Question

Inverse hyperbolic sine The inverse of hyperbolic sine is defined in several ways; among them are $$ \sinh ^{-1} x=\ln (x+\sqrt{x^{2}+1})=\int_{0}^{x} \frac{d t}{\sqrt{1+t^{2}}} $$ Find the first four terms of the Taylor series for \(\sinh ^{-1} x\) using these two definitions (and be sure they agree).

Short answer

Answer: The first four terms of the Taylor series for the inverse hyperbolic sine function are: $$\sinh^{-1} x = x - \frac{1}{6}x^3 + \mathcal{O}(x^5)$$

Step-by-step solution

Use the first definition to find the Taylor series

To find the Taylor series using the first definition, we first need to find the derivatives of the function \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\). Let's find the first few derivatives: - First derivative: $$ (\sinh^{-1} x)' = \frac{1}{x + \sqrt{x^2 + 1}} \cdot \frac{d (x + \sqrt{x^2 + 1})}{dx} $$ $$ (\sinh^{-1} x)' = \frac{1 + \frac{x}{\sqrt{x^2 + 1}}}{x + \sqrt{x^2 + 1}} = \frac{1}{\sqrt{x^2 + 1}} $$ - Second derivative: $$ (\sinh^{-1} x)'' = \frac{-x}{(x^2 + 1)^{\frac{3}{2}}} $$ - Third derivative: $$ (\sinh^{-1} x)''' = \frac{3x^2 - 1}{(x^2 + 1)^{\frac{5}{2}}} $$ - Fourth derivative: $$ (\sinh^{-1} x)^{(4)} = \frac{-15x(x^2 - 2)}{(x^2 + 1)^{\frac{7}{2}}} $$ Now, we can use the Taylor series formula to find the first four terms of the series: $$ \sinh^{-1} x = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n $$ Since \(\sinh^{-1}(0) = 0\), the first term is 0. Now, let's find the other terms: - Term 1: \(\frac{(\sinh^{-1} x)'}{1!} x = \frac{1}{\sqrt{0^2 + 1}}x = x\) - Term 2: \(\frac{(\sinh^{-1} x)''}{2!} x^2 = \frac{-0}{(0^2 + 1)^{\frac{3}{2}}}\frac{1}{2}x^2 = 0\) - Term 3: \(\frac{(\sinh^{-1} x)'''}{3!} x^3 = \frac{3(0)^2 - 1}{(0^2 + 1)^{\frac{5}{2}}}\frac{1}{6}x^3 = -\frac{1}{6}x^3\) - Term 4: As the exercise asks for the first four terms, we will stop at this point. So, from the first definition, the first four terms of the Taylor series for \(\sinh^{-1} x\) are: $$ \sinh^{-1} x = x - \frac{1}{6}x^3 + \mathcal{O}(x^5) $$

Use the second definition to verify the Taylor series

To find the Taylor series using the second definition, we have \(\sinh^{-1} x = \int_0^x \frac{dt}{\sqrt{1 + t^2}}\). Let's expand the integrand in a Taylor series: $$ \frac{1}{\sqrt{1+t^2}} = 1 - \frac{1}{2}t^2 + \frac{3}{8}t^4 + \mathcal{O}(t^6) $$ Now, let's integrate term-by-term: $$ \sinh^{-1} x = \int_0^x (1 - \frac{1}{2}t^2 + \frac{3}{8}t^4) dt $$ $$ \sinh^{-1} x = \int_0^x dt - \int_0^x \frac{1}{2}t^2 dt + \int_0^x \frac{3}{8}t^4 dt $$ $$ \sinh^{-1} x = (t - \frac{1}{6}t^3 + \frac{1}{40}t^5) \Big|_0^x $$ $$ \sinh^{-1} x = x - \frac{1}{6}x^3 + \mathcal{O}(x^5) $$ The first four terms of the Taylor series using the second definition agree with those from the first definition. In conclusion, the first four terms of the Taylor series for the inverse hyperbolic sine function are: $$ \sinh^{-1} x = x - \frac{1}{6}x^3 + \mathcal{O}(x^5) $$

Key concepts

Taylor Series

The Taylor Series is a way to represent functions as infinite sums of terms. Each term is calculated from the derivatives of the function at a single point. This series allows us to approximate functions that might otherwise be difficult to deal with, using simpler polynomial expressions.

For the function \(inh^{-1} x = \ln(x + \sqrt{x^2 + 1})\), we find the Taylor Series centered at 0. The goal is to express it as \(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\).

• First Term: Since \(\sinh^{-1}(0) = 0\), the constant term is 0.
• Subsequent Terms: The coefficients are derived from the function's derivatives evaluated at zero.

By carefully computing these derivatives, we construct the first few terms of the series. In this case, starting with \(inh^{-1} x \): \( x - \frac{1}{6}x^3 + \mathcal{O}(x^5)\). Each term calculated brings us closer to a polynomial that effectively represents the function near the point x=0, enhancing our ability to approximate \(\sinh^{-1} x\).

Derivatives

Derivatives are fundamental in calculus as they measure how a function changes as its input changes. For \(\sinh^{-1} x = \ln(x + \sqrt{x^2 + 1})\), calculating derivatives is crucial to finding the Taylor series.

Here's how the process unfolds:

• First Derivative: It simplifies the expression, revealing how \(\sinh^{-1} x\) grows. Here, we have \(\frac{1}{\sqrt{x^2 + 1}}\), suggesting the rate of change diminishes as \(x\) increases.
• Higher Derivatives: Successive derivatives become increasingly complex. For example, the second derivative \(\frac{-x}{(x^2 + 1)^{\frac{3}{2}}}\) shows how the function's curvature behaves. Further derivatives provide insight into oscillatory behavior and detailed approximations.
  

Calculating derivatives helps in capturing the local behavior of the inverse hyperbolic sine function. These derivatives ensure accurate Taylor expansion, allowing us to approximate \(\sinh^{-1} x\) smoothly within a particular region around the origin.

Integrals

Integrals can provide another powerful method to determine the behavior of a function, particularly when developing a Taylor series. When applying integration to \(\sinh^{-1} x = \int_0^x \frac{dt}{\sqrt{1 + t^2}}\), the integral represents the accumulation of the function's rate of change.

The process involves:

• Taylor Expansion of the Integrand: Before integrating, the integrand \(\frac{1}{\sqrt{1+t^2}}\) is expanded as a Taylor series. This simplifies the integration process, translating the problem into simpler polynomial terms.
• Term-by-Term Integration: Each polynomial term is integrated separately. This produces a series where each term corresponds to a power of \(x\).

Applying limits after integrating results in a new series that mirrors the function's behavior over a domain. The resulting series should match the manually defined Taylor series for \(\sinh^{-1} x\), confirming the equivalence of both approaches. This method of integration thus serves as a robust tool for constructing accurate function approximations.