Taking a long time to produce output
3
$begingroup$
In the following, I want to know b[1000],
but it is taking a very long time even for b[30].
Please help
a[0]=1.
a[n_]:=a[n - 1] + 1./ Sqrt[a[n - 1]];
b[n_] := (a[n])^3/n^2;
b[30]
programming recursion sequence
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
$endgroup$
add a comment |
3
$begingroup$
In the following, I want to know b[1000],
but it is taking a very long time even for b[30].
Please help
a[0]=1.
a[n_]:=a[n - 1] + 1./ Sqrt[a[n - 1]];
b[n_] := (a[n])^3/n^2;
b[30]
programming recursion sequence
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
$endgroup$
$begingroup$
Every time you callbit is in turn callingarecursively. Try the simple change ofa[n_]:=a[n]=this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…
$endgroup$
– enano9314
Aug 27 '18 at 4:32
1
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37
add a comment |
3
3
3
$begingroup$
In the following, I want to know b[1000],
but it is taking a very long time even for b[30].
Please help
a[0]=1.
a[n_]:=a[n - 1] + 1./ Sqrt[a[n - 1]];
b[n_] := (a[n])^3/n^2;
b[30]
programming recursion sequence
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
$endgroup$
In the following, I want to know b[1000],
but it is taking a very long time even for b[30].
Please help
a[0]=1.
a[n_]:=a[n - 1] + 1./ Sqrt[a[n - 1]];
b[n_] := (a[n])^3/n^2;
b[30]
programming recursion sequence
programming recursion sequence
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
edited Aug 27 '18 at 4:35
Carl Woll
68.3k390176
68.3k390176
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
asked Aug 27 '18 at 4:24
Sachin KumarSachin Kumar
356110
356110
$begingroup$
Every time you callbit is in turn callingarecursively. Try the simple change ofa[n_]:=a[n]=this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…
$endgroup$
– enano9314
Aug 27 '18 at 4:32
1
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37
add a comment |
$begingroup$
Every time you callbit is in turn callingarecursively. Try the simple change ofa[n_]:=a[n]=this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…
$endgroup$
– enano9314
Aug 27 '18 at 4:32
1
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37
$begingroup$
Every time you call
b it is in turn calling a recursively. Try the simple change of a[n_]:=a[n]= this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…$endgroup$
– enano9314
Aug 27 '18 at 4:32
$begingroup$
Every time you call
b it is in turn calling a recursively. Try the simple change of a[n_]:=a[n]= this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…$endgroup$
– enano9314
Aug 27 '18 at 4:32
1
1
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37
add a comment |
1 Answer
1
active
oldest
votes
7
$begingroup$
Use memoization:
a[0] = 1.;
a[n_] := a[n] = a[n - 1] + 1./ Sqrt[a[n - 1]]
b[n_] := (a[n])^3/n^2;
b[1000] //AbsoluteTiming
0.052607, 2.2588
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
$endgroup$
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
7
$begingroup$
Use memoization:
a[0] = 1.;
a[n_] := a[n] = a[n - 1] + 1./ Sqrt[a[n - 1]]
b[n_] := (a[n])^3/n^2;
b[1000] //AbsoluteTiming
0.052607, 2.2588
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
$endgroup$
add a comment |
7
$begingroup$
Use memoization:
a[0] = 1.;
a[n_] := a[n] = a[n - 1] + 1./ Sqrt[a[n - 1]]
b[n_] := (a[n])^3/n^2;
b[1000] //AbsoluteTiming
0.052607, 2.2588
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
$endgroup$
add a comment |
7
7
7
$begingroup$
Use memoization:
a[0] = 1.;
a[n_] := a[n] = a[n - 1] + 1./ Sqrt[a[n - 1]]
b[n_] := (a[n])^3/n^2;
b[1000] //AbsoluteTiming
0.052607, 2.2588
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
$endgroup$
Use memoization:
a[0] = 1.;
a[n_] := a[n] = a[n - 1] + 1./ Sqrt[a[n - 1]]
b[n_] := (a[n])^3/n^2;
b[1000] //AbsoluteTiming
0.052607, 2.2588
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
answered Aug 27 '18 at 4:33
Carl WollCarl Woll
68.3k390176
68.3k390176
add a comment |
add a comment |
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$begingroup$
Every time you call
bit is in turn callingarecursively. Try the simple change ofa[n_]:=a[n]=this should run quite quickly even for large n. Though be warned that it will use up memory, since you are trading speed for storage here. See this for more details. reference.wolfram.com/language/tutorial/…$endgroup$
– enano9314
Aug 27 '18 at 4:32
1
$begingroup$
@enano9314, Thanks a lot for your help
$endgroup$
– Sachin Kumar
Aug 27 '18 at 4:37