作业帮数学英语翻译有人会吗?
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数学英语翻译有人会吗?
Definition 1. A lattice path of length n is a sequence (γ0, . . . , γn) of points γi in the plan Z × Z for all 0 ≤ i ≤ n and such that γi+1 − γi = (1, 0) (east-step) or (0, 1) (north-step) for 1 ≤ i ≤ n − 1.
As shown by Arworn [1], we can encode each homomorphism f ∈ Hom1 (Pn, Pk) by a lattice path γ = (γ0, . . . , γn−1) in N × N between the lines y = x and y = x − k + 1 as follows:
• γ0 = (0, 0), and for j = 1, . . . , n − 1,
• γj+1 = γj + (1, 0) if f(j) > f(j − 1),
• γj+1 = γj + (0, 1) if f(j) < f(j − 1).
For example, if the images of successive vertices of f ∈ Hom(P15, P11) are
1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 6, 5;
then the corresponding lattice path is given by Figure 1.
Definition 2. For nonnegative integers n,m, t, s, Let L(n,m) be the set of all the lattice paths from the origin to (n,m) and L(n,m; t, s) the set of lattice paths in L(n,m) that stay between the lines y = x + t and y = x − s (being allowed to touch them), where n + t ≥ m ≥ n − s.
Lemma 4. Let K = min(⌊n+k 2 ⌋, n), then
Proof. It follows from the above correspondence that each homomorphism from Pn to Pk
is encoded by a lattice path in some L(#E,#N; 0, k − 1), where #E is the number of4
ZHICONG LIN AND JIANG ZENG
Figure 2. Reflection of the segment of the path from O to the first reaching point F with respect to the line y = x + t + 1.
east-steps and #N the number of north-steps. The path structures require that
Therefore, we must have #E ≥ (n − 1)/2, #E ≤ n − 1 and #E ≤ (k + n − 2)/2.
To evaluate the sum in (2.3), we need a formula for the cardinality of L(n,m; t, s). First of all, each lattice path in L(n,m) can be encoded by a word of length n+m on the alphabet {A,B} with n letters A and m letters B. So, the cardinality of L(n,m) is given by the binomial coefficient. Next, each lattice path in L(n,m) which passes above the line y = x + t (or reaching the line y = x + t + 1) can be mapped to a lattice path from (−t − 1, t + 1) to (n,m) by the reflection with respect to the line y = x + t + 1 (see Figure 2). Hence, there are such lattice paths. Therefore, the number of lattice paths in L(n,m) which do not pass above the line y = x + t (or not reaching the line y = x + t + 1), where m ≤ n + t, is given by
By a similar reasoning, we can prove the following known result (see [4, Lemma 4A], for
example). For the reader’s convenience, we provide a sketch of the proof.
Lemma 5. The cardinality of L(n,m; t, s) is given by
Definition 1. A lattice path of length n is a sequence (γ0, . . . , γn) of points γi in the plan Z × Z for all 0 ≤ i ≤ n and such that γi+1 − γi = (1, 0) (east-step) or (0, 1) (north-step) for 1 ≤ i ≤ n − 1.
As shown by Arworn [1], we can encode each homomorphism f ∈ Hom1 (Pn, Pk) by a lattice path γ = (γ0, . . . , γn−1) in N × N between the lines y = x and y = x − k + 1 as follows:
• γ0 = (0, 0), and for j = 1, . . . , n − 1,
• γj+1 = γj + (1, 0) if f(j) > f(j − 1),
• γj+1 = γj + (0, 1) if f(j) < f(j − 1).
For example, if the images of successive vertices of f ∈ Hom(P15, P11) are
1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 5, 6, 5, 6, 5;
then the corresponding lattice path is given by Figure 1.
Definition 2. For nonnegative integers n,m, t, s, Let L(n,m) be the set of all the lattice paths from the origin to (n,m) and L(n,m; t, s) the set of lattice paths in L(n,m) that stay between the lines y = x + t and y = x − s (being allowed to touch them), where n + t ≥ m ≥ n − s.
Lemma 4. Let K = min(⌊n+k 2 ⌋, n), then
Proof. It follows from the above correspondence that each homomorphism from Pn to Pk
is encoded by a lattice path in some L(#E,#N; 0, k − 1), where #E is the number of4
ZHICONG LIN AND JIANG ZENG
Figure 2. Reflection of the segment of the path from O to the first reaching point F with respect to the line y = x + t + 1.
east-steps and #N the number of north-steps. The path structures require that
Therefore, we must have #E ≥ (n − 1)/2, #E ≤ n − 1 and #E ≤ (k + n − 2)/2.
To evaluate the sum in (2.3), we need a formula for the cardinality of L(n,m; t, s). First of all, each lattice path in L(n,m) can be encoded by a word of length n+m on the alphabet {A,B} with n letters A and m letters B. So, the cardinality of L(n,m) is given by the binomial coefficient. Next, each lattice path in L(n,m) which passes above the line y = x + t (or reaching the line y = x + t + 1) can be mapped to a lattice path from (−t − 1, t + 1) to (n,m) by the reflection with respect to the line y = x + t + 1 (see Figure 2). Hence, there are such lattice paths. Therefore, the number of lattice paths in L(n,m) which do not pass above the line y = x + t (or not reaching the line y = x + t + 1), where m ≤ n + t, is given by
By a similar reasoning, we can prove the following known result (see [4, Lemma 4A], for
example). For the reader’s convenience, we provide a sketch of the proof.
Lemma 5. The cardinality of L(n,m; t, s) is given by
定义1.一个晶格路径的长度n是一个序列(γ0,.,γn)的分γi在Z计划为所有0×Z≤≤n和我这样γi + 1−γi =(1,0)(东步骤)或(0,1)(北一步)1≤≤n−1我.
如图所示,Arworn[1],我们可以对每个同态f Hom1∈(Pn,Pk)通过一个晶格路径γ=(γ0,.,γn−1)在N×N字里行间y = x和y = x−k + 1如下:
•γ0 =(0,0),j = 1,.,n−1,
•γj + 1 =γj +(1,0)如果f(j)> f(j−1),
•γj + 1 =γj +(0,1)如果f(j)< f(j−1).
例如,如果连续图像的顶点∈f的坎(P15,P11)
1、2、3、2、3、4、5、4、3、4、5、6、5、6、5;
那么相应的点阵路径给出了图1.
定义2.为非负整数n,m,t,s,让L(n,m)是集所有的点阵路径从原点到(n,m)和L(n,;t,s)套晶格路径在L(n,m),待字里行间y = x + t和y = x−年代(被允许碰),其中n + t m≥≥n−年代.
引理4.让K = min(⌊n + K 2⌋,n),然后
证明.它遵循从上面的对应,每个同态从Pn,Pk
是由一个晶格路径编码在一些L(# E,# N;0,k−1),其中# E是完成数量
ZHICONG林和江泽民曾庆红
图2.反映了部分的路径从O到第一个到达点F就行y = x + t + 1.
东的步骤和#北步骤的数目N.路径结构要求
因此,我们必须有# E≥(n−1)/ 2,# E≤n−1和# E≤(k + n−2)/ 2.
评价的总和(2.3),我们需要一个公式为基数的L(n,;t,s).首先,每个栅格路径在L(n,m)由一个词可以编码长度为n + m在字母表{ a、B }与n字母a和B .所以,million基数的L(n,m)是给定的二项式系数.接下来,每个栅格路径在L(n,m),通过线上方的y = x + t(或达到线y = x + t + 1)可以被映射到一个晶格路径从(−−1 t,t + 1)(n,m)的反射就行y = x + t + 1(见图2).因此,有这样的晶格路径.因此,晶格路径的数量在L(n,m),不通过线上方的y = x + t(或没达到线y = x + t + 1),m≤n + t,给出了
通过一个类似的推理,我们可以证明下面的已知的结果(参见[4、引理4 a),
例).为读者的方便,我们提供一个素描的证据.
引理5.基数的L(n,;t,s)了
如图所示,Arworn[1],我们可以对每个同态f Hom1∈(Pn,Pk)通过一个晶格路径γ=(γ0,.,γn−1)在N×N字里行间y = x和y = x−k + 1如下:
•γ0 =(0,0),j = 1,.,n−1,
•γj + 1 =γj +(1,0)如果f(j)> f(j−1),
•γj + 1 =γj +(0,1)如果f(j)< f(j−1).
例如,如果连续图像的顶点∈f的坎(P15,P11)
1、2、3、2、3、4、5、4、3、4、5、6、5、6、5;
那么相应的点阵路径给出了图1.
定义2.为非负整数n,m,t,s,让L(n,m)是集所有的点阵路径从原点到(n,m)和L(n,;t,s)套晶格路径在L(n,m),待字里行间y = x + t和y = x−年代(被允许碰),其中n + t m≥≥n−年代.
引理4.让K = min(⌊n + K 2⌋,n),然后
证明.它遵循从上面的对应,每个同态从Pn,Pk
是由一个晶格路径编码在一些L(# E,# N;0,k−1),其中# E是完成数量
ZHICONG林和江泽民曾庆红
图2.反映了部分的路径从O到第一个到达点F就行y = x + t + 1.
东的步骤和#北步骤的数目N.路径结构要求
因此,我们必须有# E≥(n−1)/ 2,# E≤n−1和# E≤(k + n−2)/ 2.
评价的总和(2.3),我们需要一个公式为基数的L(n,;t,s).首先,每个栅格路径在L(n,m)由一个词可以编码长度为n + m在字母表{ a、B }与n字母a和B .所以,million基数的L(n,m)是给定的二项式系数.接下来,每个栅格路径在L(n,m),通过线上方的y = x + t(或达到线y = x + t + 1)可以被映射到一个晶格路径从(−−1 t,t + 1)(n,m)的反射就行y = x + t + 1(见图2).因此,有这样的晶格路径.因此,晶格路径的数量在L(n,m),不通过线上方的y = x + t(或没达到线y = x + t + 1),m≤n + t,给出了
通过一个类似的推理,我们可以证明下面的已知的结果(参见[4、引理4 a),
例).为读者的方便,我们提供一个素描的证据.
引理5.基数的L(n,;t,s)了