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 Check out our web site! http://mathforum.org/dr.math/ |
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