Hello,
I looked at the manual on interpolation, and I can do the examples,
but I don't understand why one needs an apostrophe to plot.
For instance,
(%i1) load(interpol);
(%o1) /usr/local/share/maxima/5.12.0/share/numeric/interpol.mac
(%i2) xx: makelist([i*.1,(i*.1)^2],i,1,10);
(%o2) [[0.1, 0.01], [0.2, .04000000000000001], [0.3,
.09000000000000002],
[0.4, 0.16], [0.5, 0.25], [.6000000000000001, .3600000000000001],
[.7000000000000001, .4900000000000001], [0.8, .6400000000000001],
[0.9, 0.81],
[1.0, 1.0]]
(%i3) linearinterpol(xx);
(%o3) (0.3 x - 0.02) charfun2(x, minf, 0.2)
+ (1.9 x - .8999999999999999) charfun2(x, 0.9, inf)
+ (1.7 x - .7199999999999995) charfun2(x, 0.8, 0.9)
+ (1.5 x - .5600000000000003) charfun2(x, .7000000000000001, 0.8)
+ (1.3 x - .4200000000000002) charfun2(x, .6000000000000001,
.7000000000000001) + (1.1 x - 0.3) charfun2(x, 0.5, .6000000000000001)
+ (.8999999999999998 x - .1999999999999999) charfun2(x, 0.4, 0.5)
+ (.7000000000000002 x - .1200000000000001) charfun2(x, 0.3, 0.4)
+ (0.5 x - 0.06) charfun2(x, 0.2, 0.3)
(%i4) f(x):= ''%;
(%o4) f(x) := (0.3 x - 0.02) charfun2(x, minf, 0.2)
+ (1.9 x - .8999999999999999) charfun2(x, 0.9, inf)
+ (1.7 x - .7199999999999995) charfun2(x, 0.8, 0.9)
+ (1.5 x - .5600000000000003) charfun2(x, .7000000000000001, 0.8)
+ (1.3 x - .4200000000000002) charfun2(x, .6000000000000001,
.7000000000000001) + (1.1 x - 0.3) charfun2(x, 0.5, .6000000000000001)
+ (.8999999999999998 x - .1999999999999999) charfun2(x, 0.4, 0.5)
+ (.7000000000000002 x - .1200000000000001) charfun2(x, 0.3, 0.4)
+ (0.5 x - 0.06) charfun2(x, 0.2, 0.3)
(%i5) f(2);
(%o5) 2.9
(%i6) plot2d('(f(x)),[x,0,1]);
Why does one need the apostrophe and the parentheses in the expression
in (%i6), and what does it mean?
TIA,
-sen
--
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| Sheldon E. Newhouse | e-mail: sen1 at math.msu.edu |
| Mathematics Department | |
| Michigan State University | telephone: 517-355-9684 |
| E. Lansing, MI 48824-1027 USA | FAX: 517-432-1562 |
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