Expansion for Tan x about Pi/2

[ghotra]

Is there an "prettier" way to obtain a taylor series for tan x about pi/2? Currently, I expand sin and cos and do polynomial long division. The typical application of the formula doesn't seem to work because of the singularity about pi/2.

Thanks.

 

[HallsofIvy]

tan x does not exist, as you say, at pi/2. Therefore it can't be "analytic" at pi/2 and does not have a Taylor series at pi/2!

 

[cjellison]

Hmm...

Sorry perhaps I should have spoken correctly.

How do I find the series expansion of tan x about pi/2?

Mathematica input "Series[Tan[x],{x,Pi/2,3}]" gives:

[tex]
-\frac{1}{x-\pi/2} + \frac{1}{3}\left(x - \frac{\pi}{2}\right) + \frac{1}{45}\left(x - \pi/2\right)^3 + \cdots
[/itex]

Again, I can obtain this series if I write sin and cos as a Taylor series about pi/2 and perform the corresponding long division. I am looking for a "prettier" way of obtaining this series.

Why am I doing this? I am trying to find the residue of tan z about pi/2.

 

[cjellison]

Additional Question

When I see a function: tan z

How can determine, by inspection only, what order the pole is? It is simple to do for polynomials.

This particular example would be enlightening...I had to find the series to determine the order of the pole.

For something like 1/(1-cos z) I think of the series expansion of cos z about 0 and know that the 1 will cancel...so my first term, when I do the division, will be z^{-2}. Thus, I know this is a second order pole. Is this _the_ method for determining the order of a pole, or is there a better way?

 

[krab]

Do a Taylor expansion of (pi/2-x)tan(x).

 

[krab]

For f(x) having a pole at x=h, just evaluate the limit with x going to h of (x-h)^n f(x). If there is no limit for any n, it is an essential singularity. If there is no limit for all n<N, and the limit is zero for n>N, then the order of the pole is N.

 

[ghotra]

You mention taking the Taylor expansion of (z-pi/2)tan(z). I assume you want me to expand about the point pi/2. Even so, this does not take away the complication. The first non-zero term is:

$$\frac{\mathrm{d}}{\mathrm{d}z}\left[(z-\pi/2)\tan z \right]_{z=\pi/2} (z-\pi/2)$$

But, when I take the derivative I will get a $\sec^2 z$...I cannot evaluate that at pi/2. This method does not work. Please correct me if I have misunderstood.

Thanks.

 

[lurflurf]

You mention taking the Taylor expansion of (z-pi/2)tan(z). I assume you want me to expand about the point pi/2. Even so, this does not take away the complication. The first non-zero term is:

$$\frac{\mathrm{d}}{\mathrm{d}z}\left[(z-\pi/2)\tan z \right]_{z=\pi/2} (z-\pi/2)$$

But, when I take the derivative I will get a $\sec^2 z$...I cannot evaluate that at pi/2. This method does not work. Please correct me if I have misunderstood.

Thanks.

you need to interpet z=pi/2 as a limit z->pi/2
It is probably easier to find series for sin and cos and do formal division of series.

 

[krab]

But, when I take the derivative I will get a $\sec^2 z$...I cannot evaluate that at pi/2. This method does not work.

But the sec^2 is multiplied by x-pi/2, so if e=x-pi/2, it converges to 1/e as e->0. The other term is tan(x), which converges to -1/e. The two cancel, so the limit is zero. Now go to the second derivative. It's messy, and you have to use tricks like l'Hopital's rules, but still better than dividing two infinite series.