What is the minimum of an ordinary function including a given function? [on hold]
Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.
analysis
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Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.
analysis
asked yesterday
Fei Xu
11
New contributor
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday
add a comment |
Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.
analysis
asked yesterday
Fei Xu
11
New contributor
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let begin{align}g(lambda)=int_{Omega}(u(x)-lambda)^2dx,end{align} where $lambdainmathbb R$ and $uin L^2(Omega)$. Then to find the minimum of $g$.
analysis
analysis
asked yesterday
Fei Xu
11
New contributor
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
Fei Xu
11
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Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
asked yesterday
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asked yesterday
Fei Xu
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asked yesterday
Fei Xu
11
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Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, amWhy, hardmath, Eevee Trainer, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday
add a comment |
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday
add a comment |
1 Answer
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Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
answered yesterday
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
answered yesterday
Fei Xu
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Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
answered yesterday
Fei Xu
11
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Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
answered yesterday
Fei Xu
11
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Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Method one: By the definition of $g$, one can obtain that
begin{align}g(lambda)=|Omega|cdotlambda^2-2int_{Omega}udxcdotlambda+int_{Omega}u^2dx.end{align}
Hence begin{align}g_{min}=gleft(frac{1}{|Omega|}int_{Omega}udxright)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
Method two: Let $lambda_0inmathbb R$ be the minimal point. For any $epsilon>0$, consider the auxiliary function
begin{align}h(epsilon)=g(lambda_0+epsilon)=int_{Omega}(u-lambda_0-epsilon)^2dx.end{align}
Hence by the definition of $lambda_0$,
begin{align}0=frac{d}{depsilon}h(epsilon)|_{epsilon=0}=-2int_{Omega}u-lambda_0Longrightarrow lambda_0=frac{1}{|Omega|}int_{Omega}udx,end{align} and the minimum of $g$ is begin{align}g(lambda_0)=int_{Omega}u^2dx-frac{1}{|Omega|}left(int_{Omega}xdxright)^2.end{align}
answered yesterday
Fei Xu
11
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answered yesterday
Fei Xu
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answered yesterday
Fei Xu
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answered yesterday
Fei Xu
11
11
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Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Fei Xu is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Welcome to MSE. Please include what you tried too
– Ankit Kumar
yesterday
I'm not getting a full picture of your problem (setup and goal) from the current wording in the body of the Question. Perhaps the unknown to control for the minimum of $g$ is $lambda$, but there is scarcely enough information provided about $u$ or $Omega$ to make a reasoned mathematical argument. In many settings the minimum is achieved by taking $lambda$ to be the average value of $u(x)$, but the little information given does not justify that average value is meaningful.
– hardmath
yesterday