Matlab/Octave finty - Revision history

Balogh at 14:24, 25 February 2012

2012-02-25T14:24:19Z

Balogh at 08:18, 22 February 2012

2012-02-22T08:18:21Z

Balogh at 08:05, 22 February 2012

2012-02-22T08:05:15Z

Balogh: 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...

2012-02-17T10:30:36Z

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...

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 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)

← Older revision		Revision as of 14:24, 25 February 2012
Line 15:		Line 15:
	x = (x + 2/x) / 2;		x = (x + 2/x) / 2;
	fprintf('%18.15f\n', x)		fprintf('%18.15f\n', x)
		+
		+	Este iny postup
		+
		+	printf ("%.15f\n", sqrt(2));
		+	produces
		+	1.414213562373095
		+
		+	With
		+
		+	function x = approxroot2(n)
		+	x = 2;
		+	while n
		+	printf ("%.15f\n", x - 1);
		+	x = 2 + 1/x;
		+	n--;
		+	endwhile
		+	x -= 1;
		+	printf ("%.15f\n", x);
		+	endfunction
		+
		+	approxroot2(25);
		+	produces
		+	1.000000000000000
		+	1.500000000000000
		+	1.400000000000000
		+	1.416666666666667
		+	1.413793103448276
		+	1.414285714285714
		+	1.414201183431953
		+	1.414215686274510
		+	1.414213197969543
		+	1.414213624894870
		+	1.414213551646055
		+	1.414213564213564
		+	1.414213562057320
		+	1.414213562427273
		+	1.414213562363800
		+	1.414213562374690
		+	1.414213562372821
		+	1.414213562373142
		+	1.414213562373087
		+	1.414213562373096
		+	1.414213562373095
		+	1.414213562373095
		+	1.414213562373095
		+	1.414213562373095
		+	1.414213562373095
		+	1.414213562373095
		+

@@ Line 17: / Line 17: @@
-'''How do I evaluate  1^2 + 2^2 +3^2 +...+100^2'''
+'''How do I evaluate <math> 1^2 + 2^2 +3^2 + \dots +100^2</math>'''
   sumsq(1:100)
@@ Line 40: / Line 40: @@
 Or
 *101*201/6
-because (sum(n^2)) = n*(n+1)*(2*n+1)/6)
 Or

← Older revision		Revision as of 08:05, 22 February 2012
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
		+	100101201/6
		+	because (sum(n^2)) = n(n+1)(2*n+1)/6)
		+
		+	Or
		+
		+	norm(1:100)^2
		+
		+	This will also work
		+
		+	100*var(-100:100)