Mixing
Calculating the residue of $frac{10z^4-10sin(z)}{z^3}, z(0) = 0$
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
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 '18 at 12:25
Tianlalu
3,09621038
asked Oct 25 '18 at 22:49
TootsieRoll
163
add a comment |
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
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 '18 at 12:25
Tianlalu
3,09621038
asked Oct 25 '18 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 '18 at 0:15
add a comment |
$$frac{10z^4-10sin(z)}{z^3}, quad z(0) = 0.$$
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 '18 at 12:25
Tianlalu
3,09621038
asked Oct 25 '18 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 '18 at 12:25
Tianlalu
3,09621038
asked Oct 25 '18 at 22:49
TootsieRoll
163
edited Nov 21 '18 at 12:25
Tianlalu
3,09621038
asked Oct 25 '18 at 22:49
TootsieRoll
163
edited Nov 21 '18 at 12:25
Tianlalu
3,09621038
edited Nov 21 '18 at 12:25
Tianlalu
3,09621038
edited Nov 21 '18 at 12:25
Tianlalu
3,09621038
3,09621038
asked Oct 25 '18 at 22:49
TootsieRoll
163
asked Oct 25 '18 at 22:49
TootsieRoll
163
asked Oct 25 '18 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 '18 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 '18 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 '18 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 '18 at 0:15
add a comment |
2 Answers
2
active
oldest
votes
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 '18 at 20:29
Markus Scheuer
60.2k455144
add a comment |
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 '18 at 0:22
Oscar Lanzi
12.1k12036
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 '18 at 20:29
Markus Scheuer
60.2k455144
add a comment |
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 '18 at 20:29
Markus Scheuer
60.2k455144
add a comment |
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 '18 at 20:29
Markus Scheuer
60.2k455144
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 '18 at 20:29
Markus Scheuer
60.2k455144
answered Oct 26 '18 at 20:29
Markus Scheuer
60.2k455144
answered Oct 26 '18 at 20:29
Markus Scheuer
60.2k455144
answered Oct 26 '18 at 20:29
Markus Scheuer
60.2k455144
60.2k455144
add a comment |
add a comment |
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 '18 at 0:22
Oscar Lanzi
12.1k12036
add a comment |
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 '18 at 0:22
Oscar Lanzi
12.1k12036
add a comment |
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 '18 at 0:22
Oscar Lanzi
12.1k12036
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 '18 at 0:22
Oscar Lanzi
12.1k12036
answered Oct 26 '18 at 0:22
Oscar Lanzi
12.1k12036
answered Oct 26 '18 at 0:22
Oscar Lanzi
12.1k12036
answered Oct 26 '18 at 0:22
Oscar Lanzi
12.1k12036
12.1k12036
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 '18 at 0:15