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K G
Jul 08, 2021

Can you help me out?

in Math Problems

We use the infinitesimals that as x tends to 0,(1+x)^n-1 is an equivalent infinitesimal to nx .

We also extends this concept for n less than 1 , by saying that (1+x)^(1/n) -1 is equivalent to x/n

We draw these conclusion from the binomial expansion of (1+x)^n

But how can we extend the concept of binomial theorem fir n<1 or fractional values if n.

1 comment
Ian Fowler
Jul 22, 2021  ·  Edited: Jul 22, 2021

I'm not going to prove it here - I think a Maclaurin expansion ? for (1+x)^n will do it. I'm pretty sure that it will produce the Binomial Expansion. At any rate,

For REAL n: (1+x)^n = 1 + nx + [n(n-1)/2!]x^2 + [n(n-1)(n-2)/3!]x^3 + [n(n-1)(n-2)(n-3)/4!]x^4 + . . .


So the term in position R+1 is:

t(R+1) = [n(n-1)(n-2) ......(n-R+1) / R!] x^R This is an infinite series that will converge only when: -1 <x < 1. Footnote. If n is a finite whole number, then this expansion will eventually produce terms of 0 to infinity and then it essentially becomes a finite series. Another thought. Any infinite series that converges essentially behaves more and more like an infinite Geometric Series for extremely large values of n. That is, the larger the value of n, the more the ratio of successive terms becomes closer to a constant. Hence the radius of convergence

is -1 < x <1.

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