Helping Ada Lovelace with her Homework: Classroom Exercises from a Victorian Calculus Course – Solution to Problem 2

Here is a transcription of the letter containing De Morgan’s solution to Problem 2, which he included in a letter [LB 170, [Nov. 1840], ff. 27r-27v]:

My dear Lady Lovelace,

I can soon put you out of your misery about p. 206. You have shown correctly that \(\phi(x+y)=\phi(x)+\phi(y)\) can have no other solution than \(\phi(x)=ax\), but the preceding question is not of the same kind; it is not show that there can be no other solution except \(\frac{1}{2} (a^x+a^{-x})\) but show that \(\frac{1}{2} (a^x+a^{-x})\) is a solution: that is, try this solution.

\[\begin{array}{rcl}\phi(x+y)&=&\frac{1}{2} (a^{x+y}+a^{-x-y} )\\
\phi(x-y)&=&\frac{1}{2} (a^{x-y}+a^{-x+y} )\\
          \phi(x+y)+\phi(x-y)&=&\frac{1}{2} (a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y} )\end{array}\]

\[\begin{array}{ll}\mbox{[On the other hand]}\hspace{20pt}&    2\phi(x)\cdot \phi(y)=2\cdot \frac{1}{2} (a^x+a^{-x} )\cdot \frac{1}{2} (a^y+a^{-y} )\\
                                                            &     =\frac{1}{2} (a^x+a^{-x} )(a^y+a^{-y} )\\
                                                             & =\frac{1}{2} (a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y} )\end{array}\]
the same as before.

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
A De Morgan

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