9.2 Limit and complex number
1.Exercise about limit
Prove that \(a_n=(-1)^n\) has no limit
Suppose \((-1)^n\) has \(\lim L\)
NTP: \(\left|(-1)^n - L\right| < \epsilon\)
- \(n\) is even
\(\left|1 - L\right| < \epsilon \implies -\epsilon < L - 1 < \epsilon\)
\(\implies 1 - \epsilon < L < 1 + \epsilon\)
- \(n\) is odd
\(\left| -1 - L\right| < \epsilon \implies -\epsilon < L + 1 < \epsilon\)
\(\implies -1 - \epsilon < L < -1 + \epsilon\)
If we take \(\epsilon = \frac{1}{2}\)
\(L \in \left( \frac{1}{2}, \frac{3}{2} \right) \cup \left( -\frac{3}{2}, -\frac{1}{2} \right)\) Contradiction
2.Infinity
We say that \(\lim_{n\to\infty}a_{n}=+\infty\) if \(\forall M>0,\exists n_0|a_n>M, ~\forall n\geq n_0\)
\(\lim_{n\to\infty}a_{n}=-\infty\)
3.Proposition
Let \(a_n\) be a sequence, prove that
If \(a_n\) is convergent(\(\exists \lim_{n\to\infty}a_{n}=L\in R)\Rightarrow a_n\) is bounded
(why no equivalent \(a_n=(-1)^n\))
Proof
Consider \(\varepsilon=1\) in the definition of limit
Then \(\exists n_0||a_n-L|<1~~\forall n\geq n_0\)
Then \(\forall n\geq n_0,|a_n|<|L|+1\) due to
\(||a_n|-|L||\leq |a_n-L|<1\Rightarrow|a_n|<|L|+1\)
To control \(a_1,a_2,a_3...a_{n_0-1}\),pick a large \(M\) such that \(|a_j|\leq M,~\forall j=1,2,...,n_0-1\)
Then \(|a_n|\leq max\{M,|L|+1\}\)
4.Algebra of limits=9=
Consider \((a_n)_{n\in N}\),\((b_n)_{n\geq 1}\) two sequences
\(\lim_{n\to\infty}a_{n}=a\) \(\lim_{n\to\infty}b_{n}=b\)
-
\(\lim_{n\to\infty}(a_{n}+b_n)=a+b\)
To prove this, let \(\varepsilon>0\). We need to find \(n_0\in N\)
N.T.P \(|a_n+b_n-(a+b)|<\varepsilon\)
Note that \(|a_{n}+b_{n}-(a+b)|\leq|a_{n}-a|+|b_{n}-b|\)
Since \(a_n\rightarrow a\), we know that there is some \(n_0\in N||a_n-a|<\frac{\varepsilon}{2},~\forall n\geq n_0\)
Similarly, \(b_n\rightarrow b\) so we can take \(n_1\in N||b_n-b|<\frac{\varepsilon}{2},~\forall n\geq n_1\) more strict restrict
Then if \(n\geq max\{n_0,n_1\}\) we have that
\(|a_n+b_n-(a+b)|\leq |a_n-a|+|n_b+b|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)
-
\(\lim_{n\to\infty}(a_{n}b_n)=ab\)
To prove this, consider \(\varepsilon >0\)
N.T.P. \(\exists n_0\in N||a_nb_n-ab|<\varepsilon~~\forall n\geq n_0\)
Note that \(|a_nb_n-ab|=|a_nb_n+a_nb-a_nb-ab|\leq |a_n(b_n-b)|+|b||a_n-a|\leq |a_n||b_n-b|+|b||a_n-a|\)
Now since \(a_n\) is convergent, there \(M\in R||a_n|\leq M~\forall n\in N\)
Then \(|a_nb_n-ab|\leq M|b_n-b|+|b||a_n-a|\) spilit into two parts
Consider \(n_1\in N||b_n-b|<\frac{\varepsilon}{2M}~\forall n\geq n_1(b_n\rightarrow b)\)
Also, \(n_2\in N||a_n-a|<\frac{\varepsilon}{2|b|}~\forall n\geq n_2(a_n\rightarrow a)\)
Then, if \(n\geq max\{n_1,n_2\},\) we have that
\(|a_nb_n-ab|\leq M|b_n-b|+|b||a_n-a|<M\frac{\varepsilon}{2M}+|b|\frac{\varepsilon}{2|b|}=\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon\)
-
If \(b\neq 0\), \(\lim_{n\to\infty}\frac{a_{n}}{b_n}=\frac{a}{b}\)
Proof exercise
Lemma 1: If \(\lim_{n \to \infty} a_n = \ell\) and \(\ell > 0\), then there is a number \(X\) such that \(a_n > \frac{1}{2} \ell\) for all \(n > X\).
Proof: Since \(\frac{1}{2} \ell > 0\), the terms \(a_n\) must eventually lie within a distance \(\frac{1}{2} \ell\) of the limit \(\ell\).
In other words, there is a number \(X\) such that
\(|a_n - \ell| < \frac{1}{2} \ell\), for all \(n > X\).
Hence \(-\frac{1}{2} \ell < a_n - \ell < \frac{1}{2} \ell\), for all \(n > X\)
and so the left-hand inequality gives
\(\frac{1}{2} \ell < a_n\), for all \(n > X\), as required.
Proof: We assume that \(m > 0\); the proof for the case \(m < 0\) is similar. Once again, the idea is to write the required expression in terms of \(a_n - \ell\) and \(b_n - m\):
\[ \frac{a_n}{b_n} - \frac{\ell}{m} = \frac{m(a_n - \ell) - \ell(b_n - m)}{b_n m}. \]Now, however, there is a slight problem: \(\{m(a_n - \ell) - \ell(b_n - m)\}\) is certainly a null sequence, but the denominator is rather awkward. Some of the terms \(b_n\) may take the value \(0\), in which case the expression is undefined.
However, by Lemma 1, we know that for some \(X\) we have
\[ b_n > \frac{1}{2} m, \quad \text{for all } n > X. \]Thus, for all \(n > X\):
\[ \left|\frac{a_n}{b_n} - \frac{\ell}{m}\right| = \frac{|m(a_n - \ell) - \ell(b_n - m)|}{b_n m} \]\[ \leq \frac{|m(a_n - \ell) - \ell(b_n - m)|}{\frac{1}{2} m^2} \]\[ \leq \frac{|m| \times |a_n - \ell| + |\ell| \times |b_n - m|}{\frac{1}{2} m^2}. \]
5.Back to the example
\(a_n\) defined by: \(a_1>0(any)\). \(a_{n+1}=\frac{a_n^2+2}{2a_n}\)
Conjecture: \(lim_{n\to\infty}a_n=\sqrt2\) whatever \(a_n\) we choose
We will prove that 1) \(a_n\geq \sqrt2 ~~~\forall n\geq 2\) 2) \(a_{n+1}\leq a_n\) (decreasing)
Once we prove this, we will conclude that \(\exists\lim_{n\to\infty}a_{n}=l\)
Recall: decreasing and bounded below \(\Rightarrow\) \(a_n\) convergent (\(\lim_{n\to\infty}a_n=inf\{a_n:n\in N\}\))
-
Let \(n\in N\)
\(a_{n+1}=\frac{a_n^2+2}{2a_n}\Rightarrow(a_n-\sqrt2)^2>0\) if \(a_n\neq \sqrt2\)
\(\forall a_1>0,a_n>\sqrt2,~\forall n\in N\)
-
Since \(a_n>\sqrt2~~\forall n\geq 2\), we have \(a_n^2>2~~\forall n\geq 2\)
\(a_{n+1}=\frac{a_n^2+2}{2a_n}<\frac{a_n^2+a_n^2}{2a_n}=a_n\)
Then, \(\exists l\in R|lim_{n\to\infty}a_n=l\) \((l>0)\)
\(l=lim_{n\to\infty}a_{n+1}=lim_{n\to\infty}\frac{a_n^2+2}{2a_n}=\frac{l^2+2}{2l}\)
Then \(l^2=2\Leftrightarrow l=\sqrt2\)
6.Series(infinite sums)
Consider a sequence \((a_n)_{n\in N}\) of real numbers
We defined another sequence \(S_1=a_1,S_2=a_1+a_2,S_3=a_1+a_2+a_3\)
\(S_n=a_1+a_2+a_3+...+a_n\)
We write \(\begin{aligned}&\sum_{k=1}^na_k&\end{aligned}\).\((S_n)_{n\in N}\) is the sequence of partial sums.
Definition: We say that the series defined by \((a_n)_{n\in N}\) is convergent if \(\lim_{n\to\infty}S_n\) exists
In that case, we write \(\begin{aligned}&\sum_{k=1}^{\infty}a_k=\lim_{n\to\infty}S_n=\lim_{n\to\infty}\sum_{k=1}^{\infty}a_k&\end{aligned}\)
Example
\(a_n=\frac{1}{2^n},S_n=\begin{aligned}&\sum_{k=1}^n\frac{1}{2^k}=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2^n}\rightarrow2&\end{aligned}\)conjecture: convergent
\(b_n=\frac{1}{n},S_n=\begin{aligned}&\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\rightarrow&\end{aligned}\)
Each group \(\geq \frac{1}{2}\). \(S_n = (1) + \left(\frac{1}{2}\right) + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}\right) + \cdots\)
\(c_n=(-1)^n,S_n=\begin{aligned}&\sum_{k=1}^n(-1)^n=1-1+1-1+...+(-1)^n\rightarrow&\end{aligned}\)
\(d_n=\frac{1}{n^2},S_n=\begin{aligned}&\sum_{k=1}^n\frac{1}{k^2}=1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}\rightarrow&\end{aligned}\)
(\(\sum_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi^2}{6}\))
The geometric series
\(a_n=r^n\) with \(|r|<1\)
Consider \(S_n=\sum_{k=0}^nr^k=1+r+r^2+r^3+...+r^n\)
\(rS_n=r+r^2+r^3+r^4+...+r^{n+1}\)
\(S_n-rS_n=1-r^{n+1}\Rightarrow S_n=\frac{1-r^{n+1}}{1-r}\)
Now, \(lim_{n\rightarrow \infty}\frac{1-r^{n+1}}{1-r}=\frac{1}{1-r}\) Since \(lim_{n\rightarrow \infty}r^{n+1}=0\)
We check that for any \(r\in (0,1)\), \(lim_{n\rightarrow \infty}r^{n}=0\)
Take \(\varepsilon>0\), N.T.P \(\exists n_0\in N|\forall n\geq n_0~~|r^n-0|<\varepsilon\)
\(log(|r|^n)<log(\varepsilon)\), then \(nlog(|r|)<log(\varepsilon)\)
\(n>\frac{log(\varepsilon)}{log(|r|)}\) pick \(n_0>\frac{log(\varepsilon)}{log(|r|)}\)
7.Complex number
Approach: include solutions to the equation \(x^2+1=0\)
Consider the set \(C=\{a+bi:a\in R,b\in R,i^2=-1\}\)
Example
\(z_1=3+2i,z_2=-1+3i\)
Definition: If \(z=a+bi\), \(Re(z)=a\), \(Im(z)=b\). Real part imaginary part imaginary unit
Geometrical representation: \(C\sim R\times R\)
Remark: \(R\hookrightarrow C\)
Operations
Sum: \(z=a+bi,w=c+di\). \(z+w=a+c+(b+d)i\)
\(Re(z+w)=Re(z)+Re(w)\), \(Im(z+w)=Im(z)+Im(w)\)
Product: Use \(i^2=-1\),\(z=a+bi,w=c+di\)
\(z\cdot w=(a+bi)(c+di)=ac-bd+(ad+bc)i\)
Absolute value
\(z\in C,z=a+bi,|z|=\sqrt{a^2+b^2}\)
Conjugation: \(z\in C,z=a+bi\) we define \(\bar z=a-bi\)
Note that \(z\cdot \bar z=|z|^2\), \(|z|=z\Leftrightarrow z=0=0+0i\)
Inverse
Let \(z\in C\),\(z=a+bi\)
Define \(z^{-1}=\frac{\bar z}{|z|^2}\), then \(z\cdot z^{-1}=z\cdot \frac{\bar z}{|z|^2}=1\)
Polar form for complex numbers
Let \(z=a+bi.~~a,b\in R\)
We write \(z=rcos(\theta)+i\cdot rsin(\theta)=r(cos(\theta)+i\cdot sin(\theta))\) \(r=|z|,|cos(\theta)+i\cdot sin(\theta)|=1\)
Also: \(z=\sqrt2(cos(\frac{9\pi}{4})+isin(\frac{9\pi}{4}))\)
So, do not miss plus the \(2k\pi\)
Argument
Given \(z\in C\), we define \(arg(z)=\{\theta\in R|z=r(cos(\theta)+i\cdot sin(\theta))\}\)
\(z=1+i\), \(arg(z)=\{\frac{\pi}{4}+2k\pi:k\in Z\}\)
We call "Principal argument" to the unique \(\theta\in arg(z)|-\pi<\theta\leq \pi\)
Notation \(Arg(1+i)=\frac{\pi}{4}\)
