# Taylor Series Question

1. Nov 26, 2007

### ColdFusion85

[SOLVED] Taylor Series Question

I have to find the Taylor series of $$\frac{3}{z-4i}$$ about -5. Therefore, we want the series in powers of z+5. Now, following the textbook it appears that we want to get this in a form that resembles a geometric series so that we can easily express the Taylor series in the form of a geometric series...

If we get it in the form $$\frac{1}{1+t}$$, then Taylor series is $$\sum_{n=0}^\infty (-1)^n (t)^n$$

Now, if the denominator isn't in this geometric form the book says to "do some algebraic manipulation" to get it in a form suitable for a geometric series. My problem is how the hell is one supposed to do this "algebraic manipulation"?? Of course the textbook shows about 2 steps, which does nothing to indicate how one is supposed to figure out what exactly this manipulation is. My teacher did nothing to explain how to either. I understand we are supposed to be able to think a little, but this seems ridiculous that we are to somehow easily know what we need to add/subtract/multiply to get it in the correct form. Is there a way to go about it that is systematic or logical? I can't imagine how to manipulate the above problem I need to do.

Thanks for any help or insight.

2. Nov 26, 2007

### CompuChip

If you have something of the form
$$\frac{a}{z + b}$$
try dividing everything by b. Then you get
$$\frac{a/b}{1 + (z/b)} = C \cdot \frac{1}{1 + t},$$
for some constant C and for a new variable t... can you find (more like, read off) their values now?

3. Nov 26, 2007

### ColdFusion85

Yeah that makes sense I guess. However, in the book for example, they had $$\frac{2i}{4+iz}$$ about -3i. Their process was:

$$\frac{2i}{4+iz}$$ = $$\frac{2i}{4+i(z+3i)+3}$$ = $$\frac{2i}{7+i(z+3i)}$$ = $$\frac{2i}{7}\frac{1}{1+\frac{i}{7}(z+3i)}$$

I just don't see how one would come to that conclusion in any reasonable amount of time...

4. Nov 26, 2007

### Ben Niehoff

Each of those steps seems pretty clear to me...first you put it in terms of (z + 3i), and then you divide through to get the (1 + u) in the denominator...

5. Nov 26, 2007

### ColdFusion85

Alright, I think I may have gotten the answer. Can someone check my method?

Find the Taylor series of $$\frac{3}{z-4i}$$ about -5.

We want the series in powers of z+5.

$$\frac{3}{z-4i}$$ = $$\frac{3}{z+5-4i-5}$$ = $$\frac{3}{(z+5)-(4i+5)}$$ = $$(\frac{-3}{4i+5})(\frac{1}{1-\frac{z+5}{4i+5}})$$

Define $$t = \frac{z+5}{4i+5}$$. If |t|< 1, then

$$\frac{1}{1-\frac{z+5}{4i+5}} = \sum_{n=0}^\infty (\frac{z+5}{4i+5})^n$$

and therefore,

$$\frac{3}{z-4i}$$ = $$\frac{-3}{4i+5} \sum_{n=0}^\infty (\frac{z+5}{4i+5})^n$$

= $$\sum_{n=0}^\infty \frac{-3(z+5)^n}{(4i+5)^{n+1}}$$

6. Nov 26, 2007

### CompuChip

Very good! You got it.
Note that this method even gives you the radius of convergence of the series (you could in principle convert the inequality $|t| < 1$ to an inequality for z).

7. Nov 26, 2007

### Office_Shredder

Staff Emeritus
A good way to start these is to say let w= z+5 (in the example of doing a taylor series around z=-5). This works for a whole host of problems, like sine and cosine around non-zero points, etc. You find the taylor series of w around w=0, then put it back in terms of z+5.

8. Nov 26, 2007

### ColdFusion85

Nice...yes, calculating the radius of convergence for the series is the second part to the problem. Is this just |z|<4i ?

9. Nov 26, 2007

### CompuChip

Not really. First of all, |z| is a real number (|z|^2 = Re(z)^2 + Im(z)^2) You have an inequality for |t|, you should convert it to an inequality for |z + 5|.
E.g.
|z + 5|/|4i + 5| < 1
|z + 5| < 1 * |4i + 5| = sqrt(41) ( I think )

So the series converges at every point a distance smaller than |4i + 5| away from z = -5. Technically speaking, you should check the convergence for each point for which |z + 5| = |4i + 5|, and the series diverges for all points a distance greater than |4i + 5| from -5.

Last edited: Nov 26, 2007
10. Nov 26, 2007

### ColdFusion85

Oh yeah, that makes sense. I should have thought about it more. Thanks for your help.