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Folio 174: AAL to ADM

[174r] [(mostly) in AAL's hand]

 Theorem.  Page 199.
If \(N\) be a function of \(x\) and \(y\), giving \(\frac{dN}{dx}=p+q\frac{dy}{dx}\) 
then the equation \(\frac{du}{dx.dy}=V.\frac{dN}{dx.dy}\) is incongruous &
self-contradictory, except upon the assumption
that \(u\) is, as to \(x\) and \(y\), a function of \(N\);
or contains \(x\) and \(y\) only thro' \(N\) .

Let \(N=\psi(x,y)\) give \(y=\chi(N,x)\), and
suppose, if possible, that the substitution of
this value of \(y\) in \(u\) gives \(u=\beta(N,x)\), \(x\) 
not disappearing with \(y\) .  Then \(x\) and \(y\) 
varying

 \(\frac{du}{dx.dy}=\frac{d\beta}{dN}.\frac{dN}{dx}+\frac{d\beta}{dN}.\frac{dN}{dy}+\frac{d\beta}{dx}\)  [in above line, \(\frac{du}{dx.dy}\) is crossed through in pencil, and '\( 1\) ' written above; </br>'\( =\frac{du}{dx}+\frac{du}{dy}\) ' added in pencil at end of line --- in ADM's hand?]

 \(=\frac{d\beta}{dN}\cdot\left(\frac{dN}{dx}+\frac{dN}{dy}\right)+\frac{d\beta}{dx}=\frac{d\beta}{dN}\cdot\frac{dN}{dx.dy}+\frac{d\beta}{dx}=\) 

 \(=V.\frac{dN}{dx.dy}\)  , which equation being
universal\, is true on the supposition that \(x\) 
does not vary\, or that \(\frac{d\beta}{dx}=0\) .  This gives \(\frac{d\beta}{dN}=V\);

 or \(\frac{du}{dx.dy}=V\frac{dN}{dx.dy}+\frac{d\beta}{dx}=V\frac{dN}{dx.dy}\) 
because \(\frac{d\beta}{dN}\) and \(V\) being independent of the variations
&c, &c.  Hence \(\frac{d\beta}{dx}=0\) always; or \(\beta\) does not
contain \(x\) directly, &c.

I think the above is correct.  I cannot see
[174v] the use (page 200) of introducing \(t\) in
the proof there given .  Is it possible that
I have committed an error in my original
understanding of the ennunciation [sic] of the Theorem;
& that the \(du\) ['of the equation' crossed out] and the \(dN\) 
of the equation \(du=V.dN\)  do not mean
the \(du\) and \(dN\) derived from differentiating
with respect to the quantities \(x\) and \(y\) ,
already introduced ; but with respect
to ['some' crossed out] other given quantity?

I suspect so  .
[the following appears underneath in pencil --- still in Ada's hand]}

 \(u=\beta(N, x)\) 

 \(\frac{du}{dx}=\frac{d\beta}{dN}\frac{dN}{dx}+\frac{d\beta}{dx}\) 

 \(\frac{d^nu}{dx.dy}=\) 

About this document

Date of authorship: 

Sept 1841

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170