Linear approximation of non-linear equation.
$begingroup$
I am trying to understand what it means to make a linear approximation but will get more specific in a minutes. First, some context.
The problem I am working on is more of a physics but the question I have is a purely mathematical one (I think). I am trying to derive the linear equations of motion of a system. It is a 2 DOF, so two equations. I came up with:
$$
I_x*ddot{q_x} + K_1*q_x = -r*F_1*sin(q_x)+r*F_2*cos(q_y)
$$
and
$$
I_y*ddot{q_y} + K_2*q_y = -r*F_1*cos(q_x)-r*F_2*cos(q_y)
$$
From above, there are two variables, $q_x$ and $q_y$. You can ignore all other letters since I want to focus on the trigonometric parts, on the right side. The problem is that there are two non-linear elements, the cos(q) and the sin(q).
I have two questions now:
Question 1:
For the specific example (assuming that the equations are correct), the linear approximation of this system would be to use Taylor series, and as a result say $cos(q)=1$ and $sin(q)=q$ ? I have seens that many times, when using Taylor, for the first one, they would right $cos(q)=1-frac{q^2}{q!}$ because it is a better aproximation. But that is non-linear so it will not do for what I want to do, correct?
Question 2:
This is a more general question about linearization. I have heard some times, in the context of engineering, something along the lines of "this system is to non-linear for us to use a linear version of it". I especially remember that being said when having ln(). Does this basically mean that there is not good linear (Taylor) approximation for the system that would be useful?
ordinary-differential-equations taylor-expansion systems-of-equations dynamical-systems nonlinear-system
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
$endgroup$
add a comment |
$begingroup$
I am trying to understand what it means to make a linear approximation but will get more specific in a minutes. First, some context.
The problem I am working on is more of a physics but the question I have is a purely mathematical one (I think). I am trying to derive the linear equations of motion of a system. It is a 2 DOF, so two equations. I came up with:
$$
I_x*ddot{q_x} + K_1*q_x = -r*F_1*sin(q_x)+r*F_2*cos(q_y)
$$
and
$$
I_y*ddot{q_y} + K_2*q_y = -r*F_1*cos(q_x)-r*F_2*cos(q_y)
$$
From above, there are two variables, $q_x$ and $q_y$. You can ignore all other letters since I want to focus on the trigonometric parts, on the right side. The problem is that there are two non-linear elements, the cos(q) and the sin(q).
I have two questions now:
Question 1:
For the specific example (assuming that the equations are correct), the linear approximation of this system would be to use Taylor series, and as a result say $cos(q)=1$ and $sin(q)=q$ ? I have seens that many times, when using Taylor, for the first one, they would right $cos(q)=1-frac{q^2}{q!}$ because it is a better aproximation. But that is non-linear so it will not do for what I want to do, correct?
Question 2:
This is a more general question about linearization. I have heard some times, in the context of engineering, something along the lines of "this system is to non-linear for us to use a linear version of it". I especially remember that being said when having ln(). Does this basically mean that there is not good linear (Taylor) approximation for the system that would be useful?
ordinary-differential-equations taylor-expansion systems-of-equations dynamical-systems nonlinear-system
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
$endgroup$
1
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
1
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35
add a comment |
$begingroup$
I am trying to understand what it means to make a linear approximation but will get more specific in a minutes. First, some context.
The problem I am working on is more of a physics but the question I have is a purely mathematical one (I think). I am trying to derive the linear equations of motion of a system. It is a 2 DOF, so two equations. I came up with:
$$
I_x*ddot{q_x} + K_1*q_x = -r*F_1*sin(q_x)+r*F_2*cos(q_y)
$$
and
$$
I_y*ddot{q_y} + K_2*q_y = -r*F_1*cos(q_x)-r*F_2*cos(q_y)
$$
From above, there are two variables, $q_x$ and $q_y$. You can ignore all other letters since I want to focus on the trigonometric parts, on the right side. The problem is that there are two non-linear elements, the cos(q) and the sin(q).
I have two questions now:
Question 1:
For the specific example (assuming that the equations are correct), the linear approximation of this system would be to use Taylor series, and as a result say $cos(q)=1$ and $sin(q)=q$ ? I have seens that many times, when using Taylor, for the first one, they would right $cos(q)=1-frac{q^2}{q!}$ because it is a better aproximation. But that is non-linear so it will not do for what I want to do, correct?
Question 2:
This is a more general question about linearization. I have heard some times, in the context of engineering, something along the lines of "this system is to non-linear for us to use a linear version of it". I especially remember that being said when having ln(). Does this basically mean that there is not good linear (Taylor) approximation for the system that would be useful?
ordinary-differential-equations taylor-expansion systems-of-equations dynamical-systems nonlinear-system
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
$endgroup$
I am trying to understand what it means to make a linear approximation but will get more specific in a minutes. First, some context.
The problem I am working on is more of a physics but the question I have is a purely mathematical one (I think). I am trying to derive the linear equations of motion of a system. It is a 2 DOF, so two equations. I came up with:
$$
I_x*ddot{q_x} + K_1*q_x = -r*F_1*sin(q_x)+r*F_2*cos(q_y)
$$
and
$$
I_y*ddot{q_y} + K_2*q_y = -r*F_1*cos(q_x)-r*F_2*cos(q_y)
$$
From above, there are two variables, $q_x$ and $q_y$. You can ignore all other letters since I want to focus on the trigonometric parts, on the right side. The problem is that there are two non-linear elements, the cos(q) and the sin(q).
I have two questions now:
Question 1:
For the specific example (assuming that the equations are correct), the linear approximation of this system would be to use Taylor series, and as a result say $cos(q)=1$ and $sin(q)=q$ ? I have seens that many times, when using Taylor, for the first one, they would right $cos(q)=1-frac{q^2}{q!}$ because it is a better aproximation. But that is non-linear so it will not do for what I want to do, correct?
Question 2:
This is a more general question about linearization. I have heard some times, in the context of engineering, something along the lines of "this system is to non-linear for us to use a linear version of it". I especially remember that being said when having ln(). Does this basically mean that there is not good linear (Taylor) approximation for the system that would be useful?
ordinary-differential-equations taylor-expansion systems-of-equations dynamical-systems nonlinear-system
ordinary-differential-equations taylor-expansion systems-of-equations dynamical-systems nonlinear-system
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
asked Jan 9 at 19:13
Dimitris PantelisDimitris Pantelis
387
387
1
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
1
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35
add a comment |
1
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
1
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35
1
1
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
1
1
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35
add a comment |
0
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1
$begingroup$
It only makes sense, in general, to linearise about some trajectory (typically a stable equilibrium point).
$endgroup$
– copper.hat
Jan 9 at 19:18
1
$begingroup$
The answer is part physics. The well known linear equation for a pendulum works well when the amplitude of the swing is small. If it is large then the linear equation is not so good.
$endgroup$
– user121049
Jan 9 at 19:35