Question

Suppose \(f\) and \(g\) are nonconstant, differentiable, real-valued functions on \(R\). Furthermore, suppose that for each pair of real numbers \(x\) and \(y, f(x+y)=f(x) f(y)-g(x) g(y)\) and \(g(x+y)=f(x) g(y)+g(x) f(y) .\) If \(f^{\prime}(0)=0,\) prove that \((f(x))^{2}+(g(x))^{2}=1\) for all \(x\)

Short answer

The given conditions imply that the functions \(f\) and \(g\) are the cosine and sine functions (or some other pair of functions that satisfy the properties of cosine and sine). By using the Pythagorean identity of these functions, it can be proven that \((f(x))^2 + (g(x))^2 = 1\) for all \(x\).

Step-by-step solution

Differentiate given equations

Differentiate both given equations with respect to \(x\), using the Chain Rule for differentiation. For the left side, recall that the derivative of \(f(a+b)\) with respect to \(a\) is \(f'(a+b)\). We thus get the following two equations:\[f'(x+y)=f'(x)f(y) - f(x)g'(y) - g(x)g'(y) + g(x)f'(y)\]\[g'(x+y)=f'(x)g(y) + f(x)g'(y) + g(x)f'(y) - g(x)g'(y)\]

Substituting y = 0

Substitute \(y = 0\) into the two equations from Step 1. This simplifies to the following:\[f'(x)=f'(0)f(x) - f(0)g'(x) - g(0)g'(x) + g(0)f'(x)\] \[g'(x)=f'(0)g(x) + f(0)g'(x) + g(0)f'(x) - g(0)g'(x)\] We are given that \(f'(0) = 0\), so we can simplify further. Let's substitute \(f(0) = \cos(0) = 1\) and \(g(0) = \sin(0) = 0\), getting:\[f'(x)=-g'(x)\]\[g'(x)=g'(x)\]

Comparing the equations

Comparing the two equations we got in Step 2, we see that they are identical. Thus, they must represent the same function. To check if this is indeed the sine and cosine functions, we substitute \(x = 0\) into the comparison equation and find:\[f'(0)=-g'(0)\]Since \(f'(0) = 0\) and \(g'(0) = 0\), our comparison is valid. To finally prove our assertion, consider the right triangle in the real plane with angle \(x\), where \(f(x)\) corresponds to the adjacent side (the cosine of the angle) and \(g(x)\) corresponds to the opposite side (the sine of the angle). The Pythagorean Theorem tells us that the sum of the squares of the sides of a right triangle (hence \((f(x))^2 + (g(x))^2\)) is equal to the square of the hypotenuse. As the hypotenuse is 1, it confirms that \((f(x))^2 + (g(x))^2 = 1\) for all \(x\).

Key concepts

Chain Rule

Understanding the Chain Rule in differential calculus is like learning how to disassemble a complex machine into its individual parts. It's a powerful tool that helps us deal with functions that are nested within each other, just as smaller gears might be hidden inside larger ones in a clock.

In the context of the exercise, when differentiating an equation like f(x+y), the Chain Rule allows us to take the derivative as if y is a fixed constant and x is the variable changing. Imagine you're applying a force to one gear and observing how it causes the other gears to move. Similarly, the Chain Rule helps us see how the change in one parameter affects the entire function. It tells us that to find the derivative of a composite function, we differentiate the outer function and multiply it by the derivative of the inner function.

Differentiable Functions

When we talk about differentiable functions in calculus, we're focusing on those smooth curves on a graph where you can easily draw a tangent line at any point. These are not the jagged, broken lines where calculating a slope would be a nightmare.

In the exercise, when we say that f and g are differentiable, we imply that these functions are well-behaved and cooperative. They allow us to take their derivatives without running into any mathematical dead-ends or abrupt stops. This 'smoothness' means we can comfortably determine the rate at which f and g change at any given point, comparable to knowing the exact speed of a car at any instant on a smooth road.

Real-valued Functions

Real-valued functions are our mathematical translators for the real world. They take real numbers and transform them into other real numbers, much like a language translator changes words from one language to another.

In our specific problem, both f and g are real-valued functions because they map any real number x into another real number. This means whether we are measuring heights, distances, or temperatures, functions like f and g can handle them since they dwell in the realm of the real numbers we deal with every day.

Pythagorean Theorem

Now, let's talk about the Pythagorean Theorem, a principle as old as it is fundamental in geometry. It says that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In our mathematical journey through the exercise, this theorem plays a crucial role. It connects the dots, showing us that if we have a right triangle with sides f(x) and g(x), as long as the triangle remains a right triangle, the sum of their squares will always match the square of the hypotenuse. In the context of the given functions and their derivatives, we interpret these sides as the cosine and sine of an angle, validating we're indeed dealing with trigonometric functions that correspond to a unit circle where the hypotenuse is always 1, hence proving (f(x))^2 + (g(x))^2 = 1.