Question: "Let A=f2+g2, where f,g are functions of x..."

Aryan Salinas

Answered question

2022-03-21

Let

A = f^{2} + g^{2}

, where f,g are functions of x such that

f^{'} = (c - 1) (f \cos (x) \sin (x) + g \sin^{2} (x))

g^{'} = - (c - 1) (f \cos^{2} (x) + g \cos (x) \sin (x))

,
for some constant c.
How do I show that

A^{'} \leq 4 | c - 1 | A

Abdullah Avery

Beginner2022-03-22Added 19 answers

Cauchy-Schwartz' inequality tells you that

| a f + b g | \leq \sqrt{a^{2} + b^{2}} \sqrt{f^{2} + g^{2}}

Starting from your result, we obtain

A^{'} \leq | A^{'} | \leq 2 | c - 1 | 1 . \sqrt{f^{2} + g^{2}} 1 . \sqrt{f^{2} + g^{2}} = 2 | c - 1 | A

Malia Booth

Beginner2022-03-23Added 16 answers

You could also use the second to last equation to find via trigonometric theorems for the double angle

A^{'} = (c - 1) ((f^{2} - g^{2}) \sin (2 x) - 2 f g \cos (2 x))

then using that

{(f^{2} - g^{2})}^{2} + {(2 f g)}^{2} = {(f^{2} + g^{2})}^{2} = A^{2}

like in the construction of Pythagorean triples you even get

A^{'} \leq | c - 1 | A

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