Question

Consider the sequence \(\phi_{n}(x)=2 n x e^{-n x^{2}}, 0 \leq x \leq 1\) (a) Show that \(\lim _{n \rightarrow \infty} \phi_{n}(x)=0\) for \(0 \leq x \leq 1\); hence $$ \int_{0}^{1} \lim _{n \rightarrow \infty} \phi_{n}(x) d x=0 $$ (b) Show that \(\int_{0}^{1} 2 n x e^{-x x^{2}} d x=1-e^{-x}\); hence $$ \lim _{n \rightarrow \infty} \int_{0}^{1} \phi_{n}(x) d x=1 $$ Thus, in this example, $$ \lim _{n \rightarrow \infty} \int_{a}^{b} \phi_{n}(x) d x \neq \int_{a}^{b} \lim _{n \rightarrow \infty} \phi_{n}(x) d x $$ even though \(\lim _{n \rightarrow \infty} \phi_{n}(x)\) exists and is continuous.

Short answer

Question: Show that the example sequence provided does not satisfy the condition that the limit of the integral equals the integral of the limit. Answer: We showed that the limit of the sequence, \(\lim_{n \rightarrow \infty} \phi_n(x)\), is 0 for all \(0 \leq x \leq 1\). However, when we evaluate the integral, we find that \(\int_{0}^{1}\lim_{n \rightarrow \infty} \phi_n(x) dx = 0\) does not equal the limit of the integral, which is \(\lim_{n \rightarrow \infty} \int_{0}^{1} \phi_n(x) dx = 1\). This means that the given sequence does not satisfy the condition.

Step-by-step solution

Analyze the limit

Let's find the limit of the sequence \(\phi_n(x) = 2nx e^{-nx^2}\) as \(n \rightarrow \infty\). We can factor out \(n\) in the exponent and use the properties of limits to simplify the expression.

Simplify the expression

Consider the expression: $$ \lim _{n \rightarrow \infty} \phi_{n}(x) = \lim_{n \rightarrow \infty} 2nx e^{-nx^2} $$ Now, break the limit into product of two limits: $$ \lim _{n \rightarrow \infty} \phi_{n}(x) = \lim_{n \rightarrow \infty} (2nx) \cdot \lim_{n \rightarrow \infty} e^{-nx^2} $$

Calculate the limits

Now, let's find the limits separately: $$ \lim_{n \rightarrow \infty} (2nx) = 2x \cdot \lim_{n \rightarrow \infty} n = 2x \cdot \infty = \infty, \text{ for } x \neq 0 $$ $$ \lim_{n \rightarrow \infty} e^{-nx^2} = e^{\lim_{n \rightarrow \infty} -nx^2} = e^{-\infty} = 0, \text{ for } x \neq 0 $$ Note that \(x=0\) is a trivial case, when both factors equal to \(0\). Multiply the limits: $$ \lim _{n \rightarrow \infty} \phi_{n}(x) = \infty \cdot 0 = 0 \text{ for } 0 \leq x \leq 1 $$ Now let's show the integral: $$ \int_{0}^{1} \lim _{n \rightarrow \infty} \phi_{n}(x) dx = \int_{0}^{1} 0 dx = 0 $$ #b) Show that \(\int_{0}^{1} 2 n x e^{-x x^{2}} d x=1-e^{-x}\)

Perform substitution

Let's perform a substitution: $$ u = x^2 \implies du = 2x \, dx $$ Now, rewrite the integral with the substitution: $$ \int_{0}^{1} 2 n x e^{-x x^{2}} d x = n \int_{0}^{1} e^{-(nx^2)} \, du $$

Calculate the integral

Now, let's calculate the integral: $$ n \int_{0}^{1} e^{-(nx^2)} \, du = -\frac{d}{dn} \int_{0}^{1} e^{-(nx^2)} \, dx $$ By chain rule, we have: $$ -\frac{d}{dn} \int_{0}^{1} e^{-u} \, du = -\frac{d}{dn} (1 - e^{-u}) = -\frac{d}{dn} (1 - e^{-x^2}) $$ Now, take the limit as \(n \rightarrow \infty\): $$ \lim_{n \rightarrow \infty} n \int_{0}^{1} e^{-nx^2}\, dx = \lim_{n \rightarrow \infty} (-\frac{d}{dn} (1 - e^{-x^2})) = 1 - \lim_{n \rightarrow \infty} e^{-nx^{2}} = 1 - 0 = 1 $$ Thus, we have shown that: $$ \lim_{n \rightarrow \infty} \int_{0}^{1} \phi_{n}(x) dx \neq \int_{0}^{1} \lim_{n \rightarrow \infty} \phi_{n}(x) dx = 0 $$ This shows that the given sequence doesn't satisfy the given condition, even though the limit exists and is continuous.

Key concepts

Limit of a Sequence

Understanding the limit of a sequence is vital in advanced calculus. In our exercise, we are dealing with the sequence \(\phi_{n}(x) = 2nx e^{-nx^2}\) and want to find its behavior as \(n\) approaches infinity.

To determine \(\lim_{n \rightarrow \infty} \phi_{n}(x)\), we split the sequence into two parts: \(2nx\) and \(e^{-nx^2}\). As \(n\) increases, \(2nx\) grows indefinitely for any \(x > 0\). Meanwhile, \(e^{-nx^2}\) approaches zero because the exponent \(-nx^2\) becomes a large negative number.

• When multiplied together, \(2nx \cdot e^{-nx^2} = \infty \cdot 0\), which can seem undefined, but for values \(0 \leq x \leq 1\), this resolves to zero.

Understanding that the sequence tends to zero helps in exploring convergence and functions' integrals, setting the stage for the next parts of the exercise.

Improper Integrals

Improper integrals are crucial when working with limits and infinite boundaries. In this exercise, we focus on the integral\(\int_{0}^{1} \phi_n(x) \, dx\). This represents a special case where the limit as \(n\) approaches infinity and the improper integral interact.

The step-by-step solution uses substitution methods to find the definite integral and understand its behavior as \(n\) is increased. It makes us think about how an integral might behave differently under specific conditions.

• We calculated this limit differently from standard limits, using the substitution \(u = x^2\), converting the integral into a form that's easier to evaluate.
• Observing \(\lim_{n \rightarrow \infty}\) \(\int_{0}^{1}2nx e^{-x^2}\,dx\), we adjusted our approach to look closer at how limits behave in integration scenarios.Observations show \(\int\) evaluated independently may present surprising outcomes, furthering understanding of analytical techniques in calculus.

Uniform Convergence

Uniform convergence is a robust topic in advanced calculus, often interlinked with sequences and series functions. It is notable when handling the transition between integrations and limit operations. In this exercise, we're presented with a clear case where uniform convergence plays a crucial role.

Notice how while \(\phi_n(x)\) satisfies the pointwise convergence \(\phi_n(x) \rightarrow 0\) for fixed \(x\), uniform convergence isn't achieved when integrating over the interval. Hence,

• The limit of the integral, \(\lim_{n \rightarrow \infty} \int_{0}^{1} \phi_{n}(x) \, dx\), diverges from \(\int_{0}^{1} \lim_{n \rightarrow \infty} \phi_{n}(x) \,dx\).

This is quintessential in illustrating how the order of limit and integral may influence outcomes, emphasizing uniform convergence's significance when ensuring interchangeability. Uniform convergence mandates continuity and boundedness across the entire interval, which isn't fully satisfied here.

Substitution Method

The substitution method is a powerful technique for evaluating integrals, especially those involving complex exponents or odd coefficients. It is pivotal in transforming the given problem into a more standard form, simplifying evaluation for problem analysis. In the exercise, the substitution chosen was \(u = x^2\). This leads to \(du = 2x \, dx\), which turns the integral to easier bounds.

Using substitution, the integrals \(\int_{0}^{1} nx e^{-nx^2} \, dx\) become \(\int_{0}^{1} e^{-u} \, du\), making the integral limits and expressions manageable.

• The process allowed a transformation from complex to basic integrals, showcasing the substitution method's power in reducing calculation complexity and enhancing clarity.
• Substitution simplifies integration, crucially impacting how we manipulate limits and calculus expressions efficiently.
• It exemplifies the beautiful connection between algebraic manipulation and calculus, showing how analytical techniques intersect to solve complex calculus problems effectively.

Understanding substitution broadens problem-solving skills, especially when dealing with indefinite algebraic expressions in calculus contexts.