Question

If \(f\) is differentiable and is periodic with period \(T,\) show that \(f^{\prime}\) is also periodic with period \(T\), Determine whether $$F(x)=\int_{0}^{x} f(t) d t$$ is always periodic.

Short answer

#Answer# 1. If \(f\) is differentiable and periodic with period \(T\), its derivative \(f'\) is also periodic with period \(T\). 2. We cannot conclude that \(F(x) = \int_{0}^{x} f(t) dt\) is always periodic.

Step-by-step solution

Proving \(f'\) is also periodic with period \(T\)

If \(f\) is differentiable and periodic with period \(T\), then by definition, we have: $$f(x + T) = f(x)$$ for all values of \(x\). To prove that \(f'\) is also periodic with period \(T\), we will differentiate both sides of the equation with respect to \(x\). The chain rule gives us: $$\frac{d}{dx}(f(x + T)) = \frac{d}{dx}(f(x))$$ Now, using the chain rule: $$f'(x+T) = f'(x)$$ This shows that \(f'\) is also periodic with period \(T\), since it satisfies the definition of a periodic function.

Determining if \(F(x)\) is always periodic

We need to determine if the following integral is always periodic: $$F(x) = \int_{0}^{x} f(t) dt$$ Here, \(f(t)\) is a periodic function with period \(T\). To determine if \(F(x)\) is always periodic or not, we will analyze the difference between values of the integral at \(x+T\) and \(x\). We have: $$F(x+T) - F(x) = \int_{0}^{x+T} f(t) dt - \int_{0}^{x} f(t) dt$$ Using properties of integrals, we can rewrite this as: $$F(x+T) - F(x) = \int_{x}^{x+T} f(t) dt$$ Notice that the integral on the right side is equivalent to finding the area under the graph of the function \(f(t)\) over the interval \((x, x+T)\). Since \(f(t)\) is periodic with period \(T\), the area under the curve over this interval \((x, x+T)\) is always the same, no matter the value of \(x\). Therefore, \(F(x+T)-F(x)\) is a constant. However, this does not guarantee that \(F(x)\) is always periodic. A periodic function must satisfy the condition: $$F(x+T)=F(x)$$ for all values of \(x\). In our case, we have only proven that the difference \(F(x+T)-F(x)\) is a constant. Thus, we cannot conclude that \(F(x)\) is always periodic.

Key concepts

Periodic Functions

A periodic function is a function that repeats its values at regular intervals. This means that there is a number, known as the period, such that the function's output does not change if the input is increased by this period. For example, the sine function \( \sin(x) \) is periodic with a period of \( 2\pi \), since \( \sin(x + 2\pi) = \sin(x) \) for all \( x \).

A function \( f(x) \) is considered to be periodic if there exists a positive number \( T \) such that:

• \( f(x + T) = f(x) \)
• This relationship holds for all values of \( x \).

Periodic functions are common in nature, representing phenomena such as sound waves, tides, and the seasons.

In the context of differentiable periodic functions, if \( f \) is a differentiable function and periodic with period \( T \), then its derivative \( f' \) is also periodic with the same period. This is because when we differentiate the periodicity equation \( f(x + T) = f(x) \), we obtain \( f'(x + T) = f'(x) \), confirming the periodic nature of the derivative.

Derivative Rules

When dealing with differentiable functions, several derivative rules help us understand how to find the derivative of complex functions. The derivative of a function represents its rate of change or the slope of the function at any given point. Some fundamental rules of differentiation include:

• **Power Rule**: If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \)
• **Constant Multiple Rule**: If \( f(x) = c \cdot g(x) \), then \( f'(x) = c \cdot g'(x) \)
• **Sum Rule**: If \( f(x) = g(x) + h(x) \), then \( f'(x) = g'(x) + h'(x) \)
• **Chain Rule**: If \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x)) \cdot h'(x) \)

The chain rule is particularly important for understanding periodic functions' derivatives, as it allows us to differentiate composite functions. In the case of periodic functions, we used the chain rule to show that if \( f(x) \) is periodic with period \( T \), the derivative \( f'(x) \) also maintains periodicity with the same period. This is achieved by differentiating both sides of the periodic function equation \( f(x+T) = f(x) \), leading to the conclusion \( f'(x+T) = f'(x) \).

Integral Calculus

Integral calculus is centered around the concept of an integral, which is used to find the area under a curve mathematically. The function \( F(x) = \int_{0}^{x} f(t) \, dt \) represents the accumulated area under the curve from \( 0 \) to \( x \), where \( f(t) \) is a periodic function.

There are two main types of integrals:

• **Definite Integrals**: \( \int_{a}^{b} f(x) \, dx \) provides the exact area under \( f(x) \) from \( x = a \) to \( x = b \)
• **Indefinite Integrals**: Provides the antiderivative function, which represents the family of all functions \( F(x) \) whose derivative is \( f(x) \)

In the context of periodic functions, determining if the integral \( F(x) \) is periodic is crucial. We found that while \( F(x+T) - F(x) \) is a constant, indicating consistent areas under the periodic cycle, this does not ensure \( F(x) \) itself is periodic. A function is only periodic if \( F(x+T) = F(x) \) for all values, which is not always guaranteed by the constant difference observed in periodic function integrals. This highlights that while differentiation retains periodicity, integration may not.