## Folios 42-43: ADM to AAL

**[42r] ** My dear Lady Lovelace

Mr. Frend's death (which took place on Sunday Morning)

has made me answer your letter later than I should otherwise have

done. The family are all well, and have looked forward to this termination

for some time. My wife will answer your letter on a part of this.

Number and Mag^{n} pp. 75,76. The use of this theorem is shown in

what follows.

It proves that any quantity which lies between two others is

either one of a

set of mean proportionals between those two, or as near

to one as we please.

It is not self evident that the base of Napiers system, as

given by himself

is \(\varepsilon\) or \(1+1+\frac{1}{2}+\cdots\) as we learn

from the modern mode of presenting the

theory. The last sentence in the book (making

\(V\) a linear unit) would show

that Napier's notion was to take \(k\) in such a manner

that \(x\) shall

expound [?] \(1+x\) ['without' crossed out] or rather that

the smaller \(x\) is the more nearly shall

\( x\) expound \(1+x\) . If this were accurately done,

we should have

\(k^x=1+x\) or \(\frac{k^x-1}{x}=1\)

and this is to be nearer to the truth the smaller \(x\) is. Now when the

common theory is known, it is known that \(k=\varepsilon\) gives

\(\frac{\varepsilon^x-1}{x}=1+\frac{x}{2}+\frac{x^2}{2\cdot 3}+\cdots\cdot\) and limit of \(\frac{\varepsilon^x-1}{x}=1\)

while \(\frac{k^x-1}{x}=\log k+\frac{\overline{\log k}\vert^2.x}{2}+\cdots\) and limit of \(\frac{k^x-1}{x}=\log k\)

where the log.\ has this very base \(\varepsilon\) . Having proved these things, it is

then obvious that, \(\log k\) being never \(1\) except when \(k\) is the base or \(\varepsilon\),

the last paragraph cannot consist with any other value of \(k\) except

\( \varepsilon\) . In this book (Num & Mag.) I must refer you to the algebra, which

I do not in the Diff. Calc., many matters of series, until the whole

doctrine is reestablished.

Now ['as' crossed out] as to the Diff.\ Calc. You do not see that \(\theta\) is a

function of \(a\) and \(h\) . Let us take the simplest case of the

original theorem which is**[42v] ** \(\varphi(a+h)=\varphi a+h\ \varphi'(a+\theta h)\) (1)

Now 1.\ Why should \(\theta\) be independent of \(a\) and \(h\), we have never

proved it to be so : all we have proved is that __one__ of the

numerical values of \(\theta\) is \(<1\), or that this equation (1) can be

satisfied by a value of \(\theta<1\) . As to what \(\theta\) is, let \(\psi\) be the

inverse function of \(\varphi'\) so that \(\psi\varphi'x=x\) . Then

\(\frac{\varphi(a+h)-\varphi a}{h}=\varphi'(a+\theta h)\)

\(\psi\left(\frac{\varphi(a+h)-\varphi(a)}{h}\right)=\psi\varphi'(a+\theta h)=a+\theta h\)

\(\theta=\frac{\psi\left(\frac{\varphi(a+h)-\varphi a}{h}\right)-a}{h}.\)\(\left\{\begin{array}{l}\text{Say that this is not a function}\\ \text{of}~a~\text{and}~ h,~\text{if you dare}\end{array}\right.\)

For example \(\varphi x=c^x\)

\(\varphi'x=c^x.\log c\)

\(c^{a+h}=c^a+h\log c\ c^{a+\theta h}\)

\(c^{a+\theta h}=\frac{c^{a+h}-c^a}{h\log c}\)

\((a+\theta h)\log c=\log\frac{c^{a+h}-c^a}{h\log c}\)

\(\theta=\frac{\log\frac{c^{a+h}-c^a}{h\log c}-a\log c}{h\log c}\)

\(=\frac{\log (c^h-1)-\log(h\log c)}{h\log c}\)

In this particular example \(\theta\) happens to be a function

of \(h\) only, not of \(a\) : but you must remember that in every case

where we speak of a quantity as being generally a function

of \(a\), we do not mean thereby to deny that it may be in

particular case, not a function of \(a\) at all : just as**[43r] ** when we say that there is __a number__ (\( x\) ) which satisfies

certain conditions, we do not thereby exclude the extreme case in

which \(x=0\) .

Look at the question of differences in this manner. Any

thing which has been proved to be true of \(u_n\) relatively to \(u_{n-1}\) [,]

\( u_{n-2}\) &c has also been proved to be true of \(\Delta u_n\) relatively to

\( \Delta u_{n-1}\) [,] \(\Delta u_{n-2}\) &c. For in the set

\(\begin{array}{llll}

u_0 & & & \\

& \Delta u_0 & & \\

u_1 & & \Delta^2 u_0 & \\

& \Delta u_1 & & &c \\

u_2 & & \Delta^2 u_1 & \\

& \Delta u_2 & & \\

u_3 & & &

\end{array}\)

the first column may be rubbed out and the second column

becomes the first &c. It is obvious that the \(m+1\), \(m+2\), &c

columns are formed from the \(m\) th precisely as the 2nd, 3rd &c

are formed from the first. If then I show that up to

\( n=7\), for instance

[\( u\ldots\) crossed out] \(u_n=u_0+n\Delta u_0+\cdots\cdot\cdot\)

I also show (writing \(\Delta u_n\) for \(u_n\) ) that \(\Delta u_n=\Delta u_0+n\Delta(\Delta u_0)+\cdots\)

Perhaps you had better let the question of discontinuity

rest for the present, and take the result as proved for continuous

functions. You will presently see in a more natural manner

the entrance of discontinuity

The paper which I return is correct

Yours very truly

__ADeMorgan__

69 Gower St

Mond^{y} Ev^{g}

## About this document

All Ada Lovelace manuscript images on the

Clay Mathematics Institute website are

© 2015 The Lovelace Byron Papers,

reproduced by permission of

Pollinger Limited. To re-use them in

any form, please apply to

katyloffman@pollingerltd.com.

The LaTeX transcripts of the letters

were made by Christopher Hollings

(christopher.hollings@maths.ox.ac.uk).

Their re-use in any form requires his

permission, and is subject to the

rights reserved to the owner of

The Lovelace Byron Papers.

Bodleian Library, Oxford, UK

Dep. Lovelace Byron