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

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

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

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

Differentiate both sides

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

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

Differentiate both sides

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

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

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, $\displaystyle z = 0$.

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.