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The exercise

"Let suppose that $f$ is a function that admits a Taylor expansion such that $$ \forall x\in \mathbb{R}, \ f\left(x\right)=\sum_{n=0}^{+\infty}a_nx^n $$ with $a_n \in \left\{0,1\right\}$ and that $$f\left(\frac{2}{3}\right)=\frac{3}{2}$$

Prove that $\displaystyle f\left(\frac{1}{2}\right)$ is irrational."

My attempt ( which has two main problems )

I've tried proving that with such a function, $\displaystyle \left(a_n\right)_{n \in \mathbb{N}}$ cannot have a finite number of $1$. I think this statement is true, but I dont know how to prove it, that's the first problem.

I've supposed that the sequence $\displaystyle \left(a_n\right)_{n \in \mathbb{N}}$ was periodic of period $T \in \mathbb{N}^{*}$, it led me to the expression $$ f\left(x\right)=\frac{\displaystyle \sum_{n=0}^{T-1}a_nx^n}{1-x^{T}} $$ I dont how I can find a contradiction with this using this expression and the fact that $\displaystyle f\left(\frac{2}{3}\right)=\frac{3}{2}$. That's the second problem.

However, with this two "hics" solved, we would have that $\displaystyle \left(a_n\right)_{n \in \mathbb{N}}$ is infinite and cannot be periodic, hence the binary expansion of the real $\displaystyle f\left(\frac{1}{2}\right)$ would be infinite and not repreating would be irrational.

Any Idea ?

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Suppose that $$ \frac32=\sum_{n=0}^Na_n\Bigl(\frac23\Bigr)^n. $$ Then $$ 3^{N+1}=2\sum_{n=0}^Na_n\,2^n\,3^{N-n}. $$ The left hand side is odd, while the right hand side is even. This solves the first problem. The second can be dealt similarly. With your notation, if $$ \frac32=\frac{\sum_{n=0}^{T-1}a_n\bigl(\frac23\bigr)^n}{1-\bigl(\frac23\bigr)^T}, $$ then $$ 3^T-2^T=2\sum_{n=0}^{T-1}a_n\,2^n\,3^{T-1-n}. $$

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Let's see what happens when $(a_n)$ is a repeating sequence, shall we? Say $(a_n)$ starts repeating at $n=N$. Then we know the non-repeating part is

$$\sum_{i=0}^{N-1}a_i(\tfrac23)^i$$

Then say it repeats with period length $P>0$; define $(b_n)_{n=0}^{P-1}$ as the row of numbers that $a_i$ repeats with; that is, $a_{N+Pj+i}=b_i$ for all non-negative integers $j$ and for $0\leq i<P$. Then we know

$$f(\tfrac23)=\sum_{i=0}^{N-1}a_i(\tfrac23)^i+\sum_{a=0}^{P-1}\sum_{i=0}^\infty b_a(\tfrac23)^{N+Pi+a}$$

We know that

$$\sum_{i=0}^\infty b_a(\tfrac23)^{N+Pi+a}=b_a(\tfrac23)^{N+a}\sum_{i=0}^\infty (\tfrac23)^{Pi}=b_a(\tfrac23)^{N+a}\frac{1}{1-(\tfrac23)^P}=b_a\frac{2^{N+a}3^P}{3^{N+a}(3^P-2^P)}$$

So that

$$f(\tfrac23)=\sum_{i=0}^{N-1}a_i(\tfrac23)^i+\sum_{a=0}^{P-1}b_a\frac{2^{N+a}3^P}{3^{N+a}(3^P-2^P)}$$

Note that these sums are finite; a lot easier to work with!


Now we know:

$$\frac32=\sum_{i=0}^{N-1}a_i(\tfrac23)^i+\sum_{a=0}^{P-1}b_a\frac{2^{N+a}3^P}{3^{N+a}(3^P-2^P)}$$

which we can rewrite

$$3^{N+1}=2\sum_{i=0}^{N-1}a_i2^i3^{N-i}+2\sum_{a=0}^{P-1}b_a\frac{2^{N+a}3^{P-a}}{3^P-2^P}$$

Or

$$(3^P-2^P)\left(3^{N+1}-2\sum_{i=0}^{N-1}a_i2^i3^{N-i}\right)=2\sum_{a=0}^{P-1}b_a2^{N+a}3^{P-a}$$

Yay! We only have integers left. Let's now look $\mod 2$: the left-hand side is

$$(3^P-2^P)\left(3^{N+1}-2\sum_{i=0}^{N-1}a_i2^i3^{N-i}\right)\equiv (1-0)(1-0)\equiv 1\mod 2$$

However, the left-hand side clearly has a factor $2$, so can not be $1\mod2$; contradiction, hence $(a_n)$ can not be a repeating sequence.

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