## Folios 132-133: AAL to ADM

Ashley-Combe

Porlock

Somerset

Thurs^{dy} 4^{th} Nov^{r} ['1841' inserted by later reader]

Dear M^{r} 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 9^{th}of 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 f__act 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 8^{th} Chapter, & partly to

miscellaneous matters.

I hope M^{r} De Morgan & the ''__large boy__'' continue

to flourish. So M^{r} De M has __beat the Queen__ in

the race, __out & out__!

Yours most truly

A. A. L.

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