First, we will analyze both sides of the given approximation to see if it holds true under the condition that [tex]\( x \)[/tex] is a large number. This means we will use a series expansion to approximate the more complex expression.
We need to compare:
[tex]\[
\sqrt{1 - \frac{\pi^2}{2x^2}} \quad \text{and} \quad 1 - \frac{\pi^2}{4x^2}
\][/tex]
Let's start with the left side, [tex]\( \sqrt{1 - \frac{\pi^2}{2x^2}} \)[/tex]. For a large [tex]\( x \)[/tex], the term [tex]\( \frac{\pi^2}{2x^2} \)[/tex] is very small. We can use the binomial expansion for a square root function [tex]\( \sqrt{1 - y} \)[/tex] for small [tex]\( y \)[/tex]:
[tex]\[
\sqrt{1 - y} \approx 1 - \frac{y}{2} - \frac{y^2}{8} + \dots
\][/tex]
Here [tex]\( y = \frac{\pi^2}{2x^2} \)[/tex], so substituting [tex]\( y \)[/tex] into the expansion, we get:
[tex]\[
\sqrt{1 - \frac{\pi^2}{2x^2}} \approx 1 - \frac{1}{2} \left( \frac{\pi^2}{2x^2} \right) - \frac{1}{8} \left( \frac{\pi^2}{2x^2} \right)^2 + \dots
\][/tex]
Simplifying the first two terms in our expansion:
[tex]\[
\sqrt{1 - \frac{\pi^2}{2x^2}} \approx 1 - \frac{\pi^2}{4x^2} - \frac{\pi^4}{32x^4} + \dots
\][/tex]
Now, let's consider the right side of the equation given by [tex]\( 1 - \frac{\pi^2}{4x^2} \)[/tex].
Comparing our series expansion for the left side with the right side:
[tex]\[
\sqrt{1 - \frac{\pi^2}{2x^2}} \approx 1 - \frac{\pi^2}{4x^2} - \frac{\pi^4}{32x^4} + \dots
\][/tex]
[tex]\[
1 - \frac{\pi^2}{4x^2}
\][/tex]
We see that the main term [tex]\( 1 - \frac{\pi^2}{4x^2} \)[/tex] on the right side matches the first two terms of the series expansion for [tex]\( \sqrt{1 - \frac{\pi^2}{2x^2}} \)[/tex]. The higher-order terms (like [tex]\( \frac{\pi^4}{32x^4} \)[/tex]) are relatively small and tend to diminish rapidly for large [tex]\( x \)[/tex].
Thus, for large [tex]\( x \)[/tex], the approximation:
[tex]\[
\sqrt{1 - \frac{\pi^2}{2x^2}} \approx 1 - \frac{\pi^2}{4x^2}
\][/tex]
holds true to a good degree of accuracy.
