1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Taylor series expansion of tangent

  1. Sep 5, 2008 #1

    kreil

    User Avatar
    Gold Member

    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!

    [tex]tan(x)[/tex](n=0)

    [tex]sec^2(x)[/tex](n=1)

    [tex]2tan(x)sec^2(x)[/tex](n=2)

    [tex]2tan^2(x)sec^2(x)+2sec^4(x)[/tex](n=3)

    [tex]4tan^3(x)sec^2(x)+4tan(x)sec^4(x)+8sec^4(x)tan(x)[/tex](n=4)

    [tex]8tan^4(x)sec^2(x)+12tan^2(x)sec^4(x)+16tan^2(x)sec^4(x)+4sec^6(x)+32tan^2xsec^4(x)+8sec^6(x)[/tex](n=5)

    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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    kreil

    User Avatar
    Gold Member

    Thanks!
     
  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:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Taylor series expansion of tangent
  1. Taylor expansion (Replies: 8)

  2. Taylor series expansion (Replies: 15)

  3. Taylor expansion (Replies: 1)

Loading...