Periodic Conjugates

#Limits #Calculus

Problem Statement

Let $f,g: \mathbb{R} \rightarrow \mathbb{R} $ be periodic functions of periods $a$ and $b$ such that $ \underset{x \rightarrow 0}{\lim} \frac{f(x)}{x} =u \in \mathbb{R} $ and $ \underset{x \rightarrow 0}{\lim} \frac{g(x)}{x} =v \in \mathbb{R} \setminus {0}. $ Compute

$$ \underset{n \rightarrow \infty }{\lim} \frac{f((3+\sqrt{7})^n a)}{g((2 +\sqrt{2})^n b)}. $$

Solution: Of course, the argument of $f$ is terrible. However, the point is that $f$ is periodic. Therefore, the integer multiples of $a$ can be subtracted from the argument of $f$ and would not affect the value of $f.$ So let us find the closest integer. The same argument can be extended to $g.$ Let

$$ (3+\sqrt{7})^n = I + r, \ \text{where} \ \lfloor (3+\sqrt{7})^n \rfloor = I \implies 0< r <1 $$

$$ \text{Define} \ (3-\sqrt{7})^n = r' \implies 0<r'<1. $$

$$ I+ r +r' = (3+\sqrt{7})^n +(3-\sqrt{7})^n = 2 \left( 3^n + \binom{n}{2}3^{n-2}\cdot 7 + \binom{n}{4} 3^{n-4} \cdot 7^2 + \dots \right) $$

Notice that the quantity on the right is an integer, so must be the quantity on the left. Also, notice that $I$ is an integer $\implies r + r'$ is an integer. However, $ 0< r+r'< 2 \implies r+r' = 1. $

$$ \implies r = 1 - r' = 1 - (3-\sqrt{7})^n. $$

$$ \implies (3+\sqrt{7})^n = I + r = I + 1- (3-\sqrt{7})^n $$

$$ \implies f((3+\sqrt{7})^na)= f(( I + 1- (3-\sqrt{7})^n )a)= f( (I+1)a -(3-\sqrt{7})^n a) $$

$$= f(- (3-\sqrt{7})^n a)$$

$$ \text{Similarly,} \ g((2 +\sqrt{2})^n b)= g(-(2-\sqrt{2})^n b ) $$

$$ \implies \underset{n \rightarrow \infty }{\lim} \frac{f((3+\sqrt{7})^n a)}{g((2 +\sqrt{2})^n b)} = \underset{n \rightarrow \infty }{\lim} \frac{f(-(3-\sqrt{7})^n a)}{g(-(2-\sqrt{2})^n b)} $$

$$ = \underset{n \rightarrow \infty }{\lim} \frac{\frac{f(-(3-\sqrt{7})^n) a}{-(3-\sqrt{7})^n a}}{\frac{g(-(2-\sqrt{2})^n b)}{-(2-\sqrt{2})^n b}} \cdot \frac{a}{b} \cdot \left(\frac{3-\sqrt{7}}{2-\sqrt{2}} \right)^n $$

Notice that $ \underset{n \rightarrow \infty }{\lim} \frac{f(-(3-\sqrt{7})^n a)}{-(3-\sqrt{7})^n a} = u $ and $ \underset{n \rightarrow \infty }{\lim} \frac{g(-(2-\sqrt{2})^n b)}{-(2-\sqrt{2})^n b } = v. $ Also notice that $ 0< \frac{3-\sqrt{7}}{2-\sqrt{2}} < 1. $

$$ \implies \underset{n \rightarrow \infty }{\lim} \frac{\frac{f(-(3-\sqrt{7})^n a)}{-(3-\sqrt{7})^n a}}{\frac{g(-(2-\sqrt{2})^nb)}{-(2-\sqrt{2})^nb}} \cdot \frac{a}{b} \cdot \left(\frac{3-\sqrt{7}}{2-\sqrt{2}} \right)^n = \frac{u}{v} \cdot \frac{a}{b} \cdot 0 =0. $$