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c4 binomial expansion

Wondering why it must always be in the form (1+ax)^n when doing the expansion in ascending powers of x? (probably should have found this out before my exam but whatever)

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Original post by black1blade
Wondering why it must always be in the form (1+ax)^n when doing the expansion in ascending powers of x? (probably should have found this out before my exam but whatever)


Because we know the expansion of (1+ax)n(1+ax)^n very well and so it is much easier to manipulate any expression in the from (b+ax)n(b+ax)^n, b1b \neq 1, into the more comfortable form bn(1+abx)nb^n(1+\frac{a}{b}x)^n before expanding it.
But you could do the questions without factorising right? Actually I guess you can't use standard binomial formula with the combinations in it when n is negative.
Original post by black1blade
But you could do the questions without factorising right? Actually I guess you can't use standard binomial formula with the combinations in it when n is negative.


(a + b)^n is limited to integer n > 0. For negative or fractional powers we need to use (1 + ax)^n. We factorise to get the bracket in the required form.

This is valid for |ax| < 1, expansion is infinite for negative or fractional powers, but it's finite and ascending (as written) to x^n for integer n > 0.
(edited 6 years ago)
Reply 4
Original post by black1blade
But you could do the questions without factorising right? Actually I guess you can't use standard binomial formula with the combinations in it when n is negative.


What do you think (1+ax)n(1+ax)^n is? It's just the expansion of (b+ax)n(b+ax)^n with b=1b=1 and when nn is negative, we replace the binomial coefficients nCk^nC_k with n(n1)(n2)(nk+1)/k!n(n-1)(n-2)\cdots (n-k+1)/k!.
Original post by Physics Enemy
There's (a + b)^n but that's limited to integer n > 0. For negative or fractional powers we need to use binomial. We often factories for binomial to get the expanded bracket in the required form.

Binomial is valid for |ax| < 1, expansion is infinite for negative and/or fractional powers, but finite and ascending (as we write it) up to x^n for integer n > 0.

So can you find a finite expression for the expansion of (1+ax)^n? (a sum to infinity as the expansion produces a geometric progression?). Also yeah must ax be less than 1 so that successive powers of x get smaller so that expansion can converge to a finite value?
Reply 6
Original post by Physics Enemy
For negative or fractional powers we need to use (1 + ax)^n.


No we don't.
Original post by Zacken
What do you think (1+ax)n(1+ax)^n is? It's just the expansion of (b+ax)n(b+ax)^n with b=1b=1 and when nn is negative, we replace the binomial coefficients nCk^nC_k with n(n1)(n2)(nk+1)/k!n(n-1)(n-2)\cdots (n-k+1)/k!.


So replacing binomial coefficients with the formula for combinations when k=1? I feel like I don't actually understand this topic (can do past paper questions til the cows come home) because it wasn't very well taught this year and I did c2 last year so forgotten the expansion work I'd already done and tge combination formula.
Reply 8
Original post by Physics Enemy
Do it with a technique from C4 or earlier then?


(a+b)n=an+nan1b+n(n1)2an2b2+(a+b)^n = a^n + na^{n-1}b + \frac{n(n-1)}{2} a^{n-2} b^2 + \cdots just as you would do for (1+a)n(1+a)^n by setting b=1b=1.
So when ax<1 and you do an expansion in ascending powers of x, do you eventually reach a power of x with no power like positive integer expansion? If not, how do you go about finding sum to infinity of the expansion and what would such an expression look like?
Original post by Zacken
(a+b)n=an+nan1b+n(n1)2an2b2+(a+b)^n = a^n + na^{n-1}b + \frac{n(n-1)}{2} a^{n-2} b^2 + \cdots just as you would do for (1+a)n(1+a)^n by setting b=1b=1.


That's (1 + ax)^n with the factorised term carried through, you adapted the nCr formula to be valid for negative/fractional powers. I referred to (a + b)^n in terms of the standard C4 result with nCr coefficients.
(edited 6 years ago)
Reply 11
Original post by black1blade
So replacing binomial coefficients with the formula for combinations when k=1? I feel like I don't actually understand this topic (can do past paper questions til the cows come home) because it wasn't very well taught this year and I did c2 last year so forgotten the expansion work I'd already done and tge combination formula.


Essentially this isn't really a topic. It's learning the Newton Binomial equation (a+b)n=an+nC1an1b+nC2an2b2++nCn1abn1+bn(a+b)^n = a^n + ^nC_1 a^{n-1} b + ^nC_2 a^{n-2}b^2 + \cdots + ^nC_{n-1} a b^{n-1} + b^n.

That's all there is. We also note that nCk=n!(nk)!k!=n(n1)(nk+1)k!^nC_k = \frac{n!}{(n-k)! k!} = \frac{n(n-1)\cdots (n-k+1)}{k!} which works for negative and/or fractional nn. Some people prefer to learn (1+a)n=an+nC1an1+nC2an2++1(1+a)^n = a^n + ^nC_1 a^{n-1} + ^nC_2 a^{n-2} + \cdots + 1 and then do everything of the form [ tex](a+b)^n = b^n(1 + a/b)^n by factorising out, but they are mistaken when they claim that "this must be done".
Reply 12
Original post by Physics Enemy
That's the (1 + ax)^n formula with the factorised term carried through. It's adapted to be valid for negative or fractional powers. I was talking about (a + b)^n in terms of the standard result with nCr coefficients.


What do you think the nCr coefficients are defined as?
Original post by Zacken
What do you think the nCr coefficients are defined as?


n!/[r!(n - r)!], that's not going to work with n = - 2.735 and r = 2. Hence we use the (1 + ax)^n formula.
(edited 6 years ago)
Reply 14
Original post by Physics Enemy
n!/(n - r)! , that's not going to work with say, n = - 2.735 and r = 2. Hence we go to the (1 + ax)^n formula.


Actually it's n!/(n-r)!r!

Where do you think the (1+ax)^n formula comes from...?
Reply 15
Original post by black1blade
Wondering why it must always be in the form (1+ax)^n when doing the expansion in ascending powers of x? (probably should have found this out before my exam but whatever)


I don't think it does have to be in that form but if you're unsure, expand both bn(1+ab)n b^n(1+\frac{a}{b})^n and (a+b)n (a+b)^n and see if they are equivalent.
So slightly confused by second formula in terms of expressing combinations. When doing the (1+a)^n expansion, you have 1+nx+n(n-1)x^2/2!... How does that tie in with the combinations?
Original post by black1blade
So when ax<1 and you do an expansion in ascending powers of x, do you eventually reach a power of x with no power like positive integer expansion? If not, how do you go about finding sum to infinity of the expansion and what would such an expression look like?


I did say expansion terminates at x^n for integer n > 0, but is infinite for negative or fractional powers.

You have formulae to sum various converging series. Depends on the series, esp pre-uni level.
(edited 6 years ago)
Reply 18
Original post by black1blade
So slightly confused by second formula in terms of expressing combinations. When doing the (1+a)^n expansion, you have 1+nx+n(n-1)x^2/2!... How does that tie in with the combinations?


nCk=n!(nk)!k!=n(nk+1)(nk)!(nk)!k!=(nk)(nk+1)k!^nC_k = \frac{n!}{(n-k)!k!} = \frac{n\cdots (n-k+1)(n-k)!}{(n-k)! k! } = \frac{(n-k)\cdots (n-k+1)}{k!}.

So for example 1+nC1x+nC2x2=1+nx+n(n1)/2x1 + ^nC1 x + ^nC2 x^2 = 1 + nx + n(n-1)/2 x since nC2=n(n1)/2!=n(n1)/2^nC_2 = n(n-1)/2! = n(n-1)/2
Original post by Zacken
Actually it's n!/(n-r)!r!

Where do you think the (1+ax)^n formula comes from...?


I forgot r! in the denom. Still doesn't work with n = - 2.753 and r = 2, so you've adapted the formula with nCr coefficients. In C4 we just use 2 tidier formulae.

I get you want to merge them into one, but that doesn't help OP imo and isn't 'C4 like'. This is semantic, with 1 general formula split (or not) into 2 sub formulae.
(edited 6 years ago)

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