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6

You're on the right track. Note that since $0<L<1$, we have $$ L^m+L^{m+1}+\dots+L^{m+n-1}\leq \sum_{k=m}^{\infty}L^k=\frac{L^m}{1-L} $$ which can be made as small as we like by choosing $m$ large enough.


4

You are very close to a complete proof. All you need is the "final step". Here is your proof, completed with the "final step" (in details). Let $\phi(x):[a,b]\rightarrow [a,b]$ be a continuous function. Show that if $\phi(x)$ is a contraction mapping on $[a,b]$ then the sequence $\{x^{(k)}\}$ defined by $x^{(k+1)} = \phi(x^{(k)})$ is a Cauchy sequence. ...


2

Gauss-Legendre Gauss-Legendre quadrature is designed for integrals over the integration domain $[−1,+1]$, but this can be modified to accept any finite domain $[a, b]$ through the transformation $$\int_{a}^{b}f(x)dx=\frac{b-a}{2}\int_{-1}^{1}f\left(\frac{b-a}{2}x+\frac{b+a}{2}\right)dx$$ so $a=-\infty$ and $b=\infty$, we must truncate the upper and lower ...


2

That approximation just comes from an asymptotic approximation for the incomplete $\Gamma$ function. We have: $$ F(x)\,e^{-x} = e^{-x}+x\,\Gamma(0,x) = \color{red}{1+\left(-1+\gamma+\log(x)\right)x}\color{blue}{-\frac{1}{2}x^2+\frac{1}{12}x^3-\frac{1}{72}x^4+\ldots}$$ where the blue part is the Taylor series at $x=0$ of an entire function, hence a ...


2

Since $T(A)=-(D+L)^{-1}U$: $$T(cA)=-(cD+cL)^{-1}cU=-c^{-1}(D+L)^{-1}cU=-(D+L)^{-1}U=T(A).$$ Thus the iteration matrix is not changed at all by rescaling $A$. So although you can just as well solve $Ax=b$ or $(cA)x=cb$, Gauss-Seidel's convergence and convergence rate are not affected.


1

The only way to have a spatially smooth solution is for $\partial u/\partial r$ go to zero as $r$ goes to zero. This means you can use L'Hopitals's rule in evaluating that second term: $\displaystyle \frac{1}{r}\frac{\partial u}{\partial r} \rightarrow \frac{\partial_r(\partial u/\partial r)}{\partial_r(r)} = \frac{\partial^2 u}{\partial r^2}$


1

Here is an example showing it is not singular. $$A=\begin{pmatrix} 0 & 1 \\ 1 & 1\end{pmatrix}$$ The first pivot is zero, but the matrix is not singular.


1

It seems the authors might have used (I admit that the numbers do not match completely) $$ c=\frac{2e^\gamma}{3F_1(6)}\approx1.18787 $$ (where $F_1$ is also defined on page 7) instead of $$ c=\frac{2e^\gamma}{3f_1(6)}\approx 0.82396 $$ I don't have time or energy to understand which one is the correct one to put in the end. Mathematica calculates $$ F_1(s)=\...


1

From your original equation, you can do a regular perturbation expansion $w=w_0+\epsilon w_1+\epsilon^2w_2+\ldots$ to get, at $O(1)$, $$w_0^4-\frac{4}{3}w_0^3=0,$$ which has three solutions $w_0=0$ and one solution $w_0=4/3$. At $O(\epsilon)$, you get $w_0^3w_1-w_0^2w_1=1$, which is not consistent for the roots $w_0=0$, so consider $w_0=4/3$ to give, $$ \...


1

We suppose there are two polynomial $P_n(x)$ and $Q_n(x)$ such that for $i=0,1,\cdots,n$ $$P_n(x_i)=f(x_i)$$ and $$Q_n(x_i)=f(x_i)$$ let $$R_n(x)=P_n(x)-Q_n(x)$$ for $i=0,1,\cdots,n$ we have $$R_n(x_i)=P_n(x_i)-Q_n(x_i)=f(x_i)-f(x_i)=0$$ therefor $R_n(x_i)$ has $n+1$ roots whereas degree of $R_n$ is $n$, thus we can say $$R_n(x)=0$$ in the other words $...


1

Written in base 2, the smallest floating-point number larger than 7 is $7+2^{-12} = 111.000000000001$ (with 11 zeros). From this, you can conclude the mantissa has 15 bits, and so when we increment 70 by the smallest amount possible, we get $1000110.00000001$; this distance is $2^{-8}$.



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