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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: onur369 on December 21, 2010, 11:03:32 pm

Title: Binomial Theorem
Post by: onur369 on December 21, 2010, 11:03:32 pm: Can someone please explain it in year 12 standards. Im not sure what exactly I have to know about the Binomial Theorem but Im finding it hard to understand. Can you genius' explain to me exactly what this theorem is?
Title: Re: Binomial Theorem
Post by: brightsky on December 21, 2010, 11:18:03 pm: Basically just Pascal's triangle, it's used for expansion. Just say you have something like (ax + b)^6. Then the binomial theorem helps you expand this in one step without the hassle of expanding term by term. Essentially it just utilises combinations (how many of a certain term there is) and hence figures out the coefficient in front of it - the expansion is as follows $\binom{6}{0} (ax)^0 (b)^6 + \binom{6}{1} (ax)^1 (b)^5 + ... \binom{6}{6} (ax)^6 (b)^6$ . The theory behind the derivation of the theorem is quite straight forward. If you observe the Pascal's triangle, you'll notice the pattern (the pattern exists because of the way it is expanded).
Title: Re: Binomial Theorem
Post by: Greatness on December 21, 2010, 11:20:39 pm: It pretty much allows you to expand $(a+x)^n$
Title: Re: Binomial Theorem
Post by: aznxD on December 21, 2010, 11:22:32 pm: Pretty much this:
$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k$

The coefficients can be observed in pascal's triangle. The power of the first term decreases by one, and the second increases by one.

For example: $(a+b)^5 = a^5 + 5a^4 b + 10a^3 b^2 + 10a^2 b^3 + 5a b^4 + b^5$
Title: Re: Binomial Theorem
Post by: werdna on December 21, 2010, 11:26:06 pm: Oh man I'm so glad I'm not repeating this subject. ;D
Title: Re: Binomial Theorem
Post by: Greatness on December 21, 2010, 11:31:59 pm: aznxD and brightsky have got it covered. but with pascals triangle you will notice that the 6th row is 1 5 10 10 5 1 thus giving the coefficents of the x terms in descending order.
Title: Re: Binomial Theorem
Post by: onur369 on December 22, 2010, 11:54:19 pm: So I just opened the book for the first time. Example 1, in relation to the Binomial Theorem. Expand (2x-3)^4 using the Binomial Theorem. Ok I do the first step properly, but then I dont know what do to on the other part. (http://www3.wolframalpha.com/Calculate/MSP/MSP286519de5df6i0523h5h00001i72gagehd4gh59b?MSPStoreType=image/gif&s=42&w=226&h=20) thats the final answer but i dont know what do the step before the answer.

I do all (http://www.forkosh.dreamhost.com/mathtex.cgi?%20\binom{6}{0}%20%28ax%29^0%20%28b%29^6%20+%20\binom{6}{1}%20%28ax%29^1%20%28b%29^5%20+%20...%20\binom{6}{6}%20%28ax%29^6%20%28b%29^6) this correct but its the step right afterwards.
Title: Re: Binomial Theorem
Post by: Greatness on December 22, 2010, 11:59:50 pm: This part (4/0)x(2x)^4becomes 16x^4. If you look at pascals triangle you will notice that in the 5th row, the first term is 1 so you multiply 16x^4 by 1 which gives you 16x^4. You do that with each term, but you would find all the x terms first then look at the pascals triangle to determine the rest of the coefficients.
I need to learn how to use latex...
Title: Re: Binomial Theorem
Post by: aznxD on December 23, 2010, 12:11:01 am: You got to substitute it with the right numbers.

It's:

$(2x-3)^4 = \sum_{k=0}^4 {4 \choose k}(2x)^{4-k}(-3)^k$

$\binom{4}{0} (2x)^4 (-3)^0 + \binom{4}{1} (2x)^3 (-3)^1 + ... \binom{4}{4} (2x)^0 (-3)^4$

This becomes: $16x^4 - 96x^3 + 216x^2 - 216x + 81$

EDIT: I also got to learn how to use latex efficiently. This took me like 15 mins -_-
:P
Title: Re: Binomial Theorem
Post by: onur369 on December 23, 2010, 12:12:50 am: Quote from: aznxD on December 23, 2010, 12:11:01 am

$\binom{4}{0} (2x)^4 + \binom{4}{1} (2x)^3 (-3)^1 + ... \binom{4}{4} (2x)^0 (-3)^4$

This becomes: $16x^4 - 96x^3 + 216x^2 - 216x + 81$

I dont understand how we get 16x^4 :@
Title: Re: Binomial Theorem
Post by: Greatness on December 23, 2010, 12:14:23 am: (2x)^4=16x^4
Title: Re: Binomial Theorem
Post by: aznxD on December 23, 2010, 12:17:09 am: Do you know how to work out combinations?
For the first term:

$\binom{4}{0} = 1$

$(2x)^4 = 16x^4$

$1 \times 16x^4 = 16x^4$
Title: Re: Binomial Theorem
Post by: onur369 on December 23, 2010, 12:18:35 am: Woops, I actually ment the other part, 2x^3 (-3)^1= -24x^3. How do we get -96x^3?
Title: Re: Binomial Theorem
Post by: Greatness on December 23, 2010, 12:19:32 am: Pascals triangle:
   1
   1 1
   1 2 1
   1 3 3 1
   1 4 6 4 1 <<<<<<<<< This is the row you are looking at. Multiply each x term by these to give the coefficients
Title: Re: Binomial Theorem
Post by: Greatness on December 23, 2010, 12:21:35 am: you multiply the -24x^3 by 4 as you can see from pascals triangle. this is a pretty simple method - that is pascals
x^4 term gets multplied by 1
x^3 '' '' '' '' 4
x^2 '' '' '' '' 6
x^1 '' '' '' '' 4
x^0 '' '' '' '' 1
Title: Re: Binomial Theorem
Post by: aznxD on December 23, 2010, 12:21:59 am: $\binom{4}{1} (2x)^3 (-3)^1$

-----------------------------------------------------------

$\binom{4}{1} = 4$ (This can be worked out by looking at the fifth row of pascals triangle)

$(2x)^3 = 8x^3$

$(-3)^1 = -3$

$4 \times 8x^3 \times -3 = -96x^3$
Title: Re: Binomial Theorem
Post by: luken93 on December 23, 2010, 12:23:13 am: Row 1: 1
Row 2: 1 1
Row 3: 1 2 1
Row 4: 1 3 3 1
Row 5: 1 4 6 4 1

For the first number in row 5, we have $1 \times (2x)^4(-3)^0 = 16x^4$
For the second number in row 5, we have $4 \times (2x)^3(-3)^1 = 4 \times 8x^3 \times -3 = -96x^3$
For the third number in row 5, we have $6 \times (2x)^2(-3)^2 = 6 \times 4x^2 \times 9 = 216x^2$

etc
Title: Re: Binomial Theorem
Post by: Greatness on December 23, 2010, 12:25:18 am: isnt the 3rd term 6 x 4x^2 x 9? which equals 216x^2
Title: Re: Binomial Theorem
Post by: aznxD on December 23, 2010, 12:27:45 am: Quote from: swarley on December 23, 2010, 12:25:18 am
isnt the 3rd term 6 x 4x^2 x 9? which equals 216x^2

yes it is
Title: Re: Binomial Theorem
Post by: onur369 on December 23, 2010, 12:33:58 am: Thanks guys, Ill have a go at it a few times and let you know how I go. Btw, this is the first time I came across it, no foundation of it from yr11 whatsoever so I am rusty -.-
Title: Re: Binomial Theorem
Post by: aznxD on December 23, 2010, 12:36:00 am: Don't worry it wasn't in the year 11 course except for combinations
Title: Re: Binomial Theorem
Post by: Greatness on December 23, 2010, 12:37:21 am: i never learnt it as well :( but after you do a few questions you'll get the hang of it :)
Title: Re: Binomial Theorem
Post by: luken93 on December 23, 2010, 07:43:57 am: Quote from: swarley on December 23, 2010, 12:25:18 am
isnt the 3rd term 6 x 4x^2 x 9? which equals 216x^2
Haha I know I saw it as soon as I posted, all good now :)
Title: Re: Binomial Theorem
Post by: QuantumJG on December 23, 2010, 09:12:18 am: $(ax+b)^n = \sum_{k=0}^n {n \choose k}(ax)^{n-k}(b)^k$

Where:

${n \choose k} = \dfrac{ n ! }{ k ! ( n - k ) ! }$

Pascal's triangle is a lot more simpler and useful for n < 6.
Title: Re: Binomial Theorem
Post by: TrueTears on December 24, 2010, 05:26:35 pm: binominal theorem is a special case of the multinomial theorem, the proof can be done combinatorially.