MATH 221

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

Last updated: 6 September 2014

Lecture 39

There is one function that

(a)	In the Beginning, created something from nothing, and
(b)	is "unchanging", or rather, its change is itself.

Through the ages thinkers have contemplated this function and nowadays it is common to write (a) and (b) in abbreviated form:

(a')	$\text{god}\left(0\right)=1,$ and
(b')	$\frac{d\text{god}\left(t\right)}{dt}=\text{god}\left(t\right),$

but the meaning is still the same.

Two of the children of god are eve and adam: $$\text{god}\left(it\right)=\text{eve}\left(t\right)+i\text{adam}\left(t\right)\text{.}$$ If we try to understand god in normal terms $$\text{god}\left(t\right)=a_0+a_{1}t+a_2{t}^2+a_3{t}^3+\cdots$$ then we find that since $\text{god}\left(0\right)=1,$ $$a_0=1,$$ and since $\frac{d\text{god}\left(t\right)}{dt}=\text{god}\left(t\right),$ $$\begin{array}{ccc}{\displaystyle a_1}& =& {\displaystyle a_0,}\\ {\displaystyle a_2}& =& {\displaystyle \frac{1}{2}{a}_1,}\\ {\displaystyle a_3}& =& {\displaystyle \frac{1}{3!}{a}_2,}\\ {\displaystyle a_4}& =& {\displaystyle \frac{1}{4!}{a}_3,\dots }\end{array}$$ and so $$\text{god}\left(t\right)=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+\frac{t^5}{5!}+\cdots$$ from which we see that god exists everywhere and goes on forever.

One of the amazing things about god is that $$\text{god}(t+s)=\text{god}\left(t\right)\text{god}\left(s\right)\text{.}$$ You can see that god must be this way by supposing that there were a "different" function that

(a'')	is "unchanging".
(b'')	In the Beginning, was the way that god is after $s$ millenia.

Because of the chain rule, $$\frac{d\text{god}(t+s)}{dt}=\text{god}(t+s),$$ and $$\text{god}(0+s)=\text{god}\left(s\right),$$ and, since $$\frac{d\text{god}\left(t\right)\text{god}\left(s\right)}{dt}=\text{god}\left(t\right)\text{god}\left(s\right),$$ and $$\text{god}\left(0\right)\text{god}\left(s\right)=\text{god}\left(s\right),$$ we see that both $\text{god}(t+s)$ and $\text{god}\left(t\right)\text{god}\left(s\right)$ fit the job description for this "different" function job and this is why $$\text{god}(t+s)=\text{god}\left(t\right)\text{god}\left(s\right)\text{.}$$ What about eve and adam? Since $$\begin{array}{ccc}{\displaystyle \text{god}\left(it\right)}& =& {\displaystyle 1+it+\frac{{\left(it\right)}^2}{2!}+\frac{{\left(it\right)}^3}{3!}+\frac{{\left(it\right)}^4}{4!}+\frac{{\left(it\right)}^5}{5!}+\frac{{\left(it\right)}^6}{6!}+\cdots }\\ & =& {\displaystyle 1+\phantom{it}+\frac{i^2{t}^2}{2!}+\phantom{\frac{i^3{t}^3}{3!}}+\frac{i^4{t}^4}{4!}+\phantom{\frac{i^5{t}^5}{5!}}+\frac{i^6{t}^6}{6!}+\cdots }\\ & & {\displaystyle \phantom{1+}it\phantom{+\frac{i^2{t}^2}{2!}+}\frac{i^3{t}^3}{3!}\phantom{+\frac{i^4{t}^4}{4!}+}\frac{i^5{t}^5}{5!}\phantom{+\frac{i^6{t}^6}{6!}+\cdots }}\\ & =& {\displaystyle 1+\phantom{it}-\frac{t^2}{2!}\phantom{-\frac{i{t}^3}{3!}}+\frac{t^4}{4!}\phantom{+\frac{i{t}^5}{5!}}-\frac{t^6}{6!}+\cdots }\\ & & {\displaystyle \phantom{1+}it\phantom{-\frac{t^2}{2!}}-\frac{i{t}^3}{3!}\phantom{+\frac{t^4}{4!}}+\frac{i{t}^5}{5!}\phantom{-\frac{t^6}{6!}}-\cdots }\\ & =& {\displaystyle (1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots )+i(t-\frac{t^3}{3!}+\frac{t^5}{5!}-\cdots )}\end{array}$$ and since $\text{god}\left(it\right)=\text{eve}\left(t\right)+i\text{adam}\left(t\right),$ it follows that $$\begin{array}{ccc}{\displaystyle \text{eve}\left(t\right)}& =& {\displaystyle 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\cdots ,\text{and}}\\ {\displaystyle \text{adam}\left(t\right)}& =& {\displaystyle t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\cdots \text{.}}\end{array}$$ From these we see that $$\text{eve}\left(0\right)=1\text{and}\text{adam}\left(0\right)=0,$$ and $$\frac{d\text{eve}\left(t\right)}{dt}=-\text{adam}\left(t\right),\frac{d\text{adam}\left(t\right)}{dt}=\text{eve}\left(t\right)$$ and $$\text{eve}(-t)=\text{eve}\left(t\right)\text{and}\text{adam}(-t)=-\text{adam}\left(t\right),$$ all of which illustrate that eve and adam are both identical twins and opposites at the same time. Another manifestation is $$\begin{array}{ccc}{\displaystyle 1}& =& {\displaystyle \text{god}\left(0\right)}\\ & =& {\displaystyle \text{god}(it-it)}\\ & =& {\displaystyle \text{god}(it+(-it))}\\ & =& {\displaystyle \text{god}\left(it\right)\text{god}\left(i(-t)\right)}\\ & =& {\displaystyle \left(\text{eve}(t+i\text{adam}\left(t\right))\right)(\text{eve}(-t)+i\text{adam}(-t))}\\ & =& {\displaystyle (\text{eve}\left(t\right)+i\text{adam}\left(t\right))(\text{eve}\left(t\right)-i\text{adam}\left(t\right))}\\ & =& {\displaystyle {\text{eve}}^2\left(t\right)-i^2{\text{adam}}^2\left(t\right)}\\ & =& {\displaystyle {\text{eve}}^2\left(t\right)+{\text{adam}}^2\left(t\right),}\end{array}$$ i.e. $${\text{eve}}^2\left(t\right)+{\text{adam}}^2\left(t\right)=1\text{.}$$ Let $x=\text{adam}\left(t\right)$ and $y=\text{eve}\left(t\right)\text{.}$ Then, in the Beginning, the point $(x,y)$ was at $(1,0)$ $$\begin{array}{ccc}& =& {\displaystyle {\int }_{t=0}^{t=d}\sqrt{{\text{adam}}^2\left(t\right)+{\text{eve}}^2\left(t\right)}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=d}\sqrt{1}dt}\\ & =& {\displaystyle {\int }_{t=0}^{t=d}dt=}\\ & =& {\displaystyle t{|}_{t=0}^{t=d}}\\ & =& {\displaystyle d-0}\\ & =& {\displaystyle d\text{.}}\end{array}$$ So $$\text{eve}\left(t\right)=\begin{array}{}y\text{-coordinate of a point on a}\\ \text{circle of radius 1 which is}\\ \text{distance}\hspace{0.17em}d\hspace{0.17em}\text{from}\hspace{0.17em}(1,0)\end{array}$$ and $$\text{adam}\left(t\right)=\begin{array}{}x\text{-coordinate of a point on a}\\ \text{circle of radius 1 which is}\\ \text{distance}\hspace{0.17em}d\hspace{0.17em}\text{from}\hspace{0.17em}(1,0)\end{array}$$ $$\begin{array}{c} -1 1 x -1 1 1 y (x,y)=(adam(t),eve(t)) \end{array}$$ The triangle in this picture has $$\begin{array}{c} 1 t adam(t) eve(t) \end{array}$$ and so $$\begin{array}{ccc}{\displaystyle \text{eve}\left(t\right)}& =& {\displaystyle \frac{\text{opposite}}{\text{hypotenuse}},\text{and}}\\ {\displaystyle \text{adam}\left(t\right)}& =& {\displaystyle \frac{\text{adjacent}}{\text{hypotenuse}}}\end{array}$$ since the hypotenuse is length 1, the opposite edge is length $\text{eve}\left(t\right),$ and the adjacent edge is length $\text{adam}\left(t\right)\text{.}$

Note: Mathematicians are a cloistered group and prefer to study $\text{god}\left(t\right)$ in anonymity. Thus they write $$\begin{array}{c}{e}^t\text{instead of}\text{god}\left(t\right)\\ \text{cos}\hspace{0.17em}t\text{instead of}\text{eve}\left(t\right)\\ \text{sin}\hspace{0.17em}t\text{instead of}\text{adam}\left(t\right)\end{array}$$ in the mathematical literature.

Notes and References

These are a typed copy of Lecture 39 from a series of handwritten lecture notes for the class MATH 221 given on December 13, 2000.

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