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Math Help - solve equation

  1. #1
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    solve equation

    a_{n+1}=7a_n-6, and a_1=10,a_2=64
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  2. #2
    Super Member Quacky's Avatar
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    Re: solve equation

    What do you mean "solve equation"? There isn't anything to solve. You have a recurrence relation. What is the exact wording of the question?
    Last edited by Quacky; November 15th 2011 at 08:18 AM.
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  3. #3
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    Re: solve equation

    IN this case we don't want recurrence relationship, we need determine a_n using n only.
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  4. #4
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    Re: solve equation

    Hello, greg1987!

    Have you been taught anything about recurrence relations?

    It's impossible to help you if I don't know what you have learned so far
    . . and where you are having difficulty.


    \text{Given: }\:a_{n+1}\:=\:7a_n-6,\;\text{ and }\,a_1=10,\;a_2=64
    . . \text{Find the closed form: }f(n).

    By the way, the answer is:. f(n) \:=\:9\cdot7^{n-1} + 1

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  5. #5
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    Re: solve equation

    Quote Originally Posted by greg1987 View Post
    a_{n+1}=7a_n-6, and a_1=10,a_2=64
    Is a_n the n^{th} term of an arithmetic progression?
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  6. #6
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    Re: solve equation

    Quote Originally Posted by sbhatnagar View Post
    Is a_n the n^{th} term of an arithmetic progression?
    Clearly not, since the recurrence relation would then take the form a_{n+1} - a_n = d where d is a constant, and obviously this is not the case. Not to mention the fact that post #4 obviously answers your question in the negative.
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  7. #7
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    Re: solve equation

    For "linear recurrences with constant coefficients", things such as a_{n+1}= ba_n+ c where b and c are numbers, there is a very general method: first drop the number that is not multiplying and "a": a_{n+1}= ba_n and "try" a_n= r^n. Then a_{n+1}= r^{n+1} so the equation becomes r^{n+1}= br^n and, dividing both sides by r^n, r= b. That is a_n= b^n satisfies a_{n+1}= ba_n. In fact it is easy to see that any constant times that, a_n= Pb^n, is a solution. To find a solution to the entire equation, "try" a constant: if a_n= A the equation becomes A= bA+ c so (1- b)A= c and (if b is not 1), A= \frac{c}{1- b}. The solution to the entire equation is the sum of those: a_n= Pb^n+ \frac{c}{1- b} for any constant p. (If b= 1, the equation is a_{n+1}= a_n+ c and we have an arithmetic sequence.)


    This same idea can be used for higher order recurrences: If we have a_{n+2}= 3a_{n+1}- 2a_n, setting a_n= r^n gives [tex]r^{n+2}= 3r^{n+1}- 2r^n[tex] and dividing through by r^n and moving everything to the left, r^2- 3r+ 2= (r- 1)(r- 2)= 0. That is, both a_n= 1^n= 1 and a_n= 2^n satisfy the recurrence and the general solution is A(1^n)+ B(2^n)= A+ B2^n for any constants, A and B.
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