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Folios 74-76: AAL to ADM

[74r]

 Ockham

 Monday. 4th Jany   ['1841' added by later reader]

 Dear Mr De Morgan.  We have
had company ever since I
last wrote to you, so I have
been at a Stand-still, &
only yesterday was able to
read over with attention your
replies.  I am reluctant to
trouble you again with remarks
on the Series \(1+\frac{x^2}{2}+\frac{x^4}{2.3.4}+\) &c,
for it seems as if I was
determined to plague you
about it.  However I feel I
must do so.  Your added
[74v] remarks of last time, about
\( B\) &c, are quite clear in
themselves, but I felt at
once that they did not
meet my difficulty which
was that as long as \(\frac{5}{4}\) 
(which is greater than \(1\) ) is to
be added to \(\frac{1}{2}\sqrt{1000,000+\frac{1}{4}}\),
it matters not whether for
\( \sqrt{1000,000+\frac{1}{4}}\) we substitute
the whole number next above
it or ''the intermediate fraction''
alluded to by you in the line
I have marked [mark a bit like her '\( \sqrt{}\) '], but we
never can bring out \(n=\) to
anything less than \(502\), whence
\( n+1\) the required term must
[75r] be \(503\) .  This, after reading
over & over, remains in my
mind a most obstinate fact,
and I believe I have found
out the real source of the
discrepancy between the result
at the bottom of the first 
page & the top of the Second
one.  I am presumptuous
enough to think there is
certainly an error in your
writing out, in the line I
have marked X, & it is one
which is very likely to have
occurred in writing ['it' crossed out] in
a hurry.      
The \((n+1)^{\rm th}\) term divided by
the \(n\)th is I believe not
[75v] \(\frac{Z}{(2n-2)(2n-3)}\), but \(\frac{Z}{2n(2n-1)}\),
and I have re-written & now
enclose the rest of the demonstration
(exactly like yours) with this
correction.  The result comes
out as I expected, owing to
\( \frac{1}{4}\) taking the place of \(\frac{5}{4}\),
& everything appears to me
consistent. the \(n\)th term
divided by the \((n-1)^\text{th}\) term
would be, (as you have
written) \(\frac{Z}{(2n-2)(2n-3)}\), & this
correction would do instead of
the others, & be perhaps ['a' inserted] more
simple mode of making it,
as your demonstration would
[76r] then remain correct, the \(n\)th 
term being in that case the
required unknown one instead
of the \((n+1)^\text{th}\) .     
I am afraid all this is a
little complicated to explain
in letters, & perhaps I have
still not succeeded very
perfectly in doing so; but
I feel it now all very
clear in my own mind,
& am only anxious to receive
confirmation as to my being
right, both as satisfactory
to me in the present instance,
& as tending to give me
[76v] confidence in future in
my own conclusions, or, (if
I am in this case puzzle-
-headed), a due diffidence
of them.                  

 I therefore beg your indulgence
for being so teasing.               

 Believe me

 Yours most truly

 A. A. Lovelace

About this document

Date of authorship: 

4 Jan 1841

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170