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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 limit.
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.

About this document

Date of authorship: 

1 Aug 1840

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170