设f(x)在区间[a,b]上满足a≤f(x)≤b,且有|f’(x)|≤q<1,令μn=f(μn-1)(n=1,2,…),μ0∈[a,b],证明:级数(μn+1一μn)绝对收敛.

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问题 设f(x)在区间[a,b]上满足a≤f(x)≤b,且有|f(x)|≤q<1,令μn=f(μn-1)(n=1,2,…),μ0∈[a,b],证明:级数n+1一μn)绝对收敛.

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答案由|μn+1一μn|=|f(μn)一f(μn-1)|=|f1)||μn一μn-1|≤q|μn一μn-1|≤q2|μn-1-μn-2|≤…≤qn|μ1一μ0| [*]

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