Show that f^2 − g^2 = C ?

Suppose f and g are continuously differentiable functions such that f(x) = g'(x)and g(x) = f'(x) and that any product of f, f', g and g' is commutative for all
x ∈ R. Show that #f^2# − #g^2# = C for some real constant C.

1 Answer
Noah G
Mar 27, 2018

See below for proof.

Explanation:

We have:

#f^2 - g^2 = C#

#(f(x) + g(x))(f(x) - g(x)) = C#

By the product rule:

#(f'(x) + g'(x))(f(x)- g(x)) + (f(x) + g(x))(f'(x) - g'(x)) = 0#

Note that the right hand side becomes #0# because the derivative of any constant is #0#. We now use our definitions that #g'(x) = f(x)# and #f'(x) = g(x)# to rewrite in terms of #f(x)# and #g(x)#.

#(g(x) + f(x))(f(x) - g(x)) + (f(x) + g(x))(g(x) - f(x)) = 0#

#(g(x) + f(x))(f(x) - g(x) + g(x) - f(x)) = 0#

#(g(x) + f(x))(0) = 0#

#0 = 0#

As required.

Hopefully this helps!

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