[108r]

 Ockham

 Sunday.  6^th July

 Dear M^r De Morgan.  It is perhaps unfair of
me to write again with a batch of observations
& enquiries, before you have had time to reply
to the previous one.  But I am so anxious to
get the present matters off my mind, that I
cannot resist dispatching them by this post.

I have two series of observations to send,
one relating to the passage from page 107, (line
8 from the bottom), to the last line of page 108;
the other to certain former passages in pages 99,
100 & 103, concerning which some questions have
suddenly occurred to me quite recently.
I shall begin with pages 107 & 108: I enclose
you my development & explanation of \(\int\frac{x^ndx}{\sqrt{a^2-x^2}}\) up
to \(\int\frac{x^ndx}{\sqrt{a^2-x^2}}=-x^{n-1}\sqrt{a^2-x^2}+(n-1)a^2\int\frac{x^{n-2}dx}{\sqrt{a^2-x^2}}-(n-1)\int\frac{x^ndx}{\sqrt{a^2-x^2}}\) 
from which you will judge if I understand it
so far.  I should tell you that I have not yet
begun page 109.
I will now ask two or three questions : 1^stly : page 107,
[108v] (line 3 from the bottom): ''the diff. co of \(a^2-x^2\) being\(-2xdx)\) 
&c''.  This surely is incorrect; & you will see that
in my development I have written it as I fancy
it should be ''being \(=(-2x)\), &c''
2^ndly : page 108, (lines 8, 9, 10 form the top) : ''By \(\int UdV\) 
''we mean \(\cdots\cdots\cdots\cdots\cdot\) p. 102, where
''the values of \(\Delta V\) in the several terms are
''different, but comminuent.''  I do not see that
this is a case of page 102 rather than of page 100;
in other words, that the increments in this
Integration are ''unequal but comminuent''.
3^rdly : the subtraction in line 15 from the top, of
\( (n-1)x^{n-2}\times dx\) for \(d.(-x^{n-1})\) appears to me quite
inconsistent with the inseparable indivisible
nature of a diff. co.
4^thly : Lines 9, 10 from the bottom, ''We have therefore
''&c \(\cdots\cdots\cdots\cdot\cdot\) that of \(\sqrt{a^2-x^2}x^{n-2}dx\) ''.
Admitted, most fully.  But \(\int\sqrt{a^2-x^2}x^{n-2}dx\) does
not answer exactly to \(\int vdx\) or \(\int\sqrt{v}d2u\), and
therefore it appears to me that this Integration is
not strictly an example of lines 5, 6, 7 (from the bottom)
of page 107.  You will remember that \(-x^{n-1}\) was \(=2V\),
therefore the \(x^{n-2}\) of \((\sqrt{a^2-x^2}x^{n-2})\) is equal to \((-1)\times\frac{2V}{x}\) 
or \(\frac{-1}{x}.2V\) .  So that another factor \(\frac{-1}{x}\) enters into the
[109r] expression which was, as I understand it, to answer
strictly to \(\int vdu\) or \(\int\sqrt{v}d2u\) 
5^thly (line 5 from the bottom) page 108: I think there
is an Erratum.  Surely \(\int\left(\frac{a^2x^{n-2}}{\sqrt{a^2-x^2}}-\frac{x^ndx}{\sqrt{a^2-x^2}}\right)\) 
ought to be \(\int\left(\frac{a^2x^{n-2}dx}{\sqrt{a^2-x^2}}-\frac{x^ndx}{\sqrt{a^2-x^2}}\right)\) 

I don't know if my pencil Sheet enclosed
will be very intelligible, for it is as I wrote
it down at the time quite roughly, & without
any very great amplitude or method.

I now proceed to my series of observations
relating to former pages, beginning with page 102,
(line 10 from the bottom)

 ''\( +\) less than \(nC\frac{\Omega^2}{2}\), or \(Ch\frac{\Omega}{2}\) '';
now in order to ['effect' inserted] the substitution of \(Ch\frac{\Omega}{2}\) for \(nC\frac{\Omega^2}{2}\) 
the latter is resolved into \(C.n\Omega.frac{\Omega}{2}\), & ['for' inserted] \(n\Omega\) is
substituted \(h\) .  But by the hypothesis & conditions,
\( h\) must be less than \(n\Omega\) .  Therefore it does not
necessarily follow that that which is proved less than
\( nC\frac{\Omega^2}{2}\), is also less than \(Ch\frac{\Omega}{2}\) . You see
my objection.
2^ndly .  See Note to page 102 : If the ''completion of the ['first' inserted] Series''
[109v] in this page is unnecessary, surely it is equally
unnecessarily in the first Series of page 100; for the
same observation applies to the latter as to the
former, viz : that the additional term is
comminuent with \(w\) .
3^rdly .  See page 99 (line 8 from the bottom) :

 ''\( \int_a^x\varphi x.dx=(x-a)a+\frac{(x-a)^2}{2}=\frac{x^2-a^2}{2}\) ''
This is another form of \(\int_a^{a+h}xdx=ha+\frac{h^2}{2}\) 8 lines
above, & of the limit of the summation for \(\varphi x=x\) in
the previous page.  And therefore it appears to me
that it ought to be

 \(\int_a^xx.dx=(x-a)a+\frac{(x-a)^2}{2}=\frac{x^2-a^2}{2}\) 
I do not see what business \( \varphi x\) has.

Now at last, I have done troubling you.
I am very anxious on all these points.
With many apologies, believe me

 Yours very truly

 A. A. Lovelace