How can I show f(x)=x2-cosx=0 has exactly 2 solutions?

[StephenPrivitera]

How can I show f(x)=x²-cosx=0 has exactly 2 solutions? I can show it has at least two by using the fact that f is even and continuous and f(0)<0 and f(1)>0 so there's at least one solution in 0<x<1 and thus one solution in -1<x<0. I'm trying to derive a contradiction by assuming there's a third solution but I can't seem to find one.

 

[Hurkyl]

One of the cuter ways to prove a function has no zeroes in a certain interval (e.g. [1, &infin;) )is to prove that it's left endpoint is positive and that it's increasing throughout the interval

Another thing that might help is Rolle's theorem; if f(a) and f(b) are both zero, then f'(c) is zero for some c between a and b.

 

[StephenPrivitera]

Originally posted by Hurkyl
One of the cuter ways to prove a function has no zeroes in a certain interval (e.g. [1, &infin;) )is to prove that it's left endpoint is positive and that it's increasing throughout the interval

I've already shown that f(x) cannot be zero if |x|>1 by using the fact that cosx<1 or cosx=1 (I didn't write it in the post). I've identified the existence of two numbers, c and -c such that 0<c<1 where f(c)=f(-c)=0. I don't think this method will help here.

Originally posted by Hurkyl
One of the cuter ways to prove a function has no zeroes in a certain interval (e.g. [1, &infin;) )is to prove that it's left endpoint is positive and that it's increasing throughout the interval

Yes unfortunately, the fact that I should use this theorem has been haunting me. The chapter is on MVT, so I figured it should come into play somewhere. I cannot see how this helps at all. If I go by contradiction (assume a third x), then there will be 4 x's such that f(x)=0 (since when f(x)=0, f(-x)=0). So by MVT there will be 3 x such that f'(x)=0 (repeatedly applying the theorem). Thus, for at least 3 x, 2x=-sinx. So now the problem reduces to showing that this equation has 2 or less solutions. We have x=0 as a solution. Ah yes, the next derivative 2 + cosx is always positive. So f' is always increasing. So 2x=-sinx has only one solution.
I can take it from here. Thanks

 

[Hurkyl]

To simplify things a little bit...

If you assume there are more than 2 solutions, there must be more than 2 positive solutions, and zero can't lie between those two solutions.

 

[StephenPrivitera]

Originally posted by Hurkyl
If you assume there are more than 2 solutions, there must be more than 2 positive solutions

Wait, why?

 

[Hurkyl]

You already noticed that it's an even function. If the 3rd solution is negative, that means there's a 4th solution that's positive. (or, you can just play with the two negative solutions, if you prefer)


Basically, the thing I wanted to get at is saying "Well, if there are three solutions, then by Rolle's theorem 2x = -sin x has 3 solutions" is a non sequitor; in principle Rolle's theorem could return the same number three times. Given the actual problem at hand, it will give you the same number twice... but there's still that logical detail to prove that rolle's theorem doesn't give the same number a third time.

 

[StephenPrivitera]

Ok well in that case there are at least two positive solutions. I think you made a mistake when you said there will be more than 2 positive solutions.
I figured that Rolle's theorem gives distinct x by applying it to disjoint intervals. If c, c_p are distinct positive solutions with c<c_p, then Rolle's theorem guanantees an x for each interval [-c_p,-c], [-c,c], [c,c_p] and each x will not be an endpoint of the interval. I found it necessary however to consider both positive and negative solutions.

 

[Hurkyl]

Yes, that's what I meant; if there's a third solution, there must be a second positive solution!


Consider this: you know the only solution to 2x+sinx = 0 is x=0... so no solution can exist in the range [c, c_p].

 

[StephenPrivitera]

Originally posted by Hurkyl
Consider this: you know the only solution to 2x+sinx = 0 is x=0... so no solution can exist in the range [c, c_p].

Ah you're too creative!
I should have seen that myself.