1.用归纳法可证：φ∈C^∞(R)==>g∈C^∞(R)==>h∈C^∞(R),且容易验证：h(x)=1,|x|≤1/2h(x)=0,|x|≥1.==>h^((k))(x)=0,|x|≤1/2或|x|≥1.2.设函数列U(n)(x)=h(ζ(n)x)a(n)x^n/n!根据1。得：U(n)∈C^∞(R)，只要证明在任意[-r,r]上，∑{n≥0}U^((k))(n)(x)"normalment"收敛，则有f∈C^∞(R),其中U^((k))(n)为k阶导数.下面证明：在任意[-r,r]上，∑{n≥0}U^((k))(n)(x))"normalment"收敛。设M(n)=Max{|h^((k))(x)|,x∈R},L(k)=Max{M(s),s≤k}U^((k))(n)(x)==∑{k≥s≥0}[C(s,k)h^((s))(ζ(n)x)[ζ(n)]^s]**[a(n)x^(n-k+s)/(n-k+s)!]由于|h^((s))(ζ(n)x)[ζ(n)x]^s]a(n)x|≤M(s)==>当n≥k+1时，|U^((k))(n)(x)|≤ ∑{k≥s≥0}[C(s,k)M(s)]**[|x|^(n-k-1)/(n-k+s)!]≤≤[L(k)r^(n-k-1)/(n-k)!][∑{k≥s≥0}C(s,k)]==2^kL(k)r^(n-k-1)/(n-k)!==>∑{n≥2k+1}2^kL(k)r^(n-k-1)/(n-k)!收敛==》在任意[-r,r]上，∑{n≥0}U^((k))(n)(x))"normalment"收敛。==》f∈C^∞(R)。3。f^((k))(0)=∑{n≥0}U^((k))(n)(0)ⅰ.n>k,==>n-k+s>0==>U^((k))(n)(0)==∑{k≥s≥0}[C(s,k)h^((s))(ζ(n)0)[ζ(n)]^s]**[a(n)0^(n-k+s)/(n-k+s)!]=0ⅱ.n<k,U^((k))(n)(0)==∑{k≥s≥k-n}[C(s,k)h^((s))(ζ(n)0)[ζ(n)]^s]**[a(n)0^(n-k+s)/(n-k+s)!]==C(k-n,k)h^((k-n))(ζ(n)0)[ζ(n)]^(k-n)]a(n)=0ⅲ.n=kU^((k))(n)(0)=C(0,k)h(ζ(n)0)a(k)=a(k)==>f^((k))(0)=a(k).