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Finding an Expression as an Elementary Function for a Power Series
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Consider $$h(z)=sum_n=1^inftyfrac(z-2)^nn.$$
I wish to find an expression for $h$ as an elementary function.
This question has me stumped. I considered another function, $$f(z)=sum_n=1^infty n(z-2)^n.$$ This is much easier to express as an elementary function, as
$$f(z)=sum_n=1^infty n(z-2)^n=(z-2)fracddzsum_n=1^infty (z-2)^n=fracz+3(z+2)^2.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.
I would really appreciate a hint.
complex-analysis proof-verification power-series
asked 52 mins ago
JulianAngussmith
314
add a comment |Â
up vote
2
down vote
favorite
Consider $$h(z)=sum_n=1^inftyfrac(z-2)^nn.$$
I wish to find an expression for $h$ as an elementary function.
This question has me stumped. I considered another function, $$f(z)=sum_n=1^infty n(z-2)^n.$$ This is much easier to express as an elementary function, as
$$f(z)=sum_n=1^infty n(z-2)^n=(z-2)fracddzsum_n=1^infty (z-2)^n=fracz+3(z+2)^2.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.
I would really appreciate a hint.
complex-analysis proof-verification power-series
asked 52 mins ago
JulianAngussmith
314
1
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Consider $$h(z)=sum_n=1^inftyfrac(z-2)^nn.$$
I wish to find an expression for $h$ as an elementary function.
This question has me stumped. I considered another function, $$f(z)=sum_n=1^infty n(z-2)^n.$$ This is much easier to express as an elementary function, as
$$f(z)=sum_n=1^infty n(z-2)^n=(z-2)fracddzsum_n=1^infty (z-2)^n=fracz+3(z+2)^2.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.
I would really appreciate a hint.
complex-analysis proof-verification power-series
asked 52 mins ago
JulianAngussmith
314
Consider $$h(z)=sum_n=1^inftyfrac(z-2)^nn.$$
I wish to find an expression for $h$ as an elementary function.
This question has me stumped. I considered another function, $$f(z)=sum_n=1^infty n(z-2)^n.$$ This is much easier to express as an elementary function, as
$$f(z)=sum_n=1^infty n(z-2)^n=(z-2)fracddzsum_n=1^infty (z-2)^n=fracz+3(z+2)^2.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.
I would really appreciate a hint.
complex-analysis proof-verification power-series
complex-analysis proof-verification power-series
asked 52 mins ago
JulianAngussmith
314
asked 52 mins ago
JulianAngussmith
314
asked 52 mins ago
JulianAngussmith
314
asked 52 mins ago
JulianAngussmith
314
asked 52 mins ago
JulianAngussmith
314
314
1
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago
add a comment |Â
1
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago
1
1
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
Hint:
Instead of looking at $h$, first try to calculate what $$fracpartial hpartial z$$ would be.
answered 42 mins ago
5xum
85.8k388154
add a comment |Â
up vote
1
down vote
Taking the convergent part of this sum and making
$$
y = z-2
$$
we have
$$
h(y) = sum_k=1^inftyfracy^kk = intsum_k=0^inftyy^kdy = intfracdy1-y
$$
answered 25 mins ago
Cesareo
6,8872414
add a comment |Â
up vote
1
down vote
Hint: formally $h'(z)=sum (z-2)^n-1$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Hint:
Instead of looking at $h$, first try to calculate what $$fracpartial hpartial z$$ would be.
answered 42 mins ago
5xum
85.8k388154
add a comment |Â
up vote
1
down vote
Hint:
Instead of looking at $h$, first try to calculate what $$fracpartial hpartial z$$ would be.
answered 42 mins ago
5xum
85.8k388154
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint:
Instead of looking at $h$, first try to calculate what $$fracpartial hpartial z$$ would be.
answered 42 mins ago
5xum
85.8k388154
Hint:
Instead of looking at $h$, first try to calculate what $$fracpartial hpartial z$$ would be.
answered 42 mins ago
5xum
85.8k388154
answered 42 mins ago
5xum
85.8k388154
answered 42 mins ago
5xum
85.8k388154
answered 42 mins ago
5xum
85.8k388154
85.8k388154
add a comment |Â
add a comment |Â
up vote
1
down vote
Taking the convergent part of this sum and making
$$
y = z-2
$$
we have
$$
h(y) = sum_k=1^inftyfracy^kk = intsum_k=0^inftyy^kdy = intfracdy1-y
$$
answered 25 mins ago
Cesareo
6,8872414
add a comment |Â
up vote
1
down vote
Taking the convergent part of this sum and making
$$
y = z-2
$$
we have
$$
h(y) = sum_k=1^inftyfracy^kk = intsum_k=0^inftyy^kdy = intfracdy1-y
$$
answered 25 mins ago
Cesareo
6,8872414
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Taking the convergent part of this sum and making
$$
y = z-2
$$
we have
$$
h(y) = sum_k=1^inftyfracy^kk = intsum_k=0^inftyy^kdy = intfracdy1-y
$$
answered 25 mins ago
Cesareo
6,8872414
Taking the convergent part of this sum and making
$$
y = z-2
$$
we have
$$
h(y) = sum_k=1^inftyfracy^kk = intsum_k=0^inftyy^kdy = intfracdy1-y
$$
answered 25 mins ago
Cesareo
6,8872414
answered 25 mins ago
Cesareo
6,8872414
answered 25 mins ago
Cesareo
6,8872414
answered 25 mins ago
Cesareo
6,8872414
6,8872414
add a comment |Â
add a comment |Â
up vote
1
down vote
Hint: formally $h'(z)=sum (z-2)^n-1$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
add a comment |Â
up vote
1
down vote
Hint: formally $h'(z)=sum (z-2)^n-1$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint: formally $h'(z)=sum (z-2)^n-1$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
Hint: formally $h'(z)=sum (z-2)^n-1$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
edited 11 mins ago
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
answered 43 mins ago
Kavi Rama Murthy
32.4k31644
32.4k31644
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
add a comment |Â
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
Does this give $-ln|3-z|-ln|z-2|+C$?
– JulianAngussmith
16 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
No, the sum of the series is not a real valued function. Please see my revised answer.
– Kavi Rama Murthy
11 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
I believe I have calculated the sum incorrectly. I have that $$h(z)=intsum_n=1^infty (z-2)^n-1 dz=intsum_n=0^infty (z-2)^n-frac1z-2 dz.$$
– JulianAngussmith
8 mins ago
add a comment |Â
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1
Hint: See en.wikipedia.org/wiki/Logarithm#Power_series
– gammatester
46 mins ago
Thank you for the hint, I had not considered this. Despite this, I still can't yield an answer. I'm open to additional help if you are willing.
– JulianAngussmith
33 mins ago