# Tag Info

32

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: $$L^s(L^tf(x)) = L^sf(x+t) = f(x+s+t) = ... 12$$ \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 ...

7

This is a difficult question to answer. The FDM is the oldest and is based upon the application of a local Taylor expansion to approximate the differential equations. The FDM uses a topologically square network of lines to construct the discretization of the PDE. This is a potential bottleneck of the method when handling complex geometries in multiple ...

7

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 ...

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.

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 $$\Delta^k x(x - 1)\cdots(x - (n-1)) = n! {x \choose n-k} = k! {n \choose k} x(x - 1) \cdots(x - (n-k-1))$$ ...

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 $$f[x_0]:=f(x_0),$$ $$f[x_0,\dots,x_j]:=\frac{f[x_1,\dots,x_j]-f[x_0,\dots,x_{j-1}]}{x_j-x_0}\quad,\quad j\ge1. ... 5 UPDATE : Let's start by showing a solution of the difference equation :$$\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 ...

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.

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 $$\sum_{i=a}^b \sum_{j=c}^d (-1)^{N-i+j} = (-1)^N \left(\sum_{i=a}^b (-1)^i\right) ... 4 how does one make sense of exponentiating or taking the logarithm of an operator? The operator is linear, and therefore so are its positive integer powers, hence any power series in that operator has a chance of making sense. At least the series is a limit of linear operators, and the series makes perfect sense without any limiting process when applied ... 4 Actually you can use Taylor expansion to derive the formula$$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 ...

3

The interpolating polynomial is linear in the data, so all you need is $7$ basis polynomials corresponding to one of the given values being $1$ and the others being $0$. The three polynomials for $x_1$, $x_2$ and $x_4$, are straightforward; for instance \begin{align} p_4(x) &= ... 3 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 ... 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 $$for ... 3 Notice that$$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

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: $$\frac{\partial u}{\partial t} = \nabla \cdot (D \nabla u - ... 3 Just shift t:$$\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}$$In effect I’m substituting s=t+1, rewriting the summation in terms of s, and then renaming s back to t. 3 Use a Finite-Difference, Time Domain scheme, which uses centered time and space differences. You can scale your grid such that c=1. I will illustrate for an explicit scheme only. Here,$$u_i^n = u(i \Delta x,n \Delta t)$$where$$u_i^{n+1} = r^2 (u_{i+1}^n -2u_{i}^n + u_{i-1}^n) +2 u_i^n - u_i^{n-1}u_i^{n+1} = r^2 (u_{i+1}^n + u_{i-1}^n) ...

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

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. ...

3

We are given: $\tag 1 \displaystyle y'' + 4y = \cos(x), 0 \le x \le4, y(0) = 0, y\left(\frac{pi}{4}\right) = 0, h = \frac{\pi}{20}$ We know that if the linear boundary-value problem: $y'' = p(x)y' + q(x)y+r(x), a \le x \le b, y(a) = \alpha, y(b) = \beta$, satisfies: (i) $p(x), q(x)$, and $r(x)$ are continuous on $[a, b]$ (ii) $q(x) \gt 0$ on $[a, b].$ ...

2

We have $$\left(r \frac{\partial^2}{\partial r^2} +\frac{\partial}{\partial r} +r \frac{\partial^2}{\partial z^2}\right)\phi(r,z) = 0.$$ Letting $h=k=1$, Boyer makes the following natural replacement: $$\begin{eqnarray*} \frac{\partial^2\phi(r,z)}{\partial r^2} &\rightarrow& \phi(m+1,n)-2\phi(m,n)+\phi(m-1,n) \\ ... 2 Using the vector notations f=(f(i))_i and h=(h_i)_i and the matrix notation g=(g_n(i,j))_{i,j}, you impose that f=g\cdot h and you ask for an expression of h as a function of f and g. Let D=\det g. If g is not invertible, that is, if D=0, either this is impossible or the solution is not unique. If g is invertible, that is, if D\ne 0, ... 2 The trouble with your method is not when A is singular, it's when A is not diagonalizable. The solution of the initial value problem x_{n+1} = A x_n, x_0 given, is x_n = A^n x_0. Now we can write A = S^{-1} J S where S is invertible and J is in Jordan canonical form, and so x_n = S^{-1} J^n S x_0. For a d \times d Jordan block$$ J = ...

2

When you do the von Neumann analysis, I guess you end up with an amplification factor $G=U^{i+1}/U^i$ which depends both on $d=ka/h^2$ and $j$ (and $\theta$). So for stability you have to find $d$ which satisfies $|G(d,j,\theta)|\leq 1$, for $0<\theta<2\pi$ and $0<j<N$. The equation you are solving is not the same as the 1D equation, so it's not ...

2

It depends. Do A and B depend on the index $i$? If A and B are constant, then you can split the solution up into a homogeneous solution $u_{i}^{(H)}$ and an inhomogeoneous solution $u_{i}^{(I)}$. The homogeneous solution satisfies $$A u_{i+1}^{(H)} + 2 u_{i}^{(H)} + B u_{i-1}^{(H)} = 0$$ with initial conditions such as $u_{0}^{(H)} = u_0$ and ...

2

Stability in $L^\infty$ should imply $L^2$-stability (if your domain is bounded) since you have the bound $\|\cdot\|_{L^2}\leq C\|\cdot\|_{L^\infty}$. You have the continuous embedding $L^p(U)\subset L^q(U)$ for $p>q$, provided $U$ is bounded. This means that $\|u\|_{L^q}\leq C\|u\|_{L^p}$ for $u\in L^p(U)$. Generally, a good numerical scheme ought to be ...

2

well, not exactly for Neumann BCs, you don't fill in any values, you just supply a relation between the values of the points on the boundary and the points right next to them. think of this as having 2 different governing equations, one of them which is valid inside the boundary, and the other is valid only on the boundary. when you discretize your system, ...

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