Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expanding a simple cube root

  1. Oct 12, 2014 #1
    A quiz at the end of Steven Krantz's Calculus Demystified includes the following problem:
    Find
    [tex]
    \lim_{x \to \infty} [ \sqrt[3]{x+1}
    -
    \sqrt[3]{x} ]
    [/tex]
    I see how one can use the Maclaurin series to get
    [tex]
    \sqrt[3]{x+1} = 1 + \frac {x}{3} - \frac {x^2}{9} + \frac {5 x^3}{81} + . . .
    [/tex]
    but trying it with the cube root of x gives me zero plus an endless series of undefined terms.
    Is there a way to expand [itex] \sqrt[3]{x} [/itex] and solve this problem?
     
  2. jcsd
  3. Oct 12, 2014 #2
    But why do you want expand that in series?
    Isn't it by simple high school algebra :
    ac{2}{3}+%20%28x+1%29^\frac{2}{3}+x^{\frac{1}{3}}%28x+1%29^{\frac{1}{3}}%20}.gif
    for x>0?
     
  4. Oct 12, 2014 #3
    Oh yeah, you're right zoki85 - thanks!

    I also just realized that I could simply define y = x + 1, plug in y-1 for all the x's in the series above and there's the answer to my other question. :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Expanding a simple cube root
  1. Volume of a cube (Replies: 2)

  2. Expand integrand (Replies: 1)

Loading...