Results 1 to 8 of 8

Math Help - Binomial Expension

  1. #1
    Member
    Joined
    Oct 2009
    Posts
    202

    Binomial Expension

    Write down the first three term in the binomial expansion of 3\sqrt {1 + \frac{x}{125}}.Hence,deduce that 3\sqrt {126} \approx 5.013.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391

    Re: Binomial Expension

    Quote Originally Posted by mastermin346 View Post
    Write down the first three term in the binomial expansion of 3\sqrt {1 + \frac{x}{125}}.Hence,deduce that 3\sqrt {126} \approx 5.013.
    Write this as \displaystyle 3\left(1 + \frac{x}{125}\right)^{\frac{1}{2}}.

    You should know that the binomial series is \displaystyle (1 + X)^{\alpha} = 1 + \alpha X + \frac{\alpha(\alpha - 1)}{2!}X^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!}X^3 + \dots. Here you need to let \displaystyle X = \frac{x}{125} and \displaystyle \alpha = \frac{1}{2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2009
    Posts
    202

    Re: Binomial Expension

    3\left ( 1 + \frac{x}{125} \right )^\frac{1}{2} let
    a = 1
    b = \frac{x}{125} and n = \frac{1}{2}

    I get 3 + \frac{3}{250}x - \frac{3}{125000}x^2 + \frac{3}{31250000}x^3 + ...

    Then?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391

    Re: Binomial Expension

    Quote Originally Posted by mastermin346 View Post
    3\left ( 1 + \frac{x}{125} \right )^\frac{1}{2} let
    a = 1
    b = \frac{x}{125} and n = \frac{1}{2}

    I get 3 + \frac{3}{250}x - \frac{3}{125000}x^2 + \frac{3}{31250000}x^3 + ...

    Then?
    Correct.

    Now note that \displaystyle 3\sqrt{126} = 3\sqrt{1 + 125} = 3\sqrt{1 + \frac{15625}{125}}. So what do you need to let \displaystyle x equal in your series?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2009
    Posts
    202

    Re: Binomial Expension

    x = 15625
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,607
    Thanks
    1574
    Awards
    1

    Re: Binomial Expension

    Quote Originally Posted by mastermin346 View Post
    Write down the first three term in the binomial expansion of \sqrt[3] {1 + \frac{x}{125}}.Hence,deduce that \sqrt[3] {126} \approx 5.013.
    Clearly the question should be \color{blue}\sqrt[3] {126}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,487
    Thanks
    1391

    Re: Binomial Expension

    Quote Originally Posted by Plato View Post
    Clearly the question should be \color{blue}\sqrt[3] {126}
    Yes that would probably make more sense :P
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,527
    Thanks
    773

    Re: Binomial Expension

    One has to represent \sqrt[3]{126} as 5\sqrt[3]{1+\frac{1}{125}}. Taylor expansion for \sqrt[3]{1+x} works well when x\approx 0. One can check the Taylor expansion

    (1+x)^{1/3}=1+\frac{x}{3}-\frac{x^2}{9}+\frac{5x^3}{81}+\dots

    for x = 125 gives 118869.8, which is very far from the correct value 5.013. On the other hand, the Lagrange form of the remainder says that (1+x)^{1/3}=1+\frac{x}{3}+R_2(x) where |R_2(1/125)|<10^{-5}, so 5(1+x)^{1/3}=5+\frac{5x}{3}+5R_2(x) and |5R_2(1/125)|<10^{-4}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: July 15th 2010, 05:33 AM
  2. Replies: 1
    Last Post: November 12th 2009, 12:38 AM
  3. Binomial Theorem or Binomial Coefficient
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: October 2nd 2009, 01:06 PM
  4. Replies: 1
    Last Post: March 11th 2009, 11:09 PM
  5. Relation between Negative Binomial and Binomial Distros
    Posted in the Advanced Statistics Forum
    Replies: 0
    Last Post: November 5th 2007, 06:59 AM

Search Tags


/mathhelpforum @mathhelpforum