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New answers tagged na.numerical-analysis

0

I figured it out. here is the code. from scipy.interpolate import CubicSpline def P(Cm): Ps = [] Cmb = [i for i in Cm] c0 = Cmb[0] if c0 == 0.0 or c0 == 0: pass else: Cmb.insert(0, 0.0) for i in range(len(Cmb)-1): ys = [0.0 for i in range(len(Cmb))] ...

1

I'll posit that perhaps asking for a taxonomy on quantum algorithms might be overly narrowing, and searching for elusive superpolynomial speedups might be white whales. In actuality quantum computation might offer opportunities that are somewhat orthogonal to anything conceivable classically. As an example, I think of proposals for quantum money, such as by ...

0

The classical material derivative $D\varphi/Dt$ of a test function $\varphi \in C_c^\infty(\mathbb{R}_+\times D)$ is obtained by setting $$\dfrac{D\varphi}{Dt}(x) := \dfrac{\partial\tilde\varphi}{\partial t}(0,x)\;\;\mbox{with}\;\; \tilde\varphi(t,x) = \varphi(t,T_t(x))$$ for $x\in D$. Expanding out using the chain rule, we have  \dfrac{D\varphi}{Dt}(x) =...

3

Yes, this is true, and there is a proof which closely tracks your intuition. As you know, this estimate can be proved in the continuum by applying the Sobolev embedding twice, first to get $\nabla u \in L^p$ for $p<\frac{2d}{d-2}=6$, and then once more to get $u\in L^\infty$. So for simplicity let me discuss how to get discrete versions of the Sobolev ...

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