[20r] My dear Lady Lovelace

With regard to the error in Peacock you will see
that you have omitted a sign.  It is very common to suppose that
if \(\varphi x\) differentiated gives \(\psi x\), then \(\varphi(-x)\) gives \(\psi(-x)\), but
this should be \(\psi(-x)\times\text{diff.co.}(-x)\) or \(\psi(-x)\times-1\) .  Thus

 \(y=\varepsilon^x\)  \(\frac{dy}{dx}=\varepsilon^x\) 

 \(y=\varepsilon^{-x}\)  \(\frac{dy}{dx}=\varepsilon^{-x}\times(-1)=-\varepsilon^{-x}\) 

 As to the note, my copy of Peacock wants a few pages at
the beginning by reason of certain thumbing of my own and others
in 1825.  I remember however that there is a note which I did
not attend to, nor need you.  But if curiosity prompts, pray
sent it to me in writing.

As to \(du=\varphi(x).dx\), you should not have written it
\(du=\varphi(x)\) as you proposed but

 \(\frac{du}{dx}=\varphi x\) 

 The differential coeff^t is the limit of \(\frac{\Delta u}{\Delta x}\), and is a Total
symbol.  Those whose [sic] write

 \(y=x^2 \therefore  dy=2xdx\) make an error

but if \(dy=2xdx+\alpha\) be the truth,
\( \alpha\) diminishes without limit as compared with \(dx\), when
\( dx\) diminishes.  Consequently \(\alpha\) is of no use in finding
[20v] any limit, and those who use differentials, as they
are called, do not differ at the end of their process
from those who make limiting ratios as they go
along.  You can however for the present transform
Peacock's formula \(du=A\,dx\) into \(\frac{du}{dx}=A\) .

There is the erratum you mention in Alg. p. 225

As to p. 226

 \(\frac{1+b}{1-b}=\frac{1+x}{x}\) 

 \({\small (1+b)x=(1-b)(1+x)}\) 

 \({\small x+bx=1+x-b-bx}\) 

 \({\small bx=1-b-bx}\) 

 \({\small 2bx+b=1}\) 

 \({\small (2x+1)b=1~~~~b=}\frac{1}{2x+1}\) 

 Verification \(\frac{1+\frac{1}{2x+1}}{1-\frac{1}{2x+1}}=\frac{2x+1+1}{2x+1-1}=\frac{2x+2}{2x}\) 

 \(=\frac{2(x+1)}{2x}=\frac{x+1}{x}\) 

[21r] p.\,212.  To shew that for instance

\( n\,\frac{n-1}{2}\,\frac{n-2}{3}\,\frac{n-3}{4}\,\frac{n-4}{5}+m\,n\,\frac{n-1}{2}\,\frac{n-2}{3}\,\frac{n-3}{4}+m\,\frac{m-1}{2}\,n\,\frac{n-1}{2}\,\frac{n-2}{3}\) 

\( +m\,\frac{m-1}{2}\,\frac{m-2}{3}\,n\,\frac{n-1}{2}+m\,\frac{m-1}{2}\,\frac{m-2}{3}\,\frac{m-3}{4}\,n+m\,\frac{m-1}{2}\,\frac{m-2}{3}\,\frac{m-3}{4}\,\frac{m-4}{5}\) 

\( =\overline{m+n}\,\frac{m+n-1}{2}\,\frac{\overline{m+n-2}}{3}\,\frac{m+n-3}{4}\,\frac{m+n-4}{5}\)  without actual multiplication

 \(m\) and \(n\) being whole numbers

 1.\ \(m\,\frac{m-1}{2}\,\frac{m-2}{3}\cdots\cdot\cdot\frac{m-(r-1)}{r}\) is the number of ways in which \(r\) 
can be taken out of \(m\) (see chapter on combinations in the
                                               Arithmetic)

If then we denote by \((a,b)\) the number of ways in
which \(a\) can be taken out of \(b\), we have to prove that

\( (5,n)+(1,m)\times (4,n)+(2,m)\times (3,n)+(3,m)\times (2,n)\) 

 \((4,m)\times (1,n)+(5,m)=(5,m+n)\) 

 Suppose ['the' crossed out?] \(m+n\) counters to be divided into two parcels,
one containing \(m\) and the other \(n\) counters
He who would take 5 out of them must either
take

 \(0\) out of the \(m\) and \(5\) out of the \(n\) 

or  \(1\ \, \cdots\cdot\   m\ \cdots\ 4\, \cdots\cdots  n\) 

or  \(2\ \, \cdots\cdot\   m\ \cdots\ 3\, \cdots\cdots  n\) 

     \( 3\ \, \cdots\cdot\   m\ \cdots\ 2\, \cdots\cdots  n\) 

     \(4\ \, \cdots\cdot\   m\ \cdots\ 1\, \cdots\cdots  n\) 

      \(5\, \cdots\cdot\   m\ \cdots\ 0\, \cdots\cdots  n\) 

 Now if to take say the third of these cases, we
can take two of \(m\) in \(a\) ways and \(3\) out of \(n\) in \(b\) ways
[21v] we can do both together in \(a\times b\) ways.  For if
for instance there are \(12\) things in one lot and \(7\) in another,
we can take one out of each lot in \(12\times 7\) ways, since any
one of the twelve may come out with any one of the seven
Hence the number of distinct ['distinct' inserted] ways of bringing \(2\) out of \(m\) and
\( 3\) out of \(n\) together is

 \((2,m)\times (3,n)\) 

 I think you will now be able to make out that the
preceding theorem is true when \(m\) and \(n\) are whole,
whence, by the reasoning in the book it must be
true when they are fractional.

 This reasoning you do not see.  It is an appeal to the
nature of the method by which algebraical operations
are performed.  There is no difference of operation in
the fundamental rules (addition subtⁿ   multⁿ & divⁿ)
whether the symbols be whole nos or fractions.  Hence
if a theorem be true when the letters are any wh.  nos, it
remains true when they are fractions
For example, suppose it proved that for all whole
nos

 \((a+b)\times (a+b)=a\times a+2a\times b+b\times b\) 

 we should then, if we performed the operation \((a+b)\times (a+b)\) 
remembering that \(a\) and \(b\) are whole numbers find \(a\times a+\) \,&c \[\begin{array}{l} a+b\\ \underline{a+b}\\ aa+ab\\ ~~~~+ab+bb\\ \hline\\ aa+2ab+bb \end{array}\]

[22r] Now in no part of this operation are you required
to stop and do ['or omit' inserted] anything because the letters are whole
numbers which you would not do or not omit if they
were fractions.  Consequently, the reservation that the
letters are whole numbers cannot affect the result
which if true with it is true without.  This
principle requires some algebraical practice to see
the necessity of its truth.

The notation of functions is very abstract.  Can
you put your finger upon the part of Chapt. X
at which there is difficulty

The equation

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

 is supposed to be universally true for all values of
\( x\) and \(y\) .  You have hitherto had to deal with
equations in which value was the thing sought:
now it is not value, but form.  Perhaps you
are thinking of the latter when it ought to
be of the former.

With our remembrances to L.' Lovelace I am

 Yours very truly

 ADeMorgan

69 Gower St. No^r 14/40