Finally 5 may be proved as follows if λ ? l then l

Info icon This preview shows pages 198–200. Sign up to view the full content.

View Full Document Right Arrow Icon
Finally (5) may be proved as follows. If Λ = λ = l , then l - < φ ( n ) < l + for every positive value of and all sufficiently large values of n , so that φ ( n ) l . Conversely, if φ ( n ) l , then the inequalities above written hold for all sufficiently large values of n . Hence l - is inferior and l + superior, so that λ = l - , Λ 5 l + , and therefore Λ - λ 5 2 . As Λ - λ = 0, this can only be true if Λ = λ . Examples XXXII. 1. Neither Λ nor λ is affected by any alteration in any finite number of values of φ ( n ). 2. If φ ( n ) = a for all values of n , then m = λ = Λ = M = a . 3. If φ ( n ) = 1 /n , then m = λ = Λ = 0 and M = 1. 4. If φ ( n ) = ( - 1) n , then m = λ = - 1 and Λ = M = 1. 5. If φ ( n ) = ( - 1) n /n , then m = - 1, λ = Λ = 0, M = 1 2 . 6. If φ ( n ) = ( - 1) n { 1 + (1 /n ) } , then m = - 2, λ = - 1, Λ = 1, M = 3 2 . 7. Let φ ( n ) = sin nθπ , where θ > 0. If θ is an integer then m = λ = Λ = M = 0. If θ is rational but not integral a variety of cases arise. Suppose, e.g. , that θ = p/q , p and q being positive, odd, and prime to one another, and q > 1. Then φ ( n ) assumes the cyclical sequence of values sin( pπ/q ) , sin(2 pπ/q ) , . . . , sin { (2 q - 1) pπ/q } , sin(2 qpπ/q ) , . . . . It is easily verified that the numerically greatest and least values of φ ( n ) are cos( π/ 2 q ) and - cos( π/ 2 q ), so that m = λ = - cos( π/ 2 q ) , Λ = M = cos( π/ 2 q ) . The reader may discuss similarly the cases which arise when p and q are not both odd.
Image of page 198

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
[IV : 84] POSITIVE INTEGRAL VARIABLE 183 The case in which θ is irrational is more difficult: it may be shown that in this case m = λ = - 1 and Λ = M = 1. It may also be shown that the values of φ ( n ) are scattered all over the interval [ - 1 , 1] in such a way that, if ξ is any number of the interval, then there is a sequence n 1 , n 2 , . . . such that φ ( n k ) ξ as k → ∞ . * The results are very similar when φ ( n ) is the fractional part of . 83. The general principle of convergence for a bounded func- tion. The results of the preceding sections enable us to formulate a very im- portant necessary and sufficient condition that a bounded function φ ( n ) should tend to a limit, a condition usually referred to as the general principle of con- vergence to a limit. Theorem 1. The necessary and sufficient condition that a bounded func- tion φ ( n ) should tend to a limit is that, when any positive number is given, it should be possible to find a number n 0 ( ) such that | φ ( n 2 ) - φ ( n 1 ) | < for all values of n 1 and n 2 such that n 2 > n 1 = n 0 ( ) . In the first place, the condition is necessary . For if φ ( n ) l then we can find n 0 so that l - 1 2 < φ ( n ) < l + 1 2 when n = n 0 , and so | φ ( n 2 ) - φ ( n 1 ) | < (1) when n 1 = n 0 and n 2 = n 0 . In the second place, the condition is sufficient . In order to prove this we have only to show that it involves λ = Λ. But if λ < Λ then there are, however small may be, infinitely many values of n such that φ ( n ) < λ + and infinitely many such that φ ( n ) > Λ - ; and therefore we can find values of n 1 and n 2 , each greater than any assigned number n 0 , and such that φ ( n 2 ) - φ ( n 1 ) > Λ - λ - 2 , which is greater than 1 2 - λ ) if is small enough. This plainly contradicts the inequality (1). Hence λ = Λ, and so φ ( n ) tends to a limit.
Image of page 199
Image of page 200
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern