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Math Help - recurrence relation

  1. #1
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    recurrence relation

    "show x_n = \frac{n+1}{9^n} satisfies x_{n+1}\leq\frac{1}{2}x_n for all n \geq 1". i don't know how to show it. i can see that it is true but i can't see how to approach it. i have tried a few things like writing one expression in terms of the other etc. but it doesn't get me where i need to be. can someone give me a pointer please? thank you
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  2. #2
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    Re: recurrence relation

    You can't show that \frac{n+2}{9^{n+1}}\le\frac{1}{2}\cdot\frac{n+1}{9  ^n}?
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    Re: recurrence relation

    i can convince myself, but i am not sure of what is allowed for a formal proof. what you wrote is where i started from, then took out the 1/9 fraction of the lhs, cancelled down, but i wasn't sure what i was left with was a valid proof. even though it was 'obvious' that it is so. it is possible i am overcomplicating it in my mind. would it be ok to just use 'this implies...' etc. and then as n is positive the resultant inequality is always true, just state that as so?
    Last edited by learning; January 10th 2013 at 08:11 AM.
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  4. #4
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    Re: recurrence relation

    After multiplying by 2\cdot9^{n+1}, which produces an equivalent inequality since 2\cdot9^{n+1}>0, you get 2n+4<9n+9. Can you show that this is implied by n ≥ 1?
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    Re: recurrence relation

    i absolutely comprehend fully what you're saying. that is what i did. i just was unaware of how to really set it out. i don't really know how to 'say' the answer. it's not something i have experienced much submitting a problem like this, but as i say, it's not a problem of understanding it myself, rather what is the technical language to write up an answer here? for i can see that one inequality implies the other, and that the resultant inequality is 'true' always.
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  6. #6
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    Re: recurrence relation

    I would write the solution to this problem as follows.

    \begin{align*}n\ge1&\implies2n+4\le9n+9\\ &\iff\frac{n+2}{9^{n+1}}\le \frac{1}{2}\cdot\frac{n+1}{9  ^n}\\ &\iff x_{n+1}\leq\frac{1}{2}x_n\end{align*}

    Quote Originally Posted by learning View Post
    it is possible i am overcomplicating it in my mind. would it be ok to just use 'this implies...' etc. and then as n is positive the resultant inequality is always true, just state that as so?
    The level of details depends on the context, such as the course level (elementary school vs. high school), instructor's requirements and so on. If you need to derive this inequality from the ordered field axioms, then indeed there would be quite a few steps. But since the question is about sequences or possibly recurrence relations, the level seems pretty advanced, and therefore details of proving simple inequalities, which is the subject matter of middle school, are usually not required.
    Thanks from MarkFL and learning
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  7. #7
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    Re: recurrence relation

    thanks very much. i think i had nothing to worry about then because what you have written resembles what i first did. i'm very happy that you have given this response because it confirms to me there is not something else i have to look for.
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