## Folios 3-4: ADM to AAL

** [3r]** My dear Lady Lovelace

The solution of the \(t^3+t^2+t\) problem is correct

and I have no doubt from it that you fully understand the problem

of the stone.

The difficulty you meet with in the variable coefficient would

be fatal to the process if the coefficient increased __without limit__

as the value of \(k\) diminishes __without limi__t.

But in this case \(t\) which you call a variable, is really

a given constant, for it is __not__ \(t\) which varies, but the __time__

which is first \(t\) , then \(t+k\) , then \(t+2k\) , &c.

There is a little incorrectness in the phraseology of variable

quantities. A quantity varies, it is first \(x\) , and then \(x+a\) ;

here it is usual to say that \(x\) varies, whereas it is not

\( x\) , but the magnitude which \(x\) represents, which changes

and is no longer represented by \(x\) , but by \(x+a\)

Thus if we pass in thought from 10 seconds to 11 eleven *sic*

seconds, we say let 10 vary, and become 11. Now 10 is a

fixed symbol, and so is 11; it is we ourselves who vary our

supposition, and pass from one to the other.

But even if the coefficients of \(k\) __were __variable, it

would not vitiate the result, as may be thus shown

Let \(a+vk\) be an expression in which \(a\) is a constant

\( k\) diminishes without limit, and \(v\) at the same time

varies, say increases. Let \(v\) always remain finite, that

is, let it not increase without limit while \(k\) diminishes**[3v**] Suppose for instance, that it never exceeds a

certain number (no matter how great) say a

million. Then \(vk\) never exceeds \(1000,000.k\) . Now if

\( k\) may be made as small as we please, so may \(1000,000k\) ,

and still more \(vk\) , which is less, or at least not greater.

That is \(vk\) diminishes without limit, and

\( a+vk\) has the limit \(a\) .

But if \(v\) increased without limit as \(k\)

diminished, the case __might__ be altered (not necessarily__would__. For example

1. Let \(v=\) [\( \frac{1}{k}\) crossed out] \(\frac{1+k}{k}\)

\(a+vk=a+1+k\) and the limit is \(a+1\)

2. Let \(v=\frac{1+k}{k^2}\)

\(a+vk=a+\frac{1+k}{k}\,,\) and increases without limit

3. Let \(v=\frac{1+k}{\sqrt k}\)

\(a+vk=a+(1+k)\sqrt k\,,\) and the limit is \(a\) , as at first.

And \(\frac{1+k}{k}\) , \(\frac{1+k}{k^2}\) , \(\frac{1+k}{\sqrt k}\) all increase without

limit as \(k\) diminishes without limit.

**[4r]** Your correction of the press is right. On the cover

of no 12 you will see a list of errata.

[diagram here in original] When two lines at right angles

are made standards of position, and

when the position of a point is

determined by its perpendicular

distance from the lines, as \(PA\) , \(PB\) ,

then \(PA\) and \(PB\) are called __co-ordinates__ of \(P\) . One of

them, as \(BP\) is usually found by its equal \(OA\) , and called

the __abscissa__ of \(P\) , the other, \(AP\) , is called the ordinate.

I have written something on the paper which I return.

My wife desires kind remembrances, and with our

united remembrance to Lord Lovelace

I remain

Yours very truly

ADeMorgan

3 Grotes' Place Blackheath

August 1, 1840

Should you decide on supporting our old printing

Society, I shall be very happy to have Lord Lovelace's name

inserted. We are beginning with an Old Saxon treatise on

astronomy with a translation.

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