How to determine the period of composite functions?











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For exemple, take :$fleft(xright)=$cos$left(2xright)cdot :$sin$left(3xright)$. Period of cos$left(2xright)$ is $pi$ and that of sin$left(3xright)$ is $frac{2pi }{3}$. But why is the period of $fleft(xright)$ $2pi$?



For good measure, here's another example: How do I prove that the period of $fleft(xright)=frac{tanleft(xright)}{1+sinleft(xright)}$ is $2pi$?






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  • 3




    $f$ is the product of two functions, not the composition of two functions.
    – lhf
    Nov 13 at 17:22






  • 3




    $2pi$ is the LCM, the least common multiple, of the periods of the other functions.
    – Mason
    Nov 13 at 17:23










  • @Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
    – Daniel Oscar
    Nov 13 at 17:33















up vote
0
down vote

favorite












For exemple, take :$fleft(xright)=$cos$left(2xright)cdot :$sin$left(3xright)$. Period of cos$left(2xright)$ is $pi$ and that of sin$left(3xright)$ is $frac{2pi }{3}$. But why is the period of $fleft(xright)$ $2pi$?



For good measure, here's another example: How do I prove that the period of $fleft(xright)=frac{tanleft(xright)}{1+sinleft(xright)}$ is $2pi$?






real-analysis functions







share|cite|improve this question


















  • 3




    $f$ is the product of two functions, not the composition of two functions.
    – lhf
    Nov 13 at 17:22






  • 3




    $2pi$ is the LCM, the least common multiple, of the periods of the other functions.
    – Mason
    Nov 13 at 17:23










  • @Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
    – Daniel Oscar
    Nov 13 at 17:33













up vote
0
down vote

favorite









up vote
0
down vote

favorite











For exemple, take :$fleft(xright)=$cos$left(2xright)cdot :$sin$left(3xright)$. Period of cos$left(2xright)$ is $pi$ and that of sin$left(3xright)$ is $frac{2pi }{3}$. But why is the period of $fleft(xright)$ $2pi$?



For good measure, here's another example: How do I prove that the period of $fleft(xright)=frac{tanleft(xright)}{1+sinleft(xright)}$ is $2pi$?






real-analysis functions







share|cite|improve this question













For exemple, take :$fleft(xright)=$cos$left(2xright)cdot :$sin$left(3xright)$. Period of cos$left(2xright)$ is $pi$ and that of sin$left(3xright)$ is $frac{2pi }{3}$. But why is the period of $fleft(xright)$ $2pi$?



For good measure, here's another example: How do I prove that the period of $fleft(xright)=frac{tanleft(xright)}{1+sinleft(xright)}$ is $2pi$?






real-analysis functions




real-analysis functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 13 at 17:19









Daniel Oscar

1047




1047








  • 3




    $f$ is the product of two functions, not the composition of two functions.
    – lhf
    Nov 13 at 17:22






  • 3




    $2pi$ is the LCM, the least common multiple, of the periods of the other functions.
    – Mason
    Nov 13 at 17:23










  • @Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
    – Daniel Oscar
    Nov 13 at 17:33














  • 3




    $f$ is the product of two functions, not the composition of two functions.
    – lhf
    Nov 13 at 17:22






  • 3




    $2pi$ is the LCM, the least common multiple, of the periods of the other functions.
    – Mason
    Nov 13 at 17:23










  • @Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
    – Daniel Oscar
    Nov 13 at 17:33








3




3




$f$ is the product of two functions, not the composition of two functions.
– lhf
Nov 13 at 17:22




$f$ is the product of two functions, not the composition of two functions.
– lhf
Nov 13 at 17:22




3




3




$2pi$ is the LCM, the least common multiple, of the periods of the other functions.
– Mason
Nov 13 at 17:23




$2pi$ is the LCM, the least common multiple, of the periods of the other functions.
– Mason
Nov 13 at 17:23












@Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
– Daniel Oscar
Nov 13 at 17:33




@Mason. Ok, but why is it like that? I mean, why do we use the LCM? I was kind of searching to build a strong intuitive notion...
– Daniel Oscar
Nov 13 at 17:33










1 Answer
1






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$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
Since $g$ is $pi-$periodic and $h$ is $frac{3pi}{2}-$periodic and $pineqfrac{3pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $pi$ (to guaranty that the periodicity of $g$) and a multiple of $frac{3pi}{2}$ (to guaranty that the periodicity of $h$) at the same time.
Hence, $T$ is the $LCM(pi,frac{3pi}{2})=2pi$



You can use the same logic to answer the second question.






share|cite|improve this answer

















  • 1




    However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
    – Ingix
    Nov 13 at 19:51












  • Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
    – hamza boulahia
    Nov 13 at 20:27











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$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
Since $g$ is $pi-$periodic and $h$ is $frac{3pi}{2}-$periodic and $pineqfrac{3pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $pi$ (to guaranty that the periodicity of $g$) and a multiple of $frac{3pi}{2}$ (to guaranty that the periodicity of $h$) at the same time.
Hence, $T$ is the $LCM(pi,frac{3pi}{2})=2pi$



You can use the same logic to answer the second question.






share|cite|improve this answer

















  • 1




    However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
    – Ingix
    Nov 13 at 19:51












  • Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
    – hamza boulahia
    Nov 13 at 20:27















up vote
1
down vote













$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
Since $g$ is $pi-$periodic and $h$ is $frac{3pi}{2}-$periodic and $pineqfrac{3pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $pi$ (to guaranty that the periodicity of $g$) and a multiple of $frac{3pi}{2}$ (to guaranty that the periodicity of $h$) at the same time.
Hence, $T$ is the $LCM(pi,frac{3pi}{2})=2pi$



You can use the same logic to answer the second question.






share|cite|improve this answer

















  • 1




    However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
    – Ingix
    Nov 13 at 19:51












  • Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
    – hamza boulahia
    Nov 13 at 20:27













up vote
1
down vote










up vote
1
down vote









$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
Since $g$ is $pi-$periodic and $h$ is $frac{3pi}{2}-$periodic and $pineqfrac{3pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $pi$ (to guaranty that the periodicity of $g$) and a multiple of $frac{3pi}{2}$ (to guaranty that the periodicity of $h$) at the same time.
Hence, $T$ is the $LCM(pi,frac{3pi}{2})=2pi$



You can use the same logic to answer the second question.






share|cite|improve this answer












$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
Since $g$ is $pi-$periodic and $h$ is $frac{3pi}{2}-$periodic and $pineqfrac{3pi}{2}$, then it is clear that the period $T$ of $f$ is a multiple of $pi$ (to guaranty that the periodicity of $g$) and a multiple of $frac{3pi}{2}$ (to guaranty that the periodicity of $h$) at the same time.
Hence, $T$ is the $LCM(pi,frac{3pi}{2})=2pi$



You can use the same logic to answer the second question.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 13 at 17:42









hamza boulahia

931319




931319








  • 1




    However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
    – Ingix
    Nov 13 at 19:51












  • Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
    – hamza boulahia
    Nov 13 at 20:27














  • 1




    However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
    – Ingix
    Nov 13 at 19:51












  • Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
    – hamza boulahia
    Nov 13 at 20:27








1




1




However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
– Ingix
Nov 13 at 19:51






However this proves only that $2pi$ is a period, not that it is the smallest period. For example, while $sin(x)$ and $cos(x)$ have smallest period $2pi$, their product has smallest period $pi$.In this case, looking at the zeros of $f(x)$ yields that a smaller period would have to be $pi$, but we have $f(x+pi)=-f(x)$.
– Ingix
Nov 13 at 19:51














Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
– hamza boulahia
Nov 13 at 20:27




Yes that's right. to find the smallest period one must solve the equation $f(x+T)=f(x)$ for $T$.
– hamza boulahia
Nov 13 at 20:27


















 

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