Why the periodicity of solution for $theta''+gammatheta = 0$ implies $sqrt{gamma} = ninmathbb{N}$?
One solution for $$theta''+gammatheta = 0$$ is, for $gamma>0$,
$$Theta(theta) = Acos sqrt{gamma}theta + Bsin sqrt{gamma}theta$$
My book says that because of the $2pi$ periodicity of $theta$ we have that $sqrt{gamma} = ninmathbb{N}$
algebra-precalculus differential-equations trigonometry periodic-functions
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
add a comment |
One solution for $$theta''+gammatheta = 0$$ is, for $gamma>0$,
$$Theta(theta) = Acos sqrt{gamma}theta + Bsin sqrt{gamma}theta$$
My book says that because of the $2pi$ periodicity of $theta$ we have that $sqrt{gamma} = ninmathbb{N}$
algebra-precalculus differential-equations trigonometry periodic-functions
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53
add a comment |
One solution for $$theta''+gammatheta = 0$$ is, for $gamma>0$,
$$Theta(theta) = Acos sqrt{gamma}theta + Bsin sqrt{gamma}theta$$
My book says that because of the $2pi$ periodicity of $theta$ we have that $sqrt{gamma} = ninmathbb{N}$
algebra-precalculus differential-equations trigonometry periodic-functions
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
One solution for $$theta''+gammatheta = 0$$ is, for $gamma>0$,
$$Theta(theta) = Acos sqrt{gamma}theta + Bsin sqrt{gamma}theta$$
My book says that because of the $2pi$ periodicity of $theta$ we have that $sqrt{gamma} = ninmathbb{N}$
algebra-precalculus differential-equations trigonometry periodic-functions
algebra-precalculus differential-equations trigonometry periodic-functions
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
asked Nov 29 '18 at 19:47
PaprikaPaprika
61312
61312
Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53
add a comment |
Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53
Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53
Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53
add a comment |
1 Answer
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The minimal period of the solution is $frac{2pi}{sqrtγ}$. You want that $2pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$$
2pi=frac{2pi}{sqrtγ}cdot n.
$$
This is directly equivalent to $sqrtγ=n$.
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The minimal period of the solution is $frac{2pi}{sqrtγ}$. You want that $2pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$$
2pi=frac{2pi}{sqrtγ}cdot n.
$$
This is directly equivalent to $sqrtγ=n$.
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
add a comment |
The minimal period of the solution is $frac{2pi}{sqrtγ}$. You want that $2pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$$
2pi=frac{2pi}{sqrtγ}cdot n.
$$
This is directly equivalent to $sqrtγ=n$.
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
add a comment |
The minimal period of the solution is $frac{2pi}{sqrtγ}$. You want that $2pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$$
2pi=frac{2pi}{sqrtγ}cdot n.
$$
This is directly equivalent to $sqrtγ=n$.
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
The minimal period of the solution is $frac{2pi}{sqrtγ}$. You want that $2pi$ is also a period of the solution. Thus you need it to be an integer multiple of the minimal period,
$$
2pi=frac{2pi}{sqrtγ}cdot n.
$$
This is directly equivalent to $sqrtγ=n$.
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
answered Nov 29 '18 at 20:14
LutzLLutzL
56.5k42054
56.5k42054
add a comment |
add a comment |
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Τhis doesn't make a lot of sense, $gamma$ is just a positive constant, how can it be equal to a natural number ?
– Rebellos
Nov 29 '18 at 19:53