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A function of rational proportions

Give an example of a differentiable function f:\mathbb{R} \rightarrow \mathbb{R} such that for each x \in \mathbb{Q} it holds f(x) \in \mathbb{Q} and f'(x) \notin \mathbb{Q}.

Solution

Let us define

    \[F(x) = x (1-x) (1+x) \quad , \quad -1 \leq x \leq 1\]

and extend it periodically to all of \mathbb{R}. Specifically F(0)=F(-1)=F(1) and hence F has zeroes all the integers. We also note that F is differentiable and the derivative is given by F'(x)= 1-3x^2 \; , \; -1 \leq 1 \leq 1. The only thing we have to check is the differentiability of the function at the odd integers. But that is quite easy since

    \[\lim_{x \rightarrow 1^-}F'(x) = -2 = \lim_{x \rightarrow -1^+}F'(x)\]

Let us define the function under question as

(1)   \begin{equation*} f(x) = \sum_{n=0}^{\infty} \frac{F\left ( n! x \right )}{\left ( n! \right )^2} \end{equation*}

The above sum converges uniformly because the nominator is bounded. We can also differentiate term by term since the series of f' also converges uniformly.

All that remains now is to prove that f , the way it is defined , has the requested properties. Let q be a rational number. For all n that are greater or equal to the denominator of q , n!q is rational. Hence , F(n!q) . This in return means that the sum in (1) for rational q is a finite rational sum , hence rational as we wanted.

All that remains now is to prove that f'(q) \notin \mathbb{Q}. What we are using is a slightly variant of the well known proof that e is irrational.  Indeed , let f(q) = \frac{P}{Q} where P, Q are integers. Without loss of generality , let us suppose that they are positive. Thus, for a suitable large natural number R>Q we have that R! f(q) is an integer and

    \begin{align*} 0 & < R! \left( f(q) - \sum_{n=0}^{R} \frac {F ^{\prime}(n!x)}{(n!)^2} \right) \\ &< \sum_{n = R+1}^{\infty} \frac {R!}{n!} \\ &=\frac{1}{R+1} +\frac{1}{(R+1)(R+2)} + \frac{1}{(R+1)(R+2)(R+3)} + ... \\ &< \frac{1}{R+1} +\frac{1}{(R+1)^2} + \frac{1}{(R+1)^3} + ... \\ &= \frac{1}{R} \\ &< 1 \end{align*}

However , the last is a contradiction since the LHS is an integers and the RHS is less than 1. This completes the proof.

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