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4

Your approach is sound. What remains to do to complete it is to note that the matrix $$\left(\begin{array}{ccc} 1 & \cdots & 1 \\ x_0 & \cdots & x_n \\ & \vdots & \\ x_0^n & \cdots & x_n^n \end{array}\right)$$ is called a Vandermonde matrix and that its determinant is known to be $$\prod_{0\leqslant ...


4

If $y$ is twice differentiable, $$y(x+h)=y(x)+hy'(x)+O(h^2)$$ $$y(x-h)=y(x)-hy'(x)+O(h^2)$$ Subtract $$(y(x+h)-y(x-h))=2hy'(x)+O(h^2)$$ $$\frac{y(x+h)-y(x-h)}{2h}=y'(x)+O(h)$$ And the approximation is to $O(h)$ (actually, you can get $o(h)$ in this case, using Taylor-Young formula, but not $O(h^2)$). However, if $y$ has a third derivative, ...


3

Some trivial observations: The Delaunay method is not continuous in the sample locations. If you perturb one of the sample points, it can change the topology of the Delaunay triangulation, and the value of the computed integral will jump discontinuously. The areas of Voronoi regions, on the other hand, vary continuously with the sample locations. Moving ...


2

Another not complete answer, but I've actually performed some numerical studies of the two cases: TLDR; For most cases they're almost indistinguishable. I suspect that this is related to the observations in Han de Bruijn's answer - and that the Delaunay and Voronoi regions are close to the same size, since the Delaunay triangulation tries to produce "nice" ...


1

I assume that all $x_k$ are distinct. Let $f_0,\ldots,f_n$ be the Lagrange basis polynomials for $x_0,\ldots,x_n$. That is, each $f_k$ has degree $n$ and $f_k(x_m)=\delta_{km}$. Now the weights are fixed since $$\int_a^b f_k(x)dx=\sum_{m=0}^nw_mf_k(x_m)=w_k.$$ For any polynomial $f$ of degree at most $n$ we find ...


1

Since $0<x<1$, it is simplest to use the Alternating Series Test estimate for the error (although Taylor's Remainder Formula will give the same estimate): If we use the first 5 nonzero terms of the series, $\;\;\;\displaystyle\frac{x^3}{3!}-\frac{x^5}{5!}+\frac{x^7}{7!}-\frac{x^9}{9!}+\frac{x^{11}}{11!}$, the error satisfies ...


1

For differentiable functions like this, this is easy to find out: A differentiable function $f$ is (locally) Lipschitz if and only if the derivative $f'$ is (locally) bounded. To see this, first note that if $f$ is Lipschitz with constant $L$ on $(a,b)$, this implies $$ |f'(x)| = |\lim_{y \to x} \frac{f(y)-f(x)}{y-x}| \leq L, $$ because each term in the ...


1

The simplest breakdown would be this: Analytical solutions can be obtained exactly with pencil and paper; Numerical solutions cannot be obtained exactly in finite time and typically cannot be solved using pencil and paper. These distinctions, however, can vary. There are increasingly many theorems and equations that can only be solved using a computer; ...


1

Analytical approach example: Find the root of $f(x)=x-5$. Analytical solution: $f(x)=x-5=0$, add $+5$ to both sides to get the answer $x=5$ Numerical solution: let's guess $x=1$: $f(1)=1-5=-4$. A negative number. Let's guess $x=6$: $f(6)=6-5=1$. A positive number. The answer must be between them. Let's try $x=\frac{6+1}{2}$: $f(\frac{7}{2})<0$ So it ...


1

The (truncated) series is from the point $t_i$. Note that each of the polynomial terms is of the form $(t-t_i)$, where $t=t_{i+1}$. Additionally, the initial term is $y(t_i)$, so when $t=t_i$, the series evaluates to $y(t_i)$. If it were from $t_{i+1}$, then the initial term would be $y(t_{i+1})$, since that is what $y$ is at $t_{i+1}$. You might think of ...



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