青云英语翻译

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翻译结果1翻译结果2 翻译结果3翻译结果4翻译结果5

翻译结果1复制译文编辑译文朗读译文返回顶部

假定在时间t1机器人enterwhereγ()= 0。然后我们从方程(20)和(21),dξ= 0和d2ξ= 0。应用引理1(2)部分给DZD = 0和d2zd = 0.sinceð= 0,这意味着DZ = DZD = 0。此外γ()= 0意味着,我们的要求onφ(),至少是对他们的大。书面d2z = 0和d = 0组件的形式给d2zid = 0和DI = 0。

翻译结果2复制译文编辑译文朗读译文返回顶部

猜想对时间的 t1 的那机器人 enterwhere?()= 0.然后我们有从等式 (20) 和 (21) 那 d?= 0 和 d2?= 0.应用辅助定理 1 部分 (2) 给 dZd = 0 和 d2Zd = 0.Since d = 0 上,这意味着 dZ = dZd = 0。此外?()= 0 意味着,按我们的要求 onf(),至少之一 thei 是大的。写 d2Z = 0 和 d = 在组成形式中的 0 给 d2zid = 0 和二 = 0。

翻译结果3复制译文编辑译文朗读译文返回顶部

假设在时间 t1 的机器人 enterwhere (γ) = 0。然后我们有从方程 (20) 和 (21) 的 dξ = 0 和 d2ξ = 0。应用引理 1 零件 (2) 使 dZd = 0 和 d2Zd=0.Since d = 0,这意味着 dZ = dZd = 0。加法 (γ) = 0 意味着,由我们要求 onφ(),这至少是一个只管很大。写作 d2Z = 0 和 d = 0 的组件形式给 d2zid = 0 和迪 = 0。

翻译结果4复制译文编辑译文朗读译文返回顶部

假设在时间t1机器人enterwhere銄()=0。 然后我们从方程(20)和(21),dξ=0和d2阀门ξ=0)。 lemma申请1件(2)使dzd=0和d2zd=0。由于d=0,这意味着dz=dzd=0。 除了将()=0意味着,我们的要求onφ(),其中至少一个‘slow也较大。 以书面形式d2z=0,d=0在“组件”表单将为d2看见=0和di=0。

翻译结果5复制译文编辑译文朗读译文返回顶部

假设在时间t1机器人enterwhere γ () =0。 然后我们有从式(20)和(21)那dξ=0和d2ξ=0。 申请1的题词(第2部分)给dZd=0和d2Zd=0.Since d=0,这暗示dZ=dZd =0。 另外γ () =0由我们的要求onφ暗示, (),那至少一个thei是大。 写d2Z=0和d=0在组分形式给d2zid=0和di=0。
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