Mixing
Find $f$ on $[1,a]$ such that $f ge xlog x$, $f$ is strictly convex, analytic, and touches $x log x$ only $n$...
$begingroup$
I am looking for a function $f$ defined on $[1,a]$ that satisfies the following:
$f(x) ge x log x$ on $[1,a]$
$f(x)$ is strictly convex
$f(x)$ is analytic on $[1,a]$.
$f(x)-xlog(x)=0$ only $n$ times on $[1,a]$.
My attempt was to construct $f$ as follows:
$f(x)=xlog(x)+g(x)$
where $g(x)$ would have oscillatory behavior (e.g., $cos(x)$). However, every time I try to add oscillations I violate strict convexity condition.
real-analysis calculus convex-analysis
asked Jan 15 at 15:08
BobyBoby
1,0181929
$endgroup$
add a comment |
$begingroup$
I am looking for a function $f$ defined on $[1,a]$ that satisfies the following:
$f(x) ge x log x$ on $[1,a]$
$f(x)$ is strictly convex
$f(x)$ is analytic on $[1,a]$.
$f(x)-xlog(x)=0$ only $n$ times on $[1,a]$.
My attempt was to construct $f$ as follows:
$f(x)=xlog(x)+g(x)$
where $g(x)$ would have oscillatory behavior (e.g., $cos(x)$). However, every time I try to add oscillations I violate strict convexity condition.
real-analysis calculus convex-analysis
asked Jan 15 at 15:08
BobyBoby
1,0181929
$endgroup$
$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04
add a comment |
$begingroup$
I am looking for a function $f$ defined on $[1,a]$ that satisfies the following:
$f(x) ge x log x$ on $[1,a]$
$f(x)$ is strictly convex
$f(x)$ is analytic on $[1,a]$.
$f(x)-xlog(x)=0$ only $n$ times on $[1,a]$.
My attempt was to construct $f$ as follows:
$f(x)=xlog(x)+g(x)$
where $g(x)$ would have oscillatory behavior (e.g., $cos(x)$). However, every time I try to add oscillations I violate strict convexity condition.
real-analysis calculus convex-analysis
asked Jan 15 at 15:08
BobyBoby
1,0181929
$endgroup$
I am looking for a function $f$ defined on $[1,a]$ that satisfies the following:
$f(x) ge x log x$ on $[1,a]$
$f(x)$ is strictly convex
$f(x)$ is analytic on $[1,a]$.
$f(x)-xlog(x)=0$ only $n$ times on $[1,a]$.
My attempt was to construct $f$ as follows:
$f(x)=xlog(x)+g(x)$
where $g(x)$ would have oscillatory behavior (e.g., $cos(x)$). However, every time I try to add oscillations I violate strict convexity condition.
real-analysis calculus convex-analysis
real-analysis calculus convex-analysis
asked Jan 15 at 15:08
BobyBoby
1,0181929
asked Jan 15 at 15:08
BobyBoby
1,0181929
edited Jan 15 at 17:18
Boby
asked Jan 15 at 15:08
BobyBoby
1,0181929
asked Jan 15 at 15:08
BobyBoby
1,0181929
asked Jan 15 at 15:08
BobyBoby
1,0181929
1,0181929
$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04
add a comment |
$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04
$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04
add a comment |
1 Answer
1
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$begingroup$
Example, $n=3$ on the interval $[1,3]$,
$$
f(x) = xlog x+frac{(x-1)^2(x-2)^2(x-3)^2}{10}
$$
It is analytic on $(0,+infty)$.
Here is $f(x)$ on $[1,3]$
Here is $f(x) - xlog x$. It is of course nonnegative and vanishes only at $1,2,3$.
Here is $f''(x)$. It is positive, so $f$ is strictly convex.
The coefficient $1/10$ was chosen so that $f''(x)$ is strictly positive.
If we consider $f(x)= xlog x+k(x-1)^2(x-2)^2(x-3)^2$ for $k>0$, then $f''(x)$
converges uniformly to $1/x$ as $k to 0$, so of course we can choose $k$ so close to zero that $f''(x)$ is positive on $[1,3]$.
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
add a comment |
Your Answer
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1 Answer
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$begingroup$
Example, $n=3$ on the interval $[1,3]$,
$$
f(x) = xlog x+frac{(x-1)^2(x-2)^2(x-3)^2}{10}
$$
It is analytic on $(0,+infty)$.
Here is $f(x)$ on $[1,3]$
Here is $f(x) - xlog x$. It is of course nonnegative and vanishes only at $1,2,3$.
Here is $f''(x)$. It is positive, so $f$ is strictly convex.
The coefficient $1/10$ was chosen so that $f''(x)$ is strictly positive.
If we consider $f(x)= xlog x+k(x-1)^2(x-2)^2(x-3)^2$ for $k>0$, then $f''(x)$
converges uniformly to $1/x$ as $k to 0$, so of course we can choose $k$ so close to zero that $f''(x)$ is positive on $[1,3]$.
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
add a comment |
$begingroup$
Example, $n=3$ on the interval $[1,3]$,
$$
f(x) = xlog x+frac{(x-1)^2(x-2)^2(x-3)^2}{10}
$$
It is analytic on $(0,+infty)$.
Here is $f(x)$ on $[1,3]$
Here is $f(x) - xlog x$. It is of course nonnegative and vanishes only at $1,2,3$.
Here is $f''(x)$. It is positive, so $f$ is strictly convex.
The coefficient $1/10$ was chosen so that $f''(x)$ is strictly positive.
If we consider $f(x)= xlog x+k(x-1)^2(x-2)^2(x-3)^2$ for $k>0$, then $f''(x)$
converges uniformly to $1/x$ as $k to 0$, so of course we can choose $k$ so close to zero that $f''(x)$ is positive on $[1,3]$.
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
$endgroup$
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
add a comment |
$begingroup$
Example, $n=3$ on the interval $[1,3]$,
$$
f(x) = xlog x+frac{(x-1)^2(x-2)^2(x-3)^2}{10}
$$
It is analytic on $(0,+infty)$.
Here is $f(x)$ on $[1,3]$
Here is $f(x) - xlog x$. It is of course nonnegative and vanishes only at $1,2,3$.
Here is $f''(x)$. It is positive, so $f$ is strictly convex.
The coefficient $1/10$ was chosen so that $f''(x)$ is strictly positive.
If we consider $f(x)= xlog x+k(x-1)^2(x-2)^2(x-3)^2$ for $k>0$, then $f''(x)$
converges uniformly to $1/x$ as $k to 0$, so of course we can choose $k$ so close to zero that $f''(x)$ is positive on $[1,3]$.
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
$endgroup$
Example, $n=3$ on the interval $[1,3]$,
$$
f(x) = xlog x+frac{(x-1)^2(x-2)^2(x-3)^2}{10}
$$
It is analytic on $(0,+infty)$.
Here is $f(x)$ on $[1,3]$
Here is $f(x) - xlog x$. It is of course nonnegative and vanishes only at $1,2,3$.
Here is $f''(x)$. It is positive, so $f$ is strictly convex.
The coefficient $1/10$ was chosen so that $f''(x)$ is strictly positive.
If we consider $f(x)= xlog x+k(x-1)^2(x-2)^2(x-3)^2$ for $k>0$, then $f''(x)$
converges uniformly to $1/x$ as $k to 0$, so of course we can choose $k$ so close to zero that $f''(x)$ is positive on $[1,3]$.
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
answered Jan 15 at 16:30
GEdgarGEdgar
62.5k267171
62.5k267171
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
add a comment |
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Thanks. Nice example. Do you think there is a general procedure for doing this?
$endgroup$
– Boby
Jan 15 at 17:08
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Of course. You can do the same thing. You can take an interval $[1,a]$ for any $a>1$. You can choose any integer $n ge 0$. You can choose any $n$ points in $[1,a]$. Then repeat what I did. The commentary at the end about $k$ is what you need for this general case.
$endgroup$
– GEdgar
Jan 15 at 19:30
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
$begingroup$
Got it. Thanks.....
$endgroup$
– Boby
Jan 15 at 19:31
add a comment |
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$begingroup$
Can you do $n=3$? Say, $f(x) = x log x$ only at the endpoints and one point between.
$endgroup$
– GEdgar
Jan 15 at 15:54
$begingroup$
@GEdgar NO! Do you have an example?
$endgroup$
– Boby
Jan 15 at 16:04