广大-数院(大数据)-(2)
Mathematical Theorems and Operators
Question 1
This part is sometimes referred to as the second fundamental theorem of calculus or the Newton-Leibniz theorem.
Let f be a real-valued function on a closed interval [a,b] and F a continuous function on [a,b] which is an antiderivative of f in (a,b):
F′(x)=f(x)F'(x)=f(x)F′(x)=f(x)
If f is Riemann integrable on [a,b] then
∫abf(x)dx=F(b)−F(a)\int_{a}^{b} f(x)dx=F(b)-F(a)∫abf(x)dx=F(b)−F(a)
The second part is somewhat stronger than the corollary because it does not assume that f is continuous.
When an antiderivative F of f exists, then there are infinitely many antiderivatives for f, obtained by adding an arbitrary constant to F. Also, by the first part of the theorem, antiderivatives of f always exist when f is continuous.
Question 2
Taylor’s theorem - Let k≥1 be an integer and let the function f:ℝ→ℝ be k times differentiable at the point a∈ℝ. Then there exists a function hₖ:ℝ→ℝ such that
f(x)=∑i=0kf(i)(a)i!(x−a)i+hk(x)(x−a)kf(x)=\sum_{i=0}^{k}\frac{f^{(i)}(a)}{i!} (x-a)^{i} +h_{k}(x)(x-a)^{k}f(x)=i=0∑ki!f(i)(a)(x−a)i+hk(x)(x−a)k
and
limx→ahk(x)=0\lim_{x\to a} h_{k}(x)=0x→alimhk(x)=0
This is called the Peano form of the remainder.
The polynomial appearing in Taylor’s theorem is the k-th order Taylor polynomial
Pk(x)=f(a)+f′(a)(x−a)+f′′(a)2!(x−a)2+⋯+f(k)(a)k!(x−a)kP_{k}(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^{2} +\cdots+ \frac{f^{(k)}(a)}{k!}(x-a)^{k}Pk(x)=f(a)+f′(a)(x−a)+2!f′′(a)(x−a)2+⋯+k!f(k)(a)(x−a)k
of the function f at the point a. The Taylor polynomial is the unique “asymptotic best fit” polynomial in the sense that if there exists a function hₖ:ℝ→ℝ and a k-th order polynomial p such that
f(x)=p(x)+hk(x)(x−a)k,limx→ahk(x)=0f(x)=p(x)+h_{k}(x)(x-a)^{k},\quad \lim_{x\to a} h_{k}(x)=0f(x)=p(x)+hk(x)(x−a)k,x→alimhk(x)=0
then p=Pₖ. Taylor’s theorem describes the asymptotic behavior of the remainder term
Rk(x)=f(x)−Pk(x)R_{k}(x)=f(x)-P_{k}(x)Rk(x)=f(x)−Pk(x)
which is the approximation error when approximating f with its Taylor polynomial. Using the little-o notation, the statement in Taylor’s theorem reads as
Rk(x)=o(∣x−a∣k),x→aR_{k}(x)=o(|x-a|^{k}),\quad x\to aRk(x)=o(∣x−a∣k),x→a
Question 3
The Laplace operator is a second-order differential operator in the n-dimensional Euclidean space, defined as the divergence (∇⋅) of the gradient (∇f). Thus if f is a twice-differentiable real-valued function, then the Laplacian of f is the real-valued function defined by:
Δf=∇2f=∇⋅∇f(1)\Delta f=\nabla^{2} f=\nabla\cdot\nabla f \tag{1}Δf=∇2f=∇⋅∇f(1)
where the latter notations derive from formally writing:
∇=(∂∂x1,…,∂∂xn)\nabla=\left( \frac{\partial}{\partial x_{1}},\dots, \frac{\partial}{\partial x_{n}} \right)∇=(∂x1∂,…,∂xn∂)
Explicitly, the Laplacian of f is thus the sum of all the mixed second partial derivatives in the Cartesian coordinates xᵢ:
Δf=∑i=1n∂2f∂xi2(2)\Delta f=\sum_{i=1}^{n}\frac{\partial^{2} f}{\partial x_{i}^{2}} \tag{2}Δf=i=1∑n∂xi2∂2f(2)
As a second-order differential operator, the Laplace operator maps Cᵏ functions to Cᵏ⁻² functions for k≥2. It is a linear operator Δ:Cᵏ(ℝⁿ)→Cᵏ⁻²(ℝⁿ), or more generally, an operator Δ:Cᵏ(Ω)→Cᵏ⁻²(Ω) for any open set Ω⊆ℝⁿ.
Alternatively, the Laplace operator can be defined as:
∇2f(x)=limR→02nR2(f‾shell R−f(x))=limR→02nAn−1R1+n∫shell R(f(r)−f(x))drn−1\nabla^{2} f(\boldsymbol{x})=\lim_{R\to0}\frac{2n}{R^{2}} \left( \overline{f}_{\text{shell }R} -f(\boldsymbol{x}) \right)=\lim_{R\to0}\frac{2n}{A_{n-1} R^{1+n}} \int_{\text{shell }R} (f(\boldsymbol{r})-f(\boldsymbol{x}))dr^{n-1}∇2f(x)=R→0limR22n(fshell R−f(x))=R→0limAn−1R1+n2n∫shell R(f(r)−f(x))drn−1
Where n is the dimension of the space, f‾shell R\overline{f}_{\text{shell }R}fshell R is the average value of f on the surface of a sphere of radius R, ∫shell Rf(r)drn−1\int_{\text{shell }R} f(\boldsymbol{r})dr^{n-1}∫shell Rf(r)drn−1 is the integral over the shell of radius R, and Aₙ₋₁ is the hypervolume of the boundary of the unit n-sphere.
Question 4
Let us suppose we have a smooth bounded function f:ℝⁿ→ℝ and we solve the extension problem
u(x,0)=f(x) for x∈Rn(1.1)u(x,0)=f(x) \text{ for } x\in\mathbb{R}^{n} \tag{1.1}u(x,0)=f(x) for x∈Rn(1.1)
Δu(x,y)=0 for x∈Rn and y>0(1.2)\Delta u(x,y)=0 \text{ for } x\in\mathbb{R}^{n} \text{ and } y>0 \tag{1.2}Δu(x,y)=0 for x∈Rn and y>0(1.2)
to obtain a smooth bounded function u. It is well known that -uᵧ(x,0)=(-Δ)¹ᐟ²f(x), and therefore we can realize (-Δ)¹ᐟ² as the operator T:f↦-uᵧ(x,0) in the above extension problem.
This is easy to show by applying T twice. When we place -uᵧ(x,0) instead of f as the Dirichlet condition in (1.1), we obtain -uᵧ(x,y) instead of u as the solution of (1.1)-(1.2). Then T(T(f))(x)=T(-uᵧ(x,0))(x)=uᵧᵧ(x,0)=-Δₓf(x). To show T=(-Δ)¹ᐟ² it is only left to check that T is indeed a positive operator, which follows by a simple integration by parts argument.
Question 5
The Fourier series of a complex-valued P-periodic function s(x), integrable over the interval [0,P] on the real line, is defined as a trigonometric series of the form
∑n=−∞∞cnei2πnPx\sum_{n=-\infty}^{\infty} c_{n} e^{i2\pi\frac{n}{P}x}n=−∞∑∞cnei2πPnx
(or ∑m=−∞∞cme−i2πmPx\sum_{m=-\infty}^{\infty} c_{m} e^{-i2\pi\frac{m}{P}x}∑m=−∞∞cme−i2πPmx for some authors), such that the Fourier coefficients cₙ are complex numbers defined by the integral
cn=1P∫0Ps(x)e−i2πnPxdxc_{n}=\frac{1}{P}\int_{0}^{P} s(x)e^{-i2\pi\frac{n}{P}x} dxcn=P1∫0Ps(x)e−i2πPnxdx
The series does not necessarily converge (in the pointwise sense) and, even if it does, it is not necessarily equal to s(x). Only when certain conditions are satisfied (e.g. if s(x) is continuously differentiable) does the Fourier series converge to s(x), i.e.,
s(x)=∑n=−∞∞cnei2πnPxs(x)=\sum_{n=-\infty}^{\infty} c_{n} e^{i2\pi\frac{n}{P}x}s(x)=n=−∞∑∞cnei2πPnx
For functions satisfying the Dirichlet sufficiency conditions, pointwise convergence holds. However, these are not necessary conditions and there are many theorems about different types of convergence of Fourier series (e.g. uniform convergence or mean convergence). The definition naturally extends to the Fourier series of a (periodic) distribution s (also called Fourier-Schwartz series). Then the Fourier series converges to s(x) in the distribution sense.
The process of determining the Fourier coefficients of a given function or signal is called analysis, while forming the associated trigonometric series (or its various approximations) is called synthesis.
Question 6
This part is sometimes referred to as the first fundamental theorem of calculus.
Let f be a continuous real-valued function defined on a closed interval [a,b]. Let F be the function defined, for all x in [a,b], by
F(x)=∫axf(t)dtF(x)=\int_{a}^{x} f(t)dtF(x)=∫axf(t)dt
Then F is uniformly continuous on [a,b] and differentiable on the open interval (a,b), and
F′(x)=f(x)F'(x)=f(x)F′(x)=f(x)
for all x in (a,b), so F is an antiderivative of f.
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