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Calc_Chap09 - MATH1131 Mathematics 1A Calculus Chapter 09...

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MATH1131 – Mathematics 1A Calculus Chapter 09 Logarithmic and Exponential Functions Dr. Thanh Tran School of Mathematics and Statistics The University of New South Wales Sydney, Australia 1 1 The natural log function 2 The exponential function 2 Let a be a positive real number. We can define a p for all p Z . E.g. a 0 = 1, a 2 = a × a , a 3 = 1 / a 3 . We can also define a 1 / q for q = 1 , 2 , 3 , . . . as the q th of a , i.e., b = a 1 / q if b q = a . Thus we can define a p / q with p , q Z , q > 0. How do we define a 3 or a π ? We will use the logarithmic and exponential functions. (You will see later a 3 = e 3 ln a .) The logarithmic and exponential functions are used in high school mathematics, but they are never defined rigorously. We will use the fundamental theorem of calculus to define the logarithmic function as an integral . The exponential function will be defined as the inverse of the log function. 3 The natural log function Definition 1 The function ln : ( 0 , ) −→ R is defined by the formula ln ( x ) = integraldisplay x 1 1 t dt . ln is read ‘ell en’ or ‘log’. When x > 1, ln x is the area of 1 x t y = 1 / t The natural log function 4
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Properties of ln ( x ) Theorem 2 For all x , y > 0 and r Q there hold 1 ln ( xy ) = ln x + ln y 2 ln ( x / y ) = ln x ln y 3 ln ( x r ) = r ln x 4 Signs of ln ( x ) : ln x > 0 if x > 1 , ln x = 0 if x = 1 , ln x < 0 if x < 1 ; 5 ln x → ∞ as x → ∞ 6 ln x → −∞ as x 0 + The natural log function 5 Proof: 1 By letting t = xu we have ln ( xy ) = integraldisplay xy 1 dt t = integraldisplay x 1 dt t + integraldisplay xy x dt t = ln x + integraldisplay y 1 du u = ln x + ln y . 2 By letting t = x / u we have ln ( x / y ) = integraldisplay x / y 1 dt t = integraldisplay x 1 dt t + integraldisplay x / y x dt t = ln x integraldisplay y 1 du u = ln x ln y . 3 By letting t = s r we have ln ( x r ) = integraldisplay x r 1 dt t = r integraldisplay x 1 ds s = r ln x . The natural log function 6 4 Use the fact that 1 / t > 0 when t > 0. 5 Considering the lower sum of the integral integraldisplay x 1 dt t as in the diagram, we obtain t y 1 2 4 8 y = 1 t integraldisplay 2 n 1 dt t 1 × 1 2 + 2 × 1 4 + 4 × 1 8 + · · · + 2 n 1 × 1 2 n = 1 2 + 1 2 + 1 2 + · · · + 1 2
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