How do I get the power series representation of the square root of (8+x^3)?

How do I get the power series representation of the square root of (8+x^3)?

💝

Please Give me  Love react👇🙏🙏🙏

Send your opinion 👇🙏🙏🙏

To find the power series representation of $\sqrt{8+{𝑥}^3}$, we will use the binomial series expansion for non-integer exponents. The binomial series for $(1+𝑢{)}^{𝑛}$ where $𝑛$ is any real number and $\mathrm{\mid }𝑢\mathrm{\mid }<1$ is given by:

$$(1+𝑢{)}^{𝑛}=\sum _{𝑘=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{𝑛}{𝑘}\right){𝑢}^{𝑘}$$

where $\left(\genfrac{}{}{0px}{}{𝑛}{𝑘}\right)$ is the generalized binomial coefficient:

$$\left(\genfrac{}{}{0px}{}{𝑛}{𝑘}\right)=\frac{𝑛(𝑛-1)(𝑛-2)\cdots (𝑛-𝑘+1)}{𝑘!}$$

First, rewrite $\sqrt{8+{𝑥}^3}$ in a form suitable for the binomial expansion:

$$\sqrt{8+{𝑥}^3}=\sqrt{8(1+\frac{{𝑥}^3}{8})}=\sqrt{8}\sqrt{1+\frac{{𝑥}^3}{8}}=2\sqrt{2}\sqrt{1+\frac{{𝑥}^3}{8}}$$

Now, let $𝑢=\frac{{𝑥}^3}{8}$ and $𝑛=\frac{1}{2}$. We then expand $\sqrt{1+𝑢}$ as follows:

$$\sqrt{1+𝑢}=(1+𝑢{)}^{\frac{1}{2}}=\sum _{𝑘=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{\frac{1}{2}}{𝑘}\right){𝑢}^{𝑘}$$

Using the binomial coefficient for $𝑛=\frac{1}{2}$:

$$\left(\genfrac{}{}{0px}{}{\frac{1}{2}}{𝑘}\right)=\frac{\frac{1}{2}(\frac{1}{2}-1)(\frac{1}{2}-2)\cdots (\frac{1}{2}-𝑘+1)}{𝑘!}$$

Now substitute $𝑢=\frac{{𝑥}^3}{8}$:

$$\sqrt{1+\frac{{𝑥}^3}{8}}=\sum _{𝑘=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{\frac{1}{2}}{𝑘}\right){\left(\frac{{𝑥}^3}{8}\right)}^{𝑘}$$

​

 Thus:

$$\sqrt{8+{𝑥}^3}=2\sqrt{2}\sum _{𝑘=0}^{\mathrm{\infty }}\left(\genfrac{}{}{0px}{}{\frac{1}{2}}{𝑘}\right){\left(\frac{{𝑥}^3}{8}\right)}^{𝑘}$$