[146r]

 Ashley-Combe
 Wed^dy. 16^th Nov^r 

 ['1842' or '1847' added by later reader]

[CH: this might date from earlier than 1842, as it contains material that seems to fit better with the earlier letters]

Dear M^r De Morgan.  I am very much obliged
for your long letter.  The Formula in Peacock
comes out quite correct now that I have
written diff. co of \((b-x)^4=4(b-x)^3\times(-1)=-4(b-x)^3\) .
It is odd that notwithstanding the caution you
gave me in Town on this very point, I should
have fallen into the trap.  There is nothing like
one's own blunders after all for instruction.
I do not however understand why example (19)
page 4, has not come out wrong also in
my working out.  I enclose a copy of my
solution, and it appears to me it ought to
be wrong, because I surely should have had
diff.\ co of \((1-x)^4=4(1-x)^3\times(-1)=-4(1-x)^3\), whereas
I have diff.\ co of \((1-x)^4=4(1-x)^3\) .      

On looking over my development again very
carefully, I am inclined to think that my solution
[146v] \(\frac{(1+x)^2}{(1-x)^5}\times(7+x)\), comes out right only because
I have managed to make another blunder of
a sign in the course of the proofs, which has
corrected the first blunder.  I therefore now
write on the other side of the paper, what I
think it should be.        
The note in page 2 I do not imagine to be of
any consequence.  It is on ''rendering the Differentiation
''of complicated Functions sometimes much easier'' by
means of three Theorems from Maclaurin's Fluxions.

Certainly had I thought a little
more upon what I read some weeks ago, before
I wrote my last letter to you, I should not
have sent the question about \(du=\varphi(x)\times dx\) [flourishes at tops of stems of 'd's, here and after].
I ['must have' inserted] forgot exactly what a Differential Co-efficient
means, when I did so.  But how is it then
that in your 1^st Chapter of the Differential Calculus''
there is no mention of the multiplication by \(dx\) ?''
I conclude that the real Differential Co-efficient
is \(\frac{du}{dx}=\varphi(x)\), and that Peacock's solutions are''
not strictly speaking Differential Co-efficients. ''
I think pages 13 to 15 of your Elementary
[147r] Illustrations bear considerably upon the observations
in your letter, do they not?     
Your explanation of Euler's proof of the Binomial
Theorem is perfectly satisfactory to me.  Unluckily
I have not any Book here which contains the
Theory of Combinations.  I wanted to refer to
this when reading page 215, as I have forgotten
it in it's [\textit{sic}] particulars.  However this can very
well wait a short time, & I have only to take
the Formula for Combinations for granted meanwhile.
The necessity of the truth of \((1+x)^n\times(1+x)^m=(1+x)^{n+m}\) 
for all values of \(n\) and \(m\), since it is true
when they are whole numbers, I shall probably
see more clearly at some further time.             

I can explain exactly what my
difficulty is in Chapter X. ''For instance, if we
''know that \(\varphi(xy)=x\times\varphi y\), supposing this always
''true, it is true when \(y=1\), which gives \(\varphi(x)=\) 
''\( =x\times\varphi(1)\) .  But \(\varphi(1)\) is an independent quantity,
''made by writing \(1\) instead of \(y\) in \(\varphi(y)\) .  Let us
''call it \(c\) &c. ''
It is this substitution of \(1\) and of \(c\), and
consequent ascertainment of the form which will
[147v] satisfy the equation, which is all dark to me.
It is ditto in lines 12, 13, & 14 from the top.

I understand quite well I believe from
''We have seen that if \(\varphi x=c^x\) &c'', all through
the next page.
That I do not comprehend at all the means of
deducing from a Functional Equation the form
which will satisfy it, is I think clear from
my being quite unable to solve the example
at the end of the Chapter ''Shew that the equation
''\( \varphi(x+y)+\varphi(x-y)=2\varphi x\times\varphi y\) is satisfied
''by \(\varphi x=\frac{1}{2}(a^x+a^{-x})\) ''.  I have tried
several times, substituting first \(1\) for \(x\), then
\( 1\) for \(y\) .  but I can make nothing whatever
of it, and I think it is evident there is
something that has preceded, which I have
not understood. The 2^nd example given for
practice ''Shew that \(\varphi(x+y)=\varphi x+\varphi y\) can
''have no other solution than \(\varphi x=ax\) '', I
have not attempted.             
I have a question to ask upon page 229.

''By extracting a sufficiently high root of \(z\), we
[148r] ''can bring \(z^m\) as near to \(1\) as we please, or
''make \(z^m-1\) as small as we please; that is
''(page 187) \(z^m-1\) may be made as nearly equal
''to the sum of the whole series as we please''.    

I cannot find what it is that is referred
to in page 187; and Secondly, it appears to me
somewhat of a contradiction that a quantity
\( z^m-1\) which can certainly be made as small
as we please by the diminution of \(m\), should
become as near as we please to a fixed
limit or sum (the \(\log z\) I conclude is the
sum of the series, referred to), since by continued
diminution the quantity \(z^m-1\) may become a
great deal less than the sum of the Series, &
keep receding from it.
To return to Chapter X, there is one other
thing in it that I do not understand.  Page
205, lines 5, 6, 7 from the bottom.  It seems
to me fallacious to substitute first one value
\( 0\), for a letter; & then another value, let \(y=-x\),
[148v] in the same equation & in a manner at the
same time.  How can the two suppositions
consist together at all.           

I go on well with the Trigonometry, &
have nearly finished the Number & Magnitude.
I think there is another Erratum in page
34 of the Trigonometry, line 13 from the bottom

 \(=\frac{OM}{ON}\cdot\frac{ON}{OP}-\frac{NR}{NP}\cdot\frac{NP}{NO}\) &c
should be \(-\frac{NR}{NP}\cdot\frac{NP}{OP}\) 

 I am really ashamed to send you such
troublesome letters.                   

Believe me

 Yours most truly

 A. A. Lovelace