Problems with some multivariable calc

1. May 23, 2004

Hey everyone, our teacher assigned us 15 pretty tough problems to finish by the end of the year and my partner and I have gotten through all but these last two, which none of the other teachers at school can figure out either. Any help would be greatly appreciated, thanks!

In the (Epsilon , Delta) definition of the limit (formal definition), limit as x -> c of f(x) = L, let f(x) = x^3 + 3x^2 - x + 1, and let c = 2. Find the least upper bound on delta so that f(x) is bounded within epsilon of L for all sufficiently small epsilon > 0.

Evaluate the limit (bear with me because I don't know how to insert math symbols, so I'll type it out),

limit as n -> infinity of (the series from k = 1 to k = n [ (1 + (2k) / n) ^ 2 (2 / n)]).

I'm 99.99% sure the answer is 26/3 (from the graphing), but I'm not sure how to prove it. Thanks in advance everyone!

2. May 23, 2004

AKG

Let's get something we can work with:

$$f(x) = x^3 + 3x^2 - x + 1$$

$$c = 2$$

$$0 < |x - c| < \delta$$

$$|f(x) - L| < \epsilon$$

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \left [ \left (1 + \frac{2k}{n} \right )^2 \left ( \frac{2}{n} \right ) \right ]$$

You're going to have to let us know if this is the right interpretation.

Last edited: May 23, 2004
3. May 23, 2004

Perfect...both are correct.

Last edited: May 23, 2004
4. May 23, 2004

AKG

First off, this all looks like single-variable calculus, not multi, right? Anyways, this one is quite easy if you're familiar with integration and the Riemann sum.

$$\lim_{n \rightarrow \infty} \sum_{k=1}^n \left [ \left (1 + \frac{2k}{n} \right )^2 \left ( \frac{2}{n} \right ) \right ]$$

When we integrate, it can be seen as taking a function, and cutting the area underneath it into rectangles, and summing the areas of the rectangles as the number of rectangles approaches infinity (and the width approaches zero).

So, given n rectangles, the area under the curve would be the sum of the areas of all the rectangles. Each rectangle has a uniform width. Let's say we're integrating from x=a to x=b. So, with n rectangles, each rectangle will have a width of (b-a)/n. That's where the (2/n) in the sum comes from. Now, each rectangle has a height. The height is normally some function of $x_i$ (or you may have seen $x^*$) which would be something like the midpoint of the subinterval. So, the height is $f(x^*)$

It looks to me that we have the following:

$$\int_{1}^3 x^2 dx = \frac{26}{3}$$

Seems like your graphing was right. Start with n = 5, and make the rectangles such that the height of a rectangle on the interval [m,n] is f(n). This is called the Right-Endpoint Method (or something like that). Remember, the height of a rectangle is $f(x^*)$ and you should find that in general, if looking at an overall interval of [a,b], that:

$$x^* = a + \frac{k(b - a)}{n}$$

Hopefully, if you consider all this it will all come together and make sense, I don't know if I did the best job of explaining it though.

5. May 23, 2004

AKG

$$\int_a^b f(x) dx = \lim _{n \rightarrow \infty} \sum _{i=1} ^n f(x_i) \left ( \frac{b-a}{n} \right ) \mbox { where } x_i = a\ +\ \frac{k(b - a)}{n}$$

$$\mbox{when using the Right Endpoint Method}$$