Let \(I\) be an open interval containing \(0\). Suppose \(f'\) exists in a neighborhood of \(0\) and \(f''(0)\) exists. Show that \[ f(x)=f(0)+f'(0)\sin x+\frac12f''(0)\sin^2x+o(x^2) \qquad(x\to0). \]
Solution
1. As \(x\to0\),\[ f'(x) = f'(0)+f''(0)x+o(x). \]
The existence of \(f''(0)\) means precisely that \[ \frac{f'(x)-f'(0)}{x} \longrightarrow f''(0). \] Rearranging this limit gives the stated first-order expansion of \(f'\).
2. The ordinary second-order Peano expansion holds:\[ f(x) = f(0)+f'(0)x+\frac12f''(0)x^2+o(x^2). \]
Set \[ R(x) \coloneqq f(x)-f(0)-f'(0)x-\frac12f''(0)x^2. \] Then \(R(0)=0\) and, by step 1, \[ R'(x) = f'(x)-f'(0)-f''(0)x = o(x). \]
For \(x\neq0\) sufficiently close to \(0\), Cauchy’s mean value theorem applied to \(R(t)\) and \(t^2\) on the interval with endpoints \(0\) and \(x\) gives a point \(\xi\) strictly between those endpoints such that \[ \frac{R(x)}{x^2} = \frac{R'(\xi)}{2\xi}. \] As \(x\to0\), also \(\xi\to0\), and step 1 gives \[ \frac{R'(\xi)}{2\xi}\longrightarrow0. \] Hence \(R(x)=o(x^2)\).
3. As \(x\to0\),\[ x = \sin x+o(x^2) \]and\[ x^2 = \sin^2x+o(x^2). \]
The standard expansion \[ \sin x=x+O(x^3) \] gives \[ x-\sin x=O(x^3)=o(x^2). \] Moreover, \[ x^2-\sin^2x = (x-\sin x)(x+\sin x) = O(x^4) = o(x^2). \]
4. Therefore\[ \boxed{ f(x) = f(0)+f'(0)\sin x +\frac12f''(0)\sin^2x +o(x^2) }. \]
2. Multiplication by the fixed constants \(f'(0)\) and\(\frac12f''(0)\) preserves the \(o(x^2)\) error, and a finite sum of\(o(x^2)\) terms is again \(o(x^2)\).
5. Q.E.D.
Step 4 is the desired expansion.