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42v|43r

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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 Magn  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

 Mondy Evg

 

 

 

About this document

Date of authorship: 

22 Feb 1841

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170