Taylor series of $csc(x)$












2












$begingroup$


How can I compute the Taylor series of $csc(x):=frac{1}{sin(x)}$ at $0$? When I google this I can see that indeed this series can be computed, the first term being $1/x$ but how can this be if $sin(x)$ vanishes at $x=0$?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $pmb sin(x)$ vanishes (is 0) at $x=0$
    $endgroup$
    – J. W. Tanner
    Jan 24 at 2:06






  • 1




    $begingroup$
    There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
    $endgroup$
    – Travis
    Jan 24 at 2:07
















2












$begingroup$


How can I compute the Taylor series of $csc(x):=frac{1}{sin(x)}$ at $0$? When I google this I can see that indeed this series can be computed, the first term being $1/x$ but how can this be if $sin(x)$ vanishes at $x=0$?






calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $pmb sin(x)$ vanishes (is 0) at $x=0$
    $endgroup$
    – J. W. Tanner
    Jan 24 at 2:06






  • 1




    $begingroup$
    There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
    $endgroup$
    – Travis
    Jan 24 at 2:07














2












2








2


1



$begingroup$


How can I compute the Taylor series of $csc(x):=frac{1}{sin(x)}$ at $0$? When I google this I can see that indeed this series can be computed, the first term being $1/x$ but how can this be if $sin(x)$ vanishes at $x=0$?






calculus







share|cite|improve this question











$endgroup$




How can I compute the Taylor series of $csc(x):=frac{1}{sin(x)}$ at $0$? When I google this I can see that indeed this series can be computed, the first term being $1/x$ but how can this be if $sin(x)$ vanishes at $x=0$?






calculus




calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 24 at 2:09







Sak

















asked Jan 24 at 2:00









SakSak

1,7181537




1,7181537








  • 1




    $begingroup$
    $pmb sin(x)$ vanishes (is 0) at $x=0$
    $endgroup$
    – J. W. Tanner
    Jan 24 at 2:06






  • 1




    $begingroup$
    There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
    $endgroup$
    – Travis
    Jan 24 at 2:07














  • 1




    $begingroup$
    $pmb sin(x)$ vanishes (is 0) at $x=0$
    $endgroup$
    – J. W. Tanner
    Jan 24 at 2:06






  • 1




    $begingroup$
    There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
    $endgroup$
    – Travis
    Jan 24 at 2:07








1




1




$begingroup$
$pmb sin(x)$ vanishes (is 0) at $x=0$
$endgroup$
– J. W. Tanner
Jan 24 at 2:06




$begingroup$
$pmb sin(x)$ vanishes (is 0) at $x=0$
$endgroup$
– J. W. Tanner
Jan 24 at 2:06




1




1




$begingroup$
There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
$endgroup$
– Travis
Jan 24 at 2:07




$begingroup$
There's some confusion about definitions here. One can indeed expand $csc x$ in a series about $x = 0$, but since it's not defined there, it does not have a Taylor series based there.
$endgroup$
– Travis
Jan 24 at 2:07










1 Answer
1






active

oldest

votes


















5












$begingroup$

What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.



You probably found the Laurent series(*). Since
$$g(x)=x csc x=frac x{sin x}$$
is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say
$$g(x)=a_0+a_1x+a_2x^2+cdots$$
at least at some interval containing $x=0$, then
$$csc x= frac{g(x)}x=frac{a_0}x+a_1+a_2 x+a_3 x^2+cdots,$$
which gives the aforementioned Laurent series.




(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form
$$f(x)=frac{b_{-k}}{(x-a)^k}+cdots+frac{b_{-2}}{(x-a)^2}+frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+cdots.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a great explanation thank you very much! I understand now my confusion.
    $endgroup$
    – Sak
    Jan 24 at 2:16










  • $begingroup$
    The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
    $endgroup$
    – Johnathan Gross
    Jan 26 at 18:39











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









5












$begingroup$

What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.



You probably found the Laurent series(*). Since
$$g(x)=x csc x=frac x{sin x}$$
is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say
$$g(x)=a_0+a_1x+a_2x^2+cdots$$
at least at some interval containing $x=0$, then
$$csc x= frac{g(x)}x=frac{a_0}x+a_1+a_2 x+a_3 x^2+cdots,$$
which gives the aforementioned Laurent series.




(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form
$$f(x)=frac{b_{-k}}{(x-a)^k}+cdots+frac{b_{-2}}{(x-a)^2}+frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+cdots.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a great explanation thank you very much! I understand now my confusion.
    $endgroup$
    – Sak
    Jan 24 at 2:16










  • $begingroup$
    The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
    $endgroup$
    – Johnathan Gross
    Jan 26 at 18:39
















5












$begingroup$

What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.



You probably found the Laurent series(*). Since
$$g(x)=x csc x=frac x{sin x}$$
is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say
$$g(x)=a_0+a_1x+a_2x^2+cdots$$
at least at some interval containing $x=0$, then
$$csc x= frac{g(x)}x=frac{a_0}x+a_1+a_2 x+a_3 x^2+cdots,$$
which gives the aforementioned Laurent series.




(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form
$$f(x)=frac{b_{-k}}{(x-a)^k}+cdots+frac{b_{-2}}{(x-a)^2}+frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+cdots.$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This is a great explanation thank you very much! I understand now my confusion.
    $endgroup$
    – Sak
    Jan 24 at 2:16










  • $begingroup$
    The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
    $endgroup$
    – Johnathan Gross
    Jan 26 at 18:39














5












5








5





$begingroup$

What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.



You probably found the Laurent series(*). Since
$$g(x)=x csc x=frac x{sin x}$$
is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say
$$g(x)=a_0+a_1x+a_2x^2+cdots$$
at least at some interval containing $x=0$, then
$$csc x= frac{g(x)}x=frac{a_0}x+a_1+a_2 x+a_3 x^2+cdots,$$
which gives the aforementioned Laurent series.




(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form
$$f(x)=frac{b_{-k}}{(x-a)^k}+cdots+frac{b_{-2}}{(x-a)^2}+frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+cdots.$$






share|cite|improve this answer











$endgroup$



What you found in Wikipedia can't be a Taylor series. In fact there's not such thing as the Taylor series of $csc(x)$ at $x=0$ since this function is not defined at $x=0$, much less the derivatives of any order.



You probably found the Laurent series(*). Since
$$g(x)=x csc x=frac x{sin x}$$
is continuous and differentiable at $x=0$ if we define $g(0)=1$, and it's Taylor series can be found, say
$$g(x)=a_0+a_1x+a_2x^2+cdots$$
at least at some interval containing $x=0$, then
$$csc x= frac{g(x)}x=frac{a_0}x+a_1+a_2 x+a_3 x^2+cdots,$$
which gives the aforementioned Laurent series.




(*) The Laurent series of a function at $x=a$ is a generalization of the Taylor series for functions $f(x)$ with a pole at $x=a$, that is, functions that go to infinity at $x=a$ but such that $(x-a)^k f(x)$ has a removable discontinuity at $x=a$. If $k$ is the least integer such that this is true, then the Laurent series has the form
$$f(x)=frac{b_{-k}}{(x-a)^k}+cdots+frac{b_{-2}}{(x-a)^2}+frac{b_{-1}}{x-a}+b_0+b_1(x-a)+b_2(x-a)^2+cdots.$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 24 at 2:22

























answered Jan 24 at 2:08









Alejandro Nasif SalumAlejandro Nasif Salum

4,765118




4,765118












  • $begingroup$
    This is a great explanation thank you very much! I understand now my confusion.
    $endgroup$
    – Sak
    Jan 24 at 2:16










  • $begingroup$
    The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
    $endgroup$
    – Johnathan Gross
    Jan 26 at 18:39


















  • $begingroup$
    This is a great explanation thank you very much! I understand now my confusion.
    $endgroup$
    – Sak
    Jan 24 at 2:16










  • $begingroup$
    The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
    $endgroup$
    – Johnathan Gross
    Jan 26 at 18:39
















$begingroup$
This is a great explanation thank you very much! I understand now my confusion.
$endgroup$
– Sak
Jan 24 at 2:16




$begingroup$
This is a great explanation thank you very much! I understand now my confusion.
$endgroup$
– Sak
Jan 24 at 2:16












$begingroup$
The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
$endgroup$
– Johnathan Gross
Jan 26 at 18:39




$begingroup$
The Laurent series of csc(x) at x=0 is 1/x+x/6+... Might be a good idea to include that.
$endgroup$
– Johnathan Gross
Jan 26 at 18:39


















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