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There are several ways to think about approximating $\tan x$ as a rational function. Taking the log-derivative of $\sin x=x\prod_{k\ge 1}\left(1-\frac{x^2}{k^2\pi^2}\right)$ gives $$\cot x=\frac{1}{x}-2\sum_{k\ge 1}\frac{x}{k^2\pi^2-x^2}.$$Keeping only the $k=1$ term,$$\tan x\approx\frac{x(\pi^2-x^2)}{\pi^2-3x^2}.$$This is a reasonable approximation if |x|\... 4 You can also directly apply Dirichlet's test (the integral of the numerator is zero when taken over a period, therefore uniformly bounded on all finite intervals). We have $$\cos^{2 k} u = 2^{-2 k} \binom {2 k} k + 2^{1 - 2 k} \sum_{1 \leq j \leq k} \binom {2 k} {k - j} \cos 2 j u, \\ \sum_{k \geq 0} \frac {(i \alpha \cos u)^{2 k}} {(2 k)!} = J_0(\alpha) + ... 4 I don't see an elegant way to do this so we will apply brute force starting from this formula$$w_k=\frac{a_n}{a_{n-1}}\frac{\int_a^bw(x)p_{n-1 }(x)^2dx}{p_n^{\prime}(x_k)p_{n-1}(x_k)}$$In our case we write it as$$v_k=\frac{b_m}{b_{m-1}}\frac{\int_0^1\sqrt y\,q_{m-1}(y)^2dy}{q_m^{\prime}(y_k)q_{m-1}(y_k)}Now b_m is the leading coefficient of q_m(y)=\... 4 Hints: (1) f extends to a continuous function [0,\infty)\to\mathbb{R}. (2) \lim_{x\to\infty}f(x)=0. 3 You know by Taylor expansions that u(t+h)=u(t)+hu'(t+h/2)+\frac{h^3}6u'''(t+h/2)+O(h^5). Insert the ODE for the first derivative and compare with the numerical method \begin{align} u(t+h)&=u(t)+hf(t+h/2, u(t+h/2))+\frac{h^3}6u'''(t+h/2)+O(h^5)\\ U_{+1}&=U+hf(t+h/2, \tilde U)\\[1em]\hline u(t+h)-U_{+1}&=u(t)-U+h\left[f(t+h/2, u(t+h/2))-f(t+h/2, \... 3 A defining property of the outer product is its anti-symmetry a∧b = -b∧a, $$which implies that$$ 2\,a∧a=0. 3 Let us diagonalize the matrix A such that {\bf u}_t + A {\bf u}_x = {\bf 0} as follows: \begin{aligned} A &= \begin{pmatrix}0 & a\\ b & 0\end{pmatrix} \\ &= \begin{pmatrix}-\sqrt{a/b} & \sqrt{a/b}\\ 1 & 1\end{pmatrix} \begin{pmatrix}-\sqrt{a/b} & 0\\ 0 & \sqrt{a/b}\end{pmatrix} \frac12 \begin{pmatrix}-\sqrt{b/a} & 1\\ \... 3 Your approach works, because it yields another ODE: Let z(t)=x(T-t). Then \frac{\partial }{\partial t}z(t)=\frac{\partial }{\partial t}x(T-t)\\ =-x'(T-t)=-f(T-t,x(T-t))\\ =-f(T-t,z(t)). $$3 In fact, that matrix does not have a dominant eigenvalue. You have that |\lambda_1| = |\lambda_2|=|\lambda_3| =1 and so there is no reason why the method should converge, unless you take as initial approximation an eigenvector associated with \lambda_1. The exact values for \lambda_2, \lambda_3 are -\frac 12 \pm \frac{\sqrt{3}}{2}. 3 You should check out chapter 3 in Demmel's text. Let me summarize some of the results for dense least-squares, for A an m\times n matrix: A is well-conditioned: In this case, using the normal equations is around as accurate as other methods and is also the fastest, so using them is fine, even though they are not numerically stable. A is not well-... 2 Start with the Taylor expansion$$ u(x,t+k)=u(x,t)+u_t(x,t)k+\frac12u_{tt}(x,t)k^2+... Use the PDE to convert time into space derivatives, u_t=-[f(u)]_x, then \begin{align} u_{tt}&=-[f(u)]_{xt}=-[f'(u)u_t]_x\\ &=[f'(u)[f(u)]_x]_x\\ \end{align} Now realize this derivative structure using symmetric divided differences and a half-step scheme. Let ... 2 You get an automatic Newton solver by using the BVP solver scipy.integrate.solve_bvp. You program your equation as usual def eqn(x,u): return [ u, (1+u**2)**1.5*(T*u+0.5*w*x*(x-L))/K ] def bc(u0,uL): return [ u0, uL ] x = np.linspace(0,L,21); a = 0.1 u = [ a*x*(L-x), a*(L-2*x) ] res = solve_bvp(eqn, bc, x, u, tol = 1e-12); # the solution is ... 2 The interval is quite simply a consequence of following the standard proof of Picard-Lindelöf. As the Lipschitz constant is globally L=1, one does not need a restriction in the y direction. In the next step, the Picard iteration is considered on C([−ϵ,ϵ]) where it has a Lipschitz constant as a mapping on a function space of Lϵ=ϵ, \bigl|P[y_1](t)... 2 The value ofy'(a)$depends on the method you use and the step size. However, because your system is linear, solving for$y'(a)$is just a single step linear interpolation/extrapolation. Alternatively, you can keep track of$y'(a)$during your iterations (since the DE is linear, every step of the approximate solution is going to look like$y=y_0+y'(a)y_1$)... 2 This is not an identity, but just an approximation for$\tan$. If you look at their graphs, you'll notice that the two functions line up very, very well though for$|x|\leq\pi$. 2 Yes, in$c_1h^p+c_2h^{p+1}+...$, the second term will dominate the first one for$h>\frac{c_1}{c_2}. In numerical applications, the many steps required by smaller step sizes eventually accumulate floating point noise sufficient to dominate the truncation error, so that a loglog plot of error vs. step size has a V shape with a fuzzy left leg, a middle ... 2 Let me reformulate your assertion under the precise form : $$\lim_{x\to \infty}\left(\sqrt[n]{x^n+q(x)}-x\right)=\dfrac{1}{n}c_{n-1} \ \ \ \text{where} \ \ \ q(x):=c_{n-1}x^{n-1}+\cdots+c_0 \tag{1}$$ and prove it. Indeed : $$\sqrt[n]{x^n+q(x)}=\sqrt[n]{x^n\left(1+\dfrac{q(x)}{x^n}\right)}=x\sqrt[n]{1+\dfrac{q(x)}{x^n}}.\tag{2}$$ Now, let us use the first ... 2 It is easy to check that you get from the Euler method to a second order method by taking the slope at the midpoint as in the explicit midpoint or modified Euler method $$y(x+Δx)=y(x)+f(x+\tfrac12Δx, y(x)+\tfrac12f(x,y(x))Δx)Δx$$ or by combining two slopes as in the explicit trapezoidal or improved Euler or Heun's 2nd order method y(x+Δx)=y(x)+\tfrac12f(x,... 2 Calculate backwards, with partial integration: \begin{align} \int_a^b(x-a)(x-b)f''(x)dx &=[(x-a)(x-b)f'(x)]_a^b-\int_a^b(2x-a-b)f'(x)dx\\ &=0-0-[(2x-a-b)f(x)]_a^b+\int_a^b2f(x)dx\\ &=-(b-a)(f(b)+f(a))+2\int_a^bf(x)dx \end{align} so that indeed the claimed identity holds. 2 No, it is not. Take f(x)=1 for all x, and take g(x)=1 if x is rational and g(x)=-1 if x is irrational. f and f\circ g are both constant, but g is nowhere continuous. 2 Assuming IEEE754 double representation (so there is a sign bit on exponent and mantissa), the largest P is \approx 2^{2^{11-1}}=2^{1024}, so x^{64}\approx 2^{1024} and hence x\approx 2^{16}=65536. This ignores all subtleties about precision of the result depending on the machine implementation, order of operations, etc. 2 As you want to compute the inverse, let us start from that end: Considerg(x)=\begin{cases}3x^5-10x^3+30x&-1\le x\le 1\\ 15x+8&x\ge1\\ 15x-8&x\le -1\end{cases} $$Then g is C^2 and has no critical points, hence has a C^2 inverse f that looks like a line plus a sigmoidal. To match your demands for x\ll 0 and x\gg 0, we consider a ... 2 To solve 2D quasilinear systems of conservation laws$$ {\bf u}_t + {\bf A}({\bf u})\, {\bf u}_x + {\bf B}({\bf u})\, {\bf u}_y = {\bf 0} numerically, various strategies can be followed: Implement a 2D finite-volume scheme, such as the 2D Lax-Friedrichs method \begin{aligned} {\bf u}_{i,j}^{n+1} &= \frac{{\bf u}_{i-1,j}^{n} + {\bf u}_{i+1,j}^{n} + {\... 2 I think that the process you're envisioning, if you did it rigorously, becomes equivalent to Picard's iterative process: Given an initial value problem y'(t)= f(t,y), y(0) = y_0, the sequence of functions defined by \begin{align*} \phi_0(t) &= y_0 \\ \phi_{k+1}(t) &= y_0 + \int_0^t f(s,\phi_k(s))\,ds \end{align*} converges to a solution \... 2 In the last line of your code, you have h/2. It should be h/3. You also are using the trapezoid weights instead of the simpson's weights. In fact, I can't figure out why your two results are different at all, since the calculations in the last two lines are identical. 2 Another point of view is the sampling theorem, as the integrated function is periodic and integrated over 2 periods. The limit frequency of \sin^2x =\frac12(1-\cos2x) is 2, so with 4 sub-intervals you are at the minimal sampling frequency. If you write S(h)=\frac{4T(h)-T(2h)}3 as per Richardson extrapolation, then the term T(2h) is under-sampled with ... 2 You are missing an important part of the story... After you multiply the equation by a smooth test function, you extend it by density to some larger (Sobolev) space. The finite element spaces will be subspaces of that Sobolev space, not of C^{\infty}_c(\Omega), so they are not supposed to be differentiable across elements. 1 No! Note the bar on \bar{\mathcal{F}}. You rewrite every appearance of \Delta_\delta y(x)=\frac{y(x+\delta)-y(x)}{\delta}, \Delta_\delta^2 y(x)=\frac{y(x+2\delta)-2y(x+\delta)+y(x)}{\delta^2}, etc., so an equation of the form \mathcal{F}(y(x),\Delta_\delta y(x), \Delta_\delta^2 y(x),\dots;x)=0 $$becomes an equation involving only x, y(x), y(x+\... 1 The best course of action is to reduce to a system of first order differential equations and use a standard ode solver on that system.$$ \begin{cases} y' = z\\ z' = -\frac{k}{m} (y-x) +\frac fm \\ x' = \frac kb (y-x) \end{cases}\$ Edit: With the set of parameters given in the comments below, Wolfram returns a numerical solution with a maximum error in ...