Mixing
Why does this function have pole of order n=2, not 3?
$begingroup$
$frac{1-cos(z)}{z^4}$, find the order of pole for this function at $z=0$
Initially I thought if you plug in $z=0$ you get $frac{0}{0^4}$ so one removable singularity, thus pole is of order n=3.
However, if you taylor expand the cosine, you see that the pole order is n=2.
My question is, why is the first method wrong, and when does it/does it not work? (Because some problems the first method works fine)
complex-analysis
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
$endgroup$
add a comment |
$begingroup$
$frac{1-cos(z)}{z^4}$, find the order of pole for this function at $z=0$
Initially I thought if you plug in $z=0$ you get $frac{0}{0^4}$ so one removable singularity, thus pole is of order n=3.
However, if you taylor expand the cosine, you see that the pole order is n=2.
My question is, why is the first method wrong, and when does it/does it not work? (Because some problems the first method works fine)
complex-analysis
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
$endgroup$
$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
$begingroup$
because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
$endgroup$
– MinYoung Kim
Jan 24 at 7:00
add a comment |
$begingroup$
$frac{1-cos(z)}{z^4}$, find the order of pole for this function at $z=0$
Initially I thought if you plug in $z=0$ you get $frac{0}{0^4}$ so one removable singularity, thus pole is of order n=3.
However, if you taylor expand the cosine, you see that the pole order is n=2.
My question is, why is the first method wrong, and when does it/does it not work? (Because some problems the first method works fine)
complex-analysis
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
$endgroup$
$frac{1-cos(z)}{z^4}$, find the order of pole for this function at $z=0$
Initially I thought if you plug in $z=0$ you get $frac{0}{0^4}$ so one removable singularity, thus pole is of order n=3.
However, if you taylor expand the cosine, you see that the pole order is n=2.
My question is, why is the first method wrong, and when does it/does it not work? (Because some problems the first method works fine)
complex-analysis
complex-analysis
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
asked Jan 24 at 4:26
MinYoung KimMinYoung Kim
887
887
$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
$begingroup$
because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
$endgroup$
– MinYoung Kim
Jan 24 at 7:00
add a comment |
$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
$begingroup$
because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
$endgroup$
– MinYoung Kim
Jan 24 at 7:00
$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
$begingroup$
because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
$endgroup$
– MinYoung Kim
Jan 24 at 7:00
$begingroup$
because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
$endgroup$
– MinYoung Kim
Jan 24 at 7:00
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The order of the pole of $f$ at $z=0$ is the least $k$ such that
$$z^k f(z)$$
has a removable discontinuity at $z=0$. So the pole of this function is of order $2$ if
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}$$
exists and is finite but
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}$$
is infinite.
Since
$$cos(z)=1-frac{z^2}2+frac{z^4}{4!}-frac{z^6}{6!}+cdots,$$
then
$$z^2cdot frac{1-cos(z)}{z^4}=frac12-frac{z^2}{4!}+cdots,$$
so
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}=frac12.$$
But is easy to see, through the same reasoning, that
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}=infty.$$
Of course that
$$lim_{zto 0} z^3cdot frac{1-cos(z)}{z^4}$$
is also finite ($0$ indeed, and the same is true for higher exponents), but since $2$ is the least exponent such that the limit
$$lim_{zto 0} z^kcdot frac{1-cos(z)}{z^4}$$
is finite, the order of the pole is $2$.
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
add a comment |
$begingroup$
The initial problem with your first approach is that $1-cos z$ has a double zero at 0 (since $cos$ has slope 0 at 0). In general, if $f$ has a zero of order $k$ at 0 and $g$ a zero or order $j$ at 0 then $f/g$ has a zero of order $k-j$ (where a zero of order $-n$ is a pole of order $n$).
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
Because $1-cos z=0$ when $z=0$ and note that $$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots$$ so your function can be written as $$frac{(frac{z^2}{2!}-frac{z^4}{4!}+cdots)}{z^4}=frac{phi(z)}{z^2}$$ where $phi(z)=frac{1}{2!}-frac{z^2}{4!} cdots $ with $phi(0) neq0$
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
$endgroup$
add a comment |
$begingroup$
$1 - cos z= {1 over 2} z^2 + cdots $ .
So, ${1 - cos z over z^4} = {1 over 2} {1 over z^2} + cdots $
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The order of the pole of $f$ at $z=0$ is the least $k$ such that
$$z^k f(z)$$
has a removable discontinuity at $z=0$. So the pole of this function is of order $2$ if
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}$$
exists and is finite but
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}$$
is infinite.
Since
$$cos(z)=1-frac{z^2}2+frac{z^4}{4!}-frac{z^6}{6!}+cdots,$$
then
$$z^2cdot frac{1-cos(z)}{z^4}=frac12-frac{z^2}{4!}+cdots,$$
so
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}=frac12.$$
But is easy to see, through the same reasoning, that
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}=infty.$$
Of course that
$$lim_{zto 0} z^3cdot frac{1-cos(z)}{z^4}$$
is also finite ($0$ indeed, and the same is true for higher exponents), but since $2$ is the least exponent such that the limit
$$lim_{zto 0} z^kcdot frac{1-cos(z)}{z^4}$$
is finite, the order of the pole is $2$.
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
add a comment |
$begingroup$
The order of the pole of $f$ at $z=0$ is the least $k$ such that
$$z^k f(z)$$
has a removable discontinuity at $z=0$. So the pole of this function is of order $2$ if
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}$$
exists and is finite but
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}$$
is infinite.
Since
$$cos(z)=1-frac{z^2}2+frac{z^4}{4!}-frac{z^6}{6!}+cdots,$$
then
$$z^2cdot frac{1-cos(z)}{z^4}=frac12-frac{z^2}{4!}+cdots,$$
so
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}=frac12.$$
But is easy to see, through the same reasoning, that
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}=infty.$$
Of course that
$$lim_{zto 0} z^3cdot frac{1-cos(z)}{z^4}$$
is also finite ($0$ indeed, and the same is true for higher exponents), but since $2$ is the least exponent such that the limit
$$lim_{zto 0} z^kcdot frac{1-cos(z)}{z^4}$$
is finite, the order of the pole is $2$.
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
add a comment |
$begingroup$
The order of the pole of $f$ at $z=0$ is the least $k$ such that
$$z^k f(z)$$
has a removable discontinuity at $z=0$. So the pole of this function is of order $2$ if
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}$$
exists and is finite but
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}$$
is infinite.
Since
$$cos(z)=1-frac{z^2}2+frac{z^4}{4!}-frac{z^6}{6!}+cdots,$$
then
$$z^2cdot frac{1-cos(z)}{z^4}=frac12-frac{z^2}{4!}+cdots,$$
so
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}=frac12.$$
But is easy to see, through the same reasoning, that
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}=infty.$$
Of course that
$$lim_{zto 0} z^3cdot frac{1-cos(z)}{z^4}$$
is also finite ($0$ indeed, and the same is true for higher exponents), but since $2$ is the least exponent such that the limit
$$lim_{zto 0} z^kcdot frac{1-cos(z)}{z^4}$$
is finite, the order of the pole is $2$.
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
$endgroup$
The order of the pole of $f$ at $z=0$ is the least $k$ such that
$$z^k f(z)$$
has a removable discontinuity at $z=0$. So the pole of this function is of order $2$ if
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}$$
exists and is finite but
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}$$
is infinite.
Since
$$cos(z)=1-frac{z^2}2+frac{z^4}{4!}-frac{z^6}{6!}+cdots,$$
then
$$z^2cdot frac{1-cos(z)}{z^4}=frac12-frac{z^2}{4!}+cdots,$$
so
$$lim_{zto 0} z^2cdot frac{1-cos(z)}{z^4}=frac12.$$
But is easy to see, through the same reasoning, that
$$lim_{zto 0} zcdot frac{1-cos(z)}{z^4}=infty.$$
Of course that
$$lim_{zto 0} z^3cdot frac{1-cos(z)}{z^4}$$
is also finite ($0$ indeed, and the same is true for higher exponents), but since $2$ is the least exponent such that the limit
$$lim_{zto 0} z^kcdot frac{1-cos(z)}{z^4}$$
is finite, the order of the pole is $2$.
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
edited Jan 24 at 4:41
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
answered Jan 24 at 4:33
Alejandro Nasif SalumAlejandro Nasif Salum
4,765118
4,765118
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
add a comment |
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
So I realized that if you taylor expand, you get the right answer. My question was, is my original method just plain wrong? or applicable only to certain cases?
$endgroup$
– MinYoung Kim
Jan 24 at 4:39
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
If you plug in $z=0$ you only get and indetermination and you can't say anything, since you don't know how quick each part goes to zero. In fact, the $0$ you wrote in the numerator, corresponding to $1-cos z$, omits the fact that this expression goes to zero as fast as $z^2$; that's why you conclude that the order is $3$ and not $2$. You have to effectively calculate the limits, since an expression that goes to zero can do this at any rate.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:45
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
$begingroup$
How fast a function goes to 0 ($z^2$ or$z$ etc.)$ is something I haven't learned. So I guess it is just best to stick with the safe method of just expanding a taylor series when I see one
$endgroup$
– MinYoung Kim
Jan 24 at 4:48
1
1
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
$begingroup$
Actually, it is the Taylor series that tells you "how fast" some expression goes to zero. But yes, at least until you feel very comfortable with these concepts, it's better to stick to the definitions and the basic properties to decide the order of the poles.
$endgroup$
– Alejandro Nasif Salum
Jan 24 at 4:50
add a comment |
$begingroup$
The initial problem with your first approach is that $1-cos z$ has a double zero at 0 (since $cos$ has slope 0 at 0). In general, if $f$ has a zero of order $k$ at 0 and $g$ a zero or order $j$ at 0 then $f/g$ has a zero of order $k-j$ (where a zero of order $-n$ is a pole of order $n$).
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
The initial problem with your first approach is that $1-cos z$ has a double zero at 0 (since $cos$ has slope 0 at 0). In general, if $f$ has a zero of order $k$ at 0 and $g$ a zero or order $j$ at 0 then $f/g$ has a zero of order $k-j$ (where a zero of order $-n$ is a pole of order $n$).
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
$endgroup$
add a comment |
$begingroup$
The initial problem with your first approach is that $1-cos z$ has a double zero at 0 (since $cos$ has slope 0 at 0). In general, if $f$ has a zero of order $k$ at 0 and $g$ a zero or order $j$ at 0 then $f/g$ has a zero of order $k-j$ (where a zero of order $-n$ is a pole of order $n$).
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
$endgroup$
The initial problem with your first approach is that $1-cos z$ has a double zero at 0 (since $cos$ has slope 0 at 0). In general, if $f$ has a zero of order $k$ at 0 and $g$ a zero or order $j$ at 0 then $f/g$ has a zero of order $k-j$ (where a zero of order $-n$ is a pole of order $n$).
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
answered Jan 24 at 5:59
Van LatimerVan Latimer
363110
363110
add a comment |
add a comment |
$begingroup$
Because $1-cos z=0$ when $z=0$ and note that $$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots$$ so your function can be written as $$frac{(frac{z^2}{2!}-frac{z^4}{4!}+cdots)}{z^4}=frac{phi(z)}{z^2}$$ where $phi(z)=frac{1}{2!}-frac{z^2}{4!} cdots $ with $phi(0) neq0$
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
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add a comment |
$begingroup$
Because $1-cos z=0$ when $z=0$ and note that $$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots$$ so your function can be written as $$frac{(frac{z^2}{2!}-frac{z^4}{4!}+cdots)}{z^4}=frac{phi(z)}{z^2}$$ where $phi(z)=frac{1}{2!}-frac{z^2}{4!} cdots $ with $phi(0) neq0$
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
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Because $1-cos z=0$ when $z=0$ and note that $$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots$$ so your function can be written as $$frac{(frac{z^2}{2!}-frac{z^4}{4!}+cdots)}{z^4}=frac{phi(z)}{z^2}$$ where $phi(z)=frac{1}{2!}-frac{z^2}{4!} cdots $ with $phi(0) neq0$
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
$endgroup$
Because $1-cos z=0$ when $z=0$ and note that $$cos z=1-frac{z^2}{2!}+frac{z^4}{4!}-cdots$$ so your function can be written as $$frac{(frac{z^2}{2!}-frac{z^4}{4!}+cdots)}{z^4}=frac{phi(z)}{z^2}$$ where $phi(z)=frac{1}{2!}-frac{z^2}{4!} cdots $ with $phi(0) neq0$
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
answered Jan 24 at 4:34
Chinnapparaj RChinnapparaj R
5,7332928
5,7332928
add a comment |
add a comment |
$begingroup$
$1 - cos z= {1 over 2} z^2 + cdots $ .
So, ${1 - cos z over z^4} = {1 over 2} {1 over z^2} + cdots $
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
$1 - cos z= {1 over 2} z^2 + cdots $ .
So, ${1 - cos z over z^4} = {1 over 2} {1 over z^2} + cdots $
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
$1 - cos z= {1 over 2} z^2 + cdots $ .
So, ${1 - cos z over z^4} = {1 over 2} {1 over z^2} + cdots $
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
$endgroup$
$1 - cos z= {1 over 2} z^2 + cdots $ .
So, ${1 - cos z over z^4} = {1 over 2} {1 over z^2} + cdots $
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
edited Jan 24 at 5:51
J. W. Tanner
3,2701320
3,2701320
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
answered Jan 24 at 4:33
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
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$begingroup$
Why didn't you write $frac{0^2}{0^4}$??
$endgroup$
– Lord Shark the Unknown
Jan 24 at 6:01
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because if you plug in $z=0$ to $1-cos(z)$ you get $1-1=0^1$
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– MinYoung Kim
Jan 24 at 7:00