Let f(x) be a real valued function not identically zero satisfies the equation, f(x + y^n ) = f(x) + (f(y))^n for all real x & y and f’(0) = 0 where n (>1) is an odd natural number. Find f(10)

To solve the functional equation \( f(x + y^n) = f(x) + (f(y))^n \) for a real-valued function \( f \), given that \( f'(0) = 0 \) and \( n \) is an odd natural number greater than 1, we can start by exploring the implications of the conditions provided.

Analyzing the Functional Equation

First, let's substitute \( x = 0 \) into the equation:

\[ f(0 + y^n) = f(0) + (f(y))^n \] This simplifies to: \[ f(y^n) = f(0) + (f(y))^n \]

This equation suggests that \( f(y^n) \) is related to \( f(y) \) and a constant \( f(0) \). Next, we can also substitute \( y = 0 \) into the original equation:

\[ f(x + 0^n) = f(x) + (f(0))^n \] This simplifies to: \[ f(x) = f(x) + (f(0))^n \]

From this, we can conclude that \( (f(0))^n = 0 \), which implies \( f(0) = 0 \).

Exploring Further Properties

Now that we know \( f(0) = 0 \), we can rewrite our earlier equation:

\[ f(y^n) = (f(y))^n \]

This indicates that \( f(y^n) \) is the \( n \)-th power of \( f(y) \). Since \( n \) is odd, this suggests that \( f \) could be a polynomial function. Let's assume \( f(y) = k y^m \) for some constants \( k \) and \( m \). We will determine \( m \) by substituting back into the functional equation.

Finding the Form of f

Substituting \( f(y) = k y^m \) into the equation \( f(y^n) = (f(y))^n \) gives us:

\[ f(y^n) = k (y^n)^m = k y^{mn} \] and \[ (f(y))^n = (k y^m)^n = k^n y^{mn} \]

Setting these equal, we have:

\[ k y^{mn} = k^n y^{mn} \]

For this to hold for all \( y \), we need \( k = k^n \). Since \( k \neq 0 \) (as \( f \) is not identically zero), we can conclude that \( k = 1 \). Thus, we have \( f(y) = y^m \) for some \( m \).

Determining the Value of m

Next, we need to find \( m \). We know that \( f'(0) = 0 \). The derivative of \( f(y) = y^m \) is:

\[ f'(y) = m y^{m-1} \]

Evaluating this at \( y = 0 \) gives \( f'(0) = m \cdot 0^{m-1} \). For this to equal zero, either \( m = 0 \) or \( m > 1 \). Since \( m \) must be a positive integer (as \( f \) is not identically zero), we conclude that \( m = 1 \). Therefore, we have:

\[ f(y) = y \]

Calculating f(10)

Finally, we can find \( f(10) \):

\[ f(10) = 10 \]

Thus, the value of \( f(10) \) is \( 10 \).