Analysis | Part IA, 2006

Prove that if the function ff is infinitely differentiable on an interval (r,s)(r, s) containing aa, then for any x(r,s)x \in(r, s) and any positive integer nn we may expand f(x)f(x) in the form

f(a)+(xa)f(a)+(xa)22!f(a)++(xa)nn!f(n)(a)+Rn(f,a,x),f(a)+(x-a) f^{\prime}(a)+\frac{(x-a)^{2}}{2 !} f^{\prime \prime}(a)+\cdots+\frac{(x-a)^{n}}{n !} f^{(n)}(a)+R_{n}(f, a, x),

where the remainder term Rn(f,a,x)R_{n}(f, a, x) should be specified explicitly in terms of f(n+1)f^{(n+1)}.

Let p(t)p(t) be a nonzero polynomial in tt, and let ff be the real function defined by

f(x)=p(1x)exp(1x2)(x0),f(0)=0.f(x)=p\left(\frac{1}{x}\right) \exp \left(-\frac{1}{x^{2}}\right) \quad(x \neq 0), \quad f(0)=0 .

Show that ff is differentiable everywhere and that

f(x)=q(1x)exp(1x2)(x0),f(0)=0,f^{\prime}(x)=q\left(\frac{1}{x}\right) \exp \left(-\frac{1}{x^{2}}\right) \quad(x \neq 0), \quad f^{\prime}(0)=0,

where q(t)=2t3p(t)t2p(t)q(t)=2 t^{3} p(t)-t^{2} p^{\prime}(t). Deduce that ff is infinitely differentiable, but that there exist arbitrarily small values of xx for which the remainder term Rn(f,0,x)R_{n}(f, 0, x) in the Taylor expansion of ff about 0 does not tend to 0 as nn \rightarrow \infty.

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