Chapter 4: Problem 108
The theory of interference of coherent oscillators requires the limit \(\lim _{\delta \rightarrow 2 m \pi} \frac{\sin ^{2}(N \delta / 2)}{\sin ^{2}(\delta / 2)},\) where \(N\) is a positive integer and \(m\) is any integer. Show that the value of this limit is \(N^{2}\).
Short Answer
Expert verified
Answer: The limit is \(N^2\).
Step by step solution
01
Rewrite the limit expression
First, let's rewrite the limit expression in a more convenient form:
$$
\lim_{\delta \rightarrow 2 m \pi} \frac{\sin ^{2}(N \delta / 2)}{\sin ^{2}(\delta / 2)}
$$
Let \(x = \delta / 2\), then as \(\delta \rightarrow 2 m \pi\), we have \(x \rightarrow m \pi\). So the expression becomes:
$$
\lim_{x \rightarrow m \pi} \frac{\sin ^{2}(N x)}{\sin ^{2}(x)}
$$
02
Apply L'Hôpital's rule
In this step, we will apply L'Hôpital's rule to the limit. L'Hôpital's rule states that if the limit of a function is of the indeterminate form (0/0 or ∞/∞), then:
$$
\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}
$$
provided that the limit on the right-hand side exists. Our expression is in the form 0/0 as both \(\sin(Nx)\) and \(\sin(x)\) would be zero for \(x=m \pi\). Let's find the derivatives of the numerator and denominator with respect to \(x\):
1. Derivative of the numerator:
Using chain rule and double angle formula for the sine function, we have
$$
\frac{d}{dx} \sin^2(Nx) = 2 \sin(Nx) \cos(Nx) \cdot N = N \sin(2Nx)
$$
2. Derivative of the denominator:
Similarly, for the denominator
$$
\frac{d}{dx} \sin^2(x) = 2 \sin(x) \cos(x) = \sin(2x)
$$
So applying L'Hôpital's rule, our expression becomes:
$$
\lim_{x \rightarrow m \pi} \frac{N \sin(2Nx)}{\sin(2x)}
$$
03
Apply L'Hôpital's rule again
Applying L'Hôpital's rule one more time, we find the second derivatives of the numerator and denominator with respect to \(x\):
1. Second derivative of the numerator:
$$
\frac{d^2}{dx^2} \left( N \sin(2Nx) \right) = 4N^2 \cos(2Nx)
$$
2. Second derivative of the denominator:
$$
\frac{d^2}{dx^2} \sin(2x) = 4 \cos(2x)
$$
So our expression, after applying L'Hôpital's rule again, is:
$$
\lim_{x \rightarrow m \pi} \frac{4N^2 \cos(2Nx)}{4 \cos(2x)}
$$
04
Evaluate the limit
The limit can now be easily evaluated as:
$$
\lim_{x \rightarrow m \pi} \frac{4N^2 \cos(2Nx)}{4 \cos(2x)} = \frac{4N^2}{4} = N^2
$$
Hence, the value of the limit is \(N^2\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
L'Hôpital's Rule
When faced with complex limits, L'Hôpital's rule is an indispensable tool for calculus students. This rule cleverly converts a challenging limit problem involving indeterminate forms into a more straightforward derivative problem. It's reserved for situations where a limit yields an indeterminate form like 0/0 or ∞/∞. What you do is take the derivatives of both the numerator and denominator and then attempt to evaluate the limit again.
For instance, if you have a limit of the form \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \), and direct substitution results in a dreaded indeterminate form, L'Hôpital's rule allows you to explore the limit of the derivatives: \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \). Crucially, this switch is only valid if the derivatives are continuous near 'a' and the limit of the derivatives exists or leads to ±∞. It’s not uncommon to apply L'Hôpital's rule multiple times, as seen in our textbook example, where successive applications finally resolve the limit.
For instance, if you have a limit of the form \( \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \), and direct substitution results in a dreaded indeterminate form, L'Hôpital's rule allows you to explore the limit of the derivatives: \( \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)} \). Crucially, this switch is only valid if the derivatives are continuous near 'a' and the limit of the derivatives exists or leads to ±∞. It’s not uncommon to apply L'Hôpital's rule multiple times, as seen in our textbook example, where successive applications finally resolve the limit.
Indeterminate Forms
Indeterminate forms are mathematical expressions whose behavior isn't immediately clear when tend towards their limits. Common examples include 0/0, ∞/∞, 0·∞, ∞ - ∞, 1^∞, 0^0, and ∞^0. These forms arise when evaluating limits of functions and are deceptive - they don’t tell us anything about the limit’s value without further analysis.
In the context of our problem, we start with a limit that looks like 0/0 as \( \delta \) approaches \( 2m\pi \). This seemingly simple ratio gives no clues on the surface as to what the limit might be as \( \delta \) nears \( 2m\pi \). That's why L'Hôpital's rule is so valuable - it provides a methodical way to probe these cryptic forms. By examining the behavior the functions' derivatives instead, we can often pinpoint the exact value of the limit.
In the context of our problem, we start with a limit that looks like 0/0 as \( \delta \) approaches \( 2m\pi \). This seemingly simple ratio gives no clues on the surface as to what the limit might be as \( \delta \) nears \( 2m\pi \). That's why L'Hôpital's rule is so valuable - it provides a methodical way to probe these cryptic forms. By examining the behavior the functions' derivatives instead, we can often pinpoint the exact value of the limit.
Sine Function
The sine function is one of the fundamental trigonometric functions with extensive applications in various fields, including mathematics, physics, and engineering. It’s defined as the y-coordinate of a point on the unit circle as it rotates from an initial position on the positive x-axis. In calculus, the sine function is intriguing because it beautifully oscillates between -1 and 1.
When dealing with limits involving sine, it's common to see the function returning to 0 at integer multiples of \( \pi \). This cyclical nature often leads to the indeterminate form 0/0 in limit problems. The exercise we are tackling involves a squared sine function, which accentuates this behavior. By applying L'Hôpital’s rule and deciphering the derivatives of the sine function, we convert a tricky trigonometric limit into a more approachable algebraic expression, ultimately revealing that the original limit is indeed \( N^2 \).
When dealing with limits involving sine, it's common to see the function returning to 0 at integer multiples of \( \pi \). This cyclical nature often leads to the indeterminate form 0/0 in limit problems. The exercise we are tackling involves a squared sine function, which accentuates this behavior. By applying L'Hôpital’s rule and deciphering the derivatives of the sine function, we convert a tricky trigonometric limit into a more approachable algebraic expression, ultimately revealing that the original limit is indeed \( N^2 \).