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January 30th, 2016, 07:59 AM   #1
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stuck!

Hello!

I'm stuck with the following problem... maybe someone can give me ideas on how to go about it...

I have two functions f(x) and g(x) and I know that f has one zero at y. I need to show that g(y) > 0. Any suggestions on what to consider in order to prove this? I know the result should be true.

Thanks!
Natalie

Last edited by skipjack; January 30th, 2016 at 08:33 AM.
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January 30th, 2016, 08:04 AM   #2
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What's the relationship between f(x) and g(x)?
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January 30th, 2016, 08:11 AM   #3
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What type of relation ?
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January 30th, 2016, 08:14 AM   #4
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Your original post appears incomplete. f(y) = 0 does not imply g(y) > 0 unless there is some relationship between f and g.
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January 30th, 2016, 08:41 AM   #5
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yes yes they are related of course apologies for that...the expressions look like this... the constants $\displaystyle M_i, P_i , i=1,2,...6$ are related as well...I just need ideas on what to consider to prove this result...it is not true in general but for the constants i have it is.

for fixed A > 0:

$\displaystyle f(y;A) = A*[M_1 y^{\gamma_1+\gamma_3} + M_2 y^{\gamma_1+\gamma_2} + M_3 y^{\gamma_2+\gamma_3} + M_4 y^{\gamma_1 + 1} + M_5 y^{\gamma_2 + 1} + M_6 y^{\gamma_3 + 1}] - [P_1 y^{\gamma_1+\gamma_3} + P_2 y^{\gamma_1+\gamma_2} + P_3 y^{\gamma_2+\gamma_3} + P_4 y^{\gamma_1 } + P_5 y^{\gamma_2} + P_6 y^{\gamma_3 }] $

I know that for each fixed $\displaystyle A > 0, y \mapsto f(y;A) $ has a unique root.

$\displaystyle g(y;A) = A*[M_1 (\gamma_1 + \gamma_3 -1) y^{\gamma_1+\gamma_3} + M_2 (\gamma_1 + \gamma_2 -1) y^{\gamma_1+\gamma_2} + M_3 (\gamma_2 + \gamma_3 -1)y^{\gamma_2+\gamma_3} + M_4 \gamma_1 y^{\gamma_1 + 1} + M_5 \gamma_2 y^{\gamma_2 + 1} + M_6 \gamma_3 y^{\gamma_3 + 1}] - [P_1 (\gamma_1 + \gamma_3 -1)y^{\gamma_1+\gamma_3} + P_2 (\gamma_1 + \gamma_2 - 1)y^{\gamma_1+\gamma_2} + P_3 (\gamma_2 + \gamma_3 -1)y^{\gamma_2+\gamma_3} + P_4 (\gamma_1 -1) y^{\gamma_1} + P_5 (\gamma_2 - 1) y^{\gamma_2 } + P_6 (\gamma_3 -1 ) y^{\gamma_3 }] $
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