Find minimum of $frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$
$begingroup$
The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of
$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?
I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?
functions optimization convex-optimization
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add a comment |
$begingroup$
The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of
$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?
I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?
functions optimization convex-optimization
$endgroup$
2
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Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19
add a comment |
$begingroup$
The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of
$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?
I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?
functions optimization convex-optimization
$endgroup$
The question is quite straight-forward, how does one find the minimum function (depending on $lambda$) of
$frac{x^2}{2}+x-lambda x$ when $x geq 1, lambda geq 0$?
I know the answer is $frac{3}{2} - lambda$ when $0 leq lambda leq 2$ and $-frac{(1-lambda)^2}{2}$ when $lambda geq 2$, but how does one compute that?
functions optimization convex-optimization
functions optimization convex-optimization
asked Dec 21 '18 at 3:18
ProShitposterProShitposter
827
827
2
$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19
add a comment |
2
$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19
2
2
$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19
$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.
$endgroup$
add a comment |
$begingroup$
I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
begin{align*}
f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
&= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
end{align*}
Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.
$endgroup$
add a comment |
$begingroup$
Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.
$endgroup$
add a comment |
$begingroup$
Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.
$endgroup$
Note that
$$
x^2/2 +(1-lambda)x = frac{1}{2}left((x+(1-lambda))^2 - (1-lambda)^2right).
$$
From here, I believe you can derive the answer you want.
answered Dec 21 '18 at 3:30
induction601induction601
1,285314
1,285314
add a comment |
add a comment |
$begingroup$
I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
begin{align*}
f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
&= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
end{align*}
Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.
$endgroup$
add a comment |
$begingroup$
I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
begin{align*}
f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
&= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
end{align*}
Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.
$endgroup$
add a comment |
$begingroup$
I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
begin{align*}
f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
&= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
end{align*}
Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.
$endgroup$
I would not give the full detail of the answer. But you can consider the following quadratic form of the equation :
Define $f(x,lambda) = frac{x^{2}}{2}+x - lambda x$ with constraints $xgeq1$ and $lambdageq0$. Now, arrange the function as follows :
begin{align*}
f(x,lambda) &= frac{1}{2}(x^{2} +2(1-lambda)x)\
&= frac{1}{2}bigg((x-(1-lambda))^{2}-(1-lambda)^{2}bigg)
end{align*}
Hence, $f$ attains minimum whenever $x=1-lambda$ (I leave the detail for you to figure out by yourself). The rest of the arguments might not be so difficult to follow on your own.
answered Dec 21 '18 at 3:36
Evan William ChandraEvan William Chandra
626313
626313
add a comment |
add a comment |
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2
$begingroup$
Derivative and set equal to zero.
$endgroup$
– Zachary Selk
Dec 21 '18 at 3:19