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3

The Legendre polynomial $P_{k}=C_{k}\frac{d^{k}}{dx^{k}}(x^{2}-1)^{k}$ ($C_k$ is a normalization constant) is orthogonal to all polynomials of lower order with respect to the inner product $(f,g)=\int_{-1}^{1}f(x)g(x)dx$. In fact, the Legendre polynomials can be derived by applying the Graham-Schmidt procedure to $\{ 1,x,x^{2},x^{3},\cdots \}$. So your ...


2

I suffices $n=1$ and $c=0$. In this way you get $e^x=1+x+E_1(x)=1+x+\frac{1}{2}e^\xi x^2$. Note that $\frac{1}{2}e^\xi x^2>0$ for all $\xi$ provided $x\neq 0$


2

The Taylor expansion is for $$ φ(t,y(t),h)−φ(t,y(t)−s,h)=\frac{∂}{∂y}φ(t,y(t),h)s+O(s^2) $$ where $s=e_p(t)h^p$ so that $O(s^2)=O(h^{2p})$.


2

In compositional data analysis, the distance function $d(x,y) = |\ln(x) - \ln(y)| = \left| \ln\left( \frac{x}{y} \right) \right|$ is known as the log-ratio distance, and it was introduced by J. Aitchison. But also the log-normal distribution involves this distance function. If parameterized with the geometric mean $m$, the probability density function of the ...


2

Every tail of this series is bounded above by a geometric series. For example: \begin{align} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {8!} + \frac 1 {9!} + \frac 1 {10!} + \cdots} \\[10pt] \le {} & \frac 1 {0!} + \frac 1 {1!} + \cdots + \frac 1 {6!} + \overbrace{\frac 1 {7!} + \frac 1 {7!\cdot 8} + ...


1

Ignore the $k$, that's basically irrelevant. The point is really that if $a,b$ are small positive numbers, then $|\sqrt{a}-\sqrt{b}|$ is much larger than $|a-b|$. You can see this with a linear approximation: $\sqrt{b} \approx \sqrt{a} + \frac{b-a}{2\sqrt{a}}.$ So $|\sqrt{a}-\sqrt{b}| \approx \frac{|a-b|}{2 \sqrt{a}}$. Thus the error in the input basically ...


1

The error of the trapezoid method is given in terms of the second derivative of a function. Our function is smooth and fast-decaying outside any right neighbourhood of the origin, but in a right neighbourhood of the origin the second derivative of $f$ is unbounded and behaves like $\frac{1}{x^{1/2}}$. That is the reason for which $p\to 2+\frac{1}{2}$. You ...


1

You have an ODE $$ \frac{dx}{dt} = f(t,x) = (x+1)t $$ which separates to $$ \frac{dx}{x+1} = tdt\\ \log |x+1| + C = \frac{t^2}{2}\\ x = C_1 e^{t^2/2} - 1. $$ With $x(0) = 0$ that gives $C_1 = 1$ and $$ x(t) = e^{t^2/2} - 1. $$ Computing the first step we get $$\begin{aligned} k_1 &= hf(0, 0) = 0\\ k_2 &= hf\left(\frac{h}{2}, \frac{k_1}{2}\right) = ...


1

The first equation looks like the Eikonal equation in $\Bbb R^{1+1}$, it is solved by $u=\sqrt{x-t}$, you would then plug this into the wave equation to determine $f$. We see that $u_x=x/u$, $u_t=-t/u$, $u_{xx}=(u-x^2/u)/u^2$ and $u_{tt}=-(u+t^2/u)/u^2$ thus$$u_{xx}-u_{tt}=(u-x^2/u)/u^2+(u+t^2/u)/u^2$$ $$=2u/u^2+(t^2-x^2)/u^3$$ ...


1

Finally, uranix hint with preconditioning lead me to a solution. The key performance problem comes from solving lots of systems of the form $Ax = b$ with our $A$. Fortunately, $A$ has so many nice properties, that the PCG algorithm works well when using ichol as a preconditioner. Thus using eigs' capability to take $x\mapsto A^{-1}x$ as a function leads to ...


1

You've already calculated your answer: The series is given by $\sum \limits_{n = 0}^\infty \frac{e^3}{n!} (x-3)^n$. You don't ned to transform this in any way to the Taylor-series around $0$. However, you can easily deduce the Taylor series at $3$ from the one at $0$ by using the identity $e^x = e^{x-3} e^3$.


1

HINT The question wants you to expand your series for $\mathrm{e}^x$, close to $x=3$ and show that it gives the same as series for $\mathrm{e}^x$, close to $x=0$. In other words: $$\mathrm{e}^3 + \mathrm{e}^3(x-3)+\frac{1}{2!}\mathrm{e}^3(x-3)^2 + \frac{1}{3!}\mathrm{e}^3(x-3)^3 \cdots \equiv 1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots$$ Let's start by ...


1

Your code t(n+1)=t(n)-f(n).*((t(n+1)-t(n))/(f(n+1)-f(n))); is attempting to use t(n+1) and f(n+1) before they have been assigned values.


1

This doesn't work for series in general. It does work when the terms are decreasing in absolute value, have alternating signs, and approach $0$. That happens with the alternating harmonic series, for which the error is typically about half of the first excluded term. With $\sin z = \sum\limits_{n=0}^\infty (-1)^n \dfrac{x^{2n+1}}{(2n+1)!}\vphantom{\dfrac ...



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