An approach to "early transcendentals"

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

Last update: 08 July 2012

The function $\mathrm{god}\left(t\right)$

There is one function that

1. (a) in the Beginning, created something from nothing, and
2. (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, $$(a\prime )\mathrm{god}\left(0\right)=1,and(b\prime )\frac{d\mathrm{god}\left(t\right)}{dt}=\mathrm{god}\left(t\right),$$ but the meaning is still the same.

Two of the children of god are eve and adam: $$\mathrm{god}\left(it\right)=\mathrm{adam}\left(t\right)+i\mathrm{eve}\left(t\right).$$

Trying to understand $\mathrm{god}\left(t\right)$

If we try to "understand" god in "normal" terms, $$\mathrm{god}\left(t\right)=a_0+a_{1}t+a_2{t}^2+a_3{t}^3+\cdots ,$$ then $$\begin{array}{llll}since& \mathrm{god}\left(0\right)=1,& a_0=1,& and\\ since& \frac{d\mathrm{god}\left(t\right)}{dt}=\mathrm{god}\left(t\right),& a_1=a_0,& and\\ & & 2a_2=a_1,& and\\ & & 3a_3=a_2,& and\\ & & 4a_4=a_3,& and\\ & & 5a_5=a_4,& ...,\; etc.,\end{array}$$ and so $$\mathrm{god}\left(t\right)=1+t+\frac{1}{2!}{t}^2+\frac{1}{3!}{t}^3+\frac{1}{4!}{t}^4+\cdots ,$$ which illustrates that $\mathrm{god}\left(t\right)$ exists everywhere and goes on forever.

An amazing thing about $\mathrm{god}\left(t\right)$

One of the amazing things about god is that $$\mathrm{god}(t+s)=\mathrm{god}\left(t\right)\mathrm{god}\left(s\right).$$ To see why god is this why suppose that there is a "different" function such that

1. (a′′) is "unchanging" $(i.e.\frac{d\widetilde{\mathrm{god}}\left(t\right)}{dt}=\widetilde{\mathrm{god}}\left(t\right)),$ and
2. (b′′) in the Beginning, was the way that god is after $s$ millenia (i.e. $\widetilde{\mathrm{god}}\left(0\right)=\mathrm{god}\left(s\right)$).

By the chain rule, $$\frac{d\mathrm{god}(t+s)}{dt}=\mathrm{god}(t+s)and\mathrm{god}(0+s)=\mathrm{god}\left(s\right),$$ and so $$\mathrm{god}(t+s)=\widetilde{\mathrm{god}}\left(t\right).$$ Also, $$\frac{d\left(\mathrm{god}\left(t\right)\mathrm{god}\left(s\right)\right)}{\mathrm{god}\left(t\right)\mathrm{god}\left(s\right)},and\mathrm{god}\left(0\right)\mathrm{god}\left(s\right)=\mathrm{god}\left(s\right),$$ and so $$\mathrm{god}\left(t\right)\mathrm{god}\left(s\right)=\widetilde{\mathrm{god}}\left(t\right)=\mathrm{god}(t+s).$$

What about $\mathrm{adam}\left(t\right)$ and $\mathrm{eve}\left(t\right)$?

$$\begin{array}{rcl}\mathrm{god}\left(it\right)& =& 1+it+\frac{{\left(it\right)}^2}{2!}+\frac{{\left(it\right)}^3}{3!}+\frac{{\left(it\right)}^4}{4!}+\frac{{\left(it\right)}^5}{5!}+\cdots \\ & =& 1\phantom{+it}+\frac{i^2{t}^2}{2!}\phantom{+\frac{i^3{t}^3}{3!}}+\frac{i^4{t}^4}{4!}\phantom{+\frac{i^5{t}^5}{5!}}+\frac{i^6{t}^6}{6!}\phantom{+\frac{i^7{t}^7}{7!}}+\cdots \\ & & \phantom{1}+it\phantom{+\frac{i^2{t}^2}{2!}}+\frac{i^3{t}^3}{3!}\phantom{+\frac{i^4{t}^4}{4!}}+\frac{i^5{t}^5}{5!}\phantom{+\frac{i^6{t}^6}{6!}}+\frac{i^7{t}^7}{7!}\phantom{+\cdots }\\ & =& 1\phantom{+it}-\frac{t^2}{2!}\phantom{-\frac{i{t}^3}{3!}}+\frac{t^4}{4!}\phantom{+\frac{i{t}^5}{5!}}-\frac{t^6}{6!}\phantom{-\frac{i{t}^7}{7!}}+\cdots \\ & & \phantom{1}+it\phantom{-\frac{t^2}{2!}}-\frac{i{t}^3}{3!}\phantom{+\frac{t^4}{4!}}+\frac{i{t}^5}{5!}\phantom{-\frac{t^6}{6!}}-\frac{i{t}^7}{7!}\phantom{+\cdots }\\ & =& (1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\frac{t^8}{8!}-\cdots )+i(t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\cdots )\end{array}$$ and, since adam and eve are the children of god, $$i.e.\; because\mathrm{god}\left(it\right)=\mathrm{adam}\left(t\right)+i\mathrm{eve}\left(t\right),$$ we see that $$\begin{array}{rcll}\mathrm{adam}\left(t\right)& =& 1-\frac{t^2}{2!}+\frac{t^4}{4!}-\frac{t^6}{6!}+\frac{t^8}{8!}-\cdots ,& and\\ \mathrm{eve}\left(t\right)& =& t-\frac{t^3}{3!}+\frac{t^5}{5!}-\frac{t^7}{7!}+\frac{t^9}{9!}-\cdots ,& \end{array}$$ from which it follows that $$\begin{array}{rclcrcl}\mathrm{adam}\left(0\right)& =& 1,& & \mathrm{eve}\left(0\right)& =& 0,\\ \mathrm{adam}(-t)& =& \mathrm{adam}\left(t\right),& & \mathrm{eve}(-t)& =& -\mathrm{eve}\left(t\right),\\ \frac{d\mathrm{adam}\left(t\right)}{dt}& =& -\mathrm{eve}\left(t\right),& & \frac{d\mathrm{eve}\left(t\right)}{dt}& =& \mathrm{adam}\left(t\right).\end{array}$$ So, adam and eve are complete opposites and identical twins at the same time.

Complete opposites and identical twins at the same time, another manifestation

$$\begin{array}{rcl}1& =& \mathrm{god}\left(0\right)=\mathrm{god}(it-it)=\mathrm{god}(it+i(-t))=\mathrm{god}\left(it\right)\mathrm{god}\left(i(-t)\right)\\ & =& (\mathrm{adam}\left(t\right)+i\mathrm{eve}\left(t\right))(\mathrm{adam}(-t)+i\mathrm{eve}(-t))\\ & =& (\mathrm{adam}\left(t\right)+i\mathrm{eve}\left(t\right))(\mathrm{adam}\left(t\right)-i\mathrm{eve}\left(t\right))\\ & =& {\left(\mathrm{adam}\left(t\right)\right)}^2+\mathrm{eve}\left(t\right))2,\end{array}$$ i.e. $$1={\left(\mathrm{adam}\left(t\right)\right)}^2+{\left(\mathrm{eve}\left(t\right)\right)}^2.$$

Through the ages: where are we now?

Let $x=\mathrm{adam}\left(t\right)$ and $y=\mathrm{eve}\left(t\right).$

1. (A) In the Beginning the point $(x,y)$ was at $(\mathrm{adam}\left(0\right),\mathrm{eve}\left(0\right))=(1,0),$ and since $1={\left(\mathrm{adam}\left(t\right)\right)}^2+{\left(\mathrm{eve}\left(t\right)\right)}^2,x^2+y^2=1,$ and
2. (B) adam and eve travel through the ages on a circle of radius 1.

Where are they after $d$ millenia? $$\begin{array}{rcl}\genfrac{}{}{0ex}{}{The\; distance\; traveled}{afterdmillenia}& =& \underset{t=0}{\overset{t=d}{\int }}\mathrm{ds}\\ & =& \underset{t=0}{\overset{t=d}{\int }}\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}\mathrm{dt}\\ & =& \underset{t=0}{\overset{t=d}{\int }}\sqrt{{\left(\frac{d\mathrm{adam}\left(t\right)}{dt}\right)}^2+{\left(\frac{d\mathrm{eve}\left(t\right)}{dt}\right)}^2}\mathrm{dt}\\ & =& \underset{t=0}{\overset{t=d}{\int }}\sqrt{{(-\mathrm{eve}\left(t\right))}^2+{\left(\mathrm{adam}\left(t\right)\right)}^2}\mathrm{dt}\\ & =& \underset{t=0}{\overset{t=d}{\int }}\sqrt{1}\mathrm{dt}\\ & =& \underset{t=0}{\overset{t=d}{\int }}\mathrm{dt}\\ & =& {\phantom{|}\phantom{\frac{\mathrm{A}}{\mathrm{A}}}t|}_{t=0}^{t=d}\\ & =& d-0=d,\end{array}$$ and so $$\begin{array}{rcll}\mathrm{adam}\left(t\right)& =& \genfrac{}{}{0ex}{}{x-coordinate\; of\; the\; point\; on\; a\; circle\; of\; radius\; 1}{which\; is\; distancedfrom\; the\; point(1,0),}& and\\ \mathrm{eve}\left(t\right)& =& \genfrac{}{}{0ex}{}{y-coordinate\; of\; the\; point\; on\; a\; circle\; of\; radius\; 1}{which\; is\; distancedfrom\; the\; point(1,0).}& \end{array}$$

The triangle in this picture is $$\begin{array}{c} \[ \mathrm{adam}\left(d\right) \] \[ \mathrm{eve}\left(d\right) \] \[ 1 \] \[ adjacent \] \[ opposite \] \[ hypotenuse \]\end{array}$$ and so $$\mathrm{adam}\left(d\right)=\frac{opposite}{hypotenuse}and\mathrm{eve}\left(d\right)=\frac{adjacent}{hypotenuse}$$ for a right triangle with angle $d.$

Some remarks on society

1. It is interesting to note that our school system like to introduce our children to $\mathrm{adam}\left(t\right)$ and $\mathrm{eve}\left(t\right)$ but prefer to hide from my child how close they really are to $\mathrm{god}\left(t\right).$
2. Mathematicians are a cloistered group and prefer to study god, adam, and eve in anonymity. In the mathematical literature
   1. $\mathrm{god}\left(t\right)$ is usually called $e^t,$
   2. $\mathrm{adam}\left(t\right)$ is usually termed $\cos t,$ and
   3. $\mathrm{eve}\left(t\right)$ is usually called $\sin t.$

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