Aryan Salinas

2022-03-21

Let $A{\textstyle \phantom{\rule{0.222em}{0ex}}}={f}^{2}+{g}^{2}$ , where f,g are functions of x such that

${f}^{\prime}=(c-1)(f\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right)+g{\mathrm{sin}}^{2}\left(x\right))$ ,

${g}^{\prime}=-(c-1)(f{\mathrm{cos}}^{2}\left(x\right)+g\mathrm{cos}\left(x\right)\mathrm{sin}\left(x\right))$ ,

for some constant c.

How do I show that${A}^{\prime}\le 4|c-1|A$ ?

for some constant c.

How do I show that

Abdullah Avery

Beginner2022-03-22Added 19 answers

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

${A}^{\prime}\le \left|{A}^{\prime}\right|\le 2|c-1|1.\sqrt{{f}^{2}+{g}^{2}}1.\sqrt{{f}^{2}+{g}^{2}}=2|c-1|A$

Starting from your result, we obtain

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

${A}^{\prime}=(c-1)(({f}^{2}-{g}^{2})\mathrm{sin}\left(2x\right)-2fg\mathrm{cos}\left(2x\right))$

then using that

$({f}^{2}-{g}^{2})}^{2}+{\left(2fg\right)}^{2}={({f}^{2}+{g}^{2})}^{2}={A}^{2$

like in the construction of Pythagorean triples you even get

${A}^{\prime}\le |c-1|A$

then using that

like in the construction of Pythagorean triples you even get

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