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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$begingroup$

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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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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