Aryan Salinas

2022-03-21

Let $A\phantom{\rule{0.222em}{0ex}}={f}^{2}+{g}^{2}$, where f,g are functions of x such that
${f}^{\prime }=\left(c-1\right)\left(f\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right)+g{\mathrm{sin}}^{2}\left(x\right)\right)$,
${g}^{\prime }=-\left(c-1\right)\left(f{\mathrm{cos}}^{2}\left(x\right)+g\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right)\right)$,
for some constant c.
How do I show that ${A}^{\prime }\le 4|c-1|A$ ?

Abdullah Avery

Cauchy-Schwartz' inequality tells you that
$|af+bg|\le \sqrt{{a}^{2}+{b}^{2}}\sqrt{{f}^{2}+{g}^{2}}$
Starting from your result, we obtain
${A}^{\prime }\le |{A}^{\prime }|\le 2|c-1|1.\sqrt{{f}^{2}+{g}^{2}}1.\sqrt{{f}^{2}+{g}^{2}}=2|c-1|A$

Malia Booth

You could also use the second to last equation to find via trigonometric theorems for the double angle
${A}^{\prime }=\left(c-1\right)\left(\left({f}^{2}-{g}^{2}\right)\mathrm{sin}\left(2x\right)-2fg\mathrm{cos}\left(2x\right)\right)$
then using that
${\left({f}^{2}-{g}^{2}\right)}^{2}+{\left(2fg\right)}^{2}={\left({f}^{2}+{g}^{2}\right)}^{2}={A}^{2}$
like in the construction of Pythagorean triples you even get
${A}^{\prime }\le |c-1|A$

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