(1+x)^n = x^n

[Fritzo]

Originally posted by: SoulAssassin
It's true that you never use this crap when you get out of school.

I'm out of school, and apparently I'm right now using it in an internet forum.

 

[tealk]

Disprove This

 

[Random Variable]

Originally posted by: tealk
Disprove This

No, I will not have sex with you.

 

[chuckywang]

Originally posted by: Random Variable
(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1^(1/n)*(1+x) = 1^(1/n)*x

You get a contradictory statement if you let the nth root of 1 be the same number on both sides of the equation. If you let them be different numbers, then you get all the solutions. Why is that?

How did you get from the second to third line?

 

[Duddy]

Nerds!

*pulls each one of your underwear up your cracks*

 

[chuckywang]

Originally posted by: sdifox

Originally posted by: Random Variable
(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1^(1/n)*(1+x) = 1^(1/n)*x

You get a contradictory statement if you let the nth root of 1 be the same number on both sides of the equation. If you let them be different numbers, then you get all the solutions. Why is that?

? I can't think of any way (1+x)^n=x^n


Other than n=0 I mean

Well, as long as we're setting values, x=-1/2 and n is even.

 

[sdifox]

Originally posted by: chuckywang

Originally posted by: sdifox

Originally posted by: Random Variable
(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1^(1/n)*(1+x) = 1^(1/n)*x

You get a contradictory statement if you let the nth root of 1 be the same number on both sides of the equation. If you let them be different numbers, then you get all the solutions. Why is that?

? I can't think of any way (1+x)^n=x^n


Other than n=0 I mean

Well, as long as we're setting values, x=-1/2 and n is even.

and that is why I am not a math major man talk about being rusty.

 

[Random Variable]

Originally posted by: chuckywang

Originally posted by: Random Variable
(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1^(1/n)*(1+x) = 1^(1/n)*x

You get a contradictory statement if you let the nth root of 1 be the same number on both sides of the equation. If you let them be different numbers, then you get all the solutions. Why is that?

How did you get from the second to third line?

(z^n)^(1/n) = (1*z^n)^(1/n) = 1^(1/n)*z

 

[Engineer]

Originally posted by: SoulAssassin
It's true that you never use this crap when you get out of school.

That's not true. You've got to help your kids with it when they are doing their homework!

 

[chuckywang]

Originally posted by: Random Variable

Originally posted by: chuckywang

Originally posted by: Random Variable
(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1^(1/n)*(1+x) = 1^(1/n)*x

You get a contradictory statement if you let the nth root of 1 be the same number on both sides of the equation. If you let them be different numbers, then you get all the solutions. Why is that?

How did you get from the second to third line?

(z^n)^(1/n) = (1*z^n)^(1/n) = 1^(1/n)*z

Your answer lies in the fact that 1+x can never equal to x.

 

[Krazy4Real]

x = the plane taking off

 

[LordMorpheus]

Originally posted by: SoulAssassin
It's true that you never use this crap when you get out of school.

It's true that you never used this crap when you got out of school.

 

[kevinthenerd]

Originally posted by: Duddy
Nerds!

*pulls each one of your underwear up your cracks*

Ouch!

 

[kevinthenerd]

Originally posted by: Krazy4Real
x = the plane taking off

LMAO!

 

[cirthix]

x=-.5 for all even n, while all values of x are true for n=0. This statement is not true for any values of odd n.

 

[Inspector Jihad]

1+1=2

 

[Ilikepiedoyou]

how is this done?

(1*z^n)^(1/n) = 1^(1/n)*z

 

[Random Variable]

Originally posted by: cirthix
x=-.5 for all even n, while all values of x are true for n=0. This statement is not true for any values of odd n.

that's true if you restrict yourself to the reals

 

[manowar821]

http://youtube.com/watch?v=MiMWJ1xBo8w

 

[DrPizza]

Egads. Had to come back to this thread when I realized I made a stupid, careless mistake. When I was thinking about the nth root, I didn't stop to think that there were n nth roots. Ooops! There are other solutions other than n=0.

 

[silverpig]

We use it in physics for some semi-classical approximations in statistical physics. Of course the times we use it are when n = 10^26 or larger (maybe even 10^(10^26), so it's valid there.