Real Analysis

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 9 July 2014

Lecture 11

The interest sequence

If you borrow $500 on your credit card at 14% interest, find the amounts due at the end of two years if the interest is compounded

(a)	annually,
(b)	quarterly,
(c)	monthly,
(d)	daily,
(e)	hourly,
(f)	every second,
(g)	every nanosecond,
(h)	continuously.

(a) You owe $$500+500\left(.14\right)=500(1+.14)$$ after one year and $$500(1+.14)(1+.14)$$ after two years.

(c) You owe $$500+500\left(\frac{.14}{12}\right)=500(1+\frac{.14}{12})$$ after one month, $$500(1+\frac{.14}{12})(1+\frac{.14}{12})$$ after two months, and $$500{(1+\frac{.14}{12})}^{24}$$ after two years.

(f) You owe $$500+500\left(\frac{.14}{365·24·2600}\right)$$ after 1 second, and $$500{(1+\left(\frac{.14}{365·24·2600}\right))}^{2·365·24·2600}$$ after two years.

(h) You owe $$\underset{n\to \infty }{\text{lim}}500{(1+\frac{.14}{n})}^{2n}$$ after $2$ years. $$\begin{array}{ccc}{\displaystyle \underset{n\to \infty }{\text{lim}}500{(1+\frac{.14}{n})}^{2n}}& =& {\displaystyle 500\underset{n\to \infty }{\text{lim}}{e}^{\text{log}{(1+\frac{.14}{n})}^{2n}}}\\ & =& {\displaystyle 500\underset{n\to \infty }{\text{lim}}{e}^{2n\text{log}(1+\frac{.14}{n})}}\\ & =& {\displaystyle 500\underset{n\to \infty }{\text{lim}}{e}^{2·\left(.14\right)\frac{\text{log}(1+\frac{.14}{n})}{.14/4}}}\\ & =& {\displaystyle 500\underset{n\to \infty }{\text{lim}}{e}^{.28\frac{\text{log}(1+\frac{.14}{n})}{.14/n}}\text{.}}\end{array}$$ Recall $$\underset{x\to 0}{\text{lim}}\frac{\text{log}(1+x)}{x}=1\text{.}$$ So you owe $$500·e^{.28}$$ after two years.

Note: $$\begin{array}{ccc}{\displaystyle 500{(1+.14)}^2}& =& {\displaystyle 649.80}\\ {\displaystyle 500{(1+\frac{.14}{12})}^{24}}& \approx & {\displaystyle 660.49}\\ {\displaystyle 500e^{.28}}& \approx & {\displaystyle 661.56}\end{array}$$

Newton Iteration – Solving $f\left(x\right)=0$

The Taylor series of $f\left(x\right)$ at $x=a$ is $$f\left(x\right)=f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)(x-a)+\frac{1}{2!}\left(\frac{d^{2}f}{d{x}^2}{|}_{x=a}\right){(x-a)}^2+\cdots$$ Then, for nice functions $f$ (functions that don't jump around much) $$f\left(x\right)\approx f\left(a\right)+\left(\frac{df}{dx}{|}_{x=a}\right)(x-a)=f\left(a\right)+f\prime \left(a\right)(x-a)\text{.}$$ Create a sequence: $$\begin{array}{ccc}{\displaystyle z_0}& =& {\displaystyle \text{your choice}}\\ {\displaystyle z_1}& =& {\displaystyle -\frac{f\left(z_0\right)}{f\prime \left(z_0\right)}+z_0,}\\ {\displaystyle z_2}& =& {\displaystyle -\frac{f\left(z_1\right)}{f\prime \left(z_1\right)}+z_1,}\\ {\displaystyle }& =& {\displaystyle ⋮}\end{array}$$ Then, if $\underset{n\to \infty }{\text{lim}}{z}_n=z$ (the sequence $z_n$ converges) then $f\left(z_{n+1}\right)$ is very close to $$f\left(z_n\right)+f\prime \left(z_n\right)(z_{n+1}-z_n)=f\left(z_n\right)+f\prime \left(z_n\right)(-\frac{f\left(z_n\right)}{f\prime \left(z_n\right)}+z_n-z_n)=0\text{.}$$ So $f\left(z\right)=0\text{.}$

This "trick" for solving $f\left(z\right)=0$ totally fails if $f\prime \left(z_n\right)$ ever comes out $0$ (or very very small) or if $f$ jumps around wildly.

Picard iterlation – Solving $f\left(x\right)=x\text{.}$

Create a sequence $\left(a_n\right)\text{:}$ $$\begin{array}{ccc}{\displaystyle a_1}& =& {\displaystyle \text{your choice}}\\ {\displaystyle a_2}& =& {\displaystyle f\left(a_1\right),}\\ {\displaystyle a_3}& =& {\displaystyle f\left(a_2\right),}\\ & =& {\displaystyle ⋮}\end{array}$$ Then, if $\underset{n\to \infty }{\text{lim}}{a}_n=a$ (the sequences $\left(a_n\right)$ converges) then $f\left(a\right)=a$ (because $f\left(a_n\right)=a_{n+1}$ is very close to $a_n$ for large enough $n\text{).}$

Show that the equation $x^3+x-1=0$ has a solution between $0$ and $1\text{.}$ $$\begin{array}{ccc} y=x3 - 1 1 x - 1 1 y & & y=x3+1 x 1 2 y \\ y=x3-1 - 1 x - 2 - 1 y & & y=x - 1 1 x - 1 1 y \\ \multicolumn{3}{c}{ y=x3+x-1 1 x - 1 1 y }\end{array}$$ Notes:

(a)	If $x=0$ then $x^3+x-1=-1\text{.}$
(b)	If $x=1$ then $x^3+x-1=1+1-1=1\text{.}$

So $x^3+x-1$ has a zero between $x=0$ and $x=1$ (because $x^3+x-1$ is continuous).

Transform the equation $x^3+x-1=0$ to the form $x=f\left(x\right)$ and use Picard iteration to find the solution to $3$ decimal places.

Since $x^3+x-1=0$ is the same as $x(x^2+1)=1,$ $x=\frac{1}{x^2+1}$ is of the form $x=f\left(x\right)\text{.}$

Let $a_1=\frac{1}{2}\text{.}$ Then $$\begin{array}{ccc}{\displaystyle a_2}& =& {\displaystyle \frac{1}{{\left(\frac{1}{2}\right)}^2+1}=\frac{1}{\frac{1}{4}+1}=\frac{1}{\frac{5}{4}}=\frac{4}{5}=0.8}\\ {\displaystyle a_3}& =& {\displaystyle \frac{1}{{\left(\frac{4}{5}\right)}^2+1}=\frac{1}{\frac{16}{25}+1}=\frac{1}{\frac{41}{25}}=\frac{25}{41}\approx \mathrm{.}\overline{60975}\overline{60975}}\\ {\displaystyle a_4}& =& {\displaystyle \frac{1}{{\left(\frac{25}{41}\right)}^2+1}=\frac{1}{\frac{625}{1681}+1}=\frac{1}{\frac{625+1681}{1681}}=\frac{1681}{2306}\approx .728967}\\ {\displaystyle a_5}& =& {\displaystyle .6530046}\\ {\displaystyle a_6}& =& {\displaystyle .7010582}\\ {\displaystyle a_7}& =& {\displaystyle .6704737}\\ {\displaystyle a_8}& =& {\displaystyle .68987635}\\ {\displaystyle a_9}& =& {\displaystyle .67753918}\\ {\displaystyle a_{10}}& =& {\displaystyle .68537308}\\ {\displaystyle a_{11}}& =& {\displaystyle .680394233}\\ {\displaystyle a_{12}}& =& {\displaystyle .6835567}\\ {\displaystyle a_{13}}& =& {\displaystyle .681547222}\\ {\displaystyle a_{14}}& =& {\displaystyle .68282382}\\ {\displaystyle a_{15}}& =& {\displaystyle .6820126}\end{array}$$ So, to $3$ decimal places of accuracy $x=.682$ is a zero of $x^3+x-1=0\text{.}$

Note: Another expression for $x^3+x-1=0$ is $x=1-x^3\text{.}$

Let $$\begin{array}{ccc}{\displaystyle a_1}& =& {\displaystyle \frac{1}{2}}\\ {\displaystyle a_2}& =& {\displaystyle 1-{\left(\frac{1}{2}\right)}^3=1-\frac{1}{8}=\frac{7}{8}=.875}\\ {\displaystyle a_3}& =& {\displaystyle 1-{\left(\frac{7}{8}\right)}^3\approx .330078}\\ {\displaystyle a_4}& \approx & {\displaystyle 1-{\left(.330078\right)}^3\approx .964037}\\ {\displaystyle a_5}& \approx & {\displaystyle .104055,}\\ & =& {\displaystyle ⋮}\end{array}$$ and this sequence is not converging, but oscillating between close to $1$ and close to $0\text{.}$

Notes and References

These are notes from a 2010 course on Real Analysis 620-295. This page comes from 100324suggLect11.pdf and was given on 24 March 2010.

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