Maths Question

Barks · 19 May 2011 at 00:10

The answer is not between 1 or 9 this time!

I know people hate these threads but I'd honestly be extremely grateful as this has been annoying me all evening.

A = ((1 - X) ^ n) - ((1 - X) ^ (n+1))

Now apparently you can take out a common factor which is (1 - X) ^ n

And you then get A = (X(1-X) ^ n)

How on earth do you get there?

I know how to factorize it but that does not get (X(1-X) ^ n) yet plug in some values and indeed it does equal the top line.

Any ideas?

Cheers

Castiel · 19 May 2011 at 00:14

I haven't the foggiest idea what any of that even means....but I look forward to hopefully learning as much as I did in the 1 or 9 thread....

Barks · 19 May 2011 at 00:16

Shouldn't need a Berkely Ring Theorist to solve this one.

Mason- · 19 May 2011 at 00:21

going by power rules...
x^2 * x^2 = x^2+2 = x^4

(x-1)^n * (x-1)^1 = x^n+1 see where im going...

A = (1-x)^n - (1-x)^n+1

A = (1-x)^n * [1 - (1-x)]

part inside square brackets

[1 - (1-x)] = 1 - 1 - -x = 1-1+x = x
so A = (1-x)^n * [x]
or A = x(1-x)^n

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Judgeneo · 19 May 2011 at 00:24

Originally took on phone, then couldn't find cable, so had to take with my webcam, so apologies for the quality, the light isn't great in my room

cheechm · 19 May 2011 at 00:26

Because ((1 - X) ^ (n+1)) = ((1 - X) ^ (n)).((1 - X) ^ (1))

The easiest way to look at it now is : assume that (1 - X ) ^ n = the variable "y"

Your new equation is:

A = y - y x ((1 - X) ^ 1)

Obviously y is a common factor:

Therefore:

A = y (1 - (( 1 - X ) ^ 1))

Anything to the power of 1 is itself, so:

A = y (1 - 1 - ( - X ))

Giving A = y (1 + X)

Now we know that (1 - X ) ^ n =y

Therefore: A = (1 - X ) ^ n (1 + X)

Shayper · 19 May 2011 at 00:27

Mason- said:
going by power rules...
x^2 * x^2 = x^2+2 = x^4

(x-1)^n * (x-1)^1 = x^n+1 see where im going...

A = (1-x)^n - (1-x)^n+1

A = (1-x)^n * [1 - (1-x)]

part inside square brackets

[1 - (1-x)] = 1 - 1 - -x = 1-1+x = x
so A = (1-x)^n * [x]
or A = x(1-x)^n

This

Reminds me of rearranging equations when proving an infinity series by induction, evil, evil subject. But awesome feeling when you get it right

Barks · 19 May 2011 at 00:29

Mason- said:
going by power rules...
x^2 * x^2 = x^2+2 = x^4

(x-1)^n * (x-1)^1 = x^n+1 see where im going...

A = (1-x)^n - (1-x)^n+1

A = (1-x)^n * [1 - (1-x)]

part inside square brackets

[1 - (1-x)] = 1 - 1 - -x = 1-1+x = x
so A = (1-x)^n * [x]
or A = x(1-x)^n

Brilliant, thanks for that mate. It was the part in bold I didn't think of simplifying.

Edit - Cheers to everyone else too some of your replies weren't there when I wrote the above.

Mason- · 19 May 2011 at 00:32

Shayper said:
This

Reminds me of rearranging equations when proving an infinity series by induction, evil, evil subject. But awesome feeling when you get it right

Yep I hate proof by induction!

Judgeneo · 19 May 2011 at 00:35

Why did I pick the module on metric spaces and topology

Shayper · 19 May 2011 at 00:58

Mason- said:
Yep I hate proof by induction!

It's evil, but I do love a challenge

Disclaimer: as long as doesn't involve protracted physical or mental effort

sk82jack · 19 May 2011 at 01:35

Judgeneo said:
Why did I pick the module on metric spaces and topology

I'm with you there.

Beginning of the year I thought topology sounded ok so I signed myself up for Intro to Topology and Algebraic Topology. With my exams coming up in a couple of weeks I now hate myself :mad:

lol

Alex74 · 19 May 2011 at 01:37

Judgeneo said:
Why did I pick the module on metric spaces and topology

Did that last year. Was an interesting course, certainly better than the fluids and dynamics or statistical modelling options

.

Hardest exam out of the three though..