Order of pole $fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2$

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Find the order of pole at $z_0=1$



$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2$$



I have two options?



Write the Laurent series or using derivatives, all seems very tiresome, and I am may missing something is there another way?




complex-analysis




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  • 1




    Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
    – Lord Shark the Unknown
    Jul 17 at 10:29










  • $log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
    – gbox
    Jul 17 at 10:34







  • 1




    $1-cos w$ has a double zero at $w=0$.
    – Lord Shark the Unknown
    Jul 17 at 10:35










  • I agree it is obvious that the order is 5
    – Dr Peter McGowan
    Jul 17 at 10:36














up vote
0
down vote

favorite












Find the order of pole at $z_0=1$



$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2$$



I have two options?



Write the Laurent series or using derivatives, all seems very tiresome, and I am may missing something is there another way?




complex-analysis




share|cite|improve this question

















  • 1




    Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
    – Lord Shark the Unknown
    Jul 17 at 10:29










  • $log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
    – gbox
    Jul 17 at 10:34







  • 1




    $1-cos w$ has a double zero at $w=0$.
    – Lord Shark the Unknown
    Jul 17 at 10:35










  • I agree it is obvious that the order is 5
    – Dr Peter McGowan
    Jul 17 at 10:36












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Find the order of pole at $z_0=1$



$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2$$



I have two options?



Write the Laurent series or using derivatives, all seems very tiresome, and I am may missing something is there another way?




complex-analysis




share|cite|improve this question













Find the order of pole at $z_0=1$



$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2$$



I have two options?



Write the Laurent series or using derivatives, all seems very tiresome, and I am may missing something is there another way?





complex-analysis





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 17 at 10:30









Bernard

110k635103




110k635103









asked Jul 17 at 10:27









gbox

5,30851841




5,30851841







  • 1




    Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
    – Lord Shark the Unknown
    Jul 17 at 10:29










  • $log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
    – gbox
    Jul 17 at 10:34







  • 1




    $1-cos w$ has a double zero at $w=0$.
    – Lord Shark the Unknown
    Jul 17 at 10:35










  • I agree it is obvious that the order is 5
    – Dr Peter McGowan
    Jul 17 at 10:36












  • 1




    Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
    – Lord Shark the Unknown
    Jul 17 at 10:29










  • $log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
    – gbox
    Jul 17 at 10:34







  • 1




    $1-cos w$ has a double zero at $w=0$.
    – Lord Shark the Unknown
    Jul 17 at 10:35










  • I agree it is obvious that the order is 5
    – Dr Peter McGowan
    Jul 17 at 10:36







1




1




Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
– Lord Shark the Unknown
Jul 17 at 10:29




Using Laurent expansion isn't tiresome al all. But isn't it obvious that the order of the pole is $5$?
– Lord Shark the Unknown
Jul 17 at 10:29












$log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
– gbox
Jul 17 at 10:34





$log^4(z)$ is order $4$ but how can I infer on $(1-cos(z-1))^2$? can I use derivative
– gbox
Jul 17 at 10:34





1




1




$1-cos w$ has a double zero at $w=0$.
– Lord Shark the Unknown
Jul 17 at 10:35




$1-cos w$ has a double zero at $w=0$.
– Lord Shark the Unknown
Jul 17 at 10:35












I agree it is obvious that the order is 5
– Dr Peter McGowan
Jul 17 at 10:36




I agree it is obvious that the order is 5
– Dr Peter McGowan
Jul 17 at 10:36










2 Answers
2






active

oldest

votes

















up vote
1
down vote













Hint. Note that for $w=z-1$ then
$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(w)log^4(1+w)[1-cos(w)]^2$$
Now use the Taylor expansions of $sin(w)$, $log(1+w)$, $cos(w)$ at $0$. In particular recall that $cos(w)=1-fracw^22+ o(w^2).$






share|cite|improve this answer























  • @Bernard I think I have already edit that.
    – Robert Z
    Jul 17 at 10:34










  • But I need to take $cos$ to the power of $2$ so it order of $4$?
    – gbox
    Jul 17 at 10:51










  • Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
    – Robert Z
    Jul 17 at 10:53


















up vote
0
down vote













Let $u=z-1$ therefore$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(u)log^4(1+u)[1-cos(u)]^2=dfrac2sindfracu2cos^3dfracu2log^4(1+u)$$then the pole is of order $3$ since the zeros of $log^4(1+u)$ and $sindfracu2$ are of order $4$ and $1$ respectively.






share|cite|improve this answer























  • As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
    – Lord Shark the Unknown
    Jul 17 at 14:13










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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













Hint. Note that for $w=z-1$ then
$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(w)log^4(1+w)[1-cos(w)]^2$$
Now use the Taylor expansions of $sin(w)$, $log(1+w)$, $cos(w)$ at $0$. In particular recall that $cos(w)=1-fracw^22+ o(w^2).$






share|cite|improve this answer























  • @Bernard I think I have already edit that.
    – Robert Z
    Jul 17 at 10:34










  • But I need to take $cos$ to the power of $2$ so it order of $4$?
    – gbox
    Jul 17 at 10:51










  • Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
    – Robert Z
    Jul 17 at 10:53















up vote
1
down vote













Hint. Note that for $w=z-1$ then
$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(w)log^4(1+w)[1-cos(w)]^2$$
Now use the Taylor expansions of $sin(w)$, $log(1+w)$, $cos(w)$ at $0$. In particular recall that $cos(w)=1-fracw^22+ o(w^2).$






share|cite|improve this answer























  • @Bernard I think I have already edit that.
    – Robert Z
    Jul 17 at 10:34










  • But I need to take $cos$ to the power of $2$ so it order of $4$?
    – gbox
    Jul 17 at 10:51










  • Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
    – Robert Z
    Jul 17 at 10:53













up vote
1
down vote










up vote
1
down vote









Hint. Note that for $w=z-1$ then
$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(w)log^4(1+w)[1-cos(w)]^2$$
Now use the Taylor expansions of $sin(w)$, $log(1+w)$, $cos(w)$ at $0$. In particular recall that $cos(w)=1-fracw^22+ o(w^2).$






share|cite|improve this answer















Hint. Note that for $w=z-1$ then
$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(w)log^4(1+w)[1-cos(w)]^2$$
Now use the Taylor expansions of $sin(w)$, $log(1+w)$, $cos(w)$ at $0$. In particular recall that $cos(w)=1-fracw^22+ o(w^2).$







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 17 at 10:35


























answered Jul 17 at 10:30









Robert Z

84.2k955123




84.2k955123











  • @Bernard I think I have already edit that.
    – Robert Z
    Jul 17 at 10:34










  • But I need to take $cos$ to the power of $2$ so it order of $4$?
    – gbox
    Jul 17 at 10:51










  • Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
    – Robert Z
    Jul 17 at 10:53

















  • @Bernard I think I have already edit that.
    – Robert Z
    Jul 17 at 10:34










  • But I need to take $cos$ to the power of $2$ so it order of $4$?
    – gbox
    Jul 17 at 10:51










  • Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
    – Robert Z
    Jul 17 at 10:53
















@Bernard I think I have already edit that.
– Robert Z
Jul 17 at 10:34




@Bernard I think I have already edit that.
– Robert Z
Jul 17 at 10:34












But I need to take $cos$ to the power of $2$ so it order of $4$?
– gbox
Jul 17 at 10:51




But I need to take $cos$ to the power of $2$ so it order of $4$?
– gbox
Jul 17 at 10:51












Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
– Robert Z
Jul 17 at 10:53





Yes! $(1-cos(w))^2=(fracw^22+ o(w^2))^2=w^4/4+o(w^4)$
– Robert Z
Jul 17 at 10:53











up vote
0
down vote













Let $u=z-1$ therefore$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(u)log^4(1+u)[1-cos(u)]^2=dfrac2sindfracu2cos^3dfracu2log^4(1+u)$$then the pole is of order $3$ since the zeros of $log^4(1+u)$ and $sindfracu2$ are of order $4$ and $1$ respectively.






share|cite|improve this answer























  • As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
    – Lord Shark the Unknown
    Jul 17 at 14:13














up vote
0
down vote













Let $u=z-1$ therefore$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(u)log^4(1+u)[1-cos(u)]^2=dfrac2sindfracu2cos^3dfracu2log^4(1+u)$$then the pole is of order $3$ since the zeros of $log^4(1+u)$ and $sindfracu2$ are of order $4$ and $1$ respectively.






share|cite|improve this answer























  • As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
    – Lord Shark the Unknown
    Jul 17 at 14:13












up vote
0
down vote










up vote
0
down vote









Let $u=z-1$ therefore$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(u)log^4(1+u)[1-cos(u)]^2=dfrac2sindfracu2cos^3dfracu2log^4(1+u)$$then the pole is of order $3$ since the zeros of $log^4(1+u)$ and $sindfracu2$ are of order $4$ and $1$ respectively.






share|cite|improve this answer















Let $u=z-1$ therefore$$fracsin^3(z-1)log^4(z)[1-cos(z-1)]^2=fracsin^3(u)log^4(1+u)[1-cos(u)]^2=dfrac2sindfracu2cos^3dfracu2log^4(1+u)$$then the pole is of order $3$ since the zeros of $log^4(1+u)$ and $sindfracu2$ are of order $4$ and $1$ respectively.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 17 at 11:22









Bernard

110k635103




110k635103











answered Jul 17 at 10:43









Mostafa Ayaz

8,6023630




8,6023630











  • As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
    – Lord Shark the Unknown
    Jul 17 at 14:13
















  • As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
    – Lord Shark the Unknown
    Jul 17 at 14:13















As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
– Lord Shark the Unknown
Jul 17 at 14:13




As the pole has order $5$, then something has gone wrong ... possibly your trig function calculations.
– Lord Shark the Unknown
Jul 17 at 14:13












 

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