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 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(n-1)/n! . . . . I then used attempted to use the error equation |m/(n+1) (x-c)^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: 20,926 Thanks: 2205 $\dfrac{1}{1 + y} = (1 + y)^{-1} = 1 - y + y^2 - y^3 \,+ \,.\,.\,.$ (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
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 Originally Posted by skipjack $\dfrac{1}{1 + y} = (1 + y)^{-1} = 1 - y + y^2 - y^3 \,+ \,.\,.\,.$ (for |$y$| < 1). Now substitute $x^{1/3}$ for $y$, etc. That was easy. Maybe you changed the original problem too much.
latex problem - it is not displaying December 10th, 2017, 09:18 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 Without latex: 1/(1 + y) = (1 + y)^(-1) = 1 - y + y² - y³ + . . . (for |y| < 1). Now substitute x^(1/3) for y, etc. December 12th, 2017, 12:57 PM   #5
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Quote:
 Originally Posted by skipjack Without latex: 1/(1 + y) = (1 + y)^(-1) = 1 - y + y² - y³ + . . . (for |y| < 1). Now substitute x^(1/3) for y, etc.
You now have a series in $\displaystyle x^{\frac{1}{3}}$. I wonder if the original intent was to get a series in x? Tags binomial, question, theorem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post beesee Probability and Statistics 5 September 18th, 2015 01:38 PM Richard Fenton Number Theory 3 January 13th, 2015 08:29 AM SH-Rock Probability and Statistics 1 January 6th, 2011 03:08 PM rooster Applied Math 1 October 27th, 2009 08:30 AM SH-Rock Calculus 1 December 31st, 1969 04:00 PM

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