Difference between revisions of "Matlab/Octave finty"
From RoboWiki
(New page: Niektoré zaujímavé konštrukcie a postupy '''How to find square root of 2 in 15 significant figures''' octave:1> format long octave:2> sqrt(2) ans = 1.41421356237310 Alebo iný...) |
|||
| Line 15: | Line 15: | ||
x = (x + 2/x) / 2; | x = (x + 2/x) / 2; | ||
fprintf('%18.15f\n', x) | fprintf('%18.15f\n', x) | ||
| + | |||
| + | |||
| + | '''How do I evaluate 1^2 + 2^2 +3^2 +...+100^2''' | ||
| + | |||
| + | sumsq(1:100) | ||
| + | |||
| + | a=1:100; | ||
| + | b=a.^2; | ||
| + | c=sum(b) | ||
| + | |||
| + | s = 0; | ||
| + | for i = 1:100, | ||
| + | s = s + i^2; | ||
| + | endfor | ||
| + | |||
| + | or you could vectorize the loop and get: | ||
| + | |||
| + | s = sum((1:100).^2); | ||
| + | |||
| + | Or | ||
| + | |||
| + | (1:100)*(1:100)' | ||
| + | |||
| + | Or | ||
| + | 100*101*201/6 | ||
| + | because (sum(n^2)) = n*(n+1)*(2*n+1)/6) | ||
| + | |||
| + | Or | ||
| + | |||
| + | norm(1:100)^2 | ||
| + | |||
| + | This will also work | ||
| + | |||
| + | 100*var(-100:100) | ||
Revision as of 09:05, 22 February 2012
Niektoré zaujímavé konštrukcie a postupy
How to find square root of 2 in 15 significant figures
octave:1> format long octave:2> sqrt(2) ans = 1.41421356237310
Alebo iný postup
x = 1.5;
x = (x + 2/x) / 2;
x = (x + 2/x) / 2;
x = (x + 2/x) / 2;
x = (x + 2/x) / 2;
fprintf('%18.15f\n', x)
How do I evaluate 1^2 + 2^2 +3^2 +...+100^2
sumsq(1:100)
a=1:100; b=a.^2; c=sum(b)
s = 0; for i = 1:100, s = s + i^2; endfor
or you could vectorize the loop and get:
s = sum((1:100).^2);
Or
(1:100)*(1:100)'
Or
100*101*201/6
because (sum(n^2)) = n*(n+1)*(2*n+1)/6)
Or
norm(1:100)^2
This will also work
100*var(-100:100)
