# Prove f^2+g^2=1: Differentiable Functions f & g

• Janez25
In summary, two differentiable functions f and g with the given conditions are proven to satisfy the equation (f(x))^2 + (g(x))^2 = 1 for all x ∈ R. To prove this, the differential equations f'' = -f and g'' = -g are obtained and the solutions sinx and cosx are used, or alternatively, the function (f(x))^2 + (g(x))^2 is equated with 1 and the steps are reversed. It is also mentioned that f and g are twice differentiable.

## Homework Statement

Suppose f:R→R and g:R→R are both differentiable and that f'(x)=g(x) and g'(x)=-f(x) for all x ∈ R; f(0)=0 and g(0)=1.
Prove : (f(x))²+(g(x))²=1 for all x ∈ R.

## The Attempt at a Solution

I know I need the find d/dx[f(x)²+g(x)²]=d/dx[1], but I am not sure what that is going to help me find and how to use the result.

Proceed with d/dx[f(x)²+g(x)²]=d/dx[1] and use the fact that f'(x)=g(x) and g'(x)=-f(x) for all x ∈ R to get an equality that is clearly true. Then see if you can make all of your steps reversible.

Prove that both f and g are twice differentiable.

Obtain the differential equations: f'' = -f and g'' = -g

Then you know that the function sinx, with initial conditions sin(0) = 0 and d(sin0)/dx = 1 and the funtion cosx with initial condition cos(0) = 1 and d(cos0)/dx = 0 are the solutions to the differential equations.

If you aren't familiar with defining cosine and sine with a differential equation, then just do what snipez advised. Assume that
d(f^2(x) + g^2(x))/dx = d(1)/dx until you get an equality that is true and try to reverse it. Also remember that if y' = 0 then y = c, where c is some real number.

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## 1. What does it mean for a function to be differentiable?

Being differentiable means that a function is smooth and has a well-defined derivative at every point in its domain. This means that the function has a unique tangent line at every point and can be approximated by a linear function near that point.

## 2. How do you prove that f^2 + g^2 = 1 for differentiable functions f and g?

There are a few different approaches to proving this equation. One possible approach is to use the Pythagorean identity for trigonometric functions and relate f and g to sine and cosine, taking advantage of the fact that sine and cosine have derivatives that can be expressed in terms of each other. Another approach is to use the definition of differentiability and show that the derivatives of f^2 and g^2 are equal to each other.

## 3. Can this equation still hold if f and g are not differentiable?

No, this equation only holds for differentiable functions f and g. If f and/or g are not differentiable, then the derivatives of f^2 and g^2 may not be well-defined, and the equation will not hold.

## 4. How does this equation relate to the concept of orthogonality?

This equation is closely related to the concept of orthogonality because it states that the sum of the squares of two functions (f and g) is equal to 1. This is similar to the Pythagorean theorem, which states that the sum of the squares of the sides of a right triangle is equal to the square of the hypotenuse. In this case, f and g can be thought of as the sides of a right triangle, and the sum of their squares is equal to the square of the hypotenuse, which is 1.

## 5. What are some real-world applications of this equation?

This equation has many real-world applications, particularly in fields such as physics and engineering. For example, in electrical engineering, this equation is used in the calculation of impedance, which is a measure of how much a circuit resists the flow of electricity. In physics, this equation is used in the study of waves and oscillations, where f and g can represent different aspects of a wave, such as its amplitude and frequency.