Question

Taylor series for f(x)=ln(sec(x)) at a=0

Answer 1

take a few derivatives. to find a general formula.

Answer 2

the general taylor for natural log should be in your book! you can follow that to get a general formula at a=0 after finding a few derivatives and plugging in 0

Answer 3

right o, i have done that i found f(x)=ln(sec(x), f'(x)=tan(x), f''(x)=sec^2(x), f'''(x)=2tan(x)sec^2(x), f''''(x)=-2(cos(2x)-2)sec^4(x)

Answer 4

@phi @satellite73 could you try and take a look when you get a chance, please?

Answer 5

I am also having issues using my ti-83 to calculate either Cn or Sn if you can help with getting the terms to pull up in a table or just by calculating

Answer 6

i have an idea

Answer 7

because doing this by hand is really going to suck
instead, take the derivative, and you get something nice, namely
\[\frac{d}{dx}\ln(\sec(x))=\tan(x)\]

Answer 8

now the power series for \(\tan(x)\) is well known
take that one, and integrate it term by term
it will be much much easier this way i think

Answer 9

that will help, let me give it a shot. would you happen to know how to plug in f(x) into calc ti-83 to get a table showing either series or sum of series?

Answer 10

no sorry
but if you are just looking for a calculator answer (calculators are so last century) you can do this
http://www.wolframalpha.com/input/?i=ln%28sec%28x%29%29

Answer 11

well she also wants us to know how to calculate the error
Find the exact error in approximating ln(sec(−0.3)) by its fourth degree Taylor polynomial at a=0