Mixing
How to find the inverse of a function numerically
$begingroup$
This is an extension of my previous question posted in here Inverse of a function of a 3rd order
Now, I have another one which seems to be more complicated. I don't know how to solve them numerically. The function is as follow
$$z(zeta)=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
in which $z$ is a complex number and cannot be zero. $m_k$, $a$ and $b$ are constant.
How to solve this function for $zeta(z)$?
I am a Matlab user so I would appreciate if someone can refer to any built-in Matlab function that can be used.
Example
The following $z(zeta)$ is obtained using the following input:
$$a=-2.08$$
$$b = 4.08$$
$$m_1 = 0.5$$
$$m_2 = -0.03$$
$$zeta = costheta+isintheta$$ for $theta = [0,2pi]$
for
The results is:
z = [
2.9400 + 0.0000i
3.2277 + 2.1618i
2.2730 + 1.2986i
0.4557 - 2.2605i
-1.4094 - 6.6606i
-2.3950 - 9.7020i
-1.9857 -10.1025i
-0.4083 - 7.8642i
1.5321 - 3.9596i
2.9400 - 0.0000i
]
Thanks!
numerical-methods inverse-function
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
$endgroup$
add a comment |
$begingroup$
This is an extension of my previous question posted in here Inverse of a function of a 3rd order
Now, I have another one which seems to be more complicated. I don't know how to solve them numerically. The function is as follow
$$z(zeta)=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
in which $z$ is a complex number and cannot be zero. $m_k$, $a$ and $b$ are constant.
How to solve this function for $zeta(z)$?
I am a Matlab user so I would appreciate if someone can refer to any built-in Matlab function that can be used.
Example
The following $z(zeta)$ is obtained using the following input:
$$a=-2.08$$
$$b = 4.08$$
$$m_1 = 0.5$$
$$m_2 = -0.03$$
$$zeta = costheta+isintheta$$ for $theta = [0,2pi]$
for
The results is:
z = [
2.9400 + 0.0000i
3.2277 + 2.1618i
2.2730 + 1.2986i
0.4557 - 2.2605i
-1.4094 - 6.6606i
-2.3950 - 9.7020i
-1.9857 -10.1025i
-0.4083 - 7.8642i
1.5321 - 3.9596i
2.9400 - 0.0000i
]
Thanks!
numerical-methods inverse-function
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
$endgroup$
$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04
add a comment |
$begingroup$
This is an extension of my previous question posted in here Inverse of a function of a 3rd order
Now, I have another one which seems to be more complicated. I don't know how to solve them numerically. The function is as follow
$$z(zeta)=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
in which $z$ is a complex number and cannot be zero. $m_k$, $a$ and $b$ are constant.
How to solve this function for $zeta(z)$?
I am a Matlab user so I would appreciate if someone can refer to any built-in Matlab function that can be used.
Example
The following $z(zeta)$ is obtained using the following input:
$$a=-2.08$$
$$b = 4.08$$
$$m_1 = 0.5$$
$$m_2 = -0.03$$
$$zeta = costheta+isintheta$$ for $theta = [0,2pi]$
for
The results is:
z = [
2.9400 + 0.0000i
3.2277 + 2.1618i
2.2730 + 1.2986i
0.4557 - 2.2605i
-1.4094 - 6.6606i
-2.3950 - 9.7020i
-1.9857 -10.1025i
-0.4083 - 7.8642i
1.5321 - 3.9596i
2.9400 - 0.0000i
]
Thanks!
numerical-methods inverse-function
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
$endgroup$
This is an extension of my previous question posted in here Inverse of a function of a 3rd order
Now, I have another one which seems to be more complicated. I don't know how to solve them numerically. The function is as follow
$$z(zeta)=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
in which $z$ is a complex number and cannot be zero. $m_k$, $a$ and $b$ are constant.
How to solve this function for $zeta(z)$?
I am a Matlab user so I would appreciate if someone can refer to any built-in Matlab function that can be used.
Example
The following $z(zeta)$ is obtained using the following input:
$$a=-2.08$$
$$b = 4.08$$
$$m_1 = 0.5$$
$$m_2 = -0.03$$
$$zeta = costheta+isintheta$$ for $theta = [0,2pi]$
for
The results is:
z = [
2.9400 + 0.0000i
3.2277 + 2.1618i
2.2730 + 1.2986i
0.4557 - 2.2605i
-1.4094 - 6.6606i
-2.3950 - 9.7020i
-1.9857 -10.1025i
-0.4083 - 7.8642i
1.5321 - 3.9596i
2.9400 - 0.0000i
]
Thanks!
numerical-methods inverse-function
numerical-methods inverse-function
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
edited Jan 28 at 12:19
BeeTiau
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
asked Jan 28 at 11:26
BeeTiauBeeTiau
758
758
$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04
add a comment |
$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04
$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Scale your equation
$$z=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
by $zeta^2$ and obtain the equivalent form
$$z zeta^2=aleft(zeta+m_1zeta^3+m_2zeta^4 right)+bleft(zeta^3+m_1zeta+m_2 right),$$
which can be reordered into
$$ 0 = am_2 zeta^4+(am_1+b)zeta^3-zzeta^2+(bm_1+a)zeta+bm_2.$$
The right-hand side is a polynomial in $zeta$ of order at most $4$. MATLAB has a built-in function called roots which will compute all the roots of a polynomial by finding the eigenvalues of the companion matrix.
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
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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
$begingroup$
Scale your equation
$$z=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
by $zeta^2$ and obtain the equivalent form
$$z zeta^2=aleft(zeta+m_1zeta^3+m_2zeta^4 right)+bleft(zeta^3+m_1zeta+m_2 right),$$
which can be reordered into
$$ 0 = am_2 zeta^4+(am_1+b)zeta^3-zzeta^2+(bm_1+a)zeta+bm_2.$$
The right-hand side is a polynomial in $zeta$ of order at most $4$. MATLAB has a built-in function called roots which will compute all the roots of a polynomial by finding the eigenvalues of the companion matrix.
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
$endgroup$
add a comment |
$begingroup$
Scale your equation
$$z=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
by $zeta^2$ and obtain the equivalent form
$$z zeta^2=aleft(zeta+m_1zeta^3+m_2zeta^4 right)+bleft(zeta^3+m_1zeta+m_2 right),$$
which can be reordered into
$$ 0 = am_2 zeta^4+(am_1+b)zeta^3-zzeta^2+(bm_1+a)zeta+bm_2.$$
The right-hand side is a polynomial in $zeta$ of order at most $4$. MATLAB has a built-in function called roots which will compute all the roots of a polynomial by finding the eigenvalues of the companion matrix.
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
$endgroup$
add a comment |
$begingroup$
Scale your equation
$$z=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
by $zeta^2$ and obtain the equivalent form
$$z zeta^2=aleft(zeta+m_1zeta^3+m_2zeta^4 right)+bleft(zeta^3+m_1zeta+m_2 right),$$
which can be reordered into
$$ 0 = am_2 zeta^4+(am_1+b)zeta^3-zzeta^2+(bm_1+a)zeta+bm_2.$$
The right-hand side is a polynomial in $zeta$ of order at most $4$. MATLAB has a built-in function called roots which will compute all the roots of a polynomial by finding the eigenvalues of the companion matrix.
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
$endgroup$
Scale your equation
$$z=aleft(frac{1}{zeta}+m_1zeta+m_2zeta^2 right)+bleft(zeta+frac{m_1}{zeta}+frac{m_2}{zeta^2} right)$$
by $zeta^2$ and obtain the equivalent form
$$z zeta^2=aleft(zeta+m_1zeta^3+m_2zeta^4 right)+bleft(zeta^3+m_1zeta+m_2 right),$$
which can be reordered into
$$ 0 = am_2 zeta^4+(am_1+b)zeta^3-zzeta^2+(bm_1+a)zeta+bm_2.$$
The right-hand side is a polynomial in $zeta$ of order at most $4$. MATLAB has a built-in function called roots which will compute all the roots of a polynomial by finding the eigenvalues of the companion matrix.
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
answered Jan 28 at 21:09
Carl ChristianCarl Christian
5,9161723
5,9161723
add a comment |
add a comment |
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$begingroup$
It can be that there is no algebraic expression for the inverse. Does it suffice to have an algorithm that finds the input value, when given the output value $z$?
$endgroup$
– Matti P.
Jan 28 at 11:52
$begingroup$
Hi @MattiP. Yes. The input $zeta = e^{i theta}$ where $theta$ is from $0$ to $2pi$
$endgroup$
– BeeTiau
Jan 28 at 12:04