1. Find The Maclaurin Series Of Sin⁻¹x Using The Integral ∫ (1/√(1-x²)) Dx = Sin⁻¹x + C, And Provide The First Four Non-zero Terms. 2. Using The Maclaurin Series Of F(x) = X²cos2x, Determine F^(50)(0).

In the realm of calculus, Maclaurin series serve as a powerful tool for representing functions as infinite sums of terms. These series, which are special cases of Taylor series centered at zero, provide valuable insights into the behavior of functions and their derivatives. This article delves into the fascinating world of Maclaurin series, exploring their application in finding the series representation of the inverse sine function, ${\sin^{-1} x}$, and in determining the 50th derivative of the function ${f(x) = x^2 \cos 2x}$ at ${x = 0}$. We will navigate through the process of deriving Maclaurin series, emphasizing the underlying principles and techniques involved.

1. Maclaurin Series for sin⁻¹x

1.1 Utilizing the Integral Representation

The journey to find the Maclaurin series for ${\sin^{-1} x}$ begins with a fundamental integral representation: ${\int \frac{1}{\sqrt{1-x^2}} dx = \sin^{-1} x + c}$ This integral forms the cornerstone of our derivation. To proceed, we express the integrand, ${\frac{1}{\sqrt{1-x^2}}}$ as a binomial series. This transformation allows us to leverage the power of series manipulation to unveil the Maclaurin series for ${\sin^{-1} x}$. The binomial series expansion is a crucial technique in this process, enabling us to represent the integrand in a form that is amenable to integration and subsequent series representation.

1.2 Binomial Series Expansion

We can rewrite the integrand as ${(1-x^2)^{-\frac{1}{2}}}$ and apply the binomial series expansion, which states: ${(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + ...}$ where ${|u| < 1}$. In our case, ${u = -x^2}$ and ${n = -\frac{1}{2}}$. Substituting these values into the binomial series, we obtain: ${(1-x^2)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-x^2) + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}(-x^2)^2 + \frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}(-x^2)^3 + ...}$ Simplifying this expression, we get: ${(1-x^2)^{-\frac{1}{2}} = 1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2^2 \cdot 2!}x^4 + \frac{1 \cdot 3 \cdot 5}{2^3 \cdot 3!}x^6 + ...}$ This expansion represents the integrand as an infinite series, which we can then integrate term by term to find the Maclaurin series for ${\sin^{-1} x}$. The binomial series expansion is a powerful tool that allows us to express functions in terms of infinite sums, which can then be manipulated and integrated to derive new series representations.

1.3 Term-by-Term Integration

Integrating term by term, we have: ${\sin^{-1} x = \int (1-x^2)^{-\frac{1}{2}} dx = \int \left(1 + \frac{1}{2}x^2 + \frac{1 \cdot 3}{2^2 \cdot 2!}x^4 + \frac{1 \cdot 3 \cdot 5}{2^3 \cdot 3!}x^6 + ...\right) dx}$ This yields: ${\sin^{-1} x = x + \frac{1}{2} \cdot \frac{x^3}{3} + \frac{1 \cdot 3}{2^2 \cdot 2!} \cdot \frac{x^5}{5} + \frac{1 \cdot 3 \cdot 5}{2^3 \cdot 3!} \cdot \frac{x^7}{7} + ... + C}$ To find the constant of integration, ${C}$, we set ${x = 0}$. Since ${\sin^{-1} 0 = 0}$, we find that ${C = 0}$. Thus, the Maclaurin series for ${\sin^{-1} x}$ is: ${\sin^{-1} x = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7 + ...}$ This series provides a representation of the inverse sine function as an infinite sum of terms, which is valid for ${|x| < 1}$. The term-by-term integration is a crucial step in this process, allowing us to transform the series representation of the integrand into the series representation of the inverse sine function.

1.4 First Four Non-Zero Terms

The first four non-zero terms of the Maclaurin series for ${\sin^{-1} x}$ are: ${x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \frac{5}{112}x^7}$ These terms provide a good approximation of ${\sin^{-1} x}$ for values of ${x}$ close to zero. The more terms we include in the series, the more accurate the approximation becomes. The Maclaurin series provides a powerful tool for approximating the values of functions, especially when dealing with functions that are difficult to evaluate directly.

2. Maclaurin Series and Derivatives of f(x) = x²cos2x

2.1 Maclaurin Series of cos2x

To find the 50th derivative of ${f(x) = x^2 \cos 2x}$ at ${x = 0}$, we first need the Maclaurin series for ${\cos 2x}$. Recall the Maclaurin series for ${\cos x}$: ${\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ...}$ Substituting ${2x}$ for ${x}$, we get: ${\cos 2x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x)^{2n} = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + ...}$ This simplifies to: ${\cos 2x = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n}}{(2n)!}x^{2n} = 1 - \frac{4x^2}{2!} + \frac{16x^4}{4!} - \frac{64x^6}{6!} + ...}$ The Maclaurin series for ${\cos 2x}$ is a fundamental building block for finding the Maclaurin series of ${f(x) = x^2 \cos 2x}$. By understanding the series representation of ${\cos 2x}$, we can then manipulate it to derive the series for ${f(x)}$ and subsequently determine its derivatives.

2.2 Maclaurin Series of f(x) = x²cos2x

Now, we find the Maclaurin series for ${f(x) = x^2 \cos 2x}$ by multiplying the series for ${\cos 2x}$ by ${x^2}$: ${f(x) = x^2 \cos 2x = x^2 \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n}}{(2n)!}x^{2n} = \sum_{n=0}^{\infty} \frac{(-1)^n 2^{2n}}{(2n)!}x^{2n+2}}$ This gives us: ${f(x) = x^2 - \frac{4x^4}{2!} + \frac{16x^6}{4!} - \frac{64x^8}{6!} + ...}$ The Maclaurin series for ${f(x)}$ is obtained by multiplying the series for ${\cos 2x}$ by ${x^2}$, which shifts the powers of ${x}$ in the series. This manipulation allows us to express ${f(x)}$ as an infinite sum of terms, which is crucial for finding its derivatives.

2.3 Finding f^(50)(0)

The general form of the Maclaurin series is: ${f(x) = \sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!}x^k}$ Comparing the coefficient of ${x^{50}}$ in the Maclaurin series of ${f(x)}$, we need to find the term where ${2n + 2 = 50}$, which means ${2n = 48}$ and ${n = 24}$. The coefficient of ${x^{50}}$ in the series for ${f(x)}$ is: ${\frac{(-1)^{24} 2^{2(24)}}{(2(24))!} = \frac{2^{48}}{48!}}$ Therefore, we have: ${\frac{f^{(50)}(0)}{50!} = \frac{2^{48}}{48!}}$ Solving for ${f^{(50)}(0)}$, we get: ${f^{(50)}(0) = \frac{2^{48}}{48!} \cdot 50! = 2^{48} \cdot 50 \cdot 49}$ Thus, ${f^{(50)}(0) = 2^{48} \cdot 50 \cdot 49}$ This result demonstrates the power of Maclaurin series in determining higher-order derivatives of functions at a specific point. By comparing the coefficients of the series, we can directly relate the derivatives to the terms in the series, providing a powerful tool for calculus.

In this exploration of Maclaurin series, we have successfully derived the series representation for ${\sin^{-1} x}$ and determined the 50th derivative of ${f(x) = x^2 \cos 2x}$ at ${x = 0}$. These examples highlight the versatility and utility of Maclaurin series in calculus and mathematical analysis. Understanding Maclaurin series provides a solid foundation for tackling more complex problems in various fields of science and engineering. The ability to represent functions as infinite sums and to extract derivative information from these series is a testament to the power and elegance of calculus.