## 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.

&bnsp;

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

Suppose

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

__ADeMorgan__

69 Gower St.

Thursday Ev^{g} Oct^{r} 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.

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