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Old Nov 5th, 2005, 12:13 AM   #1
sparda
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Math problem

Well im trying to implement a math formula, the difficult part is understanding it.
I would like to know why:
1000 / (pi ( 500 / pi )^(2/3) ) = 2( 500/pi)^(1/3)
this is a simplification problem.
i want to know why it equals = 2(500/pi)^(1/3)
if anyone can explain this i'd greatly appreciate it. To all the cynics: This is not homework as many of you might be thinking! (i've read alot of posts) I just want to know how it works, thats all. I just want an explanation, not the answer.
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Old Nov 5th, 2005, 2:10 AM   #2
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ah,this is realy a simple math problem:
(1) (500/pi)^(2/3)=(500^(2/3))*pi^(-2/3)
(2)pi*(500/pi)^(2/3)=pi*(500^(2/3))*pi^(-2/3)=(500^(2/3))*pi^(1/3)
(3)1000 / (pi ( 500 / pi )^(2/3) )=(2*500^(1))/((500^(2/3))*pi^(1/3))=2*(500^(1-2/3))*pi^(0-1/3)
=2*(500^(1/3))*(pi^(-1/3))=2*(500/pi)^(1/3)=2( 500/pi)^(1/3)
if you write it down on a paper and deduce it ,you will get the result.; -)
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Old Nov 5th, 2005, 2:14 AM   #3
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Quote:
Originally Posted by sinix
ah,this is realy a simple math problem:
(1) (500/pi)^(2/3)=(500^(2/3))*pi^(-2/3)
(2)pi*(500/pi)^(2/3)=pi*(500^(2/3))*pi^(-2/3)=(500^(2/3))*pi^(1/3)
(3)1000 / (pi ( 500 / pi )^(2/3) )=(2*500^(1))/((500^(2/3))*pi^(1/3))=2*(500^(1-2/3))*pi^(0-1/3)
=2*(500^(1/3))*(pi^(-1/3))=2*(500/pi)^(1/3)=2( 500/pi)^(1/3)
if you write it down on a paper and deduce it ,you will get the result.; -)
I was sitting here for ages trying to work it out too.

I still don't get it. :p
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Old Nov 5th, 2005, 2:42 AM   #4
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yeah, in 2 or three steps i got it down to 1000=1000 which satisfied me.
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Old Nov 5th, 2005, 3:01 AM   #5
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Smile Basic rules

just use these rules:
A^(x)*A^(y)=A^(x+y)
A^(x)=(1/A)^(-x)
A^(x)/A^(y)=A^(x-y)
(A*B)^(x)=A^(x)*B^(x)
(A/B)^(x)=A^(x)/B^(x)
use these foundamental rules,you'll get the result.
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Old Nov 5th, 2005, 3:04 AM   #6
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Quote:
Originally Posted by sinix
ah,this is realy a simple math problem:
(1) (500/pi)^(2/3)=(500^(2/3))*pi^(-2/3)
(2)pi*(500/pi)^(2/3)=pi*(500^(2/3))*pi^(-2/3)=(500^(2/3))*pi^(1/3)
(3)1000 / (pi ( 500 / pi )^(2/3) )=(2*500^(1))/((500^(2/3))*pi^(1/3))=2*(500^(1-2/3))*pi^(0-1/3)
=2*(500^(1/3))*(pi^(-1/3))=2*(500/pi)^(1/3)=2( 500/pi)^(1/3)
if you write it down on a paper and deduce it ,you will get the result.; -)
I got the *answer* and the involved maths just fine (I have the evaluated number), but I still don't get the way you presented your reply. Mind actually explaining it, specifically, a little more coherently?

EDIT: Your second post helps
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% rc4 hexkey < input > output
#define S ,t=s[i],s[i]=s[j],s[j]=t /* rc4 hexkey <file */
unsigned char k[256],s[256],i,j,t;main(c,v,e)char**v;{++v;while(++i)s[ 
i]=i;for(c=0;*(*v)++;k[c++]=e)sscanf((*v)++-1,"%2x",&e);while(j+=s[i]
+k[i%c]S,++i);for(j=0;c=~getchar();putchar(~c^s[t+=s[i]]))j+=s[++i]S;}
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Old Nov 5th, 2005, 3:24 AM   #7
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we start with 1000/(pi(500/pi)^2/3)=s(500/pi)^1/3

multiply both sides by the denominator of the left expression and we get:

1000=2pi(500/pi)

the pi's cancel and we get:

1000=2(500)

which is 1000=1000

which is true.
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Old Nov 5th, 2005, 5:19 AM   #8
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Quote:
Originally Posted by sinix
just use these rules:
A^(x)*A^(y)=A^(x+y)
A^(x)=(1/A)^(-x)
A^(x)/A^(y)=A^(x-y)
(A*B)^(x)=A^(x)*B^(x)
(A/B)^(x)=A^(x)/B^(x)
use these foundamental rules,you'll get the result.
Yeh, I have all those rules in front of me, but I still couldn't get the answer. It's been a while since I've tried anything remotely challenging in the maths department.
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Old Nov 5th, 2005, 5:27 AM   #9
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Quote:
Originally Posted by bl00dninja
we start with 1000/(pi(500/pi)^2/3)=s(500/pi)^1/3

multiply both sides by the denominator of the left expression and we get:

1000=2pi(500/pi)

the pi's cancel and we get:

1000=2(500)

which is 1000=1000

which is true.
Sooooo much more concise, thanks.
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% rc4 hexkey < input > output
#define S ,t=s[i],s[i]=s[j],s[j]=t /* rc4 hexkey <file */
unsigned char k[256],s[256],i,j,t;main(c,v,e)char**v;{++v;while(++i)s[ 
i]=i;for(c=0;*(*v)++;k[c++]=e)sscanf((*v)++-1,"%2x",&e);while(j+=s[i]
+k[i%c]S,++i);for(j=0;c=~getchar();putchar(~c^s[t+=s[i]]))j+=s[++i]S;}
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Old Nov 5th, 2005, 8:02 PM   #10
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thaks alot for clearing this math problem up, i really apprieciate it
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