# (1.0 / 2) process repeated 5 times; what is the algrabraic formula?

by mr magoo
Tags: algrabraic, formula, process, repeated, times
 P: 23 1 / 2 = 0.5 0.5 / 2 = 0.25 0.25 / 2 = 0.125 0.125 / 2 = 0.0625 0.0625 / 2 = 0.03125 What is the algebraic formula for this?
 P: 1,384 $\frac{1}{2^5}$
 P: 23 This is a new one; 64 / 2 = 32 32 / 2 = 16 16 / 2 = 8 8 / 2 = 4 4 / 2 = 2 2 / 2 = 1 1 / 2 = 0.5 0.5 / 2 = 0.25 0.25 / 2 = 0.125 0.125 / 2 = 0.0625 0.0625 / 2 = 0.03125 $\frac{64}{2^{10}}$
P: 23

## (1.0 / 2) process repeated 5 times; what is the algrabraic formula?

Thanks.
 P: 1,384 That should actually be $\frac{64}{2^{11}}$. Edit: Enclose your "10" in { } to make it appear correctly.
 P: 23 Your right, I added one too many and thought there was only ten.
 P: 23 Thanks for the editing tip.
PF Patron
P: 1,930
 Quote by jgens $\frac{1}{2^5}$
But that's not a formula.

$$\frac{1}{2^n}$$ is a formula.
P: 1,384
 Quote by Char. Limit But that's not a formula.
I could nitpick and argue that $\frac{1}{2^n}$ is actually an expression and not a formula since it does not contain an equals sign; but the distinction is really not all that relevant. The OP wanted to know how to express "1 divided by 2 fives times" algebraically and one way is $\frac{1}{2^5}$. I really don't understand the objection.
HW Helper
P: 3,320
 Quote by mr magoo This is a new one; 64 / 2 = 32 32 / 2 = 16 16 / 2 = 8 8 / 2 = 4 4 / 2 = 2 2 / 2 = 1 1 / 2 = 0.5 0.5 / 2 = 0.25 0.25 / 2 = 0.125 0.125 / 2 = 0.0625 0.0625 / 2 = 0.03125 $\frac{64}{2^{10}}$
Also notice that since we divided 64 by 2 five times and we got to 1, so $\frac{64}{2^5}=1$ rearranging, we get $64=2^5$ so we can express the answer as

$$\frac{64}{2^{10}}=\frac{2^5}{2^{10}}$$

And if you remember the rule of indices, $$\frac{2^a}{2^b}=2^{a-b}$$ so $$\frac{2^5}{2^{10}}=2^{5-10}=2^{-5}=\frac{1}{2^5}$$

As we got in your first question.
 HW Helper P: 6,757 The formula (not sure if this is considered algebraic) or notation for a product series in the original example would be: $$\prod_{i=1}^5 \ \frac{1}{2}$$

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