New proof of rolles theorom

[Alpharup]

Can concepts of definite integrals be applied to prove rolles theorom? if so is my proof correct?

Let us consider a function y=g(x). it is continous. the range of y is(-∞,∞).let the derivative of g(x) be f(x)..ie g'(x)=f(x).so the area between the limits a and b is lim(b,a)∫g(x)dx.


So, lim(b,a)∫g(x)dx=f(b)-f(a)+c-----------{1}

We also know area between limits is (b-a)Ʃf{a+(r-1)Δx}/n.--------{2}
Let us take the term Ʃf{a+(r-1)Δx}/n=k-----{3}
We know {2}={3}
So (b-a)Ʃf{a+(r-1)Δx}/n.= lim(b,a)∫g(x)dx=f(b)-f(a)

So, (b-a)*k=f(b)-f(a) [from {3}]


So k=f(b)-f(a)/b-a. Since g(x) is continous, let's say g(c)=k..ie f'(c)=k


hence f'(c)=f(b)-f(a)/b-a

Hence proved

 

[HallsofIvy]

One problem with this- both Rolle's theorem and the mean value theorem are important in proving the "fundamental theorem of calculus"- that the derivative and integral are "inverse" operations. So there is circular reasoning involved in your proof.

 

[Alpharup]

One problem with this- both Rolle's theorem and the mean value theorem are important in proving the "fundamental theorem of calculus"- that the derivative and integral are "inverse" operations. So there is circular reasoning involved in your proof.

But many textbooks avoid the proof of rolles theorom since it is complicated. I tried to learn how the extreme value theorom came to prove rolles theorom came but i got a headache!.So i came up with my own way of proving. Many modern higher secondary textbooks introduce calculus in this order-limits,differentiation and rules,integration(introduced as reverse of differentiation) and rules,area problem by anti derivative and integral approach and then only rolle's theorom without proof.So i tried my proof.

If I had known the proof of this rolles theorom and mean value theorom before integration, I would not hav posted here. The syllabus requires me to study this way and naturally i will try to create the proof from what i hav learnt( by circular proof)!

 

[A. Bahat]

You can think of this, perhaps not as a proof, but as a consistency check, since the Fundamental Theorem of Calculus depends on Rolle's theorem.

 

[wisvuze]

If you've found such a proof independently, I say good work!
It is not decisively circular, as it is not necessarily true that the proof of the fundamental theorem of calculus requires rolle's theorem ( depending on how you prove it ). I think what is interesting is what you will need to assume ( in common ) in any configuration of proofs ( rolle's theorem first or FTC first.. et c ), to help see what is going on between the two.

 

[lavinia]

Can concepts of definite integrals be applied to prove rolles theorom? if so is my proof correct?

Let us consider a function y=g(x). it is continous. the range of y is(-∞,∞).let the derivative of g(x) be f(x)..ie g'(x)=f(x).so the area between the limits a and b is lim(b,a)∫g(x)dx.


So, lim(b,a)∫g(x)dx=f(b)-f(a)+c-----------{1}

We also know area between limits is (b-a)Ʃf{a+(r-1)Δx}/n.--------{2}
Let us take the term Ʃf{a+(r-1)Δx}/n=k-----{3}
We know {2}={3}
So (b-a)Ʃf{a+(r-1)Δx}/n.= lim(b,a)∫g(x)dx=f(b)-f(a)

So, (b-a)*k=f(b)-f(a) [from {3}]So k=f(b)-f(a)/b-a. Since g(x) is continous, let's say g(c)=k..ie f'(c)=khence f'(c)=f(b)-f(a)/b-a

Hence proved

You have mixed up your notation so it is hard to follow what you are saying. You start by saying that f is the derivative of g then integrate g to get f.

So let's fix this and say that g is the derivative of f.

- first of all the term k changes as the Riemann sum becomes increasingly refined. It is not a constant. So your last equation, k=f(b)-f(a)/b-a, is false for any particular value of k.

- but this aside, we can define k by your last equation, k=f(b)-f(a)/b-a. the whole point of Rolle's theorem is to show that the derivative of f, which I am taking to be g (note your Riemann sums have the wrong letters in them), must equal this average value at some point. But this does not follow just because g is continuous which is what you argue at the end of the post.

Maybe I don't understand what you are trying to get at. If I don't please clarify and we can go through it again.

I think though that the key fact in proving Rolle's theorem is that if the derivative of a function has constant sign then it must either be increasing or decreasing. This follows easily from the definition in terms of Riemann sums. Then you can use the continuity of the derivative to invoke the intermediate value theorem.

 

[mathwonk]

there are of course two fundamental theorems of calculus, one that the integral from a to x of a continuous g has derivative equal to g(x). The other says that a function f(x) with continuous derivative equals f(a) + the integral from a to x of that derivative.

The first one does not use Rolle, but I am curious to see how the second one can be proved without Rolle. It seems Lavinia is using the second one to deduce that a function whose continuous derivative does not change sign is monotone. I.e. certainly the integral of that derivative is monotone, but why is that integral equal to the original function?

Also the usual Rolle does not assume continuity of the derivative.

The usual proof of Rolle can hardly be simpler:
1) a differentiable function on [a,b] is also continuous, hence if f(a) = f(b), it has an extremum at some interior point.
2) A differentiable function with an extremum at an interior point has derivative zero there.

It is tempting to try to argue that the derivative must change sign and hence must equal zero in between by intermediate values, but the point is that the intermediate value theorem does not apply since the derivative is not assumed continuous. In fact one interesting point is that after proving Rolle, one can deduce that all derivatives, continuous or not, do satisfy the intermediate value theorem anyway.

Perhaps I have not understood the discussion above. In any event, any new argument is always worth giving no matter what it assumes.

 

[arildno]

In any event, any new argument is always worth giving no matter what it assumes.

My version is:
"In any event, any new valid argument is always worth giving no matter what it assumes", unless arguments are, by definition, valid.

 

[mathwonk]

boy what a stickler! Only valid arguments?!

 

[arildno]

boy what a stickler! Only valid arguments?!

In maths, yes; in politics, anything goes as long as you get the desired votes!

 

[Alpharup]

You have mixed up your notation so it is hard to follow what you are saying. You start by saying that f is the derivative of g then integrate g to get f.

So let's fix this and say that g is the derivative of f.

- first of all the term k changes as the Riemann sum becomes increasingly refined. It is not a constant. So your last equation, k=f(b)-f(a)/b-a, is false for any particular value of k.

- but this aside, we can define k by your last equation, k=f(b)-f(a)/b-a. the whole point of Rolle's theorem is to show that the derivative of f, which I am taking to be g (note your Riemann sums have the wrong letters in them), must equal this average value at some point. But this does not follow just because g is continuous which is what you argue at the end of the post.

Maybe I don't understand what you are trying to get at. If I don't please clarify and we can go through it again.

I think though that the key fact in proving Rolle's theorem is that if the derivative of a function has constant sign then it must either be increasing or decreasing. This follows easily from the definition in terms of Riemann sums. Then you can use the continuity of the derivative to invoke the intermediate value theorem.

I deifinitely know nothing about Riemann sums as iam a higher secondary student. But by the term "sum", I assume that you r mentioning about the summation of area between a and b.It was my fault that I didnt mention about 'k'

I already said k=Ʃf{a+(r-1)Δx}/n

Here Δx=b-a/n where n is the number of rectangular strips between a and b and tends to infinity. Since n tends to infinity Δx tends to 0. The small area is equal to Δxf(m) where m is a constant between a abd b. Dividing the area between a and b into infinite rectangles and adding them up we would definitely get an approximate area between a and b. I too agree with your fact that it can't give the definite area as we do by integration but as the infinite number of rectangles of small area we can approach the definite area between a and b.

We know lim sinα/α=1 when α tends to zero. By using similar fact we can say Ʃf{a+(r-1)Δx}/n is also a constant..ie k is a constant.Moreover since the area between a and b is a constant and is equal to lim(b,a)∫g(x)dx. b and a are defined values and b-a is also a constant.

Hence the term lim(b,a)∫g(x)dx/b-a is also a constant ie..k is a constant

Hope my argument is right.

 

[lavinia]

Also the usual Rolle does not assume continuity of the derivative.

The usual proof of Rolle can hardly be simpler:
1) a differentiable function on [a,b] is also continuous, hence if f(a) = f(b), it has an extremum at some interior point.
2) A differentiable function with an extremum at an interior point has derivative zero there.

It is tempting to try to argue that the derivative must change sign and hence must equal zero in between by intermediate values, but the point is that the intermediate value theorem does not apply since the derivative is not assumed continuous. In fact one interesting point is that after proving Rolle, one can deduce that all derivatives, continuous or not, do satisfy the intermediate value theorem anyway.

Thanks Mathwonk. I love that the derivative can be discontinuous but that it still must be zero at some point if f(a) = f(b).

 

[mathwonk]

yes this implies that the intermediate value theorem always holds even for discontinuous derivatives. I.e. if f'(a) > 0 and f'(b) < 0, then f is increasing at a and decreasing at b, hence not monotonic. Thus by the intermediate value theorem for f, there are two points where f has the same value and then by Rolle f' is zero somewhere between a and b. Thus IVT holds also for f'.

I think this is marvelous as well. and this simple proof is perhaps not that well known, at least i have never seen it. this is one of the arguments i discovered while trying to make an old subject fresh for myself after years of teaching it by thinking it all through again and again.

Well of course I have only to say this to find the counterexample. This is known as Darboux's theorem, and the proof in Kitchen's Calculus of one variable, page 202, is if anything even simpler than mine. I.e. he notes that the hypothesis that f'(a) > 0 and f'(b) < 0 implies again that the maximum of f on [a,b] occurs at an interior point where thus f'= 0. the same proof appears in spivak calculus chapter 11, problem 39, p.187 of my old edition.

The more complicated proof I recalled was for a different result, that a continuous function differentiable away from c, whose derivatives have a limit at c, is also differentiable at c. Even that argument, using MVT applied to the difference quotient, seems easy now.

 

[Alpharup]

Anywayzz, why are mean value theorom and rolles theorom are used?

 

[lugita15]

yes this implies that the intermediate value theorem always holds even for discontinuous derivatives. I.e. if f'(a) > 0 and f'(b) < 0, then f is increasing at a and decreasing at b, hence not monotonic. Thus by the intermediate value theorem for f, there are two points where f has the same value and then by Rolle f' is zero somewhere between a and b. Thus IVT holds also for f'.

I think this is marvelous as well. and this simple proof is perhaps not that well known, at least i have never seen it. this is one of the arguments i discovered while trying to make an old subject fresh for myself after years of teaching it by thinking it all through again and again.

Well of course I have only to say this to find the counterexample. This is known as Darboux's theorem, and the proof in Kitchen's Calculus of one variable, page 202, is if anything even simpler than mine. I.e. he notes that the hypothesis that f'(a) > 0 and f'(b) < 0 implies again that the maximum of f on [a,b] occurs at an interior point where thus f'= 0. the same proof appears in spivak calculus chapter 11, problem 39, p.187 of my old edition.

The more complicated proof I recalled was for a different result, that a continuous function differentiable away from c, whose derivatives have a limit at c, is also differentiable at c. Even that argument, using MVT applied to the difference quotient, seems easy now.

Can we say something even stronger, that if all we know the right handed derivative exists then he right-handed derivative satisfies the intermediate value theorem?