Homework Help: 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.