Violation of L'Hopital's Rule in Simplifying Rational Functions

[thomas49th]

Homework Statement

http://gyazo.com/f3161ad1a64909e84c2e033b442d7be1

I take it LHoptial is used to get

(2z+i/3)/(18z)

plugging -i/3 gives 1/18

But isn't this a violation of L'Hopital as
g'(z) does equal 0 (which is not allowed) if z is 0?
http://en.wikipedia.org/wiki/L'Hôpital's_rule

 

[clamtrox]

No, the L'Hopital applies only if both g(z) and f(z) go to zero or infinity. In this case downstairs does not.

 

[thomas49th]

so your saying I shouldn't of used L'Hopital. How else should I of solved it?

 

[Office_Shredder]

Simple poles in this form have a very easy way to calculate the residue. Consider f(z)/z. Suppose f(z) is holomorphic at z=0 (note f(z) might be, for example/ 1/(1+z), so you CAN write your problem in this form).

Then the Laurent series for f(z)/z is
f(0)/z+ f'(0) + f''(0) z /2 +...

just taking the Taylor series for f(z) and dividing it by z. Is it clear now what the residue is?

 

[D H]

Homework Statement

http://gyazo.com/f3161ad1a64909e84c2e033b442d7be1

I take it LHoptial is used to get

(2z+i/3)/(18z)

plugging -i/3 gives 1/18

But isn't this a violation of L'Hopital as
g'(z) does equal 0 is z is 0?
http://en.wikipedia.org/wiki/L'Hôpital's_rule

L'Hôpital's rule is perfectly valid here. The function f(z) has two poles, one at i/3 and the other at -i/3. Looking at the latter (your example), we need to calculate
\lim_{z\rightarrow -i/3} (z+i/3)f(z)<br /> = \lim_{z\rightarrow -i/3}\frac{(z+i/3)z}{9z^2+1}<br /> = \lim_{z\rightarrow -i/3}\frac{z^2+i/3\,z}{9z^2+1}
At the limit we have something of the form 0/0, so L'Hopital's rule applies. Taking the derivatives of the numerator and denominator yields
\lim_{z\rightarrow -i/3} (z+i/3)f(z)<br /> = \lim_{z\rightarrow -i/3}\frac{2z+i/3}{18z} = \frac 1 {18}
At the limit, the numerator and denominator are -i/3 and -18i/3, respectively, so the residue is 1/18.

The same analysis works for the other pole as well.

 

[Ray Vickson]

Homework Statement

http://gyazo.com/f3161ad1a64909e84c2e033b442d7be1

I take it LHoptial is used to get

(2z+i/3)/(18z)

plugging -i/3 gives 1/18

But isn't this a violation of L'Hopital as
g'(z) does equal 0 is z is 0?
http://en.wikipedia.org/wiki/L'Hôpital's_rule

If we have a function of the form F(z) = \frac{f(z)}{(z-a) g(z)} with g(a) \neq 0, the residue of F at a is just \text{Res}(F)(a) = \frac{f(a)}{g(a)}. There is no need to use L'Hospital's rule here, because the work performed by the rule has already been done. Here is what I mean: you could define \text{Res}(F)(a) = \lim_{z \rightarrow a} (z-a) F(z). If you do that, notice that we have
(z-a) F(z) = (z-a) \frac{f(z)}{(z-a)g(z)} = \frac{f(z)}{g(z)}, which has a perfectly nice limit equal to f(a)/g(a), with no need to use L'Hospital.

In fact, you are looking at the problem backwards: the whole point of L'Hospital is to "discover" the powers m and n (if any) in the expression \frac{N(z)}{D(z)} = <br /> \frac{(z-a)^n u(z)}{(z-a)^m v(z)}, \text{ where } v(a) \neq 0. In the current problem you don't need to do that, because it has already been done for you.

RGV

 

[thomas49th]

Ok, thanks. I understand, haven't looked at the Laurent series in detail but looks it looks ok.
While we are on the subject of limits I have got

Lim_{z \rightarrow ia} \frac{(z-ia)}{(z^2 + a^2)(z^2+b^2)}e^{imz}

Is there a quick way to find it's residue without multiplying out then taking the derivative of upstairs and downstairs until something happens?

edit: what is with this new itex. Itallic tex?

 

[Ray Vickson]

Ok, thanks. I understand, haven't looked at the Laurent series in detail but looks it looks ok.
While we are on the subject of limits I have got

Lim_{z \rightarrow ia} \frac{(z-ia)}{(z^2 + a^2)(z^2+b^2)}e^{imz}

Is there a quick way to find it's residue without multiplying out then taking the derivative of upstairs and downstairs until something happens?

edit: what is with this new itex. Itallic tex?

Your expression looks OK to me, but maybe you don't like Lim, and would rather have \lim instead? If so, write "\lim" to override the math italics (same as saying "\exp" instead of "exp", "\sin" instead of "sin", etc.)

Anyway, you can write z^2 + a^2 = (z - ia)(z + ia), then cancel out the (z - ia) factor in both the numerator and denominator. You can just put z = ia in what is left. So, basically, you can just write down the answer with hardly any work at all.

RGV

 

[thomas49th]

Thanks, I didn't spot that :)