May 18, 2021

Notes on Whittaker & Watson, Chapter XI, part 1

Whittaker & Watson, A Course of Modern Analysis. Chapter XI, Integral Equations. 11.1 Definition of Integral Equations. Examples: solving $\phi$ for $f(x)=\int e^{xt}\phi(t)dt$ (Laplace), $f(x)=\int_{-\infty}^{\infty} \cos{(xt)}\phi(t)dt$ (Fourier). 11.11 Hadamard’s inequality. For square matrix $A = (a_1, a_2, \cdots, a_n)$ with columns normalised $\|a_i\|_2=1, i=1,2,\cdots,n$, the determinant $-1\leq \det(A)\leq 1$. Collorary. For $A=(a_{ij})$, with each entry $|a_{ij}|<M$, the column vectors $\|a_i\|_2\leq n^{\frac{1}{2}}M$, therefore $|\det(A)|\leq n^{\frac{1}{2}n}M^n$. 11.2 Fredholm’s equation of the second kind is defined as $$ \phi(x)=f(x)+\lambda\int_a^bK(x,\xi)\phi(\xi)d\xi $$ in which $\phi$ is the unknown, $f$ is given, $K(x, \xi)$ is the kernel (the term nucleus was used in Whittaker & Watson) such that $K(\cdot, \cdot)$ is continuous over $[a,b]\times[a,b]$ and $K(x,y)=0$ when $y>x$.

May 10, 2021

Notes on Whittaker & Watson, Chapter X

Whittaker & Watson, A Course of Modern Analysis. Chapter X, Linear Differential Equations. 10.1 Set-up the discussion of the homogeneous second order ODE $$ \frac{d^2u}{dz^2}+p(z)\frac{du}{dz}+q(z)u=0. $$ 10.2 Series solution in the neighbourhood $S_b$ of an ordinary point $b$. With the substitution $u=v\int_b^zp(\zeta)d\zeta$, the equation becomes $$ \frac{d^2v}{dz^2}+J(z)u=0. $$ where $J(z)=q(z)-\frac{1}{2}\frac{dp}{dz}-\frac{p^2}{4}$. Series solution $v(z)=\sum_{n-0}^{\infty}v_n(z)$ is uniform convergent and analytic in $S_b$ with $$ v_0(z)=a_0+a_1(z-b),\newline v_n(z)=\int_b^z(\zeta-z)J(\zeta)v_{n-1}(\zeta)d\zeta, $$ and $a_0=v(b)$, $a_1=v'(b)$ determined by initial conditions at $z=b$.

May 5, 2021

Notes on Whittaker & Watson, Chapter VIII

Whittaker & Watson, A Course of Modern Analysis. Chapter VIII, Asymptotic Expansions and Summable Series. 8.1. An example: the function $f(x)=\int_x^\infty t^{-1}e^{x-t}$ can be expanded as $$ f(x)=\sum_{m=0}^n\frac{(-1)^mm!}{x^{m+1}} + (-1)^{n+1}(n+1)!\int_x^\infty \frac{e^{x-t}}{t^{n+2}}dt. $$ The series sum $S_n=\sum_{m=0}^n\frac{(-1)^mm!}{x^{m+1}}$ diverges, and the remainder $|f(x)-S_n(x)|<\frac{1}{2^{n+1}n^2}$ for $x\ge 2n$. 8.2 The Poincaré’s definition of asymptotic expansion: $$ f(x)\sim A_0+\frac{A_1}{z}+\frac{A_2}{z^2}+\cdots+\frac{A_n}{z^n}+\cdots $$ The divergent series on the RHS is the asymptotic expansion of $f(x)$ if the remainder $R_n(z)=z^n\left(f(z)-S_n(z)\right) \to 0$ when $|z|\to\infty$ given $n$, even if $\lim_{n\to\infty}R_n(z) = \infty$ at fixed $z$.

May 4, 2021

Notes on Whittaker & Watson, Chapter VII, part 3

Whittaker & Watson, A Course of Modern Analysis. Chapter VII, The Expansion of Function in Infinite Series. 7.8 Borel’s theorem. For the Taylor series of $f(z)=\sum_{n=0}^\infty a_nz^n$ converging when $|z|\le r$, we define $$ f_1(z)=\int_0^\infty e^{-t}\phi(zt)dt, $$ with $\phi(z)=\sum_{n=0}^\infty \frac{a_nz^n}{n!}$ being Borel’s function associated with $\sum_{n=0}^\infty a_nz^n$. The function $f_1$ thus defined agrees with $f$ when $|z|< r$. 7.81 Borel’s integral as analytic continuation to Borel’s polygon. For any $\zeta$ in the interior of the polygon, $\phi(\zeta t)=\frac{1}{2\pi i}\sum_{n=0}^\infty\frac{\zeta^nt^n}{n!

May 3, 2021

Notes on Whittaker & Watson, Chapter VII, part 2

Whittaker & Watson, A Course of Modern Analysis. Chapter VII, The Expansion of Function in Infinite Series. 7.4 When the function $f$ has poles at $a_1, a_2, \cdots$ with residues $b_1, b_2, \cdots$ located within the circles of radius $R_n\to\infty$ such that $\forall m\le n: |a_m|<R_n$ and $f$ bounded on these circles, the value of the function $f(z)$ can be expressed as the sum $$ f(z)=f(0)+\sum_{n=1}^\infty b_n\left(\frac{1}{z-a_n}+\frac{1}{a_n}\right), $$ by Cauchy’s theorem integrating on these circles.

May 3, 2021

Notes on Whittaker & Watson, Chapter VII, part 1

Whittaker & Watson, A Course of Modern Analysis. Chapter VII, The Expansion of Function in Infinite Series. 7.1 Darboux’s formula calculates $f(z)-f(a)$ via an $n$th order polynomial $\phi(t)$ using the values of the function at $a$ and its derivatives up to $n$th order at $a$ and $z$ $$ f(z)-f(a) = \frac{1}{\phi^{(n)}(0)}\sum_{m=1}^n(-1)^{m-1}(z-a)^m(\phi^{(n-m)}(1)f^{(m)}(z)-\phi^{(n-m)}(0)f^{(m)}(a)) + R_n(z), $$ where the remainder $$ R_n(z) = (-1)^{n}(z-a)^{n+1}\int_0^1\phi(t)f^{(n+1)}(a+t(z-a))dt. $$ The Taylor series can be seen as a specific case of Darboux’s formula with $\phi(t)=(t-1)^n$.