Question

Suppose that \(c^{\prime}(0)=a\) and \(s^{\prime}(0)=b\) where \(a^{2}+b^{2} \neq 0,\) and $$ \begin{array}{l} c(x+y)=c(x) c(y)-s(x) s(y) \\ s(x+y)=s(x) c(y)+c(x) s(y) \end{array} $$ for all \(x\) and \(y\). (a) Show that \(c\) and \(s\) are differentiable on \((-\infty, \infty),\) and find \(c^{\prime}\) and \(s^{\prime}\) in terms of \(c, s, a,\) and \(b\). (b) (For those who have studied differential equations.) Find \(c\) and \(s\) explicitly.

Short answer

A: Yes, both functions c(x) and s(x) are differentiable on (-∞, ∞) as we showed their derivatives are well-defined and satisfy the properties given for all x ∈ ℝ.

Step-by-step solution

Proof of Differentiability

To prove that the functions are differentiable on \((-\infty, \infty)\), we will make use of the given properties as well as the rules of differentiation. We are given that the functions \(c(x)\) and \(s(x)\) satisfy the following properties for all \(x, y \in \mathbb{R}\): $$ \begin{array}{l} c(x+y)=c(x) c(y)-s(x) s(y) \\ s(x+y)=s(x) c(y)+c(x) s(y) \end{array} $$ We will now compute the derivatives of \(c(x)\) and \(s(x)\) and show that they satisfy the above properties. Using the chain rule, we get: $$ \begin{aligned} \dfrac{d}{dx} c(x+y) &= \dfrac{d}{dx} (c(x) c(y) - s(x) s(y)) \\ \dfrac{d}{dx} s(x+y) &= \dfrac{d}{dx} (s(x) c(y) + c(x) s(y)) \end{aligned} $$

Finding the Derivatives

Applying the product rule to find the derivatives: $$ \begin{aligned} c'(x+y) &= (c'(x) c(y) - c(x) c'(y)) - (s'(x) s(y) - s(x) s'(y)) \\ s'(x+y) &= (s'(x) c(y) + c'(x) s(y)) + (s'(y) c(x) + s(y) c'(x)) \end{aligned} $$ Now compute the derivatives of the functions at \(x=0\): $$ \begin{aligned} c'(y) &= (c'(0) c(y) - c(0) c'(y)) - (s'(0) s(y) - s(0) s'(y)) \\ s'(y) &= (s'(0) c(y) + c'(0) s(y)) + (s'(y) c(0) + s(y) c'(0)) \end{aligned} $$ Using the known information that \(c'(0) = a\) and \(s'(0) = b\), we get: $$ \begin{aligned} c'(y) &= a c(y) - b s(y) \\ s'(y) &= b c(y) + a s(y) \end{aligned} $$ These equations show that \(c(x)\) and \(s(x)\) have well-defined derivatives for all \(x \in \mathbb{R}\). Hence, both functions are differentiable on \((-\infty, \infty)\).

(Optional) Step 3: Finding c(x) and s(x) via Differential Equations

If you have studied differential equations, you can find explicit expressions for \(c(x)\) and \(s(x)\). Note that we can express the derivatives \(c'(y)\) and \(s'(y)\) as a system of linear homogeneous differential equations and solve for \(c(y)\) and \(s(y)\). $$ \begin{aligned} c'(y) &= a c(y) - b s(y) \\ s'(y) &= b c(y) + a s(y) \end{aligned} $$ The differential equation can be solved using standard techniques like diagonalization of coefficients or eigenvectors of the coefficient matrix. In this case, it turns out that the solutions are: $$ \begin{aligned} c(x) = A\cos(wx) + B\sin(wx) \\ s(x) = -B\cos(wx) + A\sin(wx) \end{aligned} $$ where \(w = \sqrt{a^2 + b^2}\), and \(A\) and \(B\) are constants that depend on the initial values of the functions and their derivatives.

Key concepts

Differentiable Functions

Differentiable functions are fundamental in calculus. Essentially, a function is said to be differentiable at a point if it has a derivative there. A derivative represents how a function changes as its input changes. For a function to be differentiable across its entire domain, it has to be smooth or continuous, without any sharp bends or breaks.
The exercise gives us two functions, \(c(x)\) and \(s(x)\), and asks us to show they're differentiable on the entire number line. This involves computing their derivatives using given properties.
Differentiability, as shown in this exercise, involves verifying that the derivatives exist and that the functions smoothly oscillate, which is confirmed by applying the chain rule and product rule to find the derivatives. Thus, differentiable functions like \(c(x)\) and \(s(x)\) are essential in modeling and predicting behavior smoothly and continuously.

Chain Rule

The chain rule is a powerful method in differential calculus used to find the derivative of composite functions. It tells us how to differentiate a function inside another function. If you have two differentiable functions \(f\) and \(g\), the chain rule says that the derivative of their composite \(f(g(x))\) is \(f'(g(x)) \cdot g'(x)\).
In the exercise, the problem uses the chain rule implicitly when differentiating functions like \(c(x+y)\) and \(s(x+y)\). By breaking these into simpler parts, we can differentiate \(c(x+y)\) and \(s(x+y)\) as combinations of \(c(x), c(y), s(x),\) and \(s(y)\) using this rule.

• Differentiate the outer function keeping the inner function unchanged.
• Multiply it by the derivative of the inner function.

Understanding the chain rule is crucial because it helps in managing complex expressions that involve nested functions, making it easier to handle in calculus.

Linear Differential Equations

Linear differential equations are equations that relate a function to its derivatives. They appear in many natural phenomena modeling and are pivotal in mathematical analysis. These equations are vast, but the basic form can often be expressed as something like \(a(x)y' + b(x)y = c(x)\).
The challenge in the exercise was to find \(c(x)\) and \(s(x)\) using a system of linear differential equations derived from the derivatives: \(c'(y) = a c(y) - b s(y)\) and \(s'(y) = b c(y) + a s(y)\). By solving these equations, one can find explicit expressions for \(c(x)\) and \(s(x)\). Techniques like diagonalization of matrices or finding eigenvectors can simplify solving these systems.
They help determine the behavior of systems modeled by such equations, often leading to solutions that describe wave-like or oscillatory movements, which are applicable in various fields such as physics and engineering.