Mathematical Methods for Physics and Engineering
1. Basic sum of series [Page 117-118]:
Basic series:
$$\begin{array}{ll} Arithmetic\ series:S_N=\frac{N}{2}(first\ term+last\ term)\\ Geometric\ series:S_N=\frac{a(1-r^N)}{1-r} \end{array}$$
Especially, $S_N=\frac{a}{1-r}$ for infinite geometric series with $|r|<1$.
Arithmetico-geometric series:
$$\begin{array}{ll} S_N & =\sum\limits_{n=0}^{N-1}(a+nd)r^n\\ & =\mathbf{\frac{a-[a+(N-1)d]r^N}{1-r}+\frac{rd(1-r^{N-1})}{(1-r)^2}} \end{array}$$
Especially, $S_N=\frac{a}{1-r}+\frac{rd}{(1-r)^2}$ for infinite arithmetico-geometric series with $|r|<1$.
2. Some useful methods [Page 119-124]:
2.1 The difference method
For series $u_n$, if you can find $u_n=f(n)-f(n-1)$, then $\sum\limits_{n=1}^{N}u_n=f(N)-f(0)$.
(Try it $\sum\limits_{n=1}^{N}\frac{1}{n(n+1)(n+2)}$, $\sum\limits_{n=1}^{N}n^3$ )
Example:
I will show you how to find $\sum\limits_{n=1}^{N}n^3$ which isn’t stated clearly on page 121.
$$\begin{array}{rl} N^3-(N-1)^3 &=3N^2-3N+1\\ (N-1)^3-(N-2)^3 &=3(N-1)^2-3(N-1)+1\\ &…\\ 1^3-0^3&=3-3+1\\ \rightarrow N^3&=3\sum\limits_{n=1}^{N}n^2-3\sum\limits_{n=1}^{N}n+N\\ \rightarrow \sum\limits_{n=1}^{N}n^2&=\frac{N(N+1)(2N+1)}{6}\\ \end{array}$$
Similarly,
$$\begin{array}{rl} N^4-(N-1)^4 &=4N^3-6N^2+4N-1\\ (N-1)^4-(N-2)^4 &=4(N-1)^3-6(N-1)^2+4(N-1)-1\\ &…\\ 1^4-0^4&=4-6+4-1\\ \rightarrow N^4&=4\sum\limits_{n=1}^{N}n^3-6\sum\limits_{n=1}^{N}n^2+4\sum\limits_{n=1}^{N}n-N\\ \rightarrow \sum\limits_{n=1}^{N}n^3&=\frac{N^2(N+1)^2}{4} \end{array}{}$$
2.2 The transformation method
This is not only limited to few ways. Here’s three examples.
Example:
Find $S(x)=\frac{x^4}{3(0!)}+\frac{x^5}{4(1!)}+\frac{x^6}{5(2!)}+….$
$$\begin{array}{rl} \frac{S(x)}{x}&=\frac{x^3}{3(0!)}+\frac{x^4}{4(1!)}+\frac{x^5}{5(2!)}+…\\ \frac{d}{dx}[\frac{S(x)}{x}]&=\frac{x^2}{0!}+\frac{x^3}{1!}+\frac{x^4}{2!}+…\\ \frac{S(x)}{x}&=\int x^2e^xdx\\ \end{array}$$
With $\frac{S(x)}{x}=0$ at $x=0$, we finally find $S(x)=x^3e^x-2x^2e^x+2xe^x-2x$.
Find $S=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+….$
$$\begin{array}{rl} Set\ f(x) &= 1+2x+3x^2+4x^3+…\Rightarrow S = f(\frac{1}{2})\\ \int f(x)dx&=x+x^2+x^3+…=\frac{x}{1-x}\\ f(x)&=(\frac{x}{1-x})’=\frac{1}{(1-x)^2}\Rightarrow S=f(\frac{1}{2})=4 \end{array}$$
You can check the result by the formula in section 1: $S=\frac{a}{1-r}+\frac{rd}{(1-r)^2}=\frac{1}{1-\frac{1}{2}}+\frac{\frac{1}{2}\times1}{(1-\frac{1}{2})^2}=4$.
Find $S(\theta)=1+\cos\theta+\frac{\cos2\theta}{2!}+\frac{\cos3\theta}{3!}+….$
$$\begin{array}{rl} S(\theta)&=Re\{1+e^{i\theta}+\frac{e^{2i\theta}}{2!}+\frac{e^{3i\theta}}{3!}+… \}\\ &=Re\{e^{e^{i\theta}} \}\\ &=e^{cos\theta}cos(sin\theta) \end{array}$$
3. Convergence of infinite series [Page 124-131]:
3.1 The definition:
Absolute convergence: If the series $\sum|u_n|$ converges then $\sum u_n$ also converges, and $\sum u_n$ is said to be absolutely convergent.
Conditional convergence: if $\sum |u_n|$ diverges whilst $\sum u_n$ converges then $\sum u_n$ is said to be conditionally convergent.
3.2 Test of convergence:
3.2.1 Preliminary test:
For positive terms only series (necessary but not sufficient): $\sum u_n$ is convergent if $\lim\limits_{n\rightarrow\infty}u_n=0$.
3.2.2 Comparison test:
For positive terms only series: $\sum u_n$ is convergent if $\sum v_n$ is convergent and $u_n < v_n$ ($\forall n>N$, $N$ is finite).
3.2.3 D’Alembert’s ratio test:
For positive terms only series: define $\rho=\lim\limits_{n\rightarrow\infty}\frac{u_{n+1}}{u_n}$,
(i) if $\rho<1$, then u_n is convergent.
(ii) if $\rho>1$, then u_n is divergent.
(iii) if $\rho=1$, then u_n is undetermined.
3.2.4 Ratio comparison test:
For positive terms only series: $\sum u_n$ is convergent if $\sum v_n$ is convergent and $\frac{u_{n+1}}{u_n} \leq \frac{v_{n+1}}{v_n}$ ($\forall n>N$, $N$ is finite$)$; $\sum u_n$ is divergent if $\sum v_n$ is divergent and $\frac{u_{n+1}}{u_n} \geq \frac{v_{n+1}}{v_n}$ ($\forall n>N$, $N$ is finite$)$.
3.2.5 Quotient test:
For positive terms only series: define $\rho=\lim\limits_{n\rightarrow \infty}\frac{u_n}{v_n}$,
(i) if $\rho \neq 0$ but is finite then $\sum u_n$ and $\sum v_n$ both converge or both diverge.
(ii) if $\rho=0$ and $v_n$ converges then $u_n$ converges.
(iii) if $\rho=\infty$ and $v_n$ diverges then $u_n$ diverges.
3.2.6 Integral test:
For positive terms only series: $\sum u_n$ is convergent if $u_n = f(n)$ and $\lim\limits_{N\rightarrow \infty}\int^{N}f(x)dx$ exists.
Example:
Determine if $\sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ (Riemann zeta series) is convergent.
$$\lim_{N\rightarrow \infty}\int^N \frac{1}{x^p}dx=\lim\limits_{N\rightarrow \infty}\frac{N^{1-p}}{1-p}\\ = \begin{cases} 0&,&\ p>1\\ \infty&,&\ p\leq1 \end{cases}$$
which means $\sum\limits_{n=1}^{\infty}\frac{1}{n^p}$ is convergent for $p>1$ and divergent for $p\leq1$.
3.2.7 Cauchy’s root test:
For positive terms only series: define \rho=\lim\limits_{n\rightarrow \infty}(u_n)^{1/n},
(i) if \rho<1, then u_n is convergent.
(ii) if \rho>1, then u_n is divergent.
(iii) if \rho=1, then u_n is undetermined.
3.2.8 Grouping items:
Let’s get back to the Riemann zeta series stated before. Since
$$\begin{array}{rl} \sum\limits_{n=1}^{\infty}\frac{1}{n^p}=&\frac{1}{1^p}+(\frac{1}{2^p}+\frac{1}{3^p})+(\frac{1}{4^p}+\cdots+\frac{1}{7^p})+\cdots\\ <&\frac{1}{1^p}+\frac{2}{2^p}+\frac{4}{4^p}+\cdots\\ \end{array}$$
which converges for $p>1$, then $\frac{1}{n^p}$ must also converge for $p>1$. Since
$$\begin{array}{rl} \sum\limits_{n=1}^{\infty}\frac{1}{n^p}=&\frac{1}{1^p}+(\frac{1}{2^p}+\frac{1}{3^p})+(\frac{1}{4^p}+\cdots+\frac{1}{7^p})+\cdots\\ \geq&\frac{1}{1}+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\cdots+\frac{1}{8})+\cdots\\ >&1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots\\ &(for\ p\leq1) \end{array}$$
which diverges, then $\frac{1}{n^p}$ must also diverge for $p\leq1$.
3.2.9 Alternating series test:
$\sum\limits_{n=1}^{\infty}(-1)^{n+1}u_n$ can be shown to converge provided:
(i) \lim\limits_{n\rightarrow\infty}u_n=0.
(ii) u_n < u_{n−1} for all n > N for some finite N.
See page 131 for more details.
3.2.10 Power series test:
If $P(z)=a_0+a_1z+a_2z^2+\cdots$ where $z$ is a complex number, then $P(z)$ converges for:
$\rho=\lim\limits_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}z|=|z|\lim\limits_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|<1$
where $\rho$ is called radius of convergence.
Example:
Determine the range of z for which the following complex power series converges:
$P(z)=1-\frac{z}{2}+\frac{z^2}{4}-\frac{z^3}{8}+\cdots$
Answer:
It’s easy to know it converges for $|z|<2$. The problem is how to investigate the condition on the circle $|z|=2$. We can write $z=2e^{i\theta}$, then:
$P(z)=1-e^{i\theta}+(e^{i\theta})^2-(e^{i\theta})^3+\cdots=\frac{1}{1+e^{i\theta}}$
We find it converges at all points on the circle except $\theta=\pi$ which makes $P(z)$ infinite.
4. Cauchy product [Page 131]:
If $\sum u_n$ converges to $S$ and $\sum v_n$ converges to $T$, then the Cauchy product:
$\sum(u_1v_n+u_2v_{n-1}+\cdots+u_nv_1)$
converges to $ST$.
5. Operations with power series [Page 134-135]:
If two power series $P(x)$ and $Q(x)$ are convergent, then the sum, difference, substitution, differentiation and integration also converge to the same range where $P(x)=a_0+a_1x+a_2x^2+\cdots$, so as $Q(x)$.
6. Taylor series [Page 138-140]:
$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f”(a)+\cdots+\frac{(x-a)^{n-1}}{(n-1)!}f^{(n-1)}(a)+R_n(x)$
where $R_n(x)=\frac{(x-a)^{n}}{n!}f^{(n)}(\xi)$, $\xi \in[a,x]$ called Lagrange remainder, or $R_n(x)=o(x^{n-1})$, $\xi \in[a,x]$ called Peano remainder.
If set $a = 0$, then we obtain Maclaurin series:
$f(x)=f(0)+f'(0)x+\frac{f”(0)}{2!}x^2+\cdots+\frac{f^{(n-1)}}{(n-1)!}x^{n-1}(0)+R_n(x)$
with error estimation $|R_n(x)|\leq \frac{M}{n!}x^n$.
Here’s some important Maclaurin formulas:
$$\begin{array}{rl} e^x&=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+o(x^3)\\ sinx&=x-\frac{x^3}{3!}+\frac{x^5}{5!}+o(x^5)\\ cosx&=1-\frac{x^2}{2!}+\frac{x^4}{4!}+o(x^4)\\ \frac{1}{1-x}&=1+x+x^2+x^3+o(x^3)\\ \frac{1}{1+x}&=1-x+x^2-x^3+o(x^3)\\ ln(1+x)&=x-\frac{x^2}{2}+\frac{x^3}{3}+o(x^3)\\ arctanx&=x-\frac{x^3}{3}+\frac{x^5}{5}+o(x^5)\\ (1+x)^{\alpha}&=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+o(x^3)\\ arcsinx&=x+\frac{1}{2}\times\frac{x^3}{3}+\frac{1}{2}\times\frac{3}{4}\times\frac{x^5}{5}+o(x^5)\\ tanx&=x+\frac{x^3}{3}+\frac{2x^5}{15}+o(x^5)\\ \sqrt{1+x}&=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}+o(x^3)\\ \end{array}$$
7. Evaluation of limits [Page 142]:
7.1 Use ln:
$\lim\limits_{x\rightarrow \infty}y(x)=\lim\limits_{x\rightarrow \infty}e^{\ln[y(x)]}$.
7.2 Use l’Hopital’s rule:
$\lim\limits_{x\rightarrow \infty}\frac{f(x)}{g(x)}=\lim\limits_{x\rightarrow \infty}\frac{f'(x)}{g'(x)}$.
