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Let $V = \mathrm{Vect}( e^{ 4 \pi it } , e^{ 5 \pi i t} , e^{ 6 \pi i t} ) $ and $f \in V$.

I want to determine the common period between the functions : $t \mapsto e^{ 4 \pi it } $, $t \mapsto e^{ 5 \pi i t }$ and $t \mapsto e^{ 6 \pi i t}$ ; i.e the period of $f \in V$.

We have

  • $ t \mapsto e^{ 4 \pi i t} $ is periodic with period $\frac{1}{2}$
  • $ t \mapsto e^{ 5 \pi i t} $ is periodic with period $\frac{2}{5}$
  • $ t \mapsto e^{ 6 \pi i t} $ is periodic with period $\frac{1}{3}$

and I don't know how to pursue it. Is there a particular method to found a common period? I need a hint.

Thank you

Tohiea
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    lcm works for rational numbers too – lhf Dec 10 '20 at 13:27
  • @lhf $$ \mathrm{lcm}( \frac{1}{2}, \frac{2}{5}, \frac{1}{3} ) = \mathrm{lcm}( \frac{1}{2}, \mathrm{lcm}( \frac{2}{5}, \frac{1}{3} ) ) $$ Now $$ \mathrm{lcm}( \frac{2}{5}, \frac{1}{3} ) = \frac{ \frac{2}{5} \frac{1}{3} }{ \mathrm{gcd}( \frac{2}{5}, \frac{1}{3} ) } = 2 $$ Then $$ \mathrm{lcm}( \frac{1}{2}, \frac{2}{5}, \frac{1}{3} ) = \mathrm{lcm}( \frac{1}{2}, 2 ) = \frac{2}{2} \mathrm{lcm}( \frac{1}{2}, 2 ) = \frac{1}{2} \mathrm{lcm}( 1, 4 ) = \frac{4}{2} = 2 $$ It is correct ? – Tohiea Dec 10 '20 at 13:45

2 Answers2

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Hint: You can use that

$$\operatorname{lcm}\biggl(\frac pq,\frac rs\biggr)=\frac{\operatorname{lcm}(p,r)}{\gcd(q,s)}.$$

Bernard
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  • $$ \mathrm{lcm} (\frac{1}{2}, \frac{2}{5} , \frac{1}{3} )= \mathrm{lcm} (\frac{1}{2}, \mathrm{lcm} ( \frac{2}{5} , \frac{1}{3} ) ) $$ Now $$ \mathrm{lcm} ( \frac{2}{5} , \frac{1}{3} ) = \frac{ \mathrm{lcm} ( 2,1) }{ \mathrm{gcd}(5,3) } = \frac{2}{ \mathrm{gcd}(5,3) \mathrm{gcd} ( 2,1) } = 2 $$ Then $$ \mathrm{lcm} (\frac{1}{2}, \frac{2}{5} , \frac{1}{3} )= \mathrm{lcm} (\frac{1}{2}, 2) = 2$$ Thank you – Tohiea Dec 10 '20 at 14:04
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    It is associative, as for integers: you could do it in a single stroke: $\dfrac{\operatorname{lcm}(1,2,1)}{\gcd(2,3,5)}$ – Bernard Dec 10 '20 at 14:11
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The periods are $\frac{1}{2}, \frac{2}{5},\frac{1}{3}$.

Since $\operatorname{lcm}(2,5,3)=30$, we can write these periods as $\frac{15}{30}, \frac{12}{30},\frac{10}{30}$.

Since $\operatorname{lcm}(15,12,10)=60$, the lcm of the periods is $\frac{60}{30}=2$.

lhf
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