Math Problem Statement

Do e only

Do e only (E = {x | x = n^3 / (2n^2 + 1), n ∈ ℕ})

Solution

To analyze set E={xx=n32n2+1,nN}E = \{x \mid x = \frac{n^3}{2n^2 + 1}, n \in \mathbb{N} \}, let’s proceed step by step:


Step 1: Examine the general form of the elements of EE

Each xx in the set EE is given by: xn=n32n2+1where nN(natural numbers).x_n = \frac{n^3}{2n^2 + 1} \quad \text{where } n \in \mathbb{N} \, \text{(natural numbers)}.

Here:

  • The numerator is n3n^3,
  • The denominator is 2n2+12n^2 + 1.

Step 2: Behavior of xnx_n as nn \to \infty

To analyze the behavior of xnx_n, divide the numerator and denominator by n2n^2: xn=n32n2+1=n3/n2(2n2+1)/n2=n2+1n2.x_n = \frac{n^3}{2n^2 + 1} = \frac{n^3 / n^2}{(2n^2 + 1) / n^2} = \frac{n}{2 + \frac{1}{n^2}}.

As nn \to \infty, the 1n2\frac{1}{n^2} term in the denominator approaches 0. Thus: limnxn=n2n2(grows without bound).\lim_{n \to \infty} x_n = \frac{n}{2} \to \frac{n}{2} \, (\text{grows without bound}).

Therefore, the elements of EE increase unbounded as nn \to \infty.


Step 3: Behavior for small values of nn

To determine the smallest element of the set, substitute small values of nn (starting from n=1n = 1):

  1. For n=1n = 1: x1=13212+1=12+1=13.x_1 = \frac{1^3}{2 \cdot 1^2 + 1} = \frac{1}{2 + 1} = \frac{1}{3}.

  2. For n=2n = 2: x2=23222+1=824+1=88+1=89.x_2 = \frac{2^3}{2 \cdot 2^2 + 1} = \frac{8}{2 \cdot 4 + 1} = \frac{8}{8 + 1} = \frac{8}{9}.

  3. For n=3n = 3: x3=33232+1=2729+1=2718+1=2719.x_3 = \frac{3^3}{2 \cdot 3^2 + 1} = \frac{27}{2 \cdot 9 + 1} = \frac{27}{18 + 1} = \frac{27}{19}.

Observe that: x1=13,x2=89,x3=2719.x_1 = \frac{1}{3}, \, x_2 = \frac{8}{9}, \, x_3 = \frac{27}{19}.

We can see that xnx_n increases as nn increases.


Step 4: Smallest element and behavior of EE

  • The smallest element of EE occurs at n=1n = 1, where x1=13x_1 = \frac{1}{3}.
  • There is no largest element because xnx_n \to \infty as nn \to \infty.

Step 5: Lub (Least Upper Bound) and Glb (Greatest Lower Bound)

  • Greatest Lower Bound (glb): The smallest element of EE is 13\frac{1}{3}.
  • Least Upper Bound (lub): Since xnx_n \to \infty, the set is unbounded above, so there is no finite lub.

Final Answer:

For the set EE:

  1. Smallest element: 13\frac{1}{3},
  2. Largest element: Does not exist (unbounded above),
  3. glb (infimum): 13\frac{1}{3},
  4. lub (supremum): \infty.

Let me know if you want further explanations or details!


Related Questions:

  1. What happens if the denominator of EE is modified to 2n212n^2 - 1?
  2. Can you prove that xnx_n is strictly increasing for all nNn \in \mathbb{N}?
  3. What would the behavior of xnx_n be if nn starts at a different integer (e.g., n2n \geq 2)?
  4. How would the sequence behave if nn were allowed to take real values?
  5. How can you calculate the asymptotic growth rate of xnx_n?

Tip:

To determine bounds of sequences, analyze their behavior as nn \to \infty and test small values systematically.

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Math Problem Analysis

Mathematical Concepts

Sequences
Limits
Infimum and Supremum

Formulas

x_n = n^3 / (2n^2 + 1)
Limit behavior for large n

Theorems

Behavior of sequences as n → ∞
Greatest Lower Bound (infimum)
Least Upper Bound (supremum)

Suitable Grade Level

Undergraduate or Advanced High School