Proving 0 < θ1 < β in θ12-γθ1+β=0

  • Thread starter Thread starter Kinetica
  • Start date Start date
Click For Summary
The discussion revolves around proving the inequality 0 < θ1 < β from the quadratic equation θ12 - γθ1 + β = 0. Participants note that substituting θ1 = 0 yields a positive result, indicating that β must be greater than zero. However, they express uncertainty about demonstrating that θ1 can take values between zero and β, particularly without more information about the parameters γ and β. The conversation highlights the need for clarity on the values and relationships of these parameters to establish the required inequalities. Ultimately, the conclusion emphasizes that additional context is necessary to assert the inequalities definitively.
Kinetica
Messages
83
Reaction score
0

Homework Statement



θ12-γθ1+β=0

Show that

0<θ1


The Attempt at a Solution



I know that for θ1=0, θ12-γθ1+β>0:
Substituting, we get 02-0+β=β, which is positive.

I don't know how to show that for θ1=λ, θ12-γθ1+β<0.

I also don't know how to show that these results imply that there is zero between these two values. Which in turns means that 0<θ1<β.
 
Physics news on Phys.org
Well, θ12-γθ1+β=0 is just a quadratic equation in θ1, so there should be at most two possible values for θ1, in terms of the other parameters.

\theta_{1} = \frac{\gamma\pm\sqrt{\gamma^{2}-4\beta}}{2}

But as things stand, you haven't given nearly enough information to assert the inequalities required. For example, why shouldn't β be zero or even negative? Is γ greater than zero, greater than one?
 
consider for instance β=0 and y=θ then θ can be whatever you like as long as y is also!
 
OK so I found this on your other request, and it looks like it could apply here.
Kinetica said:
θ12
θ12
but that just means that the relative sizes of θ1 and β depend on θ2. So still no closer to that inequality - there's something you aren't telling us about this question...
 
Kinetica said:

Homework Statement



θ12-γθ1+β=0

Show that

0<θ1

The Attempt at a Solution



I know that for θ1=0, θ12-γθ1+β>0:
Substituting, we get 02-0+β=β, which is positive.

I don't know how to show that for θ1=λ, θ12-γθ1+β<0.

I also don't know how to show that these results imply that there is zero between these two values. Which in turns means that 0<θ1<β.
This thread is very similar in topic to a thread you started one day earlier: https://www.physicsforums.com/showthread.php?p=3717073#post3717073 .

Where did λ (lambda) come from, or should that be γ (gamma) ?

I suppose we can infer that β > 0 from the inequality, 0<θ1<β, and because you mentioned it in passing, "Substituting, we get 02-0+β=β, which is positive."

It makes no sense to plug values such as 0 or γ or λ in for θ1 in the quadratic polynomial θ12-γθ1+β to see if 0<θ1<β .

As Joffan said, you need more information regarding β and γ, before you can say much about θ1 .
 

Similar threads

Replies
24
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
27
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
5
Views
5K