Problem 21 - Essential Calculus

Question

In Exercises \(19-24,\) find the \(n\) th Taylor polynomial centered at \(c\). $$ f(x)=\sqrt{x}, \quad n=4, \quad c=1 $$

Step-by-Step Solution

Answer

The fourth degree Taylor polynomial for the function \( \sqrt{x} \) centered at \( c=1 \) is \( P_4(x)=1 +\frac{1}{2}(x-1) -\frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} - \frac{15}{16}\frac{(x-1)^4}{4!} \).

Step 1: Calculating Derivatives

The derivatives need to be evaluated up to the fourth order. For \( \sqrt{x} \) at \( x=1 \), they are as follows: \[ f(x)=\sqrt{x} \Rightarrow f'(x)=\frac{1}{2\sqrt{x}}, \]\[ f''(x)=-\frac{1}{4x\sqrt{x}}, \]\[ f'''(x)=\frac{3}{8x^2\sqrt{x}}, \]and \[ f''''(x)=-\frac{15}{16x^3\sqrt{x}}, \] .

Step 2: Evaluate Derivatives at \( c=1 \)

The derivatives obtained in step 1 are then evaluated at \( c=1 \). The results are: \[ f(1)=\sqrt{1}=1,\ f'(1)=\frac{1}{2},\ f''(1)=-\frac{1}{4},\ f'''(1)=\frac{3}{8}, \]and \[ f''''(1)=-\frac{15}{16} . \]

Step 3: Compose Taylor Polynomial

The nth degree Taylor polynomial is given as \[ P_n(x)=f(c) +f'(c)(x-c) +f''(c)\frac{(x-c)^2}{2!} + f'''(c)\frac{(x-c)^3}{3!} + \f''''(c)\frac{(x-c)^4}{4!}\]Substitute the values from Step 2 into this equation:\[P_4(x)=1 +\frac{1}{2}(x-1) -\frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} - \frac{15}{16}\frac{(x-1)^4}{4!} \]

Key Concepts

Derivatives

Derivatives are the building blocks for understanding how functions change. In examining the Taylor polynomial for the function \(f(x) = \sqrt{x}\) centered at a point \(c\), it is crucial to compute the series of derivatives. Derivatives provide the necessary information to construct a polynomial approximation of a function.

• The first derivative \(f'(x) = \frac{1}{2\sqrt{x}}\) tells us how fast \(f(x)\) is increasing or decreasing.


• The second derivative \(f''(x) = -\frac{1}{4x\sqrt{x}}\) gives insight into the concavity, showing how the rate of change is itself changing.


• The third derivative, \(f'''(x) = \frac{3}{8x^2\sqrt{x}}\), and the fourth, \(f''''(x) = -\frac{15}{16x^3\sqrt{x}}\), provide further nuances by detailing higher-order behavior of the function.

Each derivative evaluated at the center \(c=1\) delivers specific coefficients for our polynomial, crucial for producing an accurate approximation.

Polynomial Approximation

Polynomial approximation, particularly using Taylor series, is an effective way to estimate complex functions with polynomials. A Taylor polynomial like \(P_n(x)\) centers around a specific point \(c\) and imitates the behavior of \(f(x)\) within a region.

• Step 1 begins by finding the necessary derivatives of \(f(x)\). These derivatives handle the steepness and curvature of the original function.


• Step 3 involves creating a polynomial using these derivatives, providing a snapshot of the function's likeness over small intervals.

The npolynomial takes this form: \[P_n(x) = f(c) + f'(c)(x-c) + f''(c)\frac{(x-c)^2}{2!} + \ldots + f''''(c)\frac{(x-c)^4}{4!}\]Each term represents a deeper dive into how closely \(P_n\) can model \(f(x)\), particularly near \(c=1\) for this exercise.

Function Evaluation

Function evaluation is pivotal in determining the accuracy of a polynomial approximation. Once derivatives are calculated, evaluating them at a specific center \(c\), in this case, \(c=1\), helps refine the approximation.

• Each derivative evaluated at \(c=1\) yields a precise numerical value, crafting the coefficients for our Taylor polynomial.


• The function itself at \(x = 1\) gives \(f(1) = \sqrt{1} = 1\), a straightforward calculation.


• Subsequently, evaluating derivatives like \(f'(1) = \frac{1}{2}\) and \(f''(1) = -\frac{1}{4}\) sets the stage for forming the polynomial.

These strategically calculated values ensure that \(P_4(x)\) remains a faithful representation of \(f(x)\) in its vicinity, allowing us to gleam insights or spoil us from intensive computations.