If f and g are two functions for which f' = g and g' = f for all x, then prove that
f^2 - g^2 must be a constant.
ff' = fg
gg' = gf
ff' = gg'
2f f' - 2g g' = 0
f^2 - g^2 = c
I don't really get why f^2 - g^2 would be a constant.
take integrals across the equation
2ff' - 2gg' = 0
∫2f df - ∫2g dg = 0
(f^2 + c1) - (g^2 + c2) = 0
f^2 - g^2 = -(c1+c2) = c
To prove that f^2 - g^2 is a constant, we need to differentiate this expression and show that its derivative is zero.
Let's differentiate f^2 - g^2 with respect to x using the chain rule and the given conditions:
(d/dx)[f^2 - g^2] = 2f⋅f' - 2g⋅g'
Since f' = g and g' = f, we can substitute these values into the expression:
(d/dx)[f^2 - g^2] = 2f⋅g - 2g⋅f = 2fg - 2gf
Notice that the terms 2fg and -2gf cancel out:
(d/dx)[f^2 - g^2] = 0
This shows that the expression f^2 - g^2 has a derivative of zero. A function with a derivative of zero is a constant function. Therefore, f^2 - g^2 is a constant.
Hence, we have proven that if f' = g and g' = f for all x, then f^2 - g^2 must be a constant.