Notes on Mathematical Methods for Physicists Chapter1

Notes on Mathematical Methods for Physicists$\S1$

Chapter1 Mathematical Preliminaries

$\S1.1$Infinite Series

$\square$Comparison Test

Consider a convergent series {$a_n$} , we can use {$a_n$} to study the convergence of series {$u_n$} :

$0\leq u_n \leq a_n \Rightarrow${$u_n$} is convergent.

Similarly , consider a divergent series {$a_n$} , we can use {$a_n$} to study the convergence of series {$u_n$} :

$u_n \geq a_n \geq 0 \Rightarrow${$u_n$} is divergent.

$\square$Cauchy Root Test

$(a_n)^{1/n}\leq r < 1 \Rightarrow${$a_n$} is convergent.

$(a_n)^{1/n} \geq 1 \Rightarrow${$a_n$} is divergent.

$\square$d'Alembert (or Cauchy) Ratio Test

$\frac{a_{n+1}}{a_n}~\leq r<1 \Rightarrow${$a_n$} is convergent.

$\frac{a_{n+1}}{a_n}~\geq 1\Rightarrow${$a_n$} is divergent.

$\begin{equation}\underset{n\to\infty}{lim}~(\frac{a_{n+1}}{a_n})\left\{\begin{array}{lr}<1,\quad congvergent &\\=1,\quad divergent &\\>1,\quad indeterminate &\end{array}\right.\end{equation}$

At some crucial point , the test may fail. For example ,$a_n=\frac{1}{n}$(harmonic series) :

but we cannot find r (< 1) independent of n.

Since$\underset{n\to\infty}{lim}\frac{a_{n+1}}{a_n}~=1$, the test fails.

$\square$Cauchy (or Maclaurin) Integral Test

Consider$f(x)$a continuous , monotonic decreasing function , in which$f(n)=a_n$.

We have an equation , it writes ,

proof :

Then$RHS=LHS$.

$Q.E.D\quad\blacksquare$

Alternative of the equation :

In this kind of equation , the second part in the right hand side is a function that oscillates about zero.

$\square$More Sensitive Tests

1. Kummer Theorem

If$\underset{n\to\infty}{lim}(a_n\frac{u_N}{u_{n+1}}-a_{n+1})\geq C>0$, where$C$is a constant , we have {$u_n$} is convergent if$\sum_{n}a_n^{-1}$is convergent. And if$\sum_{n}a_n^{-1}$is divergent , the more weak it diverges , the more powerful the theorem will be.

If$\underset{n\to\infty}{lim}(a_n\frac{u_N}{u_{n+1}}-a_{n+1})\leq 0$, we have {$u_n$} is divergent if$\sum_{n}a_n^{-1}$is divergent.

proof :

2. Gauss's Test

For large n ,$\frac{u_n}{u_{n+1}}=1+\frac{h}{n}+\frac{B(n)}{n^2}$, we can know that

$\square$Alternating Series

For series of the form$\sum_{n=1}^{\infty}(-1)^{n+1}a_n$,$a_n>0$, we have Leibniz Criterion :

If$a_n$monotonically decreases , and$lim_{n\to\infty}a_n=0$, then {$a_n$} converges.

proof :

so when$n\uparrow,R_{2n}\uparrow$.

$Q.E.D\quad \blacksquare$

$\square$Absolute & Conditional Convergence

Absolute convergence : the absolute value of its terms form a convergent series.

Conditional convergence : not the situation above.

$\square$Operation on Series

$\bullet$If an infinite series is absolutely convergent , the series sum is independent of the order in which the terms are added.

$\bullet$An absolutely convergent series may be added termwise to , or subtracted termwise from , or multiplied termwise with another absolutely convergent series , and the resulting series will also be absolutely convergent.

$\bullet$The series (as a whole) may be multiplied with another absolutely convergent series. The limit of the product will be the product of the individual series limits. The product series , a double series , will also convergent absolutely.

$\square$Improvement of Convergence

The rate of convergence : to form a linear combination of our slowly converging series and one or more series whose sum is known.

For the known series , the following collection is particularly useful :

The series we want to sum and one or more known series (multiplied by coefficient) are combined term by term. The coefficients in the linear combination are chosen to cancel the most slowly converging terms.

$\square$Rearrangement of Double Series

For$S=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{n,m}$,

$\S1.2$Series of Functions

Let's extend our concept of infinite series to include series of functions :

$\square$Uniform Convergence

If for any small$\epsilon>1$, there exists a number$N$, independent of$x$in the interval [$a,b$] (that is ,$a<x<b$) such that ,

Then the series is said to be uniformly convergent.

$\square$Weierstrass M (Majorant) Test

If we can construct a series of numbers$\sum_{i=1}^{\infty}M_i$, in which$M_i\geq|u_i(x)|$for all$x$in the interval [$a,b$] , and$\sum_{i=1}^{\infty}M_i$is convergent , then our series$u_i(x)$will be uniformly convergent in [$a,b$].

proof :

Uniform convergence has nothing with absolute convergence. But M test can only establish for absolutely convergent series.

$\square$Abel's Test

A somewhat more delicate test for uniform convergence has been given by Abel. If$u_n(x)$can be written in the uniform$a_nf_n(x)$, and

1. The$a_n$form a convergent series ,$\sum_{n}a_n=A$.
2. For all$x$in [$a,b$] the functions$f_n(x)$are monotonically decreasing in$n$, that is ,$f_{n+1}(x)\leq f_{n}(x)$.
3. For all$x$in [$a,b$] , all the$f_n(x)$are bounded in the range$0\leq f_n(x)\leq M$, where$M$is independent of$x$.

Then$\sum_{n}u_n(x)$converges uniformly in [$a,b$]. This method is especially useful in analyzing the convergence of power series.

$\square$Properties of Uniformly Convergent Series

If a series$\sum_nu_n(x)$is uniformly convergent in [$a,b$] and the individual terms$u_n(x)$are continuous ,

1. The series sum$S(x)=\sum_{n=1}^{\infty}u_n(x)$is also continuous ,
2. $\int_{a}^{b}S(x)dx=\sum_{n=1}^{\infty}\int_{a}^{b} u_n(x)$,
3. if the following additional conditions are also satisfied , then$\sum_{n=1}^{\infty}\frac{du_n(x)}{dx}$is uniformly convergent in [$a,b$] :$\frac{du_n(x)}{dx}$is continuous in [$a,b$].

The first and second conditions are always right in physics , but the third is not because it is more restrictive.

$\square$Taylor's Expansion

We assume that our function$f(x)$has a continuous n-th derivative in the interval [$a,b$].

First , let's intergrate this n-th derivative n times :

Finally ， after integrating for the n-th time ,

where the remainder ,$R_n$, is given by the n-fold integral ,

We may convert$R_n$into a perhaps more pratical form by using the mean value theorem of integral calculus ,

Or applying Cauchy's mean value theorem of integral calculus ,

When you adjust n properly ,$R_n\to0$. Then we have Taylor Expansion , which writes

$\square$Power Series

When$a=0$, we have Maclaurin series ,

$\square$Properties of Power Series

$f(x)=\sum_{n=0}^{\infty}a_nx^n$, in which$a_n$is independent of$x$. If$lim_{n\to\infty}|\frac{a_{n+1}}{a_n}|=R^{-1}$, then$R$is the radius of convergence , and the series converges for$x\in(-R,R)$.

But the ratio / root test fails at endpoints ,$x=\pm R$needs special attention. In M test , the series is uniformly and absolutely convergent in ($-R,R$).

$\square$Uniqueness Theorem

The power-series representation is unique. Assume that

What we need to prove is that$a_n=b_n$, for all$n$.

When$x=0$, we have$a_0=b_0$. Then differentiate the series ,

When$x=0$, we have$a_1=b_1$. ··· ···

Repeating the process , we will get$a_n=b_n$.

$Q.E.D\quad \blacksquare$

This theorem will be a crucial point in our study of differential equations , in which we develop power series solutions (For instance , in theoretical physics , there's perturbation theory in quantum mechanics).

$\square$Indeterminate Forms

You can use the power-series representation of functions to prove$L'H \hat{o} pital's\,\,\,Rule$

(See exercise 1.2.12)

$\square$Inversion of Power Series

Consider that$y-y_0=\sum_{n=1}^{\infty}a_n(x-x_0)^n$, if we want to know$x-x_0=\sum_{n=1}^{\infty}b_n(y-y_0)^n$, we need to equate coefficients of$(x-x_0)^n$on both sides of the given equation.

$\S1.3$Binomial Theorem

Binomial series is a extremely significant application of the Maclaurin series. Let$f(x)=(1+x)^m$, in which$m\in\R$. Then ,

For this function , the remainder is

with$\xi\in(0,x)$.

When$x>0$, for$n>m$,$(1+\xi)^{m-n}$is a maximum for$\xi>0$. So for$x>0$,$|R_n|\leq\frac{x^n}{n!}|m(m-1)···(m-n+1)|$, with$\underset{n\to\infty}{lim}R_n=0$, when$0\leq x<1$.

Because the radius of convergence of a power series is the same for positive and negative$x$, the binomial series converges for$-1<x<1$. The endpoints cannot be determined.

Binomial Expansion :

If$m$is a nonnegative integer ,$R_n$for$n>m$vanishes for all$x$, corresponding to the fact that under those conditions$(1+x)^m$is a finite sum.

Binomial Coefficients :

In which , when$n=0$,$\begin{equation}\left\lgroup\begin{array}{lr}m\\0\end{array}\right\rgroup =1\end{equation}$.

1. When$m$is a positive integer ,$\begin{equation}\left\lgroup\begin{array}{lr}m\\n\end{array}\right\rgroup =\frac{m!}{n!(m-n)!}\end{equation}$. That is$C_m^n=\begin{equation}\left\lgroup\begin{array}{lr}m\\n\end{array}\right\rgroup\end{equation}$.

2. For negative integer$m$, set$m=-p$, there is

3. For nonintegral$m$, it is convenient to use the Pochhammer Symbol , defined for general a and nonnegative integer$n$, as