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yonski

Member

Registered: 2005-12-14

Posts: 67

Convergent binomial expansions

I'm learning about binomial expansions from my maths book but I don't really understand how you find the range of values of x for which the expansion is valid.

For example, take the expansion of √(1-3x). I calculate the first four terms to be:

1 - (3/2)x - (9/8)x² - (27/16)x³ + ...

Now because you're not raising the bracket to a positive integer, no coefficient will ever be equal to zero, so it goes on infinitely. I get that bit.

Then the book says that the expansion will only be convergent when modulus(x) < 1/3 . This bit i don't get.

Any help would be much appreciated!

Last edited by yonski (2007-07-03 06:42:01)

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Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

HallsofIvy

Guest

Re: Convergent binomial expansions

Have you dealt with infinite series?  In particular how an infinte series may converge if the coefficients decrease fast enough>  Oh, and what is the general term of the coefficient of x^n?
From what you give it appears to be (3^n/2^(n+1)) but that doesn't work for n= 1.

yonski

Member

Registered: 2005-12-14

Posts: 67

Re: Convergent binomial expansions

I've looked at how to find the sum of an infinite geometric series, but that's about it. I guess i'll have to look up what you said about. Thanks.

---

Student: "What's a corollary?"
Lecturer: "What's a corollary? It's like when a theorem has a child. And names it corollary."

HallsofIvy

Guest

Re: Convergent binomial expansions

In general, "power series" (infinite sums involving x^n) have a "radius of convergence": they converge for all values of x inside that radius, diverge for x outside it.

   Often the "ratio test" is the best way to determine that radius of convergence.  If you are summing a_n, the series converges absolutely if the ratio |a_{n+1}/a_n| is less than 1 in the limit, diverges if it is larger than 1.

  For this example, assuming the general term is  a_n= (3^n/2^(n+1)) x^n, then a_{n+1}= 3^(n+1)/2^(n+2)( x^(n+1)) and the ratio |a_{n+1}/a_n|= (3/2)|x|  That will be less than 1 as long as |x|< 2/3.  That is, the series converges absolutely for -3< x< 3 and diverges for x<-3 or x>3.  (at x=3 or x=-3, the series may diverge, converge absolutely, or converge conditionally)