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3

Hint: Since $x^4+4x+3 = (x^2+2x+1)(x^2-2x+3) = (x+1)^2(x^2-2x+3)$, we have: $\dfrac{|x_{n+1}+1|}{|x_n+1|^2} = \dfrac{|\tfrac{4}{3}x_n+\tfrac{1}{3}x_n^4+1|}{|x_n+1|^2} = \dfrac{|\tfrac{1}{3}(x_n+1)^2(x_n^2-2x_n+3)|}{|x_n+1|^2} = \dfrac{|x_n^2-2x_n+3|}{3}$. Now, computing the limit as $x_n \to -1$ should be easy.

3

You may notice that you are just applying Newton's method to $f(x)=1+\frac{1}{x^3}$, since: $$x-\frac{f(x)}{f'(x)} = \frac{x(4+x^3)}{3}.$$ We have that $f(x)$ is a decreasing and concave function on $\mathbb{R}^-$, with a simple zero at $x=-1$, hence quadratic (and monotonic) convergence is ensured by the properties of Netwon's method: $$\forall ... 3 If you look at this description of Hermite-Gauss quadrature on MathWorld, its error term is$$ \frac{n!\sqrt\pi}{2^n(2n)!}f^{(2n)}(\xi), $$where \xi is some point in the region of integration. When you integrate your function$$ \frac{e^{x^2}}{1+x^2}, \qquad I = \int \frac{e^{x^2}}{1+x^2}W(x)\,dx, \quad W(x) = e^{-x^2},$$its 2n-th derivate is ... 3 Let \mathbf A = \left\{a_{ij}\right\}_{i,j=0}^n \in \mathbb C^{(n+1) \times (n+1)} be defined as$$ a_{ij} = c_{i+j \operatorname{mod} (n+1)}, $$so$$ \mathbf A = \begin{pmatrix} c_0 & c_1 & \dots & c_n\\ c_1 & c_2 & \dots & c_0\\ \vdots & \vdots & \ddots & \vdots\\ c_n & c_0 & \dots & c_{n-1} \end{pmatrix}. ...

3

This is called an anticirculant matrix, which is a special case of Hankel matrix. The eigenvalue/eigenvector formula for circulant matrix does not apply.

2

Let's look how is $e^{x^2} - (1 + x^2)$ evaluated. First, $e^{x^2}$ is evaluated, and that results in $$e^{x^2} = \sum_{k=0}^\infty \frac{x^{2k}}{k!} = 1 + x^2 + \frac{x^4}{2} + \dots \approx 1 + 10^{-8} + 0.5 \cdot 10^{-16} + \dots.$$ But from machine point of view when using double precision arithmetics, $0.5 \cdot 10^{-16}$ is less than unity roundoff, ...

2

From Abramowitz and Stegun, 15.2.10 $$(c-a){}_2F_1(a-1, b; c; z) + (2a-c+(b-a)z){}_2F_1(a, b; c; z) +a(z-1){}_2F_1(a+1, b; c; z) = 0$$ Let $G(a) = {}_2F_1(a, b; c; z)$. Then we can use $$G(a+1) = \frac{2a-c+(b-a)z}{a(1-z)}G(a)+\frac{c-a}{a(1-z)}G(a-1)$$ recurrence to compute $G(a)$ for big values of $a$, starting from a pair $G(a - \lfloor a \rfloor + ... 2 You are getting bitten by the (1-x)^(m-r) term when x=1 and m=r. The sum and the add commands handle that differently. Your m is a fixed integer, and for finite summation you should be using the add command and not the sum command. The sum command is for symbolic summation. m := 20: sum( 0^(m-r), r=0..m ); 0 add( 0^(m-r), ... 2 The solution I will give is an extension of the one I provided in this question. However, it will take into account the higher$p$. We are given that$g(\alpha) = \alpha$and that$x_{n+1} = g(x_n)$is a sequence that converges to$\alpha$(i.e. to the fixed point). The limit we are interested in calculating can be viewed as the ratio of two$p$times ... 2 Use your recurrence relations and the value you found for$p$, we can write the limit you are trying to find as $$\lim_{n\to\infty} \frac{|1 + \frac{1}{3}x_n(4+x_n^3)|}{|(1 + x_n)|^2} = \lim_{n\to\infty}\left| \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + x_n)^2}\right|$$ Consider this limit first: $$\lim_{n\to\infty} \frac{1 + \frac{1}{3}x_n(4+x_n^3)}{(1 + ... 2 A quick work would be as follows. By symmetry (x\mapsto -x in the integral), we can see that x_0=0 and x_1+x_2=0. By integrating with P_0, we obtain A_0=1. With P_1, we get A_0+A_1=0. By integrating with P_2, \frac{1}{2}=2A_1x_1+2A_2x_2. With P_4, we get \frac13=4A_1x_1^3+4A_2x_2^3. Thus, A_1x_1=\frac{1}{8} and ... 1 Due to the fact that both the integration segment [-1,1] and \omega(x) are symmetric to the transformation x \to -x, one might look for a quadrature with the same symmetry. That is weights A_i should satisfy (the minus comes from the fact that replacing x with -x also negates the first derivatives):$$ A_1 = -A_2 $$and the abscissae should ... 1 Your function is piecewise linear. I suggest finding all the points where it's slope changes, seem to be \beta_n = \frac{\gamma}{a_n} (If some a_n are zero, simply kick them out of sum), sort them by value and find the interval [\beta_k, \beta_{k+1}] where the root is (binary search, O(\log n) complexity). On that interval, the root can be found by ... 1 The fastest generic (read: linear convergence) algorithm given a continuous function and an interval [a,b] with f(a) < 0 and f(b) > 0 is called bisection (sometimes binary search): Look at the sign s of f(\frac{a+b}2) If s = 0, we are done with the exact solution \frac{a+b}2. If s = 1, we know the root is in [a, \frac{a+b}2], so ... 1 W.r.t. "Do I need to know at least the interval I want to search in?": Yes, typically you do. At least for subdivision-based solvers such as Sturm's method or pretty much anything based on Descartes' rule of sign or its variations. For other algorithms like Durand-Kerner, Aberth-Ehrlich or homotopy continuation methods, you generally don't need bounds. It ... 1 Your function is called the Generalized Exponential Integral E_p(x) and is described in http://dlmf.nist.gov/8.19:$$E_p(x) = x^{p-1}\int_x^\infty \frac{e^{-t}}{t^p}\: d t \; = \int_1^\infty \frac{e^{-xt}}{t^p}\: d t\; , $$Without a specific reference to the actual open source code, I guess that the discrete recurrence formula based on E_1 or your ... 1 Perhaps for Maple:$$0^0\neq 1\text{ ?}$$I was able to replicate the problem It is odd but \mathrm{subs}(x=1,B(x))=4 in Maple. Also Z\mathrm{ := unapply (B(x),x)} gives Z(1)=4. Maple Primes is a better site for Maple questions 1 The first one is correct by definition. One can try to get a product rule as$$ \Delta_+(u(i)\cdot v(i))=\frac{u(i+1)v(i+1)-u(i)v(i+1)+u(i)v(i+1)-u(i)v(i)}{h}=\\ =\Delta_+(u(i))\cdot v(\color{red} {i+1})+u(i)\cdot\Delta_+v(i) $$but you see the difference in indexing. 1 For any i = 1, \dots, 4 matrix A_i is symmetric. Then, by spectral theorem, for any A_i there exist numbers \lambda_1^i, \dots, \lambda_4^i\in \mathbb R such that$$ x^TA_ix = \sum_{k=1}^{4}\lambda_k^i\, x_i^2, \quad \lambda_k^i \in \mathbb R \quad \forall \, k,i = 1, \dots, 4 \quad $$Therefore the system of equations can be rewritten as$$ ... 1 Only to make more obvious what was already obvious, as said in several comments : $$z=y'-\sin(x)$$ $$z'=y''-\cos(x)x'$$ $$z'=y-z=y''-\cos(x)x'=y-(y'-\sin(x))$$ $$y''+y'-y-\cos(x)x'-\sin(x)=0$$ $$x'''+x''-x'-\cos(x)x'-\sin(x)=0$$ Numerical computation of$x(t)$,$y(t)$,$z(t)$requires to state a third condition, for example$x''(0)$, or$z(0)\$, any ...

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