math4b-7 - Math 4B Lecture 7 Doug Moore MONSIEUR JOURDAIN...

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Math 4B Lecture 7 April 22, 2013 Doug Moore
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MONSIEUR JOURDAIN: By my faith! For more than forty years I have been speaking prose without knowing anything about it, and I am much obliged to you for having taught me that. (Quo- tation from Moliere, Le Bourgeois gentilhomme ) BANKER: By my faith! For more than forty years I have been calculating the Cauchy-Euler polygon without knowing anything about it, and I am much obliged to you for having taught me that.
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The sad fact of life is that most differential equations cannot be solved exactly, so one often uses approximations. Even the simplest initial-value problem dy dt = ky, y (0) = 1 , has the solution y ( t ) = e kt for its solution; to use this would require the use of the expo- nential function. Bankers, for example, are afraid to use this solution because they would have trouble explaining it to their customers. So they use an approximation, called the Cauchy- Euler polygon to find an approximate solution instead. They call this approximation compound interest .
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We imagine that y ( t ) is the amount of indebtedness of a credit card account at time t , the interest rate being k . We suppose that the bank would like to maximize its rate of return by com- pounding sufficiently frequently that the amount of indebtedness at the end of one year will be a good approximation to e k , cor- responding to an instantaneous growth rate of k . How often should the bank compound? Quarterly? Monthly? Daily? Let us suppose that the interest rate on the credit card is 18%, so we take k = . 18 to represent the instantaneous growth rate. With no compounding, $100 of debt will grow to (1+ . 18)($100) or $118 in one year.
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Quarterly compounding corresponds to n = 4. In this case, 1 + . 18 4 4 = 1 . 19252 , so a debt of $100 grows to a debt of $119.25. Monthly compounding corresponds to n = 12. In this case, 1 + . 18 12 12 = 1 . 19562 , so a debt of $100 grows to a debt of $119.56. Daily compounding corresponds to n = 365. In this case, 1 + . 18 365 365 = 1 . 19716 , so a debt of $100 grows to a debt of $119.72.
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Since e ( . 18) = lim n →∞ 1 + . 18 n n = 1 . 19722 , under instantaneous compounding a debt of $100 grows to a debt of $119.72. Compound interest gives a discrete version of exponential growth which gets better as the time interval for compounding gets smaller. Compound interest gives an easily understood example of the Cauchy polygon approximation to solutions of differential equations.
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PICARD’S THEOREM. Suppose that the real-valued function
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  • Winter '14
  • Math, Cauchy-Euler polygon, Euler polygon

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