# Thread: Terms of sec(z) series

1. ## Terms of sec(z) series

The question
Find the first three non-zero terms of the Taylor series for sec(z) about 0.

My attempt
$\displaystyle f(z) = \frac{1}{cos(z)}$
Radius of convergence depends on the zeros of cos(z), so $\displaystyle |z| < \frac{\pi}{2}$

Now, I'm not sure how to proceed. I know the series for cos(z), but I'm not sure how to use it in this context. Any advice?

2. ## Re: Terms of sec(z) series

$\displaystyle \displaystyle \frac{1}{\cos{(z)}} = c_0 + c_1x + c_2x^2 + c_3x^3 + \dots$.

At $\displaystyle \displaystyle z = 0, \frac{1}{\cos{(0)}} = c_0 + c_1(0) + c_2(0)^2 + c_3(0)^3 + \dots \implies c_0 = 1$.

Differentiate both sides

$\displaystyle \displaystyle \frac{\sin{(z)}}{\cos^2{(z)}} = c_1 + 2c_2x + 3c_3x^2 + 4c_4x^3 + \dots$

At $\displaystyle \displaystyle z = 0, \frac{\sin{(0)}}{\cos^2{(0)}} = c_1 + 2c_2(0) + 3c_3(0)^2 + 4c_4(0)^3 + \dots \implies c_1 = 0$.

Differentiate both sides

$\displaystyle \displaystyle \frac{\cos^2{(z)} + 2\sin{(z)}}{\cos^3{(z)}} = 2c_2 + 3\cdot 2c_3x + 4\cdot 3c_4x^2 + 5\cdot 4c_5x^3 + \dots$

At $\displaystyle \displaystyle z = 0, \frac{\cos^2{(0)} + 2\sin{(0)}}{\cos^3{(0)}} = 2c_2 + 3\cdot 2c_3(0) + 4\cdot 3c_4(0)^2 + 5\cdot 4c_5(0)^3 + \dots \implies c_2 = \frac{1}{2}$.

Keep going like this until you have the required number of terms.

3. ## Re: Terms of sec(z) series

How did you get that first line?

4. ## Re: Terms of sec(z) series

You have to write your function as an arbitrary polynomial centred about a point, in this case, $\displaystyle \displaystyle z = 0$.

5. ## Re: Terms of sec(z) series

Ahh I see, I think I get what you're doing. Thanks.

Say I wanted to get the series representation rather than just the coefficients, what would be the best way to go about that?

6. ## Re: Terms of sec(z) series

Originally Posted by Glitch
Ahh I see, I think I get what you're doing. Thanks.

Say I wanted to get the series representation rather than just the coefficients, what would be the best way to go about that?
Getting the coefficients gives you the series representation...

7. ## Re: Terms of sec(z) series

Forgive my ignorance, but how would I get a general representation of the series, using summation?

8. ## Re: Terms of sec(z) series

Originally Posted by Glitch
Forgive my ignorance, but how would I get a general representation of the series, using summation?
You would have to try to find a pattern in the coefficients.

9. ## Re: Terms of sec(z) series

Thanks. That's the reason I asked, by the way, just in case there was a better way. Time is of the essence in an exam, after all.

10. ## Re: Terms of sec(z) series

Originally Posted by Glitch
Thanks. That's the reason I asked, by the way, just in case there was a better way. Time is of the essence in an exam, after all.
In an exam they would either give you the series, or they would ask you to find a reasonably easy series.