Let \[ y(h)=1-2\sin^2(2\pi h), \qquad f(y)=\frac{2}{1+\sqrt{1-y}}. \] Justify the asymptotic formula \[ f(y(h))=2-4\sqrt2\,\pi |h|+O(h^2) \qquad(h\to0), \] with a remainder bounded in absolute value by a constant multiple of \(h^2\) near \(0\).
Solution
Set \[ u=2\pi h \] and \[ a=\sqrt2\,\abs{\sin u}. \]
1. One has\[ \sqrt{1-y(h)} = \sqrt2\,\abs{\sin(2\pi h)} = a. \]
By definition, \[ 1-y(h) = 2\sin^2(2\pi h). \] Taking the nonnegative square root gives \[ \sqrt{1-y(h)} = \sqrt2\,\sqrt{\sin^2(2\pi h)} = \sqrt2\,\abs{\sin(2\pi h)}. \]
2. Hence\[ f(y(h)) = \frac{2}{1+a} = 2-2a+\frac{2a^2}{1+a}. \]
Step 1 gives the first equality. For the second, \[ \begin{aligned} \frac2{1+a}-(2-2a) &= \frac{2-2(1-a)(1+a)}{1+a}\\ &= \frac{2a^2}{1+a}. \end{aligned} \]
3. For every real \(u\),\[ \abs{ \abs{\sin u}-\abs u } \leq \frac{\abs u^3}{6}. \]
The reverse triangle inequality gives \[ \abs{ \abs{\sin u}-\abs u } \leq \abs{\sin u-u}. \] Taylor’s theorem at \(0\) gives \[ \abs{\sin u-u} \leq \frac{\abs u^3}{6}, \] because the third derivative of \(\sin\) has absolute value at most \(1\).
4. One has\[ -2a = -4\sqrt2\,\pi\abs h + R_1(h), \]where\[ \abs{R_1(h)} \leq \frac{\sqrt2}{3}(2\pi)^3\abs h^3. \]
Since \[ a=\sqrt2\,\abs{\sin u}, \] one has \[ \begin{aligned} -2a &= -2\sqrt2\,\abs u -2\sqrt2 \left( \abs{\sin u}-\abs u \right)\\ &= -4\sqrt2\,\pi\abs h + R_1(h), \end{aligned} \] where \[ R_1(h) = -2\sqrt2 \left( \abs{\sin u}-\abs u \right). \] Step 3 gives \[ \abs{R_1(h)} \leq \frac{\sqrt2}{3}\abs u^3 = \frac{\sqrt2}{3}(2\pi)^3\abs h^3. \]
5. The second remainder term in step 2 satisfies\[ 0 \leq \frac{2a^2}{1+a} \leq 16\pi^2 h^2. \]
Since \(a\geq0\), \[ \frac{2a^2}{1+a} \leq 2a^2 = 4\sin^2u. \] The elementary bound \[ \abs{\sin u}\leq\abs u \] gives \[ 4\sin^2u \leq 4u^2 = 16\pi^2h^2. \]
6. There is a constant \(C>0\) such that for all sufficiently small\(h\),\[ \abs{ f(y(h)) - \left( 2-4\sqrt2\,\pi\abs h \right) } \leq Ch^2. \]
Combining steps 2, 4, and 5 gives \[ f(y(h)) - \left( 2-4\sqrt2\,\pi\abs h \right) = R_1(h) + \frac{2a^2}{1+a}. \] For \(\abs h\leq1\), step 4 gives \[ \abs{R_1(h)} \leq \frac{\sqrt2}{3}(2\pi)^3h^2, \] and step 5 gives \[ \frac{2a^2}{1+a} \leq 16\pi^2h^2. \] Thus the desired bound holds with, for example, \[ C = \frac{\sqrt2}{3}(2\pi)^3 + 16\pi^2. \]
7. Therefore\[ \boxed{ f(y(h)) = 2-4\sqrt2\,\pi\abs h + O(h^2) } \]as \(h\to0\).
Step 6 is exactly the stated \(O(h^2)\) estimate, with an explicit constant bounding the remainder divided by \(h^2\) near \(0\).
8. Q.E.D.
Step 7 is the required asymptotic formula.