(1+x)^n = x^n

[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?

 

[Lorax]

learn FOIL.

 

[IceBergSLiM]

PI

 

[zerocool1]

log (1+x)* log(n)=(log x) * log n

if i'm not mistaken.

 

[SoulAssassin]

It's true that you never use this crap when you get out of school.

 

[Random Variable]

Originally posted by: Lorax
learn FOIL.

Binomial expansion isn't very helpful when n=20.

 

[kevinthenerd]

(1+x)^n = x^n is only approximately true for very large values of x.

but

(1+x)^1 = 1 + x
(1+x)^2 = 1 + x + x + x^2 = 1 + 2x + x^2
(1+x)^3 = 1 + x + x + x^2 + x + x^2 + x^2 + x^3 = 1 + 3x + 3x^2 + x^3

You might want to read http://en.wikipedia.org/wiki/Pascal_triangle


By the way, I thought I came up with something very cool one day, but it really bothered me to discover that I wasn't the first to come up with this idea for polynomials of three or more terms: http://en.wikipedia.org/wiki/Pascal_pyramid

 

[kevinthenerd]

Originally posted by: Random Variable

Originally posted by: Lorax
learn FOIL.

Binomial expansion isn't very helpful when n=20.

but Mathcad is. ;-)

 

[kevinthenerd]

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

It depends on what you'll do. A physicist or a research engineer will use it, but your run-of-the-mill bachelor-degree CAD-crunching engineer won't.

 

[kevinthenerd]

Originally posted by: IcebergSlim
PI

42

 

[rocadelpunk]

Originally posted by: kevinthenerd
(1+x)^n = x^n is only approximately true for very large values of x.

but

(1+x)^1 = 1 + x
(1+x)^2 = 1 + x + x + x^2 = 1 + 2x + x^2
(1+x)^3 = 1 + x + x + x^2 + x + x^2 + x^2 + x^3 = 1 + 3x + 3x^2 + x^3

You might want to read http://en.wikipedia.org/wiki/Pascal_triangle


By the way, I thought I came up with something very cool one day, but it really bothered me to discover that I wasn't the first to come up with this idea for polynomials of three or more terms: http://en.wikipedia.org/wiki/Pascal_pyramid

pretty impressive you came up with that on your own...can't believe you hadn't heard of pascal triangle though if you're messing around with binomial expansion.

 

[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

 

[kevinthenerd]

Originally posted by: zerocool1
log (1+x)* log(n)=(log x) * log n

if i'm not mistaken.

No. Divide both sides by log(n). Then your expression reads

log(1+x) = log(x)

Now exponentiate both sides.

1 + x = x

This is not true.

 

[zerocool1]

woah, forgot about pascal's triangle.

 

[kevinthenerd]

Originally posted by: rocadelpunk

Originally posted by: kevinthenerd
(1+x)^n = x^n is only approximately true for very large values of x.

but

(1+x)^1 = 1 + x
(1+x)^2 = 1 + x + x + x^2 = 1 + 2x + x^2
(1+x)^3 = 1 + x + x + x^2 + x + x^2 + x^2 + x^3 = 1 + 3x + 3x^2 + x^3

You might want to read http://en.wikipedia.org/wiki/Pascal_triangle


By the way, I thought I came up with something very cool one day, but it really bothered me to discover that I wasn't the first to come up with this idea for polynomials of three or more terms: http://en.wikipedia.org/wiki/Pascal_pyramid

pretty impressive you came up with that on your own...can't believe you hadn't heard of pascal triangle though if you're messing around with binomial expansion.

I've heard of the Pascal triangle. It was the Pascal pyramid that I thought I invented. They don't teach you this stuff in mechanical engineering curriculum. It's just not that important.

 

[kevinthenerd]

I got my bachelor's in ME this last December. I figured getting a master's would only hurt my job-hunting even more, so I've decided to learn masters level material at my own pace for the fun of it. Luckily, I downloaded enough grad-level video lectures before I left to fill about 50 DVD-Rs. I get about one course on each. That should hold me over for a few years. By then I will have acquired about four masters degrees' worth of education.

(Remember my thread on teaching myself Calc 3?)

 

[ryan256]

Wait a minute... thats impossible. If you simplify by taking the n root of both sides you'd get:

(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1+x = x

There's no way that can happen. No number can be equal to itself plus 1. Or am I missing something here?

 

[kevinthenerd]

Originally posted by: ryan256
Wait a minute... thats impossible. If you simplify by taking the n root of both sides you'd get:

(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1+x = x

There's no way that can happen. No number can be equal to itself plus 1. Or am I missing something here?

the rest of the thread

 

[ryan256]

Originally posted by: kevinthenerd

Originally posted by: ryan256
Wait a minute... thats impossible. If you simplify by taking the n root of both sides you'd get:

(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1+x = x

There's no way that can happen. No number can be equal to itself plus 1. Or am I missing something here?

the rest of the thread

Enlighten me then.
Yes I know about Pascal's Triangle and binomial coefficients.
That still doesn't change the fact that (1+x)^n = x^n is totally impossible unless n=0.

 

[kevinthenerd]

Originally posted by: ryan256

Originally posted by: kevinthenerd

Originally posted by: ryan256
Wait a minute... thats impossible. If you simplify by taking the n root of both sides you'd get:

(1+x)^n = x^n
((1+x)^n)^(1/n) = (x^n)^(1/n)
1+x = x

There's no way that can happen. No number can be equal to itself plus 1. Or am I missing something here?

the rest of the thread

Enlighten me then.
Yes I know about Pascal's Triangle and binomial coefficients.
That still doesn't change the fact that (1+x)^n = x^n is totally impossible unless n=0.

I never said it was possible. n=0 is a trivial case. I already disproved it.

 

[rocadelpunk]

Originally posted by: kevinthenerd

Originally posted by: rocadelpunk

Originally posted by: kevinthenerd
(1+x)^n = x^n is only approximately true for very large values of x.

but

(1+x)^1 = 1 + x
(1+x)^2 = 1 + x + x + x^2 = 1 + 2x + x^2
(1+x)^3 = 1 + x + x + x^2 + x + x^2 + x^2 + x^3 = 1 + 3x + 3x^2 + x^3

You might want to read http://en.wikipedia.org/wiki/Pascal_triangle


By the way, I thought I came up with something very cool one day, but it really bothered me to discover that I wasn't the first to come up with this idea for polynomials of three or more terms: http://en.wikipedia.org/wiki/Pascal_pyramid

pretty impressive you came up with that on your own...can't believe you hadn't heard of pascal triangle though if you're messing around with binomial expansion.

I've heard of the Pascal triangle. It was the Pascal pyramid that I thought I invented. They don't teach you this stuff in mechanical engineering curriculum. It's just not that important.

oops, I need to read better : P

 

[sao123]

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?

X=0

 

[DrPizza]

;;;;;

Originally posted by: sao123

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?

X=0

Rethink that. (1+0)^3 = 0^3?? Nope.

The only solution (as stated above) is the trivial solution when n=0. Otherwise, there are no solutions because x would have to equal x+1


Also, someone above said something about it being approximately true for large values of x. That depends on the value of n. If n is a positive integer, the difference between the two sides increases. But, if n is a negative integer, the two sides get closer to each other as x gets very large; in fact, both sides approach zero. (not that n necessarily has to be an integer though)

 

[sao123]

Originally posted by: DrPizza
;;;;;

Originally posted by: sao123

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?

X=0

Rethink that. (1+0)^3 = 0^3?? Nope.

The only solution (as stated above) is the trivial solution when n=0. Otherwise, there are no solutions because x would have to equal x+1


Also, someone above said something about it being approximately true for large values of x. That depends on the value of n. If n is a positive integer, the difference between the two sides increases. But, if n is a negative integer, the two sides get closer to each other as x gets very large; in fact, both sides approach zero. (not that n necessarily has to be an integer though)

after i types it realized what i meant...
n=0.

 

[Random Variable]

I never stated that x must be restricted to the reals.

let's take n=4 for example

(1+x)^4 = x^4
1^(1/4)*(1+x) = 1^(1/4)*x

the 4th roots of 1 are 1, -1, i, and -i

so the possibilities are 1+x=-x, 1+x=ix, 1+x=-ix, -(1+x)=ix, -(1+x)=-ix, etc. (with some answers repeating)

If you use the binomial theorem, you'll have to solve 4x^3+6x^2+4x+1=0 (which isn't too hard to factor). But for larger n, factoring becomes impossible.

EDIT: The answers are -1/2, -1/2+i/2, and -1/2-i/2, so one of the answers is real and the other two are complex