Discover the Taylor Series for 3/(z-4i) about -5 | SOLVED

[ColdFusion85]

[SOLVED] Taylor Series Question

I have to find the Taylor series of [tex]\frac{3}{z-4i}[/tex] 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 [tex]\frac{1}{1+t}[/tex], then Taylor series is [tex]\sum_{n=0}^\infty (-1)^n (t)^n[/tex]

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.

 

[CompuChip]

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

 

[ColdFusion85]

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

[tex]\frac{2i}{4+iz}[/tex] = [tex]\frac{2i}{4+i(z+3i)+3}[/tex] = [tex]\frac{2i}{7+i(z+3i)}[/tex] = [tex]\frac{2i}{7}\frac{1}{1+\frac{i}{7}(z+3i)}[/tex]

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

 

[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...

 

[ColdFusion85]

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

Find the Taylor series of [tex]\frac{3}{z-4i}[/tex] about -5.

We want the series in powers of z+5.

[tex]\frac{3}{z-4i}[/tex] = [tex]\frac{3}{z+5-4i-5}[/tex] = [tex]\frac{3}{(z+5)-(4i+5)}[/tex] = [tex](\frac{-3}{4i+5})(\frac{1}{1-\frac{z+5}{4i+5}})[/tex]

Define [tex] t = \frac{z+5}{4i+5}[/tex]. If |t|< 1, then

[tex]\frac{1}{1-\frac{z+5}{4i+5}} = \sum_{n=0}^\infty (\frac{z+5}{4i+5})^n[/tex]

and therefore,

[tex]\frac{3}{z-4i}[/tex] = [tex]\frac{-3}{4i+5} \sum_{n=0}^\infty (\frac{z+5}{4i+5})^n[/tex]

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

 

[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 [itex]|t| < 1[/itex] to an inequality for z).

 

[Office_Shredder]

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.

 

[ColdFusion85]

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

 

[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.

 

[ColdFusion85]

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