)$$ by induction. Divide by the $\Gamma$ on the left and we're done.


4

$$ x^\underline n = \prod_{k=0}^{n-1}(x-k)= \frac{\Gamma(x+1)}{\Gamma(x+1-n)} $$


4

According to Florian Cajori, A history of mathematical notations (1928 - Dover reprint) : "A provisional, temporary notation $\Delta$ for differential coefficient or différences des fonctions was used in 1706 by Johann Bernoulli." (see Cajori, page 205 of 2nd vol). L.Euler introduced the symbolism for finite differences in Institutiones calculi ...


4

Actually you can use Taylor expansion to derive the formula $$y_{i-1}=y(x-\Delta x_i)=y(x)-\frac{dy}{dx}\Delta x_i+O(\Delta x^2)$$ $$y_{i+1}=y(x+\Delta x_{i+1})=y(x)+\frac{dy}{dx}\Delta x_{i+1}+O(\Delta x^2)$$ By neglecting higher order terms $O(\Delta x^2)$ $$y_{i+1}-y_{i-1}=\frac{dy}{dx}\Delta x_{i+1}+\frac{dy}{dx}\Delta x_i\Rightarrow ...


4

Hint: Notice that $\Delta^k (x^n)$ is a polynomial of degree $n-k$, because the highest order term in $x$ cancels out with each application of $\Delta$ (of course you should prove this). So for $\Delta^n (x^n)$ what you have to show is that all that will be left is the constant term.


4

Just shift $t$: $$\sum_{t=1}^{n-1}(t+1)^{\underline 4}=\sum_{t=2}^nt^{\underline 4}=\frac15\left((n+1)^{\underline5}-2^{\underline 5}\right)=\frac15(n+1)^{\underline 5}$$ In effect I’m substituting $s=t+1$, rewriting the summation in terms of $s$, and then renaming $s$ back to $t$.


4

Here is an old scicomp.SE question that answered some of your question: What are criteria to choose between finite-differences and finite-elements? In my humble opinion, FEM is the most flexible one in terms of dealing with complex geometry and complicated boundary conditions. FEM also allows the adaptive/local procedure to get higher order local ...


4

You are right. In the first case, for $$\frac{f(x+h)-f(x)}{h} - f'(x)$$ you only have an $o(1)$ bound. A function like $f(x) = x\cdot\lvert x\rvert^\alpha$ for $0 < \alpha < 1$ is continuously differentiable on all of $\mathbb{R}$, but at $0$ the difference quotient converges only of the order $\lvert h\rvert^{\alpha}$ to the derivative. In the ...


4

Suppose we are modeling a quantity $u$, say the concentration of a chemical, driven by a flow in some fluid in a region $\Omega$ with no source (meaning we are not adding more chemical into the fluid after starting the timer). Then the convection-diffusion pde to describe the phenomenon is: $$ \frac{\partial u}{\partial t} = \nabla \cdot (D \nabla u - ...


4

Notice that $$x^{\underline k}=\frac{x^{\underline{k+1}}}{x-k}$$ for $k\ge 0$. If we generalize this to negative $k$, we have $$\begin{align*} x^{\underline{-1}}&=\frac{x^{\underline0}}{x-(-1)}=\frac1{x+1}\\\\ x^{\underline{-2}}&=\frac1{(x+1)(x+2)}\\\\ &\;\vdots\\\\ ...


3

counterexample :set $f(x)=-1$ for $0\leq x<1$ , $f(x)=1$ for $1<x\leq 2$ and $f(1)=0$. Clearly that limit is zero on for $x\not=1$ and $\frac{f(1+h)+f(1-h)-2f(1)}{h^2}=0. $


3

While the approach of robjohn is certainly possible, it is often better to take the approach suggested by the original poster: $$ \left\{\frac{d}{dx}\left[ A(x)\frac{d\, u(x)}{dx}\right]\right\}_i = \frac{A_{i+1/2}\left[\frac{u_{i+1}-u_{i}}{h}\right] - A_{i-1/2}\left[\frac{u_{i}-u_{i-1}}{h}\right]}{h} $$ As he noted, $A$ is evaluated on half grid point. ...


3

This is really the same answer as that of Qiaochu Yuan, but I find the "binomial coefficients of $x$", much as I approve the notation, a bit distracting when next to ordinary binomial coefficients. One can do without them, using falling factorial powers instead: $x^\underline n=x(x-1)\ldots(x-n+1)$, which is of course the same as $n!\binom xn$. Elementarily ...


3

Split your $\Phi$ and $I$ into columns, so \begin{align} \Phi &= \left [ \phi_1, \phi_2, \ldots, \phi_n\right ] \\ I &= \left [e_1, e_2, \ldots, e_n \right ] \end{align} where $\phi_i, e_i \in \mathbb R^n,\ \forall i$. So, instead one matrix system you'll get $n$ linear systems $$ \frac {d \phi_i}{dt} = A(t) \phi_i(t) \\ \phi_i(t_0) = e_i $$ for ...



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