Why is this recursive code so slow?












5












$begingroup$


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]]}]









share|improve this question











$endgroup$












  • $begingroup$
    How slow? How many minutes/seconds?
    $endgroup$
    – JonyD
    Apr 12 at 8:28
















5












$begingroup$


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]]}]









share|improve this question











$endgroup$












  • $begingroup$
    How slow? How many minutes/seconds?
    $endgroup$
    – JonyD
    Apr 12 at 8:28














5












5








5





$begingroup$


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]]}]









share|improve this question











$endgroup$




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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Apr 12 at 8:27









Roman

5,56111131




5,56111131










asked Apr 12 at 4:08









morapimorapi

355




355












  • $begingroup$
    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




$begingroup$
How slow? How many minutes/seconds?
$endgroup$
– JonyD
Apr 12 at 8:28










1 Answer
1






active

oldest

votes


















12












$begingroup$

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.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you very much. I really appreciate it.
    $endgroup$
    – morapi
    Apr 12 at 6:35






  • 1




    $begingroup$
    delayed assignments definitely sound slower than immediate, 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












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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









12












$begingroup$

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.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you very much. I really appreciate it.
    $endgroup$
    – morapi
    Apr 12 at 6:35






  • 1




    $begingroup$
    delayed assignments definitely sound slower than immediate, 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
















12












$begingroup$

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.






share|improve this answer











$endgroup$













  • $begingroup$
    thank you very much. I really appreciate it.
    $endgroup$
    – morapi
    Apr 12 at 6:35






  • 1




    $begingroup$
    delayed assignments definitely sound slower than immediate, 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














12












12








12





$begingroup$

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.






share|improve this answer











$endgroup$



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.







share|improve this answer














share|improve this answer



share|improve this answer








edited Apr 12 at 8:53

























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




    $begingroup$
    delayed assignments definitely sound slower than immediate, 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






  • 1




    $begingroup$
    delayed assignments definitely sound slower than immediate, 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
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– 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


















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