Taylor series $(x+1)^{\frac{1}{3}}$

Question

Complete the Maclaurin polynomial of degree three for $(x+1)^{1/3}$.

I have completed the first two derivatives of this function and thus have coefficients but am not certain how to put them into the polynomial form.

Thank you!

Answer 1

If you have the first three derivatives correct, then the polynomial is $$ f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3 $$ where $f(x)=(1+x)^{1/3}$.

If you know the general binomial theorem, you have $$ (1+x)^\alpha=\sum_{k=0}^\infty \binom{\alpha}{k}x^k $$ where the convergence is for $|x|<1$, if $\alpha$ is not a positive integer. The binomial coefficient is defined as $$ \binom{\alpha}{k}=\frac{\alpha(\alpha-1)\dots(\alpha-k+1)}{k!}. $$ So you can check your answer against this general formula.

Answer 2

$$f(0)=(x+1)^{\frac13}$$ Taylor/Maclaurian says that: $$f(x)=f(0)+f'(0)x+\frac{f''(0)}{2!}x^2+...$$ $$f(0)=(0+1)^{\frac13}=1$$ $$f'(0)=\frac13(0+1)^{\frac13-1}=\frac13$$ $$f''(0)=\frac13\left(\frac{1}3-1\right)(0+1)^{\frac13-2}=\frac13\left(\frac{1}3-1\right)$$ $$f'''(0)=\frac13\left(\frac{1}3-1\right)\left(\frac{1}3-2\right)(0+1)^{\frac13-3}=\frac13\left(\frac{1}3-1\right)\left(\frac{1}3-2\right)$$ Now, : $$f(x)=1+\frac13\frac{x}{1!}+\frac13\left(\frac{1}3-1\right)\frac{x^2}{2!}+\frac13\left(\frac{1}3-1\right)\left(\frac{1}3-2\right)\frac{x^3}{3!}+\cdots$$ The general form to which is: $$f(x)=\sum_{k=0}^{\infty}\frac{\frac13(\frac13-1)\dots(\frac13-k+1)}{k!}x^k$$