# Tag Info

4

Here the author is using the Citardauq Formula $x=\frac{2c}{-b\mp \sqrt{b^2-4ac}}$. It can be useful in avoiding loss of significance in the computation of one of the roots. We can obtain the Citardauq Formula by multiplying top and bottom of the Quadratic Formula by $-b\mp\sqrt{b^2-4ac}$. Suppose for example that $b$ is fairly large positive, and $4ac$ ...

2

You can verify that \begin{align} \sum_{k=1}^{3\ 260\ 805}\frac{1}{p_k} &= 3.1415926\color{blue}{396}... \\ \pi &= 3.1415926\color{blue}{535}... \\ \sum_{k=1}^{3\ 260\ 806}\frac{1}{p_k} &= 3.1415926\color{blue}{579}... \end{align} so $$N=3\ 260\ 806.$$

2

$$x\tan(x)=(\epsilon^\alpha x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\cdots)^2+\frac{1}{3}(\epsilon^\alpha x_0+\epsilon^\beta x_1+\epsilon^\gamma x_2+\cdots)^4+O(\epsilon^\kappa)=\epsilon$$ and you know that $\alpha<\beta<\gamma<\kappa$ and that $\alpha$ should be chosen to make the largest term on the LHS be $O(\epsilon)$. Expanding gives ...

2

Let $a=4\sqrt k$ $$\frac{d^2\theta}{dt^2} + a\frac{d\theta}{dt}+g\sin(\theta)=0$$ Change of function : $\frac{d\theta}{dt}=z(\theta)$ Do not confuse with $\frac{d\theta}{dt}=z(t)$ which fails to reduce the order of the ODE. $\frac{d^2\theta}{dt^2}=\frac{dz}{dt}=\frac{dz}{d\theta}\frac{d\theta}{dt}=z\frac{dz}{d\theta}$ $$z\frac{dz}{d\theta} + ... 2 A reference to the Doolittle's paper is missing on Wiki: Doolittle, M.H.: Method employed in the solution of normal equations and the adjustment of a triangularization. In: U. S. Coast and Geodetic Survey, Report, pp. 115–120 (1878). For more information, see Mathematicians of Gaussian Elimination by Grcar. Apparently, Doolittle worked as a computer. 1 Use the quadratic Taylor formula, i.e., linear expansion with quadratic error term,$$ f(a+h)=f(a)+f'(a)h+\frac12 f''(a+\theta h)h^2 $$with h=-\frac{f(a)}{f'(a)}. Note that the constant and linear term cancel for the Newton update so that you get f(x)=O(f(a)^2). For x to be better than a you need that the constant in this asymptotic estimate is ... 1 It needs some more careful statement in order to become subject to proof. It is not always true that a Newton iteration produces a closer approximation to a root, though there is a basin of attraction around a simple root of a smooth (e.g. twice continuously differentiable) function wherein the iterations will converge to that root. The analysis of ... 1 Crank-Nicolson for the PDE \frac{∂u}{∂t}=L(t,u) plus boundary conditions first performs a discretization in the space dimension(s) forming a vector of sample point values \vec u(t)=\{u(t,x_k)\}_{k=0,…,N}, x_k=x_0+k·Δx, and transforming the operator L plus the boundary conditions into a function f(t,\vec u) of this vector approximating them. Then ... 1 This is the midpoint method. With some elementary transformations you find that$$ U^*=\frac{U^n+U^{n+1}}2 Consider for simplicity an autonomous equation u'=f(u). Write the midpoint as increment of the last point, U^*=U_n+\frac k2 S. Recursively inserting this in the fixed point equation gives \begin{align} S&=f(U^n+\tfrac k2 ... 1 If there are no restrictions on g apart from those mentioned by you, then clearly your claim is false as you can see by considering the function g: [0,1] \to [0,1] defined by g(x) = x^2. In this case g'(1) = 2. After your edit: This the most I can say If g is a function satisfying the mentioned property then for all x,y \in [a,b] ... 1 If 0< g'(x) < 1 everywhere on [a,b], then the following property must hold:b-a > g(b)-g(a).$$And a suitably similar condition for when -1 < g'(x) < 0. In other words, the function must grow more slowly than x over the interval. 1 If you are dealing with this numerically, use expm1(n*log1p(x)) (if those functions are available in your math library) to get high precision results. expm1(x)=exp(x)-1 and log1p(x) = ln(1+x) both for small x without cancellation. 1 You missed that the left boundary condition is a derivative u_x(0,t)=t/5. Transcribing task descriptions has its uses. You can compare your solution to the exact solution u(x,t)=(x+t)^2/10. If the method is correctly second order, the numerical value should coincide with the actual values. 1 You have to evaluate them exactly as the method prescribes. You already did this correctly for the RK2 method. This might seem to be a lot more effort for RK4. But consider that RK4 is O(h^4). Roughly, to get a accuracy of e.g. about 10^{-4} for t=1 you need h=0.1 and 10 steps netting 40 function evaluations. To get the same accuracy for the ... 1 Almost all engineering calculations involve solutions of nonlinear equations and, in a single steady state simulation, these are typically solved billions of times (in a dynamic simulation, zillions of time). Most of the time, these equations are solved using Newton-Raphson method (or higher order equivalents). Just to give you a simple one, consider ... 1 From what I understand, \Delta\theta and \Delta\tilde g are column vectors with n entries (so they are n\times 1 matrices) and A is an n\times n matrix. Therefore, \Delta\theta^T A is the product of a 1\times n matrix with a n\times n matrix (notice the transpose on the \Delta\theta), so that the result is a 1\times n matrix, also ... 1 You could start with Taylor expansion on each term:$$f(x-h)=f(x)+f'(x)(-h)+\frac{1}{2!}f''(x)(-h)^2+O(h^3)\\ f(x)=f(x)\\ f(b)=f(x)+f'(x)(b-x)+\frac{1}{2!}f''(x)(b-x)^2+O((b-x)^3)$$Then add three equations with weights w_0,w_1,w_2:$$w_0f(x-h)+w_1f(x)+w_2f(b)=w_0f(x)+w_1f(x)+w_2f(x)+w_0f'(x)(-h)+w_2 f'(x)(b-x)+w_0 \frac{1}{2!}f''(x)(-h)^2 + w_2 ...

Only top voted, non community-wiki answers of a minimum length are eligible