Let the function $f\left( x \right) = \left( {\frac{{1 - \cos x}} {{{x^2}}}} \right)\cos (3x)$ if $x\ne 0$ and $f(0)=\frac{1}{2}$. Prove that any derivative of $f$ is bounded on $\mathbb{R}$. Thank so much for helping.
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The original function is an entire function hence its derivatives are bounded on $[-1,1]$. Moreover, any derivative of $\frac{1-\cos x}{x^2}$ is bounded on $\mathbb{R}\setminus[-1,1]$ since it is a linear combination of functions of the form $\frac{f(x)}{x^k}$ where $f(x)$ is bounded trigonometric function and $k\geq 2$; so the same holds for $\frac{1-\cos x}{x^2}\cos(3x)$ over $\mathbb{R}$.
answered Sep 19 '14 at 13:11
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$\begingroup$ Can you explain more precisely for me why any derivative of $\frac{1-\cos x}{x^2}$ is a linear combination of functions of the form $\frac{f(x)}{x^k}$ ? $\endgroup$ – Cao Sep 19 '14 at 13:20
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$\begingroup$ Just use the formula for differentiating a product of two functions $n$ times. Any derivative of $1-\cos x$, as well as any derivative of $\frac{1}{x^2}$, is bounded on $\mathbb{R}\setminus[-1,1]$. $\endgroup$ – Jack D'Aurizio Sep 19 '14 at 13:31
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