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December 10th, 2017, 05:19 AM   #1
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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.
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December 10th, 2017, 07:58 AM   #2
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$\dfrac{1}{1 + y} = (1 + y)^{-1} = 1 - y + y^2 - y^3 \,+ \,.\,.\,.$ (for |$y$| < 1).
Now substitute $x^{1/3}$ for $y$, etc.

That was easy. Maybe you changed the original problem too much.
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December 10th, 2017, 05:38 PM   #3
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Quote:
Originally Posted by skipjack View Post
$\dfrac{1}{1 + y} = (1 + y)^{-1} = 1 - y + y^2 - y^3 \,+ \,.\,.\,.$ (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
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December 10th, 2017, 10:18 PM   #4
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Without latex: 1/(1 + y) = (1 + y)^(-1) = 1 - y + y² - y³ + . . . (for |y| < 1).
Now substitute x^(1/3) for y, etc.
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December 12th, 2017, 01:57 PM   #5
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Quote:
Originally Posted by skipjack View Post
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?
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