Existence of a root between 2 given points

[sara_87]

Homework Statement

Show that there exists one root int (0,2) of the following function:

f(x)=(1-x^2)^2-√((1-x^2)*(1-1/2*x^2))

Homework Equations

The Attempt at a Solution

I first found:
f(0)=0
and
f(2)=7.268

But, i don't know what to do now. I'm not sure if it has something to do with the mean value theorem or not, or if i have to differentiate or not.

Any ideas will be very much appreciated.
Thank you.

 

[HmBe]

Well you've found a root at x=0, f(0)=0, but this isn't in the interval you're interested in, so it's best to try another value.

With finding a root in an interval your best bet is to try and find an a and a b in the interval such that f(a)<0 and f(b)>0, then by the intermediate value theorem there will be a root.

You've already got your b, such that f(b)>0, so now you just need to find your a.

 

[sara_87]

thank you.
so, i can take b=0.1
f(0.1)=-0.0124

but, this show that there exists at least 1 root in the interval, but i need to show that there exists exactly one root.
how would i go from there?
doesn't have something to do with the derivative?

 

[MathematicalPhysicist]

Well if I read your function correctly (if I am not mistaken you're a PhD student, then why don't you use TeX?), it's:
f(x)=(1-x^2)^2-\sqrt{(1-x^2)(1-\frac{1}{2} x^2)}

If this is your function then we can see that at x=1 the function vanishes, and you also know that at the boundaries at x=0 it equals 0 and at x=2 you get a positive function.

Now you should prove that for x in (0,1] there isn't another zero as in f(x) is always positive in this interval (non-negative), it's just a matter of playing with inequlities.

If you prove that f had the same sign all over the interval (0,1] then now you should show that if it were negative in (0,1) then in (1,2) it's a positive.

I'll show you the second part:
for x in (1,2) we have (1-x^2)^2>(1-4)^2=9
\sqrt{(1-x^2)(1-1/2 x^2)} &lt; \sqrt{1/2}
so you get f(x)&gt;9-\sqrt{1/2}&gt;0.

Hopefully, I didn't get it wrong (with me it's always a possibility).

 

[MathematicalPhysicist]

Wait a minute, you want to show there exists exactly one, or that exists such root.

Cause I can now that there exist more than one root.

 

[sara_87]

thank you,
but I don't understand, where did you get the 4 from in:
(1-4)^2 ?

also, what do we do if we assume that we don't know that x=1 is a root?

 

[sara_87]

we want to show that there exists exactly 1 root

 

[HmBe]

also, what do we do if we assume that we don't know that x=1 is a root?

I'd follow the method I said above for finding a root, to show that there is only one root I'd just do a rough sketch of the function in the interval you're interested in, because you're right - with a trickier function (something you're more likely to use IVT on) you won't know where the root is, so the inequality thing is much harder to do.

However, without looking at your mark scheme, I wouldn't know if this is enough. A sketch usually suffices for us though.

 

[sara_87]

I have plotted the graph on MATLAB, and i can see that there is exactly one root (namely x=1)
but we have to use another method (not a graphical method). We were told that finding the derivative is a clue, but i don't see why.

 

[sara_87]

I'm not sure if Rolle's theorem comes into this content?

 

[HmBe]

Can you show the function is even, and thus symmetrical? Then it will have another root in (-2,0).

Then you could show using the derivative that the root at x=0 is repeated, and because the root is quartic can only have one more root...

 

[sara_87]

Yes, the function is even since
f(-x)=f(x)
(since x appears only as x^2).

But, I don't understand the next part:
The derivative is

f'(x)= -4 x \left(1-x^2\right)-\frac{-x \left(1-x^2\right)-2 x \left(1-\frac{x^2}{2}\right)}{2 \sqrt{\left(1-x^2\right) \left(1-\frac{x^2}{2}\right)}}

should i solve f'(x)=0?

 

[HmBe]

Well I suppose you could cheat a little, showing that f'(0)=0 and that f(0)=0 (which you've already done) shows there is a repeated root at x=0, then you don't have to solve anything.

 

[sara_87]

Yes, I think that's what we are supposed to do.

but, how does showing that f'(0)=0 and that f(0)=0 show that there is a repeated root at x=0?

 

[HmBe]

I was pretty sure the definition of a repeated root was that if there is a root at x=a, that is f(a)=0, and f'(a)=0, then the root is repeated. Thinking about it, it seems to make sense, but I couldn't find anything on the internet to verify this.

 

[Ray Vickson]

Well you've found a root at x=0, f(0)=0, but this isn't in the interval you're interested in, so it's best to try another value.

With finding a root in an interval your best bet is to try and find an a and a b in the interval such that f(a)<0 and f(b)>0, then by the intermediate value theorem there will be a root.

You've already got your b, such that f(b)>0, so now you just need to find your a.

There may be a problem with this approach in this case: as written, the function f(x) does not exist (in the reals) for 1 < x < sqrt(2). So, application of the intermediate-value method needs to be done with great care.

PS: I assume the OP means
f(x) = (1-x^2)^2 - sqrt[(1-x^2)(1 - (x^2/2))], not (1-x^2)^2 - sqrt[(1-x^2)(1 - 1/(2x^2))].

RGV

 

[HmBe]

Oh yes, just realized I made a pretty big mistake when I plotted the function, and that it's not continuous, so applying IVT would be very tricky.

 

[sara_87]

Oh right, i see.
If we assume we have a different function that IS continuous in (0,2) and that the IVT does apply, then we can use that f(0)=0 and f'(0)=0
But, what do you mean by:
''because the root is quartic can only have one more root''?

 

[MathematicalPhysicist]

thank you,
but I don't understand, where did you get the 4 from in:
(1-4)^2 ?

also, what do we do if we assume that we don't know that x=1 is a root?

Yes, I got it wrong. :-DThe sqrt there just pisses me off, I don't see any good estimate here.

Ok, I think I remembered my course I learned in ODE2.

If you prove that at the interval (0,2) you have exactly two maximum points and one minimum at x=1, you've proven that there isn't anymore roots for this equation.

Am I wrong here again?
What I am trying to say is, if you would have found a minimum point at the interval s.t f(x0)<0 then we must have crossed the x-axis somewhere, and then if we found two positive max, we should cross the x-axis again, thus getting at least two roots. So indeed you should analyse the function via differentiation twice and sorting the max and min in this interval.

 

[I like Serena]

I'm not sure if Rolle's theorem comes into this content?

I don't think so. Rolle only says something about at least one point (same thing for IVT).

I believe that what you need is the definition of a absolute monoticity, which is implied by the derivative (being not zero).

 

[sara_87]

Thank you.
but the derivative at 0 is =0.
how would absolute monoticity help?

 

[I like Serena]

Thank you.
but the derivative at 0 is =0.
how would absolute monoticity help?

On the interval (0,1) the derivative is not zero.
Between 0 and the zero of the first derivative (at x≈0.7) the function is strictly decreasing, meaning there is no zero (since there is indeed a zero at x=0).
Afterward it is strictly increasing and at x=1 you have another zero, meaning there is no zero before x=1.

 

[sara_87]

But, we have to assume that we don't know that x=1 is a root of the function.

 

[Ray Vickson]

Yes, I got it wrong. :-DThe sqrt there just pisses me off, I don't see any good estimate here.

Ok, I think I remembered my course I learned in ODE2.

If you prove that at the interval (0,2) you have exactly two maximum points and one minimum at x=1, you've proven that there isn't anymore roots for this equation.

Am I wrong here again?
What I am trying to say is, if you would have found a minimum point at the interval s.t f(x0)<0 then we must have crossed the x-axis somewhere, and then if we found two positive max, we should cross the x-axis again, thus getting at least two roots. So indeed you should analyse the function via differentiation twice and sorting the max and min in this interval.

Unfortunately, your statement "If you prove that at the interval (0,2) you have exactly two maximum points and one minimum at x=1, ..." does not apply. By plotting, we can see that for 0 < x < 1 we have f(x) < 0, so x = 0 and x = 1 are both zeros of f and *maxima* of f on [0,1]. For 1 < x < sqrt(2), f(x) does not exist (as a real value). For x > sqrt(2), f(x) > 0 always, and f has a minimum a bit to the right of sqrt(2). It is true, though, that x = 1 is the only root in (0,2).

RGV

 

[I like Serena]

But, we have to assume that we don't know that x=1 is a root of the function.

Why would we assume that?
You can readily verify that it is and after you do so, you can use that.

 

[sara_87]

Thank you all for these ideas.

I find what HmBe said very interesting:

''Well I suppose you could cheat a little, showing that f'(0)=0 and that f(0)=0 (which you've already done) shows there is a repeated root at x=0, then you don't have to solve anything.

and because the root is quartic can only have one more root''

So, if we have a symmetric function. in our case, there is a root (0,2) and (-2,0)
and since f(0)=f'(0)=0, then x=0 is a repeated root. But, I still don't understand how this shows that (since also f(2) is positive) that there is one more root in the interval (0,2).