线性代数网课代修|ENGO361代写|ENGO361英文辅导|ENGO361

如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决!

\]Since\[\sum_{i=1}^{c}\alpha_{i}\leq(q-1)\alpha,\qquad\sum_{i=1}^{c}\beta_{i}\leq(p+q- 1)\beta\qquad\text{and}\qquad\sum_{i=1}^{c}\gamma_{i}\leq(p-1)\gamma,\]from (1) we obtain\[e(G_{j}) \leq\frac{1}{xy}(x\alpha_{j}+x\beta_{j}+x\gamma)(y\alpha+y\beta_ {j}+y\gamma_{j})\] \[\leq\frac{1}{4xy}\big{(}y\alpha+x\alpha_{j}+(x+y)\beta_{j}+x \gamma+y\gamma_{j}\big{)}^{2}\] \[\leq\frac{1}{4c^{2}xy}\big{(}(yc+x(q-1))\alpha+(x+y)(p+q-1) \beta+(xc+y(p-1))\gamma\big{)}^{2}. \tag{2}\]In particular, for \(x=1\) and \(y=1\), since \(p\leq q\) and \(\alpha+\gamma=n-\beta\), this gives\[e(G_{j}) \leq\frac{1}{4c^{2}}\big{(}(c+q-1)\alpha+2(p+q-1)\beta+(c+p-1) \gamma\big{)}^{2}\] \[\leq\frac{1}{4c^{2}}\big{(}(c+q-1)(\alpha+\gamma)+2(p+q-1)\beta \big{)}^{2}\] \[=\frac{1}{4c^{2}}\big{(}(c+q-1)n+(2p+q-1-c)\beta\big{)}^{2}.\]If \(c\leq t_{1}\), then the above expression is maximized at \(\beta=n\) and we obtain\[e(G_{j})\leq\frac{(p+q-1)^{2}}{c^{2}}n^{2},\]which gives the first bound in both cases of the theorem. This bound is achieved if \(A=\emptyset\), \(C=\emptyset\) and \(B\) is divided into \(\binom{c}{p+q-1}\) equal-sized sets, one for each subset of \(p+q-1\) colors assigned to the set, with edges in \(G_{i}\), for \(i\in[c]\), between any vertices from sets having color \(i\) assigned. This is depicted in Consider now \(c\geq t_{1}\) and take \(x=c-p-q+1\) and \(y=p\). This gives\[yc+x(q-1)=(x+y)(p+q-1)=(c-q+1)(p+q-1),\]so from (2) we obtain\[e(G_{j}) \leq\frac{1}{4c^{2}p(c\!-\!p\!-\!q\!+\!1)}\big{(}(c-q+1)(p+q-1)( \alpha+\beta)+((c-p-q+1)c+p(p-1))\gamma\big{)}^{2}\] \[=\frac{1}{4c^{2}p(c\!-\!p\!-\!q\!+\!1)}\big{(}(c-q+1)(p+q-1)n-(pq-( c-p-q+1)^{2})\gamma\big{)}^{2}.\]If \(c\leq t_{3}\), then the above expression is maximized at \(\gamma=0\) and gives\[e(G_{j})\leq\frac{(c-q+1)^{2}(p+q-1)^{2}}{4c^{2}p(c-p-q+1)}n^{2}.\] This bound is achieved when \(C=\emptyset\), the set \(B\) is of size \(\frac{(q-1)(p+q-1)-c(q-p-1)}{2p(c-p-q+1)}n\) and is divided into \(\binom{c}{p+q-1}\) equal-sized sets, one for each subset of \(p+q-1\) colors assigned to the set, while set \(A\) is of size \(\frac{(p+q-1)(c-2p-q+1)}{2p(c-p-q+1)}n\) and is divided into \(\binom{c}{q-1}\) equal-sized sets, one for each subset of \(q-1\) colors assigned to the set. We put edges in \(G_{i}\), for \(i\in[c]\), between any vertex from a set in \(A\cup B\) having color \(i\) assigned to any other vertex in \(A\) or a vertex in \(B\) having color \(i\) assigned. This construction is possible only if the mentioned sizes of sets \(A\) and \(B\) are non-negative, which occurs when \(c\geq t_{1}\) and \(c\leq t_{2}\), and so it proves the second bound in both cases of the theorem. Note that if \(q\leq p+1\), then the size of \(B\) is always positive, which explains why \(t_{2}\) is defined in this way. This construction is illustrated in In order to prove the third bound in the first case of the theorem, consider \(c\) satisfying \(t_{2}\leq c\leq t_{4}\) and take \(x=c\) and \(y=q-1\). From (2) we get\[e(G_{j})\leq\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)\alpha+(c+q-1)(p+q-1)\beta+(c^{ 2}+(p-1)(q-1))\gamma\big{)}^{2}.\]The assumption \(c\leq t_{4}\) implies that \(c^{2}+(p-1)(q-1)\leq 2c(q-1)\), so\[e(G_{j}) \leq\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)(\alpha+\gamma)+(c+q-1)(p+ q-1)\beta\big{)}^{2}\] \[=\frac{1}{4c^{3}(q-1)}\big{(}2c(q-1)n-(c(q-p-1)-(q-1)(p+q-1)) \beta\big{)}^{2}.\]Since \(c\geq t_{2}\), the expression above is maximized at \(\beta=0\) and gives\[e(G_{j})\leq\frac{q-1}{c}n^{2},\] which proves the third bound in the first case of the theorem. This is the same bound as when forbidding a rainbow \(S_{0,q}\), so it is achieved when \(B=C=\emptyset\), while set \(A\) is divided into \(\binom{c}{q-1}\) equal-sized sets, one for each subset of \(q-1\) colors assigned to the set, and we put edges in \(G_{i}\), for \(i\in[c]\), between any vertex from a set having color \(i\) assigned to any other vertex. This is depicted for \(q=3\) and \(c=3\) in Finally, consider the last remaining bound, where we have \(c\geq t_{3}\) and \(c\geq t_{4}\). From (2) for \(x=c-p+1\) and \(y=c-q+1\) we obtain\[e(G_{j}) \leq\frac{1}{4c^{2}(c\!-\!p\!+\!1)(c\!-\!q\!+\!1)}\big{(}(c^{2}-(p \!-\!1)(q\!-\!1))(\alpha+\gamma)+(2c-p-q+2)(p+q-1)\beta\big{)}^{2}\] \[=\frac{1}{4c^{2}(c\!-\!p\!+\!1)(c\!-\!q\!+\!1)}\big{(}(c^{2}-(p\! -\!1)(q\!-\!1))n-((c-p-q+1)^{2}-pq)\beta\big{)}^{2}.\]Since \(c\geq t_{3}\), the above expression is maximized at \(\beta=0\) and gives\[e(G_{j})\leq\frac{(c^{2}-(p-1)(q-1))^{2}}{4c^{2}(c-p+1)(c-q+1)}图片描述

发表回复

您的电子邮箱地址不会被公开。 必填项已用 * 标注