December 10th, 2017, 04:19 AM  #1 
Newbie Joined: Dec 2017 From: australia Posts: 2 Thanks: 0  Question binomial theorem
Given 1/(1+cube root (x)) find its series using the binomial theorem. Use this series to approximate y at x = 1/2 and find the error between the real value and the approximated value for y at x = 1/2 (assuming n = 6). I was given a question similar to above which I have changed a bit as I still want to do the original myself. I haven't been able to work out how to do it. the best I could get have I changed the cube root into a 1/3 power and moved it to the top of the fraction making it 1/3. assuming N equals 6 I put the power together giving 2. i then plugged it into the equation (1+x)^n=1+1x+n(n1)/n! . . . . I then used attempted to use the error equation m/(n+1) (xc)^n+1. This error equation I thought was right but I don't get an answer that appears correct. can anyone explain where I am going wrong with this method? I have tried a few different ways but this seems to get me the closest. thanks in advance. 
December 10th, 2017, 06:58 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1529 
$\dfrac{1}{1 + y} = (1 + y)^{1} = 1  y + y^2  y^3 \,+ \,.\,.\,.$ (for $y$ < 1). Now substitute $x^{1/3}$ for $y$, etc. That was easy. Maybe you changed the original problem too much. 
December 10th, 2017, 04:38 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567  
December 10th, 2017, 09:18 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,704 Thanks: 1529 
Without latex: 1/(1 + y) = (1 + y)^(1) = 1  y + yÂ²  yÂ³ + . . . (for y < 1). Now substitute x^(1/3) for y, etc. 
December 12th, 2017, 12:57 PM  #5 
Global Moderator Joined: May 2007 Posts: 6,454 Thanks: 567  

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