Prove that this equation has at least one real root

[utkarshakash]

Homework Statement

Let f:R→R be a continuous and differentiable function, then prove that the equation f'(x)+λf(x)=0 has at least one real root between any pair of roots of f(x)=0, λ being a real number

Homework Equations

The Attempt at a Solution

All that I know from Rolle's Theorem is that between a pair of roots of f(x) there must be atleast one root of f'(x). But I can't figure out how to deal with that extra term 'λf(x)'?

 

[dirk_mec1]

Hint: f'(x) = -λf(x).

 

[shortydeb]

Do you mean f is continuously differentiable from R to R? No textbook would say a function is continuous and differentiable on a given interval because on a given interval, a differentiable function is always continuous.

 

[haruspex]

Hint: f'(x) = -λf(x).

That is not provided as a differential equation. The question could be worded better. It is defining a function g(x) = f'(x)+λf(x), and asks you to show that g has a root between each pair of roots of f.

 

[haruspex]

Hint 1: Does the form f'+λf remind you of anything?
Hint 2: If h(x) has no roots then f(x)h(x) has the same roots as f(x).

 

[Dick]

You might also imagine what the graph of log(|f|) looks like on the interval and then think about how that might relate to your problem. That might give you some intuition about the problem.

 

[utkarshakash]

Hint 1: Does the form f'+λf remind you of anything?
Hint 2: If h(x) has no roots then f(x)h(x) has the same roots as f(x).

Is it related to differential equations?

 

[haruspex]

Is it related to differential equations?

Yes.

 

[utkarshakash]

Yes.

OK then it's time for me to wait a little because my teacher hasn't started DE yet. I wonder why he gives questions which involves DE's.

 

[Dick]

OK then it's time for me to wait a little because my teacher hasn't started DE yet. I wonder why he gives questions which involves DE's.

You don't have to use differential equations. Draw a sample function and sketch the graph of log(|f|).

 

[Saitama]

Yes.

Solving the given differential equation, $f(x)=ce^{-\lambda x}$ (where c is a constant). How will this function have any roots (except $\infty$)?

 

[verty]

Solving the given differential equation, $f(x)=ce^{-\lambda x}$ (where c is a constant). How will this function have any roots (except $\infty$)?

This fooled me too. If g(x) = 0 for all x, THEN f is that function (is in that family).

 

[verty]

Hint 3: will it suffice to consider only pairs of adjacent roots of f?

 

[haruspex]

Solving the given differential equation, $f(x)=ce^{-\lambda x}$ (where c is a constant). How will this function have any roots (except $\infty$)?

As I pointed out to dirk_mec1, they have not provided us with a differential equation for f.
Let me reword the question to make it clearer:

Let f:R→R be a continuous and differentiable function, then prove that the function g(x) = f'(x)+λf(x) has at least one real root between any pair of roots of f(x), λ being a real number.

(It's wrong to talk about an equation having roots. Functions have roots, equations have solutions. A root of f(x) is a solution of f(x)=0.)
That said, you have indeed found a function that has no roots, and it is the function h(x) in my second hint.

 

[shortydeb]

I don't know if the proposition given in the OP is true if f is differentiable but not continuously differentiable, but you can use the Intermediate Value Theorem to prove the proposition if f is continuously differentiable.

 

[verty]

I was just noticing that a function like $f(x) = e^{-\frac{1}{x^2}} * e^{-\frac{1}{(x-1)^2}}, f(0) = f(1) = 0$ is a problem for this type of argument.

 

[CompuChip]

Is that differentiable at x = 0, x = 1?

 

[epenguin]

Either I am missing something or (as it seems to me) a big meal is being made of something damned obvious. The question does assume that f have two real roots, otherwise 'between' makes no sense.

So just consider what f'(x)+λf(x) is at one root of f and the next root of f.

 

[shortydeb]

Look at any pair of roots. Consider the case where f'(x)+λf(x) is positive at the first root and negative at the second root. What does the Intermediate Value Theorem tell you about f'(x)+λf(x) between the two roots? This is assuming f is continuously differentiable on R.

 

[utkarshakash]

Look at any pair of roots. Consider the case where f'(x)+λf(x) is positive at the first root and negative at the second root. What does the Intermediate Value Theorem tell you about f'(x)+λf(x) between the two roots? This is assuming f is continuously differentiable on R.

Since f'(x)+λf(x) has changed its sign from +ve to -ve there must be atleast one point where it became zero. Is this logic correct?

 

[epenguin]

Since f'(x)+λf(x) has changed its sign from +ve to -ve there must be atleast one point where it became zero. Is this logic correct?

Exactly, it is almost the same thing as your Rolle's theorem of #1 .

 

[haruspex]

epenguin and shortydeb, how are you proposing to show that f' switches sign at consecutive roots of f? It does, of course, but I don't see this as completely trivial.

 

[epenguin]

epenguin and shortydeb, how are you proposing to show that f' switches sign at consecutive roots of f? It does, of course, but I don't see this as completely trivial.

You are asking for a proof of Rolle's theorem?

It seemed the OP was assuming it as something already known.

(Mathematicians and non-mathematicians probably part company about whether they want one. )

 

[haruspex]

You are asking for a proof of Rolle's theorem?

Are we using different statements of Rolle's theorem, or are you using it i some clever way I'm not seeing?
As I understand it, it says that

a differentiable function which takes the same value at two distinct points must have a point somewhere between them where the first derivative is zero.

If α and β are consecutive roots of f then the function g(x) = f'(x) + λf(x) does not necessarily take the same value at those two roots, and anyway you're not interested in g'. You can apply Rolle's theorem to f, but that only tells you f' is zero somewhere between; it does not tell you g has a root in between.

 

[Dick]

Are we using different statements of Rolle's theorem, or are you using it i some clever way I'm not seeing?
As I understand it, it says that If α and β are consecutive roots of f then the function g(x) = f'(x) + λf(x) does not necessarily take the same value at those two roots, and anyway you're not interested in g'. You can apply Rolle's theorem to f, but that only tells you f' is zero somewhere between; it does not tell you g has a root in between.

If a and b are two consecutive roots then look at log(|f|) on the interval (a,b). It approaches -infinity at a and -infinity at b and it has a finite value in between. Sketch a graph. Think about its slope f'/f and use the MVT (which is a form of Rolle's theorem). Mustn't it take on all real values? I guess I'm not seeing what all the fuss is about.

 

[haruspex]

If a and b are two consecutive roots then look at log(|f|) on the interval (a,b). It approaches -infinity at a and -infinity at b and it has a finite value in between. Sketch a graph. Think about its slope f'/f and use the MVT (which is a form of Rolle's theorem). Mustn't it take on all real values? I guess I'm not seeing what all the fuss is about.

Sure, but epenguin and shortydeb seem to be saying it's much easier even than that.
Fwiw, the solution I posted hints for is to consider f(x)e^λx. This must have the same set of roots as f, and its derivative must have the same set of roots as f'+λf.

 

[Dick]

Sure, but epenguin and shortydeb seem to be saying it's much easier even than that.
Fwiw, the solution I posted hints for is to consider f(x)e^λx. This must have the same set of roots as f, and its derivative must have the same set of roots as f'+λf.

Oh, I see what you are up to. Apply Rolle's to f(x)e^(λx). I kind of like the MVT approach because that doesn't need too much cleverness. But that's still nice.

 

[epenguin]

Are we using different statements of Rolle's theorem, or are you using it i some clever way I'm not seeing?
As I understand it, it says that If α and β are consecutive roots of f then the function g(x) = f'(x) + λf(x) does not necessarily take the same value at those two roots, and anyway you're not interested in g'. You can apply Rolle's theorem to f, but that only tells you f' is zero somewhere between; it does not tell you g has a root in between.

Between successive roots (which we should consider, as pointed out by vertex and others) of f(x), f'(x) changes sign*. At any root of f(x) , g(x) = f'(x) . So between roots of f, g changes sign. If g changes sign in an interval, it has a root in that interval. QED.

*(There may be different versions of Rolle's theorem. I often use it but never name it - I often have to look it up when anyone mentions it. The reason is I can never quite believe that such an obvious statement got honoured with the name 'theorem' so I need to check it is what I think it is.)

 

[haruspex]

Between successive roots (which we should consider, as pointed out by vertex and others) of f(x), f'(x) changes sign*.

Sorry, but I can't find that. Would you mind providing the number of the post? I did find this from utkarshakash

Consider the case where f'(x)+λf(x) is positive at the first root and negative at the second root.

but that is assuming a change of sign, not proving one. Maybe you're thinking of another theorem.

 

[epenguin]

Sorry, but I can't find that. Would you mind providing the number of the post? I did find this from utkarshakash

Oops, verty #13.

Anyway it looks like the OP has got it, #20, not quite explicit.

 

[verty]

I gave that hint about adjacent roots but I should have thought more about the question first, because it (the question as well as my hint) was a little misleading. Dick (as usual, I'm pleased to say) led us directly to the best way to look at it, a geometric argument.

 

[haruspex]

I gave that hint about adjacent roots but I should have thought more about the question first, because it (the question as well as my hint) was a little misleading. usual, I'm pleased to say) led us directly to the best way to look at it, a geometric argument.

Thanks for clearing that up, verty. (I still like my e^λx method.)

 

[Dick]

Thanks for clearing that up, verty. (I still like my e^λx method.)

I like it too. It's elegant and does use pure Rolle's theorem, not the MVT. But I like the MVT approach because I think it's easier to discover graphically, without the extra bit of cleverness. Tough call really.