## Folios 20-22: ADM to AAL

**[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 subt^{n }mult^{n} & div^{n})

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

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