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Infinite SequenceDate: 09/28/97 at 23:32:19 From: Moni Subject: Calculus - infinite series I'm having trouble with this question: From f:n ---> 3 + (-1/2)^n where n belongs to the set of natural numbers Describe a strip that contains all but a finite number of points of the graph of f I know that you first have to find the limit, which I think is 3, and then choose E = 0.001 2 cases: L + E = 3 + 0.001 and L - E = 3 - 0.001 f(n) > 3.001 and f(n) < 2.999 But I don't know what to do now. Thanks.
Date: 09/29/97 at 11:19:30
From: Doctor Rob
Subject: Re: Calculus - infinite series
Point 1: This is an infinite *sequence*, not a *series*. An infinite
series is the sum of the terms of an infinite sequence.
Point 2: The graph of f is the set of pairs {(n, f(n)): n is a
natural number}. A strip of this graph is defined by
inequalities involving n and/or f(n).
You have made a good start. Yes, the limit L = 3, and you have chosen
E = 0.001, which is fine. Your inequalities, however, are backwards,
I think, since all but a finite number of values of f satisfy
2.999 < f(n) < 3.001
These inequalities define a strip in the plane extending to infinity
in both positive and negative n directions which contains all but a
finite number of the points of the graph of f. If you mean the strip
to extend only to infinity in the positive n direction, add the
condition n >= 1.
Perhaps you meant f(n) > 3.001 and f(n) < 2.999 to be the start of the
process of finding excluded values of n. If so, that is the way to
start:
f(n) = 3 + (-1/2)^n > 3.001
<==> (-1/2)^n > 0.001
<==> n is even and (1/4)^(n/2) > 0.001
<==> (n/2)*log(1/4) > log(0.001)
<==> -n*log(2) > -3
<==> n < 3/log(2) ~=~ 9.96578
f(n) = 3 + (-1/2)^n < 2.999
<==> (-1/2)^n < -0.001
<==> n is odd and (-1/2)*(1/4)^((n-1)/2) < -0.001
<==> (1/4)^((n-1)/2) > 0.002
<==> -(n-1)*log(2) > -3 + log(2)
<==> n < 3/log(2) ~=~ 9.96578
Thus points are excluded if and only if n < 10. Thus there are
exactly nine points of the graph of f not in your strip.
-Doctor Rob, The Math Forum
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