To obtain the binomial series for the function $\frac{x^2}{\sqrt{2} + x}$ under the condition $|x| < 1$, we can approach this problem step by step, starting with simplifying the expression and then using the binomial expansion.

Step 1: Express the function in a form suitable for the binomial expansion

We begin by rewriting the expression $\frac{x^2}{\sqrt{2} + x}$. Notice that the binomial series expansion applies when we have a term of the form $(1 + u)^n$, where $|u| < 1$.

Start by factoring out $\sqrt{2}$ from the denominator:

$\frac{x^2}{\sqrt{2} + x} = \frac{x^2}{\sqrt{2}\left(1 + \frac{x}{\sqrt{2}}\right)}$

This simplifies to:

$\frac{x^2}{\sqrt{2} + x} = \frac{1}{\sqrt{2}} \cdot \frac{x^2}{1 + \frac{x}{\sqrt{2}}}$

Now we have the function $\frac{x^2}{1 + \frac{x}{\sqrt{2}}}$, and we will apply the binomial series expansion to this part.

Step 2: Apply the binomial series

We know that for $|u| < 1$, the binomial series expansion for $(1 + u)^{-1}$ is:

$(1 + u)^{-1} = \sum_{n=0}^{\infty} (-1)^n u^n$

In our case, $u = \frac{x}{\sqrt{2}}$, so the expansion of $\frac{1}{1 + \frac{x}{\sqrt{2}}}$ is:

$\frac{1}{1 + \frac{x}{\sqrt{2}}} = \sum_{n=0}^{\infty} (-1)^n \left(\frac{x}{\sqrt{2}}\right)^n$

This can be written as:

$\frac{1}{1 + \frac{x}{\sqrt{2}}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(\sqrt{2})^n}$

Step 3: Multiply by $x^2$

Now, we multiply the above series by $x^2$ to get the final expression for $\frac{x^2}{1 + \frac{x}{\sqrt{2}}}$:

$\frac{x^2}{1 + \frac{x}{\sqrt{2}}} = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{(\sqrt{2})^n}$

This simplifies to:

$\frac{x^2}{1 + \frac{x}{\sqrt{2}}} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+2}}{(\sqrt{2})^n}$

Step 4: Include the $\frac{1}{\sqrt{2}}$ factor

Finally, remember that we factored out $\frac{1}{\sqrt{2}}$ in Step 1. Therefore, the full expression for $\frac{x^2}{\sqrt{2} + x}$ is:

$\frac{x^2}{\sqrt{2} + x} = \frac{1}{\sqrt{2}} \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+2}}{(\sqrt{2})^n}$

Final Binomial Series

Thus, the binomial series expansion for $\frac{x^2}{\sqrt{2} + x}$ is:

$\frac{x^2}{\sqrt{2} + x} = \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+2}}{(\sqrt{2})^{n+1}}$

Conclusion

We have derived the binomial series for $\frac{x^2}{\sqrt{2} + x}$ for $|x| < 1$, which converges for $|x| < \sqrt{2}$.

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Would you like further details or clarification on any part of the solution? Also, here are 5 related questions for you to explore:

1. How would the binomial series change if the denominator had a different form, such as $1 + x^2$?
2. What are the convergence conditions for a general binomial series expansion?
3. Can you apply the binomial series to compute approximations for specific values of $x$, like $x = 0.5$?
4. How do binomial series relate to Taylor expansions in general?
5. How would you expand a function like $\frac{1}{\sqrt{1 - x}}$ using the binomial series?

Tip: When using the binomial series, always check the convergence radius, which depends on the magnitude of the argument inside the series.