Mixing
Solutions to an Equation
$begingroup$
I was wondering how to find the real solutions to this equation algebraically:
$$0=left(sinpi xright)^2+sinleft(frac{n}{x}piright)^2$$ if $n$ is known.
functions
asked Jan 30 at 20:51
JamesJames
218
$endgroup$
add a comment |
$begingroup$
I was wondering how to find the real solutions to this equation algebraically:
$$0=left(sinpi xright)^2+sinleft(frac{n}{x}piright)^2$$ if $n$ is known.
functions
asked Jan 30 at 20:51
JamesJames
218
$endgroup$
$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:14
$begingroup$
Yes, I realized that but how do I solve for $x$?
$endgroup$
– James
Jan 30 at 21:15
$begingroup$
It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:19
$begingroup$
The point of the equation is that I don't have to factor $n$.
$endgroup$
– James
Jan 30 at 21:20
add a comment |
$begingroup$
I was wondering how to find the real solutions to this equation algebraically:
$$0=left(sinpi xright)^2+sinleft(frac{n}{x}piright)^2$$ if $n$ is known.
functions
asked Jan 30 at 20:51
JamesJames
218
$endgroup$
I was wondering how to find the real solutions to this equation algebraically:
$$0=left(sinpi xright)^2+sinleft(frac{n}{x}piright)^2$$ if $n$ is known.
functions
functions
asked Jan 30 at 20:51
JamesJames
218
asked Jan 30 at 20:51
JamesJames
218
asked Jan 30 at 20:51
JamesJames
218
asked Jan 30 at 20:51
JamesJames
218
asked Jan 30 at 20:51
JamesJames
218
218
$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:14
$begingroup$
Yes, I realized that but how do I solve for $x$?
$endgroup$
– James
Jan 30 at 21:15
$begingroup$
It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:19
$begingroup$
The point of the equation is that I don't have to factor $n$.
$endgroup$
– James
Jan 30 at 21:20
add a comment |
$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:14
$begingroup$
Yes, I realized that but how do I solve for $x$?
$endgroup$
– James
Jan 30 at 21:15
$begingroup$
It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:19
$begingroup$
The point of the equation is that I don't have to factor $n$.
$endgroup$
– James
Jan 30 at 21:20
$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:14
$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:14
$begingroup$
Yes, I realized that but how do I solve for $x$?
$endgroup$
– James
Jan 30 at 21:15
$begingroup$
Yes, I realized that but how do I solve for $x$?
$endgroup$
– James
Jan 30 at 21:15
$begingroup$
It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:19
$begingroup$
It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
$endgroup$
– Spasoje Durovic
Jan 30 at 21:19
$begingroup$
The point of the equation is that I don't have to factor $n$.
$endgroup$
– James
Jan 30 at 21:20
$begingroup$
The point of the equation is that I don't have to factor $n$.
$endgroup$
– James
Jan 30 at 21:20
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You can write: $$(sin πx)^2>=0$$ and $$(sin frac{n}{x}π)^2>=0$$
and when you add them up, you have:
$$(sin πx)^2 + (sin frac{n}{x}π)^2>=0$$
In the case of solving your equation, equality in the third inequality takes place - which means that equality takes place in the first two inequalities too (suppose this were not true, one of the squares would be greater than 0, which would mean that $(sin πx)^2 + (sin frac{n}{x}π)^2>0$, contradicting the equation to solve)
Therefore, you have $(sin πx)=0$ and $(sin frac{n}{x}π)=0$
$(sin πx)=0$ is equivalent to $πx=πk, k ∈ mathbb{Z} $ , which means $x ∈ mathbb{Z}$
$(sin frac{n}{x}π)=0$ is equivalent to $frac{n}{x}π = πk, k ∈ mathbb{Z} $, which means: $x=n/k, k ∈ mathbb{Z} $
Therefore, x is a whole number which satisfies $x=n/k$, where $n ∈ mathbb{N} $ (I assume from the notation) and $ k ∈ mathbb{Z} $, which means x is a divisor of n.
So, if we let $S $= set of solutions of given equation, $Ssubseteq D_n$, the set of the (both positive and negative) divisors of n.
We can check that indeed every divisor of n verifies the given equation, therefore $D_n subseteq S$, which, along with earlier-proved $Ssubseteq D_n$, gives us $S = D_n$
The set of solutions for this equation is the set of both negative and positive divisors of n.
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
$endgroup$
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
$begingroup$
You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
$endgroup$
– Parallelism Alert
Jan 30 at 21:30
$begingroup$
Hmmm, I am confused. How does this help?
$endgroup$
– James
Jan 31 at 0:26
$begingroup$
Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
$endgroup$
– James
Jan 31 at 0:32
$begingroup$
Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
$endgroup$
– Parallelism Alert
Jan 31 at 14:42
|
show 4 more comments
$begingroup$
Because of the left side, $(sin(xpi))^2=0 $ if and only if x is an integer.
Same goes for the right side, so $frac{n}{x}$ must be an integer.
for every i that satisfy $i|n$ both of them will work. so for every $x$ that $x|n$ the equation is true
answered Jan 30 at 20:55
ShaqShaq
3049
$endgroup$
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
1
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
|
show 4 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You can write: $$(sin πx)^2>=0$$ and $$(sin frac{n}{x}π)^2>=0$$
and when you add them up, you have:
$$(sin πx)^2 + (sin frac{n}{x}π)^2>=0$$
In the case of solving your equation, equality in the third inequality takes place - which means that equality takes place in the first two inequalities too (suppose this were not true, one of the squares would be greater than 0, which would mean that $(sin πx)^2 + (sin frac{n}{x}π)^2>0$, contradicting the equation to solve)
Therefore, you have $(sin πx)=0$ and $(sin frac{n}{x}π)=0$
$(sin πx)=0$ is equivalent to $πx=πk, k ∈ mathbb{Z} $ , which means $x ∈ mathbb{Z}$
$(sin frac{n}{x}π)=0$ is equivalent to $frac{n}{x}π = πk, k ∈ mathbb{Z} $, which means: $x=n/k, k ∈ mathbb{Z} $
Therefore, x is a whole number which satisfies $x=n/k$, where $n ∈ mathbb{N} $ (I assume from the notation) and $ k ∈ mathbb{Z} $, which means x is a divisor of n.
So, if we let $S $= set of solutions of given equation, $Ssubseteq D_n$, the set of the (both positive and negative) divisors of n.
We can check that indeed every divisor of n verifies the given equation, therefore $D_n subseteq S$, which, along with earlier-proved $Ssubseteq D_n$, gives us $S = D_n$
The set of solutions for this equation is the set of both negative and positive divisors of n.
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
$endgroup$
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
$begingroup$
You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
$endgroup$
– Parallelism Alert
Jan 30 at 21:30
$begingroup$
Hmmm, I am confused. How does this help?
$endgroup$
– James
Jan 31 at 0:26
$begingroup$
Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
$endgroup$
– James
Jan 31 at 0:32
$begingroup$
Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
$endgroup$
– Parallelism Alert
Jan 31 at 14:42
|
show 4 more comments
$begingroup$
You can write: $$(sin πx)^2>=0$$ and $$(sin frac{n}{x}π)^2>=0$$
and when you add them up, you have:
$$(sin πx)^2 + (sin frac{n}{x}π)^2>=0$$
In the case of solving your equation, equality in the third inequality takes place - which means that equality takes place in the first two inequalities too (suppose this were not true, one of the squares would be greater than 0, which would mean that $(sin πx)^2 + (sin frac{n}{x}π)^2>0$, contradicting the equation to solve)
Therefore, you have $(sin πx)=0$ and $(sin frac{n}{x}π)=0$
$(sin πx)=0$ is equivalent to $πx=πk, k ∈ mathbb{Z} $ , which means $x ∈ mathbb{Z}$
$(sin frac{n}{x}π)=0$ is equivalent to $frac{n}{x}π = πk, k ∈ mathbb{Z} $, which means: $x=n/k, k ∈ mathbb{Z} $
Therefore, x is a whole number which satisfies $x=n/k$, where $n ∈ mathbb{N} $ (I assume from the notation) and $ k ∈ mathbb{Z} $, which means x is a divisor of n.
So, if we let $S $= set of solutions of given equation, $Ssubseteq D_n$, the set of the (both positive and negative) divisors of n.
We can check that indeed every divisor of n verifies the given equation, therefore $D_n subseteq S$, which, along with earlier-proved $Ssubseteq D_n$, gives us $S = D_n$
The set of solutions for this equation is the set of both negative and positive divisors of n.
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
$endgroup$
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
$begingroup$
You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
$endgroup$
– Parallelism Alert
Jan 30 at 21:30
$begingroup$
Hmmm, I am confused. How does this help?
$endgroup$
– James
Jan 31 at 0:26
$begingroup$
Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
$endgroup$
– James
Jan 31 at 0:32
$begingroup$
Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
$endgroup$
– Parallelism Alert
Jan 31 at 14:42
|
show 4 more comments
$begingroup$
You can write: $$(sin πx)^2>=0$$ and $$(sin frac{n}{x}π)^2>=0$$
and when you add them up, you have:
$$(sin πx)^2 + (sin frac{n}{x}π)^2>=0$$
In the case of solving your equation, equality in the third inequality takes place - which means that equality takes place in the first two inequalities too (suppose this were not true, one of the squares would be greater than 0, which would mean that $(sin πx)^2 + (sin frac{n}{x}π)^2>0$, contradicting the equation to solve)
Therefore, you have $(sin πx)=0$ and $(sin frac{n}{x}π)=0$
$(sin πx)=0$ is equivalent to $πx=πk, k ∈ mathbb{Z} $ , which means $x ∈ mathbb{Z}$
$(sin frac{n}{x}π)=0$ is equivalent to $frac{n}{x}π = πk, k ∈ mathbb{Z} $, which means: $x=n/k, k ∈ mathbb{Z} $
Therefore, x is a whole number which satisfies $x=n/k$, where $n ∈ mathbb{N} $ (I assume from the notation) and $ k ∈ mathbb{Z} $, which means x is a divisor of n.
So, if we let $S $= set of solutions of given equation, $Ssubseteq D_n$, the set of the (both positive and negative) divisors of n.
We can check that indeed every divisor of n verifies the given equation, therefore $D_n subseteq S$, which, along with earlier-proved $Ssubseteq D_n$, gives us $S = D_n$
The set of solutions for this equation is the set of both negative and positive divisors of n.
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
$endgroup$
You can write: $$(sin πx)^2>=0$$ and $$(sin frac{n}{x}π)^2>=0$$
and when you add them up, you have:
$$(sin πx)^2 + (sin frac{n}{x}π)^2>=0$$
In the case of solving your equation, equality in the third inequality takes place - which means that equality takes place in the first two inequalities too (suppose this were not true, one of the squares would be greater than 0, which would mean that $(sin πx)^2 + (sin frac{n}{x}π)^2>0$, contradicting the equation to solve)
Therefore, you have $(sin πx)=0$ and $(sin frac{n}{x}π)=0$
$(sin πx)=0$ is equivalent to $πx=πk, k ∈ mathbb{Z} $ , which means $x ∈ mathbb{Z}$
$(sin frac{n}{x}π)=0$ is equivalent to $frac{n}{x}π = πk, k ∈ mathbb{Z} $, which means: $x=n/k, k ∈ mathbb{Z} $
Therefore, x is a whole number which satisfies $x=n/k$, where $n ∈ mathbb{N} $ (I assume from the notation) and $ k ∈ mathbb{Z} $, which means x is a divisor of n.
So, if we let $S $= set of solutions of given equation, $Ssubseteq D_n$, the set of the (both positive and negative) divisors of n.
We can check that indeed every divisor of n verifies the given equation, therefore $D_n subseteq S$, which, along with earlier-proved $Ssubseteq D_n$, gives us $S = D_n$
The set of solutions for this equation is the set of both negative and positive divisors of n.
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
edited Jan 30 at 21:24
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
answered Jan 30 at 21:14
Parallelism AlertParallelism Alert
3027
3027
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
$begingroup$
You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
$endgroup$
– Parallelism Alert
Jan 30 at 21:30
$begingroup$
Hmmm, I am confused. How does this help?
$endgroup$
– James
Jan 31 at 0:26
$begingroup$
Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
$endgroup$
– James
Jan 31 at 0:32
$begingroup$
Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
$endgroup$
– Parallelism Alert
Jan 31 at 14:42
|
show 4 more comments
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
$begingroup$
You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
$endgroup$
– Parallelism Alert
Jan 30 at 21:30
$begingroup$
Hmmm, I am confused. How does this help?
$endgroup$
– James
Jan 31 at 0:26
$begingroup$
Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
$endgroup$
– James
Jan 31 at 0:32
$begingroup$
Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
$endgroup$
– Parallelism Alert
Jan 31 at 14:42
$begingroup$
Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
$endgroup$
– James
Jan 30 at 21:16
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Yes, I understand that, but how do I solve for $x$ by transforming the whole formula, not just parts.
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– James
Jan 30 at 21:16
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You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
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– Parallelism Alert
Jan 30 at 21:30
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You could denote $a=sin (πx)$ and $b= sin (frac{n}{x}pi)$, which would reduce your equation to $a^2+b^2=0$, equivalent to $(a+b)^2-2ab=0$. Using from here $$sin(m)+sin(n)=2sin(frac{m+n}{2})cos(frac{m-n}{2})$$ and $$sin(m)*sin(n)=frac{cos(m-n)-cos(m+n)}{2}$$ might help.
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– Parallelism Alert
Jan 30 at 21:30
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Hmmm, I am confused. How does this help?
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– James
Jan 31 at 0:26
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Hmmm, I am confused. How does this help?
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– James
Jan 31 at 0:26
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Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
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– James
Jan 31 at 0:32
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Because after that I get this: $$cosleft(2pi xright)+cosleft(frac{2npi}{x}right)-2$$ which doesn't help much.
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– James
Jan 31 at 0:32
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Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
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– Parallelism Alert
Jan 31 at 14:42
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Factorize $cos(2πx)+cos(frac{2nπ}{x})$ as $2cos(frac{2πx+frac{2nπ}{x}}{2}) cos(frac{2πx-frac{2nπ}{x}}{2})$ which is clearly, less (or equal) than 2, with equality when both terms are either -1 or 1. Yet I believe this still isn't 'algebraic' enough, isn't it?
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– Parallelism Alert
Jan 31 at 14:42
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show 4 more comments
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Because of the left side, $(sin(xpi))^2=0 $ if and only if x is an integer.
Same goes for the right side, so $frac{n}{x}$ must be an integer.
for every i that satisfy $i|n$ both of them will work. so for every $x$ that $x|n$ the equation is true
answered Jan 30 at 20:55
ShaqShaq
3049
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But how can i get the zeroes algebraically.
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– James
Jan 30 at 20:56
1
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What do you mean by "algebrically" ?
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– Shaq
Jan 30 at 20:57
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I mean can you solve for x by transforming the equation.
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– James
Jan 30 at 20:58
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You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
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– Shaq
Jan 30 at 20:58
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Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
|
show 4 more comments
$begingroup$
Because of the left side, $(sin(xpi))^2=0 $ if and only if x is an integer.
Same goes for the right side, so $frac{n}{x}$ must be an integer.
for every i that satisfy $i|n$ both of them will work. so for every $x$ that $x|n$ the equation is true
answered Jan 30 at 20:55
ShaqShaq
3049
$endgroup$
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
1
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
|
show 4 more comments
$begingroup$
Because of the left side, $(sin(xpi))^2=0 $ if and only if x is an integer.
Same goes for the right side, so $frac{n}{x}$ must be an integer.
for every i that satisfy $i|n$ both of them will work. so for every $x$ that $x|n$ the equation is true
answered Jan 30 at 20:55
ShaqShaq
3049
$endgroup$
Because of the left side, $(sin(xpi))^2=0 $ if and only if x is an integer.
Same goes for the right side, so $frac{n}{x}$ must be an integer.
for every i that satisfy $i|n$ both of them will work. so for every $x$ that $x|n$ the equation is true
answered Jan 30 at 20:55
ShaqShaq
3049
answered Jan 30 at 20:55
ShaqShaq
3049
answered Jan 30 at 20:55
ShaqShaq
3049
answered Jan 30 at 20:55
ShaqShaq
3049
3049
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
1
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
|
show 4 more comments
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
1
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
$begingroup$
But how can i get the zeroes algebraically.
$endgroup$
– James
Jan 30 at 20:56
1
1
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
What do you mean by "algebrically" ?
$endgroup$
– Shaq
Jan 30 at 20:57
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
I mean can you solve for x by transforming the equation.
$endgroup$
– James
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
You can take the devirative. 0 is to lowest point of the function so you can look where the devirative is zero and this lead you to the minimum points, this is what you mean?
$endgroup$
– Shaq
Jan 30 at 20:58
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
$begingroup$
Yes, but how do you find the derivative.
$endgroup$
– James
Jan 30 at 21:00
|
show 4 more comments
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$begingroup$
You should note that you have two non-negative quantities being equal to zero, that means that they both must be zero, the roots of $sin x$ are $0+2kpi$ or $pi +2kpi$ with $k$ being just an integer (the $2kpi$ is there just because $sin$ is periodic with period $2pi$)... That being said, given a $n$ your solutions would be all the $x$ that divide $n$, or in a more "compact" way, the equation holds $forall x$ s.t. $x|n$
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– Spasoje Durovic
Jan 30 at 21:14
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Yes, I realized that but how do I solve for $x$?
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– James
Jan 30 at 21:15
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It's very simple, a way would be to factor $n$ in it's prime power decomposition, then with that you can check all the divisors of $n$, let me do an example to clear things out, consider $n=6$, it's prime power decomposition is: $6=2 cdot 3$ that means that it's divisors are $pm 1, pm 2, pm 3$ and $pm 6$, therefore your solutions are: $x=pm 1, pm 2, pm 3, pm 6$
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– Spasoje Durovic
Jan 30 at 21:19
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The point of the equation is that I don't have to factor $n$.
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– James
Jan 30 at 21:20