Binomial theorem with more than 2 terms

  • I
  • Thread starter dyn
  • Start date
  • #1
dyn
749
57
Hi.
Is the binomial theorem ##(1+x)^n = 1+nx+(n(n-1)/2)x^2 + ….## valid for x replaced by an infinite series such as ##x+x^2+x^3+...## with every x in the formula replaced by the infinite series ?

If so , does the modulus of the infinite series have to be less than one for the series to converge ?
 

Answers and Replies

  • #3
dyn
749
57
My question arises from an example I have just come across regarding finding the residue of ##1/(z^5cos(z))## about the point z=0. The cos(z) is written out as a power series and then it seems the binomial theorem is applied to it with x replaced by the power series and n= -1. Is the modulus of that power series less than one ?
 
  • #5
dyn
749
57
The example expands ##1/coz(z)## as a binomial of the form ##(1+x)^n## with ## x## represented by the infinite power series starting with ##-z^2/2!## and ##n= -1##
 
  • #6
fresh_42
Mentor
Insights Author
2022 Award
17,643
18,300
So the question is whether ##\dfrac{1}{\lim_{n\to \infty}\sum_{k=0}^n a_k} = \lim_{n \to \infty} \dfrac{1}{\sum_{k=0}^n a_k}##, so what do you know about ##\lim_{n \to \infty} \dfrac{f_n}{g_n}##?
 
  • #7
dyn
749
57
You've lost me now
 
  • #8
fresh_42
Mentor
Insights Author
2022 Award
17,643
18,300
A power series such as ##z^5\cos z## is a limit, as every infinite series. So we have the quotient ##1## divided by that limit of partial sums. You asked whether this can be calculated by as limit of ##1## divided by those partial sums. Write down what you have, with limits instead of ##\infty##. This is only a symbol. If you want to know what you can do with it, you have to use its definition.
 
  • #9
dyn
749
57
If I am using the binomial expansion to find the residue ie. the coefficient of the ##1/z## term does it even matter if the series converges ? Whether it converges or not I should get the correct coefficient ?
 

Suggested for: Binomial theorem with more than 2 terms

  • Last Post
Replies
1
Views
630
Replies
10
Views
524
  • Last Post
Replies
1
Views
371
Replies
13
Views
3K
  • Last Post
Replies
0
Views
438
Replies
3
Views
315
Replies
0
Views
345
Replies
1
Views
314
  • Last Post
Replies
3
Views
577
Replies
9
Views
443
Top