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 $2pi$?
real-analysis functions
asked Nov 13 at 17:19
Daniel Oscar
1047
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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 $2pi$?
real-analysis functions
asked Nov 13 at 17:19
Daniel Oscar
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
add a comment |
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
asked Nov 13 at 17:19
Daniel Oscar
1047
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
asked Nov 13 at 17:19
Daniel Oscar
1047
asked Nov 13 at 17:19
Daniel Oscar
1047
asked Nov 13 at 17:19
Daniel Oscar
1047
asked Nov 13 at 17:19
Daniel Oscar
1047
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
add a comment |
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
add a comment |
1 Answer
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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.
answered Nov 13 at 17:42
hamza boulahia
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered Nov 13 at 17:42
hamza boulahia
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
add a comment |
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.
answered Nov 13 at 17:42
hamza boulahia
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
add a comment |
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.
answered Nov 13 at 17:42
hamza boulahia
931319
$$ f(x)=cos(2x)sin(3x)=g(x)h(x)$$
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.
answered Nov 13 at 17:42
hamza boulahia
931319
answered Nov 13 at 17:42
hamza boulahia
931319
answered Nov 13 at 17:42
hamza boulahia
931319
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
add a comment |
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
add a comment |
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$f$ 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