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Unformatted text preview: Section 3.4: Exponential growth and decay (concluded). Interest If you put $100 in the bank and it’s compounded yearly with 12% annual interest, after 1 year you get your original $100 plus $12 interest, for a total of $100 times 1.12 = $112 . If your money is compounded quarterly (i.e., 4 times per year) with 3% interest per quarter, after 1 year you get $100 times 1.03 times 1.03 times 1.03 times 1.03, or $100 × (1+.12/4) 4 ≈ $112.55 . If your money is compounded monthly, after 1 year you get $100 × (1+.12/12) 12 ≈ $112.68 . If your money is compounded continuously, after 1 year you get lim N →∞ $100 × (1+.12/ N ) N = $100 exp .12 ≈ $112.75 . Why does lim N →∞ (1+ r / N ) N equal e r ? One way to prove this is to reduce it to our definition of the number e as lim s → (1+ s ) 1/ s . lim N →∞ (1+ r / N ) N = ..?.. ..?.. = lim N →∞ [(1+ r / N ) N / r ] r (this step is just algebra) = lim s → 0+ [(1+ s ) 1/ s ] r (see Remark 1 below) = [ lim s → 0+ (1+ s ) 1/ s ] r (see Remark 2 below) = e r . Remark 1: Exercise 55 in section 1.6 asks you to show that lim x →∞ f ( x ) = lim t → 0+ f (1/ t ) and lim x → – ∞ f ( x ) = lim t → 0– f (1/ t ) (if these limits exist). This remains true if we replace 1/ t by C / t for any positive constant C . That is the principle being used here. (We won’t prove it in 92.141 this year, but you’re allowed to use it on all homework problems.) you’re allowed to use it on all homework problems....
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This note was uploaded on 02/13/2012 for the course MATH 141 taught by Professor Staff during the Fall '11 term at UMass Lowell.
 Fall '11
 Staff
 Calculus

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