Expanding a function with a power series Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that $e^lnz=z$ from the power seriesHow to compute power series by compositionPower (Maclaurin) series of $a/x ( exp(x/a) - 1)^-1$Function that Represents Divergent Power Series?Expand a function to power seriesExpand the function $f(x)$ into a power seriesDifficulty in finding interval of convergence with power seriesExpansion in power series of $z=0$Power Series Representation of a Function, (differentiation)Representation of function as power series - unique?
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Expanding a function with a power series
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Prove that $e^lnz=z$ from the power seriesHow to compute power series by compositionPower (Maclaurin) series of $a/x ( exp(x/a) - 1)^-1$Function that Represents Divergent Power Series?Expand a function to power seriesExpand the function $f(x)$ into a power seriesDifficulty in finding interval of convergence with power seriesExpansion in power series of $z=0$Power Series Representation of a Function, (differentiation)Representation of function as power series - unique?
$begingroup$
How would I expand the following function as a power series, around $eta=0$?
$$g_0(1,eta)=fracleft(fracPVNkTright)_0-14eta$$
Note that:
$$left(fracPVNkTright)_0=1+frac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k$$
Then we have:
$$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
power-series
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
$endgroup$
add a comment |
$begingroup$
How would I expand the following function as a power series, around $eta=0$?
$$g_0(1,eta)=fracleft(fracPVNkTright)_0-14eta$$
Note that:
$$left(fracPVNkTright)_0=1+frac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k$$
Then we have:
$$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
power-series
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
$endgroup$
$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37
add a comment |
$begingroup$
How would I expand the following function as a power series, around $eta=0$?
$$g_0(1,eta)=fracleft(fracPVNkTright)_0-14eta$$
Note that:
$$left(fracPVNkTright)_0=1+frac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k$$
Then we have:
$$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
power-series
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
$endgroup$
How would I expand the following function as a power series, around $eta=0$?
$$g_0(1,eta)=fracleft(fracPVNkTright)_0-14eta$$
Note that:
$$left(fracPVNkTright)_0=1+frac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k$$
Then we have:
$$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
power-series
power-series
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
edited Mar 26 at 18:38
Jackson Hart
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
asked Mar 26 at 17:53
Jackson HartJackson Hart
5452726
5452726
$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37
add a comment |
$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37
$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Starting from your last equation: $$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
you have:
$$g_0(1,eta)=frac34(eta_c-eta)+sum_k=1^4frac k4eta_c^kA_keta^k-1$$
The terms after the + sign are ok, you just need to deal with the expansion of $(eta_c-eta)^-1$. You can give $eta_c$ as a factor, then you have $$frac33eta_cleft(1-fracetaeta_cright)^-1$$
For $|x|<1$ you have $$frac 11-x=x+x^2+x^3+...=sum_n=1^infty x^n$$
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
add a comment |
$begingroup$
Hint:
The sum is a polynomial with no constant term and its handling is not a problem.
Then
$$fracetaeta_c-eta=1-frac11-dfracetaeta_c=fracetaeta_c+fraceta^2eta_c^2+fraceta^3eta_c^3+cdots$$
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
$endgroup$
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Starting from your last equation: $$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
you have:
$$g_0(1,eta)=frac34(eta_c-eta)+sum_k=1^4frac k4eta_c^kA_keta^k-1$$
The terms after the + sign are ok, you just need to deal with the expansion of $(eta_c-eta)^-1$. You can give $eta_c$ as a factor, then you have $$frac33eta_cleft(1-fracetaeta_cright)^-1$$
For $|x|<1$ you have $$frac 11-x=x+x^2+x^3+...=sum_n=1^infty x^n$$
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
add a comment |
$begingroup$
Starting from your last equation: $$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
you have:
$$g_0(1,eta)=frac34(eta_c-eta)+sum_k=1^4frac k4eta_c^kA_keta^k-1$$
The terms after the + sign are ok, you just need to deal with the expansion of $(eta_c-eta)^-1$. You can give $eta_c$ as a factor, then you have $$frac33eta_cleft(1-fracetaeta_cright)^-1$$
For $|x|<1$ you have $$frac 11-x=x+x^2+x^3+...=sum_n=1^infty x^n$$
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
$endgroup$
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
add a comment |
$begingroup$
Starting from your last equation: $$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
you have:
$$g_0(1,eta)=frac34(eta_c-eta)+sum_k=1^4frac k4eta_c^kA_keta^k-1$$
The terms after the + sign are ok, you just need to deal with the expansion of $(eta_c-eta)^-1$. You can give $eta_c$ as a factor, then you have $$frac33eta_cleft(1-fracetaeta_cright)^-1$$
For $|x|<1$ you have $$frac 11-x=x+x^2+x^3+...=sum_n=1^infty x^n$$
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
$endgroup$
Starting from your last equation: $$g_0(1,eta)=fracfrac3etaeta_c-eta+sum_k=1^4kA_kleft(fracetaeta_cright)^k4eta$$
you have:
$$g_0(1,eta)=frac34(eta_c-eta)+sum_k=1^4frac k4eta_c^kA_keta^k-1$$
The terms after the + sign are ok, you just need to deal with the expansion of $(eta_c-eta)^-1$. You can give $eta_c$ as a factor, then you have $$frac33eta_cleft(1-fracetaeta_cright)^-1$$
For $|x|<1$ you have $$frac 11-x=x+x^2+x^3+...=sum_n=1^infty x^n$$
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
answered Mar 26 at 19:15
AndreiAndrei
13.9k21330
13.9k21330
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
add a comment |
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I believe it should $frac34eta_c$ right? Then is the total answer: $sum_n=1^infty left(fracetaeta_cright)^n+sum_k=1^4frac k4eta_c^kA_keta^k-1$. I need everything to be in terms of $fracetaeta_c$
$endgroup$
– Jackson Hart
Mar 27 at 18:12
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I need everything to be in terms of $fracetaeta_c$ I need to multiply this function by another polynomial: $C_01r^0 left(fracetaeta_cright)+C_11r^1left(fracetaeta_cright)+C_21r^2left(fracetaeta_cright)+C_31r^3left(fracetaeta_cright)+C_41r^4left(fracetaeta_cright)+C_51r^5left(fracetaeta_cright)+C_61r^6left(fracetaeta_cright)$ If I take your solution and multiply by this polynomial, how do I get an expansion about $eta=0$? I need to get the coefficients in front of each term
$endgroup$
– Jackson Hart
Mar 27 at 18:19
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
$begingroup$
I accepted this answer since it answered my question. I posted a related question on another thread if you are interested in looking at it.
$endgroup$
– Jackson Hart
Mar 27 at 18:30
add a comment |
$begingroup$
Hint:
The sum is a polynomial with no constant term and its handling is not a problem.
Then
$$fracetaeta_c-eta=1-frac11-dfracetaeta_c=fracetaeta_c+fraceta^2eta_c^2+fraceta^3eta_c^3+cdots$$
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
$endgroup$
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
add a comment |
$begingroup$
Hint:
The sum is a polynomial with no constant term and its handling is not a problem.
Then
$$fracetaeta_c-eta=1-frac11-dfracetaeta_c=fracetaeta_c+fraceta^2eta_c^2+fraceta^3eta_c^3+cdots$$
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
$endgroup$
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
add a comment |
$begingroup$
Hint:
The sum is a polynomial with no constant term and its handling is not a problem.
Then
$$fracetaeta_c-eta=1-frac11-dfracetaeta_c=fracetaeta_c+fraceta^2eta_c^2+fraceta^3eta_c^3+cdots$$
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
$endgroup$
Hint:
The sum is a polynomial with no constant term and its handling is not a problem.
Then
$$fracetaeta_c-eta=1-frac11-dfracetaeta_c=fracetaeta_c+fraceta^2eta_c^2+fraceta^3eta_c^3+cdots$$
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
answered Mar 26 at 18:49
Yves DaoustYves Daoust
133k676232
133k676232
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
add a comment |
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
I'm not sure how to get the full answer. I'm just trying to get a different representation of my equation, just checking to see if series expansion would work.
$endgroup$
– Jackson Hart
Mar 26 at 19:01
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
$begingroup$
@JacksonHart: my hint explains that it would indeed work, but converge for $|eta|<|eta_c|$.
$endgroup$
– Yves Daoust
Mar 27 at 7:45
add a comment |
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$begingroup$
What does the subscript $0$ mean on the right hand side of the first equation?
$endgroup$
– Andrei
Mar 26 at 18:22
$begingroup$
Sorry I forgot to put it on 2nd equation. It is corrected now.
$endgroup$
– Jackson Hart
Mar 26 at 18:37