Patrick B. answered • 07/02/20
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Math and computer tutor/teacher
1a)
YES,
Let x = 0.99999....
Then 10x = 9.99999.....
Subtracting the equations, 10x - x = 9.99999... - 0.99999....
the repeating digits of 9 cancel:
9x = 9
x=1
1b) for this reason
1.9999.... = 2
2.9999.... = 3
etc
2a) A(n) = 3 + 5(n-1)
= 3 + 5n - 5
= 5n-2
A(n) = A(n-1) + 5 where A1=3
2b) A(1000) = 5(1000)-2
= 5000-2
= 4998
2c-2d) the explicit rule is faster;
Mark M.
Some texts refer to A(0) as the first term while others refer to A(1) as the first term. In the former case, A(n) = 3 + 5n. In the latter case, A(n) = 5n - 2. Thus, the nomenclature of the 1000th term can be confusing. Thankfully most books use A(1), but in computer programming, A(0) is often understand is the first term if you don't have control over the iteration process.
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07/03/20
Mark M.
1b) is a bit of arithmetic chicanery. If 0.9 repeats without end then how can 10 times that number have one more 9? That is how many 9's is one more than an unlimited number of 9's? 0.9 repeating = 1 in the theory of limits. You tell me how close to get to 1 and I shall tell you how many 9's are necessary.
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07/03/20
Tom K.
1b) 1.0000000000000 (the 0 is repeated) 2d) I would also add that the explicit rule also shows the long-term trend. With computers, the recursive rule for reasonable n often is faster than the time it takes to figure out the closed form in many cases.07/02/20