Calculating the residue of $frac{10z^4-10sin(z)}{z^3}, z(0) = 0$
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$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
I've gotten that $$operatorname{Res} = 0$$ but I'm not quite sure if that is correct, or if I have even used a correct pathway towards it. How should one work around sine (or cosine, for that matter) during residue calculations?
complex-analysis residue-calculus
edited Nov 21 at 12:25
Tianlalu
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
add a comment |
up vote
3
down vote
favorite
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
I've gotten that $$operatorname{Res} = 0$$ but I'm not quite sure if that is correct, or if I have even used a correct pathway towards it. How should one work around sine (or cosine, for that matter) during residue calculations?
complex-analysis residue-calculus
edited Nov 21 at 12:25
Tianlalu
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
I've gotten that $$operatorname{Res} = 0$$ but I'm not quite sure if that is correct, or if I have even used a correct pathway towards it. How should one work around sine (or cosine, for that matter) during residue calculations?
complex-analysis residue-calculus
edited Nov 21 at 12:25
Tianlalu
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
I've gotten that $$operatorname{Res} = 0$$ but I'm not quite sure if that is correct, or if I have even used a correct pathway towards it. How should one work around sine (or cosine, for that matter) during residue calculations?
complex-analysis residue-calculus
complex-analysis residue-calculus
edited Nov 21 at 12:25
Tianlalu
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
edited Nov 21 at 12:25
Tianlalu
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
edited Nov 21 at 12:25
Tianlalu
2,709632
edited Nov 21 at 12:25
Tianlalu
2,709632
edited Nov 21 at 12:25
Tianlalu
2,709632
2,709632
asked Oct 25 at 22:49
TootsieRoll
163
asked Oct 25 at 22:49
TootsieRoll
163
asked Oct 25 at 22:49
TootsieRoll
163
163
Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15
add a comment |
Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15
Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15
Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.
We obtain
begin{align*}
color{blue}{[z^{-1}]}&color{blue}{frac{10z^4-10sin z}{z^3}}\
&=[z^{-1}]left(10z-frac{10}{z^3}left(z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsright)right)\
&=-10[z^{-1}]left(frac{1}{z^2}-frac{1}{3!}+frac{z^2}{5!}-cdotsright)\
&,,color{blue}{=0}
end{align*}
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
add a comment |
up vote
0
down vote
One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.
We obtain
begin{align*}
color{blue}{[z^{-1}]}&color{blue}{frac{10z^4-10sin z}{z^3}}\
&=[z^{-1}]left(10z-frac{10}{z^3}left(z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsright)right)\
&=-10[z^{-1}]left(frac{1}{z^2}-frac{1}{3!}+frac{z^2}{5!}-cdotsright)\
&,,color{blue}{=0}
end{align*}
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
add a comment |
up vote
1
down vote
Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.
We obtain
begin{align*}
color{blue}{[z^{-1}]}&color{blue}{frac{10z^4-10sin z}{z^3}}\
&=[z^{-1}]left(10z-frac{10}{z^3}left(z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsright)right)\
&=-10[z^{-1}]left(frac{1}{z^2}-frac{1}{3!}+frac{z^2}{5!}-cdotsright)\
&,,color{blue}{=0}
end{align*}
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
add a comment |
up vote
1
down vote
up vote
1
down vote
Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.
We obtain
begin{align*}
color{blue}{[z^{-1}]}&color{blue}{frac{10z^4-10sin z}{z^3}}\
&=[z^{-1}]left(10z-frac{10}{z^3}left(z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsright)right)\
&=-10[z^{-1}]left(frac{1}{z^2}-frac{1}{3!}+frac{z^2}{5!}-cdotsright)\
&,,color{blue}{=0}
end{align*}
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
Your answer is correct. The residue of a function is the coefficient of $z^{-1}$ of the Laurent series expansion of the function.
We obtain
begin{align*}
color{blue}{[z^{-1}]}&color{blue}{frac{10z^4-10sin z}{z^3}}\
&=[z^{-1}]left(10z-frac{10}{z^3}left(z-frac{z^3}{3!}+frac{z^5}{5!}-cdotsright)right)\
&=-10[z^{-1}]left(frac{1}{z^2}-frac{1}{3!}+frac{z^2}{5!}-cdotsright)\
&,,color{blue}{=0}
end{align*}
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
answered Oct 26 at 20:29
Markus Scheuer
59.3k454141
59.3k454141
add a comment |
add a comment |
up vote
0
down vote
One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
add a comment |
up vote
0
down vote
One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
add a comment |
up vote
0
down vote
up vote
0
down vote
One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
One way to work with sines and cosines in this problem is to reckon with their power series expansions found in many places, for instance here. You might also note that an even function has only even powers in its Laurent series whereas the residue comes from the $(-1)$ power, therefore even functions will have a residue of zero. Hmm, can you see why I would mention that last fact?
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
answered Oct 26 at 0:22
Oscar Lanzi
11.6k11935
11.6k11935
add a comment |
add a comment |
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Can you show us the pathway you chose so we can compare it with what we would do?
– Oscar Lanzi
Oct 26 at 0:15