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Taylor expansion of scalar fields
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Starting of with electrodynamics I have to compute the taylor expansion around $vecr = 0$ of
$psi (vecr) = |vecr - vecr_0|^frac32$ where $vecr_0$ is a constant vector up to second order and
$psi(vecr) = e^iveckvecr$ where $veck$ is a constant vector up to arbitrary order.
I don't have problems with multidimensional taylor expansions as long as there are no vectors involved and functions look like $f(x,y,z) = y cdot sin(xz) + xz^2$ for example. Therefore, in the cases above I feel lost.
Can someone explain how to solve the exercise?
homework-and-exercises electromagnetism differentiation approximations
edited 42 mins ago
Qmechanic♦
98.2k121731066
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up vote
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Starting of with electrodynamics I have to compute the taylor expansion around $vecr = 0$ of
$psi (vecr) = |vecr - vecr_0|^frac32$ where $vecr_0$ is a constant vector up to second order and
$psi(vecr) = e^iveckvecr$ where $veck$ is a constant vector up to arbitrary order.
I don't have problems with multidimensional taylor expansions as long as there are no vectors involved and functions look like $f(x,y,z) = y cdot sin(xz) + xz^2$ for example. Therefore, in the cases above I feel lost.
Can someone explain how to solve the exercise?
homework-and-exercises electromagnetism differentiation approximations
edited 42 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
offline
165
By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
That's what I mean
– offline
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Starting of with electrodynamics I have to compute the taylor expansion around $vecr = 0$ of
$psi (vecr) = |vecr - vecr_0|^frac32$ where $vecr_0$ is a constant vector up to second order and
$psi(vecr) = e^iveckvecr$ where $veck$ is a constant vector up to arbitrary order.
I don't have problems with multidimensional taylor expansions as long as there are no vectors involved and functions look like $f(x,y,z) = y cdot sin(xz) + xz^2$ for example. Therefore, in the cases above I feel lost.
Can someone explain how to solve the exercise?
homework-and-exercises electromagnetism differentiation approximations
edited 42 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
offline
165
Starting of with electrodynamics I have to compute the taylor expansion around $vecr = 0$ of
$psi (vecr) = |vecr - vecr_0|^frac32$ where $vecr_0$ is a constant vector up to second order and
$psi(vecr) = e^iveckvecr$ where $veck$ is a constant vector up to arbitrary order.
I don't have problems with multidimensional taylor expansions as long as there are no vectors involved and functions look like $f(x,y,z) = y cdot sin(xz) + xz^2$ for example. Therefore, in the cases above I feel lost.
Can someone explain how to solve the exercise?
homework-and-exercises electromagnetism differentiation approximations
homework-and-exercises electromagnetism differentiation approximations
edited 42 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
offline
165
edited 42 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
offline
165
edited 42 mins ago
Qmechanic♦
98.2k121731066
edited 42 mins ago
Qmechanic♦
98.2k121731066
edited 42 mins ago
Qmechanic♦
98.2k121731066
98.2k121731066
asked 1 hour ago
offline
165
asked 1 hour ago
offline
165
asked 1 hour ago
offline
165
165
By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
That's what I mean
– offline
1 hour ago
add a comment |Â
By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
That's what I mean
– offline
1 hour ago
By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
That's what I mean
– offline
1 hour ago
That's what I mean
– offline
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $vecr=xhatx+yhaty+zhatz$, $veck=k_xhatx+k_yhaty+k_zhatz$, and $vecr_0=x_0hatx+y_0haty+z_0hatz$) and write the expressions out that way, it might be clearer:
$$psi(vecr)topsi(x,y,z)=left(sqrt(x-x_0)^2+(y-y_0)^2+(z-z_0)^2right)^3/2$$
$$psi(vecr)topsi(x,y,z)=e^i(xk_x+yk_y+zk_z)$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$psi(vecr)topsi(r,theta)=left(sqrtr^2+r_0^2-2rr_0costhetaright)^3/2$$
$$psi(vecr)topsi(r,theta)=e^irkcostheta$$
where $theta$ is, in the first example, the angle between $vecr$ and $vecr_0$ (in other words, the axis of rotational symmetry of the function is located along $vecr_0$), and is, in the second example, the angle between $vecr$ and $veck$ (so that the axis of rotational symmetry is along $veck$).
answered 1 hour ago
probably_someone
14.5k12451
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $vecr=xhatx+yhaty+zhatz$, $veck=k_xhatx+k_yhaty+k_zhatz$, and $vecr_0=x_0hatx+y_0haty+z_0hatz$) and write the expressions out that way, it might be clearer:
$$psi(vecr)topsi(x,y,z)=left(sqrt(x-x_0)^2+(y-y_0)^2+(z-z_0)^2right)^3/2$$
$$psi(vecr)topsi(x,y,z)=e^i(xk_x+yk_y+zk_z)$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$psi(vecr)topsi(r,theta)=left(sqrtr^2+r_0^2-2rr_0costhetaright)^3/2$$
$$psi(vecr)topsi(r,theta)=e^irkcostheta$$
where $theta$ is, in the first example, the angle between $vecr$ and $vecr_0$ (in other words, the axis of rotational symmetry of the function is located along $vecr_0$), and is, in the second example, the angle between $vecr$ and $veck$ (so that the axis of rotational symmetry is along $veck$).
answered 1 hour ago
probably_someone
14.5k12451
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
add a comment |Â
up vote
2
down vote
accepted
Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $vecr=xhatx+yhaty+zhatz$, $veck=k_xhatx+k_yhaty+k_zhatz$, and $vecr_0=x_0hatx+y_0haty+z_0hatz$) and write the expressions out that way, it might be clearer:
$$psi(vecr)topsi(x,y,z)=left(sqrt(x-x_0)^2+(y-y_0)^2+(z-z_0)^2right)^3/2$$
$$psi(vecr)topsi(x,y,z)=e^i(xk_x+yk_y+zk_z)$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$psi(vecr)topsi(r,theta)=left(sqrtr^2+r_0^2-2rr_0costhetaright)^3/2$$
$$psi(vecr)topsi(r,theta)=e^irkcostheta$$
where $theta$ is, in the first example, the angle between $vecr$ and $vecr_0$ (in other words, the axis of rotational symmetry of the function is located along $vecr_0$), and is, in the second example, the angle between $vecr$ and $veck$ (so that the axis of rotational symmetry is along $veck$).
answered 1 hour ago
probably_someone
14.5k12451
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $vecr=xhatx+yhaty+zhatz$, $veck=k_xhatx+k_yhaty+k_zhatz$, and $vecr_0=x_0hatx+y_0haty+z_0hatz$) and write the expressions out that way, it might be clearer:
$$psi(vecr)topsi(x,y,z)=left(sqrt(x-x_0)^2+(y-y_0)^2+(z-z_0)^2right)^3/2$$
$$psi(vecr)topsi(x,y,z)=e^i(xk_x+yk_y+zk_z)$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$psi(vecr)topsi(r,theta)=left(sqrtr^2+r_0^2-2rr_0costhetaright)^3/2$$
$$psi(vecr)topsi(r,theta)=e^irkcostheta$$
where $theta$ is, in the first example, the angle between $vecr$ and $vecr_0$ (in other words, the axis of rotational symmetry of the function is located along $vecr_0$), and is, in the second example, the angle between $vecr$ and $veck$ (so that the axis of rotational symmetry is along $veck$).
answered 1 hour ago
probably_someone
14.5k12451
Don't let the vector notation confuse you - it's the same situation as an ordinary multidimensional Taylor expansion. After all, in each case, the end result is a scalar, not a vector. If you split each vector into its components (for example, $vecr=xhatx+yhaty+zhatz$, $veck=k_xhatx+k_yhaty+k_zhatz$, and $vecr_0=x_0hatx+y_0haty+z_0hatz$) and write the expressions out that way, it might be clearer:
$$psi(vecr)topsi(x,y,z)=left(sqrt(x-x_0)^2+(y-y_0)^2+(z-z_0)^2right)^3/2$$
$$psi(vecr)topsi(x,y,z)=e^i(xk_x+yk_y+zk_z)$$
Of course, these may not be (and in fact probably aren't) the easiest coordinate systems to use for performing these Taylor expansions, but the same idea applies - pick a coordinate system and expand as you would with a normal multidimensional function (but be careful with periodic coordinates, as they make things more complicated to interpret if you choose to Taylor-expand in that direction). The neat thing is that Taylor expansions in different coordinates will tell you somewhat different things about a function's behavior, since different coordinates become small in different regions of space. For example, you might want to expand only in the magnitudes $r$, $r_0$, $k$ of each of the vectors, in which case you would get:
$$psi(vecr)topsi(r,theta)=left(sqrtr^2+r_0^2-2rr_0costhetaright)^3/2$$
$$psi(vecr)topsi(r,theta)=e^irkcostheta$$
where $theta$ is, in the first example, the angle between $vecr$ and $vecr_0$ (in other words, the axis of rotational symmetry of the function is located along $vecr_0$), and is, in the second example, the angle between $vecr$ and $veck$ (so that the axis of rotational symmetry is along $veck$).
answered 1 hour ago
probably_someone
14.5k12451
edited 52 mins ago
answered 1 hour ago
probably_someone
14.5k12451
answered 1 hour ago
probably_someone
14.5k12451
answered 1 hour ago
probably_someone
14.5k12451
14.5k12451
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
add a comment |Â
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
That is helpful as I can now do what I usually do. Nevertheless that seems to be not the fastest way. Could you explain how to get the result quicker. Other examples found in the internet usually just use $vecr$ and $vecr_0$ in order to compute the solution, which seems to be quite elegant but hard to understand if you are new to the topic.
– offline
57 mins ago
add a comment |Â
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By $e^iveckvecr$ do you mean $e^iveckcdotvecr$?
– probably_someone
1 hour ago
That's what I mean
– offline
1 hour ago