Mixing
Is $ x^3 $ convex , on positive x-axis
$begingroup$
Intuitively its epigraph looks so convex to me. But when I tried to prove it is convex by proving
$ f(x) ge f(y) +f'(y)(x-y) $ for $ x,y ge 0$
I've got
$ x^3 ge y^3 +6xy -6 y^2 $ which is wrong when $x$ is 0 and $y$ is 10.
Is it indeed not convex and I mistakenly thought its epigraph is convex?
functions
asked Feb 1 at 1:53
Anson NGAnson NG
23829
$endgroup$
|
show 1 more comment
$begingroup$
Intuitively its epigraph looks so convex to me. But when I tried to prove it is convex by proving
$ f(x) ge f(y) +f'(y)(x-y) $ for $ x,y ge 0$
I've got
$ x^3 ge y^3 +6xy -6 y^2 $ which is wrong when $x$ is 0 and $y$ is 10.
Is it indeed not convex and I mistakenly thought its epigraph is convex?
functions
asked Feb 1 at 1:53
Anson NGAnson NG
23829
$endgroup$
1
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23
|
show 1 more comment
$begingroup$
Intuitively its epigraph looks so convex to me. But when I tried to prove it is convex by proving
$ f(x) ge f(y) +f'(y)(x-y) $ for $ x,y ge 0$
I've got
$ x^3 ge y^3 +6xy -6 y^2 $ which is wrong when $x$ is 0 and $y$ is 10.
Is it indeed not convex and I mistakenly thought its epigraph is convex?
functions
asked Feb 1 at 1:53
Anson NGAnson NG
23829
$endgroup$
Intuitively its epigraph looks so convex to me. But when I tried to prove it is convex by proving
$ f(x) ge f(y) +f'(y)(x-y) $ for $ x,y ge 0$
I've got
$ x^3 ge y^3 +6xy -6 y^2 $ which is wrong when $x$ is 0 and $y$ is 10.
Is it indeed not convex and I mistakenly thought its epigraph is convex?
functions
functions
asked Feb 1 at 1:53
Anson NGAnson NG
23829
asked Feb 1 at 1:53
Anson NGAnson NG
23829
asked Feb 1 at 1:53
Anson NGAnson NG
23829
asked Feb 1 at 1:53
Anson NGAnson NG
23829
asked Feb 1 at 1:53
Anson NGAnson NG
23829
23829
1
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23
|
show 1 more comment
1
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23
1
1
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
To start, the actual definition of convexity is as follows:
$$fleft(tx+(1-t)yright)leq tf(x)+(1-t)f(y)$$
A better way, as suggested by Theo Bendit, is to prove $f''(x)geq0$.
answered Feb 1 at 1:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
add a comment |
$begingroup$
If you'd really like to use the definition directly, try showing that
$$tx^3+uy^3-(tx+uy)^3=tu(x-y)^2(t(2x+y)+u(x+2y))$$
whenever $t+u=1$.
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
To start, the actual definition of convexity is as follows:
$$fleft(tx+(1-t)yright)leq tf(x)+(1-t)f(y)$$
A better way, as suggested by Theo Bendit, is to prove $f''(x)geq0$.
answered Feb 1 at 1:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
add a comment |
$begingroup$
To start, the actual definition of convexity is as follows:
$$fleft(tx+(1-t)yright)leq tf(x)+(1-t)f(y)$$
A better way, as suggested by Theo Bendit, is to prove $f''(x)geq0$.
answered Feb 1 at 1:55
obscuransobscurans
1,152311
$endgroup$
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
add a comment |
$begingroup$
To start, the actual definition of convexity is as follows:
$$fleft(tx+(1-t)yright)leq tf(x)+(1-t)f(y)$$
A better way, as suggested by Theo Bendit, is to prove $f''(x)geq0$.
answered Feb 1 at 1:55
obscuransobscurans
1,152311
$endgroup$
To start, the actual definition of convexity is as follows:
$$fleft(tx+(1-t)yright)leq tf(x)+(1-t)f(y)$$
A better way, as suggested by Theo Bendit, is to prove $f''(x)geq0$.
answered Feb 1 at 1:55
obscuransobscurans
1,152311
edited Feb 1 at 2:14
answered Feb 1 at 1:55
obscuransobscurans
1,152311
answered Feb 1 at 1:55
obscuransobscurans
1,152311
answered Feb 1 at 1:55
obscuransobscurans
1,152311
1,152311
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
add a comment |
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
$begingroup$
I started with the definition but could not prove it. Too many terms are there...
$endgroup$
– Anson NG
Feb 1 at 2:07
add a comment |
$begingroup$
If you'd really like to use the definition directly, try showing that
$$tx^3+uy^3-(tx+uy)^3=tu(x-y)^2(t(2x+y)+u(x+2y))$$
whenever $t+u=1$.
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
$endgroup$
add a comment |
$begingroup$
If you'd really like to use the definition directly, try showing that
$$tx^3+uy^3-(tx+uy)^3=tu(x-y)^2(t(2x+y)+u(x+2y))$$
whenever $t+u=1$.
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
$endgroup$
add a comment |
$begingroup$
If you'd really like to use the definition directly, try showing that
$$tx^3+uy^3-(tx+uy)^3=tu(x-y)^2(t(2x+y)+u(x+2y))$$
whenever $t+u=1$.
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
$endgroup$
If you'd really like to use the definition directly, try showing that
$$tx^3+uy^3-(tx+uy)^3=tu(x-y)^2(t(2x+y)+u(x+2y))$$
whenever $t+u=1$.
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
answered Feb 1 at 2:29
K B DaveK B Dave
3,687317
3,687317
add a comment |
add a comment |
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1
$begingroup$
It is convex; try showing $f''(x) ge 0$ for $x ge 0$.
$endgroup$
– Theo Bendit
Feb 1 at 1:56
$begingroup$
Do you think $f'(y)=6y$? Isn't it $f'(x)=3y^2$ ?
$endgroup$
– kimchi lover
Feb 1 at 1:57
$begingroup$
You are right, I made a mistake.
$endgroup$
– Anson NG
Feb 1 at 2:04
$begingroup$
But then I will get, $ 2y^3+x^3 ge 3xy^2 $. Is it indeed true for all$ x, y, > 0 $ ?
$endgroup$
– Anson NG
Feb 1 at 2:05
$begingroup$
You can check this inequality: $x^3+2y^3-3xy^2=(x-y)^2(x+2y)ge 0$ for $x,yge0$.
$endgroup$
– kimchi lover
Feb 1 at 2:23