Why is this recursive code so slow? Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?What are the hidden specifications for FindRootWhy does this function inside FindRoot fail to evaluate?Very slow mathematica finite differencesUsing Mathematica to solve a recursive system of differential equationsImproving the speed on an iterated differential systemForward iterations of coupled recursion equationsManipulate+FindRoot+Plot3D very slow/crashAttacking a “Mathematica can't solve” problemErrors using FindRoot on slow numerical functionAvoiding a for loop to create a list
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Why is this recursive code so slow?
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?What are the hidden specifications for FindRootWhy does this function inside FindRoot fail to evaluate?Very slow mathematica finite differencesUsing Mathematica to solve a recursive system of differential equationsImproving the speed on an iterated differential systemForward iterations of coupled recursion equationsManipulate+FindRoot+Plot3D very slow/crashAttacking a “Mathematica can't solve” problemErrors using FindRoot on slow numerical functionAvoiding a for loop to create a list
$begingroup$
This code for the first five iterations the speed is okay, but after that the speed is very slow, I cannot understand what is wrong with this? Would you please help me fix it?
Clear[A, r, x, s, e]
s := 0.3405
e := 1.6539*10^-21
u[0] := 0.
u[1] := 0.1
A[r_] := A[r] =
Piecewise[r - 2.5 s - 48*e *s^12*r^-13 + 24*e*s^6*r^-7,
r > 2.5 s, -48*e*s^12*r^-13 + 24*e*s^6*r^-7,
s <= r <= 2.5 s, r - s -
24*e*s^-1, r < s]
For[i = 2, i < 101,
i++, u[i_] :=
x /. FindRoot[
u[i - 1] +
1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) -
0.9 A[x] == x , x, 1.]; Print[u[i]]]
equation-solving recursion
edited Apr 12 at 8:27
Roman
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
$endgroup$
add a comment |
$begingroup$
This code for the first five iterations the speed is okay, but after that the speed is very slow, I cannot understand what is wrong with this? Would you please help me fix it?
Clear[A, r, x, s, e]
s := 0.3405
e := 1.6539*10^-21
u[0] := 0.
u[1] := 0.1
A[r_] := A[r] =
Piecewise[r - 2.5 s - 48*e *s^12*r^-13 + 24*e*s^6*r^-7,
r > 2.5 s, -48*e*s^12*r^-13 + 24*e*s^6*r^-7,
s <= r <= 2.5 s, r - s -
24*e*s^-1, r < s]
For[i = 2, i < 101,
i++, u[i_] :=
x /. FindRoot[
u[i - 1] +
1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) -
0.9 A[x] == x , x, 1.]; Print[u[i]]]
equation-solving recursion
edited Apr 12 at 8:27
Roman
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
$endgroup$
$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28
add a comment |
$begingroup$
This code for the first five iterations the speed is okay, but after that the speed is very slow, I cannot understand what is wrong with this? Would you please help me fix it?
Clear[A, r, x, s, e]
s := 0.3405
e := 1.6539*10^-21
u[0] := 0.
u[1] := 0.1
A[r_] := A[r] =
Piecewise[r - 2.5 s - 48*e *s^12*r^-13 + 24*e*s^6*r^-7,
r > 2.5 s, -48*e*s^12*r^-13 + 24*e*s^6*r^-7,
s <= r <= 2.5 s, r - s -
24*e*s^-1, r < s]
For[i = 2, i < 101,
i++, u[i_] :=
x /. FindRoot[
u[i - 1] +
1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) -
0.9 A[x] == x , x, 1.]; Print[u[i]]]
equation-solving recursion
edited Apr 12 at 8:27
Roman
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
$endgroup$
This code for the first five iterations the speed is okay, but after that the speed is very slow, I cannot understand what is wrong with this? Would you please help me fix it?
Clear[A, r, x, s, e]
s := 0.3405
e := 1.6539*10^-21
u[0] := 0.
u[1] := 0.1
A[r_] := A[r] =
Piecewise[r - 2.5 s - 48*e *s^12*r^-13 + 24*e*s^6*r^-7,
r > 2.5 s, -48*e*s^12*r^-13 + 24*e*s^6*r^-7,
s <= r <= 2.5 s, r - s -
24*e*s^-1, r < s]
For[i = 2, i < 101,
i++, u[i_] :=
x /. FindRoot[
u[i - 1] +
1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) -
0.9 A[x] == x , x, 1.]; Print[u[i]]]
equation-solving recursion
equation-solving recursion
edited Apr 12 at 8:27
Roman
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
edited Apr 12 at 8:27
Roman
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
edited Apr 12 at 8:27
Roman
5,56111131
edited Apr 12 at 8:27
Roman
5,56111131
edited Apr 12 at 8:27
Roman
5,56111131
5,56111131
asked Apr 12 at 4:08
morapimorapi
355
asked Apr 12 at 4:08
morapimorapi
355
asked Apr 12 at 4:08
morapimorapi
355
355
$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28
add a comment |
$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28
$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28
$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I recommend you learn the distinction between immediate (=) and delayed (:=) assignments. They make the difference between slow and fast code here. Start with this tutorial or this book chapter, then look at memoization.
s = 0.3405;
e = 1.6539*10^-21;
u[0] = 0.;
u[1] = 0.1;
A[r_] = Piecewise[r - 2.5 s - 48*e*s^12*r^-13 + 24*e*s^6*r^-7, r > 2.5 s,
-48*e*s^12*r^-13 + 24*e*s^6*r^-7, s <= r <= 2.5 s,
r - s - 24*e*s^-1, r < s];
u[i_] := u[i] = x /. FindRoot[
u[i - 1] + 1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) - 0.9 A[x] == x, x, 1.]
Array[u, 100]
0.1, 1.77164, 1.37065, 1.04259, 0.887781, 0.708344, 0.59461,
0.457228, 0.367364, 0.296071, 0.256104, 0.20463, 0.208487, 1.20917,
1.04197, 0.939331, 0.879865, 0.827963, 0.774591, 0.72775, 0.67934,
0.63666, 0.592369, 0.553172, 0.512352, 0.476112, 0.438261, 0.404563,
0.369277, 0.339073, 0.321616, 0.301118, 0.296195, 0.224688, 0.273538,
0.31357, 0.33593, 0.366902, 0.38813, 0.417572, 0.437777, 0.465834,
0.48511, 0.511907, 0.530336, 0.55598, 0.573633, 0.598219, 0.615159,
0.638772, 0.655054, 0.677768, 0.693441, 0.715321, 0.73043, 0.751535,
0.766118, 0.786503, 0.800596, 0.820306, 0.833941, 0.852182, 0.85901,
0.874152, 0.871531, 0.78396, 0.781416, 0.696402, 0.693931, 0.611329,
0.608927, 0.528603, 0.526267, 0.448099, 0.445825, 0.369701, 0.367485,
0.315658, 0.325798, 0.341207, 0.351098, 0.366134, 0.375788, 0.390468,
0.399897, 0.414237, 0.42345, 0.437466, 0.446473, 0.46018, 0.46899,
0.4824, 0.491022, 0.504149, 0.51259, 0.525444, 0.533712, 0.546306,
0.554408, 0.56675
(takes about 1.3 seconds)
Alternatively, use
Table[u[i], i, 1, 100]
(same result). Your combination of For and Print shows the results but doesn't let you keep using them for more calculations.
answered Apr 12 at 4:43
RomanRoman
5,56111131
$endgroup$
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
$begingroup$
delayedassignments definitely sound slower thanimmediate, even if I have never worked with Mathematica
$endgroup$
– Roland
Apr 12 at 10:07
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I recommend you learn the distinction between immediate (=) and delayed (:=) assignments. They make the difference between slow and fast code here. Start with this tutorial or this book chapter, then look at memoization.
s = 0.3405;
e = 1.6539*10^-21;
u[0] = 0.;
u[1] = 0.1;
A[r_] = Piecewise[r - 2.5 s - 48*e*s^12*r^-13 + 24*e*s^6*r^-7, r > 2.5 s,
-48*e*s^12*r^-13 + 24*e*s^6*r^-7, s <= r <= 2.5 s,
r - s - 24*e*s^-1, r < s];
u[i_] := u[i] = x /. FindRoot[
u[i - 1] + 1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) - 0.9 A[x] == x, x, 1.]
Array[u, 100]
0.1, 1.77164, 1.37065, 1.04259, 0.887781, 0.708344, 0.59461,
0.457228, 0.367364, 0.296071, 0.256104, 0.20463, 0.208487, 1.20917,
1.04197, 0.939331, 0.879865, 0.827963, 0.774591, 0.72775, 0.67934,
0.63666, 0.592369, 0.553172, 0.512352, 0.476112, 0.438261, 0.404563,
0.369277, 0.339073, 0.321616, 0.301118, 0.296195, 0.224688, 0.273538,
0.31357, 0.33593, 0.366902, 0.38813, 0.417572, 0.437777, 0.465834,
0.48511, 0.511907, 0.530336, 0.55598, 0.573633, 0.598219, 0.615159,
0.638772, 0.655054, 0.677768, 0.693441, 0.715321, 0.73043, 0.751535,
0.766118, 0.786503, 0.800596, 0.820306, 0.833941, 0.852182, 0.85901,
0.874152, 0.871531, 0.78396, 0.781416, 0.696402, 0.693931, 0.611329,
0.608927, 0.528603, 0.526267, 0.448099, 0.445825, 0.369701, 0.367485,
0.315658, 0.325798, 0.341207, 0.351098, 0.366134, 0.375788, 0.390468,
0.399897, 0.414237, 0.42345, 0.437466, 0.446473, 0.46018, 0.46899,
0.4824, 0.491022, 0.504149, 0.51259, 0.525444, 0.533712, 0.546306,
0.554408, 0.56675
(takes about 1.3 seconds)
Alternatively, use
Table[u[i], i, 1, 100]
(same result). Your combination of For and Print shows the results but doesn't let you keep using them for more calculations.
answered Apr 12 at 4:43
RomanRoman
5,56111131
$endgroup$
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
$begingroup$
delayedassignments definitely sound slower thanimmediate, even if I have never worked with Mathematica
$endgroup$
– Roland
Apr 12 at 10:07
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
add a comment |
$begingroup$
I recommend you learn the distinction between immediate (=) and delayed (:=) assignments. They make the difference between slow and fast code here. Start with this tutorial or this book chapter, then look at memoization.
s = 0.3405;
e = 1.6539*10^-21;
u[0] = 0.;
u[1] = 0.1;
A[r_] = Piecewise[r - 2.5 s - 48*e*s^12*r^-13 + 24*e*s^6*r^-7, r > 2.5 s,
-48*e*s^12*r^-13 + 24*e*s^6*r^-7, s <= r <= 2.5 s,
r - s - 24*e*s^-1, r < s];
u[i_] := u[i] = x /. FindRoot[
u[i - 1] + 1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) - 0.9 A[x] == x, x, 1.]
Array[u, 100]
0.1, 1.77164, 1.37065, 1.04259, 0.887781, 0.708344, 0.59461,
0.457228, 0.367364, 0.296071, 0.256104, 0.20463, 0.208487, 1.20917,
1.04197, 0.939331, 0.879865, 0.827963, 0.774591, 0.72775, 0.67934,
0.63666, 0.592369, 0.553172, 0.512352, 0.476112, 0.438261, 0.404563,
0.369277, 0.339073, 0.321616, 0.301118, 0.296195, 0.224688, 0.273538,
0.31357, 0.33593, 0.366902, 0.38813, 0.417572, 0.437777, 0.465834,
0.48511, 0.511907, 0.530336, 0.55598, 0.573633, 0.598219, 0.615159,
0.638772, 0.655054, 0.677768, 0.693441, 0.715321, 0.73043, 0.751535,
0.766118, 0.786503, 0.800596, 0.820306, 0.833941, 0.852182, 0.85901,
0.874152, 0.871531, 0.78396, 0.781416, 0.696402, 0.693931, 0.611329,
0.608927, 0.528603, 0.526267, 0.448099, 0.445825, 0.369701, 0.367485,
0.315658, 0.325798, 0.341207, 0.351098, 0.366134, 0.375788, 0.390468,
0.399897, 0.414237, 0.42345, 0.437466, 0.446473, 0.46018, 0.46899,
0.4824, 0.491022, 0.504149, 0.51259, 0.525444, 0.533712, 0.546306,
0.554408, 0.56675
(takes about 1.3 seconds)
Alternatively, use
Table[u[i], i, 1, 100]
(same result). Your combination of For and Print shows the results but doesn't let you keep using them for more calculations.
answered Apr 12 at 4:43
RomanRoman
5,56111131
$endgroup$
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
$begingroup$
delayedassignments definitely sound slower thanimmediate, even if I have never worked with Mathematica
$endgroup$
– Roland
Apr 12 at 10:07
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
add a comment |
$begingroup$
I recommend you learn the distinction between immediate (=) and delayed (:=) assignments. They make the difference between slow and fast code here. Start with this tutorial or this book chapter, then look at memoization.
s = 0.3405;
e = 1.6539*10^-21;
u[0] = 0.;
u[1] = 0.1;
A[r_] = Piecewise[r - 2.5 s - 48*e*s^12*r^-13 + 24*e*s^6*r^-7, r > 2.5 s,
-48*e*s^12*r^-13 + 24*e*s^6*r^-7, s <= r <= 2.5 s,
r - s - 24*e*s^-1, r < s];
u[i_] := u[i] = x /. FindRoot[
u[i - 1] + 1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) - 0.9 A[x] == x, x, 1.]
Array[u, 100]
0.1, 1.77164, 1.37065, 1.04259, 0.887781, 0.708344, 0.59461,
0.457228, 0.367364, 0.296071, 0.256104, 0.20463, 0.208487, 1.20917,
1.04197, 0.939331, 0.879865, 0.827963, 0.774591, 0.72775, 0.67934,
0.63666, 0.592369, 0.553172, 0.512352, 0.476112, 0.438261, 0.404563,
0.369277, 0.339073, 0.321616, 0.301118, 0.296195, 0.224688, 0.273538,
0.31357, 0.33593, 0.366902, 0.38813, 0.417572, 0.437777, 0.465834,
0.48511, 0.511907, 0.530336, 0.55598, 0.573633, 0.598219, 0.615159,
0.638772, 0.655054, 0.677768, 0.693441, 0.715321, 0.73043, 0.751535,
0.766118, 0.786503, 0.800596, 0.820306, 0.833941, 0.852182, 0.85901,
0.874152, 0.871531, 0.78396, 0.781416, 0.696402, 0.693931, 0.611329,
0.608927, 0.528603, 0.526267, 0.448099, 0.445825, 0.369701, 0.367485,
0.315658, 0.325798, 0.341207, 0.351098, 0.366134, 0.375788, 0.390468,
0.399897, 0.414237, 0.42345, 0.437466, 0.446473, 0.46018, 0.46899,
0.4824, 0.491022, 0.504149, 0.51259, 0.525444, 0.533712, 0.546306,
0.554408, 0.56675
(takes about 1.3 seconds)
Alternatively, use
Table[u[i], i, 1, 100]
(same result). Your combination of For and Print shows the results but doesn't let you keep using them for more calculations.
answered Apr 12 at 4:43
RomanRoman
5,56111131
$endgroup$
I recommend you learn the distinction between immediate (=) and delayed (:=) assignments. They make the difference between slow and fast code here. Start with this tutorial or this book chapter, then look at memoization.
s = 0.3405;
e = 1.6539*10^-21;
u[0] = 0.;
u[1] = 0.1;
A[r_] = Piecewise[r - 2.5 s - 48*e*s^12*r^-13 + 24*e*s^6*r^-7, r > 2.5 s,
-48*e*s^12*r^-13 + 24*e*s^6*r^-7, s <= r <= 2.5 s,
r - s - 24*e*s^-1, r < s];
u[i_] := u[i] = x /. FindRoot[
u[i - 1] + 1/(i^2 (u[i - 1] - u[i - 2])^2) (u[i - 1] - u[i - 2]) - 0.9 A[x] == x, x, 1.]
Array[u, 100]
0.1, 1.77164, 1.37065, 1.04259, 0.887781, 0.708344, 0.59461,
0.457228, 0.367364, 0.296071, 0.256104, 0.20463, 0.208487, 1.20917,
1.04197, 0.939331, 0.879865, 0.827963, 0.774591, 0.72775, 0.67934,
0.63666, 0.592369, 0.553172, 0.512352, 0.476112, 0.438261, 0.404563,
0.369277, 0.339073, 0.321616, 0.301118, 0.296195, 0.224688, 0.273538,
0.31357, 0.33593, 0.366902, 0.38813, 0.417572, 0.437777, 0.465834,
0.48511, 0.511907, 0.530336, 0.55598, 0.573633, 0.598219, 0.615159,
0.638772, 0.655054, 0.677768, 0.693441, 0.715321, 0.73043, 0.751535,
0.766118, 0.786503, 0.800596, 0.820306, 0.833941, 0.852182, 0.85901,
0.874152, 0.871531, 0.78396, 0.781416, 0.696402, 0.693931, 0.611329,
0.608927, 0.528603, 0.526267, 0.448099, 0.445825, 0.369701, 0.367485,
0.315658, 0.325798, 0.341207, 0.351098, 0.366134, 0.375788, 0.390468,
0.399897, 0.414237, 0.42345, 0.437466, 0.446473, 0.46018, 0.46899,
0.4824, 0.491022, 0.504149, 0.51259, 0.525444, 0.533712, 0.546306,
0.554408, 0.56675
(takes about 1.3 seconds)
Alternatively, use
Table[u[i], i, 1, 100]
(same result). Your combination of For and Print shows the results but doesn't let you keep using them for more calculations.
answered Apr 12 at 4:43
RomanRoman
5,56111131
edited Apr 12 at 8:53
answered Apr 12 at 4:43
RomanRoman
5,56111131
answered Apr 12 at 4:43
RomanRoman
5,56111131
answered Apr 12 at 4:43
RomanRoman
5,56111131
5,56111131
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
$begingroup$
delayedassignments definitely sound slower thanimmediate, even if I have never worked with Mathematica
$endgroup$
– Roland
Apr 12 at 10:07
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
add a comment |
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
$begingroup$
delayedassignments definitely sound slower thanimmediate, even if I have never worked with Mathematica
$endgroup$
– Roland
Apr 12 at 10:07
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
$begingroup$
thank you very much. I really appreciate it.
$endgroup$
– morapi
Apr 12 at 6:35
1
1
$begingroup$
delayed assignments definitely sound slower than immediate, even if I have never worked with Mathematica$endgroup$
– Roland
Apr 12 at 10:07
$begingroup$
delayed assignments definitely sound slower than immediate, even if I have never worked with Mathematica$endgroup$
– Roland
Apr 12 at 10:07
2
2
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
$begingroup$
@Roland it's not just that one is necessarily faster or slower than the other, it's more that they are completely different things with very different applications. For some reason this point is often overlooked by beginners in Mathematica.
$endgroup$
– Roman
Apr 12 at 10:14
add a comment |
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How slow? How many minutes/seconds?
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– JonyD
Apr 12 at 8:28