Find $lim _ { z rightarrow z _ { 0 } } frac { f ( z ) } { g ( z ) }$ with f and g holomorphic and $z_0$ a...












0












$begingroup$


Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$



My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$



In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$



My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$




  1. Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.


  2. $G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?






complex-analysis complex-numbers holomorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
    $endgroup$
    – Did
    Dec 29 '18 at 11:53
















0












$begingroup$


Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$



My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$



In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$



My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$




  1. Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.


  2. $G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?






complex-analysis complex-numbers holomorphic-functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
    $endgroup$
    – Did
    Dec 29 '18 at 11:53














0












0








0





$begingroup$


Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$



My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$



In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$



My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$




  1. Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.


  2. $G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?






complex-analysis complex-numbers holomorphic-functions







share|cite|improve this question











$endgroup$




Problem : Let $f$ and $g$ holomorphic within the neighbourhood of $z_0$. Knowing that $z_0$ is a zero of order $k$ of $f$, and a zero of order $l$ of $g$ with $l>k$



My Answer : Since f is holomorphic in a neighbourhood of $z_0$ we have :
$$
begin{aligned} f ( z ) & = f left( z _ { 0 } right) + f ^ { prime } left( z _ { 0 } right) left( z - z _ { 0 } right) + frac { f ^ { prime prime } left( z _ { 0 } right) } { 2 ! } left( z - z _ { 0 } right) ^ { 2 } + cdots \ & = left( z - z _ { 0 } right) ^ { l } left[ frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { f ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } F ( z ) end{aligned}
$$



In the same way we obtain :
$$
begin{aligned} g ( z ) & = left( z - z _ { 0 } right) ^ { l } left[ frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + sum _ { n = l + 1 } ^ { + infty } frac { g ^ { ( n ) } left( z _ { 0 } right) } { n ! } left( z - z _ { 0 } right) ^ { n - l } right] \ & = left( z - z _ { 0 } right) ^ { l } frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + left( z - z _ { 0 } right) ^ { l } G ( z ) end{aligned}
$$



My issue : To conclude I would like to write the following steps :
$$
lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = lim _ { z rightarrow z _ { 0 } } left[ frac { frac { f ^ { ( l ) } left( z _ { 0 } right) } { l ! } + F ( z ) } { frac { g ^ { ( l ) } left( z _ { 0 } right) } { l ! } + G ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right) }
$$




  1. Since $z_0$ is a zero of order $l$ of $g$ I know that ${ g ^ { ( l ) } left( z _ { 0 } right) } neq 0$.


  2. $G(z_0)$ and $F(z_0)$ need to be equal to $0$ but I'm not sure how to get this property. Where this property would come from?






complex-analysis complex-numbers holomorphic-functions




complex-analysis complex-numbers holomorphic-functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 12:11









Oscar Lanzi

13.7k12136




13.7k12136










asked Dec 29 '18 at 11:47









NotAbelianGroupNotAbelianGroup

20911




20911












  • $begingroup$
    But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
    $endgroup$
    – Did
    Dec 29 '18 at 11:53


















  • $begingroup$
    But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
    $endgroup$
    – Did
    Dec 29 '18 at 11:53
















$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53




$begingroup$
But you have an explicit formula for $F(z)$ and $G(z)$, right? So you just need to plug in $z=z_0$ in these formulas.
$endgroup$
– Did
Dec 29 '18 at 11:53










1 Answer
1






active

oldest

votes


















2












$begingroup$

A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
    $endgroup$
    – NotAbelianGroup
    Dec 29 '18 at 13:38














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1 Answer
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2












$begingroup$

A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
    $endgroup$
    – NotAbelianGroup
    Dec 29 '18 at 13:38


















2












$begingroup$

A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
    $endgroup$
    – NotAbelianGroup
    Dec 29 '18 at 13:38
















2












2








2





$begingroup$

A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?






share|cite|improve this answer









$endgroup$



A better approach: we can write $f(z)=(z-z_0)^{l} f_1(z)$ where $f_1$ is anlytic in some disk around $z_0$ with $f_1(z_0) neq 0$ and $g(z)=(z-z_0)^{k} g_1(z)$ where $g_1$ is analytic in some disk around $z_0$ and $g_1(z_0) neq 0$. This immediately gives the result since $frac {f(z)} {g(z)} to 0$ if $l >k$ and $frac {f(z)} {g(z)} to frac {f_1(z_0)} {g_1(z_0)}$ if $l=k$. Can you identify the limit in the second case by differentiating the functions $l$ times and setting $z=z_0$?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 29 '18 at 12:03









Kavi Rama MurthyKavi Rama Murthy

74.9k53270




74.9k53270












  • $begingroup$
    Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
    $endgroup$
    – NotAbelianGroup
    Dec 29 '18 at 13:38




















  • $begingroup$
    Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
    $endgroup$
    – NotAbelianGroup
    Dec 29 '18 at 13:38


















$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38






$begingroup$
Thank for your answer. Would the limit in the second case be $frac { f_1 ^ { ( l ) } left( z _ { 0 } right) } { g_1 ^ { ( l ) } left( z _ { 0 } right) }$? However I don't seem to understand how your second case relate to my problem. In my exercise $l>k$, so is the solution $0$ since the $l$-th derivative of $f$ equal to $0$? $$lim _ { z rightarrow z _ { 0 } } left[ frac { f ( z ) } { g ( z ) } right] = frac { f ^ { ( l ) } left( z _ { 0 } right) } { g ^ { ( l ) } left( z _ { 0 } right)} = 0$$
$endgroup$
– NotAbelianGroup
Dec 29 '18 at 13:38




















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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
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