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Math Help - To reject or not?

  1. #1
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    To reject or not?

    Solve |\frac{1-(\frac{2}{3})^n}{1-\frac{2}{3}}-3|<0.01

     -0.01<\frac{1-(\frac{2}{3})^n}{1-\frac{2}{3}}-3<0.01

    so -0.01<\frac{1-(\frac{2}{3})^n}{1-\frac{2}{3}}-3 and \frac{1-(\frac{2}{3})^n}{1-\frac{2}{3}}-3<0.01

    simplifying: -0.01<3(\frac{2}{3})^n and 3(\frac{2}{3})^n<0.01

    I solved the right hand equation but had to reject the left hand side equation since i could not logarithm a negative number. So do i reject the whole solution or just write the right hand side equation solution?
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  2. #2
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    Why not simplify the stuff in the modulus first?

    \displaystyle \frac{1 - \left(\frac{2}{3}\right)^n}{1 - \frac{2}{3}} - 3 = \frac{1 - \left(\frac{2}{3}\right)^n}{\frac{1}{3}} - 3

    \displaystyle = 3\left[1 - \left(\frac{2}{3}\right)^n\right] - 3

    \displaystyle = 3 - 3\left(\frac{2}{3}\right)^n - 3

    \displaystyle = -3\left(\frac{2}{3}\right)^n.


    So if \displaystyle \left|\frac{1 - \left(\frac{2}{3}\right)^n}{1 - \frac{2}{3}} - 3\right| < 0.01

    then \displaystyle -\frac{1}{100} < -3\left(\frac{2}{3}\right)^n < \frac{1}{100}.

    Go from here.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Punch View Post
    simplifying: -0.01<3(\frac{2}{3})^n and 3(\frac{2}{3})^n<0.01

    I solved the right hand equation but had to reject the left hand side equation since i could not logarithm a negative number. So do i reject the whole solution or just write the right hand side equation solution?

    Take into account that

    -0.01<0<3\left(\dfrac{2}{3}\right)^n

    So, the right hand equation is satisfied for all n.


    Edited: Sorry, I didn't see Prove It's post.
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  4. #4
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    Quote Originally Posted by Prove It View Post

    then \displaystyle -\frac{1}{100} < -3\left(\frac{2}{3}\right)^n < \frac{1}{100}.

    Go from here.
    -\frac{1}{100} < -3\left(\frac{2}{3}\right)^n and -3\left(\frac{2}{3}\right)^n < \frac{1}{100}.


    Left hand side: -\frac{1}{100}<-3(\frac{2}{3})^n

    \frac{1}{100}>3(\frac{2}{3})^n

    \frac{1}{300}>(\frac{2}{3})^n

     \frac{log(\frac{1}{300})}{log(\frac{2}{3})}>n

     14.07>n

    Right hand side: -3\left(\frac{2}{3}\right)^n < \frac{1}{100}

    (\frac{2}{3})^n>-\frac{1}{300}

     n>\frac{log(-\frac{1}{300})}{log(\frac{2}{3})} (REJECT)

    I rejected the right hand solution because I can't logarithm a negative number. So do I have to reject the left hand side solution too? Since in
    -\frac{1}{100} < -3\left(\frac{2}{3}\right)^n and -3\left(\frac{2}{3}\right)^n < \frac{1}{100}, and means they come together, since i cant find a solution for one of the parts, i should reject the whole solution, am i right?
    Last edited by Punch; May 8th 2011 at 07:23 AM.
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