Using Mean Value Theorem to Prove a Summation Equation?

[Yagoda]

Homework Statement

Use the mean value theorem to show that $\frac{b^3-a^3}{b-a} = \sum_{j=1}^{n}d_j^2 (x_j - x_{j-1}) \text{where} x_{j-1} < d_j < x_j .$

Homework Equations

The mean value theorem states that if f is continuous on [a,b] and differentiable on (a,b) then there exists a c in (a,b) such that $f(c) = \frac{f(b)-f(a)}{b-a}$.

The Attempt at a Solution

I think I'm approaching this wrong, but I am trying to think of $\sum_{j=1}^{n}d_j^2 (x_j - x_{j-1})$ as the derivative of a function $\sum_{j=1}^{n}\frac{d_j^3}{3} (x_j - x_{j-1})$. This way $\frac{f(b)-f(a)}{b-a}$ resembles $\frac{b^3-a^3}{b-a}$, but then it's unclear where the function is being evaluated at since you have several different d's. It seems clear that the d's are analogous to c in the MVT, but maybe I'm confused because there is a single c in the MVT and several d's.

 

[Office_Shredder]

Does it say anything about what the x_js are?

 

[Yagoda]

The xj's are a partition of the interval [a,b]. Sorry I forgot to mention that.

 

[pasmith]

Homework Statement

Use the mean value theorem to show that $\frac{b^3-a^3}{b-a} = \sum_{j=1}^{n}d_j^2 (x_j - x_{j-1}) \text{where} x_{j-1} < d_j < x_j .$

I think you mean
$$\frac{b^3-a^3}{3} = \sum_{j=1}^{n}d_j^2 (x_j - x_{j-1})$$
(The question wants you to find a telescoping Riemann sum for $\int_a^b x^2\,dx$).

Homework Equations

The mean value theorem states that if f is continuous on [a,b] and differentiable on (a,b) then there exists a c in (a,b) such that $f(c) = \frac{f(b)-f(a)}{b-a}$.

No, it states that under those conditions there exists $c \in (a,b)$ such that
$$f'(c) = \frac{f(b)-f(a)}{b-a}$$

The Attempt at a Solution

I think I'm approaching this wrong, but I am trying to think of $\sum_{j=1}^{n}d_j^2 (x_j - x_{j-1})$ as the derivative of a function $\sum_{j=1}^{n}\frac{d_j^3}{3} (x_j - x_{j-1})$. This way $\frac{f(b)-f(a)}{b-a}$ resembles $\frac{b^3-a^3}{b-a}$, but then it's unclear where the function is being evaluated at since you have several different d's. It seems clear that the d's are analogous to c in the MVT, but maybe I'm confused because there is a single c in the MVT and several d's.

You need to apply the MVT to $f(x) = x^3$ on the subinterval $[x_{j-1}, x_j]$.

 

[Yagoda]

Sorry for the typos. I see how to use the MVT now. Not sure if I should post this is a new thread, but I'm now trying to use this result to prove that integral of f(x)=x^2 on [a,b] both exists and equals $\frac{b^3-a^3}{3}$.

I know that for a quadratic, the point that satisfies the MVT is the midpoint of the interval, $\hat{c_i} = \frac{x_j + x_{j-1}}{2}$. So to generalize for any point c in any partition we know that any point in $[x_{j-1}, x_j ]$ is at most $\frac{||P||}{2}$ units away from $\hat{c_i}$, where ||P|| is the norm of the partition. Call the actual distance from the midpoint $\partial_j$.

So I want to show that $\left| \sum_{j=1}^{n} \left(\frac{x_j + x_{j-1}}{2} + \partial_j\right)^3 (x_j - x_{j-1}) - \frac{b^3-a^3}{3} \right| < \epsilon$ when $||P||< \delta$. I was hoping to bound the d's by ||P||/2 and use that to find a delta, but this sum turns out very messy and the terms don't cancel nicely like I'd hoped. Is this the right approach?

 

[pasmith]

Sorry for the typos. I see how to use the MVT now. Not sure if I should post this is a new thread, but I'm now trying to use this result to prove that integral of f(x)=x^2 on [a,b] both exists and equals $\frac{b^3-a^3}{3}$.

I know that for a quadratic, the point that satisfies the MVT is the midpoint of the interval, $\hat{c_i} = \frac{x_j + x_{j-1}}{2}$. So to generalize for any point c in any partition we know that any point in $[x_{j-1}, x_j ]$ is at most $\frac{||P||}{2}$ units away from $\hat{c_i}$, where ||P|| is the norm of the partition. Call the actual distance from the midpoint $\partial_j$.

So I want to show that $\left| \sum_{j=1}^{n} \left(\frac{x_j + x_{j-1}}{2} + \partial_j\right)^3 (x_j - x_{j-1}) - \frac{b^3-a^3}{3} \right| < \epsilon$ when $||P||< \delta$. I was hoping to bound the d's by ||P||/2 and use that to find a delta, but this sum turns out very messy and the terms don't cancel nicely like I'd hoped. Is this the right approach?

That is not the right approach.

It is enough if you can find a Riemann sum whose value is independent of the norm of the partition. Then the function is integrable and its integral is the value of that sum. The MVT allows you to do that.

Thus, by the MVT applied to $x^3$ on $[x_{j-1},x_j]$ there exists $z_j \in [x_{j-1},x_j]$ such that
$$3z_j^2 = \frac{x_j^3 - x_{j-1}^3}{x_j - x_{j-1}}.$$
Hence, after multiplying by $(x_j - x_{j-1})/3$ and summing over j,
$$\sum_{j=1}^n z_j^2 (x_j - x_{j-1}) = \frac13 \sum_{j=1}^n (x_j^3 - x_{j-1}^3)$$
The sum on the right can now be evaluated.

 

[Yagoda]

The MVT only says that there exists a $z_j$ such that $\sum_{j=1}^n z_j^2 (x_j - x_{j-1}) = \frac13 \sum_{j=1}^n (x_j^3 - x_{j-1}^3)$, but not that this is necessarily true of any point in the interval.

We were shown a similar example where $f(x) = x$ and we chose to evaluate the Riemann sum at the midpoint of the interval $c_j = \frac{x_{j-1}+ x_j}{2}$ so it turned out to be that $\sigma = \frac12 \sum_{j=1}^n (x_{j-1}+ x_j) (x_j - x_{j-1}) = \frac12 (b-a)$, which does not appear to depend on ||P||, but we were told then that we had to show that this answer was independent of the choice of $c_j$, which we did by bounding the distance of any point in the interval $(x_{j-1}, x_j)$ from the midpoint by ||P||/2.

I am curious why this last step is not necessary with the $f(x) = x^2$ problem because it seems like using the MVT to find the sum at a certain point is similar to evaluating the sum at the midpoint.

 

[pasmith]

The MVT only says that there exists a $z_j$ such that $\sum_{j=1}^n z_j^2 (x_j - x_{j-1}) = \frac13 \sum_{j=1}^n (x_j^3 - x_{j-1}^3)$, but not that this is necessarily true of any point in the interval.

We were shown a similar example where $f(x) = x$ and we chose to evaluate the Riemann sum at the midpoint of the interval $c_j = \frac{x_{j-1}+ x_j}{2}$ so it turned out to be that $\sigma = \frac12 \sum_{j=1}^n (x_{j-1}+ x_j) (x_j - x_{j-1}) = \frac12 (b-a)$, which does not appear to depend on ||P||, but we were told then that we had to show that this answer was independent of the choice of $c_j$, which we did by bounding the distance of any point in the interval $(x_{j-1}, x_j)$ from the midpoint by ||P||/2.

I am curious why this last step is not necessary with the $f(x) = x^2$ problem because it seems like using the MVT to find the sum at a certain point is similar to evaluating the sum at the midpoint.

I appear to have been implicitly assuming that you have already proven that $x^2$ is integrable by some other means.

I think what you want to do to show that the sum is independent of the choice of point in each subinterval is to show that for all $\epsilon > 0$ there exists $\delta > 0$ such that if $\|P\| < \delta$ then
$$\left| \sum_{j} (z_j + \eta_j)^2(x_j - x_{j-1}) - \frac{b^3 - a^3}3 \right| < \epsilon$$
where $|\eta_j| < \frac12(x_j - x_{j-1})$ and $z_j$ is given by the MVT.