The given function is $f(x) = x^3$, and we are tasked with finding the Taylor series expansion around $x = 1$.

Step 1: Taylor Series Definition

The Taylor series of a function $f(x)$ about $x = a$ is given by: $f(x) = \sum_{n=0}^{\infty} c_n (x - a)^n$ where $c_n$ represents the coefficients: $c_n = \frac{f^{(n)}(a)}{n!}$ Here, $a = 1$ and $f(x) = x^3$.

Step 2: Calculate Derivatives

1. $f(x) = x^3$
2. First derivative: $f'(x) = 3x^2$
3. Second derivative: $f''(x) = 6x$
4. Third derivative: $f^{(3)}(x) = 6$
5. Fourth derivative and higher: $f^{(n)}(x) = 0$ for $n \geq 4$

Step 3: Evaluate Derivatives at $x = 1$

We evaluate each derivative at $x = 1$:

1. $f(1) = 1^3 = 1$
2. $f'(1) = 3(1)^2 = 3$
3. $f''(1) = 6(1) = 6$
4. $f^{(3)}(1) = 6$
5. $f^{(n)}(1) = 0$ for $n \geq 4$

Step 4: Find Coefficients

Using the formula $c_n = \frac{f^{(n)}(1)}{n!}$, we compute the coefficients:

1. $c_0 = \frac{f(1)}{0!} = \frac{1}{1} = 1$
2. $c_1 = \frac{f'(1)}{1!} = \frac{3}{1} = 3$
3. $c_2 = \frac{f''(1)}{2!} = \frac{6}{2} = 3$
4. $c_3 = \frac{f^{(3)}(1)}{3!} = \frac{6}{6} = 1$
5. $c_4 = \frac{f^{(4)}(1)}{4!} = \frac{0}{24} = 0$

Conclusion

The first few coefficients are:

• $c_0 = 1$
• $c_1 = 3$
• $c_2 = 3$
• $c_3 = 1$
• $c_4 = 0$

Would you like more details on how the Taylor series expansion is derived?

Related Questions

1. How do we compute higher-order coefficients in Taylor series?
2. Can the Taylor series be used to approximate $f(x) = x^3$ near other points besides $x = 1$?
3. How does the radius of convergence affect the accuracy of a Taylor series?
4. What happens if we expand $f(x) = x^3$ around a different point, say $x = 0$?
5. What is the geometric interpretation of the Taylor series for a cubic function?

Tip:

To approximate functions using Taylor series effectively, focus on finding the terms up to the degree where further terms don't significantly contribute to the result within your desired range of accuracy.