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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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