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Post by mondo on Jun 15, 2023 19:21:08 GMT
At the bottom of page 435 author gives a simple example $f = \frac{P}{Q}$ and says that this has a pole at $a$ due to $Q$ having a root in this place. This makes sense however I don't understand why we calculate a limit, which is later transformed to a derivative of only $Q(z)$? For instance, $Res[f(z),a] = \lim_{z \to a} (z-a)f(z)$ - why this way? Why not a limit of whole $P(z)/Q(z)$? Also why in the next step only denominator is transformed to a derivative, namely $\frac{Q(z) - Q(a)}{z -a}$, why we don't care about the nominator function $P(z)$?
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Post by mondo on Jun 15, 2023 19:31:45 GMT
Also, please note that in $\lim_{z \to a} (z-a)f(z)$ there is no $m$ power on $(z-a)$. At first I thought this limit is actually on a function $\phi(z) = (z-a)^m f(z)$ but this is not the case it looks like.
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Post by mondo on Jun 18, 2023 4:46:45 GMT
Corrected a typo in derivative in my first post. I still don't understand why there is a limit calculation and especially why only of the dempminator, which is later transformed to a derivative of $Q$ at $a$? Vasco, can you help here?
I see that this is a further simplification of the formula (10), and since here we consider a "simple" pole (order 1), then terms like $\frac{1}{(m-1)!}$ or $[\frac{d}{dz}]^{m-1}$ becomes $1$ and we are left with $Res[f(z), a] = (z-a)^m f(z)$ evaluated at $a$, but why is that $Res[f(z), a] = \lim_{z \to a} (z-a) f(z)$ ?
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Post by Admin on Jun 18, 2023 13:07:28 GMT
Mondo
Given that $\text{Res}[f(z),a]=[(z-a)f(z)]|_{z=a}$ and $f(z)=P(z)/Q(z)$
We can expand $Q$ as a Taylor series at $z=a$ as follows
$Q(z)=Q(a)+(z-a)Q'(a)+...$
We know that $Q(a)=0$ since $Q$ has a simple root at $a$, and so $Q(z)=(z-a)Q'(a)+...$
So $\text{Res}[f(z),a]=\bigg[(z-a)\frac{P(z)}{Q(z)}\bigg]\bigg|_{z=a}=\frac{P(z)}{Q'(a)}\bigg|_{z=a}$
From the definition of derivative we know that $\displaystyle Q'(a)=\lim_{z\rightarrow a}\frac{Q(z)-Q(a)}{z-a}$
Substituting for $Q'(a)$ we have
$\displaystyle\text{Res}[f(z),a]=\lim_{z\rightarrow a}\frac{P(z)}{\bigg[\frac{Q(z)-Q(a)}{z-a}\bigg]}=\frac{P(a)}{Q'(a)}$
Vasco
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Post by mondo on Jun 18, 2023 18:29:46 GMT
Mondo Given that $\text{Res}[f(z),a]=[(z-a)f(z)]|_{z=a}$ and $f(z)=P(z)/Q(z)$ We can expand $Q$ as a Taylor series at $z=a$ as follows $Q(z)=Q(a)+(z-a)Q'(a)+...$ We know that $Q(a)=0$ since $Q$ has a simple root at $a$, and so $Q(z)=(z-a)Q'(a)+...$ So $\text{Res}[f(z),a]=\bigg[(z-a)\frac{P(z)}{Q(z)}\bigg]\bigg|_{z=a}=\frac{P(z)}{Q'(a)}\bigg|_{z=a}$ From the definition of derivative we know that $\displaystyle Q'(a)=\lim_{z\rightarrow a}\frac{Q(z)-Q(a)}{z-a}$ Substituting for $Q'(a)$ we have $\displaystyle\text{Res}[f(z),a]=\lim_{z\rightarrow a}\frac{P(z)}{\bigg[\frac{Q(z)-Q(a)}{z-a}\bigg]}=\frac{P(a)}{Q'(a)}$ Vasco Thank you Vasco. This helps but there is one things that is still not clear, in the second to last equation on page 435 author writes $Res[f(z),a] = lim_{z \to a}(z-a)f(z)$ but then the derivative is only calculated for $Q(z)$ why not $P(z)$? |
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Post by Admin on Jun 18, 2023 18:38:35 GMT
Mondo
Your question means you have not understood my explanation.
Vasco
PS: There are a couple of small errors in my explanation which I intend to correct but they should not prevent you understanding my argument.
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Post by mondo on Jun 19, 2023 3:15:02 GMT
Mondo Your question means you have not understood my explanation. Vasco PS: There are a couple of small errors in my explanation which I intend to correct but they should not prevent you understanding my argument. That's right I still don't understand it. In (10) there is no use of a limit, only derivative, yet for this example a limit pops up. At the time author writes a formula for $Res[f(z),a]$ he only knows: - $f(z) = P/Q$ - there is a simple pole of order $1$ at $a$ - $Q$ has a root at $a$ Adding all these facts together, and using (10) I would expect, $Res[f(z),a] = (z-a)\frac{P(z)}{Q(z)}\bigg|_{z=a}$. But we know this limit leads to a pole at $a$ so we can't evaluate this limit. I guess that is the reason why we replace $Q(z)$ with its derivative but why are we allowed to do so? You said Taylor series, but Taylor series has more than one term right? Sure the first one is zero, but all the other may not be zero. |
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Post by Admin on Jun 20, 2023 15:35:55 GMT
Mondo
If $f(z)$ has an $m^{\text{th}}$ order pole at $a$ then using (10) when $m=1$ we have
$\text{Res}[f(z),a]=[(z-a)f(z)]\bigg|_{z=a}$
Now since $f(z)$ has a pole at $z=a$ then we cannot substitute $z=a$ into this expression for the residue, because $(z-a)\rightarrow 0$ and $f(a)\rightarrow\infty$ and the product is undefined.
Instead we must find
$\displaystyle \lim_{z\rightarrow a}[(z-a)f(z)]$
So now we write
$\displaystyle\text{Res}[f(z),a]=[(z-a)f(z)]\bigg|_{z=a}=\lim_{z\rightarrow a}[(z-a)f(z)]$
We consider the special case where $f=P/Q$ has a pole of order 1 at $a$ due to $Q$ having a simple root at $a$.
We can now write
$\displaystyle (z-a)f(z)=(z-a)\frac{P(z)}{Q(z)}=\frac{P(z)}{\bigg[\frac{Q(z)}{(z-a)}\bigg]}=\frac{P(z)}{\bigg[\frac{Q(z)-Q(a)}{(z-a)}\bigg]}$
since we know that $Q(a)=0$.
So we can now write
$\displaystyle\text{Res}[f(z),a]=\lim_{z\rightarrow a}[(z-a)f(z)]=\lim_{z\rightarrow a}\frac{P(z)}{\bigg[\frac{Q(z)-Q(a)}{(z-a)}\bigg]}=\frac{P(a)}{Q'(a)}$
Vasco
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Post by mondo on Jun 21, 2023 20:09:09 GMT
Yes, this explanation makes it all clear. It's neat how it lead to a derivative on $Q()$.
So this method can always be used if we recognize $P()\Q()$ form and the pole is of order $1$.
Thank you Vasco!
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