Question

Find the first few terms of the Maclaurin series for the following function: secx=1/(cosx) I know the series for cosx: 1-(x^2/2!)+(x^4/4!)-(x^6/6!)...+... would i just take that series and 1 by that?

Answer 1

i rewrot 1/cosx as a binomial series of (cosx)^-1....this is the method my professor used, but dont know where to go from here

Answer 2

f(0)=1 f'(0)=0 f""(0)= x^2/2!

Answer 3

well i know thats the long way..but im tying to figure out the other way to do it

Answer 4

i know the binomial formuals is (1+x)^p

Answer 5

and p in this case is -1

Answer 6

do i have to make the formula look like (cosx +1)^-1 ?

Answer 7

no. i'm just debating with myself whether you take 1/(every term) or if its 1/(whole thing)

Answer 8

http://www.wolframalpha.com/input/?i=maclaurin+series+1%2F%28cos%28x%29%29

Answer 9

i just new the first three terms of the series, but dont know how....im trying to follow how they do the binomial series in my book but but am not getting the same answer as them

Answer 10

need*

Answer 11

I was wrong so i'll delete what I wrote.
i'm wondering what class this is because the answer includes something I ironically haven't personally dealt with yet: Euler Numbers: http://en.wikipedia.org/wiki/Euler_number

if you scroll down here: http://mathworld.wolfram.com/MaclaurinSeries.html you'll see a lot of mclaurin series representations, look for sec(x):

The only thing that's different is the En part, where En is the nth Euler number.

I can try to answer questions

Answer 12

this class is called math methods in the physical sciences

Answer 13

sec(x) as explicit form not expanded form [with sigma]

Answer 14

i guess i could just take each successive derivative and solve it that way but it takes alot longer

Answer 15

Arguably. Because it's a known expansion, irl you don't need to solve them every time.

but for the purposes of this class, I guess you need to learn to do it from scratch

Answer 16

help!

Answer 17

I wouldn't know how to rearrange the cosine series to find the series for secant... I suggest finding a general pattern in the derivatives of \(\sec x\).

For example, the first derivative is \(\sec x \tan x\). You have \(f'(0)=0\).

The second derivative is \(\sec x\tan^2x+\sec^3x\), so \(f''(0)=1\).

You would find that all the terms of the odd order derivatives contain a factor of \(\tan x\), so \(f^{(2n+1)}(0)=0\), whereas all the even order derivatives contain a term of the form \(\sec^{n+1}x\), which give you \(f^{(2n)}(0)=1\cdot \text{(some constant that comes from myriad chain/power/product rules)}\).

The only problem is that constant. I don't think there's a simple pattern to it.