Question

Find all differentiable functions \(f : \mathbb{R} \rightarrow \mathbb{R}\) such that $$f^{\prime}(x)=\frac{f(x+n)-f(x)}{n}$$ for all real numbers \(x\) and all positive integers \(n .\)

Short answer

All differentiable functions \(f : \mathbb{R} \rightarrow \mathbb{R}\) for which \(f^{\prime}(x)=\frac{f(x+n)-f(x)}{n}\) for all real numbers \(x\) and all positive integers \(n\) are constant functions, i.e., \(f(x) = C\), where \(C\) is a real constant.

Step-by-step solution

Reinterpret the equation

Notice that the given equation \(f^{\prime}(x)=\frac{f(x+n)-f(x)}{n}\) is the definition of the derivative with a slightly modified difference ratio, i.e., instead of letting the distance between \(x+n\) and \(x\) tend towards zero, it is kept as an integer \(n\). Thus, for this derivative to be consistent for different \(n\), \(f(x+n)-f(x)\) must be equal to zero for any \(n\).

Formulate the condition

As established before, \(f(x+n)-f(x)\) = 0, which means that the function remains constant over the interval of \(n\). Thus, for any pair of integers \(m\) and \(n\), the function must satisfy: \(f(x + m) = f(x + n)\).

Solve for the function

Given that the function stays constant from one integer to the next, it can be concluded that the only differentiable function \(f : \mathbb{R} \rightarrow \mathbb{R}\) that satisfies the condition from Step 2 is a constant function. Therefore, \(f(x) = C\), where \(C\) is any real constant.

Key concepts

Definition of Derivative

The definition of a derivative is essential for understanding many concepts in calculus. At its core, it represents how a function changes as its input changes. More technically, the derivative of a function at a point is the limit of the average rate of change of the function over an interval, as the interval becomes infinitesimally small.
This is usually expressed as:

• \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

Here, \(f'(x)\) is the derivative of \(f\) at \(x\), \(h\) is the change in \(x\), and the expression inside the limit is the difference quotient.
In this problem, however, the interval is not getting smaller but is rather fixed at an integer \(n\). This means the function's values over integers give us insights into its nature. If the difference quotient \(\frac{f(x+n)-f(x)}{n}\) remains constant for every integer \(n\), we can explore how the function behaves over its domain.

Constant Function

A constant function is a function that always returns the same value no matter the input. Mathematically, it is expressed as \(f(x) = C\), where \(C\) is a real number.
This means that for any input \(x\), the function's output does not change and remains \(C\).
In the context of this problem, the conclusion that the only solution is a constant function arises from the equation \(f(x+n) = f(x)\) for all integers \(n\). If the value of the function remains unchanged between any intervals separated by \(n\), it must mean that the function does not vary at all. This stability suggests that \(f\) is not only differentiable, but also constantly returning the same output for all inputs.
This is a key insight because differentiable functions often exhibit varying value trends, but when the derivative remains zero or a condition leads to no change, it pinpoints a constant function nature.

Real Numbers

Real numbers, denoted by \(\mathbb{R}\), represent all possible numbers along the continuous number line. This includes:

• Integers like 1, 2, and -3
• Rational numbers like \(\frac{1}{2}\) and \(-\frac{3}{4}\)
• Irrational numbers like \(\pi\) and \(\sqrt{2}\)

Real numbers are crucial because they include every possible value that could be an input or output of a function that is defined over \(\mathbb{R}\).
In the exercise, the function \(f : \mathbb{R} \rightarrow \mathbb{R}\) indicates that both the domain (inputs) and codomain (outputs) are all real numbers. This suggests that our function \(f\) can map any real number to another real number. The challenge is to find out what kind of mapping could meet the given conditions.
Real number constraints are fundamental in proving that the function must be constant. Since any real \(x + n\) still results in the same function output as \(x\), it emphasizes that every real input leads to the same real output, pointing towards the function being a constant function. This ensures consistent behavior across all real numbers.