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Problems with some multivariable calc

  1. May 23, 2004 #1
    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. jcsd
  3. May 23, 2004 #2

    AKG

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    Let's get something we can work with:

    [tex]f(x) = x^3 + 3x^2 - x + 1[/tex]

    [tex]c = 2[/tex]

    [tex]0 < |x - c| < \delta[/tex]

    [tex]|f(x) - L| < \epsilon[/tex]


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

    You're going to have to let us know if this is the right interpretation.
     
    Last edited: May 23, 2004
  4. May 23, 2004 #3
    Perfect...both are correct.
     
    Last edited: May 23, 2004
  5. May 23, 2004 #4

    AKG

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    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.

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

    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 [itex]x_i[/itex] (or you may have seen [itex]x^*[/itex]) which would be something like the midpoint of the subinterval. So, the height is [itex]f(x^*)[/itex]

    It looks to me that we have the following:

    [tex]\int_{1}^3 x^2 dx = \frac{26}{3}[/tex]

    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 [itex]f(x^*)[/itex] and you should find that in general, if looking at an overall interval of [a,b], that:

    [tex]x^* = a + \frac{k(b - a)}{n}[/tex]

    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.
     
  6. May 23, 2004 #5

    AKG

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    [tex]\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} [/tex]

    [tex]\mbox{when using the Right Endpoint Method}[/tex]
     
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