Verification of Laurent series for the function $f(z)=frac{2}{(z+2)^2}-frac{5}{z-4}$
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Suppose that $$f(z)=frac{2}{(z+2)^2}-frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
begin{align}
f(z)&=-2 frac{d}{dz}left(frac{1}{z+2}right)-frac{5}{(z-2)-2} \
&=-2 frac{d}{dz}left(frac{1}{4}left(frac{1}{1+frac{(z-2)}{4}}right)right)+frac{5}{2}left(frac{1}{1-frac{(z-2)}{2}}right) \
&=-2 frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty} (-1)^nleft(frac{z-2}{4}right)^nright)+frac{5}{2}left(sum_{n=0}^{infty} left(frac{z-2}{2}right)^nright) \
&=-2sum_{n=1}^{infty} (-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}+5sum_{n=0}^{infty} frac{(z-2)}{2^{n+1}}^n.
end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=sum_{n=0}^{infty}c_n(z-2)^n, c_n=(-1)^nfrac{2(n+1)}{4^{n+3}}+frac{5}{2^{n+1}}.$$
complex-analysis proof-verification laurent-series
asked Nov 17 at 7:06
JulianAngussmith
3810
add a comment |
up vote
3
down vote
favorite
Suppose that $$f(z)=frac{2}{(z+2)^2}-frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
begin{align}
f(z)&=-2 frac{d}{dz}left(frac{1}{z+2}right)-frac{5}{(z-2)-2} \
&=-2 frac{d}{dz}left(frac{1}{4}left(frac{1}{1+frac{(z-2)}{4}}right)right)+frac{5}{2}left(frac{1}{1-frac{(z-2)}{2}}right) \
&=-2 frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty} (-1)^nleft(frac{z-2}{4}right)^nright)+frac{5}{2}left(sum_{n=0}^{infty} left(frac{z-2}{2}right)^nright) \
&=-2sum_{n=1}^{infty} (-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}+5sum_{n=0}^{infty} frac{(z-2)}{2^{n+1}}^n.
end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=sum_{n=0}^{infty}c_n(z-2)^n, c_n=(-1)^nfrac{2(n+1)}{4^{n+3}}+frac{5}{2^{n+1}}.$$
complex-analysis proof-verification laurent-series
asked Nov 17 at 7:06
JulianAngussmith
3810
2
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
1
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Suppose that $$f(z)=frac{2}{(z+2)^2}-frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
begin{align}
f(z)&=-2 frac{d}{dz}left(frac{1}{z+2}right)-frac{5}{(z-2)-2} \
&=-2 frac{d}{dz}left(frac{1}{4}left(frac{1}{1+frac{(z-2)}{4}}right)right)+frac{5}{2}left(frac{1}{1-frac{(z-2)}{2}}right) \
&=-2 frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty} (-1)^nleft(frac{z-2}{4}right)^nright)+frac{5}{2}left(sum_{n=0}^{infty} left(frac{z-2}{2}right)^nright) \
&=-2sum_{n=1}^{infty} (-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}+5sum_{n=0}^{infty} frac{(z-2)}{2^{n+1}}^n.
end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=sum_{n=0}^{infty}c_n(z-2)^n, c_n=(-1)^nfrac{2(n+1)}{4^{n+3}}+frac{5}{2^{n+1}}.$$
complex-analysis proof-verification laurent-series
asked Nov 17 at 7:06
JulianAngussmith
3810
Suppose that $$f(z)=frac{2}{(z+2)^2}-frac{5}{z-4}.$$ Find the Laurent series for $f$ in powers of $z-2$ that converges when $z=1$.
This is how I approached the problem:
begin{align}
f(z)&=-2 frac{d}{dz}left(frac{1}{z+2}right)-frac{5}{(z-2)-2} \
&=-2 frac{d}{dz}left(frac{1}{4}left(frac{1}{1+frac{(z-2)}{4}}right)right)+frac{5}{2}left(frac{1}{1-frac{(z-2)}{2}}right) \
&=-2 frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty} (-1)^nleft(frac{z-2}{4}right)^nright)+frac{5}{2}left(sum_{n=0}^{infty} left(frac{z-2}{2}right)^nright) \
&=-2sum_{n=1}^{infty} (-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}+5sum_{n=0}^{infty} frac{(z-2)}{2^{n+1}}^n.
end{align}
I believe this is the right answer, yet the answer provided (using a different method) states the answer is $$f(z)=sum_{n=0}^{infty}c_n(z-2)^n, c_n=(-1)^nfrac{2(n+1)}{4^{n+3}}+frac{5}{2^{n+1}}.$$
complex-analysis proof-verification laurent-series
complex-analysis proof-verification laurent-series
asked Nov 17 at 7:06
JulianAngussmith
3810
asked Nov 17 at 7:06
JulianAngussmith
3810
asked Nov 17 at 7:06
JulianAngussmith
3810
asked Nov 17 at 7:06
JulianAngussmith
3810
asked Nov 17 at 7:06
JulianAngussmith
3810
3810
2
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
1
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04
add a comment |
2
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
1
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04
2
2
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
1
1
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
After the last $=$ sign, what you got was:$$-2sum_{n=1}^infty(-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$sum_{n=1}^infty(-1)^{n-1}frac{2n(z-2)^{n-1}}{4^{n+1}}=sum_{n=0}^infty(-1)^nfrac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
After the last $=$ sign, what you got was:$$-2sum_{n=1}^infty(-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$sum_{n=1}^infty(-1)^{n-1}frac{2n(z-2)^{n-1}}{4^{n+1}}=sum_{n=0}^infty(-1)^nfrac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
add a comment |
up vote
2
down vote
accepted
After the last $=$ sign, what you got was:$$-2sum_{n=1}^infty(-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$sum_{n=1}^infty(-1)^{n-1}frac{2n(z-2)^{n-1}}{4^{n+1}}=sum_{n=0}^infty(-1)^nfrac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
After the last $=$ sign, what you got was:$$-2sum_{n=1}^infty(-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$sum_{n=1}^infty(-1)^{n-1}frac{2n(z-2)^{n-1}}{4^{n+1}}=sum_{n=0}^infty(-1)^nfrac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
After the last $=$ sign, what you got was:$$-2sum_{n=1}^infty(-1)^nfrac{n(z-2)^{n-1}}{4^{n+1}}.$$Since you have that $-2$ outside the sum, this is equal to$$sum_{n=1}^infty(-1)^{n-1}frac{2n(z-2)^{n-1}}{4^{n+1}}=sum_{n=0}^infty(-1)^nfrac{2(n+1)(z-2)^n}{4^{n+2}}.$$So, the provided answer is not correct.
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
edited Nov 17 at 7:57
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
answered Nov 17 at 7:30
José Carlos Santos
142k20111207
142k20111207
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
add a comment |
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
I'm unsure of how to obtain $4^{n+2}$. I have that begin{align}-2frac{d}{dz}left(frac{1}{4}sum_{n=0}^{infty}(-1)^nleft(frac{z-2}{4}right)^nright) &=-2cdot frac{1}{4}sum_{n=1}^{infty}(-1)^ncdotfrac{n}{4}cdotleft(frac{z-2}{4}right)^{n-1} \ &=sum_{n=1}^{infty} (-1)^{n+1}frac{2n(z-2)^{n-1}}{4^{n+1}}. end{align} Where have I missed the factor of $1/4$?
– JulianAngussmith
Nov 17 at 7:45
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
My mistake. I've edited my answer. The provided answer is not correct. I've checked with Mathematica.
– José Carlos Santos
Nov 17 at 7:58
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
Thank you for checking. I was scratching my head for a few seconds haha
– JulianAngussmith
Nov 17 at 8:03
add a comment |
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2
It is the same if you rearrange the summation index.
– Apocalypse
Nov 17 at 7:15
1
@Apocalypse No, it is not.
– José Carlos Santos
Nov 17 at 8:04