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Solutions of $cos(ax^c + bx) = 0$
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As per title, I would like to find the zeros of
$$ f(x) = cos(ax^c + bx)$$
where $0leq x leq K$, $a in mathbb R$, $b in mathbb R$, and $c in (0, 2]$.
I have that
$$
f(x) = 0 Leftrightarrow ax^c + bx = pi left(n - frac{1}{2} right)
$$
Now, since I am not aware of any analytical methods (except for the cases $c=1$, $c=2$, obviously) for solving equations like
$$
ax^c + bx - pi left(n - frac{1}{2} right) = 0
$$
the only method I can think of is to solve this last equation numerically for each choice of $n$ until the solutions reach outside $[0, K]$. Am I correct in assuming there is not a more efficient method? I have considered various things like series expansions, transformations of the variables etc, but without any luck.
roots
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up vote
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As per title, I would like to find the zeros of
$$ f(x) = cos(ax^c + bx)$$
where $0leq x leq K$, $a in mathbb R$, $b in mathbb R$, and $c in (0, 2]$.
I have that
$$
f(x) = 0 Leftrightarrow ax^c + bx = pi left(n - frac{1}{2} right)
$$
Now, since I am not aware of any analytical methods (except for the cases $c=1$, $c=2$, obviously) for solving equations like
$$
ax^c + bx - pi left(n - frac{1}{2} right) = 0
$$
the only method I can think of is to solve this last equation numerically for each choice of $n$ until the solutions reach outside $[0, K]$. Am I correct in assuming there is not a more efficient method? I have considered various things like series expansions, transformations of the variables etc, but without any luck.
roots
asked yesterday
Slug Pue
2,14911020
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
As per title, I would like to find the zeros of
$$ f(x) = cos(ax^c + bx)$$
where $0leq x leq K$, $a in mathbb R$, $b in mathbb R$, and $c in (0, 2]$.
I have that
$$
f(x) = 0 Leftrightarrow ax^c + bx = pi left(n - frac{1}{2} right)
$$
Now, since I am not aware of any analytical methods (except for the cases $c=1$, $c=2$, obviously) for solving equations like
$$
ax^c + bx - pi left(n - frac{1}{2} right) = 0
$$
the only method I can think of is to solve this last equation numerically for each choice of $n$ until the solutions reach outside $[0, K]$. Am I correct in assuming there is not a more efficient method? I have considered various things like series expansions, transformations of the variables etc, but without any luck.
roots
asked yesterday
Slug Pue
2,14911020
As per title, I would like to find the zeros of
$$ f(x) = cos(ax^c + bx)$$
where $0leq x leq K$, $a in mathbb R$, $b in mathbb R$, and $c in (0, 2]$.
I have that
$$
f(x) = 0 Leftrightarrow ax^c + bx = pi left(n - frac{1}{2} right)
$$
Now, since I am not aware of any analytical methods (except for the cases $c=1$, $c=2$, obviously) for solving equations like
$$
ax^c + bx - pi left(n - frac{1}{2} right) = 0
$$
the only method I can think of is to solve this last equation numerically for each choice of $n$ until the solutions reach outside $[0, K]$. Am I correct in assuming there is not a more efficient method? I have considered various things like series expansions, transformations of the variables etc, but without any luck.
roots
roots
asked yesterday
Slug Pue
2,14911020
asked yesterday
Slug Pue
2,14911020
edited yesterday
asked yesterday
Slug Pue
2,14911020
asked yesterday
Slug Pue
2,14911020
asked yesterday
Slug Pue
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2,14911020
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1 Answer
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I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=frac{1}{2},frac{1}{3},frac{1}{5}$. So,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. You will have an elementary solution. The same can be done for $c=frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.
Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.
answered 23 hours ago
Jon
4,31511022
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
I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=frac{1}{2},frac{1}{3},frac{1}{5}$. So,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. You will have an elementary solution. The same can be done for $c=frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.
Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.
answered 23 hours ago
Jon
4,31511022
add a comment |
up vote
1
down vote
I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=frac{1}{2},frac{1}{3},frac{1}{5}$. So,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. You will have an elementary solution. The same can be done for $c=frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.
Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.
answered 23 hours ago
Jon
4,31511022
add a comment |
up vote
1
down vote
up vote
1
down vote
I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=frac{1}{2},frac{1}{3},frac{1}{5}$. So,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. You will have an elementary solution. The same can be done for $c=frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.
Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.
answered 23 hours ago
Jon
4,31511022
I think that your strategy is a proper one. This can be seen by exploiting some simple cases beyond the trivial ones with $c=1,2$. E.g., let us consider the cases $c=frac{1}{2},frac{1}{3},frac{1}{5}$. So,
$$
asqrt{x}+bx-pileft(n-frac{1}{2}right)=0
$$
that can be solved with the substitution $x=y^2$ and $xge 0$. You will have an elementary solution. The same can be done for $c=frac{1}{3}$, provided you will use the Cardano formula. Things worsens for $c=frac{1}{5}$. Here you can only find a solution by numerical methods. In this case you will have to solve the equation
$$
by^5+ay-pileft(n-frac{1}{2}right)=0
$$
with no elementary solution.
Therefore, aside from very simple particular solutions, you will generally find elementarily unsolvable algebraic equations whose only solution is numerical.
answered 23 hours ago
Jon
4,31511022
answered 23 hours ago
Jon
4,31511022
answered 23 hours ago
Jon
4,31511022
answered 23 hours ago
Jon
4,31511022
4,31511022
add a comment |
add a comment |
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