r0 = 1, s0 = 0, and rn+1 = rn/2, sn+1 = sn + rn+1 for n ≥ 0.
(a) What are the formulas for the nth terms rn and sn of these sequences?
(b) What is the floating point binary representation of sn?
(c) Write a MATLAB program which generates the sequences (rn) and (sn) recursively, and
run it on your computer until the computed values satisfy sn+1 = sn. To do this you can
use a for loop with a break statement. Display the sequence of values of n and sn in two
columns.
(d) What does this tell you about the storage of real numbers in your computer (assuming it
uses binary representation)?
(a) rn = 2^(-n), sn = 1 - 2^(-n)
(b) 001111111110 (1) n-1 times and (0) 53-n times
example for n = 4
0011111111101110000000000000000000000000000000000000000000000000
(c)
>> r = 1;
s = 0;
for n = 0:100
fprintf('n = %g, sn = %24.20g\n',n,s)
r = r/2;
s1 = s + r;
if s1 ~=s
s = s1;
else
break;
end;
end;
n = 0, sn = 0
n = 1, sn = 0.5
n = 2, sn = 0.75
n = 3, sn = 0.875
n = 4, sn = 0.9375
n = 5, sn = 0.96875
n = 6, sn = 0.984375
n = 7, sn = 0.9921875
n = 8, sn = 0.99609375
n = 9, sn = 0.998046875
n = 10, sn = 0.9990234375
n = 11, sn = 0.99951171875
n = 12, sn = 0.999755859375
n = 13, sn = 0.9998779296875
n = 14, sn = 0.99993896484375
n = 15, sn = 0.999969482421875
n = 16, sn = 0.9999847412109375
n = 17, sn = 0.99999237060546875
n = 18, sn = 0.999996185302734375
n = 19, sn = 0.9999980926513671875
n = 20, sn = 0.99999904632568359375
n = 21, sn = 0.99999952316284179688
n = 22, sn = 0.99999976158142089844
n = 23, sn = 0.99999988079071044922
n = 24, sn = 0.99999994039535522461
n = 25, sn = 0.9999999701976776123
n = 26, sn = 0.99999998509883880615
n = 27, sn = 0.99999999254941940308
n = 28, sn = 0.99999999627470970154
n = 29, sn = 0.99999999813735485077
n = 30, sn = 0.99999999906867742538
n = 31, sn = 0.99999999953433871269
n = 32, sn = 0.99999999976716935635
n = 33, sn = 0.99999999988358467817
n = 34, sn = 0.99999999994179233909
n = 35, sn = 0.99999999997089616954
n = 36, sn = 0.99999999998544808477
n = 37, sn = 0.99999999999272404239
n = 38, sn = 0.99999999999636202119
n = 39, sn = 0.9999999999981810106
n = 40, sn = 0.9999999999990905053
n = 41, sn = 0.99999999999954525265
n = 42, sn = 0.99999999999977262632
n = 43, sn = 0.99999999999988631316
n = 44, sn = 0.99999999999994315658
n = 45, sn = 0.99999999999997157829
n = 46, sn = 0.99999999999998578915
n = 47, sn = 0.99999999999999289457
n = 48, sn = 0.99999999999999644729
n = 49, sn = 0.99999999999999822364
n = 50, sn = 0.99999999999999911182
n = 51, sn = 0.99999999999999955591
n = 52, sn = 0.99999999999999977796
n = 53, sn = 0.99999999999999988898
n = 54, sn = 1
(d) double-precision format can be used to store 53 binary digits or approximately 16 decimal digits of a number in decimal format.
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