Inequality
Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
$ |pe^(iqx) + q e^(-ipx) | <= e^(-cx^2) $
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
$ |pe^(iqx) + q e^(-ipx) | <= e^(-cx^2) $
Risposte
Hint, taking advantage of triangle inequality :
$| pe^(iqx) +qe^(-ipx) | <= |pe^(iqx) | +| qe^(-ipx)| = p+q = 1 .........................
$| pe^(iqx) +qe^(-ipx) | <= |pe^(iqx) | +| qe^(-ipx)| = p+q = 1 .........................
"Thomas":
I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...
But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...
done that, I observede that the minimum is positive and this proved the existence of such a $c$...
Also I have made the same demonstration and the conclusion is
$c<=1/2*(ln(1-2*p*q*(1-cosx))/(-x^2))$
Now because $0
Then also in $x=0$ there aren't problems because $lim_(x->0)1/2*(ln(1-2*p*q*(1-cosx))/(-x^2))=(pq)/2$
"Thomas":
I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...
But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...
done that, I observede that the minimum is positive and this proved the existence of such a $c$...
OK correct, I misunderstood your explanation .
Try also with the triangle inequality, it is very nice
I didn't say there is a minimum in 0, but that there is a minimum (positive) in $[-\pi,\pi]$... it can be in zero (anyway, in 0 the function is not defined but can be extended by continuity), but we don't care...
But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...
done that, I observede that the minimum is positive and this proved the existence of such a $c$...
But I had to check that the limit in zero existed!! ... in fact Weiestrass theorem tells us that there is a minimum only if the function is continuos...
done that, I observede that the minimum is positive and this proved the existence of such a $c$...
Are you sure that the function $log[(1-q)^2+q^2+2q(1-q)cosx]/(-x^2) $ has a minimum in $ x = 0 $ ?
Anyhow there is a quicker way using triangle inequality....
Anyhow there is a quicker way using triangle inequality....
First of all, sorry for the big number of mistakes I'm going to write 
...
As a firt step, it's easy to check that we can rewrite the first member this way:
$ (|pe^(iqx) + q e^(-ipx) |)^2 =(1-q)^2+q^2+2q(1-q)cosx$
with $0
so the equation, elevating both members (the quantities are all positive):
$ (1-q)^2+q^2+2q(1-q)cosx<=e^(-(2cx^2))$
taking the logaritm of both sides we need to find $c/2$ that respects:
$(log( (1-q)^2+q^2+2q(1-q)cosx))/(-x^2)>=2c$
remembering that $x$ was in $[ -pi; pi]$, if we show that the function at the left member is limited, we have finished.
But this is true because it is obviously continuos in $[-pi,pi]\0$ and in $0$ the limit exists and is finished (with de l'hopital theorem it's also easy to evaluate it and the condition $0
is it right???
...As a firt step, it's easy to check that we can rewrite the first member this way:
$ (|pe^(iqx) + q e^(-ipx) |)^2 =(1-q)^2+q^2+2q(1-q)cosx$
with $0
so the equation, elevating both members (the quantities are all positive):
$ (1-q)^2+q^2+2q(1-q)cosx<=e^(-(2cx^2))$
taking the logaritm of both sides we need to find $c/2$ that respects:
$(log( (1-q)^2+q^2+2q(1-q)cosx))/(-x^2)>=2c$
remembering that $x$ was in $[ -pi; pi]$, if we show that the function at the left member is limited, we have finished.
But this is true because it is obviously continuos in $[-pi,pi]\0$ and in $0$ the limit exists and is finished (with de l'hopital theorem it's also easy to evaluate it and the condition $0
is it right???
"nicasamarciano":[/quote]
[quote="Camillo"]Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
avevo commesso un errore, perciò ho tolto la mia precedente soluzione
Ok we wait for your updated solution.
Ok, thanks.
"Luca.Lussardi":
Ma lo scopo di questi post non era quello di rispondere in inglese?
excuse me. I have edited
Ma lo scopo di questi post non era quello di rispondere in inglese?
[quote=Camillo]Let be : $p,q > 0 $ such that $ p+q = 1 $ .
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
avevo commesso un errore, perciò ho tolto la mia precedente soluzione
Prove that exists $ c > 0 $ such that $AA x in [ -pi; pi]$ :
avevo commesso un errore, perciò ho tolto la mia precedente soluzione
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