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Math Help - Is it possible to solve this inequality for n?

  1. #1
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    Arrow Is it possible to solve this inequality for n?

    If it is, how would you go about doing it?

    \frac{\left(\frac{1}{2}\right)^{(2n+3)}+\left(\fra  c{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}
    Last edited by leverin4; April 30th 2009 at 05:24 PM.
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  2. #2
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    Lack of a response tells me no. Can anyone give me advice on approximating n?
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  3. #3
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    I tried using Mathematica, but it couldn't do it. I don't remember the exact error message, but I'll try it again tomorrow and post it. Meanwhile, isn't there anyone that can give me some advice on how to simplify it?
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  4. #4
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    It is old thread , i know

    but am interesting to know the answer *_*
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  5. #5
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    Quote Originally Posted by leverin4 View Post
    If it is, how would you go about doing it?

    \frac{\left(\frac{1}{2}\right)^{(2n+3)}+ \left(\frac{1}{3}\right)^{(2n+3)}}{2n+3}~<~10^{-1000001}
    It would be hard to find a necessary and sufficient condition on n for this inequality to hold. However, problems like this usually occur in a first course on analysis, and the aim there is not to find the smallest value of n that will satisfy the inequality, but to find some number N such that all values of n gretaer than N will satisfy the inequality. In that case, you can make the problem much easier by getting a rough upper bound for the left side of the inequality.

    For example, \textstyle\frac{(\frac12)^{2n+3}+ (\frac13)^{2n+3}}{2n+3} < (\frac12)^{2n+3}+ (\frac13)^{2n+3} < 2(\frac12)^{2n+3} < (\frac12)^{2n}. If you can find a value of N such that \textstyle(\frac12)^{2n} < 10^{-K} whenever n>N, it will follow that the original inequality will hold whenever n>N. (Here, K stands for some given large number, such as 1000001.)

    But \textstyle(\frac12)^{2n} < 10^{-K}\ \Rightarrow\ 2^{2n}>10^K\ \Rightarrow\ 2n\ln2>K\ln10\ \Rightarrow\ n>\frac{K\ln10}{2\ln2}, so take N to be any integer greater than \frac{1000001\ln10}{2\ln2}\approx 1.661\times 10^6 and your inequality will hold for all n>N.

    Just to emphasise, 1.661\times 10^6 is not the smallest number with that property. If you want to find that number, you can use a computer to check all the values of N from 1 to 1.661\times 10^6, and it will tell you which is the first one to make the inequality true.
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