# To reject or not?

• May 7th 2011, 11:37 PM
Punch
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?
• May 7th 2011, 11:58 PM
Prove It
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.
• May 8th 2011, 12:00 AM
FernandoRevilla
Quote:

Originally Posted by Punch
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.
• May 8th 2011, 04:45 AM
Punch
Quote:

Originally Posted by Prove It

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?