## Folios 24-26: ADM to AAL

**[24r] ** My dear Lady Lovelace

I send back your worked question.

The second is right the first wrong in two places

I should recommend you to get out of the

habit of writing \(d\) thus ['\( d\) ' with flourish at top of stem] or thus ['\( d\) ' with flourish and double stem]. If you have

much to do with the Diff.\ Calculus, it will make a good

deal of difference in time. The best way is to form all the

letters as like those in the book as you can.

Pages 13--15 of Elementary Illustrations bear closely

on the distinction of \(\frac{du}{dx}=a\) and \(du=adx\)

It is not because \(\overline{1+x}^m\times\overline{1+x}^n=\overline{1+x}^{m+n}\) when \(m\) &

\( n\) are whole numbers that the same is true for fractions

but because a certain other property which is therefore

true makes it necessary in the case of fractions.

For instance, the logic is as follows

\(A\) is true when \(m\) is a whole number

Whenever \(A\) is true, \(B\) is true

\(B\) is of that nature, that if true when \(m\) is a whole

number, it is also true when \(m\) is a fraction.

When \(B\) is true \(C\) is true

\(\therefore C\) is true if \(A\) be true when \(m\) is a whole number

Thus, when \(m\) and \(n\) are whole numbers,

\(\overline{1+x}^m=1+mx+m\,\frac{m-1}{2}\,x^2+\) &c \(\overline{1+x}\vert^n=1+nx+\) &c

**[24v] ** But \( \overline{1+x}^m\times\overline{1+x}^n=\overline{1+x}^{m+n}\) always

Therefore when \(m\) and \(n\) are whole numbers

\( (1+mx+\text{&c})\times (1+nx+\text{&c})=1+\overline{m+n}x+\text{&c}\)

but this last property (without any reference to its

mode of derivation) is true of \(m\) and \(n\) fractional

or negative if true of whole numbers.

Hence, if \(1+mx+\text{&c}\) be called \(\varphi m\)

\(\varphi m\times\varphi n=\varphi(m+n)\)

But this (again by an independent process)

is shown to be never universally true unless

\(\varphi m=c^m\), \(c\) being independent of \(m\)

whence \(c^m=1+mx+m\,\frac{m-1}{2}\,x^2+\cdots\cdot\)

But since \(c\) is independent of \(m\), what it is when

\( m=1\), it is always. Therefore

\(c^1=1+1.x+1.\frac{1-1}{2}\,x^2+\cdots\) or \(c=1+x\)

__ __

Let \(\varphi(xy)=x\times\varphi y\) be always true

required \(\varphi x\)

Make \(y=1\), then \(\varphi(x)=x\times\varphi(1)\)

\( \varphi(1)\) is not yet determined; let it be \(c\) . The equation

is either true for all values of \(c\), or for some (not all)

\(\varphi x=cx\)

If any particular value were needed for \(c\), it would

be found by making \(c=1\) . Do this and we have

\(\varphi(1)=c\), or \(c=c\), which is true for

all values of \(c\) .

__ __

Suppose

\(\varphi(x)=\varphi(1).x+2\varphi(1)-\overline{\varphi(1)}^2\)

to be true for all values of \(x\) . It is then true when

\( x=1\), giving

\(\varphi(1)=\varphi(1)+2\varphi(1)-\overline{\varphi(1)}\vert^2\)

or \( \overline{\varphi(1)}\vert^2-2\varphi(1)=0\)

or \(\varphi(1)=2\) only

\(\varphi(x)=2x+2.2-2^2=2x\)

\(\varphi(1)\, =2\)

which agree with each other.

__ __

You must remember than [\textit{sic}] when a form is __universally__

true, it is true in all particular cases. Then

in \(2x=x+x\), I have a perfect right to say

this is true when \(x=1\) and \(\therefore 2=1+1\), and true

__when__ \(x=20\), or \(40=20+20\) : though I do ['do' crossed out] not

thereby say that \(x\) can be \(10\) and \(1\) both at once.

__ __

\( \frac{1}{m}\log z=\left(z^{\frac{1}{m}}-1\right)-\frac{1}{2}\left(z^{\frac{1}{m}}-1\right)^2+\cdots\cdot\)

when \(m\) increases \(z^{\frac{1}{m}}-1\) diminishes

It is not \(\log z\) which \(z^{\frac{1}{m}}-1\) approaches to, but

\( \frac{1}{m}\,\log z\), which also diminishes as \(m\) increases

I see that in page 219 it is thus

\(\log z=\frac{1}{m}\left\{\overline{z^m-1}-\frac{1}{2}\,\overline{z^m-1}^n+\cdots\right\}\)

and \(m\) __diminishes__ without limit. Now as \(m\) diminishes

**[25v] ** \(\frac{1}{m}\) increases, and the assertion is that as

\(m\) diminishes, and therefore \(z^m-1\), the

product \(\frac{1}{m}\left(z^m-1\right)\)

has one factor continually increasing & the other

diminishing, so that their product approaches

without limit to \(\log z\) .

In page 187, it is shown that if \(x\)

be made sufficiently small, any term of

\( a+bx+cx^2+\cdots\) may be made to contain all the

rest as often as we please, that is may be made

as great as we please compared with the sum

of all the rest. Consequently \(m\) being small

\(z^m-1\) may be made as great as we please

compared with the sum of all the terms of the

series which follow, even if all were positive,

still more when they counterbalance each other

by difference of sign.

__ __ p.205

If \(\varphi(x+y)=\varphi x+\varphi y\) be __always__ true (hypothesis)

It is true when \(x=0\)

It is also true when \(y=-x\)

This equation being always true, is the representation

of a collection of an infinite number of truths

I do not say that these truths coexist

Put it thus. Let \(\varphi\) be such a function

that, if \(a\), \(b\), \(c\), \(d\), &c be any quantities whatever

**[26r]** \(\varphi(a+b)=\varphi a\times\varphi b\)

\(\varphi(b+c)=\varphi b\times\varphi c\)

\(\varphi(d+e)=\varphi d\times\varphi e\) &c &c

That is let \(\varphi(x+y)=\varphi(x)\times\varphi(y)\)

for all values of \(x\) and \(y\)

1.\ Let \(a=0\), then \(\varphi b=\varphi(0)\times\varphi(b)\)

or \(\varphi(0)=1\)

Let \(b=-c\), then \(\varphi(0)=\varphi(b)\times\varphi(c)\)

or \(1=\varphi(b)\times\varphi(-b)\)

and so on.

There is a want of distinction between

an equation made true by choice of values

and one which is true of itself, independently of

all values

\(x=3-x\) \(x=\frac{3}{2}\), and then only

\(x=3x+a-(2x+a)\) is true for __all__ values of \(x\),

though it cannot have more than one at a time.

There is the erratum in the Trigonometry,

as you say

Yours very truly

__ADeMorgan__

69 Gower St.

Friday Ev^{g}

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