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Trying to expand (1+x)^n using Taylor Expansion

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data

    I need to use Taylor Expansion to show that:

    (1+x)^n = 1 + nx + n(n-1)(x^2)/2! + ...


    2. Relevant equations

    y(x0 + dx) = y(x0) + dx(dy/dx) + [(dx)^2/2!](d^2y/dx^2) + ...

    3. The attempt at a solution

    I've only just begun Taylor Expansion, according to my textbook I need the above equation

    (1+x)^n
    So: x0 = 1
    and dx = x

    I'm not sure about this next part:

    y(1+x) = (1+x)^n
    So: y(x) = x^n
    dy/dx = nx^n-1
    d^2y/dx^2 = (n)(n-1)x^n-2

    However putting all of this into the equation I get:

    y(1+x) = y(1) + (x)(nx^n-1) + [x^2/2!][(n)(n-1)x^n-2] + ...

    (1+x)^n = 1^n + (nx)(x^n-1) + (n)(n-1)[x^2/2!](x^n-2) + ...

    (1+x)^n = 1 + (nx)(x^n-1) + (n)(n-1)[x^2/2!](x^n-2) + ...

    Which I get as my final answer.

    As you can see, there is a (x^n-1) in the second term, and a (x^n-2) in the third term, that shouldn't be there.

    So I'm trying to see where I went wrong, please help me out!
     
  2. jcsd
  3. Jun 23, 2012 #2

    Dick

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    The function you want to Taylor expand is y(x)=(1+x)^n. Then your Taylor expansion is y(x)=y(0)+y'(0)x+y''(0)x^2/2!+... Start by finding the derivatives of y evaluated at 0. What are y(0), y'(0), y''(0) etc etc?
     
  4. Jun 24, 2012 #3
    Thanks, here's what I get now:

    y(0) = 1^n

    y'(0) = n(1^n-1) = n

    y''(0) = (n-1)(n)(1^n-2) = (n)(n-1)

    So putting that into the equation I now get:

    (1+x)^n = 1^n + (n)(x) + (n)(n-1)(x^2)/2! + ...

    Is that the right method?
     
  5. Jun 24, 2012 #4

    Dick

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    Ok, that works. Try using more parentheses in expressions. (1^n-1) is unclear. I thought it meant (1-1)=0. It should be (1^(n-1)).
     
  6. Jun 24, 2012 #5

    HallsofIvy

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    By the way:
    [tex]1= \begin{pmatrix}n \\ 0 \end{pmatrix}[/tex]
    [tex]n= \begin{pmatrix}n \\ 1 \end{pmatrix}[/tex]
    [tex]n(n-1)= \begin{pmatrix}n \\ 2 \end{pmatrix}[/tex]
    etc.
     
  7. Dec 20, 2012 #6
    HallsofIvy, you forget multiply the left side by 1/2 in last formula:

    n choose 2 = n(n-1)/2
     
  8. Dec 20, 2012 #7

    HallsofIvy

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    Yes, thanks.
     
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