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In the context of Taylor series, derivatives play a crucial role. They help us describe how the function behaves near a specific point, known as the center. For the function \( f(x) = \frac{1}{\sqrt{1-x}} \), we compute derivatives up to the fourth degree to construct the Taylor polynomial of degree 4. The key to finding these derivatives starts with recognizing the function as \( (1-x)^{-0.5} \). This makes it easier to apply the power rule for differentiation.

• The first derivative \( f'(x) = 0.5(1-x)^{-1.5} \).
• Second derivative \( f''(x) = 0.75(1-x)^{-2.5} \).
• Third derivative \( f'''(x) = 1.875(1-x)^{-3.5} \).
• Fourth derivative \( f^{(4)}(x) = 6.5625(1-x)^{-4.5} \).

Each derivative follows a pattern where the power decreases by one with each differentiation and the coefficient is updated by multiplying with the decreasing negative exponent.

A Taylor polynomial is a polynomial approach to approximating a function. It's based on the derivatives of the function at a single point. For any function \( f(x) \), the Taylor polynomial of degree \( n \) at \( a = 0 \) is given as a sum of terms that involve derivatives evaluated at the point \( a \).This polynomial can provide a close approximation of \( f(x) \) for values of \( x \) near \( a \). Here, our calculated Taylor polynomial at \( a = 0 \) for the fourth-degree approximation is:\[ T_4(x) = 1 + 0.5x + 0.375x^2 + 0.3125x^3 + 0.2734375x^4 \]

• Each term in the polynomial involves a fraction of a derivative evaluation.
• It multiplies by \( x^n \) where \( n \) is the degree of the derivative.
• Divided by \( n! \) (factorial of \( n \)) to adjust the scale of each term.

As better approximations demand more terms, the Taylor polynomial provides increasingly accurate representations of the function near \( a \).

The remainder term \( R_n(x) \) in a Taylor series serves as a measure of the error between the actual function and the Taylor polynomial approximation. When the polynomial is of degree \( n \), the remainder involves terms that would appear in degrees higher than \( n \). It assesses how much the approximation differs from \( f(x) \).There are different ways to represent and calculate the remainder term, including:

• Lagrange Form: \[ R_n(x) = \frac{f^{(n+1)}(c)(x-a)^{n+1}}{(n+1)!} \], where \( c \) is some value between \( a \) and \( x \)
• Integral Form: \[ R_n(x) = \int_a^x \frac{(x-t)^n}{n!}f^{(n+1)}(t)dt \]

Understanding the remainder term is important for knowing the limits of approximations and ensuring the desired accuracy when using a Taylor polynomial in practice.

When we say a Taylor series is "centered at \( a = 0 \)," it indicates that \( a = 0 \) is the point around which we are approximating the function. This specific case is often referred to as a Maclaurin series, which is a Taylor series centered at zero.For the function \( f(x) = \frac{1}{\sqrt{1-x}} \), centered at \( a = 0 \), the derivatives were specifically evaluated at zero to build the Taylor polynomial. This choice simplifies calculations and often aligns well with functions defined in terms of \( (1-x) \) or similar expressions because:

• The function and its derivatives simplify more neatly at \( x = 0 \).
• Each derivative contributes directly to the polynomial terms without altering \( x \).
• It gives rise to a polynomial that fits well for small values of \( x \).

Centering a series at zero can be very helpful, especially in calculations where simplicity is desired or where function behavior is primarily analyzed near zero.