ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: Hielly on January 02, 2010, 02:33:07 pm

Title: Binomial theorem
Post by: Hielly on January 02, 2010, 02:33:07 pm: State the coefficient of $x^2$

$(x^2-3/x)^6$

$6Cr1 (x^2)^5 (-3/x)^1$

and then it becomes
$6*x^(10)* -3/x = ...$

Not sure how to go about this question, that was my attempt :S

Thanks

Edit:im still learning how to use latex, sorry about the equations
Title: Re: Binomial theorem
Post by: brightsky on January 02, 2010, 02:56:07 pm: $(x^2 - \frac{3}{x})^6$
$\implies 6C0(x^2)^6(\frac{3}{x})^0 + 6C1 (x^2)^5(\frac{3}{x})^1 + 6C2 (x^2)^4(\frac{3}{x})^2 + 6C3 (x^2)^3(\frac{3}{x})^3 + 6C4 (x^2)^2(\frac{3}{x})^4 + 6C5 (x^2)^1(\frac{3}{x})^5 + 6C6 (x^2)^0(\frac{3}{x})^6$

$\implies (1)x^{12}(1) + (6)x^10(-\frac{3}{x}) + (15)x^8(-\frac{9}{x^2}) + (20)x^6(-\frac{27}{x^3}) + (15)x^4(-\frac{81}{x^4}) + (6)x^2(\frac{243}{x^5}) + (1)1(\frac{729}{x^6})$

$\implies x^{12} - 18 x^9 + 135^6 + \frac{729}{x^6} -540x^3 - \frac{1458}{x^3} + 1215$

From the previous step to this step requires a whole heap of careful arithmetic and algebraic evaluation, too long to post the whole thing up here.

Hope this helps, but I'm not sure what you mean by the coefficient of $x^2$ .
Title: Re: Binomial theorem
Post by: Hielly on January 02, 2010, 03:00:27 pm: Quote from: brightsky on January 02, 2010, 02:56:07 pm
$(x^2 - \frac{3}{x})^6$
$\implies 6C0(x^2)^6(\frac{3}{x})^0 + 6C1 (x^2)^5(\frac{3}{x})^1 + 6C2 (x^2)^4(\frac{3}{x})^2 + 6C3 (x^2)^3(\frac{3}{x})^3 + 6C4 (x^2)^2(\frac{3}{x})^4 + 6C5 (x^2)^1(\frac{3}{x})^5 + 6C6 (x^2)^0(\frac{3}{x})^6$

yeah i understand how to do that, but is there a shortcut? the questions is just asking for the $x^2$ , do we have to go through all that ?
Title: Re: Binomial theorem
Post by: brightsky on January 02, 2010, 03:08:14 pm: Yeah, I think there is, TrueTears posted something on a maths forum about finding coefficients, I'll try and dig it out.
Title: Re: Binomial theorem
Post by: GerrySly on January 02, 2010, 03:08:36 pm: There is no co-efficient of $x^2$ ?

$\left ( x^2-\frac{3}{x}\right ) ^6=x^{12}-18 x^9+135 x^6+\frac{729}{x^6}-540 x^3-\frac{1458}{x^3}+1215$

Are you sure you wrote the question in correctly?
Title: Re: Binomial theorem
Post by: Hielly on January 02, 2010, 03:10:42 pm: Yeah i did, the answer says 0. so there is no x^2, but i want to learn the shortcut method to apply it to these sort of questions.

In mathsquest they showed you how to work it out, but its a bit vague. The method worked for 2 questions, but when i came to this question, where it has no coefficient it started turning ugly.

This is a mathsquest example,
State the coefficient of $x^2 in (3-2x)^8$
$x^0,x^1,x^2$

the third term gives a power of x^2

Third term $=8Cr2 (3)^6 (-2x)^2<br />=81648x^2$
Title: Re: Binomial theorem
Post by: Lighties on January 02, 2010, 03:11:25 pm: I think this is how I was taught

n+1th term;
$6Crn (x^2)^{6-n}(3/x)^n$

Then move things around so you separate numbers by themselves and x.

$6Crn 3^n$ times $x^{12-2n}/x^n$ , or $6Crn 3^n$ times $x^{12-3n}$

Since the coefficient of x you're trying to find is 2, then 12-3n =2. 3n=10. As n has to be an integer, then no solution exists.

Unless I did something wrong somewhere. o.o

But if n did exist, then you just sub it into the number part of the term to get the coefficient. (:

*Since the 3/x is negative, then put the negative with the 3. Sorry, forgot about that. ^^;
Title: Re: Binomial theorem
Post by: GerrySly on January 02, 2010, 03:19:23 pm: The shortcut Hielly is to get rid of that fraction

$\left (x^2-\frac{3}{x}\right )^6=\left (\frac{x^3-3}{x}\right )^6=\frac{(x^3-3)^6}{x^6}$

Now if we just take a look at the top expansion, $(x^3-3)^6$ we notice that all the $x$ terms would be $(x^3)^6, (x^3)^5, (x^3)^4, (x^3)^3, (x^3)^2, (x^3)^1, (x^3)^0$ which would never leave a $x^2$ term right? Now if we look at all those terms $x^{18}, x^{15}, x^{12}, x^9, x^6, x^3, x$ and we subtract 6 from all the exponents ( $18-6, 15-6, 12-6, 9-6, 6-6, 3-6, 1-6$ ) we'll never get $2$ , therefore there will never be an $x^2$ term.
Title: Re: Binomial theorem
Post by: brightsky on January 02, 2010, 03:34:39 pm: Now I'm confused >< Can you explain the method or post up the identity, I'm interested. :)
Title: Re: Binomial theorem
Post by: tolga on January 02, 2010, 03:45:26 pm: yeah im also interested how you did the method above
Title: Re: Binomial theorem
Post by: brightsky on January 02, 2010, 03:48:56 pm: Ok Hielly. Now I understand. It is exactly the same idea as finding the nth term in the polynomial.

Consider this example:
Find the 9th term in the expansion $(x - 2y)^{13}$

Since we start counting with 0, the 9th term is actually going to be when k=8. That is, the power on the x will 13-8=5 and the power on the -2y will be 8. The initial coefficient (that is, without taking into account the values of x and y) is either C(13,8) or C(13,5), combinations are symmetric, so it doesn't matter.

$C (13,8)(x)^5(-2y)^8 = 1287(x^5) (256y^8) = 329472x^5y^8$

Now, the only difference between finding out the nth term and finding the coefficient is you need to know which term it is that $x^2$ lies. Hope this helps.
Title: Re: Binomial theorem
Post by: Lighties on January 02, 2010, 04:01:40 pm: The method I used is what I usually use for finding coefficients, or the nth term, because it gives you both.

The first latex thingie is the general solution to any term once expanded. If you want the nth term, you just sub the right number in. Similarly, if you want to find the coefficient of a particular term knowing the power of the x, then you just sub the power of the x in to find n. But to do this you need to first separate out all the things with x, which is what I did in the second latex thingie (it might be easier to see if you do it on paper).

And what brightsky said. (:
Title: Re: Binomial theorem
Post by: TrueTears on January 02, 2010, 04:19:38 pm: For anyone who are interested, the binomial theorem comes from the more generalise multinomial theorem, hopefully this will enhance your understandings about coefficients etc.

So we basically want to find a generalised 'formula' for $(x_1+x_2+...+x_q)^n$ (Where as the binomial theorem is just for $(x_1+x_2)^n$ , ie when $q = 2$ )

So let's start off by doing some experimentation. Let's look at $(x+y+z)^7$ . How do we expand this? Let's consider the first 2 brackets, namely:

$(x+y+z)(x+y+z)(x+y+z)^5 = (x \cdot x + xy + xz +yx + y \cdot y + yz + zx +zy+ z \cdot z)(x+y+z)^5$

Now leaving them unsimplified we can see that the 2 brackets expanded into 9 'unsimplified' terms. Which is expected since we have 3 choices from the first bracket and another 3 choices from the 2nd bracket.

This means that if we expanded all 7 brackets we would get a total of $3 \times 3 \times 3 ... \times 3 = 3^7$ unsimplified terms since there are 7 brackets and each bracket has 3 'choices'.

Now let's assume we have expanded all 7 brackets and we want to find the coefficient of the $x^4y^2z^1$ term.

We notice a few things: the powers all add up to 7 and we realise that this is just a direct application of the Mississippi 'formula'.

Just imagine we have a lot $x^4y^2z^1$ terms lying around to be collected as like terms after the expansion of the 7 brackets. However each would have a different permutation.

As we list some: $xxxxyyz, xyyxxxz ...$

Thus the total 'amount' of $x^4y^2z^1$ terms lying around would be $\frac{7!}{4!2!1!}$ .

Let's try another experiment, let's say we want to find how many of $x^2y^2z^3$ terms are lying around uncollected after the expansion of the 7 brackets.

We realise after undergoing the same process as before we get $\frac{7!}{2!2!3!}$ $x^2y^2z^3$ terms are lying around.

A pattern can be seen: The numerator is always 7! (As we expect since there are always 7 terms to permute).

The denominator's factorials are correspondent to the power of each term.

Therefore we can generalise this a bit and say the coefficient of any term in the expansion of $(x+y+z)^7$ is $\frac{7!}{p(x)!p(y)!p(z)!}$ where $p(x,y,z)$ denotes the power of $x,y,z$ respectively of the term.

Now that we have generalised the result for working out the coefficients of any term we need to generalise what $(x+y+z)^7$ will be expanded into.

Let's try work out how many terms $(x+y+z)^7$ when all like terms are collected.

However $(x+y+z)^7$ seems too tedious to work with, so let's try an easier example $(x+y+z)^3$

To work out how many different like-terms there are in total when $(x+y+z)^3$ is expanded let's consider what we discovered before with the exponents. We found that the exponent must add up to $\mbox{n}$ . So all exponents of the terms of $(x+y+z)^3$ when expanded must add up to 3.

How many different combinations can we get? Certainly there can be all the different permutations of $(0,0,3)$ , $(0,1,2)$ and $(1,1,1)$ .

Thus the total number of terms we should expect should be $\frac{3!}{2!1!} + 3! + 1 = 10$ Which when we expand $(x+y+z)^3$ we certainly do get 10 terms!

Now we can try to find a more general formula for the expansion of $(x+y+z)^3$

$\implies (x+y+z)^3 = \sum_{p_1,p_2,p_3}\left[\left(\frac{3!}{p_1!p_2!p_3!}\right)\left(x^{p_1}y^{p_2}z^{p_3}\right)\right]$

What exactly does this mean?

Basically it means that we take the summation of all permutations of non-negative integer indices $p_1$ through to $p_3$ such that $p_1+p_2+p_3 = 3$

$\therefore (x+y+z)^3 = \sum_{p_1,p_2,p_3}\left[\left(\frac{3!}{p_1!p_2!p_3!}\right)\left(x^{p_1}y^{p_2}z^{p_3}\right)\right] = \sum_{0,0,3}\left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{0}z^{3}\right)\right] + \sum_{0,1,2}\left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{1}z^{2}\right)\right] + \sum_{1,1,1}\left[\left(\frac{3!}{1!1!1!}\right)\left(x^{1}y^{1}z^{1}\right)\right]$

$= \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{0}z^{3}\right)\right] + \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{3}y^{0}z^{0}\right)\right] + \left[\left(\frac{3!}{0!0!3!}\right)\left(x^{0}y^{3}z^{0}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{1}z^{2}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{0}y^{2}z^{1}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{1}y^{0}z^{2}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{1}y^{2}z^{0}\right)\right] + \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{2}y^{1}z^{0}\right)\right]+ \left[\left(\frac{3!}{0!1!2!}\right)\left(x^{2}y^{0}z^{1}\right)\right] + \left[\left(\frac{3!}{1!1!1!}\right)\left(x^{1}y^{1}z^{1}\right)\right]$

$= \left(\frac{3!}{0!0!3!}\right)\left[z^3+x^3+y^3\right] + \left(\frac{3!}{0!1!2!}\right)\left[yz^2+y^2z+xz^2+xy^2+x^2y+x^2z\right] + \left(\frac{3!}{1!1!1!}\right)\left[xyz\right]$

$= z^3+x^3+y^3 + 3(yz^2+y^2z+xz^2+xy^2+x^2y+x^2z) + 6xyz$

Now we are ready to play around with our general statement $(x_1+x_2+...+x_q)^n$

Using the same format: $(x_1+x_2+...+x_q)^n = \sum_{p_1,p_2...p_q}\left[\left(\frac{n!}{p_1!,p_2!...p_q!}\right)\left(x_1^{p_1}x_2^{p_2}...x_q^{p_q}\right)\right]$

Which basically means that we take the summation of all permutations of positive integer indices $p_1$ through to $p_q$ such that $p_1+p_2+...+p_q = 3$
Title: Re: Binomial theorem
Post by: Stroodle on January 02, 2010, 04:52:38 pm: Wow, you really love maths don't you ;D
Title: Re: Binomial theorem
Post by: brightsky on January 02, 2010, 06:06:45 pm: Genius at work, TT!! That was a better explanation than any teacher in the state could've given me!
Title: Re: Binomial theorem
Post by: Aqualim on January 05, 2010, 07:26:26 pm: Just to shed some light on the question you were asking about whether there was an eaiser way of solving this kind of equation, well there is! :)

It's probably already been explained to you in some form, but I'll share it with you anyway.

So basically we know that when finding coefficents and the 'x' value is on the left side of the question e.g. $(x-5)^5$ we count the terms in a descending order, meaning from $x^5 \to x^0$ and when the 'x' value is on the right side we count the terms in an ascending order $x^0 \to x^5$ .

Using this idea, we can do questions having two 'x' values. E.g.

$\left (x^2 \right )^6 \left ( -\frac{3}{x} \right )^0 = x^{12}$

$\left (x^2 \right )^5 \left ( -\frac{3}{x} \right )^1 = x^9$

$\left (x^2 \right )^4 \left ( -\frac{3}{x} \right )^2 = x^6$

$\left (x^2 \right )^3 \left ( -\frac{3}{x} \right )^3 = x^3$

$\left (x^2 \right )^2 \left ( -\frac{3}{x} \right )^4 = x^0$

$\left (x^2 \right )^1 \left ( -\frac{3}{x} \right )^5 = \frac{1}{x^3}$

$\left (x^2 \right )^0 \left ( -\frac{3}{x} \right )^6 = \frac{1}{x^6}$

NOTE: We are only interested in the 'x' values $\therefore$ when subracting the right side from the left, we don't include them, hence ending up with an 'x' value.

I hope that made some sense :)