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The Special Number - e

The Special Number - e. Suppose that f(n) = ( (n+1) / n ) n. Then f(1) = ( 2 / 1 ) 1 = 2 1 = 2. f(2) = ( 3 / 2 ) 2 = (1.5) 2 = 2.25. f(3) = ( 4 / 3 ) 3 = (1.33..) 3 = 2.370…. f(10) = ( 11 / 10 ) 10 = (1.1) 10 = 2.5937….

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The Special Number - e

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  1. The Special Number - e Suppose that f(n) = ((n+1)/n)n Then f(1) = (2/1)1 = 21 = 2 f(2) = (3/2)2 = (1.5)2 = 2.25 f(3) = (4/3)3 = (1.33..)3 = 2.370… f(10) = (11/10)10 = (1.1)10 = 2.5937… f(100) = (101/100)100 = (1.01)100 = 2.7048… f(1000) = (1001/1000)1000 = (1.001)1000 = 2.7169… f(100000) = (100001/100000)100000 = (1.00001)100000 = 2.7183… f(1000000) = (1000001/1000000)1000000 = (1.000001)1000000 = 2.7183…

  2. As n   f(n)  2.7183…. This value is given the name e ! Many real life growth and decay functions can be modelled mathematically using exponential functions with base e. This can be found on your calculator either by itself as ex or above the Inx key.

  3. Example1 e10 = 22026.5 e-0.3 = 0.741 etc Example2 Solve ex = 10 to 2 decimal places. ******** Using trial & error n = 2 e2 = 7.39 too small n = 3 e3 = 20.09 too large n = 2.3 e2.3 = 9.97 too small (***) n = 2.4 e2.4 = 11.02 too large n = 2.32 e2.32 = 10.18 too large n = 2.31 e2.31 = 10.07 too large To 2 dec places if ex = 10 then x = 2.30

  4. Example3 A radioactive substance decays according to the formula At = A0e(-2.1t) where A0 is the original amount and At is the amount remaining after t hours. (a) Find how much of 530g remains after 31/2 hours. (b) Show that the half-life is 20 mins ! ******** (a)At = A0e(-2.1t) = 530 Xe(-2.1 X 3.5) = 530 Xe(-7.35) = 0.34g

  5. (b) After one half-life amount left = 530g  2 = 265g If t = 20mins = 1/3hr A(1/3) = 530 x e(-2.1  3) = 530 x e(-0.7) = 263g (sufficiently close) ALTERNATIVELY We need At = 0.5A0 and At = A0e(-2.1t) So A0e(-2.1t) = 0.5A0 So e(-2.1t) = 0.5 Taking t = 1/3 we get e(-2.1  3) = 0.4966 Again sufficiently close to 0.5 so half-lifeis 20 mins.

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