Math Problem Statement

Over the given interval I determine the sets of uniform convergence for: f n (x)= sin(nx) n

,I=R.

Question 1 Select one:

a. f ⟶ uniform 0 on [0,1].

b. f ⟶ uniform 0 on R.

c. f ⟶ uniform 0 on [0,∞).

d. f ⟶ uniform 0 on

Solution

Let's analyze the sequence of functions fn(x)=sin(nx)nf_n(x) = \frac{\sin(nx)}{\sqrt{n}} over the interval I=RI = \mathbb{R} and determine the set on which it converges uniformly.

Step 1: Pointwise Convergence

First, let's determine the pointwise limit of fn(x)f_n(x) as nn \to \infty.

Given: fn(x)=sin(nx)nf_n(x) = \frac{\sin(nx)}{\sqrt{n}}

As nn increases, sin(nx)\sin(nx) oscillates between 1-1 and 11, but the amplitude of the function is scaled by 1n\frac{1}{\sqrt{n}}. This means that: limnfn(x)=limnsin(nx)n=0\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{\sin(nx)}{\sqrt{n}} = 0 for all xRx \in \mathbb{R}.

So, fn(x)f_n(x) converges pointwise to 00 on R\mathbb{R}.

Step 2: Uniform Convergence

Next, we need to check whether this convergence is uniform over different intervals.

To determine uniform convergence, consider the supremum of fn(x)|f_n(x)| over any interval AA:

supxAfn(x)=supxAsin(nx)n=1nsupxAsin(nx)\sup_{x \in A} |f_n(x)| = \sup_{x \in A} \left|\frac{\sin(nx)}{\sqrt{n}}\right| = \frac{1}{\sqrt{n}} \sup_{x \in A} |\sin(nx)|

Since sin(nx)1|\sin(nx)| \leq 1 for all xx and nn, we have: supxAfn(x)1n\sup_{x \in A} |f_n(x)| \leq \frac{1}{\sqrt{n}}

As nn \to \infty, 1n\frac{1}{\sqrt{n}} approaches 0. This indicates that fn(x)f_n(x) converges uniformly to 0 over any bounded interval in R\mathbb{R}.

However, for unbounded intervals like R\mathbb{R} or [0,)[0, \infty), the supremum over xx could potentially involve values of xx where sin(nx)\sin(nx) does not behave uniformly well, so we need to be more careful.

Step 3: Analyze Options

  • Option a: fn(x)f_n(x) converges uniformly to 0 on [0,1][0, 1].

    • On the interval [0,1][0, 1], which is bounded, we can apply our earlier result. supx[0,1]fn(x)1n\sup_{x \in [0, 1]} |f_n(x)| \leq \frac{1}{\sqrt{n}} uniformly converges to 0 as nn increases.
  • Option b: fn(x)f_n(x) converges uniformly to 0 on R\mathbb{R}.

    • On the entire real line, supxRfn(x)\sup_{x \in \mathbb{R}} |f_n(x)| could be problematic because xx could be large, causing oscillations that do not die out uniformly. Hence, this option is less likely to be true.
  • Option c: fn(x)f_n(x) converges uniformly to 0 on [0,)[0, \infty).

    • Similar to the argument above for the entire real line, the interval [0,)[0, \infty) is unbounded, and uniform convergence over an unbounded interval is generally more challenging to establish.
  • Option d: fn(x)f_n(x) converges uniformly to 0 on some unspecified set.

    • This is too vague to be a valid answer.

Conclusion

The correct choice is:

a. fn(x)f_n(x) converges uniformly to 0 on [0,1][0, 1].

Would you like further details or have any questions?


Here are five relative questions to expand your understanding:

  1. How do we determine the pointwise limit of a sequence of functions?
  2. What is the difference between pointwise convergence and uniform convergence?
  3. Why is uniform convergence more difficult to achieve on unbounded intervals?
  4. Can you think of an example where a function converges pointwise but not uniformly?
  5. How does the behavior of sin(nx)\sin(nx) impact the convergence properties of fn(x)f_n(x)?

Tip: Uniform convergence implies that the functions in the sequence get uniformly close to the limit function, regardless of the specific point in the domain.

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Math Problem Analysis

Mathematical Concepts

Uniform convergence
Pointwise convergence
Supremum

Formulas

$f_n(x) = \frac{\sin(nx)}{\sqrt{n}}$

Theorems

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Suitable Grade Level

Advanced undergraduate