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Ratio simplification using ratio series.

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data

    here is the problem: http://img155.imageshack.us/img155/5175/15399391yy7.png

    how do i simplify the fraction?

    2. Relevant equations



    3. The attempt at a solution

    i got the correct answer but i'm not sure how to simplify the fraction correctly, especially with the n+1 with a exponent 4.

    please help.
     
  2. jcsd
  3. Jan 26, 2008 #2

    Defennder

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    Have you learnt L'Hospital's rule yet? If you have then all you need to do is to evaluate [tex]\lim_{n\to \infty} \frac{n^4+16}{(n+1)^4 +16}[/tex] separately, then put that result back into the expression then you'll get the answer.
     
  4. Jan 27, 2008 #3
    Okay, ya after applying l'hopitals rule, i get 1. also, i believe i could change this (n+1)^4 to this n^4 + 1^4 i believe?

    thanks.
     
  5. Jan 28, 2008 #4

    Defennder

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    I don't think that's allowed. What rule are you following there?
     
  6. Jan 28, 2008 #5

    VietDao29

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    No, it's not correct at all.

    Here's a simple counter example. If n = -1, then your LHS will be 0, whereas your RHS is 2, they are not equal.

    To expand the terms that have the form: (a + b)n (where n is a natural number), one should use Binomial Theorem.

    Or, you can just devide both numerator, and denominator by n4 (the greatest power), like this:

    [tex]\lim_{n \rightarrow \infty} \left| (2x - 1) \frac{n ^ 4 + 16}{(n + 1) ^ 4 + 16} \right|[/tex]

    [tex]= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{\frac{n ^ 4 + 16}{n ^ 4}}{\frac{(n + 1) ^ 4 + 16}{n ^ 4}} \right|[/tex] (since 2x - 1 is a constant, independent of n, we can "pull" it out)

    [tex]= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\frac{(n + 1) ^ 4}{n ^ 4} + \frac{16}{n ^ 4}} \right|[/tex]

    [tex]= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right|[/tex]

    [tex]= |2x - 1| \lim_{n \rightarrow \infty} \left| \frac{1 + \frac{16}{n ^ 4}}{\left( \frac{n + 1}{n} \right) ^ 4 + \frac{16}{n ^ 4}} \right| = ...[/tex]

    Can you go from here? :)
     
    Last edited: Jan 28, 2008
  7. Jan 28, 2008 #6
    Viet Dao, thankyou for that thorough explanation! That was exactly my question, and it helped me a great deal!
     
  8. Jan 29, 2008 #7
    just skipping to the end, i get the interval of convergence to be [0, 1] where i plug these into the original equation and they both converge, correct? just need a confirmation.

    also the radius of convergence came to 1/2

    thanks.
     
  9. Jan 29, 2008 #8

    VietDao29

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    What is the series you are working on? You haven't shown us any series at all. :bugeye:
     
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