# Taylor's theorem with field of rational numbers

Let $\mathbb{Q}$ be the field of rational numbers. Let $\alpha \in \mathbb{R}$ with $1<\alpha<2$ be such that $\alpha^i$ is irrational for every odd integer $i$. Define $f:\mathbb{Q} \rightarrow \mathbb{R}$ by $f(0)=0$ and $f(q)=\alpha^{4i}$ where $i \in \mathbb{Z}$ with $\alpha^{2i-1}<|q|<\alpha^{2i+1}$.

(a) Show that $\frac{q^2}{4} \leq f(q) \leq 4q^2$ for every $q \in \mathbb{Q}$.

Do I need to consider cases where $i$ is odd and even?

(b) Show that $f(q)=\lim_{v \to 0} f(q+v)\ \forall q \in \mathbb{Q}$ and $f'(q)=\lim_{v \to 0} \frac{f(q+v)-f(q)}{v}=0\ \forall q \in \mathbb{Q}$.

Are these not trivial? Or maybe I am missing the point.

(c) Write the second-order Taylor expansion of $f$ around $0$.

a): No, you don't need to treat the cases $i$ even or odd differently. First think about why the function is well-defined. Then use the inequality $$\alpha^{2i-1}<|q|<\alpha^{2i+1}$$ to obtain an inequality that contains $f(q) = \alpha^{4i}$.
• For (a) I have: $a^{2i-1}<|q|<a^{2i+1}$ so $\frac{a^{4i}}{a^2}<q^2<a^{4i}.a^2$ and so $f(q)<a^2.q^2<2^2q^2=4q^2$ and $\frac{q^2}{4}<\frac{q^2}{a^2}<f(q)$ buy why equality? Is that because of $f(0)=0$? – Polp Feb 29 '16 at 18:50
• And the derivative follows by definition and since $f(q+v)-f(q)=0$ – Polp Feb 29 '16 at 18:54
• For (c) $f(0)=0$, $f'(0)=0$ and $f''(0)=0$ so I get $0$? – Polp Feb 29 '16 at 19:00