math4b-11 - Math 4B Lecture 11 May 6 2013 Doug Moore...

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Math 4B Lecture 11 May 6, 2013 Doug Moore
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Einstein is wrong (but not too wrong) because actually log 2 = . 693147 ..... not . 72 . A good approximation, however, is log 2 . = . 7 So it should be the law of sevens, not the slightly misstated rule of 72 appearing in books on personal finance, and if T is doubling time, ce kT = 2 c e kT = 2 kT = log 2 T = log 2 k . = . 7 k .
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For example, 14% interest corresponds to k = . 14 and leads to a doubling time of T = . 7 . 14 = 5 years . If Afghanistan has a population growth rate of 3 . 5% per year, its population will double every T = . 7 . 035 = 20 years . The “law of sevens” can be done in your head. And it can be explained to the man in the street, even though the reason behind the rule requires an understanding of the equation of exponential growth and decay. If a mutual fund takes 2% of your balance in fees each year, how long does it take your account to fall to half the value it would have had without the fees?
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Let V = { functions f : R R which have continuous derivatives of all orders } . Fir f and g are elements of V , we can define their sum f + g V by ( f + g )( t ) = f ( t ) + g ( t ) . If f V and c R , we can define the scalar product cf V by ( cf )( t ) = cf ( t ) . These operations make V into what is called a vector space .
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A function L : V V is said to be a linear operator if it satisfies the conditions L ( c 1 f 1 + c 2 f 2 ) = c 1 L ( f 1 ) + c 2 L ( f 2 ) , for f 1 , f 2 V, c 1 , c 2 R . An example of a linear operator is the ordinary differentiation operator d/dt —it is a familiar fact from first-year calculus that this operator satisfies the linear operator condition. ( d/dt )( c 1 f 1 + c 2 f 2 ) = c 1 ( d/dt )( f 1 ) + c 2 ( d/dt )( f 2 ) . We will denote this linear operator by D , and will use D 2 to denote the linear operator which differentiates twice in succes- sion; thus D 2 ( f ) = ( d 2 /dt 2 )( f ). More linear operators can be constructed from these—for example, L = 3 D 2 + 5 D + 4 , L ( f ) = 3 d 2 f dt 2 + 5 df dt + 4 f.
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The key point about linear operators is that if L is a linear opera- tor, solutions to the equation L ( x ) = 0 must satisfy the principle of superposition, because L ( x 1 ) = 0 and L ( x 2 ) = 0 L ( c 1 x 1 + c 2 x 2 ) = 0 . 1. Which of the following are linear operators? I. L ( f ) = 3 df dt , II. L ( f ) = df dt 2 a. Both I and II b. I but not II c. II but not I d. Neither I nor II Correct answer: b
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Homogeneous linear differential equations can be expressed quite nicely in operator notation. For example, the equation d 2 x dt 2 - dx dt - 6 x = 0 is expressed as ( D 2 - D - 6)( x ) = 0 .
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