Clay Mathematics Institute

Dedicated to increasing and disseminating mathematical knowledge
132v|133r

Click banner for images

 

Folios 132-133: AAL to ADM

[132r]

 Ashley-Combe
         Porlock
                 Somerset
Thursdy 4th Novr ['1841' inserted by later reader]

 Dear Mr De Morgan.  As I find my journey to
Town is extremely uncertain, & may possibly even not
take place at all, I will trouble you without
further delay on the more important of my present
points of difficulty.
I will begin with those relating to Chapter 9thof the
Calculus, which I am now studying.  I have arrived
at page 156.
page 132 : (at the bottom).  I make \(u=\cos^{-1}\left(\frac{1-\varepsilon^{2(C-x)}}{\varepsilon^{2(C-x)}+1}\right)\) 

 instead of \(u=\cos^{-1}\left(\frac{\varepsilon^{2(C-x)}-1}{\varepsilon^{2(C-x)}+1}\right)\) 
I enclose a paper with my version of it.
page 153 : ''For instance, we should not recommend
''the student to write the preceding thus, \(d^2.du+d^2x.du=0\),
''tho' is it certainly true that upon the implicit
''suppositions with regard to the successive Increments,
''\( \Delta^2 u.\Delta x+\Delta^2x.\Delta u\) diminishes without limit as compared
''with \((\Delta x)^3\) .''  Why this comparison with \((\Delta x)^3\) ?
[132v] Had the expression been \(\frac{\Delta^2 u.\Delta x+\Delta^2x.\Delta u}{(\Delta x)^3}\) instead
of \(\Delta^2 u.\Delta x+\Delta^2x.\Delta u\), it would then be clear that if
the Numerator diminished without limit with respect
to the Denominator, the fraction itself would approach
without limit to \(0\) .  But as it is, I see no purpose
answered by a comparison with \((\Delta x)^3\) .
Also, I not only do not see the object of this comparison,
but I do not perceive the fact itself either.
Where is the proof that \(\Delta^2 u.\Delta x+\Delta^2x.\Delta u\) does
diminish without limit with respect to \((\Delta x)^3\) ?
Page 135 : (at the top) : There is a slight misprint
\( C=K^2+K^{12}\) instead of \(C=K^2+K'^2\) 
Page 156 : (line 9 from the top) : \(u=C.\sin\theta+C'.\cos\theta+\frac{1}{2}\theta.\sin\theta\) 

 (Explain this step?)
Now I cannot ''explain this step''.
In the previous line, we have :
(1)\( \ldots\)  \(u=C\sin\theta+C'\cos\theta+\frac{1}{2}\theta.\sin\theta+\frac{1}{4}\cos\theta\) (quite clear)
(2)\( \ldots\) And \(u=\cos\theta-\frac{d^2u}{d\theta^2}\) (by hypothesis)
                            \(=\frac{1}{4}\cos\theta+\left(\frac{3}{4}\cos\theta-\frac{d^2u}{d\theta^2}\right)\) 

whence one may conclude that

 \(C.\sin\theta+C'\cos\theta+\frac{1}{2}\theta.\sin\theta=\frac{3}{4}\cos\theta-\frac{d^u}{d\theta^2}\) 
But how \(u=C\sin\theta+C'\cos\theta+\sin\theta.\frac{1}{2}\theta\) is to be deduced
[133r] I do not discover : By subtracting \(\frac{1}{4}\cos\theta\) from both
sides of (1), we get

 \(u-\frac{1}{4}\cos\theta=C\sin\theta+C'.\cos\theta+\frac{1}{2}\theta.\sin\theta\) 

But unless \(\frac{1}{4}\cos\theta=0\), (which would only be the case
I conceive if \(\theta=\frac{w}{2}\) ), I do not see how to derive
the equation in line 9 of the book.
Page 156 : Show that \(\frac{d^2u}{dx^2}-u=X\) (a function of \(x\) )

 gives \(u=C\varepsilon^x+C'\varepsilon^{-x}+\frac{1}{2}\varepsilon^x\int\varepsilon^{-x}X.du-\frac{1}{2}\varepsilon^{-x}\int\varepsilon^xX.dx\) 

 I have,  \(\frac{d^2u}{dx^2}-u=X\)  \(u=K\varepsilon^x+K'\varepsilon^{-x}\) 

 \(\frac{du}{dx}=K\varepsilon^x+K'\varepsilon^{-x}+\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}\) 

 Assume \(\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}=0\) 

 Then \(\frac{du}{dx}=K\varepsilon^x+K'\varepsilon^{-x}\), and \(\frac{d^2u}{du^2}=K\varepsilon^x+K'\varepsilon^{-x}+\) 

  \(+\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}\) 

\( \left.\begin{matrix}
X=\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x} \\
~\\
0=\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}
\end{matrix}\right\}\) \(\begin{matrix}\text{which tell nothing at all as}\\ \text{to the values of} ~\frac{dK}{dx}, ~\frac{dK'}{dx},\\ \text{of}~ K, \text{& of} ~K'\end{matrix}\)

If we had \(\left.\begin{matrix} X=\frac{dK}{dx}\varepsilon^x-\frac{dK'}{dx}\varepsilon^{-x}\\ 0=\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x} \end{matrix}\right\}\) \(\begin{matrix} \text{the expression in the}\\ \text{book will be then}\\ \text {at once deduced.}\end{matrix}\)

[133v] But I do not see how to get these two latter equations
co-existent.
I enclose an attempt of mine, making the assumed
to be \(\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}=x^2\) instead of \(=0\);
and also ['one' inserted] making this relation to be \(K+K'=x^3\),
but which latter I found led to such very complicated
results that I proceeded but a little way, thinking
it a probable loss of time to go on.
With the relationship \(\frac{dK}{dx}\varepsilon^x+\frac{dK'}{dx}\varepsilon^{-x}=x^2\), I am
as unsuccessful as with \(=0\) .
I defer to another letter some other difficulties of
mine not relating to this Chapter, but partly to 
some remaining points in the 8th Chapter, & partly to
miscellaneous matters.

I hope Mr De Morgan & the ''large boy'' continue
to flourish.  So Mr De M has beat the Queen in
the race, out & out!

 Yours most truly

 A. A. L.

About this document

Date of authorship: 

4 Nov 1841

Holding institution: 

Bodleian Library, Oxford, UK

Collection: 

Dep. Lovelace Byron

Shelfmark: 
Box 170