A Function Composition Problem

A topic that has resurfaced during my undertaking of calculus this year has been function composition. During the year, I have been drawn toward the concept of function composition, something that was evident to my father. One day, as we sat down to begin our class, he posed this problem to me:

f(x)=3x+2=\underbrace{(g\circ g\circ g\circ ... \circ g)}_\text{100  times}(x).
Find g(x).

I believe this is a very interesting problem. You should take a pencil and paper and sit down to solve this problem—it is slightly difficult to solve in your head. Here is my solution:

First, I realized that g(x) must be a linear function. Since the functions are composed, the degrees of the polynomials will multiply because the function is raised to the degree of the polynomial. For example, if g(x) = x^3, g(g(x)) = (x^3)^3 = x^{3\cdot 3} = x^9. Hence, the function must be a first degree polynomial. As such, the function g(x) can be represented by the expression ax+b, where a and b are real numbers. Next, I listed g(x), g(g(x)), and so forth, a few times, to try and find a pattern. To clarify, the notation g^n(x) means g(g(g(...(g(x) n times.

g^1(x) = ax+b

g^2(x) = a(ax+b)+b = a^2x+ab+b

g^3(x) = a(a^2x+ab+b)+b = a^3x+a^2b+ab+b

g^4(x) = a(a^3x+a^2b+ab+b)+b = a^4x+a^3b+a^2b+ab+b

From these four iterations, a pattern started to emerge:

g^n(x) = a^nx+a^{n-1}b + a^{n-2}b+...+ab+b = a^nx+b(a^{n-1}+a^{n-2}+...+a+1) = a^nx + b(\frac{a^n}{a-1})

We can equate g^{100}(x) to 3x+2 and find a and b.

3x+2 = g^{100}(x) = a^{100}x+b(\frac{a^{100}}{a-1})

We can equate the degree 1 terms to find a:

3x = a^{100}x

3 = a^{100}

a = 3^{1/100}

Using a, we can find b.

2 = b(\frac{a^{100}}{a-1}) = b(\frac{{(3^{1/100})}^{100}}{3^{1/100}-1})

b = \frac{2}{3}(3^{1/100}-1)

Hence, g(x)=ax+b=3^{1/100}x+\frac{2}{3}(3^{1/100}-1).

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