Question: Let $f$ be a real-valued function with domain $R$ satisfying $f(x+k)=1+\left[2-5 f(x)+10\{f(x)\}^{2}...

Question

Let $f$ be a real-valued function with domain $R$ satisfying $f(x+k)=1+\left[2-5 f(x)+10\{f(x)\}^{2}-10\{f(x)\}^{3}+5\{f(x)\}^{4}-\{f(x)\}^{5}\right]^{\frac{1}{5}}$ for all real $x$ and some positive constant $k$ , then the period of the function $f(x)$ is

Answer 1

Solution

The given functional equation is $f(x+k)=1+\left[2-5 f(x)+10\{f(x)\}^{2}-10\{f(x)\}^{3}+5\{f(x)\}^{4}-\{f(x)\}^{5}\right]^{\frac{1}{5}}$ .

Step 1: Simplify the expression inside the square bracket.

Let $y = f(x)$ . The expression inside the bracket is $2-5y+10y^2-10y^3+5y^4-y^5$ . We can rewrite this expression by separating out a $1$ :
$1 + (1-5y+10y^2-10y^3+5y^4-y^5)$ .

Recall the binomial expansion of $(a-b)^5$ :
$(a-b)^5 = a^5 - 5a^4b + 10a^3b^2 - 10a^2b^3 + 5ab^4 - b^5$ .

Comparing this with $1-5y+10y^2-10y^3+5y^4-y^5$ , we can see that it matches the expansion of $(1-y)^5$ .
So, the expression inside the bracket simplifies to $1 + (1-f(x))^5$ .

Step 2: Substitute the simplified expression back into the functional equation.

The given equation becomes:
$f(x+k) = 1 + [1 + (1-f(x))^5]^{\frac{1}{5}}$

Step 3: Define a new function to simplify the equation.

Let $g(x) = f(x) - 1$ . This implies $f(x) = g(x) + 1$ .
Substitute this into the equation:
$g(x+k) + 1 = 1 + [1 + (1 - (g(x)+1))^5]^{\frac{1}{5}}$
$g(x+k) = [1 + (-g(x))^5]^{\frac{1}{5}}$
$g(x+k) = [1 - (g(x))^5]^{\frac{1}{5}}$

Step 4: Find $g(x+2k)$ using the derived relation.

Let $Y = g(x)$ . Then the relation is $g(x+k) = (1-Y^5)^{1/5}$ .
Now, we apply the relation again by replacing $x$ with $x+k$ :
$g(x+2k) = [1 - (g(x+k))^5]^{\frac{1}{5}}$
Substitute the expression for $g(x+k)$ :
$g(x+2k) = [1 - ((1-Y^5)^{1/5})^5]^{\frac{1}{5}}$
$g(x+2k) = [1 - (1-Y^5)]^{\frac{1}{5}}$
$g(x+2k) = [1 - 1 + Y^5]^{\frac{1}{5}}$
$g(x+2k) = (Y^5)^{\frac{1}{5}}$
Since $f(x)$ is a real-valued function, $g(x)$ is also real. For any real number $Y$ , $(Y^5)^{1/5} = Y$ .
Therefore, $g(x+2k) = Y = g(x)$ .

Step 5: Conclude the period of $f(x)$ .

Since $g(x+2k) = g(x)$ , and we defined $f(x) = g(x) + 1$ :
$f(x+2k) = g(x+2k) + 1 = g(x) + 1 = f(x)$ .
This shows that $f(x)$ is periodic with period $2k$ .

Unless $f(x)$ is a constant function (which would imply $g(x)$ is constant, leading to $g(x)=(1/2)^{1/5}$ ), $2k$ is the fundamental period. In typical problems of this nature, it is assumed that the function is not constant unless stated otherwise.