Answer :
Let's solve each part of the question step-by-step.
### Part (1)
We want to express \( f^n(x) \) in terms of \( a, b, x, n \).
Given the function \( f(x) = ax + b \), we can construct the first few iterations of \( f \):
- For \( n = 1 \), \( f^1(x) = f(x) = ax + b \).
- For \( n = 2 \), \( f^2(x) = f(f(x)) = f(ax + b) = a(ax + b) + b = a^2x + ab + b \).
- For \( n = 3 \), \( f^3(x) = f(f^2(x)) = f(a^2x + ab + b) = a(a^2x + ab + b) + b = a^3x + a^2b + ab + b \).
To find a general form, we notice a pattern. Each iteration adds another factor of \( a \) to the term involving \( x \) and accumulates the constants \( b \) influenced by the power of \( a \):
[tex]\[ f^n(x) = a^n x + b(1 + a + a^2 + \cdots + a^{n-1}) \][/tex]
The sum \( 1 + a + a^2 + \cdots + a^{n-1} \) is a geometric series with sum given by \(\frac{1-a^n}{1-a}\) (since \( |a| < 1 \)):
[tex]\[ f^n(x) = a^n x + b \cdot \frac{1-a^n}{1-a} \][/tex]
So,
[tex]\[ \boxed{f^n(x) = a^n x + \frac{b(1-a^n)}{1-a}} \][/tex]
### Part (2)
We want to express \(\frac{f^n(x) - f^{n-1}(x)}{a^n}\) in terms of \( a, b, x, n \).
From part (1), we have:
[tex]\[ f^n(x) = a^n x + \frac{b(1-a^n)}{1-a} \][/tex]
[tex]\[ f^{n-1}(x) = a^{n-1} x + \frac{b(1-a^{n-1})}{1-a} \][/tex]
The difference is:
[tex]\[ f^n(x) - f^{n-1}(x) = \left( a^n x + \frac{b(1-a^n)}{1-a} \right) - \left( a^{n-1} x + \frac{b(1-a^{n-1})}{1-a} \right) \][/tex]
[tex]\[ = a^n x - a^{n-1} x + b \left( \frac{1-a^n}{1-a} - \frac{1-a^{n-1}}{1-a} \right) \][/tex]
[tex]\[ = a^{n-1}(ax - x) + b \left( \frac{a^{n-1}(a-1)}{1-a} \right) \][/tex]
[tex]\[ = a^{n-1}x(a - 1) + b \left( -a^{n-1} \right) \][/tex]
[tex]\[ = a^{n-1}\cdot (a-1)x - a^{n-1}b \][/tex]
Dividing by \( a^n \):
[tex]\[ \frac{f^n(x) - f^{n-1}(x)}{a^n} = \frac{a^{n-1}(ax - x) - b a^{n-1}}{a^n} \][/tex]
[tex]\[ = \frac{a^{n-1}(a - 1)x - b a^{n-1}}{a^n} \][/tex]
[tex]\[ = \frac{(a - 1)x - b}{a} \][/tex]
So,
[tex]\[ \boxed{\frac{f^n(x) - f^{n-1}(x)}{a^n} = \frac{(a - 1)x - b}{a}} \][/tex]
### Part (3)
Consider the curve \( y = \frac{f^n(x) - f^{n-1}(x)}{a^n} \) and the line \( y = ax + b \). We need to find their intersection point \( Q(x_n, y_n) \).
From part (2), the curve is \( y = \frac{(a-1)x - b}{a} \).
Setting these equal:
[tex]\[ ax + b = \frac{(a-1)x - b}{a} \][/tex]
Multiply both sides by \( a \):
[tex]\[ a^2x + ab = (a-1)x - b \][/tex]
Collecting like terms:
[tex]\[ a^2 x - (a-1)x = - b - ab \][/tex]
[tex]\[ (a^2 - a + 1)x = - b(1 + a) \][/tex]
[tex]\[ x_n = \frac{- b(1 + a)}{a^2 - a + 1} \][/tex]
Substitute \( x_n \) into \( y = ax + b \):
[tex]\[ y_n = a\left( \frac{- b(1+a)}{a^2 - a + 1} \right) + b \][/tex]
Simplifying \( y_n \):
[tex]\[ y_n = \frac{- ab(1 + a)}{a^2 - a + 1} + b \][/tex]
[tex]\[ = \frac{- ab(1 + a) + b(a^2 - a + 1)}{a^2 - a + 1} \][/tex]
[tex]\[ = \frac{b(a^2 - ab - ab - a + b)}{a^2 - a + 1} \][/tex]
[tex]\[ y_n = \frac{b(a^2 - 2ab - a + 1)}{a^2 - a + 1} \][/tex]
Thus,
[tex]\[ \boxed{x_n = \frac{- b(1 + a)}{a^2 - a + 1}} \][/tex]
[tex]\[ \boxed{y_n = \frac{b(a^2 - 2ab - a + 1)}{a^2 - a + 1}} \][/tex]
### Part (4)
Calculate the limit \(\lim_{n \rightarrow \infty} f^n(x)\).
From part (1), \( f^n(x) = a^n x + \frac{b(1-a^n)}{1-a} \).
As \( n \rightarrow \infty \), since \( -1 < a < 1 \), \( a^n \rightarrow 0 \):
[tex]\[ \lim_{n \to \infty} f^n(x) = 0 \cdot x + \frac{b(1-0)}{1-a} \][/tex]
[tex]\[ = \frac{b}{1-a} \][/tex]
Thus,
[tex]\[ \boxed{\lim_{n \rightarrow \infty} f^n(x) = \frac{b}{1-a}} \][/tex]
### Part (1)
We want to express \( f^n(x) \) in terms of \( a, b, x, n \).
Given the function \( f(x) = ax + b \), we can construct the first few iterations of \( f \):
- For \( n = 1 \), \( f^1(x) = f(x) = ax + b \).
- For \( n = 2 \), \( f^2(x) = f(f(x)) = f(ax + b) = a(ax + b) + b = a^2x + ab + b \).
- For \( n = 3 \), \( f^3(x) = f(f^2(x)) = f(a^2x + ab + b) = a(a^2x + ab + b) + b = a^3x + a^2b + ab + b \).
To find a general form, we notice a pattern. Each iteration adds another factor of \( a \) to the term involving \( x \) and accumulates the constants \( b \) influenced by the power of \( a \):
[tex]\[ f^n(x) = a^n x + b(1 + a + a^2 + \cdots + a^{n-1}) \][/tex]
The sum \( 1 + a + a^2 + \cdots + a^{n-1} \) is a geometric series with sum given by \(\frac{1-a^n}{1-a}\) (since \( |a| < 1 \)):
[tex]\[ f^n(x) = a^n x + b \cdot \frac{1-a^n}{1-a} \][/tex]
So,
[tex]\[ \boxed{f^n(x) = a^n x + \frac{b(1-a^n)}{1-a}} \][/tex]
### Part (2)
We want to express \(\frac{f^n(x) - f^{n-1}(x)}{a^n}\) in terms of \( a, b, x, n \).
From part (1), we have:
[tex]\[ f^n(x) = a^n x + \frac{b(1-a^n)}{1-a} \][/tex]
[tex]\[ f^{n-1}(x) = a^{n-1} x + \frac{b(1-a^{n-1})}{1-a} \][/tex]
The difference is:
[tex]\[ f^n(x) - f^{n-1}(x) = \left( a^n x + \frac{b(1-a^n)}{1-a} \right) - \left( a^{n-1} x + \frac{b(1-a^{n-1})}{1-a} \right) \][/tex]
[tex]\[ = a^n x - a^{n-1} x + b \left( \frac{1-a^n}{1-a} - \frac{1-a^{n-1}}{1-a} \right) \][/tex]
[tex]\[ = a^{n-1}(ax - x) + b \left( \frac{a^{n-1}(a-1)}{1-a} \right) \][/tex]
[tex]\[ = a^{n-1}x(a - 1) + b \left( -a^{n-1} \right) \][/tex]
[tex]\[ = a^{n-1}\cdot (a-1)x - a^{n-1}b \][/tex]
Dividing by \( a^n \):
[tex]\[ \frac{f^n(x) - f^{n-1}(x)}{a^n} = \frac{a^{n-1}(ax - x) - b a^{n-1}}{a^n} \][/tex]
[tex]\[ = \frac{a^{n-1}(a - 1)x - b a^{n-1}}{a^n} \][/tex]
[tex]\[ = \frac{(a - 1)x - b}{a} \][/tex]
So,
[tex]\[ \boxed{\frac{f^n(x) - f^{n-1}(x)}{a^n} = \frac{(a - 1)x - b}{a}} \][/tex]
### Part (3)
Consider the curve \( y = \frac{f^n(x) - f^{n-1}(x)}{a^n} \) and the line \( y = ax + b \). We need to find their intersection point \( Q(x_n, y_n) \).
From part (2), the curve is \( y = \frac{(a-1)x - b}{a} \).
Setting these equal:
[tex]\[ ax + b = \frac{(a-1)x - b}{a} \][/tex]
Multiply both sides by \( a \):
[tex]\[ a^2x + ab = (a-1)x - b \][/tex]
Collecting like terms:
[tex]\[ a^2 x - (a-1)x = - b - ab \][/tex]
[tex]\[ (a^2 - a + 1)x = - b(1 + a) \][/tex]
[tex]\[ x_n = \frac{- b(1 + a)}{a^2 - a + 1} \][/tex]
Substitute \( x_n \) into \( y = ax + b \):
[tex]\[ y_n = a\left( \frac{- b(1+a)}{a^2 - a + 1} \right) + b \][/tex]
Simplifying \( y_n \):
[tex]\[ y_n = \frac{- ab(1 + a)}{a^2 - a + 1} + b \][/tex]
[tex]\[ = \frac{- ab(1 + a) + b(a^2 - a + 1)}{a^2 - a + 1} \][/tex]
[tex]\[ = \frac{b(a^2 - ab - ab - a + b)}{a^2 - a + 1} \][/tex]
[tex]\[ y_n = \frac{b(a^2 - 2ab - a + 1)}{a^2 - a + 1} \][/tex]
Thus,
[tex]\[ \boxed{x_n = \frac{- b(1 + a)}{a^2 - a + 1}} \][/tex]
[tex]\[ \boxed{y_n = \frac{b(a^2 - 2ab - a + 1)}{a^2 - a + 1}} \][/tex]
### Part (4)
Calculate the limit \(\lim_{n \rightarrow \infty} f^n(x)\).
From part (1), \( f^n(x) = a^n x + \frac{b(1-a^n)}{1-a} \).
As \( n \rightarrow \infty \), since \( -1 < a < 1 \), \( a^n \rightarrow 0 \):
[tex]\[ \lim_{n \to \infty} f^n(x) = 0 \cdot x + \frac{b(1-0)}{1-a} \][/tex]
[tex]\[ = \frac{b}{1-a} \][/tex]
Thus,
[tex]\[ \boxed{\lim_{n \rightarrow \infty} f^n(x) = \frac{b}{1-a}} \][/tex]
