[24r] My dear Lady Lovelace

I send back your worked question.
The second is right the first wrong in two places

I should recommend you to get out of the
habit of writing \(d\) thus ['\( d\) ' with flourish at top of stem] or thus ['\( d\) ' with flourish and double stem].  If you have
much to do with the Diff.\ Calculus, it will make a good
deal of difference in time.  The best way is to form all the
letters as like those in the book as you can.

Pages 13--15 of Elementary Illustrations bear closely
on the distinction of \(\frac{du}{dx}=a\) and \(du=adx\) 

It is not because \(\overline{1+x}^m\times\overline{1+x}^n=\overline{1+x}^{m+n}\) when \(m\) &
\( n\) are whole numbers that the same is true for fractions
but because a certain other property which is therefore
true makes it necessary in the case of fractions.

For instance, the logic is as follows

 \(A\) is true when \(m\) is a whole number

 Whenever \(A\) is true, \(B\) is true

 \(B\) is of that nature, that if true when \(m\) is a whole

 number, it is also true when \(m\) is a fraction.

 When \(B\) is true \(C\) is true

 \(\therefore C\) is true if \(A\) be true when \(m\) is a whole number

 Thus, when \(m\) and \(n\) are whole numbers,

 \(\overline{1+x}^m=1+mx+m\,\frac{m-1}{2}\,x^2+\) &c     \(\overline{1+x}\vert^n=1+nx+\) &c

[24v] But \( \overline{1+x}^m\times\overline{1+x}^n=\overline{1+x}^{m+n}\) always
Therefore when \(m\) and \(n\) are whole numbers
\( (1+mx+\text{&c})\times (1+nx+\text{&c})=1+\overline{m+n}x+\text{&c}\) 
but this last property (without any reference to its
mode of derivation) is true of \(m\) and \(n\) fractional
or negative if true of whole numbers.

Hence, if \(1+mx+\text{&c}\) be called \(\varphi m\) 

 \(\varphi m\times\varphi n=\varphi(m+n)\) 

 But this (again by an independent process)
is shown to be never universally true unless

 \(\varphi m=c^m\), \(c\) being independent of \(m\) 

 whence \(c^m=1+mx+m\,\frac{m-1}{2}\,x^2+\cdots\cdot\) 
But since \(c\) is independent of \(m\), what it is when
\( m=1\), it is always.  Therefore

 \(c^1=1+1.x+1.\frac{1-1}{2}\,x^2+\cdots\)  or \(c=1+x\) 

                                                             

Let \(\varphi(xy)=x\times\varphi y\)  be always true
required \(\varphi x\) 

 Make \(y=1\), then \(\varphi(x)=x\times\varphi(1)\) 

\( \varphi(1)\) is not yet determined; let it be \(c\) .  The equation
is either true for all values of \(c\), or for some (not all)

 \(\varphi x=cx\) 

 If any particular value were needed for \(c\), it would
be found by making \(c=1\) .  Do this and we have

 \(\varphi(1)=c\), or \(c=c\), which is true for

 all values of \(c\) .

                                                          

Suppose
\(\varphi(x)=\varphi(1).x+2\varphi(1)-\overline{\varphi(1)}^2\) 
to be true for all values of \(x\) .  It is then true when

\( x=1\), giving

 \(\varphi(1)=\varphi(1)+2\varphi(1)-\overline{\varphi(1)}\vert^2\) 

 or \( \overline{\varphi(1)}\vert^2-2\varphi(1)=0\) 

 or \(\varphi(1)=2\) only

 \(\varphi(x)=2x+2.2-2^2=2x\) 

 \(\varphi(1)\,  =2\) 
which agree with each other.

                                                             

You must remember than [\textit{sic}] when a form is universally
true, it is true in all particular cases.  Then
in \(2x=x+x\), I have a perfect right to say
this is true when \(x=1\) and \(\therefore 2=1+1\), and true
when \(x=20\), or \(40=20+20\) : though I do ['do' crossed out] not
thereby say that \(x\) can be \(10\) and \(1\) both at once.

                                                             

\( \frac{1}{m}\log z=\left(z^{\frac{1}{m}}-1\right)-\frac{1}{2}\left(z^{\frac{1}{m}}-1\right)^2+\cdots\cdot\) 
when \(m\) increases \(z^{\frac{1}{m}}-1\) diminishes
It is not \(\log z\) which \(z^{\frac{1}{m}}-1\) approaches to, but
\( \frac{1}{m}\,\log z\), which also diminishes as \(m\) increases
I see that in page 219 it is thus

 \(\log z=\frac{1}{m}\left\{\overline{z^m-1}-\frac{1}{2}\,\overline{z^m-1}^n+\cdots\right\}\) 
and \(m\) diminishes without limit.  Now as \(m\) diminishes
[25v] \(\frac{1}{m}\) increases, and the assertion is that as

 \(m\) diminishes, and therefore \(z^m-1\), the

 product \(\frac{1}{m}\left(z^m-1\right)\) 
has one factor continually increasing & the other
diminishing, so that their product approaches
without limit to \(\log z\) .

In page 187, it is shown that if \(x\) 
be made sufficiently small, any term of
\( a+bx+cx^2+\cdots\) may be made to contain all the
rest as often as we please, that is may be made
as great as we please compared with the sum
of all the rest.  Consequently \(m\) being small

 \(z^m-1\) may be made as great as we please
compared with the sum of all the terms of the
series which follow, even if all were positive,
still more when they counterbalance each other
by difference of sign.

                                                               p.205

 

 If \(\varphi(x+y)=\varphi x+\varphi y\)   be always true (hypothesis)

 It is true when \(x=0\) 

 It is also true when \(y=-x\) 

 This equation being always true, is the representation
of a collection of an infinite number of truths

I do not say that these truths coexist

Put it thus.  Let \(\varphi\) be such a function
that, if \(a\), \(b\), \(c\), \(d\), &c be any quantities whatever

[26r] \(\varphi(a+b)=\varphi a\times\varphi b\) 

 \(\varphi(b+c)=\varphi b\times\varphi c\) 

 \(\varphi(d+e)=\varphi d\times\varphi e\)  &c &c

 That is let \(\varphi(x+y)=\varphi(x)\times\varphi(y)\) 
for all values of \(x\) and \(y\) 
1.\ Let \(a=0\), then \(\varphi b=\varphi(0)\times\varphi(b)\) 

 or \(\varphi(0)=1\) 

Let \(b=-c\), then \(\varphi(0)=\varphi(b)\times\varphi(c)\) 

 or \(1=\varphi(b)\times\varphi(-b)\) 
and so on.

 There is a want of distinction between
an equation made true by choice of values
and one which is true of itself, independently of
all values

 \(x=3-x\)  \(x=\frac{3}{2}\), and then only

 \(x=3x+a-(2x+a)\) is true for all values of \(x\),
though it cannot have more than one at a time.

There is the erratum in the Trigonometry,
as you say

 Yours very truly
     ADeMorgan
 

69 Gower St.
        Friday Ev^g