## Folios 16-17: ADM to AAL

**[16r] ** My dear Lady Lovelace

You have taken a proper time to begin with

Incommensurables and if the subject interests you, I should recommend

you to continue. You understand of course that your Diff^{l}

Calculus must be delayed from time to time while you make up

those points of Algebra and Trigonometry which you have

left behind.

D.C. p. 53. As in page 22 refers to the method of proving that

if \(P=2Q\), lim. of \(P=2.\) lim. of \(Q\)

In similar way it may be shown that if

\(\frac{\Delta u}{\Delta x}\cdot\frac{\Delta x}{\Delta u}=1\) lim.\ of \(\frac{\Delta u}{\Delta x}\times\) lim.\ of \(\frac{\Delta x}{\Delta u}=1\)

With reference to your remark remember that

\(\frac{\Delta u}{\Delta x}\cdot\frac{\Delta x}{\Delta u}=1\) and \(\frac{a}{b}\times\frac{b}{a}=1\) __are__ the same proposition

But \(\frac{du}{dx}\times\frac{dx}{du}=1\) and \(\frac{a}{b}\times\frac{b}{a}=1\) are __not__ the same

\( \frac{\Delta u}{\Delta x}\times\frac{\Delta x}{\Delta u}=1\) by common algebra \(\frac{a}{b}\times\frac{b}{a}=\frac{ab}{ab}=1\)

__ __

But we cannot say \(\frac{du}{dx}\times\frac{dx}{du}=\frac{du\ dx}{dx\ du}=1\)

because \(\frac{du}{dx}\) is a mere symbol to denote limit of \(\frac{\Delta u}{\Delta x}\) and \(du\) and \(dx\)

have no separate meaning

**[16v] ** N. & M. p. 17

The erratum exists ['but the misprint is' crossed out] and must

be set right as you propose

['for \(\frac{q_1}{p_1}-\frac{q_2}{p_2}\) ' crossed out]

The lengthiness of the proof arises from the necessity

of adapting a very common algebraical theory to Euclid's

method.

You should try some of the examples of differentiation

in Peacock's book. Remember that there are some

misprints in it. You will not have to go __through__

it to try a little of everything.

When the article __Proportion__ appears in the Penny Cycl.

which it will in a few weeks, I recommend your

attention to it

With remembrances to Lord Lovelace I am

Yours truly

__ADeMorgan__

69 Gower St.

Sunday M^{g} Sept^{r} 27/40

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