# Tag Info

7

Unfortunately, there is no single approach that will lead to robust, accurate, and high-performance implementations across the large universe of special functions. Often, two or more methods must be used for different parts of the input domain, and the necessary research and implementation work may take weeks for elementary functions and months for higher ...

3

Inspired by this question of mine, we can approximate the solution "quite" easily using Padé approximants. Let the equation be $$\frac{(1+i)^{13}\cdot i}{(1+i)^{13}-1}\cdot \frac{1}{1+i}-r=0$$ Building the simplest $[1,1]$ Padé approximant around $i=0$, we have $$0=\frac{\frac{2}{39} i (26 r+7)+\frac{1}{13} (1-13 r)}{1-\frac{4 i}{3}}$$ Canceling the ...

2

Hint : You have to substitute : We need a function $t(x)$ with $t(a)=-1$ and $t(b)=1$. The function $t(x)=\frac{2x-a-b}{b-a}$ does the job. This gives $x(t)=\frac{b-a}{2}t+\frac{a+b}{2}$ and $dx=\frac{b-a}{2}dt$, so we have $$\int_{a}^b f(x)\ dx=\frac{b-a}{2} \int_{-1}^1 f(\frac{b-a}{2}t+\frac{a+b}{2})\ dt$$

2

An affine transform may not preserve orthogonality. For example, the points $(0,0),(1,0),(0,1)$ form a right angle. By the transformation $(x,y)\to(x,x+y)$, they become $(0,0),(1,1),(0,1)$. These do no form a right angle. An affine transform has six degrees of freedom (independent coefficients), and can map any triangle onto any other triangle. All ...

2

If $f$ is continuous everywhere (for simplicity) and not differentiable at $x_0$, then $F(x)=\int_a^x f(y) dy$ is differentiable only once at $x_0$. For a concrete example, the function $$F(x)=\int_{-1}^x |y| dy$$ is differentiable only once at $0$. Thus the Newton iteration for $$G(x)=\int_{-1}^x |y| dy - \frac{1}{2}$$ for initial conditions close to ...

1

You can use both $|x_k-n| < \epsilon$ and $|x_{k + 1}-x_k| < \epsilon'$ as stopping criteria. Although you must be aware that there exists some sequence $(x_k)$ where for any $\epsilon > 0$, it exists $k$ such that $|x_{k + 1}-x_k| < \epsilon$ but $(x_k)$ does not converge. But $x\mapsto x^2$ is convex and you can prove Newton's method converge ...

1

I know this question is old, but I thought I would post some extra info around this. One reason your numerical solution to the Ordinary Differential Equation (ODE) is exploding, kind of regardless of the numerical method used, is because of the ODE itself. If you go and study some theory on numerical integration of ODEs, you will find a common model problem ...

1

What mathematical software tools do you have available? It usually doesn't pay to reinvent the wheel. In Maple you could do something like this: pde:= diff(u(x,t),t,t) = diff(u(x,t),x,x) + u(x,t)^2; ibc:= {u(0,t) = 1, D[2](u)(x,0) = 0, u(x,0)=1,u(1,t) = 1}; solution:= pdsolve(pde,ibc, numeric,time=t,range = 0..1);

1

$$g(x,T)\,\mathrm{d}x=\sqrt{1-u^2}\,\mathrm{d}u$$ where $u=(1-T)x+T\sin(x)$, so since $u=1$ when $x=x^\ast$, we have $$\int_0^{x^\ast}g(x,T)\,\mathrm{d}x=\int_0^1\sqrt{1-u^2}\,\mathrm{d}u$$

1

The underlying problem is that the plate is subjected to distributed load with density $f$, and is fixed along the boundary in such a way that the boundary can neither mode nor rotate (clamping condition). The function $u$ represents vertical displacement of the plate when it has reached an equilibrium. Since an equilibrium is being studied, no time is ...

1

Iterate: $$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a}$$ $$x_0=\frac {-b}{a}$$ I will try to edit my answer and put bounds that indicate the rate of convergence

1

Regarding the fishy solutions, when doing the discretization it can be important to regularize the solution or really strange things can happen when you try and solve the matrix-vector equation. As I have not investigated very many real problems in physics but mostly mimicked these types of equations for applications to other fields, I can't really say ...

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