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Find the Maclaurin polynomial

  1. Nov 30, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the Maclaurin polynomial for P(6)X for 1/(1+2x2) about x=0

    2. Relevant equations

    I'm pretty sure I need to use the Maclaurin polynomial for 1/(1-x)= 1+x+x2+x3+....+xn +O(xn+1)


    3. The attempt at a solution

    I simply rewrote as 1/(1-(-2x2)) then proceeded to sub and expanded. The answer was not what I expected and not what I got. Apparently it's suppose to be 1-2x2+4x4-8x6. Can anyone expain to me why is there only four terms?
     
    Last edited: Nov 30, 2009
  2. jcsd
  3. Nov 30, 2009 #2

    Hootenanny

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    Why would you expect it to be more than four terms? Don't forget that you are expanding in powers of x2.

    Perhaps it would be more obvious if we first take the expansion,

    [tex]\frac{1}{1-\left(-\theta\right)} \sim 1 - \theta + \theta^2 - \theta^3 + \theta^4[/tex]

    Now simply make the substitution [itex]\theta = 2x^2[/itex],

    [tex]\frac{1}{1-\left(-\theta\right)} \sim 1 - 2x^2 + 4x^4 - 8x^6 + 16x^8[/tex]

    Note that the fact that you are expanding in powers of x2, means that the powers of each term in the Maclaurin series doubles.
     
  4. Nov 30, 2009 #3
    I understand that..but it ask for of order 6..P6(x) yet the answer given is 1-2x2+4x2-8x6...and not 1-2x2+4x2-8x6+16x8-32x10. Is the answer given in the book just wrong?
     
  5. Nov 30, 2009 #4

    Hootenanny

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    What does order six mean?
     
  6. Nov 30, 2009 #5
    The notation Pn means the n-th order Taylor polynomial for f about a. We have the general taylor formula f(a)+ f'(a)(x-a) + f''(a)/2! ((x-a)1) +...+fn (a)/n! ((x-a)n) Which basically means P6 would be taking the formula to the 6th derivative, hence order 6.
     
  7. Nov 30, 2009 #6

    Hootenanny

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    No. P6 means a ploynomial of order six, that is a polynomial where the greatest power is six.
     
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