Mixing
Interpolator surface
$begingroup$
Writing:
Clear[u, v];
{a0, b0, c0} = {1, 2, 3};
f = ((x/a)^2 + (y/b)^2) c;
g = {a0 Sqrt[u] Cos[v], b0 Sqrt[u] Sin[v], c0 u};
h = {};
For[u = 0, u <= 1, u = u + 0.05,
For[v = 0, v <= 2 Pi, v = v + 0.05,
h = Join[h, {g}]
]
];
FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
I get:
{a -> 1., b -> 2., c -> 3.}
that's exactly how much you want.
Unfortunately a0, b0, c0 in reality I do not know them, so I should write:
FindFit[h, f, {a, b, c}, {x, y}]
I get:
{a -> 0.533886, b -> 1.06777, c -> 0.855104}
which is a result very far from what is expected. How could I fix it?
calculus-and-analysis fitting interpolation
asked Jan 20 at 14:45
TeMTeM
2,027621
$endgroup$
add a comment |
$begingroup$
Writing:
Clear[u, v];
{a0, b0, c0} = {1, 2, 3};
f = ((x/a)^2 + (y/b)^2) c;
g = {a0 Sqrt[u] Cos[v], b0 Sqrt[u] Sin[v], c0 u};
h = {};
For[u = 0, u <= 1, u = u + 0.05,
For[v = 0, v <= 2 Pi, v = v + 0.05,
h = Join[h, {g}]
]
];
FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
I get:
{a -> 1., b -> 2., c -> 3.}
that's exactly how much you want.
Unfortunately a0, b0, c0 in reality I do not know them, so I should write:
FindFit[h, f, {a, b, c}, {x, y}]
I get:
{a -> 0.533886, b -> 1.06777, c -> 0.855104}
which is a result very far from what is expected. How could I fix it?
calculus-and-analysis fitting interpolation
asked Jan 20 at 14:45
TeMTeM
2,027621
$endgroup$
add a comment |
$begingroup$
Writing:
Clear[u, v];
{a0, b0, c0} = {1, 2, 3};
f = ((x/a)^2 + (y/b)^2) c;
g = {a0 Sqrt[u] Cos[v], b0 Sqrt[u] Sin[v], c0 u};
h = {};
For[u = 0, u <= 1, u = u + 0.05,
For[v = 0, v <= 2 Pi, v = v + 0.05,
h = Join[h, {g}]
]
];
FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
I get:
{a -> 1., b -> 2., c -> 3.}
that's exactly how much you want.
Unfortunately a0, b0, c0 in reality I do not know them, so I should write:
FindFit[h, f, {a, b, c}, {x, y}]
I get:
{a -> 0.533886, b -> 1.06777, c -> 0.855104}
which is a result very far from what is expected. How could I fix it?
calculus-and-analysis fitting interpolation
asked Jan 20 at 14:45
TeMTeM
2,027621
$endgroup$
Writing:
Clear[u, v];
{a0, b0, c0} = {1, 2, 3};
f = ((x/a)^2 + (y/b)^2) c;
g = {a0 Sqrt[u] Cos[v], b0 Sqrt[u] Sin[v], c0 u};
h = {};
For[u = 0, u <= 1, u = u + 0.05,
For[v = 0, v <= 2 Pi, v = v + 0.05,
h = Join[h, {g}]
]
];
FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
I get:
{a -> 1., b -> 2., c -> 3.}
that's exactly how much you want.
Unfortunately a0, b0, c0 in reality I do not know them, so I should write:
FindFit[h, f, {a, b, c}, {x, y}]
I get:
{a -> 0.533886, b -> 1.06777, c -> 0.855104}
which is a result very far from what is expected. How could I fix it?
calculus-and-analysis fitting interpolation
calculus-and-analysis fitting interpolation
asked Jan 20 at 14:45
TeMTeM
2,027621
asked Jan 20 at 14:45
TeMTeM
2,027621
asked Jan 20 at 14:45
TeMTeM
2,027621
asked Jan 20 at 14:45
TeMTeM
2,027621
asked Jan 20 at 14:45
TeMTeM
2,027621
2,027621
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are no differences in the predictions. What you have is an over-parameterized model. You have attempted to fit too many parameters. The basic model is
$$f=a_x x^2+a_y y^2$$
There are really only 2 parameters to fit.
sol0 = FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
(* {a -> 1., b -> 2., c -> 3.} *)
f /. sol0 // Expand
(* 3. x^2 + 0.75 y^2 *)
sol1 = FindFit[h, f, {a, b, c}, {x, y}]
(* {a -> 0.533886, b -> 1.06777, c -> 0.855104} *)
f /. sol1 // Expand
(* 3. x^2 + 0.75 y^2 *)
answered Jan 20 at 16:05
JimBJimB
17.7k12763
$endgroup$
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
add a comment |
$begingroup$
This is not an answer, but a hint on how to improve your coding by coding in a more functional style. Functional code is almost always more concise and more efficient than procedural code.
{a0, b0, c0} = {1, 2, 3};
g[a_, b_, c_] = {a Sqrt[#1] Cos[#2], b Sqrt[#1] Sin[#2], c #1} &;
h = Table[g[a0, b0, c0][u, v], {u, 0, 1, .05}, {v, 0, 2 π, .05}] // Catenate;
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
$endgroup$
$begingroup$
Actually this was my first approach, but not knowingCatenateI left it. Thank you, very useful!
$endgroup$
– TeM
Jan 21 at 21:37
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are no differences in the predictions. What you have is an over-parameterized model. You have attempted to fit too many parameters. The basic model is
$$f=a_x x^2+a_y y^2$$
There are really only 2 parameters to fit.
sol0 = FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
(* {a -> 1., b -> 2., c -> 3.} *)
f /. sol0 // Expand
(* 3. x^2 + 0.75 y^2 *)
sol1 = FindFit[h, f, {a, b, c}, {x, y}]
(* {a -> 0.533886, b -> 1.06777, c -> 0.855104} *)
f /. sol1 // Expand
(* 3. x^2 + 0.75 y^2 *)
answered Jan 20 at 16:05
JimBJimB
17.7k12763
$endgroup$
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
add a comment |
$begingroup$
There are no differences in the predictions. What you have is an over-parameterized model. You have attempted to fit too many parameters. The basic model is
$$f=a_x x^2+a_y y^2$$
There are really only 2 parameters to fit.
sol0 = FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
(* {a -> 1., b -> 2., c -> 3.} *)
f /. sol0 // Expand
(* 3. x^2 + 0.75 y^2 *)
sol1 = FindFit[h, f, {a, b, c}, {x, y}]
(* {a -> 0.533886, b -> 1.06777, c -> 0.855104} *)
f /. sol1 // Expand
(* 3. x^2 + 0.75 y^2 *)
answered Jan 20 at 16:05
JimBJimB
17.7k12763
$endgroup$
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
add a comment |
$begingroup$
There are no differences in the predictions. What you have is an over-parameterized model. You have attempted to fit too many parameters. The basic model is
$$f=a_x x^2+a_y y^2$$
There are really only 2 parameters to fit.
sol0 = FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
(* {a -> 1., b -> 2., c -> 3.} *)
f /. sol0 // Expand
(* 3. x^2 + 0.75 y^2 *)
sol1 = FindFit[h, f, {a, b, c}, {x, y}]
(* {a -> 0.533886, b -> 1.06777, c -> 0.855104} *)
f /. sol1 // Expand
(* 3. x^2 + 0.75 y^2 *)
answered Jan 20 at 16:05
JimBJimB
17.7k12763
$endgroup$
There are no differences in the predictions. What you have is an over-parameterized model. You have attempted to fit too many parameters. The basic model is
$$f=a_x x^2+a_y y^2$$
There are really only 2 parameters to fit.
sol0 = FindFit[h, f, {{a, a0}, {b, b0}, {c, c0}}, {x, y}]
(* {a -> 1., b -> 2., c -> 3.} *)
f /. sol0 // Expand
(* 3. x^2 + 0.75 y^2 *)
sol1 = FindFit[h, f, {a, b, c}, {x, y}]
(* {a -> 0.533886, b -> 1.06777, c -> 0.855104} *)
f /. sol1 // Expand
(* 3. x^2 + 0.75 y^2 *)
answered Jan 20 at 16:05
JimBJimB
17.7k12763
answered Jan 20 at 16:05
JimBJimB
17.7k12763
answered Jan 20 at 16:05
JimBJimB
17.7k12763
answered Jan 20 at 16:05
JimBJimB
17.7k12763
17.7k12763
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
add a comment |
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
$begingroup$
Sometimes I think I should take a break. Thank you so much for opening my eyes!
$endgroup$
– TeM
Jan 20 at 16:09
1
1
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
$begingroup$
Same here. You might consider just using $(x/a)^2 + (y/b)^2$ which is just setting $c=1$ if the parameters $a$ and $b$ are easier to interpret. Also, using the mean of the $x$ and $y$ values for the starting values for $a$ and $b$ might be more numerically stable if the $x$'s and $y$'s are very large or very small.
$endgroup$
– JimB
Jan 20 at 16:15
add a comment |
$begingroup$
This is not an answer, but a hint on how to improve your coding by coding in a more functional style. Functional code is almost always more concise and more efficient than procedural code.
{a0, b0, c0} = {1, 2, 3};
g[a_, b_, c_] = {a Sqrt[#1] Cos[#2], b Sqrt[#1] Sin[#2], c #1} &;
h = Table[g[a0, b0, c0][u, v], {u, 0, 1, .05}, {v, 0, 2 π, .05}] // Catenate;
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
$endgroup$
$begingroup$
Actually this was my first approach, but not knowingCatenateI left it. Thank you, very useful!
$endgroup$
– TeM
Jan 21 at 21:37
add a comment |
$begingroup$
This is not an answer, but a hint on how to improve your coding by coding in a more functional style. Functional code is almost always more concise and more efficient than procedural code.
{a0, b0, c0} = {1, 2, 3};
g[a_, b_, c_] = {a Sqrt[#1] Cos[#2], b Sqrt[#1] Sin[#2], c #1} &;
h = Table[g[a0, b0, c0][u, v], {u, 0, 1, .05}, {v, 0, 2 π, .05}] // Catenate;
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
$endgroup$
$begingroup$
Actually this was my first approach, but not knowingCatenateI left it. Thank you, very useful!
$endgroup$
– TeM
Jan 21 at 21:37
add a comment |
$begingroup$
This is not an answer, but a hint on how to improve your coding by coding in a more functional style. Functional code is almost always more concise and more efficient than procedural code.
{a0, b0, c0} = {1, 2, 3};
g[a_, b_, c_] = {a Sqrt[#1] Cos[#2], b Sqrt[#1] Sin[#2], c #1} &;
h = Table[g[a0, b0, c0][u, v], {u, 0, 1, .05}, {v, 0, 2 π, .05}] // Catenate;
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
$endgroup$
This is not an answer, but a hint on how to improve your coding by coding in a more functional style. Functional code is almost always more concise and more efficient than procedural code.
{a0, b0, c0} = {1, 2, 3};
g[a_, b_, c_] = {a Sqrt[#1] Cos[#2], b Sqrt[#1] Sin[#2], c #1} &;
h = Table[g[a0, b0, c0][u, v], {u, 0, 1, .05}, {v, 0, 2 π, .05}] // Catenate;
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
answered Jan 20 at 18:15
m_goldbergm_goldberg
87.3k872198
87.3k872198
$begingroup$
Actually this was my first approach, but not knowingCatenateI left it. Thank you, very useful!
$endgroup$
– TeM
Jan 21 at 21:37
add a comment |
$begingroup$
Actually this was my first approach, but not knowingCatenateI left it. Thank you, very useful!
$endgroup$
– TeM
Jan 21 at 21:37
$begingroup$
Actually this was my first approach, but not knowing
Catenate I left it. Thank you, very useful!$endgroup$
– TeM
Jan 21 at 21:37
$begingroup$
Actually this was my first approach, but not knowing
Catenate I left it. Thank you, very useful!$endgroup$
– TeM
Jan 21 at 21:37
add a comment |
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