Physics Forums

Physics Forums (http://www.physicsforums.com/index.php)
-   General Math (http://www.physicsforums.com/forumdisplay.php?f=73)
-   -   (1.0 / 2) process repeated 5 times; what is the algrabraic formula? (http://www.physicsforums.com/showthread.php?t=592196)

mr magoo Mar31-12 02:27 PM

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

jgens Mar31-12 02:35 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
[itex]\frac{1}{2^5}[/itex]

mr magoo Mar31-12 02:44 PM

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

[itex]\frac{64}{2^{10}}[/itex]

mr magoo Mar31-12 02:49 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
Thanks.

jgens Mar31-12 02:50 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
That should actually be [itex]\frac{64}{2^{11}}[/itex].

Edit: Enclose your "10" in { } to make it appear correctly.

mr magoo Mar31-12 02:52 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
Your right, I added one too many and thought there was only ten.

mr magoo Mar31-12 02:59 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
Thanks for the editing tip.

Char. Limit Mar31-12 04:31 PM

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

Quote by jgens (Post 3843560)
[itex]\frac{1}{2^5}[/itex]

But that's not a formula.

[tex]\frac{1}{2^n}[/tex] is a formula.

jgens Mar31-12 05:08 PM

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

Quote by Char. Limit (Post 3843727)
But that's not a formula.

I could nitpick and argue that [itex]\frac{1}{2^n}[/itex] 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 [itex]\frac{1}{2^5}[/itex]. I really don't understand the objection.

Mentallic Apr1-12 09:10 PM

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

Quote by mr magoo (Post 3843573)
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

[itex]\frac{64}{2^{10}}[/itex]

Also notice that since we divided 64 by 2 five times and we got to 1, so [itex]\frac{64}{2^5}=1[/itex] rearranging, we get [itex]64=2^5[/itex] so we can express the answer as

[tex]\frac{64}{2^{10}}=\frac{2^5}{2^{10}}[/tex]

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

As we got in your first question.

rcgldr Apr2-12 06:06 PM

Re: (1.0 / 2) process repeated 5 times; what is the algrabraic formula?
 
The formula (not sure if this is considered algebraic) or notation for a product series in the original example would be:

[tex]\prod_{i=1}^5 \ \frac{1}{2} [/tex]


All times are GMT -5. The time now is 09:07 PM.

Powered by vBulletin Copyright ©2000 - 2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums