Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Finding Taylor(x) of sec(x) via sec(x)*cos(x)=1

  1. May 18, 2010 #1
    1. The problem statement, all variables and given/known data
    Calculate the taylor polynomial T7(x) of order 7 about x=0 for the function f(x)=sec(x)
    Do not take derivatives of secx and sub into taylor formula. Let the required polynomial be T7(x)=1+a_2*x^2+a_4*x^4+a_6*x^6 As degree6 is the same as degree 7. Sub this into (cosx)(secx)=1 and solve for unknown coefficients. You can quote the taylor poly of cosx without proof. There should be no appearances of X^8 or higher powers in your work.

    BTW those _ are subscripts XD so a subscript 2,4,6



    2. Relevant equations
    Given above (general Taylor poly eqn not allowed in proof)



    3. The attempt at a solution
    First of all I multiplied the taylor poly of cos x which is simply 1-x^2/2 +x^4/4! -x^6/6!

    by the eqn given for T7 secx. This resulted in a string of x terms with powers multiples of 2 from 2 to 12. I then factorised in terms of the unknown coefficients. So i ended up with a polynomial =0 as the 1 term on the left and right cancelled. (secx*cosx=1 ) Then i made myself 3 eqns by subbing x=1,2,3 into the eqn. Resulting in 3 horrendous eqns with fraction coeffs of a_2 to a_6. I solved these simultaneously over the course of 2 hours to find that a_6=0 which is clearly wrong if you have seen the solution to secx elsewhere. (I used wolfram alpha) I also noticed that in the orignal eqn from multiplying secx and cos x that the coeffs of a_2 a_4 and a_4 were x^2*cosx x^4*cosx and x^6cosx respectively. Neither method has worked. I hope i have showed i have spent a depressingly long time on this problem, any sort of guide to the solution would be fantastic.
     
  2. jcsd
  3. May 18, 2010 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Just multiply (1-x^2/2 +x^4/4! -x^6/6!)*(1+a_2*x^2+a_4*x^4+a_6*x^6) and set it equal to 1, ignoring all terms with powers higher than 6. Then equate the total coefficient of x^2 to 0. What do you get for a2? Use that to solve for a4 by setting the coefficient of x^4 to zero. Finally find a6.
     
  4. May 18, 2010 #3

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Hmmm... why not show us what you got when you expanded

    [tex](1+a_2x^2+a_4x^4+a_6x^6)\left(1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}\right))[/tex]
     
  5. May 19, 2010 #4
    thanks a lot for the tip on finding the coefficient of x and not the coeficcient. I was misinterpreting the questions hint i suspect :D
     
  6. May 20, 2010 #5

    Gib Z

    User Avatar
    Homework Helper

    Please refrain from helping this user in future. This is an ASSIGNMENT question from the MATH1901 course at the University of Sydney.

    decisive has violated the Physicsforums guidelines.
     
  7. May 20, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I thought the whole purpose of the Homework Help section was to help students with assignment questions. Am I missing something?
     
  8. May 20, 2010 #7

    Gib Z

    User Avatar
    Homework Helper

    Sorry I should have mentioned that this is not a normal homework assignment, but a hand in assignment that counts towards the final mark in the course. The student also signs a form stating that they did not receive help from other people, plagiarize, etc, although collaboration with other students doing the assignment is allowed.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook