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1) Compute the product$$\frac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.$$

2) Let $$f(x) = \frac{3}{9^x + 3}.$$ Find $$f \left( \frac{1}{1001} \right) + f \left( \frac{2}{1001} \right) + f \left( \frac{3}{1001} \right) + \dots + f \left( \frac{1000}{1001} \right).$$

3) Let $$f(x) = \frac{x + 6}{x}.$$ The sequence $$(f_n)$$ of functions is defined by $$f_1 = f$$ and $$f_n = f \circ f_{n - 1}$$for all $$n \ge 2.$$

For example,

$$f_2(x) = f(f(x)) = \frac{\frac{x + 6}{x} + 6}{\frac{x + 6}{x}} = \frac{7x + 6}{x + 6}$$

and

$$f_3(x) = f(f_2(x)) = \frac{\frac{7x + 6}{x + 6} + 6}{\frac{7x + 6}{x + 6}} = \frac{13x + 42}{7x + 6}.$$

Let S be the set of all real numbers x such that $$f_n(x) = x$$for some positive integer n. Find the number of elements in S.

Thanks :)

Jul 8, 2019

#1
+22892
+3

1)

Compute the product

$$\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}.$$

$$\begin{array}{|rcll|} \hline && \mathbf{\dfrac{(1998^2 - 1996^2)(1998^2 - 1995^2) \dotsm (1998^2 - 0^2)}{(1997^2 - 1996^2)(1997^2 - 1995^2) \dotsm (1997^2 - 0^2)}} \\\\ &=& \dfrac{(1998 - 1996)(1998 + 1996)(1998 - 1995)(1998 + 1995) \dotsm (1998 - 0)(1998 + 0)} {(1997 - 1996)(1997 + 1996)(1997 - 1995)(1997 + 1995) \dotsm (1997 - 0)(1997 + 0)} \\\\ &=& \dfrac{2\cdot 3994 \cdot 3 \cdot 3993 \dotsm 1998\cdot 1998 } {1\cdot 3993 \cdot 2 \cdot 3992 \dotsm 1997\cdot 1997 } \\\\ &=& \dfrac{2\cdot 3 \dotsm 1998 } {1 \cdot 2 \dotsm 1997 } \times \dfrac{3994 \cdot 3993 \dotsm 1998 } {3993 \cdot 3992 \dotsm 1997 } \\\\ &=& 1998 \times \dfrac{3994 } {1997 } \\\\ &=& 1998 \times \dfrac{2\cdot 1997} {1997 } \\\\ &=& 2\cdot 1998 \\ &=& \mathbf{3996 } \\ \hline \end{array}$$

Jul 8, 2019
#2
+7709
+3

2)

$$f(x) = \dfrac{3}{9^x + 3}\\ f(1-x) = \dfrac{3}{9^{1-x}+3}\\ f(1-x) = \dfrac{3\cdot 9^x}{9+3\cdot 9^x}\\ f(1-x) = \dfrac{9^x}{3+9^x}\\ \therefore f(x) + f(1-x) = 1 \quad\forall x\in \mathbb R$$

$$\quad\text{Answer}\\=\displaystyle \sum_{k=1}^{1000}f\left(\dfrac{k}{1001}\right)\\ = \displaystyle \sum_{k=1}^{500} \left(f\left(\dfrac{k}{1001}\right) + f\left(1-\dfrac{k}{1001}\right)\right)\\ = \displaystyle \sum_{k=1}^{500} 1\\ = 500$$

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Jul 8, 2019
#3
+7709
+2

3)

$$\text{When }f_1 = x,\\ f_2 = f_1(f_1) = f(x) = f_1(x) = x\\ f_3 = f_2(f_1) = f(x) = x\\ \text{When } f_k = x, \\f_{k+1} = f_k(f_1) = f(x) = f_1 (x) = x\\ \implies \text{When }f_1(x) = x, f_n (x) = x \quad \forall x \in \mathbb Z^+\\ \implies \text{The set }S \text{ only contains the solutions to the equation } f_1(x) = x.\\ \text{When }f_1(x) = x,\\ \dfrac{x+6}{x} = x\\ x^2 - x - 6 = 0\\ (x-3)(x + 2) = 0\\ x = 3 \vee x = -2\\ card(S) = 2$$

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Jul 8, 2019
edited by MaxWong  Jul 8, 2019