A student John starts with the
familiar series 1 + 1/2 + 1/4 + 1/8 + . . . .
He then takes the average
of each adjacent pair of terms and
inserts it between the terms to
obtain the new series 1 + 3/4 + 1/2 + 3/8 + 1/4 + . . . .
He divides this by
two, because there are now twice as
many terms as before. That gives
1/2 + 3/8 + 1/4 + 3/16 + 1/8 + ...
He repeats
the process indefinitely. For example, the next pair of steps gives
1/2 + 7/16 + 3/8 + 5/16 + 1/4 + 7/32 + 3/16 + 5/32 + 1/8 + ...
then
1/4 + 7/32 + 3/16 + 5/32 + 1/8 + 7/64 + 3/32 + 5/64 + 1/16 + ...
What exact
limit will the series approach?
ser=sym(1); term=sym(1);
for i=1:50
term=term/2;
ser=[ser term];
end
for iter=1:10
avgs=(ser(1:end-1)+ser(2:end))/2;
stack=[ser;avgs 0];
newser=reshape(stack,[1,size(stack,1)*size(stack,2)]);
% disp(newser(1:10))
fprintf('%s + ',newser(1:9))
fprintf('%s\n',newser(10))
ser=newser(1:2*length(ser)-1)/2;
% disp(ser(1:10))
fprintf('%s + ',ser(1:9))
fprintf('%s\n',ser(10))
fprintf('%s %f %s\n\n',[sym(eval(sum(ser))) sum(eval(ser)) sym(eval(sum(ser)))-3/2])
end
In the groupings below, the first line gives the result of interposing the means between previously successive terms. The second line shows the result of halving each term. In each case, only the first 10 terms are shown.
The third line specifies the sum of the infinite series after that halving, first as a fraction and then the decimal. The last item on that line shows the difference between the sum and 1.5. For each successive iteration, this difference is halved, showing the limit is 3/2 or 1.5.
1 + 3/4 + 1/2 + 3/8 + 1/4 + 3/16 + 1/8 + 3/32 + 1/16 + 3/64
1/2 + 3/8 + 1/4 + 3/16 + 1/8 + 3/32 + 1/16 + 3/64 + 1/32 + 3/128
7/4 1.750000 1/4
1/2 + 7/16 + 3/8 + 5/16 + 1/4 + 7/32 + 3/16 + 5/32 + 1/8 + 7/64
1/4 + 7/32 + 3/16 + 5/32 + 1/8 + 7/64 + 3/32 + 5/64 + 1/16 + 7/128
13/8 1.625000 1/8
1/4 + 15/64 + 7/32 + 13/64 + 3/16 + 11/64 + 5/32 + 9/64 + 1/8 + 15/128
1/8 + 15/128 + 7/64 + 13/128 + 3/32 + 11/128 + 5/64 + 9/128 + 1/16 + 15/256
25/16 1.562500 1/16
1/8 + 31/256 + 15/128 + 29/256 + 7/64 + 27/256 + 13/128 + 25/256 + 3/32 + 23/256
1/16 + 31/512 + 15/256 + 29/512 + 7/128 + 27/512 + 13/256 + 25/512 + 3/64 + 23/512
49/32 1.531250 1/32
1/16 + 63/1024 + 31/512 + 61/1024 + 15/256 + 59/1024 + 29/512 + 57/1024 + 7/128 + 55/1024
1/32 + 63/2048 + 31/1024 + 61/2048 + 15/512 + 59/2048 + 29/1024 + 57/2048 + 7/256 + 55/2048
97/64 1.515625 1/64
1/32 + 127/4096 + 63/2048 + 125/4096 + 31/1024 + 123/4096 + 61/2048 + 121/4096 + 15/512 + 119/4096
1/64 + 127/8192 + 63/4096 + 125/8192 + 31/2048 + 123/8192 + 61/4096 + 121/8192 + 15/1024 + 119/8192
193/128 1.507812 1/128
1/64 + 255/16384 + 127/8192 + 253/16384 + 63/4096 + 251/16384 + 125/8192 + 249/16384 + 31/2048 + 247/16384
1/128 + 255/32768 + 127/16384 + 253/32768 + 63/8192 + 251/32768 + 125/16384 + 249/32768 + 31/4096 + 247/32768
385/256 1.503906 1/256
1/128 + 511/65536 + 255/32768 + 509/65536 + 127/16384 + 507/65536 + 253/32768 + 505/65536 + 63/8192 + 503/65536
1/256 + 511/131072 + 255/65536 + 509/131072 + 127/32768 + 507/131072 + 253/65536 + 505/131072 + 63/16384 + 503/131072
769/512 1.501953 1/512
1/256 + 1023/262144 + 511/131072 + 1021/262144 + 255/65536 + 1019/262144 + 509/131072 + 1017/262144 + 127/32768 + 1015/262144
1/512 + 1023/524288 + 511/262144 + 1021/524288 + 255/131072 + 1019/524288 + 509/262144 + 1017/524288 + 127/65536 + 1015/524288
1537/1024 1.500977 1/1024
1/512 + 2047/1048576 + 1023/524288 + 2045/1048576 + 511/262144 + 2043/1048576 + 1021/524288 + 2041/1048576 + 255/131072 + 2039/1048576
1/1024 + 2047/2097152 + 1023/1048576 + 2045/2097152 + 511/524288 + 2043/2097152 + 1021/1048576 + 2041/2097152 + 255/262144 + 2039/2097152
3073/2048 1.500488 1/2048
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Posted by Charlie
on 2024-06-11 21:47:00 |
solution
