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Thread: To reject or not?

  1. #1
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    To reject or not?

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

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

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

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

    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?

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

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

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

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


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

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

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

    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

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

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

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


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

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

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

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

    $\displaystyle 14.07>n$

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

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

    $\displaystyle n>\frac{log(-\frac{1}{300})}{log(\frac{2}{3})}$ (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
    $\displaystyle -\frac{1}{100} < -3\left(\frac{2}{3}\right)^n$ and $\displaystyle -3\left(\frac{2}{3}\right)^n < \frac{1}{100}$, 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 06:23 AM.
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