Question:
Find the Taylor series expansion (up to the first 4 non-zero terms) of the function $f(x) = \sqrt[3]{x+27}$ about the point $a=0$. Use the series to estimate $\sqrt[3]{27.1}$, up to 4 significant figures.
My Attempt:
Given the function $f(x) = \sqrt[3]{x+27}$ and the point $a=0$, we want to find the Taylor series expansion of $f(x)$ up to the first 4 non-zero terms.
First, let's find the derivatives of $f(x)$ at $a=0$:
$$f'(x) = \frac{1}{3(x+27)^{2/3}}, \quad f'(0) = \frac{1}{27}$$ $$f''(x) = -\frac{2}{9(x+27)^{5/3}}, \quad f''(0) = -\frac{2}{2187}$$ $$f'''(x) = \frac{10}{27(x+27)^{8/3}}, \quad f'''(0) = \frac{10}{177147}$$
Now, we can write the Taylor series expansion of $f(x)$ around $a=0$:
$$f(x) = f(0) + xf'(0) + \frac{x^2f''(0)}{2!} + \frac{x^3f'''(0)}{3!} + \cdots$$
Substituting the values we found earlier, we get:
$$f(x) = 3 + \frac{x}{27} - \frac{x^2}{2187} + \frac{5x^3}{531441} + \cdots$$
To estimate $\sqrt[3]{27.1}$ using the series, we can plug in $x = 0.1$:
$$\sqrt[3]{27.1} \approx 3 + \frac{0.1}{27} - \frac{(0.1)^2}{2187} + \frac{5(0.1)^3}{531441} \approx 3.0037$$
Therefore, $\sqrt[3]{27.1}$ is approximately $3.0037$ up to 4 significant figures.
Am I right?
Yes you are correct. You can use a CAS software such as the open source and free software maxima to check your computation.
(%i1) taylor((x+27)^(1/3), x, 0, 5);
(%o1) $3+\frac{x}{27}-\frac{x^2}{2187}+\frac{5\,x^3}{531441}-\frac{10\,x^4}{43046721}+\frac{22\,x^5}{3486784401}+\cdots $
Note that you can start by factoring 3 as $$\sqrt[3]{x+27} = 3 \sqrt[3]{1+ {x\over 27}}$$
Then the binomial formula works for non-integer exponents as long as $|x|<1$. $(1+x)^\alpha = 1 + \pmatrix{1 \cr \alpha}x + \pmatrix{2 \cr \alpha}x +\pmatrix{3 \cr \alpha}x ...$