## Folios 149-151: AAL to ADM

Ashley-Combe

Nov^{r }27^{th}

Dear M^{r} De Morgan. I have I believe made

some little progress towards the comprehension of

the Chapter on Notation of Functions, & I enclose

you my Demonstration of one of the Exercises at

the end of it : ''Show that the equation \(\varphi(x+y)=\)

''\( =\varphi x+\varphi y\) can be satisfied by no other solution

''than \(\varphi x=ax\) .'' At the same time I am

by no means satisfied that I do understand

these Functional Equations perfectly well, because

I am completely baffled by the other Exercise :

''Shew that the equation \(\varphi(x+y)+\varphi(x-y)=\)

''\( =2\varphi x\times\varphi y\) is satisfied by \(\varphi x=\frac{1}{2}(a^x+a^{-x})\)

''for every value of \(a\) ''.

I do not know when I have been so tantalized

by anything, & should be ashamed to say __how__

much time I have spent upon it, in vain.**[149v] ** These Functional Equations are complete Will-o'-

-the-Wisps to me. The moment I fancy I have

really at last got hold of something tangible

& substantial, it all recedes further & further

& vanishes again into thin air.

But now for this perplexing \(\varphi x=\frac{1}{2}(a^x+a^{-x})\) .

I be__liev__e I have left no method untried; but

I cannot get further than as below, with any

certainty :

\(\varphi(x+y)+\varphi(x-y)=2\varphi x\times\varphi y\)

\(\because 2\varphi x=\frac{\varphi(x+y)+\varphi(x-y)}{\varphi y}\)

Since \(x\) and \(y\) may have a__n__y values whatever,

(at least such I conclude is of course intended), let

\( y=0\) . We have then

\(2\varphi x=\frac{\varphi(x)+\varphi(x)}{\varphi(0)}\)

\(\because 2\varphi x\times\varphi(0)=\varphi(x)+\varphi(x)\)

or \(2\varphi x\times\varphi(0)=2\varphi x\)

\(\varphi(0)\) must \(=1\), or \(=a^0\), since \(a^0\) is

the only function of \(0\) which can \(=1\)

I think so far is correct in itself, but whether**[150r] ** it be the ['right' inserted] road to the rest is another question.

At any rate, __I__ have not succeeded in proving

it such. To assume that since \(\varphi(0)=a^0\),

\( \varphi(x+y)=a^{x+y}\), \(\varphi(x-y)=a^{x-y}\), \(\varphi y=a^y\)

appears to me scarcely warrantable; and

besides in that case it must be equally

assumed that \(\varphi x=a^x\), (there being the same

ground for the one assumption as for the others),

and we should then have,

\(a^x=\frac{a^{x+y}+a^{x-y}}{a^y}\)

\(\because a^x=\frac{a^{x+y}}{a^y}+\frac{a^{x-y}}{a^y}\)

\(\because a^x=a^x+a^{x-2y}\)

most clearly absurd, independent of it's [\textit{sic}] being

discordant with the book.

Once I thought I had hit on something very

clever indeed, and wrote as follows :

\(\varphi(x+y)=\varphi(\overline{x+y}.1)\)

\(=\{\varphi(1)\}^{x+y}\) by equation \((\varphi a)^n=\varphi(na)\)

page 205, entirely forgetting that the \(\varphi\) of that

equation had nothing whatever to do with the__\( \varphi\) __ of any other equations; (a disagreeable truth**[150v] ** which did not occur to me until 24 hours

later). I then had \(\varphi(x-y)=\varphi(\overline{x-y}.1)=\{\varphi(1)\}^{x-y}\)

\( \varphi(y)=\varphi(y.1)=\{\varphi(1)\}^y\), and

\(2\varphi x=\frac{\{\varphi(1)\}^{x+y}+\{\varphi(1)\}^{x-y}}{\{\varphi(1)\}^y}\)

\(=\frac{\{\varphi(1)\}^{x+y}}{\{\varphi(1)\}^y}+\frac{\{\varphi(1)\}^{x-y}}{\{\varphi(1)\}^y}\)

\(=\{\varphi(1)\}^x+\{\varphi(1)\}^{x-2y}\), and supposing

\( x=y\), then \(2\varphi x=\{\varphi(1)\}^x+\frac{1}{2}\{\varphi(1)\}^{-x}\)

or calling \(\varphi(1)=a\), \(\varphi x=\frac{1}{2}(a^x+a^{-x})\)

But besides my unwarrantable assumption of

\( \varphi(\overline{x+y}.1)=\{\varphi(1)\}^{x+y}\), there was t__hi__s in the

result which was unsatisfactory, that it was

necessary to assume \(x=y\), and the result seemed

to hold good in t__hat__ case a__lone__. Also, when to

verify, I tried \(x=1\), \(\because\varphi(1)=\frac{1}{2}(a^1+a^{-1})\),

which ought to have come out \(a=a\), I could

make neither head or tail of it. Well, I

abandoned this, & tried all sorts of other resources.**[151r] ** I understand to work out something by means

similar to those in page 205 and in the

Problem I send; but equally unsuccessfully.

I also in equation (A) page 204, changed

\( \varphi(x+y)\) into \(\varphi(x-y)\) and investigated this,

thinking I might derive a hint possibly from it.

[something crossed out]} In short, many & various are the experiments

I have made, but I will not detail any

more. Indeed I think you may be possibly

heartily sick of what I h__ave__ detailed. But

I wished to show you that I have not failed

from want of t__rying__, at least; & also to give

you the chance of smiling at my expence [*sic*]. __ __

I shall have to trouble you with another

letter shortly, on other knotty points. Really I

do not give you a Sinecure. Your letters are

however well bestowed, in as far as the use

they are of to me, can make them so, and

the great encouragement that such assistance

is to me to continue my Studies with zeal

& spirit. We are to return to Surrey very**[151v] ** soon. I expect to have occasion to trouble

you again before we go, & after that I shall

hope to see you & M^{r} De Morgan in Town,

where I intend to be for two or three days

in about a fortnight. __ __

Yours very truly

A. A. L

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