Simplify \( \sqrt{\frac{1-x}{1+x^{2}}} \) using binomal theorem

To simplify \( \sqrt{\frac{1-x}{1+x^{2}}} \), we can start by rewriting the expression inside the square root. Let's factor out \( \sqrt{1+x^2} \) using the binomial expansion for \( (1+x^2)^{-1/2} \). Using the Binomial Theorem, we know that for small \( x \), \[ (1+x)^{n} \approx 1 + nx + \frac{n(n-1)}{2}x^2 + \cdots \] Here, we need to use that: \[ (1+x^2)^{-1/2} \approx 1 - \frac{1}{2}x^2 + \text{higher order terms} \] Thus, \[ \sqrt{1+x^2} \approx 1 + \frac{1}{2}x^2 \] Now, replacing \( \sqrt{1+x^2} \) in our original expression gives: \[ \frac{1-x}{\sqrt{1+x^2}} \approx \frac{1-x}{1 + \frac{1}{2}x^2} \] We can now simplify: \[ 1 - x \approx \left( 1 - x \right) \left( 1 - \frac{1}{2}x^2 \right) \approx 1 - x - \frac{1}{2}x^2 + \cdots \] Next, by simplifying \( \frac{1-x}{1 + \frac{1}{2}x^2} \) we multiply the numerator and denominator by \( 2 \): \[ \frac{2(1-x)}{2 + x^2} = \frac{2 - 2x}{2 + x^2} \] Next, we can use Taylor expansion or substitutive entries again if necessary. Thus, \[ \sqrt{\frac{1-x}{1+x^2}} \approx \sqrt{\frac{2 - 2x}{2 + x^2}} \] This can be a bit more complicated based on how much simplification is desired and how \( x \) may behave under various ranges. In conclusion, while it’s complex due to \( x \) altering the balance, we’ve utilized a binomial theme to explore the root of the quotient.