A) Start a sequence with any two real numbers, for the third and successive terms take the mean of all of the previous terms.
B) Same as part A) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one less than the number of terms.
C) Same as part A) and B) but take the mean incorrectly as follows: add the numbers together but instead of dividing by the number of terms, divide by one more than the number of terms.
What happens to the sequence in each case? Generalize further.
As expected, when the true means are used, the first mean found is the mean of the two initial numbers, and thereafter the mean stays the same:
DO
INPUT a, b
tot = a + b
FOR n = 2 TO 45
newt = tot / (n)
PRINT newt
tot = tot + newt
NEXT
LOOP
? 2,5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
3.5
What is less readily understood is that the means are also stable in the other two conditions:
DO
INPUT a, b
tot = a + b
FOR n = 2 TO 45
newt = tot / (n-1)
PRINT newt
tot = tot + newt
NEXT
LOOP
? 2,5
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
so in part B, the "mean" is the sum of the two numbers, and remains such after each generation of "mean".
In part C the first "mean" and all subsequent ones is 1/3 the sum of the first two:
DO
INPUT a, b
tot = a + b
FOR n = 2 TO 45
newt = tot / (n + 1)
PRINT newt
tot = tot + newt
NEXT
LOOP
? 2,5
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333333
2.333334
2.333334
2.333334
2.333334
2.333334
2.333334
2.333334
2.333334
2.333334
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Posted by Charlie
on 2011-05-27 02:04:40 |