Analytic inequalities by Nicholas D. Kazarinoff

By Nicholas D. Kazarinoff

Acceptable for undergraduates or even highschool scholars, this article introduces a couple of historic theorems, explores inequalities and calculus, and covers smooth theorems, together with Bernstein's evidence of the Weierstrass approximation theorem and the Cauchy, Bunyakovskii, Hölder, and Minkowski inequalities. 1961 variation. comprises 28 figures.

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3 # b a 7 64. 7 5 a a 4 3 63. + a a 65. 8 3 + 10 15 66. 5 7 + 8 12 67. 9 2 7 7 68. 2 12 5 5 69. 13 4 18 9 70. 13 8 15 45 71. 2 20 30 3 72. 5 5 7 21 73. 7 3 , 6 5 74. 7 3 , 5 4 75. 8 4 , 9 15 76. 1 1 , 8 4 3 7 78. 10 , 10 9 79. 7 7 , 13 13 5 17 , 8 6 81. 82. 3 8 1 5 77. 12 , 80. 83. 9 1 2 84. 2 7 5 3 3 7 ! Aha Fraction Notation 91. In the table below, the top number can be factored in such a way that the sum of the factors is the bottom number. For example, in the first column, 56 is factored as 7 # 8, since 7 + 8 = 15, the bottom number.

A Student Notes The following rules can help you quickly determine whether 2, 3, or 5 is a factor of a number. EXAMPLE 5 Simplify: (a) 36 15 ; (b) . 40 24 SOLUTION a) Note that 5 is a factor of both 15 and 40: 15 3 = 40 8 3 = 8 3 = 8 2 is a factor of a number if the number is even (the ones digit is 0, 2, 4, 6, or 8). 3 is a factor of a number if the sum of its digits is divisible by 3. 5 is a factor of a number if its ones digit is 0 or 5. b) 36 2 = 24 2 3 = 2 3 = 2 #5 #5 Factoring the numerator and the denominator, using the common factor, 5 #5 Rewriting as a product of two fractions; 55 ‫ ؍‬1 5 # 1 = 3.

In fact, 0 cannot be the denominator of any fraction. Later in this chapter, we will discuss why denominators cannot be 0. MULTIPLICATION AND SIMPLIFICATION Recall from arithmetic that fractions are multiplied as follows. a b c d Multiplication of Fractions For any two fractions and , ac a # c = . b d bd (The numerator of the product is the product of the two numerators. ) EXAMPLE 4 Multiply: (a) 2 # 5 4 # 8 ; (b) . x y 3 7 We multiply numerators as well as denominators. 2#5 10 2 # 5 = # = a) 3 7 3 7 21 4 # 8 4#8 32 = # = b) x y x y xy SOLUTION Try Exercise 53.

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