Question

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

Short 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-by-step solution

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 \[ f''''(x)=-\frac{15}{16x^3\sqrt{x}}, \] .

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 \[ f''''(1)=-\frac{15}{16} . \]

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:\[P_4(x)=1 +\frac{1}{2}(x-1) -\frac{1}{4}\frac{(x-1)^2}{2!} + \frac{3}{8}\frac{(x-1)^3}{3!} - \frac{15}{16}\frac{(x-1)^4}{4!} \]