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Folios 121-122: AAL to ADM



 Friday.  [something crossed out] 21st Augst

 Dear Mr De Morgan.  You have received safely I
hope my packet of yesterday, & my packet sent on
I now re-enclose you the paper marked 1.  There is
another Integral added at the bottom.  Also I have
altered one or two little minutiae in the development
of \(\int\frac{dx}{\sqrt{2ax+x^2}}\) above, which you had omitted to correct.

I quite understand your observations upon it, & see
the mistake I had made; & which related to the
Differential \(dy\), and \(d(\varphi x.x)\) 
If \(y=\varphi x.x\), then \(dy=d(\varphi x.x)=\frac{d(\varphi x.x)}{dx}dx=\) 

  \(=\left\{x.\frac{d(\varphi x)}{dx}+\varphi x.\frac{dx}{dx}\right\}dx=x.\left(\frac{d(\varphi x)}{dx}dx\right)+\varphi x.dx=\) 

  \(=x.d\varphi x+\varphi x.dx\) 
Or if \(y^2=\varphi x.x\), then \(dy^2=x.d.\varphi x+\varphi x.dx\) 

 or \(\frac{d(y^2)}{dy}dy=x.d.\varphi x+\varphi x.dx\) 

 or \(2ydy=x.d.varphi x+\varphi x.dx\), and \(ydy=\frac{1}{2}x.d.\varphi x+\frac{1}{2}\varphi x.dx\) 
[121v] This is all now right in my head.
In \(\int\frac{dx}{\sqrt{2ax+x^2}}\) we arrive then in my corrected paper,
at \(\int\frac{dx}{\sqrt{2ax+x^2}}=\log(x+a+\sqrt{2ax+x^2})\) 

 \(=\log\left(\frac{x}{2}+\frac{a}{2}+\frac{\sqrt{2ax+x^2}}{2}\right)+\log 2\) 
which, as you observe ''again with the book all but
''the \(\log.2\), which being a Constant, matters nothing''.
Very true; but why did you then insist the
\( \log 2\) in page 116? it seems as if put in on purpose
to be effaced in the parenthesis (Omit the Constant).
And it might just as well have been \(\log 3\),
\( \log 4\), \(\log\,\)(anything in the world) .

As to my two papers marked 2 (& which I
again return, merely for the convenience of reference),
I see that in order to make them valid, as applying
each to two separate & different velocities, they should
be re-written (which is not worth while), & the
terms of the enunications altered as follows :
''If two quantities \(V\), \(V'\) be respectively equal to the
''Ratios \(\frac{S}{T}\), \(\frac{S'}{T'}\), and if \(V:V'=T:T'\), then the values
‘’\( S\), \(S'\) must be to each other as the squares of \(T\), \(T'\)
''are to one another'' &c, &c
[122r] At last I believe I have it quite correctly.
As for \(\frac{dy}{dx}=\frac{y}{x}\), I see my fallacy about \(\frac{y}{x}\) being
a fixed quantity.
About page 113, ''The first form becomes impossible
''when \(x\) is greater than \(\sqrt{c}\), for &c'', I fancy I
[something crossed out] had a little misunderstood
the mathematical meaning of the words impossible
quantity.  I have loosely interpreted it as being
equivalent to ''an absurdity'', or at least to
''an absurdity, unless an extension be made in the
''ordinary meaning of words''. And in this
instance I perceived that if the Logarithm be
an odd number, there would be no absurdity
even without extension in the meaning of terms;
because that it would then merely imply a
negative Base; which negative Base, would I think
be admitted theoretically (tho' inconvenient practically)
on the common beginner's instruction on the Theory of
Logarithms. Am I right?
By the bye this subject reminds me that I think
I find a mistake in page 117, line 13 (from the top)
''(\( n\) an integer) \(\int_{-a}^{+a}x^ndx=0\) when \( n\) is odd, \(=\frac{2a^{n+1}}{n+1}\) when \(n\) is even''
[122v] It seems to me just the reverse, thus :
\( =0\) when \( n\) is even, \(=\frac{2a^{n+1}}{n+1}\) when \( n\) is odd}
I have it as follows :
\( \int_{-a}^{+a}x^ndx=\frac{a^{n+1}}{n+1}-\frac{(-a)^{n+1}}{n+1}=\frac{a^{n+1}-(-a)^{n+1}}{n+1}=\) 

 \(=\frac{a^{n+1}-a^{n+1}}{n+1}\) or \(0\) if \(n+1\) be even

 (I now see it while working; for if

 \(n+1\) be even, \( n\) must be odd.)

 and vice versa.
So I need not trouble you upon this; as I have
solved my difficulty whilst stating it.  I had only
looked at this Integral in ['a' inserted] great hurry, this morning.

I hope on Sunday to send you two
remaining papers I have to make out, on the
Accelerating Force subject;

upon \(f=\frac{dv}{dt}\), and \(v=\int fdt\) 
I think I have been encouraged by your great
kindness, so as to give you really no Sinecure
just at this moment.

 Yours most truly

 A. A. Lovelace

About this document

Date of authorship: 

21 Aug 1841

Holding institution: 

Bodleian Library, Oxford, UK


Dep. Lovelace Byron

Box 170