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Binomial theorem - finding coefficients - HSN forum

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Binomial theorem - finding coefficients


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#1 YIC

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Posted 21 January 2006 - 02:12 PM

prelim revision you know wink.gif

1. Find the coefficient of x power9.gif and the term independent of x in the expansion (1/x power2.gif - x) power1.gifpower8.gif

What I don't understand is that the two terms in the bracket are both x terms, whereas usually I see (x+y) powern.gif

2. Just a quick guassian elimination question

is the following applicable for upper triangle form?

n n 0 n
0 n n n
0 0 n n

(where "n" is non-zero number)

it is just that 0 in the first row that is annoying me. I tried a question with the above being the final upper triangular form that I got, but the answer was wrong for me.


I would appreciate any help smile.gif

#2 Paul

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Posted 21 January 2006 - 11:48 PM

For the Binomial Expansion I got the coefficient to be -816, any1 else get this?
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#3 YIC

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Posted 22 January 2006 - 01:46 AM

could you please show me the method you used?

thank you for offering help

#4 Dave

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Posted 22 January 2006 - 02:29 AM


0nnn
00nn
000n

that is what

If i am not here i am somewhere else



#5 Vixus

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Posted 22 January 2006 - 02:15 PM

QUOTE(YIC @ Jan 22 2006, 01:46 AM)
could you please show me the method you used?

thank you for offering help

View Post


It's ususally just a case of solving the binomial.
Then find the coefficients.

But first I think you need to simplify that 1/x2 - x

Let's see... maybe if you get it out to be:

(x(\fract{1}{x^3} - 1))^18

Did you actually write out Pascal's triangle row for ^18 or did you do it by calculator?

#6 George

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Posted 22 January 2006 - 05:30 PM

We have (x^{-2} - x)^{18} so using the binomial theorem, the general term is:



\begin{align*}
& {18 \choose r}(x^{-2})^r(-x)^{18-r} \\
&= {18 \choose r}x^{-2r}(-1)^{18-r}x^{18-r} \\
&= {18 \choose r}(-1)^{18-r}x^{18-3r}
\end{align*}

So the term in x9 must satisfy 18 - 3r = 9, i.e. r = 3.

You can then find the coefficient by replacing r with 3 in the expression for the general term. I agree with -816 for that.

The method is similar for the term independent of x (this satisfies 18 - 3r = 0). I got 18564.

#7 YIC

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Posted 22 January 2006 - 06:09 PM

thanks for the help everyone smile.gif



#8 YIC

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Posted 22 January 2006 - 06:36 PM

I have another question of you don't mind biggrin.gif

it is to do with graph sketching and asymptotes

to start with the function is f(x) = x power3.gif / (x power2.gif - 1)

I used long division to get x + x / (x power2.gif - 1). So far so good (unless I am wrong rolleyes.gif )

I worked out the vertical asymptote (x=1) but it is the approach to the graph that I am unsure about. I usually do a table of signs where the top row shows the value of x (values above and below the x=1 -> the asymptote). The next two rows show the numerator and denominator (respectively) of the function. The next row shows the value of the y.
It is the middle two rows that I am unsure about. I can fit the numerator and denominator in, but there is also a ( + x ) part to it and I am unsure where to place this....or should I ignore it?

If you don't understand what I am saying, I will try and explain it more

smile.gif

#9 Steve

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Posted 22 January 2006 - 06:44 PM

If you are using a table of values, you could always just do it for \displaystyle \frac{x^3}{x^2  - 1}.

I think this will make it easier to see what's happening.
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#10 YIC

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Posted 22 January 2006 - 08:08 PM

oh god thanks steve I am such an idiot! I don't know why I needed the function to stay separated for the table. thanks again smile.gif

#11 Paul

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Posted 22 January 2006 - 10:51 PM

QUOTE(YIC @ Jan 22 2006, 01:46 AM)
could you please show me the method you used?

thank you for offering help

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Yeh, soz I never included the method. Twas because I didnt know how to write it all in these messages!
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#12 YIC

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Posted 22 January 2006 - 11:29 PM

don't worry, it's fine now smile.gif





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