Taylor series of complex multivariate function Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)First four Taylor Series ExpansionsTaylor series of $1/|r - R - x|$Is there a faster convergence series than the Taylor series?Power series proof without TaylorTaylor Expansion Multivariate analysisUnivariate Taylor expansion of a multivariate function?Relation between Taylor Expansion and multivariate Gaussian DistributionWhat is the difference between a Taylor series, Taylor polynomial, analytic function and a quadradic approximation?Taylor Series expansion of an implicitly defined function $x^2 +y^2=y, y(0)=0$Taylor series of a complex Gaussian function
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Taylor series of complex multivariate function
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)First four Taylor Series ExpansionsTaylor series of $1/|r - R - x|$Is there a faster convergence series than the Taylor series?Power series proof without TaylorTaylor Expansion Multivariate analysisUnivariate Taylor expansion of a multivariate function?Relation between Taylor Expansion and multivariate Gaussian DistributionWhat is the difference between a Taylor series, Taylor polynomial, analytic function and a quadradic approximation?Taylor Series expansion of an implicitly defined function $x^2 +y^2=y, y(0)=0$Taylor series of a complex Gaussian function
$begingroup$
I am looking for a reference or some literature on Taylor series of complex multivariate functions. I found material for complex functions and material for multivariate functions, but not for both.
Is there an expression for at least the 2 or 3 first terms of the Taylor expansion of a function $f: mathbbC^n rightarrow mathbbC$?
taylor-expansion several-complex-variables
asked Apr 2 at 17:31
OpticAlOpticAl
959
$endgroup$
add a comment |
$begingroup$
I am looking for a reference or some literature on Taylor series of complex multivariate functions. I found material for complex functions and material for multivariate functions, but not for both.
Is there an expression for at least the 2 or 3 first terms of the Taylor expansion of a function $f: mathbbC^n rightarrow mathbbC$?
taylor-expansion several-complex-variables
asked Apr 2 at 17:31
OpticAlOpticAl
959
$endgroup$
$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50
add a comment |
$begingroup$
I am looking for a reference or some literature on Taylor series of complex multivariate functions. I found material for complex functions and material for multivariate functions, but not for both.
Is there an expression for at least the 2 or 3 first terms of the Taylor expansion of a function $f: mathbbC^n rightarrow mathbbC$?
taylor-expansion several-complex-variables
asked Apr 2 at 17:31
OpticAlOpticAl
959
$endgroup$
I am looking for a reference or some literature on Taylor series of complex multivariate functions. I found material for complex functions and material for multivariate functions, but not for both.
Is there an expression for at least the 2 or 3 first terms of the Taylor expansion of a function $f: mathbbC^n rightarrow mathbbC$?
taylor-expansion several-complex-variables
taylor-expansion several-complex-variables
asked Apr 2 at 17:31
OpticAlOpticAl
959
asked Apr 2 at 17:31
OpticAlOpticAl
959
asked Apr 2 at 17:31
OpticAlOpticAl
959
asked Apr 2 at 17:31
OpticAlOpticAl
959
asked Apr 2 at 17:31
OpticAlOpticAl
959
959
$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50
add a comment |
$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50
$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50
$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you're looking for a graduate course: http://www.jirka.org/scv/
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
$endgroup$
add a comment |
$begingroup$
I'll take a stab at it blending the two concepts together.
In a single complex variable we have $f(z)=a(x,y)+ib(x,y)$ where $z=x+iy$.
Suppose we have two complex variables.
Then I'd guess: $f(z_1,z_2)=a(x_1,x_2,y_1,y_2)+ib(x_1,x_2,y_1,y_2)$
Before we worry about a Taylor series, we need some condition for testing analysity.
In the single complex variable case, we have some conditions. We want the derivative tweaking just the real component to give us the derivative tweaking just the imaginary component.
So: $fraca(x+Delta x,y)+ib(x+Delta x, y)-a(x,y)-ib(x,y)Delta x=fraca(x,y+Delta y) + i b(x, y+ Delta y)-a(x,y)-ib(x,y)iDelta y$
Then we equat real and imaginary parts:
$fracpartial apartial x+ifracpartial bpartial x=(-i)(fracpartial apartial y+ifracpartial bpartial y)$
So we get :
$fracpartial apartial x=fracpartial bpartial y$
$fracpartial bpartial x=-fracpartial apartial y$
By similar reasoning, I think we need:
$fracpartial apartial x_1=fracpartial bpartial y_1$
$fracpartial bpartial x_1=-fracpartial apartial y_1$
$fracpartial apartial x_2=fracpartial bpartial y_2$
$fracpartial bpartial x_2=-fracpartial apartial y_2$
$fracpartial apartial x_1=fracpartial apartial x_2$
$fracpartial bpartial x_1=fracpartial bpartial x_2$
$fracpartial bpartial y_1=fracpartial bpartial y_2$
$fracpartial apartial y_1=fracpartial apartial y_2$
If those criteria are satisfied, then I think we can proceed with a Taylor series.
$f(z_1,z_2)=f(x_10,x_20,y_10, y_20)+fracpartial apartial x_1Delta x_1 +ifracpartial bpartial x_1Delta x_1+fracpartial apartial x_2Delta x_2+ fracpartial bpartial y_1Delta y_1 - i fracpartial apartial y_1+... + fracpartial^2 apartial x_1^2Delta x_1^2+... $
I believe you proceed in the usual way for multivariable functions just keeping in mind which terms need the imaginary unit. An x derivative of the real part of the function and a y derivative of the imaginary part do not need the imaginary units. The others do.
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
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
$begingroup$
If you're looking for a graduate course: http://www.jirka.org/scv/
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
$endgroup$
add a comment |
$begingroup$
If you're looking for a graduate course: http://www.jirka.org/scv/
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
$endgroup$
add a comment |
$begingroup$
If you're looking for a graduate course: http://www.jirka.org/scv/
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
$endgroup$
If you're looking for a graduate course: http://www.jirka.org/scv/
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
answered Apr 3 at 15:52
Jiri LeblJiri Lebl
1,683412
1,683412
add a comment |
add a comment |
$begingroup$
I'll take a stab at it blending the two concepts together.
In a single complex variable we have $f(z)=a(x,y)+ib(x,y)$ where $z=x+iy$.
Suppose we have two complex variables.
Then I'd guess: $f(z_1,z_2)=a(x_1,x_2,y_1,y_2)+ib(x_1,x_2,y_1,y_2)$
Before we worry about a Taylor series, we need some condition for testing analysity.
In the single complex variable case, we have some conditions. We want the derivative tweaking just the real component to give us the derivative tweaking just the imaginary component.
So: $fraca(x+Delta x,y)+ib(x+Delta x, y)-a(x,y)-ib(x,y)Delta x=fraca(x,y+Delta y) + i b(x, y+ Delta y)-a(x,y)-ib(x,y)iDelta y$
Then we equat real and imaginary parts:
$fracpartial apartial x+ifracpartial bpartial x=(-i)(fracpartial apartial y+ifracpartial bpartial y)$
So we get :
$fracpartial apartial x=fracpartial bpartial y$
$fracpartial bpartial x=-fracpartial apartial y$
By similar reasoning, I think we need:
$fracpartial apartial x_1=fracpartial bpartial y_1$
$fracpartial bpartial x_1=-fracpartial apartial y_1$
$fracpartial apartial x_2=fracpartial bpartial y_2$
$fracpartial bpartial x_2=-fracpartial apartial y_2$
$fracpartial apartial x_1=fracpartial apartial x_2$
$fracpartial bpartial x_1=fracpartial bpartial x_2$
$fracpartial bpartial y_1=fracpartial bpartial y_2$
$fracpartial apartial y_1=fracpartial apartial y_2$
If those criteria are satisfied, then I think we can proceed with a Taylor series.
$f(z_1,z_2)=f(x_10,x_20,y_10, y_20)+fracpartial apartial x_1Delta x_1 +ifracpartial bpartial x_1Delta x_1+fracpartial apartial x_2Delta x_2+ fracpartial bpartial y_1Delta y_1 - i fracpartial apartial y_1+... + fracpartial^2 apartial x_1^2Delta x_1^2+... $
I believe you proceed in the usual way for multivariable functions just keeping in mind which terms need the imaginary unit. An x derivative of the real part of the function and a y derivative of the imaginary part do not need the imaginary units. The others do.
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
add a comment |
$begingroup$
I'll take a stab at it blending the two concepts together.
In a single complex variable we have $f(z)=a(x,y)+ib(x,y)$ where $z=x+iy$.
Suppose we have two complex variables.
Then I'd guess: $f(z_1,z_2)=a(x_1,x_2,y_1,y_2)+ib(x_1,x_2,y_1,y_2)$
Before we worry about a Taylor series, we need some condition for testing analysity.
In the single complex variable case, we have some conditions. We want the derivative tweaking just the real component to give us the derivative tweaking just the imaginary component.
So: $fraca(x+Delta x,y)+ib(x+Delta x, y)-a(x,y)-ib(x,y)Delta x=fraca(x,y+Delta y) + i b(x, y+ Delta y)-a(x,y)-ib(x,y)iDelta y$
Then we equat real and imaginary parts:
$fracpartial apartial x+ifracpartial bpartial x=(-i)(fracpartial apartial y+ifracpartial bpartial y)$
So we get :
$fracpartial apartial x=fracpartial bpartial y$
$fracpartial bpartial x=-fracpartial apartial y$
By similar reasoning, I think we need:
$fracpartial apartial x_1=fracpartial bpartial y_1$
$fracpartial bpartial x_1=-fracpartial apartial y_1$
$fracpartial apartial x_2=fracpartial bpartial y_2$
$fracpartial bpartial x_2=-fracpartial apartial y_2$
$fracpartial apartial x_1=fracpartial apartial x_2$
$fracpartial bpartial x_1=fracpartial bpartial x_2$
$fracpartial bpartial y_1=fracpartial bpartial y_2$
$fracpartial apartial y_1=fracpartial apartial y_2$
If those criteria are satisfied, then I think we can proceed with a Taylor series.
$f(z_1,z_2)=f(x_10,x_20,y_10, y_20)+fracpartial apartial x_1Delta x_1 +ifracpartial bpartial x_1Delta x_1+fracpartial apartial x_2Delta x_2+ fracpartial bpartial y_1Delta y_1 - i fracpartial apartial y_1+... + fracpartial^2 apartial x_1^2Delta x_1^2+... $
I believe you proceed in the usual way for multivariable functions just keeping in mind which terms need the imaginary unit. An x derivative of the real part of the function and a y derivative of the imaginary part do not need the imaginary units. The others do.
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
add a comment |
$begingroup$
I'll take a stab at it blending the two concepts together.
In a single complex variable we have $f(z)=a(x,y)+ib(x,y)$ where $z=x+iy$.
Suppose we have two complex variables.
Then I'd guess: $f(z_1,z_2)=a(x_1,x_2,y_1,y_2)+ib(x_1,x_2,y_1,y_2)$
Before we worry about a Taylor series, we need some condition for testing analysity.
In the single complex variable case, we have some conditions. We want the derivative tweaking just the real component to give us the derivative tweaking just the imaginary component.
So: $fraca(x+Delta x,y)+ib(x+Delta x, y)-a(x,y)-ib(x,y)Delta x=fraca(x,y+Delta y) + i b(x, y+ Delta y)-a(x,y)-ib(x,y)iDelta y$
Then we equat real and imaginary parts:
$fracpartial apartial x+ifracpartial bpartial x=(-i)(fracpartial apartial y+ifracpartial bpartial y)$
So we get :
$fracpartial apartial x=fracpartial bpartial y$
$fracpartial bpartial x=-fracpartial apartial y$
By similar reasoning, I think we need:
$fracpartial apartial x_1=fracpartial bpartial y_1$
$fracpartial bpartial x_1=-fracpartial apartial y_1$
$fracpartial apartial x_2=fracpartial bpartial y_2$
$fracpartial bpartial x_2=-fracpartial apartial y_2$
$fracpartial apartial x_1=fracpartial apartial x_2$
$fracpartial bpartial x_1=fracpartial bpartial x_2$
$fracpartial bpartial y_1=fracpartial bpartial y_2$
$fracpartial apartial y_1=fracpartial apartial y_2$
If those criteria are satisfied, then I think we can proceed with a Taylor series.
$f(z_1,z_2)=f(x_10,x_20,y_10, y_20)+fracpartial apartial x_1Delta x_1 +ifracpartial bpartial x_1Delta x_1+fracpartial apartial x_2Delta x_2+ fracpartial bpartial y_1Delta y_1 - i fracpartial apartial y_1+... + fracpartial^2 apartial x_1^2Delta x_1^2+... $
I believe you proceed in the usual way for multivariable functions just keeping in mind which terms need the imaginary unit. An x derivative of the real part of the function and a y derivative of the imaginary part do not need the imaginary units. The others do.
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
$endgroup$
I'll take a stab at it blending the two concepts together.
In a single complex variable we have $f(z)=a(x,y)+ib(x,y)$ where $z=x+iy$.
Suppose we have two complex variables.
Then I'd guess: $f(z_1,z_2)=a(x_1,x_2,y_1,y_2)+ib(x_1,x_2,y_1,y_2)$
Before we worry about a Taylor series, we need some condition for testing analysity.
In the single complex variable case, we have some conditions. We want the derivative tweaking just the real component to give us the derivative tweaking just the imaginary component.
So: $fraca(x+Delta x,y)+ib(x+Delta x, y)-a(x,y)-ib(x,y)Delta x=fraca(x,y+Delta y) + i b(x, y+ Delta y)-a(x,y)-ib(x,y)iDelta y$
Then we equat real and imaginary parts:
$fracpartial apartial x+ifracpartial bpartial x=(-i)(fracpartial apartial y+ifracpartial bpartial y)$
So we get :
$fracpartial apartial x=fracpartial bpartial y$
$fracpartial bpartial x=-fracpartial apartial y$
By similar reasoning, I think we need:
$fracpartial apartial x_1=fracpartial bpartial y_1$
$fracpartial bpartial x_1=-fracpartial apartial y_1$
$fracpartial apartial x_2=fracpartial bpartial y_2$
$fracpartial bpartial x_2=-fracpartial apartial y_2$
$fracpartial apartial x_1=fracpartial apartial x_2$
$fracpartial bpartial x_1=fracpartial bpartial x_2$
$fracpartial bpartial y_1=fracpartial bpartial y_2$
$fracpartial apartial y_1=fracpartial apartial y_2$
If those criteria are satisfied, then I think we can proceed with a Taylor series.
$f(z_1,z_2)=f(x_10,x_20,y_10, y_20)+fracpartial apartial x_1Delta x_1 +ifracpartial bpartial x_1Delta x_1+fracpartial apartial x_2Delta x_2+ fracpartial bpartial y_1Delta y_1 - i fracpartial apartial y_1+... + fracpartial^2 apartial x_1^2Delta x_1^2+... $
I believe you proceed in the usual way for multivariable functions just keeping in mind which terms need the imaginary unit. An x derivative of the real part of the function and a y derivative of the imaginary part do not need the imaginary units. The others do.
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
answered Apr 3 at 17:24
TurlocTheRedTurlocTheRed
1,034411
1,034411
add a comment |
add a comment |
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$begingroup$
Fix all but one of the variables. Expand a taylor polynomial in that variable. Each of its coefficients is now a function of $n-1$ variables. Expand each of the coefficient functions about the next variable, holding the others constant. Keep this up and you end up with the multivariate taylor polynomial.
$endgroup$
– Paul Sinclair
Apr 3 at 0:50