An inequality with constraints.
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I came across a result in a control theory book (without proof), which states that:
Given two variables $x,z in mathbb{R}$ and four parameters $c_{1}, c_{2}, k_{1}, k_{2}$ with $c_{1}, c_{2} > 0$. A function $f(x,z)$ is defined as
begin{align}
f(x,z) &= -c_{1}x^{2} - x^{4} + z[x-k_{1}z - k_{2}x^{2}z + (c_{1}-sin x)(-c_{1}x - x^{3} + z)]
end{align}
The book reads that if we choose $k_{1} > c_{2} + c_{1} + 1 + dfrac{(c_{1}^{2}+c_{1}+1)^{2}}{2c_{1}}$ and $k_{2} ge dfrac{(c_{1}+1)^2}{4}$, then $f(x,z) le -dfrac{1}{2}c_{1}x^{2} - c_{2}z^{2}$.
I am trying to prove such result, but no luck. Now, I am trying to expand every term of $f(x,z)$ to apply the AM-GM inequality, but it does not go anywhere. Do you have any suggestion/hint on this problem? Thanks a lot!
analysis inequality control-theory
edited 12 hours ago
user10354138
6,124523
asked 13 hours ago
lyhuehue01
133
add a comment |
up vote
0
down vote
favorite
I came across a result in a control theory book (without proof), which states that:
Given two variables $x,z in mathbb{R}$ and four parameters $c_{1}, c_{2}, k_{1}, k_{2}$ with $c_{1}, c_{2} > 0$. A function $f(x,z)$ is defined as
begin{align}
f(x,z) &= -c_{1}x^{2} - x^{4} + z[x-k_{1}z - k_{2}x^{2}z + (c_{1}-sin x)(-c_{1}x - x^{3} + z)]
end{align}
The book reads that if we choose $k_{1} > c_{2} + c_{1} + 1 + dfrac{(c_{1}^{2}+c_{1}+1)^{2}}{2c_{1}}$ and $k_{2} ge dfrac{(c_{1}+1)^2}{4}$, then $f(x,z) le -dfrac{1}{2}c_{1}x^{2} - c_{2}z^{2}$.
I am trying to prove such result, but no luck. Now, I am trying to expand every term of $f(x,z)$ to apply the AM-GM inequality, but it does not go anywhere. Do you have any suggestion/hint on this problem? Thanks a lot!
analysis inequality control-theory
edited 12 hours ago
user10354138
6,124523
asked 13 hours ago
lyhuehue01
133
1
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I came across a result in a control theory book (without proof), which states that:
Given two variables $x,z in mathbb{R}$ and four parameters $c_{1}, c_{2}, k_{1}, k_{2}$ with $c_{1}, c_{2} > 0$. A function $f(x,z)$ is defined as
begin{align}
f(x,z) &= -c_{1}x^{2} - x^{4} + z[x-k_{1}z - k_{2}x^{2}z + (c_{1}-sin x)(-c_{1}x - x^{3} + z)]
end{align}
The book reads that if we choose $k_{1} > c_{2} + c_{1} + 1 + dfrac{(c_{1}^{2}+c_{1}+1)^{2}}{2c_{1}}$ and $k_{2} ge dfrac{(c_{1}+1)^2}{4}$, then $f(x,z) le -dfrac{1}{2}c_{1}x^{2} - c_{2}z^{2}$.
I am trying to prove such result, but no luck. Now, I am trying to expand every term of $f(x,z)$ to apply the AM-GM inequality, but it does not go anywhere. Do you have any suggestion/hint on this problem? Thanks a lot!
analysis inequality control-theory
edited 12 hours ago
user10354138
6,124523
asked 13 hours ago
lyhuehue01
133
I came across a result in a control theory book (without proof), which states that:
Given two variables $x,z in mathbb{R}$ and four parameters $c_{1}, c_{2}, k_{1}, k_{2}$ with $c_{1}, c_{2} > 0$. A function $f(x,z)$ is defined as
begin{align}
f(x,z) &= -c_{1}x^{2} - x^{4} + z[x-k_{1}z - k_{2}x^{2}z + (c_{1}-sin x)(-c_{1}x - x^{3} + z)]
end{align}
The book reads that if we choose $k_{1} > c_{2} + c_{1} + 1 + dfrac{(c_{1}^{2}+c_{1}+1)^{2}}{2c_{1}}$ and $k_{2} ge dfrac{(c_{1}+1)^2}{4}$, then $f(x,z) le -dfrac{1}{2}c_{1}x^{2} - c_{2}z^{2}$.
I am trying to prove such result, but no luck. Now, I am trying to expand every term of $f(x,z)$ to apply the AM-GM inequality, but it does not go anywhere. Do you have any suggestion/hint on this problem? Thanks a lot!
analysis inequality control-theory
analysis inequality control-theory
edited 12 hours ago
user10354138
6,124523
asked 13 hours ago
lyhuehue01
133
edited 12 hours ago
user10354138
6,124523
asked 13 hours ago
lyhuehue01
133
edited 12 hours ago
user10354138
6,124523
edited 12 hours ago
user10354138
6,124523
edited 12 hours ago
user10354138
6,124523
6,124523
asked 13 hours ago
lyhuehue01
133
asked 13 hours ago
lyhuehue01
133
asked 13 hours ago
lyhuehue01
133
133
1
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago
add a comment |
1
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago
1
1
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago
add a comment |
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1
I noticed that in the function definition, no $c_2$ arises, so it seems strange that the result should depend on $c_2$. Could you please check?
– Andreas
12 hours ago
Thanks @Andreas. I have checked it and it was correct. Although $c_{2}$ is not explicitly present in $f(x,z)$ but it is included in $k_{1}$, which is inside $f(x,z)$. Therefore, the result should depend on $c_{2}$.
– lyhuehue01
12 hours ago