Evaluate:
n
∫ ⌊x*⌊x*⌊x⌋⌋⌋ dx
0
where ⌊p⌋ is the floor function. ie. the greatest integer ≤ p.
For each of n=1, 2, 3, 4.
Using
clearvars,clc
prev=0;
for x=1:.0000001:5
v=floor(x * floor(x * floor(x)));
if v~=prev
fprintf('%7.5f %4d\n',x,v)
end
prev=v;
end
to explore the domain (and a little beyond):
x f(x) beginning at
that x value
1.00000 1
2.00000 8
2.25000 9
2.50000 12
2.60000 13
2.80000 14
3.00000 27
3.11111 28
3.22222 29
3.33333 33
3.40000 34
3.50000 35
3.60000 36
3.66667 40
3.72727 41
3.81818 42
3.90909 43
4.00000 64
4.06250 65
4.12500 66
4.18750 67
4.25000 72
4.29412 73
4.35294 74
4.41176 75
4.47059 76
4.50000 81
4.55556 82
4.61111 83
4.66667 84
4.72222 85
4.75000 90
4.78947 91
4.84211 92
4.89474 93
4.94737 94
5.00000 125
The breaks in value of the function itself appear to be at fractions that are multiples of 1/4, 1/10, 1/9 and 1/11 (up to x=4). Therefore doing this in symbolic (exact) values should be done in increments of 1/1980 to hit all these exactly:
tot=sym(0);
prev=sym(0);prevx=sym(0);
for u=1:4*1980
x=sym(u/1980);
v=floor(x * floor(x * floor(x)));
if v~=prev
incr=(x-prevx)*(prev);
tot=tot+incr;
% disp([x v incr tot eval(tot)])
fprintf('%7s %3s %6s %12s %17.13f\n',x, v, incr, tot, eval(tot))
prev=v; prevx=x;
end
end
resulting in the following table:
new prev f(x) accumulated sum
x f(x) * delta x exact decimal
1 1 0 0 0.0000000000000
2 8 1 1 1.0000000000000
9/4 9 2 3 3.0000000000000
5/2 12 9/4 21/4 5.2500000000000
13/5 13 6/5 129/20 6.4500000000000
14/5 14 13/5 181/20 9.0500000000000
3 27 14/5 237/20 11.8500000000000
28/9 28 3 297/20 14.8500000000000
29/9 29 28/9 3233/180 17.9611111111111
10/3 33 29/9 1271/60 21.1833333333333
17/5 34 11/5 1403/60 23.3833333333333
7/2 35 17/5 1607/60 26.7833333333333
18/5 36 7/2 1817/60 30.2833333333333
11/3 40 12/5 1961/60 32.6833333333333
41/11 41 80/33 23171/660 35.1075757575758
42/11 42 41/11 25631/660 38.8348484848485
43/11 43 42/11 28151/660 42.6530303030303
4 64 43/11 30731/660 46.5621212121212
The requested integrals are therefore
n integral
exact decimal
1 0 0.0000000000000
2 1 1.0000000000000
3 237/20 11.8500000000000
4 30731/660 46.5621212121212
Wolfram Alpha verifies
integral_0^4 floor(x floor(x floor(x))) dx = 30731/660˜46.5621
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Posted by Charlie
on 2022-09-20 10:54:05 |
computer solution; computer varification