Hi I have been told to look at ncr and do some cancelling but would like to see how the binomial expansion of(1+x)^n is derived. i.e derivation of 1+nx+ n(n-1)x^2/2!... Thanks!
You must be the king of unintentionally offensive lol. I wasn't really looking for this but I just figured it out for myself . ncr= n choose r =n!/[(n-r)!r!]. The formula just is the same as the normal expansion just with the coefficient being n choose r- for example nc2= n(n-1)/2!. I get it now lol. #epicbrainderp How would one go about it with Taylor series? (I do understand Fp2 stuff)
You must be the king of unintentionally offensive lol. I wasn't really looking for this but I just figured it out for myself . ncr= n choose r =n!/[(n-r)!r!]. The formula just is the same as the normal expansion just with the coefficient being n choose r- for example nc2= n(n-1)/2!. I get it now lol. #epicbrainderp How would one go about it with Taylor series? (I do understand Fp2 stuff)
Apologies; I assumed that this proof was taught alongside binomial theorem in C2, and as a result I thought that you were still learning it. I wasn't trying to be rude .
Let y = (1 + x)^n, and expand using Taylor's expansion (or Maclaurin's expansion - pick your poison). The result will be what you get from the binomial theorem.
Apologies; I assumed that this proof was taught alongside binomial theorem in C2, and as a result I thought that you were still learning it. I wasn't trying to be rude .
Let y = (1 + x)^n, and expand using Taylor's expansion (or Maclaurin's expansion - pick your poison). The result will be what you get from the binomial theorem.
Hi I have been told to look at ncr and do some cancelling but would like to see how the binomial expansion of(1+x)^n is derived. i.e derivation of 1+nx+ n(n-1)x^2/2!... Thanks!
If you want a different way, you can do it by induction: either directly, if you're careful about keeping track of the coefficients, or (slightly more weirdly) by differentiating, expanding inductively, and then integrating again.
If you want a different way, you can do it by induction: either directly, if you're careful about keeping track of the coefficients, or (slightly more weirdly) by differentiating, expanding inductively, and then integrating again.
If you already have the expansion for (1+x)^n, you can get the expansion for (1+x)^(n+1) by multiplying (that expansion) by (1+x).
Edit: (well, that's what I would expect it to mean, but reading the post I'm not so sure that's what he *does* mean..)
That is what I mean: (1+x)n−1(1+x)=(xn−1+(n−1)xn−2+2(n−1)(n−2)xn−3+⋯+(n−1)x+1)(1+x)=1+x(1+n−1)+x2(2(n−1)(n−2)+(n−1))+….
The differentiation way: dxd[(1+x)n]=n(1+x)n−1=n(1+(n−1)x+⋯+(n−1)xn−2+xn−1), which we may integrate if we're careful to keep track of our coefficients.
That is what I mean: ...Yes, that's what I expected the "expanding inductively" to mean.
But when you said "expanding inductively" it seemed to be in relation to the differentiation method, which was what I didn't follow. (I think you perhaps put those words in slightly the wrong place).
Yes, that's what I expected the "expanding inductively" to mean.
But when you said "expanding inductively" it seemed to be in relation to the differentiation method, which was what I didn't follow. (I think you perhaps put those words in slightly the wrong place).