Clay Mathematics Institute

Dedicated to increasing and disseminating mathematical knowledge

Click banner for images


Folios 106-107: AAL to ADM


 Ockham Park




 Dear Mr De Morgan.  You are perhaps surprised
that I have not sooner troubled you again.
And you may think it a very bad reason to
give, that I have done nothing.  We returned
here on Tuesday, & now I am working away
famously, & hope I have before me 7 or 8 months
of ditto.  You left me at page 106.  I remember
your enquiry if I were sure that I understood
\( \int_b^{b+k}fx\times\frac{dx}{dt}\) as developped [\textit{sic}] in pages 102, 103.  I
answered confidently, that I did.  I now enclose
you my own development of this Integration, that
we may be quite certain of my comprehension of
[something crossed out]} it.  On the other page of my
sheet, is the application of it to \(\int udv=uv-\int vdu\) 
(page 105); & to \(\int_a^x\frac{1}{v}\frac{dv}{dx}dx\) (page 107).

I have now two questions to propose.
I differ from you in my development of \(\int\frac{1}{1-x}dx\) 

 (see page 107)
[106v] I cannot see why the Constant \(C\) is omitted
in this more than in \(\int\frac{1}{1+x}dx\) .
I subjoin my development: Let \(v=1-x\) 
\( \int\frac{1}{1-x}dx=\int\frac{1}{1-x}\times-(-1)dx\) (which is only
another way of writing \(\int\frac{1}{1-x}.1.dx)\) 
And as \(\frac{dv}{dx}\) or \(\frac{d(1-x)}{dx}=-1\), we may in the
above substitute \(\int\frac{1}{1-x}dx=\int\frac{1}{1-x}\times-\left(\frac{d(1-x)}{dx}\right)dx\) 

 Or \(\int\frac{1}{v}dx=\int\frac{1}{v}\times-\frac{dv}{dx}dx\) 

     \(=\int\frac{1}{v}\frac{dv}{dx}.(-1)dx\) which by
\( \int budx=b\int udx\) (see page 105) is \(=(-1)\int\frac{1}{v}\frac{dv}{dx}dx\) 

 or \(=-\int\frac{1}{v}\frac{dv}{dx}dx\) 
Now since by line 4, \(\int\frac{1}{v}\frac{dv}{dx}dx=\int\frac{1}{v}dv=\) 

 \(=\log v+C\), it follows that
(--- this same expression) must \(=-(\log v+C)\) 

 \(=-(\log (1-x)+C)=-\log (1-x)-C\) 

[107r] Now how do you get rid of \((-C)\) ?

My second question is unconnected with any
of your books. But I think I may venture to
trouble you with it. In the two equations,

 \(V=gT\)   (1)

 \(S=\frac{1}{2}g.T^2\)   (2)
which you will at once recognise, I want to
know how (2) is derived from (1).
Will you refer to Mechanics (in the Useful
Knowledge Library), page 10, Note, which is
as follows, ''Let \(S\) be the space described by the
''falling body.  \(V=\frac{dS}{dT}=gT\) .  Hence \(dS=gT\,dT\),
''which being integrated gives \(S=\frac{1}{2}g.T^2\) .''
Now can I ['as yet' inserted] understand this application of
Differentiation & Integration?
I conclude that \(\frac{dS}{dT}\) here means

Diff. co of \(S\) with respect to \(T\), \(S\) being (by
Definition & Hypothesis) a function of \(T\), & of \(V\) 
I know that \(V=gT\) 

And that \(V=\frac{S}{T}\)  But I neither see how
\( V=\frac{dS}{dT}\), nor how the subsequent Integration applies.
[107v] The object, I need not say, is the solution of
\( S\) .
I mean to work very hard at my Chapter on
Integration &c, now. And I hope this
summer & autumn will see me progressing
at no small rate.
How is the Baby? And does Mr De
Morgan enjoy Highgate?  I ['am' inserted] enjoying the country
not a little, I assure you.

 Yours most truly

 A. A. Lovelace

About this document

Date of authorship: 

4 Jul 1841

Holding institution: 

Bodleian Library, Oxford, UK


Dep. Lovelace Byron

Box 170