Let's try this exercise
Find all functions $ f : RR rarr RR $ s.t. (so that) $EE F ,F' = f $, and $[xf(x)-2F(x)]*[F(x)-x^2] = 0 $, $AA x in RR.$
I know it's not the right place here, but let's see what happens.
I know it's not the right place here, but let's see what happens.
Risposte
The second equation doesn't provide $f(x)=bx, " " b!=2$ as solution...
I have two questions concerning the exercise :
* The solution indicated by Fireball, i.e. $ f(x) = 2x $ is correct but also the more general solution $ f(x) = bx (b in RR)$ is correct as can be easily checked . Where does such solution come from ?
*The two equations :
$x f -2F = 0 $
and
$ F -x^2 = 0 $
provide the same solution .
Is it correct ??
* The solution indicated by Fireball, i.e. $ f(x) = 2x $ is correct but also the more general solution $ f(x) = bx (b in RR)$ is correct as can be easily checked . Where does such solution come from ?
*The two equations :
$x f -2F = 0 $
and
$ F -x^2 = 0 $
provide the same solution .
Is it correct ??
Be careful, Fireball, something must be wrong since : $ f= (x/2)e^ (x^2/4) $ doesn't satisfy the initial relation.
Furtherly the exercise requires to find all functions that ....etc.
Furtherly the exercise requires to find all functions that ....etc.
Let $[a,b]$ be a close and limitated subset of $RR$, and $f in ccR(a,b)$,
that is, $f$ belongs to the Riemann-integrables in $[a,b]$ functions subset.
Let $c in [a,b]$. Then, $F(x)=int_c^x f(x) dx " " AAc,x in [a,b]$.
We observe that $(xf(x)-2F(x))*(F(x)-x^2) = 0 $
implies that $xf(x)-2F(x)=0 vv F(x)-x^2=0$.
First, let's consider the second relation, from which
we can immediately say: $F(x)=x^2=>f(x)=2x$.
Now, the first relation is a differential equation:
$xf(x)=2F(x)<=>(2F(x))/(f(x))=x<=>(2F(x))/(F'(x))=x
and, integrating both the members of the equation in $[0,x]$, with $F(0)-=1$:
$2logF(x)=(x^2)/2<=>F(x)=e^((x^2)/4)=>f(x)=(x/2)e^((x^2)/4)$.
This concludes the proof of my results.
that is, $f$ belongs to the Riemann-integrables in $[a,b]$ functions subset.
Let $c in [a,b]$. Then, $F(x)=int_c^x f(x) dx " " AAc,x in [a,b]$.
We observe that $(xf(x)-2F(x))*(F(x)-x^2) = 0 $
implies that $xf(x)-2F(x)=0 vv F(x)-x^2=0$.
First, let's consider the second relation, from which
we can immediately say: $F(x)=x^2=>f(x)=2x$.
Now, the first relation is a differential equation:
$xf(x)=2F(x)<=>(2F(x))/(f(x))=x<=>(2F(x))/(F'(x))=x
and, integrating both the members of the equation in $[0,x]$, with $F(0)-=1$:
$2logF(x)=(x^2)/2<=>F(x)=e^((x^2)/4)=>f(x)=(x/2)e^((x^2)/4)$.
This concludes the proof of my results.
Remark: let $f_1$ be an integrable function which satisfy $xf(x)-2F(x)=0$ and let $f_2$ integrable satisfying the second one $F(x)-x^2=0$. Fixed $A$ and $B$ disjoint subsets of $\RR$ with $A\cup B=\RR$; then the function $f=f_1$ on $A$ and $f=f_2$ on $B$ is a solution of the problem (since other regularity properties are not required), and this show that there are infinite solutions of this equation.
Pls explain in detail how you arrived to the results indicated in your post.
Congratulations to Fireball for being the first to reply in english .
Hope many others will follow .
Congratulations to Fireball for being the first to reply in english .
Hope many others will follow .
I have found two functions $f:RR->RR$ which
satisfy this property:
$f(x)=(x/2) e^(x^2/4)
and
$f(x)=2x
Notice that they are both onto and one-to-one functions.
satisfy this property:
$f(x)=(x/2) e^(x^2/4)
and
$f(x)=2x
Notice that they are both onto and one-to-one functions.
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