(adsbygoogle = window.adsbygoogle || []).push({}); 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.

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

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