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Thread: Sums involving functions of the form y=a^x

  1. #1
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    Sums involving functions of the form y=a^x

    I have my calculus 2 midterm today, and after going through the chapters I have found the only reccuring issue I have is with functions of the form $\displaystyle y=a^x$.

    If someone here could help explain how to solve these function for areas without using integral calculus I would greatly appreciate it.

    I'll give an example and my steps involved:

    Solve for the area under the function $\displaystyle y=e^x$ on the interval $\displaystyle [0, b]$.

    Let $\displaystyle x_i=0+(b/n)i$, and thus $\displaystyle y_i=e^{(bi)/n$

    $\displaystyle \displaystyle\sum_{i=1}^{n}{y_i\Delta{x}$

    $\displaystyle {\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}(b/n)$

    $\displaystyle \displaystyle\sum_{i=1}^{n}{e^{(bi)/n}(b/n)$

    $\displaystyle (b/n)\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}$

    $\displaystyle (b/n)\frac{e^{b/n}-e^{b+1}}{1-e^{b/n}}$

    It's at this point that I become flustered. I understand there is manipulation that must occur, but can someone give me tips or show me how, with ease? Keep in mind I am not a math wizard.
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  2. #2
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    I think you performed the sum incorrectly (last line):

    $\displaystyle (b/n)\displaystyle\sum_{i=1}^{n}{e^{(bi)/n}=(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}.$

    You now need to compute the limit:

    $\displaystyle \displaystyle\lim_{n\to\infty}(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}.$

    Can you do that?
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  3. #3
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    Sorry, that was a typo. The issue I have is calculating the limit, because it's obvious the answer is not zero. I'm not quite sure how to manipulate it such that I can calculate it.
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  4. #4
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    Are you allowed to use L'Hopital's Rule?
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  5. #5
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    I would assume so. It doesn't really state. How would you use this rule?
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  6. #6
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    Quote Originally Posted by quantoembryo View Post
    I would assume so. It doesn't really state. How would you use this rule?
    Well, you have to get your fraction either into the form $\displaystyle 0/0$ or $\displaystyle \infty/\infty.$ Try this:

    $\displaystyle \displaystyle\lim_{n\to\infty}(b/n)\frac{e^{b/n}-e^{b+b/n}}{1-e^{b/n}}=b(1-e^{b})\lim_{n\to\infty}(1/n)\frac{e^{b/n}}{1-e^{b/n}}$

    $\displaystyle =\displaystyle b(1-e^{b})\left[\lim_{n\to\infty}e^{b/n}\right]\left[\lim_{n\to\infty}\frac{1/n}{1-e^{b/n}}\right].$

    The limit on the left is just 1. The limit on the right is ready for L'Hopital's rule, which states that, under certain conditions, you can compute

    $\displaystyle \displaystyle\lim_{n\to\infty}\frac{f(n)}{g(n)}=\l im_{n\to\infty}\frac{f'(n)}{g'(n)}.$

    So what do you get when you apply the rule?
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  7. #7
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    Would the answer be 1-e^b?
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  8. #8
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    Nope, not quite. Can you show your work?
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  9. #9
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    $\displaystyle \frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{f'(n)}{g' (n)}$

    $\displaystyle \frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{-1/n^2}{e^{b/n}b/n^2}$

    $\displaystyle \frac{f(n)}{g(n)}=\lim_{n\to\infty}\frac{-1/n^2}{e^{b/n}b/n^2}=\frac{-1}{e^{b/n}b}$

    and therefore,
    $\displaystyle =b(1-e^{b})(1)(\frac{-1}{e^{b/n}b})$

    $\displaystyle =\frac{e^b-1}{e^{b/n}}$
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  10. #10
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    The way you're writing it is a bit off. Keep the limit sign there! Your final result should not have an n in it, because you're taking the limit as n approaches infinity.
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