Website Title: How Can Limits Help Solve This Word Problem with Intervals?

[Justabeginner]

Homework Statement

Show that \frac{a_0}{1} + \frac{a_1}{2} + ... \frac{(a_n)}{(n+1)} = 0

then a_0 + a_1x + ... + a_nx^n = 0

for some x in the interval [0, 1].

Homework Equations

The Attempt at a Solution

I thought at first the easiest value to find for a, would be 0, but it is not included on the interval and the question wants a more theoretical approach, rather than solving for a itself. I think I would be able to use limits in order to solve this question, but the most puzzling part is how to set it up (what would the function be on which the limit is being tested)? Is it the general form: \frac{a_n}{(n+1)} and a_nx^n ? Thank you.

 

[LCKurtz]

Homework Statement

Show that [IF]\frac{a_0}{1} + \frac{a_1}{2} + ... \frac{(a_n)}{(n+1)} = 0

then a_0 + a_1x + ... + a_nx^n = 0

for some x in the interval [0, 1].

Homework Equations

The Attempt at a Solution

I thought at first the easiest value to find for a, would be 0, but it is not included on the interval and the question wants a more theoretical approach, rather than solving for a itself. I think I would be able to use limits in order to solve this question, but the most puzzling part is how to set it up (what would the function be on which the limit is being tested)? Is it the general form: \frac{a_n}{(n+1)} and a_nx^n ? Thank you.

Hint: Let$$
F(x) = \int_0^x a_0+a_1t+...+a_nt^n\, dt$$and think about Rolle's theorem with $F(x)$.

 

[Justabeginner]

When I integrate the above function I get:

a_0x + \frac{a_1x^2}{2} + ... + \frac{a_nx^(n+1)}{n+1}

Rolle's theorem guarantees that there will be some c on the closed interval (a,b) such that f'(c)=0.

I remember reading somewhere in the textbook that \frac{1}{b-a} ∫_a^b f(x) dx (the average value) can be found using Rolle's Theorem?

 

[LCKurtz]

When I integrate the above function I get:

a_0x + \frac{a_1x^2}{2} + ... + \frac{a_nx^(n+1)}{n+1}

Rolle's theorem guarantees that there will be some c on the closed interval (a,b) such that f'(c)=0.

I remember reading somewhere in the textbook that \frac{1}{b-a} ∫_a^b f(x) dx (the average value) can be found using Rolle's Theorem?

Rolle's theorem has hypotheses and conclusions. State the theorem completely. Then try applying it to F(x) and show what you get. Skip the guessing about extraneous non-related issues.

 

[Justabeginner]

I was mentioning it because I thought it could be useful in this problem, but I guess not.

The derivative of f(x) is a_0 + a_1x ... + a_nx^n
f'(0)= a_0
So f'(x) is never equal to zero.
By Rolle's Theorem, this means that this equation cannot have 2 or more solutions.

 

[LCKurtz]

I asked you to state Rolle's theorem completely. Then try applying it to F(x) and show what you get. If you are going to ignore my suggestions, how do you expect me to help you? And please use consistent notation. Don't use f(x) if you mean F(x).

 

[Justabeginner]

I amn't ignoring your suggestions Professor. I thought that by Rolle's Theorem, I should find the critical points through the first derivative, and that is why I did that.

I will use F(x) from now on to indicate the function in this problem.

The Rolle's Theorem states exactly this: "Let a < b. If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)= f(b) then there is a c in (a,b) with f'(c)=0. That is, under these hypotheses F has a horizontal tangent somewhere between a and b".

 

[LCKurtz]

I amn't ignoring your suggestions Professor. I thought that by Rolle's Theorem, I should find the critical points through the first derivative, and that is why I did that.

I will use F(x) from now on to indicate the function in this problem.

The Rolle's Theorem states exactly this: "Let a < b. If F is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)= f(b) then there is a c in (a,b) with f'(c)=0. That is, under these hypotheses F has a horizontal tangent somewhere between a and b".

OK, that's better but you are still mixing notation. Use either F or f, but not both in the statement of the theorem. Now you know what the theorem says.

Now, what does it say about the $F(x)$ I suggested, and which you calculated correctly in post #3. What are a and b in your problem and does $F(x)$ satisfy the hypotheses (show this)? If so, what does the conclusion tell you?

 

[Justabeginner]

A and B are 0 and 1 in this problem. A < B and f'(x)=0 because the problem states that some x on the interval of [a,b] gives a y-value of 0. Rolle's Theorem tells me that f(a) will have to equal f(b) since A < B, and f'(c)= 0?

 

[LCKurtz]

A and B are 0 and 1 in this problem. A < B

Yes. We are dealing with the interval [0,1]

and f'(x)=0 because the problem states that some x on the interval of [a,b] gives a y-value of 0. Rolle's Theorem tells me that f(a) will have to equal f(b) since A < B, and f'(c)= 0?

I thought we were talking about your problem with $F(x)$. Rolle's theorem has hypotheses which the function $F(x)$ must satisfy before you can draw the conclusion.

1. Is F(x) continuous on the closed interval?
2. Is it differentiable on the open interval?
3 Does F(0) = F(1)?

These are the hypotheses, and you need to verify them or at least state why they are true. When you have done that, then what does the conclusion of the theorem tell you for our F(x)?

 

[Justabeginner]

I don't think I know that f(0) is equal to f(1). I do know that the function is continuous on the interval, and differentiable. How can you come to such a conclusion? Thank you.

 

[LCKurtz]

I don't think I know that f(0) is equal to f(1). I do know that the function is continuous on the interval, and differentiable. How can you come to such a conclusion? Thank you.

PLEASE use F(x) and not f(x) if you are talking about the function I suggested. I gave you a hint for F(x) and you have a formula for F(x) that you figured out in post #3. You don't have to guess whether F(0) = F(1); use the formula and check it.

 

[Justabeginner]

I must be doing something wrong because I get F(0) = 0 and F(1)≠ 0.

 

[LCKurtz]

I must be doing something wrong because I get F(0) = 0 and F(1)≠ 0.

Show your work. What do you get for F(1)? And at this point it might be a good idea for you to read the original problem again so you haven't forgotten what you are given and what you are trying to prove.

 

[Justabeginner]

a_0x + \frac{a_1x^2}{2} + ... + \frac{a_n(x^(n+1))}{n+1}

Using this formula I substituted in 0 and 1.

F(0)= a_0(0) + \frac{a_1*(0^2)}{2} + ... + \frac{a_n(0^(n+1))}{n+1} = 0

F(1)= a_0(1) + \frac{a_1*(1^2)}{2} + ... + \frac{a_n(1^(n+1))}{n+1} = a_0 + \frac{a_1}{2} + ... + \frac{a_n*(1^(n+1)}{n+1}

 

[LCKurtz]

a_0x + \frac{a_1x^2}{2} + ... + \frac{a_n(x^(n+1))}{n+1}

Using this formula I substituted in 0 and 1.

F(0)= a_0(0) + \frac{a_1*(0^2)}{2} + ... + \frac{a_n(0^(n+1))}{n+1} = 0

F(1)= a_0(1) + \frac{a_1*(1^2)}{2} + ... + \frac{a_n(1^(n+1))}{n+1} = a_0 + \frac{a_1}{2} + ... + \frac{a_n*(1^(n+1)}{n+1}

Since $1^{n+1}=1$ you hardly need to leave it in there. Note the correction I edited in post #2 to your statement of the problem. Does that help you with F(1)?

 

[Justabeginner]

I would think that means the last quantity is negligible so F(1)= a_0 + \frac {a_1}{2} ? But I still don't see how they're equal. I'm sorry I'm so slow to catch on.

 

[LCKurtz]

Since $1^{n+1}=1$ you hardly need to leave it in there. Note the correction I edited in post #2 to your statement of the problem. Does that help you with F(1)?

I would think that means the last quantity is negligible so F(1)= a_0 + \frac {a_1}{2} ?

Why would you think that? You calculated F(1) and it isn't just two terms. What does the original problem say about that quantity? You did read my post #2 showing the IF that you left out, right?

 

[Justabeginner]

Yes, I did read that post (it was a slip-up). The original problem states that the second equation is contingent upon the truth of the first equation.

 

[LCKurtz]

Yes, I did read that post (it was a slip-up). The original problem states that the second equation is contingent upon the truth of the first equation.

But the problem is giving you the first equation by saying IF it is true then prove...

So what is the value of F(1)?

Also, to satisfy my curiosity, what grade level are you in and what math courses have you completed?

 

[Justabeginner]

I'm a junior in high school and have completed everything up to Calculus, including Precalculus.

F(1)= a_0 + \frac{a_1}{2} + a^n ?

I got a^n for the last term because I presumed that if 1^(n+1) = 1, then n + 1= 1?

 

[LCKurtz]

a_0x + \frac{a_1x^2}{2} + ... + \frac{a_n(x^(n+1))}{n+1}

Using this formula I substituted in 0 and 1.

F(0)= a_0(0) + \frac{a_1*(0^2)}{2} + ... + \frac{a_n(0^(n+1))}{n+1} = 0

F(1)= a_0(1) + \frac{a_1*(1^2)}{2} + ... + \frac{a_n(1^(n+1))}{n+1} = a_0 + \frac{a_1}{2} + ... + \frac{a_n*(1^(n+1)}{n+1}

This is just $F(1)=\color{red}{a_0 + \frac{a_1}{2} + ... + \frac{a_n}{n+1}}$, leaving out the $1$'s.

I'm a junior in high school and have completed everything up to Calculus, including Precalculus.

F(1)= a_0 + \frac{a_1}{2} + a^n ?

Why are you changing your answer for F(1)?

I got a^n for the last term because I presumed that if 1^(n+1) = 1, then n + 1= 1?

That last sentence is just silly.

One last time: What does your problem tell you about the quantity F(1) highlighted above?

 

[Justabeginner]

It tells you that F(1)= F(0).

 

[LCKurtz]

It tells you that F(1)= F(0).

When you respond like that it makes me wonder if you really get it. What it tells you is that F(1) = 0. That is why F(1) = F(0). Now the last question. You have that F(x) satisfies all the hypotheses of Rolle's theorem on [0,1].

What is the conclusion of Rolle's theorem? Write it like this:

Therefore, by Rolle's theorem we have ____________. Then explain how that solves the problem.

 

[Justabeginner]

Therefore, by Rolle's Theorem we have that there is at least one point c in (a,b) where f'(c)= 0.

By integrating F(x) and finding the values of the function at x=0 and x=1, you are able to prove the three parts to Rolle's Theorem: 1) f is differentiable on (a,b) 2) f is continuous on [a,b] and 3) f(a)=f(b)=0. When you differentiate the equation derived in Post 3, you will get the original function F(x) and f'(c)= 0.

 

[LCKurtz]

Therefore, by Rolle's Theorem we have that there is at least one point c in (a,b) where f'(c)= 0.

There you go again with the lower case f's and the a and b. We are talking about our particular F(x) on [0,1] What is the conclusion for THIS FUNCTION? How does it solve the problem?

 

[Justabeginner]

The theorem tells us for this problem, that F'(x) = 0, when 0 < x < 1.

 

[LCKurtz]

The theorem tells us for this problem, that F'(x) = 0, when 0 < x < 1.

No it doesn't. Use the exact wording of the theorem. It tells us that THERE EXISTS a c such that...

Write it down exactly. And use your formula to see what it tells you. Remember, your problem asked you to prove something. Until you see what all this has to do with that, you aren't finished.

 

[Justabeginner]

In the context of this problem, Rolle's Theorem tells us that there exists a c in (a,b) such that F'(c)=0. When you plug in the values of a and b into the formula, you get a y value of zero for both, thereby indicating that the function is differentiable and continuous on the interval (a,b)

 

[LCKurtz]

In the context of this problem, Rolle's Theorem tells us that there exists a c in (a,b) such that F'(c)=0.

No. It tells us that there is a c in (0,1) such that F'(c) = 0. We are applying Rolle's theorem to THIS PROBLEM. What is so hard about that?

When you plug in the values of a and b into the formula, you get a y value of zero for both, thereby indicating that the function is differentiable and continuous on the interval (a,b)

No, no. That is a jumbled mess. You are mixing up hypotheses and conclusions. The conclusion of Rolle's theorem is we have a c in (0,1) where F'(c) = 0. That's it. That's all Rolles theorem tells you.

Now, for the last time I am going to ask you once again. You have a formula for F(x). Use it. What does F'(c) = 0 tell you? Calculate it and write it down. How does it relate to what this problem asks you to prove?

 

[Justabeginner]

F'(c)= 0 tells you that some c, when plugged into this equation: a_0 + a_1x + ... + a_nx^n is a zero. The problem is asking you to prove that a_0 + a_1c + ... + a_nc^n is equal to zero.

 

[LCKurtz]

OK, so it is done. Now, do you think you understand it well enough that you could write an organized argument that's about 5 lines long that works the problem from start to finish?

 

[Justabeginner]

I think I can try.

In the context of this problem, Rolle's Theorem tells us that there exists a c in (0,1) such that F'(c)=0. The integral of a0 + a1x + … + anxn is a0n + a1x^2/2 + … + anxn^2/2. When you plug in zero and one (a and b) into the equation, f(a) and f(b) are equal to zero (and equal to each other), thereby supporting one of the three hypotheses of the Rolle’s Theorem. The other hypotheses states that the function is differentiable on the interval which it is because the derivative of the integrated equation is the original equation. Subsequently, the function is continuous on that interval, supporting the last of the hypotheses of the Rolle’s Theorem.

^Sorry for the lack of LaTeX in that summary- I hope it is still clear enough.

 

[LCKurtz]

I think I can try.

In the context of this problem, Rolle's Theorem tells us that there exists a c in (0,1) such that F'(c)=0. The integral of a0 + a1x + … + anxn is a0n + a1x^2/2 + … + anxn^2/2. When you plug in zero and one (a and b) into the equation, f(a) and f(b) are equal to zero (and equal to each other), thereby supporting one of the three hypotheses of the Rolle’s Theorem. The other hypotheses states that the function is differentiable on the interval which it is because the derivative of the integrated equation is the original equation. Subsequently, the function is continuous on that interval, supporting the last of the hypotheses of the Rolle’s Theorem.

^Sorry for the lack of LaTeX in that summary- I hope it is still clear enough.

If this were a real class and you were to hand in this problem, you would begin with a statement of the problem including what you are given and what you are to prove, like this:

Show that if \frac{a_0}{1} + \frac{a_1}{2} + ... \frac{a_n}{(n+1)} = 0

then a_0 + a_1x + ... + a_nx^n = 0 for some x in the interval [0, 1]. Then your argument would begin like this. Let:
$$
F(x) = \int_0^x a_0+a_1t+...+a_nt^n\, dt =
a_0x + \frac{a_1x^2}{2} + ... + \frac{a_nx^{n+1}}{n+1}$$

At this point you would observe that since $F$ is a polynomial, both it and its derivative exist and are continuous on [0,1]. Now you observe that obviously $F(0)=0$ and$$
F(1) =a_0 + \frac{a_1}{2} + ... + \frac{a_n}{n+1}$$which is given to be $0$. So you have $F(1)=F(0)$.

That completes checking the hypotheses of Rolle's theorem so now you state the conclusion: There exists a $c$ in (0,1) such that $F'(c) = 0$. Since$$
F'(x) = a_0 + a_1x + ... + a_nx^n$$this says that$$
a_0 + a_1c + ... + a_nc^n=0$$which is what we were trying to prove (we have found the required value of x).

I would normally not give a complete writeup such as this, but since you are learning the subject on your own, I thought you might learn something from a proper writeup.

 

[Justabeginner]

Professor Kurtz, thank you so much for going over this with me in detail. I truly appreciate it. I didn't know the format for a 'proof' of this kind (or that it is required to write it in this way). I can say that I understand the material now, and it is all thanks to you.