- #1

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In which case depending on the value of the parameter "a" ,the function [tex]\phi_{a}(x,y) \geq 0 [/tex].

Thanks

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- Thread starter BoMa
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- #1

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In which case depending on the value of the parameter "a" ,the function [tex]\phi_{a}(x,y) \geq 0 [/tex].

Thanks

- #2

uart

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For a>1 however then your function may be negative if x and y are sufficiently close to zero.

- #3

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Could you please give me some details , how did you proceed ?

Tks

Tks

- #4

uart

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Actually I made a mistake on the [itex]a<1[/itex] case. That is no guarantee that the function is positive.

Anyway I started by writing the last term explicitly as a product, as in

[tex]\phi_{a}(x,y)=\frac{1}{a}\,x^{(1-a)/a}+\frac{1}{a}\,y^{(1-a)/a}- (\frac{1}{a}\,x^{(1-a)/a} \, \frac{1}{a}\,y^{(1-a)/a})[/tex] .

I was using the fact that for positive numbers less than one that their sum is always greater than their product (but I forgot that after dividing by "a" that they're no longer necessarily less than one).

I'll take another look at it later.

Anyway I started by writing the last term explicitly as a product, as in

[tex]\phi_{a}(x,y)=\frac{1}{a}\,x^{(1-a)/a}+\frac{1}{a}\,y^{(1-a)/a}- (\frac{1}{a}\,x^{(1-a)/a} \, \frac{1}{a}\,y^{(1-a)/a})[/tex] .

I was using the fact that for positive numbers less than one that their sum is always greater than their product (but I forgot that after dividing by "a" that they're no longer necessarily less than one).

I'll take another look at it later.

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- #5

uart

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BoMa, I wasn't looking at the full conditions on x,y,a for which the function is non-negative, I was only looking at what restrictions you need to place on "a" such that the function is non-negative over the full given domain of x,y.

First look at what is required for the product of two positive numbers to exceed their sum. It's very easy to show that this occurs if both numbers are greater than 2.

So if x=y and**1/a * x^(1/a-1)** is greater than 2 then phi is negative.

Now if**a>1** then 1/a-1 is negative and so x^(1/a-1) is a reciprocal of a power of a positive number less than one. This means that if "x" and "y" are close enough to zero then **1/a * x^(1/a-1)** can be made arbitrarily large and hence phi can be made negative. This was the rational for the second of the observations I made in my first reply.

As mention previously I made a mistake with my cursory analysis of the a<1 case. Here the powers are positive so large values of x,y are where we are likely to get a product larger than the sum of the two terms mentioned above.

So solve [itex]1/a \times x^{1/a-1} \leq 2[/itex] for the case where x=1 (worst case) which gives the result [itex]a \geq 1/2[/itex].

So you require [itex]1/2 \leq a \leq 1[/itex] to make [itex]\phi(x,y)[/itex] non-negative for all the given x,y domain.

First look at what is required for the product of two positive numbers to exceed their sum. It's very easy to show that this occurs if both numbers are greater than 2.

So if x=y and

Now if

As mention previously I made a mistake with my cursory analysis of the a<1 case. Here the powers are positive so large values of x,y are where we are likely to get a product larger than the sum of the two terms mentioned above.

So solve [itex]1/a \times x^{1/a-1} \leq 2[/itex] for the case where x=1 (worst case) which gives the result [itex]a \geq 1/2[/itex].

So you require [itex]1/2 \leq a \leq 1[/itex] to make [itex]\phi(x,y)[/itex] non-negative for all the given x,y domain.

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- #6

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Thanks for your reply , but I check for example the case

[itex] x=y \, \,\,\text{and}\,\,\, a=1/3 , [/itex]

and I see that [itex]\phi(x,y) [/itex] can be non-negative for this

value of "a" which is not in the interval , you gave [itex]1/2 \leq a \leq 1 [/itex]

And the other problem, as x and y are in the interval [0,1].

I didn't understand , when you say "... what is required for the product of two positive numbers to exceed their sum. It's very easy to show that this occurs if both numbers are greater than 2".

Could you show it to me ?

Thanks

- #7

uart

Science Advisor

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Hi Uart ,

Thanks for your reply , but I check for example the case

[itex] x=y \, \,\,\text{and}\,\,\, a=1/3 , [/itex]

and I see that [itex]\phi(x,y) [/itex] can be non-negative for this

value of "a" which is not in the interval , you gave [itex]1/2 \leq a \leq 1 [/itex]

Yes that's correct. The only claim that I'm making is that if [itex]1/2 \leq a \leq 1 [/itex] then [itex]\phi[/tex] is non-negative on the full domain, [itex]0\leq x,y\leq1[/itex].

For [itex]0 < a < 1/2[/itex] and for a > 1 then [itex]\phi[/tex] may be positive or negative.

Take your example of a=1/3 (or any positive a<1/2) and check that [itex]\phi(1,1) < 0[/itex].

Similarly take any a>1 and check that the limit [itex]\epsilon \rightarrow 0[/itex] of [itex]\phi(\epsilon,\epsilon) \rightarrow - \infty [/itex]

- #8

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Can I get the same solution, or prove your claim if I let [tex]X =\frac{1}{a}\,x^{(1-a)/a} [/tex] and [tex]Y =\frac{1}{a}\,y^{(1-a)/a} [/tex] and find in which case [tex] \phi_{a}(X,Y)=X+Y-XY [/tex] is positive ?

Thanks again

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