To approximate $f(x) = \ln(1 + 2x)$ around $a = 3$ using a Taylor polynomial of degree 3, we need to calculate the Taylor polynomial centered at $x = 3$. The Taylor polynomial $T_3(x)$ of degree 3 is given by:

$T_3(x) = f(3) + f'(3)(x - 3) + \frac{f''(3)}{2!}(x - 3)^2 + \frac{f'''(3)}{3!}(x - 3)^3$

Step 1: Calculate $f(3)$

$f(3) = \ln(1 + 2 \cdot 3) = \ln(7)$

Step 2: Calculate $f'(x)$

Using the chain rule, we find: $f'(x) = \frac{2}{1 + 2x}$ Then, $f'(3) = \frac{2}{1 + 2 \cdot 3} = \frac{2}{7}$

Step 3: Calculate $f''(x)$

Differentiating $f'(x)$ with respect to $x$: $f''(x) = -\frac{2 \cdot 2}{(1 + 2x)^2} = -\frac{4}{(1 + 2x)^2}$ Then, $f''(3) = -\frac{4}{(1 + 2 \cdot 3)^2} = -\frac{4}{49}$

Step 4: Calculate $f'''(x)$

Differentiating $f''(x)$ with respect to $x$: $f'''(x) = \frac{8 \cdot 2}{(1 + 2x)^3} = \frac{16}{(1 + 2x)^3}$ Then, $f'''(3) = \frac{16}{(1 + 2 \cdot 3)^3} = \frac{16}{343}$

Step 5: Construct $T_3(x)$

Now we substitute these values into the Taylor polynomial formula: $T_3(x) = \ln(7) + \frac{2}{7}(x - 3) - \frac{4}{49} \frac{(x - 3)^2}{2} + \frac{16}{343} \frac{(x - 3)^3}{6}$

Simplifying the coefficients: $T_3(x) = \ln(7) + \frac{2}{7}(x - 3) - \frac{2}{49}(x - 3)^2 + \frac{8}{1029}(x - 3)^3$

So, the Taylor polynomial of degree 3 is: $T_3(x) = \ln(7) + \frac{2}{7}(x - 3) - \frac{2}{49}(x - 3)^2 + \frac{8}{1029}(x - 3)^3$

Would you like further explanation on any step?

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Related Questions

1. How do Taylor series approximations change with increasing degrees?
2. What is the significance of each derivative in the Taylor series?
3. How does the choice of the center $a$ impact the approximation?
4. Can we approximate $f(x)$ for any $x$ or are there limitations?
5. What are common applications of Taylor polynomials in real-world scenarios?

Tip:

Remember, each term in the Taylor series improves accuracy near the center point $a$. Adding higher degrees can improve approximation over a broader interval around $a$.