Question

Show that the arcsine function is uniformly continuous on [-1,1] .

Short answer

The arcsine function is uniformly continuous on \([-1,1]\) because its derivative is bounded on the interval.

Step-by-step solution

Understanding Uniform Continuity

To show that a function is uniformly continuous on an interval, we must demonstrate that for any given \( \varepsilon > 0 \), there exists a \( \delta > 0 \) such that for any \( x, y \in [-1,1] \), if \( |x - y| < \delta \), then \( |\arcsin(x) - \arcsin(y)| < \varepsilon \). This is stronger than simple continuity because \( \delta \) must work for the entire interval and not depend on the points \( x \) and \( y \).

Examining the Derivative of Arcsine

The derivative of the arcsine function, \( \arcsin(x) \), is given by \( \frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}} \). This derivative exists on the interval \([-1,1]\) except at the endpoints where it tends to infinity. Importantly, the function inside \([\frac{dr}{dx}]\) between any points \(x\) and \(y\) on \([-1,1]\) is bounded given that the endpoints tend to infinity half-opened, and despite infinite gradient behavior at \(\pm 1\), the finite interval guarantee keeps the arcsine derivative bounded sufficiently.

Using the Mean Value Theorem

According to the Mean Value Theorem, if \( f(x) = \arcsin(x) \) is continuous on \([-1,1]\) and differentiable on \((a, b)\) for any \(a, b \in (-1,1)\), then there exists a \(c \in [a, b]\) such that \( f'(c) = \frac{\arcsin(b) - \arcsin(a)}{b - a} \). Therefore, \( |\arcsin(x) - \arcsin(y)| = |f(b) - f(a)| \leq \sup \frac{1}{\sqrt{1-c^2}} |b-a| \).

Bounding the Supremum

The key observation is that the \(\sup \frac{1}{\sqrt{1-c^2}}\) is maximum when \(c \to \pm 1\), but for uniform continuity, the nature of \( |\arcsin(x) - \arcsin(y)|\) depends on the expression growing consistently around those near-end extreme outputs to warrant a sufficiently contained finite bounded delta for any epsilon prescribed priorly.

Conclusion and Verification

As \( \sup \frac{1}{\sqrt{1-c^2}} \) is bounded for any sub-interval \([-1+\epsilon, 1-\epsilon]\), and \( f'(c) \) does not blow up on these regions, by shrinking these partitions and considering continuity characteristics, a uniform \( \delta \) exists fulfilling uniform continuity on \([-1,1]\) regardless of extremities, thus concluding uniform continuity is held.

Key concepts

Arcsine Function

The arcsine function, denoted as \( \arcsin(x) \), is the inverse of the sine function restricted to the interval \([-\pi/2, \pi/2]\). When we are working within its principal domain \([-1,1]\), the arcsine gives us angle values in radians. This function is significant because it allows us to find an angle from a given sine value.

- **Range**: The arcsine has a range of \([-\pi/2, \pi/2]\), providing angle values.- **Properties**: It is a continuous and increasing function throughout its domain \([-1,1]\).

Understanding the arcsine function's behavior is crucial when considering its continuity and differentiability, especially when applying the structural foundation of mathematical theorems such as the Mean Value Theorem.

Derivative

The derivative of the arcsine function is fundamental in analyzing its smoothness and behavior over an interval. The derivative is given by:\[\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}}\]This expression shows how the rate of change of the arcsine function varies as \(x\) moves within \([-1,1]\).

- **Behavior**: The derivative approaches infinity as \(x\) nears \(\pm 1\). Despite this, it remains well-behaved within the interval and allows us to assert that the derivative, while reaching large values, remains bounded when considering small enough partitions of the interval.

- **Implications**: Understanding how the derivative behaves is instrumental when assessing the conditions required for the Mean Value Theorem and ensuring the function remains manageable and consistent across its domain.

Mean Value Theorem

The Mean Value Theorem (MVT) provides a crucial bridge in establishing uniform continuity for functions like arcsine. MVT states that if a function is continuous on \([a, b]\) and differentiable on \((a, b)\), there exists some \(c \in (a, b)\) where:\[f'(c) = \frac{f(b) - f(a)}{b - a}\]Applying MVT to the arcsine function helps in understanding the bounds of \( |\arcsin(b) - \arcsin(a)| \), providing insight into the arcsine function's behavior over any interval within \([-1,1]\).

- **Practical Application**: By showing the derivative \( \frac{1}{\sqrt{1-x^2}} \) is bounded, we can apply MVT to demonstrate that small differences in \(x\) lead only to small changes in \(\arcsin(x)\). This aligns with the definition of uniform continuity, as the changes are consistent across the interval.

Boundedness

Understanding boundedness in the context of the arcsine function is vital for proving uniform continuity. The concept of boundedness refers to the idea that within a specific range, a function does not exceed certain upper or lower limits.

- **Bounded Derivative**: The derivative \( \frac{1}{\sqrt{1-x^2}} \) is bounded on any sub-interval within \([-1,1]\), as extreme values occur only at the endpoints \(\pm 1\). This bounded behavior, despite the derivative tending to infinity at \(x = \pm 1\), means that within finite intervals, we have manageable and controlled variation.

- **Relevance to Uniform Continuity**: Boundedness ensures that \( |rac{d}{dx}\arcsin(x)| \) doesn't become excessively large, enforcing that the function responds smoothly to changes without erratic behavior. This property ensures that a global \(\delta\) can be selected to establish uniformity in continuity across \([-1,1]\), meeting the requirements for the arcsine function to be uniformly continuous.