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Simple silly binomial expansion question!

So I saw that the 2nd term of the sequence 1/(1-x) has is (-1)(-X)

But!!! I keep getting 1/2 seee n=-1 right and r=1 so n!/n(n-r)! =

n!= -1

r!(n-r)! = 1!((-1)-(1))! = 1!(-2!)= 2

so the 2nd term is -1/2 (-x)


Whyyyy it says everywhere that r!(n-r)! = r! whyyyyyyyyyyyyyyyyyyyyyy
Come on !!!!!!!! 21 views and not even one reply? :frown:
Original post by Destroyviruses
So I saw that the 2nd term of the sequence 1/(1-x) has is (-1)(-X)

But!!! I keep getting 1/2 seee n=-1 right and r=1 so n!/n(n-r)! =

n!= -1

r!(n-r)! = 1!((-1)-(1))! = 1!(-2!)= 2

so the 2nd term is -1/2 (-x)


Whyyyy it says everywhere that r!(n-r)! = r! whyyyyyyyyyyyyyyyyyyyyyy


You're approaching the expansion in the wrong way - As far as I know, the factorial of a negative number is undefined - think about what the definition of n! is (n.(n-1)!), and what that would mean for n < 0.

The binomial expansion of a function in the form 1/(1+ax) is:

(1+ax)−n=1+nax+n(n−1)2!(ax)2+n(n−1)(n−2)3!(ax)3+n(n−1)(n−2)(n−3)4!(ax)4+...(1+ax)^{-n} = 1 + nax + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \frac{n(n-1)(n-2)(n-3)}{4!}(ax)^4 + ...

It shouldn't be difficult to adjust the above and get the series you've seen with the second term "(-1)(-x)".

Hope that helped.
(edited 13 years ago)
Original post by EEngWillow
You're approaching the expansion in the wrong way - As far as I know, the factorial of a negative number is undefined - think about what the definition of n! is (n.(n-1)!), and what that would mean for n < 0.

The binomial expansion of a function in the form 1/(1+ax) is:

(1+ax)−n=1+nax+n(n−1)2!(ax)2+n(n−1)(n−2)3!(ax)3+n(n−1)(n−2)(n−3)4!(ax)4+...(1+ax)^{-n} = 1 + nax + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \frac{n(n-1)(n-2)(n-3)}{4!}(ax)^4 + ...

It shouldn't be difficult to adjust the above and get the series you've seen with the second term "(-1)(-x)".

Hope that helped.


stil confused. I know how to use the equation but not how to derive it!!!! why isnt it r!(n-r)! instead of r! !!!!
Original post by EEngWillow
You're approaching the expansion in the wrong way - As far as I know, the factorial of a negative number is undefined - think about what the definition of n! is (n.(n-1)!), and what that would mean for n < 0.

The binomial expansion of a function in the form 1/(1+ax) is:

(1+ax)−n=1+nax+n(n−1)2!(ax)2+n(n−1)(n−2)3!(ax)3+n(n−1)(n−2)(n−3)4!(ax)4+...(1+ax)^{-n} = 1 + nax + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \frac{n(n-1)(n-2)(n-3)}{4!}(ax)^4 + ...

It shouldn't be difficult to adjust the above and get the series you've seen with the second term "(-1)(-x)".

Hope that helped.


Ohh are you sure its undefined? because the calculator doenst seem to think so
Original post by Destroyviruses
Ohh are you sure its undefined? because the calculator doenst seem to think so


If you type -1! into your calculator, it evaluates -(1!).

Try typing (-1)! instead, and you'll see it gives a math error.
Reply 6
Avatar for noobynoo
noobynoo
6
(1−x)−1=1+x+x2+x3+...(1-x)^{-1} = 1+x+x^2+x^3+...

Does this answer your question?
Original post by EEngWillow
If you type -1! into your calculator, it evaluates -(1!).

Try typing (-1)! instead, and you'll see it gives a math error.


Ohhhh silly me. I just read the page on my math book properly and saw that

n(n-1)/r! etc

is an alternative one to the one I learnt!!! Silly me and my selective reading.

Thankyou for your help :smile:
Original post by noobynoo
(1−x)−1=1+x+x2+x3+...(1-x)^{-1} = 1+x+x^2+x^3+...

Does this answer your question?


I got it, thanks for trying. I didn't realise there were two equations for this expansion business!:colondollar:
Reply 9
Avatar for anshul95
anshul95
11
Original post by EEngWillow
You're approaching the expansion in the wrong way - As far as I know, the factorial of a negative number is undefined - think about what the definition of n! is (n.(n-1)!), and what that would mean for n < 0.

The binomial expansion of a function in the form 1/(1+ax) is:

(1+ax)−n=1+nax+n(n−1)2!(ax)2+n(n−1)(n−2)3!(ax)3+n(n−1)(n−2)(n−3)4!(ax)4+...(1+ax)^{-n} = 1 + nax + \frac{n(n-1)}{2!}(ax)^2 + \frac{n(n-1)(n-2)}{3!}(ax)^3 + \frac{n(n-1)(n-2)(n-3)}{4!}(ax)^4 + ...

It shouldn't be difficult to adjust the above and get the series you've seen with the second term "(-1)(-x)".

Hope that helped.

If I remember correctly you can using the gamma function -gamma(2)
Original post by anshul95
If I remember correctly you can using the gamma function -gamma(2)


Yes, you can, but gamma functions aren't in the A Level syllabus. (Or are they?!)
Reply 11
Avatar for anshul95
anshul95
11
Original post by EEngWillow
Yes, you can, but gamma functions aren't in the A Level syllabus. (Or are they?!)

they aren't I was just telling you that you can.
Original post by anshul95
If I remember correctly you can using the gamma function -gamma(2)

What part are you suggesting using the gamma function for?
Well the formula is n!(n−r)!r! \frac{n!}{(n-r)!r!} but for the first few terms they have used r = 1, 2, 3... etc.

So the the terms become:

n!(n−0)!0!=1, \frac{n!}{(n-0)!0!} = 1, n!(n−1)!1!=n, \frac{n!}{(n-1)!1!} = n, n!(n−2)!2!=n(n−1)2! \frac{n!}{(n-2)!2!} = \frac{n(n-1)}{2!}.
Reply 14
Avatar for anshul95
anshul95
11
Original post by Farhan.Hanif93
What part are you suggesting using the gamma function for?


I wasn't suggesting it be used in the question, but that EEngWillow said the factorial of a negative number was undefined. Which using the gamma function it isnt.
Original post by anshul95
I wasn't suggesting it be used in the question, but that EEngWillow said the factorial of a negative number was undefined. Which using the gamma function it isnt.

I'm afraid you're wrong there. The gamma function isn't defined for negative integers.
Reply 16
Avatar for anshul95
anshul95
11
Original post by Farhan.Hanif93
I'm afraid you're wrong there. The gamma function isn't defined for negative integers.
ah yes you are right....just tried it on wolframalpha. You are just too sharp. I probably got carried away because you can do it for negative fractions. The correct answer looks quite cool though - complex infinity!!!

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