{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

AnalysisExam2 - x n-x n 1 2 = x 2 n 1-10 = ⇒ x 2 n 1-10...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Real Analysis: Exam II Extra Credit David Joseph Stith 3. Problem. Let ( x n ) be the sequence defined recursively by x 1 = 2 and x n +1 = x n 2 + 5 x n for all n 1 . Show that ( x n ) converges and find the limit of the sequence. To show that ( x n ) converges, we will show that ( x n ) is bounded below and that ( x n ) is ultimately decreasing. It will then follow that ( x n ) is bounded above as well and hence, by the Monotone Convergence Theorem, converges. To show that ( x n ) is bounded below, we will show by Mathematical Induction that x n > 0 for all n N . We have that x 1 = 2 > 0. Suppose x k > 0 for some k N . Then x k / 2 > 0 and 5 /x k > 0 so that x k +1 = x k 2 + 5 x k > 0. Therefore x k > 0 = x k +1 > 0. Therefore by Mathematical Induction, x n > 0 for all n N . Therefore ( x n ) is bounded below. Furthermore we can easily see that x n R for all n N . Now we will show that x n +1 x n +2 for all n N and hence that the one-tail of ( x n ) is decreasing. Let n N . Then x n +1 = x n 2 + 5 x n = x n +1 = x 2 n + 10 2 x n = x 2 n - 2 x n x n +1 + 10 = 0 = x 2 n - 2 x n x n +1 = - 10 = x 2 n - 2 x n x n +1 + x 2 n +1 = - 10 + x 2 n +1
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ( x n-x n +1 ) 2 = x 2 n +1-10 = ⇒ x 2 n +1-10 ≥ = ⇒ x 2 n +1 2 ≥ 5 = ⇒ x n +1 2 ≥ 5 x n +1 since ∀ n ∈ N ,x n > = ⇒ x n +1 ≥ x n +1 2 + 5 x n +1 = ⇒ x n +1 ≥ x n +2 Therefore x n +1 ≥ x n +2 for all n ∈ N . Therefore x ( n ) is ultimately decreasing. Therefore < x n < x 2 for all n ∈ N so that ( x n ) is bounded as well as monotone. Therefore by the Monotone Convergence Theorem, ( x n ) is convergent. 1 To fnd lim( x n ), let x = lim( x n ) = lim( x n +1 ) by virtue oF the Fact that any tail oF ( x n ) also converges to x . Then, x = lim( x n +1 ) = lim( x n ) 2 + 5 lim( x n ) = ⇒ x = x 2 + 5 x = ⇒ 2 x 2 = x 2 + 10 = ⇒ x 2 = 10 = ⇒ x = √ 10 since ∀ n ∈ N ,x n > ThereFore lim( x n ) = √ 10. Q.E.D. 2...
View Full Document

{[ snackBarMessage ]}