Question

) Let α be a fixed positive real number, α > 0. For a sequence {xn}, let x1 > √ α, and define x2, x3, x4, · · · by the following recurrence relation xn+1 = 1 2 xn + α xn (a) Prove that {xn} decreases monotonically (in other words, xn+1 − xn ≤ 0 for all n). (b) Prove that {xn} is bounded from below. (Hint: use proof by induction to show xn > √ α for all n) (c) Show that limn→∞ xn = √ α

Answer #1

Rudin Ch 3 No 16. Fix a positive number α. Choose x1 > √ α,
and define a sequence x2, x3,. . . by the recursion formula x n+1 =
1 2 (xn + α /xn ). (a) Prove that xn decreases monotonically and
that lim n→∞ xn = √ α. Explain how we know xn decreases. Explain
the term monotonically

If (xn) ∞ to n=1 is a convergent sequence with limn→∞ xn = 0
prove that
lim n→∞ (x1 + x2 + · · · + xn)/ n = 0 .

Define a sequence (xn)n≥1 recursively by x1 = 1, x2 = 2, and xn
= ((xn−1)+(xn−2))/ 2 for n > 2. Prove that limn→∞ xn = x exists
and find its value.

Define a sequence (xn)n≥1 recursively by x1 = 1 and
xn = 1 + 1 /(xn−1) for n > 0. Prove that limn→∞ xn = x exists
and find its value.

Given that xn is a sequence of real numbers. If (xn) is a
convergent sequence prove that (xn) is bounded. That is, show that
there exists C > 0 such that |xn| less than or equal to C for
all n in N.

Let n ≥ 2 and x1, x2, ..., xn > 0 be such that x1 + x2 + · ·
· + xn = 1. Prove that √ x1 + √ x2 + · · · + √ xn /√ n − 1 ≤ x1/ √
1 − x1 + x2/ √ 1 − x2 + · · · + xn/ √ 1 − xn

Let the LFSR be xn+5 = xn +
xn+3, where the initial values are x0=0,
x1=1, x2=0, x3=0,
x4=0
(a) Compute first 24 bits of the following LFSR.
(b) What is the period?

Let 0 < θ < 1 and let (xn) be a sequence where
|xn+1 − xn| ≤ θn for n
= 1, 2, . . ..
a) Show that for any 1 ≤ n < m one has |xm −
xn| ≤ (θn/ 1-θ )*(1 − θ m−n ).
Conclude that (xn) is Cauchy
b)If lim xn = x* , prove the following error in
approximation (the "error in approximation" is the same as error
estimation in Taylor Theorem) in t:...

Consider a sequence defined recursively as X0=
1,X1= 3, and Xn=Xn-1+
3Xn-2 for n ≥ 2. Prove that Xn=O(2.4^n) and
Xn = Ω(2.3^n).
Hint:First, prove by induction that 1/2*(2.3^n) ≤ Xn
≤ 2.8^n for all n ≥ 0
Find claim, base case and inductive step. Please show step and
explain all work and details

Question about the Mathematical Real Analysis Proof
Show that if xn → 0 then √xn → 0.
Proof. Let ε > 0 be arbitrary. Since xn → 0 there is some N
∈N such that |xn| < ε^2 for all n > N. Then for all n > N
we have that |√xn| < ε
My question is based on the sequence convergence
definition it should be absolute an-a<ε but here
why we can take xn<ε^2 rather than ε?...

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