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Folios 115-118: AAL to ADM

[115r]

 Ockham Park

 Sunday.  15th Augst 

 ['1841' added by later reader]

 Dear Mr De Morgan.  You must be beginning
to think me lost.  I have been however hard at
work, with the exception of 10 days complete
interruption from company. I have now many
thing to enquire. First of all; can I spend
an evening with Mr De Morgan & yourself on
Tuesday the 24th?  On that day we go to Town to
remain till Friday, when we move down to Ashley
for 2 months at least. I would endeavour to
be early in Gower St; before eight or not later
than eight.  And I feel as if I should have
many mathematical things to discuss.
Now to my business :
1stly: I send you a paper marked 1, containing my
development of two Integrals in page 116,
\( \int\frac{dx}{\sqrt{2ax-x^2}}=\sin^{-1}\left(\frac{x-a}{a}\right)\) 
And \(\int\frac{dx}{\sqrt{2ax+x^2}}=\log(x+a+\sqrt{2ax+x^2})+\log 2\) 
The former one I think is plain enough, & I and
the book are quite agreed upon it.  Not so with
[115v] the latter one, & I begin to suspect the book.
I cannot make it anything but

 \(\int\frac{dx}{\sqrt{2ax+x^2}}=\log(x+a+\sqrt{2ax+x^2})\) 

or else\ \   \(=\log(\frac{x}{2}+a+\frac{\sqrt{2ax+x^2}}{2})+\log 2\) 
I have tried various methods; but the
only one which I find hold good [\textit{sic}] at all, is
that applied in page 115 to \(\int\frac{dx}{\sqrt{a^2+x^2}}\), & which
seems clearly to bring out
my result above.  By the bye I have a remark
to make on the Integration of \(\int\frac{dx}{\sqrt{a^2+x^2}}\) as
developped [\textit{sic}] in page 115.
Line 10 (from the bottom), you have \(xdx=ydy\) :
This is obvious, & similarly I deduce in my paper
No 1, \((2a+x)dx=ydy\) . But I see no use in
what follows, ''and \(ydx+xdx=ydx+ydy\) ''.
It is equally obvious with the former equation,
but seems to me to have no purpose in bringing
out the results, which I deduce as follows :

Since \(xdx=ydy\), we have \(\frac{dx}{y}=\frac{dy}{x}\) 
Therefore by the Theorem of page 48, or at least ['by' inserted] a
Corollary of it, we have \(\frac{dx+dy}{x+y}=\frac{dx}{y}\), whence &c, &c.
And this is the method also which I have used
[116r] in developping [sic] \(\int\frac{dx}{\sqrt{2ax+x^2}}\) .  
2ndly : Page 113, lines 16 ['&c 17' inserted] from the bottom, you say ''The
''first form becomes impossible when \(x\) is greater than
''\( \sqrt{c}\), for in that case the Integral becomes the
''Logarithm of a Negative Quantity''.  Now there are
surely certain cases in which negative quantities
may be powers, & therefore may have Logarithms.
All the odd whole numbers may surely be the
Logarithms of Negative Quantities.
\( (-a)\times(-a)=a^2\)   But \((-a)\times(-a)\times(-a)=-a^3\) 
or \((+a)\times(+a)=a^2\)  \((+a)\times(+a)\times(+a)=a^3\) 
\( 3\) is here surely the Logarithm of a Negative Quantity.
Similarly a negative quantity multiplied into itself
any odd number of times will give a negative result.
3rdly : In the Paper marked 3, which I return
again ['for reference' inserted]; I perfectly understand the proof by means of the
Logarithms (added by you), why \(\frac{dy}{dx}\) can only \(=\frac{y}{x}\) 
when \(y\) is either \(=x\), or \(=ax\) (\) a\) being Constant)
Your proof is perfect, but still I do not see that
mine was not sufficient, tho' derived from much
more general grounds.
My argument was as follows : Given us \(\frac{dy}{dx}=\frac{y}{x}\),
what conditions must be fulfilled in order
to make this equation possible?  Firstly : I see that
[116v] since \(\frac{dy}{dx}\) means a Differential Co-efficient, which
from it's [sic] nature (being a Limit) is a constant &
fixed thing, \(\frac{y}{x}\) must also be a constant & fixed
quantity.  That is \(y\) must have to \(x\) a constant
Ratio which we may call \(a\) .
This seems to me perfectly valid.  And surely a
Differential Co-efficient is as fixed & invariable
in it's [\textit{sic}] nature as anything under the sun can be.
To be sure you may say that there is a different
Differential Co-efficient for every different initial
value of \(x\) taken to start from, thus :

 \(\frac{d(x^2)}{dx}=2x\)  if \(x=a\), \(\frac{d(x^2)}{dx}=2a\) 

  if \(x=b\), \(\frac{d(x^2)}{dx}=2b\) 
And this is perhaps what invalidates my argument
above.
4thly : In the two papers folded together & marked 2,
which I also again return for reference, I perfectly see
that tho' mathematically correct.  I was completely
wrong in my application.  But my proofs do apply
to any two different & independent velocities, whatever
of two different bodies, or of the same body moving
at two different uniform ratio [\textit{sic}] at different epochs.

Thus my paper (marked upon it 1st Paper proves
[117r] the following : that the Spaces moved over at two
different times, in virtue of the Velocity acquired at
the end of each of those times, (the impelling cause
being supposed to cease at the end respectively of each
time fixed on), would be to each other as the
squares of the times fixed on. But I perfectly
see that this is quite a different & independent
consideration from that of the Space actually moved
over by a body impelled by an accelerating force,
& how wholly inapplicable my ['former' inserted] view of it was.

I have been especially studying this
subject of [something crossed out] Accelerating Force, & believe that I now understand
it very completely.  I found I could not rest upon
it at all, until I made the whole of the subject out
entirely to my satisfaction : I enclose you (marked
4) the first of a Series of papers I am making out
in the different parts ['of' inserted] it.  This one is the more
general development of the particular case of Gravitation
in pages 27, 28; & my more especial object in it
has been the identification of the results arrived at
in this real application, with the Mathematical
Differential Co-efficient.
I have worked most earnestly & incessantly at the
Application of the Differential & Integral Calculus to the
[117v] subject of Accelerating Force, & Accelerated Motion,
during the last 2 or 3 weeks.  It has interested me
beyond everything.  After making out (according to
my own notions) the two papers on \(v=\frac{ds}{dt}\), and
\( s=\int vdt\), (the first of which I now send, & the
Second you will have in a day or two), I attacked
your Chapter 8, pages 144, 145, worked out all the
Formulae there; & had excessive trouble with my
third paper on \(t=\int\frac{ds}{v}\), (now successfully terminated);
and I am now on \(f=\frac{dv}{dt}\), page 146.
You will perhaps not approve my having thus run
a little riot, & anticipated.  But I think it has
done me great good.  And I am anxious to know
if I may read the rest of this Chapter 8, before
reading Chapter 7 on Trigonometrical Analysis; & if
I am likely to understand it all without having
read Chapter 7.
I shall probably write again tomorrow; or if not,
certainly I shall on Tuesdy .
We are very anxious to know if there is no time
between the 1st Novr & the middle of Feby, when you &
Mr De Morgan (& family) would come & stay here
for as long as you can & would like.  We should
be delighted if you would remain 2 or 3 weeks.
[118r] And if this should be impossible for you, perhaps still
you would bring Mr De Morgan & the children here,
& remain a few days; having them to stay longer.

We both of us assure you that it would be
no inconvenience whatever to us; but rather contrary
the greatest pleasure.  And I am certain it
would do Mr De Morgan good to be here for
a time. Pray consider my proposal;
at any rate for her & the children, if your own
avocations should make it impossible for you even.
Believe me

 Yours most truly

 A. A. Lovelace

 

About this document

Date of authorship: 

15 Aug 1841

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170