5

The question was Is the converse true? That is, for a fixed continuous function $f$ on $[-1, 1]$ such that there is a natural number $n$ such that $G_n(f)=G_m(f)$ for all $m\geq n$, is it true that $f$ must be a polynomial? The answer is trivially negative. Recall that the symmetric quadrature on the interval symmetric wrt. zero has the form $$Q(f)=\...


5

The analogue of the equation of the parabola is given by Apollonius in I.11 (Proposition 1 in Heath's translation) and restated in I.20. The tangent and bisecting properties are given by Apollonius in I.17 and I.32 (Prop. 11 in Heath). Because, as explained at the beginning of the treatise, a diameter is a line (parallel to the axis of a parabola) bisecting ...


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)=\...


3

With 3 points and 3 weights, you have 6 unknowns allowing you to satisfy 6 equations. There are 6 monomials $x^j$ for $0\leq j\leq 5$, so we should be able to solve for the weights and nodes that yield the exact values for the integrals for polynomials up to the fifth degree. The standard way to do this is by using orthogonal polynomials. This is explained ...


3

HINT If $f(x) = ax^2 +bx+c$, you have $$ \int_{-1}^1 f(x) dx = \frac{2a}{3} + 2c, $$ can you compute $f(0)$ and $f''(1)$ and complete the problem?


3

Clenshaw-Curtis is interpolatory but the trapezoidal rule for the transformed function is not. Thus for $f(x)=1$ Clenshaw-Curtis will always spit out $$\int_{-1}^1f(x)dx\approx2$$ But trapezoidal rule applied to the transformed function will result in $$\begin{align}\int_{-1}^1f(x)dx&=\int_0^{\pi}1\cdot\sin\theta d\theta\\ &\approx\frac{\pi}{2N}\left[...


3

You get the coefficients just be demanding that the rule is exact (gives you the exact result) for polynomials up to a certain degree. Since the rule is linear in $f$, you just need to check for the polynomials $1, x, x^2, \cdots$ Since you have three coefficients to determine, you can use the system $$ \begin{cases} A_0+A_1+A_2 = \int_0^1 x \, dx\\ x_0 A_0 +...


2

First off, you forgot the factor of $2$ in the transformation, i.e. $$\int_{0}^{4} x^2 dx = 2\int_{-1}^{1}(2x+2)^2 dx$$ Secondly your calculation is wrong, you should get $$\left(\frac{2}{\sqrt{3}} + 2\right)^2 + \left(-\frac{2}{\sqrt{3}} + 2\right)^2 = \frac{32}{3}$$ Finally, $4^3 = 64$. If you correct all your mistakes you'll get the correct answer of $...


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

In what way has it "not worked"? Have you checked your quadrature method on $1$-variable integrals? How does it do on $\int_0^1 \int_0^1 \sin(x^2)\; dx\; dy$ and $\int_0^1 \int_0^1 \cos(y^2)\; dx\; dy$?


2

First, let us simplify the integral, eliminating the absolute value that makes the problem more difficult: $$\begin{align}\int_{-1}^1|x|f(x)dx&=\int_{-1}^0-xf(x)dx+\int_0^1xf(x)dx\\ &=\int_1^0-(-y)f(-y)(-dy)+\int_0^1yf(y)dy\\ &=\int_0^1yf(-y)dy+\int_0^1yf(y)dy\\ &=\int_0^1\left(f(x)+f(-x)\right)dx\end{align}$$ Where we made the substitution $...


2

With Matlab you need some kind of toolbox to get jacobiP; not having this, we look at some stuff about Jacobi Polynomials and we find that $$P_0^{(\alpha,\beta)}(x)=1$$ $$P_1^{(\alpha,\beta)}(x)=\frac12(\alpha+\beta+2)x+\frac12(\alpha-\beta)$$ And then $$P_{n+1}^{(\alpha,\beta)}(x)=\frac{(2n+\alpha+\beta+1)\left[(2n+\alpha+\beta)(2n+\alpha+\beta+2)x+(\alpha+\...


2

Whether or not one (or more than one) of the coefficients is zero is not related to the degree of precision (or exactness if you prefer) of the given quadrature rule. Instead, it depends on how you found these coefficients. Assuming you found those coefficients by substituting $f(x) = 1, x , x^2, x^3$ into the quadrature rule and solving the $4\times 4$ ...


2

Split the interval $[0,1]$ into two $[0,1/2)\cup [1/2,1]$, so that $$ \int_0^1{\rm d}x~f(x) = \int_0^{1/2}{\rm d}x~f(x) + \int_{1/2}^1{\rm d}x~f(x) = \int_0^{h}{\rm d}x~f(x) + \int_{h}^{2h}{\rm d}x~f(x) \tag{1} $$ with $h=1/2$. The second integral in the "problematic" one, to solve it, use the change of variables $y = x - h$, ${\rm d}x = {\rm d}y$ it then ...


2

Split the computation in the middle at $1/2$ and use partial integration to remove the singularity. $$ \int_0^{1/2}(z-z^2)^{-2/3}dz = \Bigl[3z^{1/3}(1-z)^{-2/3}\Bigr]_0^{1/2}+2\int_0^{1/2}z^{1/3}(1-z)^{-5/3}dz $$ The first term has a finite value, and the second has a continuous and bounded integrand, so that standard numerical methods should give a reliable ...


2

The potential energy of the spring is $\frac{1}{2}x^2$ so, $$\frac{1}{2}x^2+\frac{1}{2} \left(\frac{dx}{dt} \right)^2=E$$ It follows, $$(\frac{dx}{dt})^2=2E-x^2$$ Supposing that $\frac{dx}{dt} \geq 0$ (ie spring is moving in positive direction), we get: $$\frac{dx}{dt}=\sqrt{2E-x^2}$$ $$dt=\frac{dx}{\sqrt{2E-x^2}}$$ $$t+c=\arcsin(\frac{x}{\sqrt{2E}})$$...


1

The answer to your first question is yes, you have to transfer the interval to $[-1,1]$ in order to apply the Gaussian Quadrature. The transformation $x=\frac {t+1}{2}$ would do the trick. For your second question you need to consider $n=4$ so you get $2n-1=7$ which covers the polynomials of degree 7 as well as polynomials of degree 6. Clearly $n=3$ is not ...


1

you could say $x'' = kx\\ x = C_1 e^{(\sqrt k)t} + C_2 e^{-(\sqrt k)t}$ $k = -1\\ \sqrt k = i\\ x = C_1 e^{it} + C_2 e^{-it}\\ e^{it} = \cos t+ i\sin t\\ x = A\cos t + B\sin t$ Or you could say let $v = x', v' =x'' = -x$ Giving us the system $v' = -x\\ x' = v$ or $\begin{bmatrix} x\\v \end{bmatrix}' = \begin{bmatrix} 0&1\\-1&0\end{bmatrix}\...


1

Clenshaw-Curtis quadrature doesn't actually use the Chebyshev polynomials of the first kind $T_n(x)$ to perform the integration, rather it uses them as a basis set to derive an interpolatory quadrature formula. Let's review briefly how it works. We have a function $$f(x)=\sum_{k=0}^{\infty}b_kT_k(x)\approx\sum_{k=0}^{n+1}b_kT_k(x)\approx\sum_{k=0}^{n+1}...


1

Use a CAS that allows arbitrary precision, e.g. Maple or Mathematica. In Maple, for example, here are the roots of the Legendre polynomial of order $38$, to $20$ digits accuracy: > Digits:= 20: fsolve(orthopoly[P](38, x); $$- 0.99804993053568761981,\; - 0.98973945426638557194,\;- 0.97484632859015350764,\;- 0.95346633093352959567,\;- 0....


1

This proof is based closely on Exercise 5.6 in Chapter II of Theodore S. Chihara, An Introduction to Orthogonal Polynomials (Gordon and Breach 1978, reprinted by Dover 2011). Given $n \geqslant 1$, and $i$ with $1 \leqslant i \leqslant n$, we prove below that there exists a polynomial $f$ such that: $f$ has degree $2n - 2$; $f(x) \geqslant 0$ for all $x$; $...


1

I would like to add a couple remarks to the good answer provided by James and to the comment of LutzL. First remark is that this falls under the problem of the trigonometric interpolation; we are requiring to find an interpolating trigonometric polynomial that equals the given one. The solution given by James can be interpreted in terms of the discrete ...


1

Perhaps the complex form is easier to work with. In that case, \begin{eqnarray*} p\left(t\right) & = & \sum_{k=-M}^{M}\hat{p}\left(k\right)e^{ikt} \end{eqnarray*} where $\hat{p}(k)$ is the Fourier coefficients. Then \begin{eqnarray*} \hat{p}\left(k\right) & = & \frac{1}{2M+1}\sum_{n=-M}^{M}p\left(nh\right)e^{-inh},\quad h=\frac{2\pi}{2M+1}\\...


1

Short Version: $dz\ne d\theta$. In Detail: Forget $r$ for the moment - it's irrelevant to the problem. The notation $\int_{|z|=1}f(e^{i\theta})\,d\theta$ doesn't really make sense, because there's no $z$ in the integrand. Assumng $f$ is hoolomorphic in a neighborhood of the closed unit disk both of the following hold: $$\int_{|z|=1}f(z)\,dz=0$$(Cauchy's ...


1

Nevermind, I got it haha. ${dx \over d \theta} = -\sin \theta$ Hence: $dx = -\sin \theta d \theta$ Rookie mistake.


1

It is not too difficult to derive the coefficients from scratch. The general idea is outlined here. Up to order 5, the result is: n=3: $$\begin{array}{cc} x_i & w_i\\\hline 0 & \frac{8}{75}\\ \pm \sqrt{\frac{5}7} & \frac{7}{25} \end{array}$$ n=4: $$\begin{array}{cc} x_i & w_i\\\hline \pm\frac{1}{3} \sqrt{5-2 \sqrt{\frac{10}{7}}} & \...


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