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Homework Help: Taylor series expansion of tangent

  1. Sep 5, 2008 #1


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    1. The problem statement, all variables and given/known data
    find the first four nonzero terms in the power series expansion of tan(x) about a=0

    2. Relevant equations
    [tex]\Sigma_{n=0}^{\infty} \frac{f^n (a)}{n!}(x-a)^n[/tex]

    3. The attempt at a solution

    Well the series has a zero term at each even n (0,2,4 etc)

    for n=1 I got x, for n=3 I got [itex]\frac{x^3}{3}[/itex]

    i ran into a problem while trying to compute n=5, since I got [itex]\frac{x^5}{10}[/itex] rather than [itex]\frac{2x^5}{15}[/itex] which is the answer. I assume that I must have differentiated wrong, so if someone could check the below derivatives I would appreciate it!







    As you can see in the n=5 expression plugging in x=0 reduces it to equaling 12; if it were instead equal to 16 I would have the correct expression.

    Also, this method is very tedious so if anyone knows of an easier way please let me know!
  2. jcsd
  3. Sep 5, 2008 #2


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    This is incorrect. Differentiating the "sec^2(x)" part of the first term of the previous derivative gives 2sec(x)(sec(x) tan(x)) so the multiplying by 2tan(x) gives 4tan^2(x)sec^2(x). This should be 4 tan^2(x)sec^2(x)+ 2sec^4(x).

  4. Sep 5, 2008 #3
    4tan^2 sec^2 + 2sec^4 (n=3)
  5. Sep 6, 2008 #4


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  6. Sep 6, 2008 #5
    tan (n=0)
    sec^2 (n=1)
    2 tan sec^2 (n=2)
    4 tan^2 sec^2 + 2 sec^4 (n=3)
    16 tan sec^4 + 8 tan^3 sec^2 (n=4)
    16 sec^6 + 88 sec^4 tan^2 + 16 sec^2 tan^4 (n=5)
    272 sec^6 tan + 416 sec^4 tan^3 + 32 sec^2 tan^5 (n=6)
    272 sec^8 + 2880 sec^6 tan^2 + 1824 sec^4 tan^4 + 64 sec^2 tan^6 (n=7)
    7936 sec^8 tan + 24576 sec^6 tan^3 + 7680 sec^4 tan^5 + 128 sec^2 tan^7 (n=8)

    See attachment,
    Ref: Mathematica by Example, (By Martha L. Abell, James P.Braselton), page 188.

    Attached Files:

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