[27r] My dear Lady Lovelace

I can soon put you out of your
misery about p. 206.

You have shown correctly that \(\varphi(x+y)=\varphi(x)+\varphi(y)\) 
can have no other solution than \(\varphi x=ax\), but the
preceding question is not of the same kind; it
is not show that there can be no other solution
except \(\frac{1}{2}\left(a^x+a^{-x}\right)\) but show that \(\frac{1}{2}\left(a^x+a^{-x}\right)\) is
a solution: that is, try this solution

 \(\varphi(x+y)=\frac{1}{2}\left(a^{x+y}+a^{-x-y}\right)\) 

 \(\varphi(x-y)=\frac{1}{2}\left(a^{x-y}+a^{-x+y}\right)\) 

\( \varphi(x+y)+\varphi(x-y)=\frac{1}{2}\left(a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y}\right)\) 

 \(2\varphi x.\varphi y=2.\frac{1}{2}\left(a^x+a^{-x}\right).\frac{1}{2}\left(a^y+a^{-y}\right)\) 

                      \(=\frac{1}{2}\left(a^x+a^{-x}\right)\left(a^y+a^{-y}\right)\) 

                      \(=\frac{1}{2}\left(a^{x+y}+a^{-x+y}+a^{x-y}+a^{-x-y}\right)\) 
the same as before.

[27v] To prove that this can be the only solution
would be above you

I think you have got all you were meant
to get from the chapter on functions.
The functional equations which can
be fully solved are few in number

                Yours very truly
                      ADeMorgan

 69 G.S.
Mon^y M^g