diff --git "a/data_all_eng_slimpj/shuffled/split2/finalzzfpxm" "b/data_all_eng_slimpj/shuffled/split2/finalzzfpxm" new file mode 100644--- /dev/null +++ "b/data_all_eng_slimpj/shuffled/split2/finalzzfpxm" @@ -0,0 +1,5 @@ +{"text":"\\section{Introduction}\n\\ \\ We have recently developed methods for obtaining exact two-point resistance of certain circulant graph namely, the complete graph minus $N$ edges \\cite {Chair1}. In this paper, using similar techniques and ideas, we consider trigonometrical sums that arise in the computation of the two-point resistance of the finite resistor networks \\cite{Wu}, in the work of McCoy and Orrick on the chiral Potts model \\cite {McCoy}, and in the Verlinde dimension formula of the twisted \/untwisted space of conformal blocks of the $SO(3)$\/$SU(2)$ WZW model \\cite{Verlinde}.\n\n\\ \\ Before considering these trigonometrical sums, we test the techniques used in \\cite{Chair1}, by first deriving the Green's function of the one-dimensional lattice graphs with free boundaries, and the two-point resistance of the $N$-cycle graph \\cite{Wu}. The same techniques is then used to evaluate a trigonometrical sum that played a crucial role to prove R. F Scott's conjecture on the permanent of the Cauchy matrix \\cite{Minc,Todd}. Having tested these techniques, an alternative derivation is then given for certain trigonometrical sum that appeared in the perturbative chiral Potts model \\cite {McCoy}, \\cite {Mehta}.\n\n\\ \\ We have also considered the general case studied by Gervois and Mehta \\cite {Mehta}, Berndt and Yeap \\cite {Berndt}, here, our results agree with those in \\cite {Mehta}. It turns out that the Verlinde's dimension formulas for the untwisted space of conformal blocks, may be obtained simply by summing over certain parameter of a trigonometrical sum considered in \\cite{Mehta}. For the twisted space of conformal blocks, however, the parameter is restricted to take some value. It is shown that the dimension of the conformal blocks on a genus $g\\geq 2$ Riemann surface may be obtained through a recursion formula that relates different genera. Mathematically speaking, the dimension of the space of conformal blocks is obtained by expanding certain generating function order by order, or using the Hirzebruch-Rlemann-Roch theorem \\cite{Zagier}.\n\n\\ \\ By using the method given in \\cite {Chair1}, we are able to obtain closed form formula for the two-point resistance of a $2\\times N $ resistor network \\cite {Chair2}. In this paper, an exact computation of the corner-to-corner resistance as well as the total effective resistance of a $2\\times N$ will be given. The total effective effective resistance, also called the Kirchhoff index \\cite{Randic}, this is an invariant quantity of the resistor network or graph.\n\n\\ \\ The exact two-point resistance of an $M\\times N$ resistor network is given in terms of a double sum and not in a closed form \\cite {Wu}. Therefore, our computation carried out in this paper, represents the first non-trivial exact results for the two-point resistance of a two-dimensional resistor network.\n\n\\ \\ This method is then used to evaluate variant of trigonometrical sums, some of which are related to number theory, we hope that these trigonometrical sums will have some physical applications. It is interesting to point out that all the computations of the trigonometrical sums in this paper are based on a formula by Schwatt \\cite{ Schwatt} on trigonometrical power sums, and the representation of the binomial coefficients by the residue operator. The Schwatt's formula is modified slightly, only in the case of the corner-to-corner resistance, the Kirchhoff index and trigonometrical sum given by $ F_{1}(N,l,2) $, and $ F_{1}(N,l,2)$, see Section 6, this was also the case in our previous paper \\cite{Chair1}.\n\n\\ \\ This paper is organized as follows; in section 2, we give an explicit computations of the two-point resistance of the $N$-cycle graph and the Green's function of the one-dimensional lattice, and in Section 3, we give a simple derivation of a trigonometrical sum connected with Scott's conjecture on the permanent of the Cauchy matrix. In section 4, we consider trigonometrical sums arising in the chiral Potts model, and in the Verlinde formula of the dimension of the conformal blocks. Exact computations of the corner-to-corner and the Kirchhoff index of $2\\times N$ resistor network will be given in section 5. In Section 6, we consider other class of trigonometrical sums some of which are related to number theory, and finally, in Section 7, our conclusions are given.\n\\section{ The two-point resistance of one-dimensional lattice using the residue operator }\n\n\\ \\ In this Section, we first start with the two-point resistance of the $N$-cycle graph computations, then, move to the trigonometrical sum related to the two-point resistance of the one-dimensional lattice with free boundaries, that is, the path graph. The two-point resistance of the $N$-cycle graph between any two nodes $\\alpha$ and $\\beta$ is given by the following simple closed formula \\cite{Wu},\n\\begin{equation} \n\\label{t1}\nR(l)=\\frac{1}{N}\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^2(n\\pi\/N)}=\\frac{l(N-l)}{N},\n\\end{equation}\nwhere $l=|\\alpha-\\beta|$, and $1\\leq \\alpha,\\beta \\leq N. $\nOur derivation for the two-point resistance starts with the following trigonometrical identity $$\\cos(2ln\\pi\/N)=\\sum_{s=0}^{l}(-1)^s \\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sin^{2s}(n\\pi\/N), $$ from which the above trigonometrical sum may be rewritten as\n\\begin{equation} \n\\label{t2}\nR(l)=\\frac{1}{2N}\\sum_{s=1}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^{N-1} \\sin^{2(s-1)}(n\\pi\/N).\n\\end{equation} \nOn the other hand, Schwatt's formula for trigonometrical power sums \\cite{Schwatt}, gives\n\\begin{equation} \n\\label{t3}\n\\sum_{n=1}^{N-1} \\sin^{2(s-1)}(n\\pi\/N)=\\frac{1}{2^{2(s-1)-1}}\\sum_{t=1}^{s-1}(-1)^{t +1}\\binom {2(s-1)}{s-1-t} +\\frac{N-1}{2^{2(s-1)}}\\binom {2(s-1)} {s-1}.\n\\end{equation}\n\n\\ \\ \n Therefore, the two-point resistance may be obtained by evaluating the binomial sums in the expression of $R(l)$, based on the residue operator. This operator played a crucial role in evaluating combinatorial sums and proving combinatorial identities \\cite{Egorychev}. First, let us recall the definition of the residue operator $\\hbox{res}$. To that end, let $G(w)= \\sum_{k=0}^{\\infty}a_{k}w^{k}$ be a generating function for a sequence $\\{a_{k}\\}$. Then the k-th coefficient of $G(w)$ may be represented by the formal residue as follows\n$$a_{k}= \\hbox{res}_w G(w){w^{-k-1}}.$$\nThis is equivalent to the Cauchy integral representation of $a_k$,\n\\[a_k=\\frac{1}{2\\pi i}\\oint _{|z|=\\rho}\\frac{G(w)}{w^{k+1}}dw ,\\] for coefficients of the Taylor series in a punctured neighborhood of zero. In particular, the generating function of the binomial coefficient sequence $\\binom {n}{k}$ for a fixed $n$ is given by $$G(w)= \\sum_{k=0}^{n}\\binom {n}{k} w^{k} =(1+w)^{n},$$ \nand hence $$\\binom {n}{k}=\\hbox{res}_w (1+w){^n}{w^{-k-1}}.$$ The other binomial coefficient that we need in this paper is the following, $$\\binom {2n}{n}=\\hbox{res}_w (1-4w){^{-1\/2}}{w^{-n-1}}.\n$$\n\n\\ \\ Before finishing this brief summary, we should mention one important property of the residue operator $\\hbox{res}$, namely linearity, this is a crucial in doing computations, linearity states that; given some contants $\\alpha$ and $\\beta$, then $$\\alpha \\hbox{res}_w G_{1}(w){w^{-k-1}}+ \\beta\\hbox{res}_w G_{2}(w){w^{-k-1}}=\\hbox{res}_w(\\alpha G_{1}(w)+ \\beta G_{2}(w)) {w^{-k-1}}.$$ \nLet us now evaluate the first term in Eq.(\\ref{t2}), using the residue operator, namely the following term\n\\begin{eqnarray}\n\\label{t4}\nR_{1}(l):&=&\\frac{2}{N}\\sum_{s=1}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-1}(-1)^{t }\\binom {2(s-1)}{s-1-t}\\nonumber\\\\&=&\\frac{2}{N}\\sum_{s=1}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-1}(-1)^{t }\\hbox{res} \\frac{(1+w)^{2(s-1)}}{w^{s-t}}\\nonumber\\\\&=&\\frac{2}{N}\\sum_{s=1}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\hbox{res}_{w=0}\\frac{(1+w)^{2(s)}}{(1+w)^{3}w^{s-1}}.\n\\end{eqnarray}\nIn obtaining Eq.(\\ref{t4}), we discarded an analytic term at the pole $w=0$ of order $s-1$. By making a change of variable $l-s=k $, then, Eq. (\\ref{t4}) may be rewritten as \n\\begin{eqnarray}\n\\label{t5}\nR_{1}(l)&=&(-1)^{l+1}\\frac{2}{N}\\hbox{res}_{w=0}\\frac{w}{(1+w)^{3}}\\sum_{k=1}^{l-1}(-1)^{ k}\\frac{2l}{2l-k}\\binom {2l-k} {k}\\bigg(\\frac{1+w}{\\sqrt{w}}\\bigg)^{2(l-k)}\\nonumber\\\\&=& (-1)^{l+1}\\frac{2}{N}\\hbox{res}_{w=0}\\frac{w}{(1+w)^{3}}\\bigg(C_{2l}\\big(\\frac{1+w}{\\sqrt{w}}\\big)-(-1)^{l}\\bigg),\n\\end{eqnarray}\nwhere $$ C_{2l}(x)= 2T_{2l}(x\/2)=\\sum_{k=0}^{l}(-1)^{k} \\frac{2l}{2l-k}\\binom {2l-k} {k}x^{2l-2k},$$\nis the normalized Chebyshev polynomial of the first kind\\cite{Rivlin}, and\n$$T_{2l}(x\/2)=\\frac{1}{2}\\Bigg\\lbrack\\Bigg(\\frac{x}{2}+\\sqrt{(x\/2)^2-1}\\bigg)^{2l} +\\Bigg(\\frac{x}{2}-\\sqrt{(x\/2)^2-1}\\bigg)^{2l} \\Bigg\\rbrack.$$\nUsing the fact that $C_{2l}\\big(\\frac{1+w}{\\sqrt{w}}\\big)= \\frac{1}{w^{l}}+w^{l}$, then the final expression for the first term $R_{1}(l) $, reads\n\\begin{eqnarray}\n\\label{t6}\nR_{1}(l)&=&(-1)^{l+1}\\frac{2}{N}\\hbox{res}_{w=0}\\frac{1}{(1+w)^{3}w^{l-1}}\\nonumber\\\\&=&-\\frac{l(l-1)}{N}\n\\end{eqnarray}\nSimilarly, the second term may written as \n\\begin{eqnarray}\n\\label{t7}\nR_{2}(l):&=&\\frac{(N-1)}{N}\\sum_{s=1}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-1)} {s-1}\\nonumber\\\\&=&(-1)^{l+1}\\frac{(N-1)}{N}\\hbox{res}_{w=0}\\frac{1}{(1+w)^{2}w^{l}}\\nonumber\\\\&=& \\frac{l(N-1)}{N}.\n\\end{eqnarray}\n Adding the contributions given by Eqs. (\\ref{t6}), and (\\ref{t7}), then, we get \n\\begin{equation}\n\\label{t8}\nR(l)=\\frac{1}{N}\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^2(n\\pi\/N)}=\\frac{l(N-l)}{N}.\n\\end{equation}\n\n\\ \\ Now, we want to evaluate the following trigonometrical sum $$ F_N(l ) = \\frac {1} {N}\\sum_{n=1}^{N-1} \\frac {1-\\cos nl\\pi\/N} {1-\\cos n\\pi\/N},$$ this sum arises in connection with the two-point resistance of a path graph \\cite{Wu}. The evaluation of this term may be done as follows; \n\\begin{eqnarray}\n\\label{t9}\nF_N(l ) &= &\\frac {1} {N}\\sum_{n=1}^{N-1} \\frac {1-\\cos nl\\pi\/N} {1-\\cos n\\pi\/N}\\nonumber\\\\&=&\\frac {1} {N}\\sum_{n=1}^\\frac{N}{2} \\frac {1-\\cos(2n-1)l\\pi\/N} {1-\\cos(2n-1)\\pi\/N}+\\frac {1} {N}\\sum_{n=1}^{\\frac{N}{2}-1} \\frac {1-\\cos 2nl\\pi\/N} {1-\\cos2n\\pi\/N},\n\\end{eqnarray}\nhere, $N$ is assumed to be even, similar steps may be used for $N$ odd. It is interesting to note that in evaluating $F_N(l )$, we only need to compute the first term since the second term is related to the two-point resistance of the $N$-cycle graph given by Eq. (\\ref{t8}). Then, the first term may be written as \n\\begin{eqnarray}\n\\label{t10}\n\\frac {1} {N}\\sum_{n=1}^\\frac{N}{2} \\frac {1-\\cos(2n-1)l\\pi\/N} {1-\\cos(2n-1)l\\pi\/N}&=&\\frac {1} {N} \\sum_{n=1}^\\frac{N}{2} \\frac {\\sin^{2}(2n-1)l\\pi\/2N} {\\sin^{2}(2n-1)l\\pi\/2N}\\nonumber\\\\&=&\\frac{1}{2N}\\sum_{s=1}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^\\frac{N}{2}\\sin^{2(s-1)}\\frac{(2n-1)\\pi}{2N}.\\nonumber\\\\.\n \\end{eqnarray}\n By using the identity $$\\sum_{n=1}^\\frac{N}{2}\\sin^{2(s-1)}(2n-1)\\pi\/2N= \\frac{1}{2}\\bigg(\\sum_{n=1}^{N-1}\\sin^{2(s-1)}n\\pi\/2N+\\sum_{n=1}^{N-1}(-1)^{n-1}\\sin^{2(s-1)}n\\pi\/2N\\bigg), $$ \n and the formulas for trigonometrical power sums given in \\cite{Schwatt}, then, one can show\n\\begin{eqnarray}\n\\label{t11}\n\\sum_{n=1}^\\frac{N}{2}\\sin^{2(s-1)}(2n-1)\\pi\/2N=\\frac{2N}{2^{2s}}\\binom {2(s-1)} {s-1},\n\\end{eqnarray}\nwhich in turn, implies that the formula for the first term should be\n \\begin{eqnarray}\n\\label{t12}\n\\frac {1} {N}\\sum_{n=1}^\\frac{N}{2} \\frac {1-\\cos(2n-1)l\\pi\/N} {1-\\cos(2n-1)\\pi\/N}&=&\\frac{1}{2}\\sum_{s=1}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-1)} {s-1}\\nonumber\\\\&=&\\frac{l}{2}.\n\\end{eqnarray}\nTo compute the second term given in Eq. (\\ref{t9}), we use the following symmetry \nenjoyed by the two-point resistance of the $N$-cycle graph, $N$ even\n\\begin{eqnarray}\n\\label{t13}\n\\frac {1} {N}\\sum_{n=1}^{N-1} \\frac {1-\\cos 2nl\\pi\/N} {1-\\cos2n\\pi\/N}&=&\\frac {2} {N}\\sum_{n=1}^{\\frac{N}{2}-1} \\frac {1-\\cos 2nl\\pi\/N} {1-\\cos2n\\pi\/N}+\\frac {1} {2N}\\big(1-(-1)^{l}\\big).\n\\end{eqnarray}\nTherefore, the second term may be obtained to give the following closed formula for $F_N(l ) $\n\\begin{equation}\n\\label{t14}\nF_N(l ) = \\frac {1} {N}\\sum_{n=1}^{N-1} \\frac {1-\\cos nl\\pi\/N} {1-\\cos n\\pi\/N}= l-\\frac{1}{N}\\bigg(\\frac{l^2}{2}+\\frac{1}{4}(1-(-1)^{l})\\bigg)\n\\end{equation}\nThis is in a complete agreement with the formula for the Green's function for the path graph in \\cite{Wu}\n\\section{Trigonometrical sum connected with Scott's conjecture}\n\n\\ \\ \n In proving R. F Scott's conjecture on the permanent of the Cauchy matrix, Minc in \\cite{Minc} needed to evaluate the following trigonometrical sum;\n \\begin{equation}\n\\label{sc0}\n \\sum_{n=1}^{N} \\frac {\\cos (2n-1)l\\pi\/N} {1-\\cos (2n-1)\\pi\/N}.\n \\end{equation}\n He obtained a closed-form formula for this sum using induction,\n and the sum turns out to be equal to $\\frac{N}{2}(N-2l)$. A short time later, Stembridge and Todd \\cite{Todd}, gave a proof for the evaluation for this sum, based on linear algebra. Here, we give a short derivation for this sum using our formula given by Eq. (\\ref{t12}), and the well-known identity \n \\begin{equation}\n\\label{sc1}\n \\sum_{n=1}^{N-1}\\frac{1}{\\sin^2(n\\pi\/N)}=\\frac{N^{2}-1}{3}.\n\\end{equation}\nOur derivation follows easily by realizing that Eq. (\\ref{t12}) is symmetric under the shift $n\\rightarrow N-n$, and as a consequence one gets \n\\begin{equation}\n\\label{sc2}\n\\sum_{n=1}^{N} \\frac {1-\\cos(2n-1)l\\pi\/N} {1-\\cos(2n-1)\\pi\/N}=N l.\n\\end{equation}\nIn order to evaluate the sum in Eq. (\\ref{sc0}), we need a formula for the sum $$\\sum_{n=1}^{N} \\frac {1} {1-\\cos(2n-1)\\pi\/N} =\\frac{1}{2}\\sum_{n=1}^{N} \\frac {1} {\\sin^{2}(2n-1)\\pi\/2N }.$$\nThe latter may be evaluated as follows\n\\begin{eqnarray}\n\\label{sc3}\n\\frac{1}{2}\\sum_{n=1}^{N} \\frac {1} {\\sin^{2}(2n-1)\\pi\/2N}&=&\\frac{1}{2}\\sum_{n=1}^{2N-1}\\frac{1}{\\sin^2(n\\pi\/2N)}-\\frac{1}{2}\\sum_{n=1}^{N-1}\\frac{1}{\\sin^2(2n\\pi\/2N)}\\nonumber\\\\&=&\\frac{N^{2}}{2}.\n\\end{eqnarray}\nIn obtainnig Eq. (\\ref{sc3}), we used the identity given in Eq. (\\ref{sc1}), thus, using Eq. (\\ref{sc2}) and Eq. (\\ref{sc3}), we may write\n\\begin{equation}\n\\label{sc4}\n \\sum_{n=1}^{N} \\frac {\\cos (2n-1)l\\pi\/N} {1-\\cos (2n-1)\\pi\/N}=\\frac{N^2}{2}-Nl.\n\\end{equation}\nThis is exactly the result obtained by Minc, Stembridge and Todd \\cite{Minc,Todd}.\n\\section{Trigonometrical sums arising in the chiral Potts model and in the Verlinde's formula}\n\n\\ \\ In this section the trigonometrical sum $T_{4}(l):=\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^4(n\\pi\/N)} $ is evaluated in a closed form using the residue operator. We also give an almost closed formula for the general case $T_{2m}(l):=\\sum_{n=1}^{N-1}\\frac{\\sin^{2}(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)} $, for $m\\geq1$. The first trigonometrical sum arises in the work of McCoy and Orrick on the chiral Potts model \\cite {McCoy}, this sum including other trigonometrical identities were proved by Gervois and Mehta \\cite {Mehta}. The second sum namely the sum $T_{2m}(l)$, was considered by Gervois and Mehta \\cite {Mehta} using a recursion formula. Here, we will obtain recursion formulas for both $T_{2m}(l)$ and $$T_{2m}:=\\sum_{n=1}^{N-1}\\frac{1}{\\sin^{2m}(n\\pi\/N)}.$$\n\n\\ \\ If, we set $N=k+2$ and $m =g-1$, $k,g$ being the level of the $su(2)$ Kac-Moody algebra, and the genus of the Riemann surface respectively. Then, the sum $ T_{2m}$ up to to some normalization factor is nothing but the dimension of the space of the conformal blocks of the $SU(2)$ WZW model. As a consequence, the recursion formula derived for $ T_{2m}$, may be used to obtain the expression for the dimension of the space of the conformal blocks for a given genus $g$. Similar computations are carried out for the twisted trigonometrical sum $$T_{2m}^{t}:=\\sum_{n=1}^{N-1}(-1)^{n+1}\\frac{1}{\\sin^{2m}(n\\pi\/N)}.$$ This is related to the dimension of the space of the conformal blocks of the $SO(3)$ WZW model.\n \\subsection {Trigonometrical sums and the perturbative chiral Potts model}\n \n \\ \\ Let us first start with the trigonometrical sums arising in the perturbative treatment of the the chiral Potts model \\cite {McCoy}. Techniques of the previous section, may be used to evaluate the sum $T_{4}(l)$, as follows\n\\begin{eqnarray}\n\\label{t15}\nT_{4}(l)&=&\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^4(n\\pi\/N)}\\nonumber\\\\&=&l^{2}\\sum_{n=1}^{N-1}\\frac{1}{\\sin^2(n\\pi\/N)}+\\frac{1}{2}\\sum_{s=2}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^{N-1} \\sin^{2(s-2)}(n\\pi\/N)\\nonumber\\\\&=&\\frac{l^2}{3}(N^{2}-1)+8\\sum_{s=2}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-2}(-1)^{t }\\binom {2(s-2)}{s-2-t}\\nonumber\\\\&+& 4(N-1)\\sum_{s=2}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-2)}{s-2},\n\\end{eqnarray}\nthe first term in the above equation follows from the well-known identity $$\\sum_{n=1}^{N-1}\\frac{1}{\\sin^2(n\\pi\/N)}=\\frac{N^{2}-1}{3},$$ while the second and the third terms may computed using the residue operator as in the previous section to give,\n\\begin{eqnarray}\n\\label{t16}\n\\sum_{s=2}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-2}(-1)^{t }\\binom {2(s-2)}{s-2-t}&=&(-1)^{l+1} \\hbox{res}_{w=0}\\frac{1}{(1+w)^{5}w^{l-2}}\\nonumber\\\\&=&\\frac{1}{4!}(l+1)l(l-1)(l-2),\n\\end{eqnarray}\nand\n\\begin{eqnarray}\n\\label{t17}\n\\sum_{s=2}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-2)}{s-2}&=& (-1)^{l+1}\\hbox{res}_{w=0}\\frac{1}{(1+w)^{4}w^{l-1}}\\nonumber\\\\&=&-\\frac{1}{3!}(l+1)l(l-1).\n\\end{eqnarray}\nTherefore, the closed formula for the sum given in Eq. (\\ref{t15}), reads\n\\begin{eqnarray}\n\\label{t18}\n\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^4(n\\pi\/N)}&=&\\frac{l^2}{3}(N^{2}-1)+\\frac{1}{3}(l+1)l(l-1)(l-2)-\\frac{2(N-1)}{3}(l+1)l(l-1)\\nonumber\\\\&=&\\frac{l^2}{3}(N-l)(N-l)+\\frac{2}{3}(N-l).\n\\end{eqnarray}\nThis is exactly the result obtained by Gervois and Mehta using a recursion formula satisfied by $T_{2m}(l)$ \\cite {Mehta}. Next, we will give another recursion formula for the sum $T_{2m}(l)$. Now, $T_{2m}(l)$, may be written as \n\\begin{eqnarray}\n\\label{t19}\nT_{2m}(l)&=&\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{2}\\sum_{n=1}^{N-1}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\frac{1}{\\sin^{2(m-s)}(n\\pi\/N)}\\nonumber\\\\&+&\\frac{1}{2}\\sum_{s=m}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^{N-1} \\sin^{2(s-m)}(n\\pi\/N). \n\\end{eqnarray}\nThe first term on the right-hand side, is written in terms of the sum $$T_{2k}=:\\sum_{n=1}^{N-1}\\frac{1}{\\sin^{2k}(n\\pi\/N)}. $$ This may be computed \\cite {Mehta}, using $$T_{2k}=\\sum_{n=1}^{N-1}\\Big(\\cot^{2}(\\frac{n\\pi}{N})+1\\Big)^{k}=\\sum_{l=1}^{k} \\binom{k}{l}S_{l}, $$ where $S_{l}=\\sum_{n=1}^{N-1}\\Big(\\cot^{2}(\\frac{n\\pi}{N}) \\Big)^{2l}$ and a recurrence relation satisfied by the power sums $S_{l} $. It turns out that $T_{2k}$, may also be obtained using a recursion formula, this will be shown shortly. Now, the second term may be written as follows \n\\begin{eqnarray}\n\\label{t20}\n\\tilde{T}_{2m}(l):&=&\\frac{1}{2}\\sum_{s=m}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^{N-1} \\sin^{2(s-m)}(n\\pi\/N)\\nonumber\\\\&=&2^{2m-1}\\sum_{s=m}^{l}(-1)^{s }\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-m}(-1)^{t }\\binom {2(s-m)}{s-m-t}\\nonumber\\\\&+&2^{2m-2}(N-1)\\sum_{s=m}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-m)}{s-m}\\nonumber\\\\&=&(-1)^{l+1}2^{2m-1}\\hbox{res}_{w=0}\\frac{1}{(1+w)^{2m+1}w^{l-m}}\\nonumber\\\\&+&(-1)^{l+1}2^{2m-2}(N-1)\\hbox{res}_{w=0}\\frac{1}{(1+w)^{2m}w^{l+1-m}}\\nonumber\\\\&=&(-1)^{m}2^{2m-1}\\frac{(l+m-1)!}{(l-m-1)!(2m)!}+ (-1)^{m+1}(N-1)2^{2m-2}\\frac{(l+m-1)!}{(l-m)!(2m-1)!}\\nonumber\\\\.\n\\end{eqnarray}\nTherefore, we succeeded in writing $\\tilde{T}_{2m}(l)$ in a closed form formula, One can check easily that our results agree with those given in \\cite {Mehta}, and so the formula for $T_{2m}(l)$ becomes\n\\begin{eqnarray}\n\\label{t21}\nT_{2m}(l)&=&\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{2}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2(m-s)}\\nonumber\\\\&+&(-1)^{m+1}2^{2m-1}\\frac{(l+m-1)!}{(l-m)!(2m)!}(mN-l).\n\\end{eqnarray}\nsetting $m=1,2$, then, our previous results given by Eq's. (\\ref{t8}) and (\\ref{t18}) respectively are recovered. From Eq. (\\ref{t21}), it is clear that in order to have a closed formula for $T_{2m}(l)$, one needs also, the exact expressions for $T_{2k}$, $ k=1\\cdots, m-1$. Next, we will show that $ T_{2k}$ satisfies a recursion formula that involves the $T_{2k}$'s. The expression for $T_{2m}$ may be obtained from $T_{2m}(l)$ from the following simple formula;\n\\begin{eqnarray}\n\\label{t22}\n\\sum_{l=1}^{N-1}T_{2m}(l)&=&\\sum_{l=1}^{N-1}\\sum_{n=1}^{N-1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{N}{2}\\sum_{n=1}^{N-1}\\frac{1}{\\sin^{2m}(n\\pi\/N)},\n\\end{eqnarray}\nTherefore, we may write\n\\begin{eqnarray}\n\\label{t23}\nT_{2m}&=&\\sum_{n=1}^{N-1}\\frac{1}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{N}\\sum_{l=1}^{N-1}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2(m-s)}\\nonumber\\\\&+&\\frac{2}{N}\\sum_{l=1}^{N-1}(-1)^{l+1}2^{2m-1}\\hbox{res}_{w=0}\\frac{1}{(1+w)^{2m+1}w^{l-m}}\\nonumber\\\\&+&\\frac{2}{N}\\sum_{l=1}^{N-1}(-1)^{l+1}2^{2m-2}(N-1)\\hbox{res}_{w=0}\\frac{1}{(1+w)^{2m}w^{l+1-m}}\\nonumber\\\\&=&\\frac{1}{N}\\sum_{l=1}^{N-1}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2(m-s)}\\nonumber\\\\&+&(-1)^{m+1}2^{2m-1}\\frac{(N+m-1)!}{N (N-m-1)!(2m+1)!}(2mN+1-N).\n\\end{eqnarray}\nAs a result, from our recursion formula, the different $T_{2m}$'s may be obtained directly, we do not have to use the recurrence relation satisfied by the power sum $S_{l}$ \\cite {Mehta}. For $m=1$ the first term in Eq. (\\ref{t23}) does not contribute and the second term gives the well known formula $ T_{2}=\\frac{N^{2}-1}{3}$. Now, for $m=2$, the first term may be computed to give $ \\frac{{(N^2-1)(N-1)}(2N-1)}{9}$, while the second term gives $ -\\frac{{(N^2-1)(N-2)}(3N+1)}{15}$, and hence, $T_{4}= \\frac{{(N^2-1)(N^2+11)}}{45}$ in a full agreement with \\cite {Mehta}, \\cite { Berndt}.\n\\subsection{The Verlinde dimension formula}\n\n\\ \\ The verlinde dimension formula may be obtained simply by setting $m=g-1$, $N=k+2$ in the expression of $T_{2m}$, where $g\\ge2$, $ k$ are the genus of the Riemann surface, and the level of the lie algebra $SU(2)$, respectively. Then, $T_{2(g-1)}$ up to some normalization factor, is the dimension of the space of conformal blocks $V_{g}$ of the $SU(2)$ WZW model \\cite{Verlinde},\n \\begin{eqnarray}\n\\label{t24}\ndimV_{g,k}&=&\\Big(\\frac{k+2}{2}\\Big)^{g-1}\\sum_{n=1}^{k+1}\\frac{1}{\\sin^{2g-2}n\\pi\/(k+2)}\\nonumber\\\\&=&\\Big(\\frac{k+2}{2}\\Big)^{g-1} \\frac{1}{(k+2)}\\sum_{l=1}^{k+1}\\sum_{s=1}^{g-2}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2g-2 -2s} \\nonumber\\\\&+&(-1)^{g}2^{2g-3}\\Big(\\frac{k+2}{2}\\Big)^{g-1} \\frac{1}{(k+2)}\\frac{1}{(2g-1)}\\binom{k+g}{2g-2}\\big((k+2)(2g-3)+1\\big).\n\\end{eqnarray}\n As a consequence, the dimension of the space of the conformal blocks of the $SU(2)$ WZW model, may be computed using our recursion formula for $T_{2k}$. Our formula Eq. (\\ref{t24}) may be used to give \n\\begin{eqnarray}\n\\label{t25}\ndimV_{2,k}&=&\\frac{(k+1)(k+2)(k+3)}{6}\\nonumber\\\\dimV_{3,k}&=&\\frac{1}{5}\\frac{(k+1)(k+2)(k+3)}{6}\\Big[\\frac{(k+1)(k+2)(k+3)}{6}+2(k+2)\\Big]\\nonumber\\\\dimV_{4,k}&=&\\frac{1}{7}.\\frac{1}{5}\\frac{(k+1)(k+2)(k+3)}{6}\\nonumber\\\\&.&\\Big[\\frac{(k+1)(k+2)(k+3)}{6}\\Big[\\frac{2(k+1)(k+2)(k+3)+27(k+2)}{6}\\Big]+6(k+2)^{2}\\Big].\\nonumber\\\\\n\\end{eqnarray}\nOur first two expressions for dimension of the space of conformal blocks agree with those computed using conformal field theory \\footnote{The last term $k+2$ in the expression of $dimV_{3,k}$ should be corrected in \\cite{Piunikhin} in order to make it a positive integer.} \\cite{Piunikhin}. For $g=4$, our formula for $dimV_{4,k} $ is identical to the formula given by Zagier \\cite{Zagier} provided the shift $k\\rightarrow k+2$ is taken. This shift is natural, since Zagier defined the dimension of the space conformal blocks as $dimV_{g,k-2}$.\n\n\\ \\ For the WZW model based on $SO(3)$, the level $k$ must be even \\cite {Thaddeus}, and the formula for the dimension of the twisted space of the conformal blocks $V_{g,k}^{t}$, may be written as \n$$dimV_{g,k}^{t}=\\Big(\\frac{k+2}{2}\\Big)^{g-1}\\sum_{n=1}^{k+1}(-1)^{n+1}\\frac{1}{\\sin^{2g-2}n\\pi\/(k+2)}.$$ In order to derive a recursion formula for the dimension $ dimV_{g,k}^{t}$, we first note that the expression for the twisted version of the trigonometrical sum $ T_{2m}(l)$ given by Eq. (\\ref{t19}) is\n\\begin{eqnarray}\n\\label{t26}\nT_{2m}^{t}(l)&=&\\sum_{n=1}^{N-1}(-1)^{n+1} \\frac{\\sin^2(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{2}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2(m-g)}^{t}\\nonumber\\\\&+&\\frac{1}{2}\\sum_{s=m}^{l}(-1)^{s +1}\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\sum_{n=1}^{N-1} (-1)^{n+1}\\sin^{2(s-m)}(n\\pi\/N), \n\\end{eqnarray}\nwhere $$T_{2m}^{t}=\\sum_{n=1}^{N-1}(-1)^{n+1}\\frac{1}{\\sin^{2m}(n\\pi\/N)} $$\nThe trigonometrical sum $$ \\sum_{n=1}^{N-1} (-1)^{n+1}\\sin^{2(s-m)}(n\\pi\/N),$$ is non-vanishing only if $ N$ is even\\cite{Schwatt} and is given by\n \\begin{eqnarray}\n\\label{t27}\n \\sum_{n=1}^{N-1} (-1)^{n+1}\\sin^{2(s-m)}(n\\pi\/N)&=&2^{2(m-s)+1}\\sum_{t=1}^{s-m}(-1)^{t }\\binom {2(s-m)}{s-m-t}\\nonumber\\\\&+&2^{2(m-s)} \\binom {2(s-m)}{s-m}\n\\end{eqnarray}\nThis formula shows clearly that the dimension of the twisted space of the conformal blocks is is non-vanishing only if $k$ is even, $N=k+2$ in agreement with the algebro-geometrical argument \\cite {Thaddeus}. Then, the expression for $T_{2m}^{t}$ may be written as\n\\begin{eqnarray}\n\\label{t28}\nT_{2m}(l)^{t}&=&\\sum_{n=1}^{N-1}(-1)^{n+1}\\frac{\\sin^2(nl\\pi\/N)}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{2}\\sum_{n=1}^{N-1}(-1)^{n+1}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}\\frac{1}{\\sin^{2(m-s)}(n\\pi\/N)}\\nonumber\\\\&+&2^{2m-1}\\sum_{s=m}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\sum_{t=1}^{s-m}(-1)^{t }\\binom {2(s-m)}{s-m-t}\\nonumber\\\\&+&2^{2m-2}\\sum_{s=m}^{l}(-1)^{s +1}\\frac{2l}{l+s}\\binom {l+s} {l-s}\\binom {2(s-m)}{s-m}\\nonumber\\\\&=&\\frac{1}{2}\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{l}{l+s}\\binom {l+s} {l-s}2^{2s}T_{2(m-s)}^{t}\\nonumber\\\\&+&(-1)^{m+1}2^{2m-1}\\frac{(l+m-1)!}{(l-m)!(2m)!}l. \n\\end{eqnarray}\nThe recursion formula for the twisted trigonometrical sum $ T_{2m}^{t}$, may be derived by setting $ l= N\/2$ in Eq. (\\ref{t28})\n\\begin{eqnarray}\n\\label{t29}\nT_{2m}^{t}&=&\\sum_{n=1}^{N-1}(-1)^{n+1}\\frac{1}{\\sin^{2m}(n\\pi\/N)}\\nonumber\\\\&=&\\sum_{s=1}^{m-1}(-1)^{s+1 }\\frac{N\/2}{N\/2+s}\\binom {N\/2+s} {N\/2-s}2^{2s}T_{2(m-s)}^{t}\\nonumber\\\\&+&(-1)^{m+1}2^{2m}\\frac{(N\/2+m-1)!}{(N\/2-m)!(2m)!}N\/2 \\nonumber\\\\&-&\\sum_{n=1}^{N-1}\\frac{1}{\\sin^{2m}(n\\pi\/N)}.\n\\end{eqnarray}\nFor $m=1$ and $m=2$ the twisted trigonometrical sums are\n\\begin{equation}\n\\label{30}\nT_{2}^{t}=\\frac{N^{2}+2}{6}\n\\end{equation}\nand\n\\begin{equation}\n\\label{31}\nT_{4}^{t}=\\frac{7N^{4}+40N^{2}+88}{360},\n\\end{equation}\nrespectively. In obtaining these results we used the expressions for $T_{2}$ and $T_{4}$. These twisted trigonometrical sums appeared earlier as coefficients of certain generating function \\cite{ Zagier}. Using the recursion formula Eq. (\\ref{t29}), the twisted trigonometrical sum $ T_{6}^{t}$ is\n\\begin{equation}\n\\label{32}\nT_{4}^{t}=\\frac{31N^{6}+294N^{4}+1344N^{2}+3056}{15120}. \n\\end{equation}\nThe twisted trigonometrical sum formula given by Eq. (\\ref{t29}), implies that the dimension of the twisted space of the conformal blocks is may be deduced for any genus $g\\geq2$, through the following formula \n\\begin{eqnarray}\n\\label{t33}\ndimV_{g,k}^{t}&=&\\Big(\\frac{k+2}{2}\\Big)^{g-1}\\sum_{n=1}^{k+1}(-1)^{n+1}\\frac{1}{\\sin^{2g-2}n\\pi\/(k+2)}\\nonumber\\\\&=&\\Big(\\frac{k+2}{2}\\Big)^{g-1}\\sum_{s=1}^{g-2}(-1)^{s+1 }\\frac{(k+2)\/2}{(k+2)\/2+s}\\binom {(k+2)\/2+s} {(k+2)\/2-s}2^{2s}T_{2g-2-2s}^{t}\\nonumber\\\\&+&(-1)^{g}2^{2g-2}\\Big(\\frac{k+2}{2}\\Big)^{g-1}\\frac{((k+2)\/2+g-2)!}{((k+2)\/2-g+1)!(2g-2)!}(k+2)\/2 \\nonumber\\\\&-&dimV_{g,k}.\n\\end{eqnarray}\nNote that, the relation between $dimV_{g,k}^{t} $ and $ dimV_{g,k}$ is expected from the simple identity $$dimV_{g,k-2}^{t}= dimV_{g,k-2}-2^{g}dimV_{g,k\/2-1},$$ where $k$ even. The formula by Zagier \\cite{Zagier} for $dimV_{g,k-2}$, may be obtained using the following generating function \n$$ \\sum_{g=1}^{\\infty}dimV_{g,k-2}\\Big(\\frac{2}{k}\\sin^{2}x\\Big)^{g-1}=\\frac{k\\sin (k-1)x}{\\sin kx \\cos x }.$$\n\\section{The corner-to-corner resistance and the Kirchhoff index of a $ 2\\times N $ resistor network}\n\n\\ \\ In general, it is hard to have a closed-form expression for the two-point resistance of a resistor network, however, if the latter has certain symmetries like circulant resistor network, then this may be possible \\cite{Chair1}. The situation gets more and more complicated in two and three dimensional resistor networks \\cite{Wu}, as the exact two-point resistance are expressed in terms of the double and triple summations. It turns out that the recently developed techniques by the author \\cite{Chair1}, may be used to obtain exact formula for the two-point resistance of the first non-trivial $ 2\\times N $ resistor network \\cite{Chair2}. In this section, we derive an exact formula for the corner-to-corner resistance and the total effective resistance of a $ 2\\times N $ resistor network. The total effective resistance is also known as the Kirchhoff index, this index was introduced in chemistry as a molecular structure descriptor, it is used for discriminating among different molecules with similar shapes and structures \\cite{Randic}. At the moment, the only exact two-point resistance not written as a double summation of an $ M\\times N $ resistor network is the asymptotic expansion of the corner-to-corner resistance \\cite{Essam, Huang}. It is known that the value of the asymptotic expansion of the corner-to-corner resistance of a rectangular resistor network provides a lower bound to the resistance of compact percolation clusters \\cite{ Domany}.\n\\subsection{The exact evaluation of the corner-to-corner resistance }\n\n\\ \\ The exact expression for the resistance between two nodes of a rectangular network of resistors with\nfree boundary conditions was given by Wu \\cite{Wu}. Suppose that the resistances in the two spatial directions\nare r = s = 1, then the resistance ${R}_{\\,\\rm free}$ between two nodes ${\\bf r}_1=(x_1, y_1)$\nand ${\\bf r}_2=(x_2, y_2)$ is \n\\begin{eqnarray}\n&&R_{\\{M\\times N\\}}^{\\,\\rm free}({\\bf r}_1,{\\bf r}_2) = \\frac {1} {N} \\Big| x_1 -x_2 \\Big| + \\frac 1 {M} \\Big| y_1 - y_2 \\Big| +\\frac 2 {MN} \\nonumber\\\\\n&&\\times{\\sum_{m=1}^{M-1}\\sum_{n=1}^{N-1}\n\\frac {\\Big[\\cos\\Big(x_1+\\frac 1 2\\Big)\\theta_m \\cos\\Big(y_1+\\frac 1 2\\Big)\\phi_n\n - \\cos\\Big(x_2+\\frac 1 2\\Big)\\theta_m \\cos\\Big(y_2+\\frac 1 2\\Big)\\phi_n\n\\Big]^2 } \n{ (1-\\cos \\theta_m ) +(1-\\cos \\phi_n ) } } ,\\nonumber \\\\\n\\label{cc1}\n\\end{eqnarray}\nwhere\n\\begin{eqnarray}\n\\theta_m= \\frac{m\\pi} M, \\hskip1cm \\phi_n= \\frac{n\\pi} N.\\nonumber\n\\end{eqnarray} \nIn order to compute the corner-to-corner resistance of a $2 \\times N$ resistor network, we set $ M = 2$, ${\\bf r}_1=(0, 0)$\nand ${\\bf r}_2=(1, N-1)$ into Eq. (\\ref{cc1}), and so\nthe double sum of the above equation is reduced to a single sum, and the corner-to-corner\nresistance may be written as\n\\begin{eqnarray}\nR_{\\{2\\times N\\}}^{\\,\\rm free}((0,0),(1,N-1))&=&\\frac{1}{N}+\\frac{N-1}{2} \\nonumber\\\\&+& \\frac{1}{3N}\\sum_{n=1}^{N-1} \\frac{(1+(-1)^{n})(1+\\cos n\\pi\/N)}{2(1-2\/3\\cos^{2}n\\pi\/2N)}.\n\\label{cc2}\n\\end{eqnarray}\nFor $N$ even, the sum over $n$ may be reduced to\n\\begin{eqnarray}\n\\frac{2}{3N}\\sum_{n=1}^{N\/2-1} \\frac{\\cos^{2} n\\pi\/N}{(1-2\/3\\cos^{2}n\\pi\/N)}&=&\\frac{1}{3N}\\sum_{n=1}^{N-1} \\frac{\\cos^{2} n\\pi\/N}{(1-2\/3\\cos^{2}n\\pi\/N)}\\nonumber\\\\&=&\\frac{1}{3N}\\sum_{j=0}^{\\infty}(2\/3)^{j}\\sum_{n=1}^{N-1}\\cos^{2(j+1)} n\\pi\/N,\n\\label{cc3}\n\\end{eqnarray}\nto evaluate this sum, we follow closely the method developed by the author in \\cite{Chair1}. As explained in \\cite{Chair1},\nthe formula for the sum $\\sum_{n=1}^{N-1} \\cos^{2J} n\\pi\/N $, given by Schwatt \\cite{Schwatt} is not the right one to use, we use instead the formula \n\\begin{eqnarray}\n\\sum_{n=1}^{N-1} \\cos^{2J} n\\pi\/N =-1+\\frac{N}{2^{2j-1}}\\sum_{p=1}^{[J\/N]}\\binom {2J}{J-pN} +\\frac{1}{2^{2j}}\\binom {2J}{J},\n\\label{cc4}\n\\end{eqnarray}\nthus, the sum contribution to the corner-to-corner resistance using the residue representation of binomials is\n\\begin{eqnarray}\n\\frac{1}{3N}\\sum_{j=0}^{\\infty}(2\/3)^{j}\\sum_{n=1}^{N-1}\\cos^{2(j+1)} n\\pi\/N&=&-\\frac{1}{N}+\\sum_{j=0}^{\\infty}\\hbox{res}_{w}\\frac{(1+w)^{2j}}{(6w)^{j}}\\frac{w^{N}}{w(1-w^{N})}\\nonumber\\\\&+&\\frac{1}{2}\\Big[\\hbox{res}_w (1-4w){^{-1\/2}}\\sum_{j=0}^{\\infty}(1\/6w)^{j}{w^{-1}}-1\\Big]\\nonumber\\\\&=&-\\frac{1}{N}+\\sqrt{3}\\frac{(2-\\sqrt{3})^{N}}{1-(2-\\sqrt{3})^{N}}+\\frac{1}{2}(\\sqrt{3}-1).\n\\label{cc5}\n\\end{eqnarray}\nFinally, the corner-to-corner resistance of $ 2\\times N $ resistor network becomes,\n\\begin{eqnarray}\nR_{\\{2\\times N\\}}^{\\,\\rm free}((0,0),(1,N-1))&=&\\frac{N-1}{2} +\\sqrt{3}\\frac{(2-\\sqrt{3})^{N}}{1-(2-\\sqrt{3})^{N}}+\\frac{1}{2}(\\sqrt{3}-1).\n\\label{cc6}\n\\end{eqnarray}\n It is not difficult to see that this formula is also valid for $N$ odd.\n \n \\ Examples. For $N=2,3, 4$ our formula Eq. (\\ref{cc6}), gives\n \\begin{eqnarray}\n R_{\\{2\\times 2\\}}^{\\,\\rm free}((0,0),(1,1))&=&1\\nonumber\\\\R_{\\{2\\times 3\\}}^{\\,\\rm free}((0,0),(1,2))&=&1.4\\nonumber\\\\R_{\\{2\\times 4\\}}^{\\,\\rm free}((0,0),(1,3))&=&1.875,\n \\label{cc7}\n \\end{eqnarray} \nthese results are in a full agreement with Eq. (\\ref{cc2}).\n\\subsection{The Kirchhoff index}\n\n\\ \\ The computation of the total effective resistance of a $ 2\\times N $ resistor network, that is, the Kirchhoff index, may be computed in two ways. It may be evaluated by summing over all effective resistances between nodes of a given resistor network, or alternatively by summing over all eigenvalues of a Laplacian associated with resistor network \\cite{Gutman}. So, we do not have to know the effective resistance between each node to compute the total effective resistance of a resistor network. The formula that gives the Kirchhoff index of a resistor network in terms of the eigenvalues is\n $$ Kf(G)=N\\sum_{n=1}^{N-1}\\frac{1}{\\lambda_n},$$ where $\\lambda_{n}$ are the eigenvalues of the Laplacian of the network, or the graph $G$ made of nodes and edges considered as unit resistors. Our network is given by the cartesian product $ 2\\times N$, that is, made of two path lines with $N$ nodes, and $N$ path lines with two nodes. Now, the Kirchhoff index of a path line is $$Kf(P_{n})=N\\sum_{n=1}^{N-1}\\frac{1}{4\\sin^2(n\\pi\/2N)}=\\frac{N}{8}\\Big[ \\sum_{n=1}^{2N-1}\\frac{1}{\\sin^2(n\\pi\/2N)}-1\\Big]=\\frac{N^{3}-N}{6}.$$ \n Thus, the contribution from these path lines is $N+\\frac{N^{3}-N}{3}$, by connecting the system together, then the corresponding eigenvalues of the laplacian are $\\lambda_{1,n}= 3(1-2\/3\\cos^{2}n\\pi\/2N)$. As a consequence, the Kirchhoff index of a $ 2\\times N$, resistor network can be written as\n\\begin{eqnarray} \nKf(2\\times N)=N+\\frac{N^{3}-N}{3}+N\\sum_{n=1}^{N-1}\\frac{1}{3(1-2\/3\\cos^{2}n\\pi\/2N)},\n\\label{Kf1}\n\\end{eqnarray}\nNote that, our simple deduction of this expression gives the same value of the Kirchhoff index given by theorem 4.1 in \\cite{Yang}. The above sum seems difficult to evaluate, however, using a simple trick, we will be able to get a closed form for the Kirchhoff index. To that end, let us write\n\\begin{eqnarray}\n\\label{Kf2}\n\\sum_{n=1}^{N-1}\\frac{1}{(1-2\/3\\cos^{2}n\\pi\/2N)}&=&\\sum_{n=1}^{N\/2-1}\\frac{1}{(1-2\/3\\cos^{2}n\\pi\/N)}\\nonumber\\\\&+&\\sum_{n=1}^{N\/2}\\frac{1}{(1-2\/3\\cos^{2}(2n-1)\\pi\/2N)},\n\\end{eqnarray}\nwhere $N$ is assumed to be even. The first sum may be \ncarried out using the following trick;\n\\begin{eqnarray}\n\\sum_{j=0}^{\\infty}(2\/3)^{j}\\sum_{n=1}^{N-1}\\cos^{2(j+1)} n\\pi\/N&=&\\frac{3}{2}\\Big[\\sum_{j=0}^{\\infty}(2\/3)^{j}\\sum_{n=1}^{N-1}\\cos^{2j} n\\pi\/N-(N-1)\\Big].\n\\label{Kf3}\n\\end{eqnarray}\nNow, the sum on the left-hand side was computed before, see Eq. (\\ref{cc5}), then one may deduce\n\\begin{eqnarray}\n\\sum_{n=1}^{N-1}\\frac{1}{(1-2\/3\\cos^{2}n\\pi\/N)}&=&\\sum_{j=0}^{\\infty}(2\/3)^{j}\\sum_{n=1}^{N-1}\\cos^{2j} n\\pi\/N \\nonumber\\\\&=&-3+\\frac{6N(2-\\sqrt{3})^{N}}{\\sqrt{3}(1-(2-\\sqrt{3})^{N})}+\\sqrt{3}N\n\\label{Kf5}.\n\\end{eqnarray}\nand so,\n\\begin{eqnarray}\n\\sum_{n=1}^{N\/2-1}\\frac{1}{(1-2\/3\\cos^{2}n\\pi\/N)}&=&\\frac{1}{2}\\Big[\\sum_{n=1}^{N-1}\\frac{1}{(1-2\/3\\cos^{2}n\\pi\/N)}-1\\Big]\\nonumber\\\\&=&-2+\\frac{3N(2-\\sqrt{3})^{N}}{\\sqrt{3}(1-(2-\\sqrt{3})^{N})}+\\frac{\\sqrt{3}}{2}N.\n\\label{Kf6}\n\\end{eqnarray}\nUsing the identity \\cite{Chair1},\n\\begin{eqnarray}\n\\label{Kf7}\n\\sum_{n=1}^{N\/2}\\cos^{2j}(2n-1)l\\pi\/N&=&\n\\frac{N}{2^{2j+1}}\\binom {2j}{j}+\\frac{N}{2^{2j}}\\sum_{p=1}^{[j\/2N]}\\binom {2j}{j-2pN}\\nonumber\\\\&-&\\frac{N}{2^{2j}}\\sum_{p=1}^{[j\/2N]}\\binom {2j}{j-(2p-1)N},\n\\end{eqnarray}\n and by following similar steps as in the above computations, then, one may show\n \\begin{eqnarray}\n \\sum_{n=1}^{N\/2}\\frac{1}{(1-2\/3\\cos^{2}(2n-1)\\pi\/2N)}&=&\\frac{3N(2-\\sqrt{3})^{2N}}{\\sqrt{3}(1-(2-\\sqrt{3})^{2N})}-\\frac{3N(2-\\sqrt{3})^{N}}{\\sqrt{3}(1-(2-\\sqrt{3})^{2N})}\\nonumber\\\\&+&\\frac{\\sqrt{3}}{2}N.\n \\label{Kf8}\n \\end{eqnarray}\n Finally, the exact expression of the Kirchhoff index of a $2\\times N$ reeds\n\\begin{eqnarray}\nKf(2\\times N)=N+\\frac{N^{3}-N}{3}+\\frac{N}{3}\\Big[-2+\\frac{6N(2-\\sqrt{3})^{2N}}{\\sqrt{3}(1-(2-\\sqrt{3})^{2N})}+\\sqrt{3}N\\Big].\n\\label{Kf9}\n\\end{eqnarray} \nOne can show that, the above formula for the Kirchhoff formula is valid for $N$ odd as well.\n\nExample, For $N=1,2,3,4,5$, the Kirchhoff indices are respectively, \n \\begin{eqnarray}\nKf(2\\times 2)&=&5\\nonumber\\\\Kf(2\\times 3)&=&14.2 \\nonumber\\\\Kf(2\\times 4)&=&30.57142857\n\\nonumber\\\\Kf(2\\times 5)&=&56.10047847\n\\end{eqnarray} \n These results are in complete agreement with those obtained using formula given by Eq. (\\ref{Kf1}), or theorem 4.1 of reference \\cite{Yang}.\n \\section{Some trigonometrical sums related to number theory}\n \n \\ \\ In this section other class of trigonometrical sums will be evaluated using similar techniques as in the previous sections. Some of these trigonometrical sums are related to number theory. We will start with the following sum\n$$S(l):=\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin^2(nl\\pi\/N)}{\\sin^2(n\\pi\/N)}, $$ which is the alternating sum associated with the sum $R(l)$ given in Eq. (\\ref{t1}). This sum has the following closed formula\n\\begin{proposition}\n\\begin{equation}\n\\label{s}\nS(l)=\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin^2(nl\\pi\/N)}{\\sin^2(n\\pi\/N)}=-l^2\n\\end{equation}\n\\end{proposition}\nTo derive the above formula, we follow similar computations carried out for $R(l)$, except that this time the sum over $n$ is non-vanishing only if $N$ is even\n\\begin{equation} \n\\label{a1}\n\\sum_{n=1}^{N-1}(-1)^n\\sin^{2(s-1)}(n\\pi\/N)=\\frac{1}{2^{2(s-1)-1}}\\sum_{t=1}^{s-1}(-1)^{t +1}\\binom {2(s-1)}{s-1-t} +\\frac{-1}{2^{2(s-1)}}\\binom {2(s-1)} {s-1},\n\\end{equation}\n Comparing Eq. (\\ref{a1}) and Eq. (\\ref{t3}), and using the previous results, then, without any further computations, the formula for the trigonometrical sum $S(l)$ is obtained. \n Due to the the symmetry enjoyed by $ S(l)$, $S(l)=S(N-l)$, the right hand side of equation (\\ref{s}) should be read with this constraint, that is for both $l$, and $N-l$, $ S(l)=-l^2$ . Next, let us consider the sums $ S_{1}(l):=\\sum_{n=1}^{N-1}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}$ and $S_{2}(l):=\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}$ that are closely related. We will prove that closed formulas for the sums $ S_{1}(l)$ and $ S_{2}(l)$ are given by\n \\begin{theorem}\n\\begin{equation} \n\\label{s1}\nS_{1}(l)=\\sum_{n=1}^{N-1}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}=\\left\\{\\begin{array}{cl}N-l & \\text{for } l \\text{odd }\\\\\n0 & \\text{for } l \\text{even },\n\\end{array} \\right.\n\\end{equation}\n\\begin{eqnarray} \n\\label{s2}\nS_{2}(l)=\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}=\\left\\{\\begin{array}{cl}-(2l-1) & \\text{for } l \\text{odd } \\text{and } N \\text{even} \\\\\n0 & \\text{for } l \\text{odd } \\text{and } N \\text{odd}\\\\\n-2l& \\text{for } l \\text{even } \\text{and } N \\text{odd}\\\\\n0& \\text{for } l \\text{even } \\text{and } N \\text{even}.\n\\end{array} \\right.\n\\end{eqnarray}\n\\end{theorem}\n\nTo prove the first formula, we use the following trigonometrical identity \\cite{Hobson},\n \\begin{equation} \n\\label{Hobson}\n\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}=\\sum_{s\\geq0}(-1)^{s}\\binom {l-s-1}{s}2^{l-2s-1}\\cos^{l-2s-1}(n\\pi\/N), \n \\end{equation}\n thus, for $l$ odd, one has\n \\begin{eqnarray} \n\\label{n1}\nS_{1}(l)=\\sum_{n=1}^{N-1}\\frac{\\sin(2l-1)n\\pi\/N}{\\sin n\\pi\/N}&=&\\sum_{s\\geq0}(-1)^{s}\\binom {2l-2-s}{s}2^{2l-2-2s}\\sum_{n=1}^{N-1}\\cos^{2l-2-2s}(n\\pi\/N)\\nonumber\\\\.\n\\end{eqnarray}\nThe sum over $n$, may be computed from Schwatt's book \\cite{Schwatt}, see Eq. (107), page $221$ to give \n\\begin{eqnarray} \n\\label{n2}\n\\sum_{n=1}^{N-1}\\cos^{2l-2-2s}(n\\pi\/N)= \\frac{-2}{2^{2l-2-2s}}\\sum_{t=1}^{l-1-s}\\binom {2l-2-2s}{l-1-s-t} +\\frac{N-1}{2^{2l-2-2s}}\\binom {2l-2-2s} {l-1-s},\n\\end{eqnarray}\nthen, the first contribution to the sum given in Eq. (\\ref{n1}) is\n\\begin{eqnarray} \n\\label{n3}\nS_{1}(l)^{'}=-2\\sum_{s=0}^{l-1}(-1)^{s}\\binom {2l-2-s}{s}\\sum_{t=1}^{l-1-s}\\binom {2l-2-2s}{l-1-s-t}=\\nonumber\\\\-2\\hbox{res}_{w=0}\\sum_{s=0}^{l-1}(-1)^{s}\\binom {2l-2-s}{s}\\Big(\\frac{1+w}{\\sqrt {w}}\\Big)^{2l-2-2s}\\frac{1}{1-w}\\nonumber\\\\=-2\\hbox{res}_{w=0}U_{2l-2}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)\\frac{1}{1-w},\n\\end{eqnarray}\nin obtaining the last line of the above equation we used the expresion for the normalized Chebyshev polynomial of the second kind $U_{n}(\\frac{x}{2})= \\sum_{k=0}^{[n\/2]}(-1)^{k}\\binom {n-k}{k}x^{n-2k}$. The residue may be evaluated using $$U_{n}(\\frac{x}{2})= \\frac{(x+\\sqrt{x^2-1})^{n+1}-(x-\\sqrt{x^2-1})^{n+1}}{2\\sqrt{x^2-1}}, $$ to obtain \n\\begin{eqnarray} \n\\label{n4}\nS_{1}(l)^{'}=-2\\hbox{res}_{w=0} \\frac{1}{w^{l-1}}\\frac{1}{(1-w)^2}=-2(l-1).\n\\end{eqnarray}\nSimilarly, the second contribution reads\n\\begin{eqnarray} \n\\label{ns3}\nS_{1}(l)^{''}&=&(N-1)\\hbox{res}_{w=0}U_{2l-2}\\Big(\\ \\frac{1+w}{2\\sqrt {w}}\\Big)\\frac{1}{w}\\nonumber\\\\&=& N-1.\n\\end{eqnarray}\nTherefore, combining these contributions, the closed formula for $ S_{1}(l)$ is \n\\begin{eqnarray} \n\\label{n00}\nS_{1}(l)=\\sum_{n=1}^{N-1}\\frac{\\sin(2l-1)n\\pi\/N}{\\sin n\\pi\/N}=N-(2l-1).\n\\end{eqnarray}\nIt is not difficult to show that there is no contribution to the sum $ S_{1}(l)$ for $l$ even. In proving the second formula for $S_{2}(l)$ Eq. (\\ref{s2}), one notes that in evaluting the sum over $n$ in $$S_{2}(l)=\\sum_{s\\geq0}(-1)^{s}\\binom {l-s-1}{s}2^{l-2s-1}\\sum_{n=1}^{N-1}(-1)^n\\cos^{l-2s-1}(n\\pi\/N),$$ turns out to depend on both $l$, $N$ unlike the previous case. By using formulas given by Eq (113), and Eq (114) in \\cite{ Schwatt}, we have\n\\begin{eqnarray} \n\\label{n5}\n\\sum_{n=1}^{N-1}(-1)^n\\cos^{l-2s-1}(n\\pi\/N)&=&\\frac{-2}{2^{2l-2-2s}}\\sum_{t=1}^{l-1-s}\\binom {2l-2-2s}{l-1-s-t} -\\frac{1}{2^{2l-2-2s}}\\binom {2l-2-2s} {s-l-1},\\nonumber\\\\\n\\end{eqnarray}\nfor $l$ odd, $ N$ even, and the sum vanishes for $l$ odd, $ N$ odd. If $l$ is even, then, the above sum is non-vanishing only for $N$ odd\n\nTherefore, for $l$ odd it follows from Eq. (\\ref{n5}) that we have \n\\begin{eqnarray} \n\\label{n7} \nS_{2}(l)&=&\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin (2l-1)n\\pi\/N}{\\sin(n\\pi\/N)}\\nonumber\\\\&=&-2\\hbox{res}_{w=0} \\frac{1}{w^{l-1}}\\frac{1}{(1-w)^2}-\\hbox{res}_{w=0} \\frac{1}{w^l}\\frac{1}{(1-w)}=-(2l-1),\n\\end{eqnarray}\n for $l$ even and $N$ odd, one has the following identity\n\\begin{eqnarray} \n\\label{n6}\n\\sum_{n=1}^{N-1}(-1)^n\\cos^{l-2s-1}(n\\pi\/N)&=&\\frac{-2}{2^{2l-1-2s}}\\sum_{t=0}^{l-1-s}\\binom {2l-1-2s}{l-1-s-t},\n\\end{eqnarray}\nfrom which\n\\begin{eqnarray} \n\\label{n0} \nS_{2}(l)=\\sum_{n=1}^{N-1}(-1)^n\\frac{\\sin(2nl\\pi\/N)}{\\sin(n\\pi\/N)}=-2\\hbox{res}_{w=0} \\frac{1}{w^{l}}\\frac{1}{(1-w)^2}=-2l,\n\\end{eqnarray}\n these results were recently verified by simulations without proof in connection with number theory \\footnote{Anonymous author working on characters of a finite field and the Polya-Vinogradov inequality.}. It is interesting to note that the closed formula for $ S_{2}(l)$ may be expected since if we let $l$ go to $ N-l$ in $ S_{1}(l)$, then, $S_{2}(l)=- l$ . Now, we will consider the following non-trivial and interesting trigonometrical sums, $$ F_{1}(N,l,2):= \\sum_{n=1}^{N-1}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2nl\\pi\/N)}{\\sin(2n\\pi\/N)},$$ and \n $$ F_{2}(N,l,2):= \\sum_{n=1}^{N-1}(-1)^{n}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2nl\\pi\/N)}{\\sin(2n\\pi\/N)},$$\n where $ F_{1}(N,N-l,2)= F_{2}(N,l,2)$, and $ N$ is assumed to be odd. So, if $ F_{2}(N,l,2)$ is known, then, $ F_{2}(N,l,2)$ may be obtained and vice-versa. In the rest of this paper, we show that both the trigonometrical sums may be evaluated to give the following closed formulas\n\\begin{theorem}\n\\begin{eqnarray} \n\\label{f1}\nF_{1}(N,l,2)& = &\\sum_{n=1}^{N-1}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2nl\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&-\\frac{1}{2}(3l-2)(3l-3)+\\frac{1}{2}(l-1)(l-2)-l+\\frac{1}{2}(1-(-1)^l)\\nonumber\\\\&+&\\frac{N-1}{2}\\Big(2l-1-(-1)^l\\Big)+N\\Big(3l-2-N+\\frac{1}{2}(1-(-1)^{l-N})\\Big)\\nonumber\\\\\n\\end{eqnarray}\nwhere $l$ is odd and the sum vanishes for even l, also, note that the last term namely the coefficient of $N$ is different from zero only if $ 3l-2>N$. The closed formula for $ F_{2}(N,l,2)$ reads\n\\begin{eqnarray} \n\\label{f2}\n F_{2}(N,l,2)&=& \\sum_{n=1}^{N-1}(-1)^{n}\\frac{\\sin(nl\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2nl\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&-\\frac{1}{2}3l(3l-1)+\\frac{1}{2}l(l-1)-l+N\\Big(3l-\\frac{(N+1)}{2}+\\frac{1}{2}(1-(-1)^{l-\\frac{(N+1)}{2}})\\Big),\\nonumber\\\\,\n\\end{eqnarray}\nwhere $l$ is even and the sum vanishes for odd l, also, note that the last term whose coefficient is $N$ is different from zero only if $ 3l>\\frac{N+1}{2}$\n\\end{theorem} \n To prove the first formula, we note that $ F_{1}(N,2l,2)=0$, and hence the only sum to consider is the sum $ F_{1}(N,2l-1,2)$. The latter may be written as \n \\begin{eqnarray} \n\\label{n8}\n F_{1}(N,2l-1,2)&=&\\sum_{n=1}^{N-1}\\frac{\\sin(n(2l-1)\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2n(2l-1)\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&\\sum_{s,k\\geq0}^{l-1}(-1)^{s+k}\\binom {2l-2-s}{s}\\binom {2l-2-k}{k}2^{2(2l-2)-2(s+k)}\\nonumber\\\\&\\times&\\sum_{j=0}^{2l-2-2k}(-1)^{j}2^j\\binom {2l-2-2k}{k}\\sum_{n=1}^{N-1}\\cos^{2l-2-2(s-j)}(n\\pi\/N).\n\\end{eqnarray}\nThe sum over $n$, formally looks like that given in Eq. (\\ref{n2}), however, the variable $t$, may be a multiple of $N$ and in that case the Schwatt's formula given by Eq (107), does not work it has to be modified slightly. The formula that takes into account this fact may be shown to be given by\n\\begin{eqnarray} \n\\label{n9}\n\\sum_{n=1}^{N-1}\\cos^{2l-2-2(s-j)}(n\\pi\/N)&= &\\frac{-2}{2^{2l-2-2(s-j)}}\\sum_{t=1}^{l-1-s}\\binom {2l-2-2(s-j)}{l-1-(s-j)-t}\\nonumber\\\\& +&\\frac{N-1}{2^{2l-2-2(s-j)}}\\binom {2l-2-2(s-j)} {l-1-(s-j)}\\nonumber\\\\&+&\\frac{2N}{2^{2l-2-2(s-j)}}\\sum_{p=1}^{[l-1-(s-j)\/N]}\\binom {2l-2-2(s-j)} {l-1-(s-j)-pN},\n\\end{eqnarray}\nwhere the first two terms in the above formula are those expected from Eq. (\\ref{n2}), while the third term is precisely the correction to the formula Eq. (\\ref{n2}) for $t$ congruent to $N$. Therefore, there are three contributions to the sum given in Eq. (\\ref{n8}), the first of which reads\n\\begin{eqnarray} \n\\label{n10}\nF_{1}^{'}(N,2l-1,2)&=&-2\\sum_{s,k\\geq0}^{l-1}(-1)^{s+k}\\binom {2l-2-s}{s}\\binom {2l-2-k}{k}2^{2l-2-2k}\\nonumber\\\\&\\times&\\sum_{j=0}^{2l-2-2k}(-1)^{j}\\frac{1}{2^j}\\binom {2l-2-2k}{k}\\sum_{t=1}^{l-1-s}\\binom {2l-2-2(s-j)}{l-1-(s-j)-t}\\nonumber\\\\&=&-2\\hbox{res}_{w=0}U_{2l-2}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)U_{2l-2}\\Big(\\frac{1+w^2}{2w}\\Big)\\frac{1}{1-w}\\nonumber\\\\&=&-2\\hbox{res}_{w=0}\\Big(\\frac{1}{w^{3l-3}}-\\frac{1}{w^{l-2}}\\Big)\\frac{1}{(1-w)^2}\\frac{1}{1-w^2}\\nonumber\\\\&=&-\\frac{1}{2}(3l-2)(3l-3)+\\frac{1}{2}(l-1)(l-2)-l+\\frac{1}{2}(1-(-1)^l)\n\\end{eqnarray}\nwhile the second contribution is\n\\begin{eqnarray} \n\\label{n11}\nF_{1}^{''}(N,2l-1,2)&=&(N-1)\\sum_{s,k\\geq0}^{l-1}(-1)^{s+k}\\binom {2l-2-s}{s}\\binom {2l-2-k}{k}2^{2l-2-2k}\\nonumber\\\\&\\times&\\sum_{j=0}^{2l-2-2k}(-1)^{j}\\frac{1}{2^j}\\binom {2l-2-2k}{k}\\binom {2l-2-2(s-j)}{l-1-(s-j)}\\nonumber\\\\&=&(N-1)\\hbox{res}_{w=0}U_{2l-2}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)U_{2l-2}\\Big(\\frac{1+w^2}{2w}\\Big)\\frac{1}{w}\\nonumber\\\\&=& (N-1)\\hbox{res}_{w=0}\\Big(\\frac{1}{w^{3l-2}}-\\frac{1}{w^{l-1}}\\Big)\\frac{1}{1-w^2}\\frac{1}{1-w}\\nonumber\\\\&=&\\frac{N-1}{2}\\Big(2l-1-(-1)^l\\Big).\n\\end{eqnarray}\n To obtain the last contribution we write the sum over $p$, in Eq. (\\ref{n9}), as $$ \\sum_{p=1}^{[l-1-(s-j)\/N]}\\binom {2l-2-2(s-j)} {l-1-(s-j)-pN}=\\hbox{res}_{w=0} \\Big( \\frac{(1+w)^{2l-2-2(s-j)}w^N}{w^{l-(s-j)}(1-w^{N})}\\Big),$$ and using the fact that $l\\leq N-1$, then, the third contribution may be computed to give\n \\begin{eqnarray} \n\\label{n12}\nF_{1}^{'''}(N,2l-1,2)&=&2N\\sum_{s,k\\geq0}^{l-1}(-1)^{s+k}\\binom {2l-2-s}{s}\\binom {2l-2-k}{k}2^{2l-2-2k}\\nonumber\\\\&\\times&\\sum_{j=0}^{2l-2-2k}(-1)^{j}\\frac{1}{2^j}\\binom {2l-2-2k}{k}\\hbox{res}_{w=0} \\Big( \\frac{(1+w)^{2l-2-2(s-j)}w^N}{w^{l-(s-j)}(1-w^{N})}\\Big)\\nonumber\\\\&=&2N\\hbox{res}_{w=0}\\Big(\\frac{1}{1-w^N}U_{2l-2}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)U_{2l-2}\\Big(\\frac{1+w^2}{2w}\\Big)w^{N-1}\\Big)\\nonumber\\\\&=&2N\\hbox{res}_{w=0}\\frac{1}{1-w^N}\\frac{1}{w^{3l-2-N}}\\frac{1}{(1-w)(1-w^2)}\\nonumber\\\\&=&N\\big(3l-2-N+\\frac{1}{2}(1-(-1)^{l-N}\\big).\n\\end{eqnarray}\nNote that this will contribute only for $3l-2>N $, and as a result the formula for the sum $ F_{1}(N,2l-1,2)$ is\n\\begin{eqnarray} \n\\label{n13}\nF_{1}(N,2l-1,2)&=&\\sum_{n=1}^{N-1}\\frac{\\sin(n(2l-1)\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin(2n(2l-1)\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&-\\frac{1}{2}(3l-2)(3l-3)+\\frac{1}{2}(l-1)(l-2)-l+\\frac{1}{2}(1-(-1)^l)\\nonumber\\\\&+&\\frac{N-1}{2}\\Big(2l-1-(-1)^l\\Big)+N\\Big(3l-2-N+\\frac{1}{2}(1-(-1)^{l-N})\\Big).\n\\end{eqnarray}\nHaving obtained a closed formula for the sum $ F_{1}(N,2l-1,2)$, we now wish to prove the formula for the alternating sum $ F_{2}(N,l,2) $. \nFirst, we note that for $N$ odd, the sum is non-vanishing only for $l$ is even. Therefore, the formula for $ F_{2}(N,l,2) $ becomes\n\\begin{eqnarray} \n\\label{n14}\n F_{2}(N,2l,2&)=&\\sum_{n=1}^{N-1}(-1)^{n}\\frac{\\sin((2l)n\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin((2l)2n\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&\\sum_{s,k\\geq0}^{[(2l-1)\/2]}(-1)^{s+k}\\binom {2l-1-s}{s}\\binom {2l-1-k}{k}2^{2(2l-1)-2(s+k)}\\nonumber\\\\&\\times&\\sum_{j=0}^{2l-1-2k}(-1)^{j}2^j\\binom {2l-1-2k}{k}\\sum_{n=1}^{N-1}(-1)^n\\cos^{2l-1-2(s-j)}(n\\pi\/N).\n\\end{eqnarray}\nThe sum over $n$ may carried out using Eq. (114), in \\cite{ Schwatt} with the slight modification as explained before, then, it is not difficult to show\n\\begin{eqnarray} \n\\label{n15}\n\\sum_{n=1}^{N-1}(-1)^n\\cos^{2l-1-2(s-j)-1}(n\\pi\/N)&=&\\frac{-2}{2^{2l-1-2(s-j)}}\\sum_{t=0}^{l-1-s}\\binom {2l-1-2(s-j)}{l-1-(s-j)-t}\\nonumber\\\\&+&\\frac{2N}{2^{2l-1-2(s-j)}}\\sum_{p\\geq1}\\binom {2l-1-2(s-j)} {l-1-(s-j)-\\frac{(2p-1)N-1}{2}},\\nonumber\\\\\n\\end{eqnarray}\nwhere the sum over $p$, may be written as\n$$\\sum_{p\\geq1}\\binom {2l-1-2(s-j)} {l-1-(s-j)-\\frac{(2p-1)N-1}{2}}=\\hbox{res}_{w=0} \\Big( \\frac{(1+w)^{2l-1-2(s-j)}w^{N\/2}}{w^{l-(s-j)+1\/2}(1-w^{N})}\\Big).$$\nBy using Eq. (\\ref{n14}), computations show that the closed formula for the sum $F_{2}(N,2l,2)$ is\n\\begin{eqnarray} \n\\label{n16}\n F_{2}(N,2l,2&)=&\\sum_{n=1}^{N-1}(-1)^{n}\\frac{\\sin((2l)n\\pi\/N)}{\\sin(n\\pi\/N)}\\frac{\\sin((2l)2n\\pi\/N)}{\\sin(2n\\pi\/N)}\\nonumber\\\\&=&-2\\hbox{res}_{w=0}\\Big(U_{2l-2}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)U_{2l-2}\\Big(\\frac{1+w^2}{2w}\\Big)\\frac{1}{\\sqrt{w}(1-w)}\\Big)\\nonumber\\\\&+&2N\\hbox{res}_{w=0}\\frac{1}{1-w^N}\\Big(U_{2l-1}\\Big(\\frac{1+w}{2\\sqrt {w}}\\Big)U_{2l-1}\\Big(\\frac{1+w^2}{2w}\\Big)\\frac{w^{N\/2}}{w}\\Big)\\nonumber\\\\&=&-2\\hbox{res}_{w=0}\\Big(\\frac{1}{w^{3l-1}}-\\frac{1}{w^{l-1}}\\Big)\\frac{1}{(1-w)^2}\\frac{1}{1-w^2}\\nonumber\\\\&+&2N\\hbox{res}_{w=0}\\frac{1}{1-w^N}\\frac{1}{w^{3l-(N+1)\/2}}\\frac{1}{(1-w)(1-w^2)}\\nonumber\\\\&=&-\\frac{1}{2}3l(3l-1)+\\frac{1}{2}l(l-1)-l+N\\Big(3l-\\frac{(N+1)}{2}+\\frac{1}{2}(1-(-1)^{l-\\frac{(N+1)}{2}})\\Big),\\nonumber\\\\\n \\end{eqnarray}\n where the last term whose coefficient is $N$, contributes only for $3l> \\frac{(N+1)}{2} $. Let us now, check that the formulas $ F_{1}(N,2l-1,2)$, $F_{2}(N,2l,2)$ are consistent with symmetry discussed earlier, that is, $ F_{1}(N,N-l,2)= F_{2}(N,l,2)$, this in turns implies that the correctness of the formulas. To do so, we will give some explicit examples, from the expression of $ F_{1}(N,2l-1,2) $ given in Eq. (\\ref{n13}), it is clear that the sum should be $N-1$, for $l=1$ and to check this, one has to take into account that when substituting $l=1$ in the formula, the last term of Eq. (\\ref{n13}) does not contribute. From the symmetry that relates the two sums, we should have $ F_{2}(N,N-1,2)=N-1$. Indeed, this is the case, we simply let $l=\\frac{N-1}{2} $ into Eq. (\\ref{n16}), this time, however, the last term of this equation does contribute. An explicit computation shows that $ F_{2}(N,2,2)= -4$, for $N> 3$, and $ F_{2}(N,2,2)= 2$, for $N= 3$, it is interesting to note that these two cases for $l=1$ are contained in the last term of Eq. (\\ref{n15}), since for $N> 3$, the last term is equal to $0$, and hence $ F_{2}(N,2,2)= -4$, while for $N= 3$, the last term is equal to $6$, that is, our formula gives the right answer. Using the symmetry, we obtain $ F_{1}(N,N-2,2)=-4$, this can be easily checked using our formula given by Eq. (\\ref{n13}), and $l=\\frac{N-1}{2} $.\n \\section{Conclusion}\n \n \\ \\ To conclude, in this paper we used our method in \\cite{ Chair1}, to give alternative derivations to closed formulas for trigonometrical sums that appear in one-dimensional lattice, and in the proof of the conjecture of F. R Scott on Permanent of the Cauchy matrix. A new derivation of certain trigonometrical sum of the perturbative chiral Potts model is given as well as new recursion formulas of certain trigonometrical sums \\cite{Mehta}. By using these recursion formulas, then, one is able deduce the Verlinde dimension formulas for the untwisted (twisted) space of conformal blocks of $SU(2)$ ($SO(3)$)WZW. In this paper, we reported closed-form formulas for the corner-to-corner resistance and the Kirchhoff index of the first non-trivial two-dimensional resistor network, $2\\times N$. We have also, considered other class of trigonometrical sums, some of which appear in number theory. Here, we followed similar formalism as in \\cite{ Chair1}, as a consequence the non-trivial circulant electrical networks (the cycle and complete graphs are not included) are related to non-trivial trigonometrical sums in number theory. For example in\\cite{ Chair1}, we had to introduce certain numbers that we called the Bejaia and Pisa numbers with well known properties so that the trigonometrical sums that arise in the computation of the two-point resistance are written in terms of these numbers nicely. By using the well known connection between the electrical networks and the random walks \\cite{Doyle}, one may hope to give interpretations to some of the trigonometrical sums in number theory other than those associated with the two-point resistance of a given electrical network, since the latter provides an alternative way to compute the basic quantity relevant to random walks known as the first passage time, the expected time to hit a target node for the first time for a walker starting from a source node \\cite{Tetali}.\n \\vspace{7mm}\n\n{\\bf Acknowledgment:}\n\nI would like to thank Professor Bruce Berndt for reading and making comments on the manuscript. Also, I would like to thank the Abdus Salam centre for Theoretical Physics for supports and\nhospitality throughout these years \n\\newpage\n\\bibliographystyle{phaip}\n","meta":{"redpajama_set_name":"RedPajamaArXiv"}} +{"text":"\\section{Introduction\\label{intro}}\nWe consider the Cauchy problem of the fourth order nonlinear Schr\\\"odinger type equations:\n\\begin{equation}\\label{D4NLS}\n\\begin{cases}\n\\displaystyle (i\\partial_{t}+\\Delta ^2)u=\\partial P_{m}(u,\\overline{u}),\\hspace{2ex}(t,x)\\in (0,\\infty )\\times \\R^{d} \\\\\nu(0,x)=u_{0}(x),\\hspace{2ex}x\\in \\R^{d}\n\\end{cases}\n\\end{equation}\nwhere $m\\in \\N$, $m\\geq 2$, $P_{m}$ is a polynomial which is written by\n\\[\nP_{m}(f,g)=\\sum_{\\substack{\\alpha ,\\beta \\in \\Z_{\\geq 0}\\\\ \\alpha +\\beta=m}}f^{\\alpha}g^{\\beta}, \n\\]\n$\\partial$ is a first order derivative with respect to the spatial variable, for example a linear combination of \n$\\frac{\\partial}{\\partial x_1} , \\, \\dots , \\, \\frac{\\partial}{\\partial x_d}$ or $|\\nabla |= \\mathcal{F}^{-1}[|\\xi | \\mathcal{F}]$\nand the unknown function $u$ is $\\C$-valued. \nThe fourth order Schr\\\"{o}dinger equation with $P_{m}(u,\\overline{u})=|u|^{m-1}u$ appears in the study of deep water wave dynamics \\cite{Dysthe}, solitary waves \\cite{Karpman}, \\cite{KS}, vortex filaments \\cite{Fukumoto}, and so on.\nThe equation (\\ref{D4NLS}) is invariant under the following scaling transformation:\n\\[\nu_{\\lambda}(t,x)=\\lambda^{-3\/(m-1)}u(\\lambda^{-4}t,\\lambda^{-1}x), \n\\]\nand the scaling critical regularity is $s_{c}=d\/2-3\/(m-1)$. \nThe aim of this paper is to prove the well-posedness and the scattering for the solution of (\\ref{D4NLS}) \nin the scaling critical Sobolev space. \n\n\nThere are many results for the fourth order nonlinear Schr\\\"{o}dinger equation \nwith derivative nonlinearities (see \\cite{S1}, \\cite{S2}, \\cite{HJ1}, \\cite{HHW}, \\cite{HHW2}, \\cite{HJ3}, \\cite{S3}, \\cite{HJ2}, \\cite{Y12}, \\cite{HN15_1}, \\cite{HN15_2}, and references cited therein).\nEspecially, the one dimensional case is well studied.\nWang (\\cite{Y12}) considered (\\ref{D4NLS}) for the case $d=1$, $m=2l+1$, $l\\ge 2$, $P_{2l+1}(u,\\overline{u})=|u|^{2l}u$ \nand proved the small data global in time well-posedness for $s=s_{c}$ by using Kato type smoothing effect. \nBut he did not treat the cubic case.\nActually, a technical difficulty appears in this case (see Theorem \\ref{notC3} below).\n\n\nHayashi and Naumkin (\\cite{HN15_1}) considered (\\ref{D4NLS}) for $d=1$ with the power type nonlineality $\\partial_{x}(|u|^{\\rho -1}u)$ ($\\rho >4$) \nand proved the global existence of the solution and the scattering in the weighted Sobolev space.\nMoreover, they (\\cite{HN15_2}) also proved that the large time asymptotics is determined by the self similar solution in the case $\\rho =4$.\nTherefore, derivative quartic nonlinearity in the one spatial dimension is the critical in the sense of the asymptotic behavior of the solution.\n\nWe firstly focus on the quartic nonlinearity $\\partial _x (\\overline{u}^4)$ in one space dimension.\nSince this nonlinearity has some good structure, the global solution scatters to a free solution in the scaling critical Sobolev space.\nOur argument does not apply to \\eqref{D4NLS} with $P (u,\\overline{u}) = |u|^3 u$ because we rely on the Fourier restriction norm method.\nNow, we give the first results in this paper. \nFor a Banach space $H$ and $r>0$, we define $B_r(H):=\\{ f\\in H \\,|\\, \\|f\\|_H \\le r \\}$. \n\\begin{thm}\\label{wellposed_1}\nLet $d=1$, $m=4$ and $P_{4}(u,\\overline{u})=\\overline{u}^{4}$. Then the equation {\\rm (\\ref{D4NLS})} is globally well-posed for small data in $\\dot{H}^{-1\/2}$. \nMore precisely, there exists $r>0$ such that for any $T>0$ and all initial data $u_{0}\\in B_{r}(\\dot{H}^{-1\/2})$, there exists a solution\n\\[\nu\\in \\dot{Z}_{r}^{-1\/2}([0,T))\\subset C([0,T );\\dot{H}^{-1\/2})\n\\]\nof {\\rm (\\ref{D4NLS})} on $(0, T )$. \nSuch solution is unique in $\\dot{Z}_{r}^{-1\/2}([0,T))$ which is a closed subset of $\\dot{Z}^{-1\/2}([0,T))$ {\\rm (see Definition~\\ref{YZ_space} and (\\ref{Zr_norm}))}. \nMoreover, the flow map\n\\[\nS^{+}_{T}:B_{r}(\\dot{H}^{-1\/2})\\ni u_{0}\\mapsto u\\in \\dot{Z}^{-1\/2}([0,T))\n\\]\nis Lipschitz continuous. \n\\end{thm}\n\\begin{rem}\nWe note that $s=-1\/2$ is the scaling critical exponent of (\\ref{D4NLS}) for $d=1$, $m=4$. \n\\end{rem}\n\\begin{cor}\\label{sccat}\nLet $r>0$ be as in Theorem~\\ref{wellposed_1}. \nFor all $u_{0}\\in B_{r}(\\dot{H}^{-1\/2})$, there exists a solution \n$u\\in C([0,\\infty );\\dot{H}^{s_{c}})$ of (\\ref{D4NLS}) on $(0,\\infty )$ and the solution scatters in $\\dot{H}^{-1\/2}$. \nMore precisely, there exists \n$u^{+}\\in \\dot{H}^{-1\/2}$ \nsuch that \n\\[\nu(t)-e^{it\\Delta^2}u^{+}\n\\rightarrow 0\n\\ {\\rm in}\\ \\dot{H}^{-1\/2}\\ {\\rm as}\\ t\\rightarrow + \\infty. \n\\]\n\\end{cor}\n\nMoreover, we obtain the large data local in time well-posedness in the scaling critical Sobolev space.\nTo state the result, we put\n\\[\nB_{\\delta ,R} (H^s) := \\{ u_0 \\in H^s | \\ u_0=v_0+w_0 , \\, \\| v_0 \\| _{\\dot{H}^{-1\/2}} < \\delta, \\, \\| w_0 \\| _{L^2} 0$ such that for all $R \\ge \\delta$ and $u_0 \\in B_{\\delta ,R} (H^{-1\/2})$ there exists a solution\n\\[\nu \\in Z^{-1\/2}([0,T]) \\subset C([0,T); H^{-1\/2})\n\\]\nfor $T=\\delta ^{8} R^{-8}$ of \\eqref{D4NLS}.\n\nFurthermore, the same statement remains valid if we replace $H^{-1\/2}$ by $\\dot{H}^{-1\/2}$ as well as $Z^{-1\/2}([0,T])$ by $\\dot{Z}^{-1\/2}([0,T])$.\n\\end{thm}\n\n\\begin{rem}\nFor $s>-1\/2$, the local in time well-posedness in $H^s$ follows from the usual Fourier restriction norm method, which covers for all initial data in $H^s$.\nIt however is not of very much interest.\nOn the other hand, since we focus on the scaling critical cases, which is the negative regularity, we have to impose that the $\\dot{H}^{-1\/2}$ part of initial data is small.\nBut, Theorem \\ref{large-wp} is a large data result because the $L^2$ part is not restricted.\n\\end{rem}\n\n\nThe main tools of the proof are the $U^{p}$ space and $V^{p}$ space which are applied to prove \nthe well-posedness and the scattering for KP-II equation at the scaling critical regularity by Hadac, Herr and Koch (\\cite{HHK09}, \\cite{HHK10}).\n\nWe also consider the one dimensional cubic case and the high dimensional cases. \nThe second result in this paper is as follows.\n\\begin{thm}\\label{wellposed_2}\n{\\rm (i)}\\ Let $d=1$ and $m=3$. Then the equation {\\rm (\\ref{D4NLS})} is locally well-posed in $H^{s}$ for $s\\ge 0$. \\\\\n{\\rm (ii)}\\ Let $d\\geq 2$ and $(m-1)d\\geq 4$. Then the equation {\\rm (\\ref{D4NLS})}\n is globally well-posed for small data in $\\dot{H}^{s_{c}}$ (or $H^{s}$ for $s\\ge s_{c}$)\n and the solution scatters in $\\dot{H}^{s_{c}}$ (or $H^{s}$ for $s\\ge s_{c}$).\n\\end{thm}\n\nThe smoothing effect of the linear part recovers derivative in higher dimensional case.\nTherefore, we do not use the $U^p$ and $V^p$ type spaces.\nMore precisely, to establish Theorem \\ref{wellposed_2}, we only use the Strichartz estimates\nand get the solution in $C([0,T);H^{s_c})\\cap L^{p_m}([0,T); W^{q_m,s_{c}+1\/(m-1)})$ \nwith $p_m =2(m-1)$, $q_m =2(m-1)d\/\\{ (m-1)d-2\\}$.\nAccordingly, the scattering follows from a standard argument.\nSince the condition $(m-1)d\\geq 4$ is equivalent to $s_{c}+1\/(m-1)\\ge 0$, \nthe solution space $L^{p_m}([0,T); W^{q_m,s_{c}+1\/(m-1)})$ has nonnegative regularity even if the data belongs to $H^{s_{c}}$ with $-1\/(m-1)\\le s_c <0$. \nOur proof of Thorem~\\ref{wellposed_2} {\\rm (ii)} cannot applied for $d=1$ \nsince the Schr\\\"odingier admissible $(a,b)$ in {\\rm (\\ref{admissible_ab})} does not exist. \n\\begin{rem}\nFor the case $d=1$, $m=4$ and $P_{4}(u,\\overline{u})\\ne \\overline{u}^{4}$, \nwe can obtain the local in time well-posedness of {\\rm (\\ref{D4NLS})} in $H^{s}$ for $s\\ge 0$ \nby the same way of the proof of Theorem~\\ref{wellposed_2}. \nActually, we can get the solution in $C([0,T];H^s)\\cap L^4 ([0,T];W^{s+1\/2,\\infty })$ \nfor $s\\ge 0$ by using the iteration argument \nsince the fractional Leibnitz rule (see \\cite{CW91}) and the H\\\"older inequality imply\n\\[\n\\left\\| |\\nabla |^{s+\\frac{1}{2}}\\prod_{j=1}^{4}u_j \\right\\|_{L^{4\/3}_{t}([0,T);L_{x}^{1})}\n\\lesssim T^{1\/4}\\| |\\nabla |^{s+\\frac{1}{2}}u_1 \\|_{L^{4}_{t}L_{x}^{\\infty}}\\| u_2 \\|_{L^{4}_{t}L_{x}^{\\infty}}\n\\| u_3 \\|_{L^{\\infty}_{t}L_{x}^{2}}\\| u_4 \\|_{L^{\\infty}_{t}L_{x}^{2}}.\n\\]\n\\end{rem}\n\nWe give a remark on our problem, which shows that the standard iteration argument does not work.\n\\begin{thm}\\label{notC3}\n{\\rm (i)}\\ Let $d=1$, $m=3$, $s<0$ and $P_{3}(u,\\overline{u})=|u|^{2}u$. Then the flow map of {\\rm (\\ref{D4NLS})} from $H^s$ to $C(\\R ; H^s)$ is not smooth. \\\\\n{\\rm (ii)}\\ Let $m\\ge 2$, $s1}$ and $P_{<1}$ as\n\\[\nP_{>1}:=\\sum_{N\\ge 1}P_N,\\ P_{<1}:=Id-P_{>1}. \n\\]\n\n\\begin{defn}\\label{YZ_space}\nLet $s <0$.\\\\\n{\\rm (i)} We define $\\dot{Z}^{s}:=\\{u\\in C(\\R ; \\dot{H}^{s}(\\R^{d}))\\cap U^{2}_{S}|\\ \\| u \\| _{\\dot{Z}^{s}}<\\infty\\}$ with the norm\n\\[\n \\| u \\| _{\\dot{Z}^{s}}:=\\left(\\sum_{N}N^{2s} \\| P_{N}u \\| ^{2}_{U^{2}_{S}}\\right)^{1\/2}.\n\\]\n{\\rm (ii)} We define $Z^{s}:=\\{u\\in C(\\R ; H^{s}(\\R^{d})) |\\ \\| u \\| _{Z^{s}}<\\infty\\}$ with the norm\n\\[\n \\| u \\| _{Z^{s}}:= \\| P_{<1} u \\| _{\\dot{Z}^{0}}+ \\| P_{>1} u \\| _{\\dot{Z}^{s}}. \n\\]\n{\\rm (iii)} We define $\\dot{Y}^{s}:=\\{u\\in C(\\R ; \\dot{H}^{s}(\\R^{d}))\\cap V^{2}_{S}|\\ \\| u \\| _{\\dot{Y}^{s}}<\\infty\\}$ with the norm\n\\[\n \\| u \\| _{\\dot{Y}^{s}}:=\\left(\\sum_{N}N^{2s} \\| P_{N}u \\| ^{2}_{V^{2}_{S}}\\right)^{1\/2}.\n\\]\n{\\rm (iv)} We define $Y^{s}:=\\{u\\in C(\\R ; H^{s}(\\R^{d})) |\\ \\| u \\| _{Y^{s}}<\\infty\\}$ with the norm\n\\[\n \\| u \\| _{Y^{s}}:= \\| P_{<1} u \\| _{\\dot{Y}^{0}}+ \\| P_{>1 }u \\| _{\\dot{Y}^{s}}.\n\\]\n\\end{defn}\n\\section{Multilinear estimate for $P_{4}(u,\\overline{u})=\\overline{u}^{4}$ in $1d$ \\label{Multi_est}}\nIn this section, we prove multilinear estimates for the nonlinearity $\\partial_{x}(\\overline{u}^{4})$ in $1d$, which plays a crucial role in the proof of Theorem \\ref{wellposed_1}.\n\\begin{lemm}\\label{modul_est}\nWe assume that $(\\tau_{0},\\xi_{0})$, $(\\tau_{1}, \\xi_{1})$, $\\cdots$, $(\\tau_{4}, \\xi_{4})\\in \\R\\times \\R^{d}$ satisfy \n$\\sum_{j=0}^{4}\\tau_{j}=0$ and $\\sum_{j=0}^{4}\\xi_{j}=0$. Then, we have \n\\begin{equation}\\label{modulation_est}\n\\max_{0\\leq j\\leq 4}|\\tau_{j}-|\\xi_{j}|^{4}|\n\\geq \\frac{1}{5}\\max_{0\\leq j\\leq 4}|\\xi_{j}|^{4}. \n\\end{equation}\n\\end{lemm}\n\\begin{proof}\nBy the triangle inequality, we obtain (\\ref{modulation_est}). \n\\end{proof}\n\n\\subsection{The homogeneous case}\n\n\\begin{prop}\\label{HL_est_n}\nLet $d=1$ and $01}u_{j} \\| _{\\dot{Y}^{-1\/2}}\n\\end{split}\n\\]\nTherefore, we obtain\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,1}'(N_{1})} \\int_{0}^{T}\\int_{\\R}\\left(N_{0}\\prod_{j=0}^{4}P_{N_{j}}u_{j}\\right)dxdt\\right| \\\\\n&\\lesssim T^{1\/2}N_0 \\| u_{0,N_{0}} \\| _{V^{2}_{S}} \\| u_{1,N_{1}} \\| _{V^{2}_{S}}\\prod_{j=2}^{3}\\| P_{>1}u_{j} \\| _{\\dot{Y}^{-1\/2}}\\|P_{<1}u_4\\|_{\\dot{Y}^{0}}\n\\end{split}\n\\]\nand note that $T^{1\/2}N_0\\le T^{1\/6}$.\n\nIn the case $T \\ge N_0^{-3}$, we divide the integrals on the left-hand side of (\\ref{hl}) into $10$ pieces of the form \\eqref{piece_form_hl} in the proof of Proposition \\ref{HL_est_n}.\nThanks to Lemma~\\ref{modul_est}, let us consider the case that $Q_{j}^{S}=Q_{\\geq M}^{S}$ for some $0\\leq j\\leq 4$.\nFirst, we consider the case $Q_{0}^{S}=Q_{\\geq M}^{S}$. \nBy the same way as in the proof of Proposition \\ref{HL_est_n} and using\n\\[\n\\|Q_{4}^{S}P_{<1}u_{4,T}\\|_{L^{12}_{t}L^{6}_{x}}\\lesssim \\|Q_{4}^{S}P_{<1}u_{4,T}\\|_{V^{2}_{S}}\\lesssim \\|P_{<1}u_{4,T}\\|_{\\dot{Y}^{0}}\n\\]\ninstead of (\\ref{L12L6_est}), we obtain\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,1}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{0}Q_{\\geq M}^{S}u_{0,N_{0},T}\\prod_{j=1}^{4}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| Q_{\\geq M}^{S}u_{0,N_{0},T} \\| _{L^{2}_{tx}} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}}\n\\prod_{j=2}^{3} \\left\\|\\sum_{1 \\le N_{j}\\lesssim N_{1}}Q_{j}^{S}u_{j,N_{j},T}\\right\\|_{L^{12}_{t}L^{6}_{x}} \\|Q_{4}^{S}P_{<1}u_{4,T}\\|_{L^{12}_{t}L^{6}_{x}}\\\\\n& \\lesssim N_0^{-\\frac{1}{2}} \\| P_{N_0} u_0 \\| _{V^2_S} \\| P_{N_1} u_1 \\| _{V^2_S} \\prod_{j=2}^{3} \\left\\| P_{>1} u_j \\right\\| _{\\dot{Y}^{-1\/2}} \\| P_{<1} u_{4} \\| _{\\dot{Y}^0}\n\\end{split}\n\\]\nand note that $N_0^{-1\/2}\\le T^{1\/6}$. \nSince the cases $Q_j^S = Q_{\\ge M}^S$ ($j=1,2,3$) are similarly handled, we omit the details here.\n\n\nWe focus on the case $Q_4^S = Q_{\\ge M}^S$.\nBy the same way as in the proof of Proposition \\ref{HL_est_n} and using\n\\[\n\\|Q_{\\ge M}^{S}P_{<1}u_{4,T}\\|_{L^{2}_{tx}}\\lesssim N_{0}^{-2} \\|P_{<1}u_{4,T}\\|_{V^{2}_{S}}\\lesssim N_{0}^{-2}\\|P_{<1}u_{4,T}\\|_{\\dot{Y}^{0}}\n\\]\ninstead of (\\ref{hi_mod_234}) with $j=4$, we obtain\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,1}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{0}Q_{\\geq M}^{S}u_{4,N_{4},T}\\prod_{j=0}^{3}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| u_{0,N_{0},T} \\| _{L^{12}_{t}L_x^6} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}}\n\\prod_{j=2}^{3} \\left\\|\\sum_{1 \\le N_{j}\\lesssim N_{1}}Q_{j}^{S}u_{j,N_{j},T}\\right\\|_{L^{12}_{t}L^{6}_{x}} \n\\|Q_{\\geq M}^{S} P_{<1}u_{4,T}\\|_{L^{2}_{tx}}\\\\\n& \\lesssim N_{0}^{-1\/2}\\| P_{N_0} u_0 \\| _{V^2_S} \\| P_{N_1} u_1 \\| _{V^2_S} \\prod_{j=2}^{3} \\left\\| P_{>1} u_j \\right\\| _{\\dot{Y}^{-1\/2}} \\| P_{<1} u_4 \\| _{\\dot{Y}^0}\n\\end{split}\n\\]\nand note that $N_0^{-1\/2}\\le T^{1\/6}$. \n\n\nWe secondly consider the case $A_{1,2}'(N_1)$.\nIn the case $T \\le N_0^{-3}$, the H\\\"older inequality implies\n\\[\n\\begin{split}\n& \\left|\\sum_{A_{1,2}'(N_{1})} \\int_{0}^{T}\\int_{\\R}\\left(N_{0}\\prod_{j=0}^{4}P_{N_{j}}u_{j}\\right)dxdt\\right| \\\\\n& \\le N_0 \\| \\ee_{[0,T)}\\|_{L^{2}_{t}}\n\\| u_{0,N_0} \\| _{L_t^4 L_x^{\\infty}} \\| u_{1,N_1} \\| _{L_t^4 L_x^{\\infty}} \\left\\| \\sum _{1 \\le N_2 \\lesssim N_1} u_{2,N_2} \\right\\| _{L_t^{\\infty} L_x^2}\n\\prod_{j=3}^{4}\\| P_{<1} u_{j} \\| _{L_t^{\\infty} L_x^4} .\n\\end{split}\n\\]\nBy the same estimates as in the proof for the case $A_{1,1}'(N_1)$ and\n\\[\n\\| P_{<1} u_{j} \\| _{L_t^{\\infty} L_x^4}\\lesssim \\| P_{<1} u_{j} \\| _{L_t^{\\infty} L_x^{2}}\n\\lesssim \\left(\\sum_{N\\le 2}\\|P_{N}P_{<1}u_{j}\\|_{V^{2}_{S}}^{2}\\right)^{1\/2}\n\\le \\|P_{<1}u_j\\|_{\\dot{Y}^{0}}\n\\]\nfor $j=3,4$, we obtain\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,2}'(N_{1})} \\int_{0}^{T}\\int_{\\R}\\left(N_{0}\\prod_{j=0}^{4}P_{N_{j}}u_{j}\\right)dxdt\\right| \\\\\n&\\lesssim T^{1\/2}N_0^{1\/2} \\| u_{0,N_{0}} \\| _{V^{2}_{S}} \\| u_{1,N_{1}} \\| _{V^{2}_{S}}\\| P_{>1}u_{2} \\| _{\\dot{Y}^{-1\/2}}\\prod_{j=3}^{4}\\|P_{<1}u_j\\|_{\\dot{Y}^{0}}\n\\end{split}\n\\]\nand note that $T^{1\/2}N_0^{1\/2}\\le T^{1\/3}$. \n\n\nIn the case $T \\ge N_0^{-3}$, we divide the integrals on the left-hand side of (\\ref{hl}) into $10$ pieces of the form \\eqref{piece_form_hl} in the proof of Proposition \\ref{HL_est_n}.\nThanks to Lemma~\\ref{modul_est}, let us consider the case that $Q_{j}^{S}=Q_{\\geq M}^{S}$ for some $0\\leq j\\leq 4$.\nBy the same argument as in the proof for the case $A_{1,1}'(N_1)$, we obtain \n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,2}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{0}Q_{\\geq M}^{S}u_{0,N_{0},T}\\prod_{j=1}^{4}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| Q_{\\geq M}^{S}u_{0,N_{0},T} \\| _{L^{2}_{tx}} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}} \\left\\|\\sum_{1 \\le N_{2}\\lesssim N_{1}}Q_{2}^{S}u_{2,N_{2},T}\\right\\|_{L^{12}_{t}L^{6}_{x}} \\prod_{j=3}^{4} \\| Q_{j}^{S}P_{<1}u_{j,T}\\|_{L^{12}_{t}L^{6}_{x}}\\\\\n& \\lesssim N_0^{-1} \\| P_{N_0} u_0 \\| _{V^2_S} | P_{N_1} u_1 \\| _{V^2_S} \\left\\| P_{>1} u_2 \\right\\| _{\\dot{Y}^{-1\/2}} \\prod _{j=3}^4 \\| P_{<1} v_j \\| _{\\dot{Y}^0}\n\\end{split}\n\\]\nif $Q_0 = Q_{\\ge M}^S$ and \n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,2}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{4}Q_{\\geq M}^{S}u_{4,N_{4},T}\\prod_{j=0}^{3}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| u_{0,N_{0},T} \\| _{L^{12}_{t}L_x^6} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}} \\left\\|\\sum_{1 \\le N_{2}\\lesssim N_{1}}Q_{2}^{S}u_{2,N_{2},T}\\right\\|_{L^{12}_{t}L^{6}_{x}} \\\\\n&\\hspace{21ex}\\times \\|Q_{3}^{S} P_{<1}u_{3,T}\\|_{L^{12}_{t}L^{6}_{x}} \\| Q_{\\geq M}^{S} P_{<1}u_{4,T}\\|_{L^{2}_{tx}}\\\\\n& \\lesssim N_0^{-1} \\| P_{N_0} u_0 \\| _{V^2_S} \\| P_{N_1} u_1 \\| _{V^2_S} \\left\\| P_{>1} u_2 \\right\\| _{\\dot{Y}^{\\frac{1}{2}}} \\prod_{j=3}^{4}\\| P_{<1} u_j \\| _{\\dot{Y}^0}\n\\end{split}\n\\]\nif $Q_4 = Q_{\\ge M}^S$\nNote that $N_0^{-1}\\le T^{1\/3}$. \nThe remaining cases follow from the same argument as above.\n\n\nWe thirdly consider the case $A_{1,3}'(N_1)$.\nIn the case $T \\le N_0^{-3}$, the H\\\"older inequality implies\n\\[\n\\begin{split}\n& \\left|\\sum_{A_{1,3}'(N_{1})} \\int_{0}^{T}\\int_{\\R}\\left(N_{0}\\prod_{j=0}^{4}P_{N_{j}}u_{j}\\right)dxdt\\right| \\\\\n& \\le N_0 \\| \\ee_{[0,T)}\\|_{L^{2}_{t}}\\| u_{0,N_0} \\| _{L_t^4 L_x^{\\infty}} \\| u_{1,N_1} \\| _{L_t^4 L_x^{\\infty}}\n\\prod_{j=2}^{4} \\| P_{<1}u_{2} \\| _{L_t^{\\infty} L_x^3}.\n\\end{split}\n\\]\nBy the same estimates as in the proof for the case $A_{1,1}'(N_1)$ and\n\\[\n\\| P_{<1} u_{j} \\| _{L_t^{\\infty} L_x^3}\\lesssim \\| P_{<1} u_{j} \\| _{L_t^{\\infty} L_x^{2}}\n\\lesssim \\left(\\sum_{N\\le 2}\\|P_{N}P_{<1}u_{j}\\|_{V^{2}_{S}}^{2}\\right)^{1\/2}\n\\le \\|P_{<1}u_j\\|_{\\dot{Y}^{0}}\n\\]\nfor $j=2, 3,4$, we obtain\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,3}'(N_{1})} \\int_{0}^{T}\\int_{\\R}\\left(N_{0}\\prod_{j=0}^{4}P_{N_{j}}u_{j}\\right)dxdt\\right| \n\\lesssim T^{1\/2}\\| u_{0,N_{0}} \\| _{V^{2}_{S}} \\| u_{1,N_{1}} \\| _{V^{2}_{S}}\\prod_{j=2}^{4}\\| P_{<1}u_{j} \\| _{\\dot{Y}^{0}}. \n\\end{split}\n\\]\n\n\nIn the case $T \\ge N_0^{-3}$, we divide the integrals on the left-hand side of (\\ref{hl}) into $10$ pieces of the form \\eqref{piece_form_hl} in the proof of Proposition \\ref{HL_est_n}.\nThanks to Lemma~\\ref{modul_est}, let us consider the case that $Q_{j}^{S}=Q_{\\geq M}^{S}$ for some $0\\leq j\\leq 4$.\nBy the same argument as in the proof for the case $A_{1,1}'(N_1)$, we obtain \n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,3}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{0}Q_{\\geq M}^{S}u_{0,N_{0},T}\\prod_{j=1}^{4}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| Q_{\\geq M}^{S}u_{0,N_{0},T} \\| _{L^{2}_{tx}} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}} \n\\prod_{j=2}^{4} \\|Q_{j}^{S}P_{<1}u_{j,T}\\|_{L^{12}_{t}L^{6}_{x}}\\\\\n& \\lesssim N_0^{-3\/2} \\| P_{N_0} u_0 \\| _{V^2_S} \\| P_{N_1} u_1 \\| _{V^2_S} \\left\\| P_{<1} u_2 \\right\\| _{Y^{-1\/2}} \\prod _{j=3}^4 \\| P_{<1} v_j \\| _{\\dot{Y}^0} \n\\end{split}\n\\]\nif $Q_0 = Q_{\\ge M}^S$ and\n\\[\n\\begin{split}\n&\\left|\\sum_{A_{1,3}'(N_{1})}\\int_{\\R}\\int_{\\R}\\left(N_{4}Q_{\\geq M}^{S}u_{4,N_{4},T}\\prod_{j=0}^{3}Q_{j}^{S}u_{j,N_{j},T}\\right)dxdt\\right|\\\\\n&\\leq N_{0} \\| u_{0,N_{0},T} \\| _{L^{12}_{t}L_x^6} \\| Q_{1}^{S}u_{1,N_{1},T} \\| _{L^{4}_{t}L^{\\infty}_{x}} \\prod _{j=2}^3 \\|Q_{j}^{S} P_{<1}u_{j,T}\\|_{L^{12}_{t}L^{6}_{x}} \n\\|Q_{\\geq M}^{S} P_{<1}u_{4,T}\\|_{L^2_{tx}}\\\\\n& \\lesssim N_0^{-3\/2} \\| P_{N_0} u_0 \\| _{V^2_S} \\| P_{N_1} u_1 \\| _{V^2_S} \\prod _{j=2}^4 \\left\\| P_{<1} u_j \\right\\| _{Y^{0}}\n\\end{split}\n\\]\nif $Q_4 = Q_{\\ge M}^S$.\nNote that $N_0^{-3\/2}\\le T^{1\/2}$. \nThe cases $Q_j^S = Q_{\\ge M}^S$ ($j=1,2,3$) are the same argument as above. \n\n\\end{proof}\n\n\nFurthermore, we obtain the following estimate.\n\n\\begin{prop}\\label{HH_est-inh}\nLet $d=1$ and $00$, we define \n\\begin{equation}\\label{Zr_norm}\n\\dot{Z}^{s}_{r}(I)\n:=\\left\\{u\\in \\dot{Z}^{s}(I)\\left|\\ \\| u \\| _{\\dot{Z}^{s}(I)}\\leq 2r \\right.\\right\\}\n\\end{equation}\nwhich is a closed subset of $\\dot{Z}^{s}(I)$. \nLet $T>0$ and $u_{0}\\in B_{r}(\\dot{H}^{-1\/2})$ are given. For $u\\in \\dot{Z}^{-1\/2}_{r}([0,T))$, \nwe have\n\\[\n \\| \\Phi_{T,u_{0}}(u) \\| _{\\dot{Z}^{-1\/2}([0,T))}\\leq \\| u_{0} \\| _{\\dot{H}^{-1\/2}} +C \\| u \\| _{\\dot{Z}^{-1\/2}([0,T))}^{4}\\leq r(1+ 16 Cr^{3})\n\\]\nand\n\\[\n\\begin{split}\n \\| \\Phi_{T,u_{0}}(u)-\\Phi_{T,u_{0}}(v) \\| _{\\dot{Z}^{-1\/2}([0,T))}\n&\\leq C( \\| u \\| _{\\dot{Z}^{-1\/2}([0,T))}+ \\| v \\| _{\\dot{Z}^{-1\/2}([0,T))})^{3} \\| u-v \\| _{\\dot{Z}^{-1\/2}([0,T))}\\\\\n&\\leq 64Cr^{3} \\| u-v \\| _{\\dot{Z}^{-1\/2}([0,T))}\n\\end{split}\n\\]\nby Proposition~\\ref{Duam_est} and\n\\[\n \\| S(\\cdot )u_{0} \\| _{\\dot{Z}^{-1\/2}([0,T))}\\leq \\| \\ee_{[0,T)}S(\\cdot )u_{0} \\| _{\\dot{Z}^{-1\/2}}\\leq \\| u_{0} \\| _{\\dot{H}^{-1\/2}}, \n\\] \nwhere $C$ is an implicit constant in (\\ref{Duam_est_1}). Therefore if we choose $r$ satisfying\n\\[\nr <(64C)^{-1\/3},\n\\]\nthen $\\Phi_{T,u_{0}}$ is a contraction map on $\\dot{Z}^{-1\/2}_{r}([0,T))$. \nThis implies the existence of the solution of (\\ref{D4NLS}) and the uniqueness in the ball $\\dot{Z}^{-1\/2}_{r}([0,T))$. \nThe Lipschitz continuously of the flow map is also proved by similar argument. \n\\end{proof} \nCorollary~\\ref{sccat} is obtained by the same way as the proof of Corollaty\\ 1.2 in \\cite{Hi}. \n\n\\subsection{The large data case}\n\nIn this subsection, we prove Theorem \\ref{large-wp}.\nThe following is the key estimate.\n\n\\begin{prop}\\label{Duam_est-inh}\nLet $d=1$. We have\n\\begin{equation}\\label{Duam_est_1-inh}\n \\| I_{1}(u_{1},\\cdots u_{4}) \\| _{\\dot{Z}^{-1\/2}} \\lesssim \\prod_{j=1}^{4} \\| u_{j} \\| _{Y^{-1\/2}}.\n\\end{equation}\n\\end{prop}\n\n\\begin{proof}\nWe decompose $u_j = v_j +w_j$ with $v_j = P_{>1}u_j \\in \\dot{Y}^{-1\/2}$ and $w_j = P_{<1} u_j \\in \\dot{Y}^0$. \n>From Propositions \\ref{HL_est_n-inh}, \\ref{HH_est-inh}, and the same way as in the proof of Proposition~\\ref{Duam_est}, \nit remains to prove that\n\\[\n\\| I_{1}(w_{1},w_2,w_3,w_{4}) \\| _{\\dot{Z}^{-1\/2}} \\lesssim \\prod_{j=1}^{4} \\| u_{j} \\| _{\\dot{Y}^0}.\n\\]\nBy Theorem \\ref{duality}, the Cauchy-Schwartz inequality, the H\\\"older inequality and the Sobolev inequality, we have\n\\[\n\\| I_{1}(w_{1},w_2,w_3,w_{4}) \\| _{\\dot{Z}^{-1\/2}}\n\\lesssim \\left\\| \\prod_{j=1}^{4}\\overline{w_{j}} \\right\\|_{L^1([0,1];L^2)}\n\\lesssim \\prod _{j=1}^4 \\| w_j \\| _{L_t^{\\infty} L_x^2}\n\\lesssim \\prod_{j=1}^{4} \\| u_{j} \\| _{\\dot{Y}^{0}},\n\\]\nwhich completes the proof.\n\\end{proof}\n\n\\begin{proof}[\\rm{\\bf{Proof of Theorem \\ref{large-wp}}}]\nLet $u_0 \\in B_{\\delta ,R}(H^{-1\/2})$ with $u_0=v_0+w_0$, $v_0 \\in \\dot{H}^{-1\/2}$, $w_0 \\in L^2$.\nA direct calculation yields\n\\[\n\\| S(t) u_0 \\| _{Z^{-1\/2}([0,1))} \\le \\delta +R.\n\\]\nWe start with the case $R=\\delta = (4C+4)^{-4}$, where $C$ is the implicit constant in \\eqref{Duam_est_1-inh}.\nProposition \\ref{Duam_est-inh} implies that for $u \\in Z^{-1\/2}_r([0,1])$ with $r=1\/(4C+4)$\n\\begin{align*}\n\\| \\Phi_{1,u_{0}}(u) \\| _{Z^{-1\/2}([0,1))} & \\leq \\| S(t) u_0 \\| _{Z^{-1\/2}([0,1))} +C \\| u \\| _{Z^{-1\/2}([0,1))}^{4} \\\\\n& \\leq 2r^4 + 16C r^4\n= r^4 (16C+2)\n\\le r\n\\end{align*}\nand\n\\begin{align*}\n\\| \\Phi_{1,u_{0}}(u)-\\Phi_{1,u_{0}}(v) \\| _{Z^{-1\/2}([0,1))}\n&\\leq C( \\| u \\| _{Z^{-1\/2}([0,1))}+ \\| v \\| _{Z^{-1\/2}([0,1))})^{3} \\| u-v \\| _{Z^{-1\/2}([0,1))}\\\\\n&\\leq 64Cr^{3} \\| u-v \\| _{Z^{-1\/2}([0,1))}\n< \\| u-v \\| _{Z^{-1\/2}([0,1))}\n\\end{align*}\nif we choose $C$ large enough (namely, $r$ is small enough).\nAccordingly, $\\Phi_{1,u_{0}}$ is a contraction map on $\\dot{Z}^{-1\/2}_{r}([0,1))$.\n\nWe note that \nall of the above remains valid if we exchange $Z^{-1\/2}([0,1))$ by the smaller space $\\dot{Z}^{-1\/2}([0,1))$ since $\\dot{Z}^{-1\/2}([0,1)) \\hookrightarrow Z^{-1\/2}([0,1))$ and the left hand side of \\eqref{Duam_est_1-inh} is the homogeneous norm.\n\nWe now assume that $u_0 \\in B_{\\delta ,R}(H^{-1\/2})$ for $R \\ge \\delta = (4C+4)^{-4}$.\nWe define $u_{0, \\lambda}(x) = \\lambda ^{-1} u_0 (\\lambda ^{-1}x)$.\nFor $\\lambda = \\delta ^{-2} R^{2}$, we observe that $u_{0,\\lambda} \\in B_{\\delta ,\\delta}(H^{-1\/2})$.\nWe therefore find a solution $u_{\\lambda} \\in Z^{-1\/2}([0,1))$ with $u_{\\lambda}(0,x) = u_{0,\\lambda}(x)$.\nBy the scaling, we find a solution $u \\in Z^{-1\/2}([0, \\delta ^8 R^{-8}))$.\n\nThanks to Propositions \\ref{HL_est_n-inh} and \\ref{HH_est-inh}, the uniqueness follows from the same argument as in \\cite{HHK10}.\n\\end{proof}\n\n\\section{Proof of Theorem~\\ref{wellposed_2}}\\label{pf_wellposed_2}\\kuuhaku\nIn this section, we prove Theorem~\\ref{wellposed_2}. \nWe only prove for the homogeneous case since the proof for the inhomogeneous case is similar. \nWe define the map $\\Phi_{T, \\varphi}^{m}$ as \n\\[\n\\Phi_{T, \\varphi}^{m}(u)(t):=S(t)\\varphi -iI_{T}^{m}(u,\\cdots, u)(t),\n\\] \nwhere\n\\[\nI_{T}^{m}(u_{1},\\cdots u_{m})(t):=\\int_{0}^{t}\\ee_{[0,T)}(t')S(t-t')\\partial \\left(\\prod_{j=1}^{m}u_{j}(t')\\right)dt'.\n\\]\nand the solution space $\\dot{X}^{s}$ as\n\\[\n\\dot{X}^{s}:=C(\\R;\\dot{H}^{s})\\cap L^{p_{m}}(\\R;\\dot{W}^{s+1\/(m-1),q_{m}}),\n\\] \nwhere $p_{m}=2(m-1)$, $q_{m}=2(m-1)d\/\\{(m-1)d-2\\}$ for $d \\ge 2$ and $p_3=4$, $q_3=\\infty$ for $d=1$. \nTo prove the well-posedness of (\\ref{D4NLS}) in $L^{2}(\\R )$ or $H^{s_{c}}(\\R^{d})$, we prove that $\\Phi_{T, \\varphi}$ is a contraction map \non a closed subset of $\\dot{X}^{s}$. \nThe key estimate is the following:\n\\begin{prop}\\label{Duam_est_g}\n{\\rm (i)}\\ Let $d=1$ and $m=3$. For any $0From $\\xi _j \\in [N-N^{-1}, N+N^{-1}]$ for $j=1,2,3$, we get\n\\[\n|-(\\xi _1-\\xi _2+\\xi _3)^4+\\xi _1^4-\\xi _2^4+\\xi _3^4|\n\\lesssim 1.\n\\]\nWe therefore obtain for sufficiently small $t>0$\n\\begin{align*}\n|\\widehat{u^{(3)}_{N}} (t,\\xi ) |\n& \\gtrsim t N^{-3s+5\/2} \\left| \\int _{\\xi _1-\\xi _2+\\xi _3 =\\xi} \\ee _{[N-N^{-1}, N+N^{-1}]} (\\xi _1) \\ee _{[N-N^{-1}, N+N^{-1}]} (\\xi _2) \\ee _{[N-N^{-1}, N+N^{-1}]} (\\xi _3) \\right| \\\\\n& \\gtrsim t N^{-3s+1\/2} \\ee _{[N-N^{-1},N+N^{-1} ]} (\\xi ) .\n\\end{align*}\nHence,\n\\[\n\\| u^{(3)}_{N} \\| _{L^{\\infty}([0,1]; H^s)} \\gtrsim N^{-2s}.\n\\]\nThis lower bound goes to infinity as $N$ tends to infinity if $s<0$, which concludes the proof.\n\\end{proof}\n\n\nSecondly, we show that absence of a smooth flow map for $d \\ge 1$ and $m \\ge 2$.\nPutting\n\\[\ng_N := N^{-s-d\/2} \\mathcal{F}^{-1}[ \\ee _{[-N,N]^d}] ,\n\\]\nwe set $u_N^{(m)} := u^{(m)} [g_N]$.\nNote that $\\| g_N \\| _{H^s} \\sim 1$.\nAs above, we show the following.\n\n\\begin{prop}\nIf $s