{"schema_version":"ocme-site-results-v1","locale":"en","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-discrete-logistic-growth"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-discrete-logistic-growth-01","title":{"zh-Hant":"離散 logistic 增量模型：[0,K] 的不變性、固定點與一條短軌道","en":"Discrete logistic increment model: invariance of [0,K], fixed points and a short orbit"},"summary":{"zh-Hant":"對模型 G(x)=x+r x(1−x/K)（K>0、0≤r≤1），證明：若 0≤x≤K，則 0≤G(x)≤K，因此從 [0,K] 出發的整條軌道都留在 [0,K]；0 與 K 是固定點，且 r>0 時它們是僅有的固定點。再以 K=8、r=1、x0=2 精確算出 x1、x2、x3，並以 r=3 的例子說明條件 r≤1 不可省。","en":"For the model G(x)=x+r x(1−x/K) with K>0 and 0≤r≤1, prove: if 0≤x≤K then 0≤G(x)≤K, so the whole orbit starting in [0,K] stays in [0,K]; 0 and K are fixed points, and for r>0 they are the only ones. Then compute x1, x2, x3 exactly for K=8, r=1, x0=2, and use an example with r=3 to show that the condition r≤1 cannot be dropped."},"mko_ids":["mko-discrete-logistic-growth"]}],"title":"Concept derivation guides"}