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Folios 18-19: ADM to AAL

[18r] My dear Lady Lovelace

First as to your non mathematical question: I do not think
an English jury would have found Mad. Laffarge guilty; but the presump-
tions of guilt and innocence can only be perfectly made by those who have
the same national opinions and feelings as the accused.  I think it very
possible that a Frenchman may know a Frenchman to be guilty upon
grounds which an Englishman would not understand; for instance, a
particular act may be such as a Frenchman may know a Frenchman
would not do, unless he had committed a murder before; and such acts
may have been proved for aught I (who have not read much of the trial)
can tell.  It is the same thing in our courts of justice: judges and counsel
can by experience make things which would appear to you or me almost
indifferent, carry very positive conclusions to their own minds.

Now as to the part of your difficulty contained between X X in the
remarks.  It matters nothing (p. 210) whether \(\frac{n-p}{p+1}x\) is negative or
positive.  If you perfectly understand why I neglect the sign in the fractional
case, the same reason applies to the negative one.  When \(n\) is negative
then \(n-p\) (\( p\) being essentially \(+\) ) is necessarily negative.  Consequently,
(\( x\) being positive) the terms alternate in sign from the very beginning,
whereas when \(n\) is positive and fractional, they do not begin to
alternate until \(p\) passes \(n\) .  But our matter is to determine conver-
gency, and if a series of positive terms be convergent, so will be the
series of similar terms alternating.  It is on the absolute magnitude
of \(\frac{n-p}{p+1}x\), independent of sign, that it depends whether the terms
shall ultimately diminish so as to create convergence, or not.
Now the limit of this is \(x\), whence follows as in the book
[18v] It is important to remember in results which depend
entirely on limits that they have nothing to do with any
vagaries which the quantity tending to a limit chooses to
play, provided that, when it has sown its wild oats, it
settles down into a steady approach to its limit.  The
sins of its youth are not to be remembered against it.
Now \(\frac{n-p}{p+1}x\) when \(n\) is positive, remains positive until
\( p\) passes \(n\) and then settles into incurable negativeness.  But
when \(x\) is ['positive' crossed out] negative, it is negative from the beginning
Now this matters nothing as to a result which depends only
on the limit to which \(\frac{n-p}{p+1}x\) approaches as \(p\) is increased
without limit.


 As to the point marked \(B\), remember that placing this
doubtful assumption, namely the expansibility of functions
of \(x\) in whole powers of \(x\), out of doubt by instances,
has been a prevailing vice of algebraical writers, and
one which is to be carefully avoided.  It was once thought, by
instance, that \(x^2+x+41\) must be a prime number, whenever
\( x\) is a whole number, for \[\begin{array}{lrl}x=0 &x^2+x+41=41 &{\small\text{a prime no}}\\ ~ =1 &=43& \cdots\cdots\cdots\\ ~ =2 &=47 &\cdots\cdots\cdots\\ ~ =3 &=53 &\cdots\cdots\cdots\\ ~ =4 &=61 &\cdots\cdots\cdots\\ ~ =5 &=71 &\cdots\cdots\cdots\\ ~ =6 &=83 &\cdots\cdots\cdots\\ ~ =7 &=97 &\cdots\cdots\cdots \end{array}\]

 [19r] Now \(x^2+x+41\), though it gives nothing but prime numbers
up to \(x=39\) inclusive, yet gives a composite number
when \(x=40\); for it then is

 \(40\times 40+40+41\)   or  \(41\times 40+41\) or \(41\times 41\) 

 and when \(x=41\) it is

 \(41\times 42+41\)   or  \(43\times 41\) 

 and for higher values it gives sometimes prime numbers
sometimes not, like other functions.  So much for instances.

 C.  The supposition as to the meaning of the non-arithmetical
roots is right (Read from p. 109 ''We shall now proceed'' to 
p.\,113 inclusive).  When we use \(\sqrt{-1}\), which we must do at
present, if at all, without full explanation, it is to be 
remembered that we say two expressions are equal when they
are algebraically the same, that is, when each side has
all the algebraical properties of the other.  ['M' crossed out] Numerical
accordance must not be looked for when one or both sides
are numerically unintelligible, and algebraical accordance
merely means that everything which is true of one side is
true of the other.  It is then unnecessary to consider any 
restrictions which may be necessary when numerical
accordance is that which is denoted by \(=\) .
But this is touching on even a higher algebra than
the one before you

 \(\frac{1}{1-x}=1+x+x^2+x^3+\cdots\cdot\) ad inf.

 which is certainly true in the arithmetical sense when
[19v] \(x<1\) .  But if \(x>1\), say \(x=2\), we have

 \(\frac{1}{1-2}\) or \(-1=1+2+4+8+16+\) &c

 which, arithmetically considered is absurd.  But nevertheless
\( -1\) and \(1+2+4+8+\) &c have the same properties

This point is treated in the chapter on the meaning
of the sign \(=\) .

My wife desires to be kindly remembered

 I remain

 Yours very truly


69 Gower St.

Thursday Evg Octr 15/40

 It is fair to tell you that the use of divergent
series is condemned altogether by some modern names of
very great note.  For myself I am fully satisfied
that they have an algebraical truth wholly independent
of arithmetical considerations; but I am also satisfied
that this is the most difficult question in mathematics.

About this document

Date of authorship: 

15 Oct 1840

Holding institution: 

Bodleian Library, Oxford, UK


Dep. Lovelace Byron

Box 170