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

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The discussion revolves around proving that if the sum \(\frac{a_0}{1} + \frac{a_1}{2} + ... + \frac{a_n}{(n+1)} = 0\), then the polynomial \(a_0 + a_1x + ... + a_nx^n = 0\) holds for some \(x\) in the interval \([0, 1]\). Participants explore using limits and Rolle's theorem to establish the relationship between the function \(F(x)\) and its derivative. They confirm that \(F(x)\) is continuous and differentiable on the interval, with \(F(0) = 0\) and \(F(1) = 0\), satisfying the conditions of Rolle's theorem. Ultimately, they conclude that there exists a point \(c\) in \((0, 1)\) where \(F'(c) = 0\), demonstrating the required relationship.
  • #31
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.
 
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  • #32
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?
 
  • #33
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.
 
  • #34
Justabeginner said:
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.
 
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  • #35
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.
 

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