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Table of Contents

Ch4-Series and limits

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2024-08-19 / 0 Comments / 0 Liked / 37 Read / 22468 Words

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:
  1. 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.

  1. 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.

  1. 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 dinvergent 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 converges 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 diverges 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)}.

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