hep-th/yymmnnn

Extremal black string with Kalb-Ramond field via $\alpha^{\prime}$ corrections

Shuxuan Ying

Department of Physics, Chongqing University

Chongqing, 401331, China

<EMAIL_ADDRESS>
Abstract

In this paper, we obtain the three-dimensional regular extremal black string
solution incorporating $\alpha^{\prime}$ corrections and a non-trivial Kalb-
Ramond field. The difficulty in considering the Kalb-Ramond field lies in the
fact that it transforms the original equations of motion into an infinite
summation form involving matrices, making it difficult to calculate the matrix
differential equations. To solve this problem, we introduce a new method that
transforms the infinite summation of matrix differential equations into a
simple trace of the matrix. As a result, we are able to obtain a non-
perturbative and non-singular extremal black string solution. Indeed, this
work serves as a good example for studying more complicated non-perturbative
solutions that incorporate the Kalb-Ramond field via complete
$\alpha^{\prime}$ corrections.

###### Contents

  1. 1 Introduction
  2. 2 Three-dimensional extremal black string at tree-level
  3. 3 Regular extremal black string via $\alpha^{\prime}$ corrections
  4. 4 Conclusion

##  1 Introduction

In recent studies [1, 2, 3], Hohm and Zwiebach made notable contributions to
understanding the $\alpha^{\prime}$ corrections of closed string theory at all
orders and derived the complete action by utilizing the $O\left(d,d\right)$
symmetry. The key point that complete action can be obtained by
$O\left(d,d\right)$ symmetry is due to the fact that when massless closed
string fields are independent of the $m$ coordinates, the action manifests an
$O\left(m,m\right)$ symmetry [4, 5]. This result has been verified in the
tree-level string effective action as well as its first-order
$\alpha^{\prime}$ correction. Consequently, it suggests that the low energy
effective action with complete $\alpha^{\prime}$ corrections can be expressed
by utilizing the standard $O\left(d,d\right)$ matrix in terms of
$\alpha^{\prime}$ corrected fields for time-dependent backgrounds [6, 4, 5, 7,
8]. Based on this observation, Hohm and Zwiebach conjectured that this result
is always true for all orders in $\alpha^{\prime}$. Conjectured this
assumption, they successfully demonstrated that the $\alpha^{\prime}$
corrections can be classified based on even powers of the Hubble parameter
within the framework of the FLRW cosmological background. In addition, it is
worth noting that the dilaton field only includes first-order time
derivatives. These features make the equations of motion (EOM) to solely
include two derivatives of the metric and be exactly solvable. This remarkable
progress introduces a new action that make the study of black holes and
cosmology incorporating non-perturbative string effects possible. By using
this action, numerous new regular black hole solutions and cosmological
solutions were studied. Refs. [9, 10, 11, 12, 13, 14] have considered the FLRW
cosmological ansatz, revealing that the $\alpha^{\prime}$ corrections can
eliminate the big bang singularity. In addition, the resolution of curvature
singularities in two-dimensional string black holes has been discussed in
refs. [15, 16, 17, 18]. However, these solutions impose significant
constraints on the form of the spacetime ansatz and require the Kalb-Ramond
field to vanish. It looks impossible to get the non-perturbative solutions
with a non-constant Kalb-Ramond field. The challenge arises from the fact that
considering the Kalb-Ramond field makes the classification of
$\alpha^{\prime}$ corrections complicated, which can no longer be simply
written by even powers of the Hubble parameter. Instead, an infinite summation
of $O\left(d,d\right)$ matrices must be considered. The previous discussion
about Kalb-Ramond field in Hohm-Zwiebach action can be found in a ref. [19].
To demonstrate the difficulty including Kalb-Ramond field, let us recall the
Hohm-Zwiebach action for the black hole background:

$I_{HZ}=-\int
dx^{3}e^{-\Phi}\left(-\dot{\Phi}^{2}+\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}^{2k}\right)\right),$
(1.1)

where all fields depend on $y$, and
$\dot{f}\left(y\right)\equiv\partial_{y}f\left(y\right)$. The coefficients are
$\bar{c}_{1}=c_{1}=-\frac{1}{8}$, $\bar{c}_{2}=-c_{2}=-\frac{1}{64}$,
$\bar{c}_{3}=c_{3}=-\frac{1}{3.2^{7}}$,
$\bar{c}_{4}=-c_{4}=-\frac{1}{2^{15}}+\frac{1}{2^{12}}\zeta\left(3\right)$ and
$\bar{c}_{k>4}$’s are unknown coefficients for the bosonic case [20]. The
$O\left(d,d\right)$ invariant dilaton and $O\left(d,d\right)$ matrix are
defined as

$e^{-\Phi}\equiv\sqrt{g}e^{-2\phi},\qquad\mathcal{S}=\left(\begin{array}[]{cc}BG^{-1}&G-BG^{-1}B\\\
G^{-1}&-G^{-1}B\end{array}\right),$ (1.2)

where $\phi$ is the physical dilaton, and $g$ does not include the direction
of $dy$. The EOM derived from the action are given by:

$\displaystyle\ddot{\Phi}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}k\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0,$ (1.3)
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4k\bar{c}_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}\right)$
$\displaystyle=$ $\displaystyle 0,$ (1.4)
$\displaystyle\dot{\Phi}^{2}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\left(2k-1\right)\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0.$ (1.5)

If we consider the three-dimensional black hole ansatz without the Kalb-Ramond
field:

$ds^{2}=a\left(y\right)^{2}\left(-dt^{2}+dx^{2}\right)+dy^{2}.$ (1.6)

The EOM (1.3) to (1.5) can be significantly simplified as follows:

$\displaystyle\dot{\Phi}+\frac{1}{2}\bar{H}\bar{f}\left(\bar{H}\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\bar{f}\left(\bar{H}\right)\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\dot{\Phi}+\bar{g}\left(\bar{H}\right)$ $\displaystyle=$
$\displaystyle 0,$ (1.7)

where the Hubble parameter is defined as
$\bar{H}\left(y\right)=\frac{\dot{a}\left(y\right)}{a\left(y\right)}$ and

$\displaystyle\bar{f}\left(\bar{H}\right)$ $\displaystyle=$ $\displaystyle
d\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(-\alpha^{\prime}\right)^{k-1}2^{2\left(k+1\right)}k\bar{c}_{k}\bar{H}^{2k-1},$
$\displaystyle\bar{g}\left(\bar{H}\right)$ $\displaystyle=$ $\displaystyle
d\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(-\alpha^{\prime}\right)^{k-1}2^{2k+1}\left(2k-1\right)\bar{c}_{k}\bar{H}^{2k}.$
(1.8)

In these EOM (1.7), if we start with the non-perturbative and non-singular
dilaton solution $\Phi$ obtained from the perturbative solution, we can obtain
$\bar{f}\left(\bar{H}\right)$ and $\bar{g}\left(\bar{H}\right)$ directly.
Since the non-perturbative expansions (1.8) introduces an extra constraint
$\dot{\bar{g}}=\bar{H}\dot{\bar{f}}$, we can also determine $\bar{H}$ in the
subsequent step. In other words, all solutions can be derived from $\Phi$.
However, when considering a non-trivial Kalb-Ramond field, we can only utilize
the complicated EOM (1.3) to (1.5). The solution $\Phi$ in these EOM cannot
determine the solution of the matrix $\mathcal{S}$ due to equation (1.4),
which includes an infinite summation of matrices, and we cannot extract any
constraints such as $\dot{\bar{g}}=\bar{H}\dot{\bar{f}}$ from equations (1.3)
to (1.5). The solution $\Phi$ in these EOM cannot determine the solution of
the matrix $\mathcal{S}$ due to the equation (1.4) includes infinite summation
of matrices, and we can not extract any constraint such as
$\dot{\bar{g}}=\bar{H}\dot{\bar{f}}$ from the (1.3) to (1.5). Therefore, it
seems impossible to obtain the non-perturbative solution including the Kalb-
Ramond field.

In this paper, our aim is to develop a method for calculating the non-
perturbative and non-singular solutions of the Hohm-Zwiebach action, including
a non-trivial Kalb-Ramond field. We begin with the three-dimensional black
string solution with axion charge [21, 22]. When $\left|Q\right|=M$ and we set
$\lambda\rightarrow\infty$ of cosmological constant $8/\lambda$, the metric
becomes extremal, which corresponds to the fields outside the fields outside a
fundamental macroscopic string [23]:

$ds^{2}=\frac{1}{y}\left(-dt^{2}+dx^{2}\right)+dy^{2},$ (1.9)

This extremal metric includes a non-trivial Kalb-Ramond field and exhibits a
naked singularity. Our method aims to remove this singularity, and it can be
summarized as follows: At first, we extract the ansatz based on the extremal
metric (1.9) and derive the corresponding Hohm-Zwiebach action. Secondly, we
calculate the perturbative solution up to the first order of $\alpha^{\prime}$
correction of the action. The key point in this step is that, when we
integrate out $d/dy$ from the matrix different equation (1.4), the constant
matrix solution may also receive $\alpha^{\prime}$ corrections. Furthermore,
we observe that in our perturbative solution, the dilaton does not receive
$\alpha^{\prime}$ corrections; only the metric and the Kalb-Ramond field are
affected. Thirdly, we use a trick to transform the infinite summation of the
matrix differential equation (1.4) into a simple trace equation:

$4e^{-\Phi}\left[\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}kc_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}\right]=\mathrm{Tr}\left(\mathbb{C}\mathcal{S}^{-1}\dot{\mathcal{S}}\right),$
(1.10)

where $\mathbb{C}$ is $4\times 4$ constant matrix. Reconsidering the EOM
(1.3), (1.4) and (1.10), we find that all fields depend on $\Phi$, allowing us
to determine the solution of $\mathcal{S}$ through equation (1.10). Finally,
we obtain the regular solution of these EOM, and it implies that the naked
curvature singularity of three-dimensional extremal black string solution can
be successfully removed. As expected, the regular solution matches the
perturbative solution in the perturbative limit $\alpha^{\prime}\rightarrow
0$. While our regular solution currently matches only the first two orders of
the perturbative solution, it can be easily generalized to arbitrary orders
using our previous method [10].

The remainder of this paper is outlined as follows: In section 2, we briefly
review the three-dimensional extremal black string at the tree-level. In
section 3, we calculate the non-perturbative and non-singular extremal black
hole solution, including Kalb-Ramond field. Section 4 is a conclusion.

##  2 Three-dimensional extremal black string at tree-level

In this section, we demonstrate how to obtain the three-dimensional extremal
black string solution from the tree-level low energy effective action of
bosonic closed string theory. We start with the three-dimensional low energy
effective action given by:

$I_{0}=\int
d^{3}x\sqrt{-g}e^{-2\phi}\left(R+4\left(\partial\phi\right)^{2}-\frac{1}{12}H^{2}\right),$
(2.11)

where $R$ is the Ricci scalar for the metric $g_{\mu\nu}$, $\phi$ represents
the dilaton, and
$H_{\mu\nu\rho}=3\partial_{\left[\mu\right.}b_{\left.\nu\rho\right]}$ is the
field strength of the anti-symmetric Kalb-Ramond field $b$. Next, we consider
the following ansatz:

$ds^{2}=a\left(y\right)^{2}\left(-dt^{2}+dx^{2}\right)+dy^{2}.$ (2.12)

This ansatz can also be written as:

$g_{\mu\nu}=\left(\begin{array}[]{ccc}-a^{2}\left(y\right)&0&0\\\
0&a^{2}\left(y\right)&0\\\ 0&0&1\end{array}\right),\qquad
b_{\mu\nu}=\left(\begin{array}[]{ccc}0&b_{tx}\left(y\right)&0\\\
b_{xt}\left(y\right)&0&0\\\
0&0&0\end{array}\right),\qquad\phi=\phi\left(y\right).$ (2.13)

It is important to note that $b_{tx}\left(y\right)=-b_{xt}\left(y\right)$.
Since the ansatz (2.12) is independent of $y$, we can introduce the following
notation to manifest the $O\left(2,2\right)$ symmetry:

$g_{\mu\nu}=\left(\begin{array}[]{cc}G_{ij}\left(y\right)&0\\\
0&1\end{array}\right),\qquad
b_{\mu\nu}=\left(\begin{array}[]{cc}B_{ij}\left(y\right)&0\\\
0&0\end{array}\right),$ (2.14)

where

$G_{ij}\left(y\right)\equiv\left(\begin{array}[]{cc}-a^{2}\left(y\right)&0\\\
0&a^{2}\left(y\right)\end{array}\right),\qquad
B_{ij}\left(y\right)\equiv\left(\begin{array}[]{cc}0&b_{tx}\left(y\right)\\\
b_{xt}\left(y\right)&0\end{array}\right).$ (2.15)

With this notation, the action (2.11) can be rewritten as:

$I_{0}=-\int
dx^{3}e^{-\Phi}\left(-\dot{\Phi}^{2}-\frac{1}{8}\mathrm{Tr}\left(\dot{\mathcal{S}}^{2}\right)\right),$
(2.16)

where the dot denotes a derivative with respect to $y$, i.e.,
$\dot{f}\left(y\right)\equiv\partial_{y}f\left(y\right)$, and

$\mathcal{S}=\left(\begin{array}[]{cc}BG^{-1}&G-BG^{-1}B\\\
G^{-1}&-G^{-1}B\end{array}\right).$ (2.17)

The $O\left(d,d\right)$ invariant dilaton $\Phi$ in action (2.16) is defined
in the action as:

$\Phi=2\phi-\frac{1}{2}\ln\left|\det G_{ij}\right|.$ (2.18)

The EOM for the action (2.16) are as follows:

$\displaystyle\ddot{\Phi}+\frac{1}{8}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2}$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\mathcal{S}\left(\dot{\mathcal{S}}\right)\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\dot{\Phi}^{2}+\frac{1}{8}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2}$
$\displaystyle=$ $\displaystyle 0.$ (2.19)

The simplest solution that includes a non-trivial Kalb-Ramond field is:

$\Phi=0,\qquad a\left(y\right)=\frac{1}{\sqrt{y}},\qquad
b_{tx}\left(y\right)=\frac{1}{y}.$ (2.20)

The corresponding Kretschmann scalar is given by:

$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=\frac{4\left(2a^{2}\ddot{a}^{2}+\dot{a}^{4}\right)}{a^{4}}=\frac{19}{4y^{4}}.$
(2.21)

It implies that the solution possesses a curvature singularity at $y=0$ and no
event horizon. This solution describes the fields outside a fundamental
macroscopic string, which corresponds to a three-dimensional extremal black
string. To understand the relationship between this tree-level singular
solution (2.20) and the extremal black string, let us recall the three-
dimensional black string with an axion charge:

$ds^{2}=-\left(1-\frac{M}{r}\right)dt^{2}+\left(1-\frac{Q^{2}}{Mr}\right)dx^{2}+\left(1-\frac{M}{r}\right)^{-1}\left(1-\frac{Q^{2}}{Mr}\right)^{-1}\frac{\lambda
dr^{2}}{8r^{2}},$ (2.22)

with

$H_{rtx}=\frac{Q}{r^{2}},\qquad\phi=\ln r+\frac{1}{2}\ln\frac{\lambda}{2},$
(2.23)

where $8/\lambda$ is related to the cosmological constant. The extremal metric
is obtained by setting $\left|Q\right|=M$, which gives:

$ds^{2}=\left(1-\frac{M}{r}\right)\left(-dt^{2}+dx^{2}\right)+\left(1-\frac{M}{r}\right)^{-2}\frac{\lambda
dr^{2}}{8r^{2}}.$ (2.24)

Using the coordinate transformations:

$\tilde{y}=-\left(\frac{\lambda}{8}\right)^{1/2}\frac{r}{M},\qquad\tilde{t}=\left(\frac{\lambda}{8}\right)^{1/4}x,\qquad\tilde{x}=\left(\frac{\lambda}{8}\right)^{1/4}t,$
(2.25)

and taking the limit $\lambda\rightarrow\infty$, which corresponds to the
vanishing cosmological constant in the action (2.11), the extremal metric
(2.24) becomes:

$ds^{2}=\frac{1}{\tilde{y}}\left(-d\tilde{t}^{2}+d\tilde{x}^{2}\right)+d\tilde{y}^{2},$
(2.26)

which matches the solution (2.20). In this ansatz, we can employ the isotropic
Hohm-Zwiebach action without considering the multitrace terms.

##  3 Regular extremal black string via $\alpha^{\prime}$ corrections

In this section, our aim is to remove the naked singularity of the solution
(2.20) by using the Hohm-Zwiebach action. Hohm and Zwiebach demonstrated that
the following low energy effective action with complete $\alpha^{\prime}$
corrections can be expressed as

$\displaystyle I_{HZ}$ $\displaystyle=$ $\displaystyle\int
d^{3}x\sqrt{-g}e^{-2\phi}\left(R+4\left(\partial\phi\right)^{2}-\frac{1}{12}H^{2}+\frac{1}{4}\alpha^{\prime}\left(R^{\mu\nu\rho\sigma}R_{\mu\nu\rho\sigma}+\ldots\right)+\alpha^{\prime
2}\left(\ldots\right)+\ldots\right)$ (3.27) $\displaystyle=$
$\displaystyle-\int
dx^{3}e^{-\Phi}\left(-\dot{\Phi}^{2}+\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}^{2k}\right)\right),$

where $\bar{c}_{1}=c_{1}=-\frac{1}{8}$, $\bar{c}_{2}=-c_{2}=-\frac{1}{64}$,
$\bar{c}_{3}=c_{3}=-\frac{1}{3.2^{7}}$,
$\bar{c}_{4}=-c_{4}=-\frac{1}{2^{15}}+\frac{1}{2^{12}}\zeta\left(3\right)$ and
$\bar{c}_{k>4}$’s are unknown coefficients for the bosonic case [20]. The EOM
of the Hohm-Zwiebach action are given by:

$\displaystyle\ddot{\Phi}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}k\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4k\bar{c}_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\dot{\Phi}^{2}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\left(2k-1\right)\bar{c}_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0.$ (3.28)

To calculate the perturbative solution, let us introduce a new variable
$\Omega\equiv e^{-\Phi}$, where $\dot{\Omega}=-\dot{\Phi}\Omega$ and
$\ddot{\Omega}=\left(-\ddot{\Phi}+\dot{\Phi}^{2}\right)\Omega$. The EOM can be
rewritten as:

$\displaystyle\frac{\left(\dot{\Omega}\right)^{2}}{\Omega^{2}}-\frac{\ddot{\Omega}}{\Omega}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}kc_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4kc_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\frac{\dot{\Omega}^{2}}{\Omega^{2}}-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\left(2k-1\right)c_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}$
$\displaystyle=$ $\displaystyle 0,$ (3.29)

Next, we make the assumption that the perturbative solutions of the EOM given
by (3.30) take the following forms:

$\displaystyle\Omega\left(y\right)$ $\displaystyle=$
$\displaystyle\Omega_{0}\left(y\right)+\alpha^{\prime}\Omega_{1}\left(y\right)+\alpha^{\prime
2}\Omega_{2}\left(y\right)+\ldots,$ $\displaystyle a\left(r\right)$
$\displaystyle=$ $\displaystyle
a_{0}\left(r\right)+\alpha^{\prime}a_{1}\left(r\right)+\alpha^{\prime
2}a_{2}\left(r\right)+\ldots.$ $\displaystyle b\left(r\right)$
$\displaystyle=$ $\displaystyle
b_{0}\left(r\right)+\alpha^{\prime}b_{1}\left(r\right)+\alpha^{\prime
2}b_{2}\left(r\right)+\ldots,$ (3.30)

where we define $b_{tx}\left(y\right)\equiv b\left(y\right)$. Substituting the
perturbative forms (3.30) into the EOM (3.30) and considering the zeroth order
in $\alpha^{\prime}$, we obtain the following equations:

$\displaystyle\ddot{\Omega}_{0}\left(y\right)$ $\displaystyle=$ $\displaystyle
0,$
$\displaystyle\dot{\Omega}_{0}^{2}+\frac{\Omega_{0}^{2}\left(\dot{b}_{0}^{2}-4a_{0}^{2}\dot{a}_{0}^{2}\right)}{2a_{0}^{4}}$
$\displaystyle=$ $\displaystyle 0,$ (3.31)

and the second equation of the EOM (3.30) leads to the following expression:

$\frac{d}{dy}\left(\begin{array}[]{cccc}\mathbb{M}_{11}^{0}&0&0&\mathbb{M}_{14}^{0}\\\
0&\mathbb{M}_{11}^{0}&-\mathbb{M}_{14}^{0}&0\\\
0&-\mathbb{M}_{41}^{0}&\mathbb{M}_{44}^{0}&0\\\
\mathbb{M}_{41}^{0}&0&0&\mathbb{M}_{44}^{0}\end{array}\right)=0,$ (3.32)

where the entries of the matrix are given by

$\displaystyle\mathbb{M}_{11}^{0}$ $\displaystyle=$
$\displaystyle\frac{\Omega_{0}\left(2a_{0}^{3}\dot{a}_{0}-b_{0}\dot{b}_{0}\right)}{2a_{0}^{4}},$
$\displaystyle\mathbb{M}_{41}^{0}$ $\displaystyle=$
$\displaystyle\frac{\Omega_{0}\dot{b}_{0}}{2a_{0}^{4}},$
$\displaystyle\mathbb{M}_{14}^{0}$ $\displaystyle=$
$\displaystyle\frac{\Omega_{0}\left(4a_{0}^{3}\dot{a}_{0}b-\left(a_{0}^{4}+b_{0}^{2}\right)\dot{b}_{0}\right)}{2a_{0}^{4}},$
$\displaystyle\mathbb{M}_{44}^{0}$ $\displaystyle=$
$\displaystyle\frac{\Omega_{0}\left(b_{0}\dot{b}_{0}-2a_{0}^{3}\dot{a}_{0}\right)}{2a_{0}^{4}}.$
(3.33)

The solution is

$\Omega_{0}=1,\qquad a_{0}=\frac{1}{\sqrt{y}},\qquad b_{0}=\frac{1}{y},$
(3.34)

which is consistent with the solution (2.20). The integration of $d/dy$ in the
second equation of EOM (3.29) yields the constant matrix

$\left(\begin{array}[]{cccc}\mathbb{M}_{11}^{0}&0&0&\mathbb{M}_{14}^{0}\\\
0&\mathbb{M}_{11}^{0}&-\mathbb{M}_{14}^{0}&0\\\
0&-\mathbb{M}_{41}^{0}&\mathbb{M}_{44}^{0}&0\\\
\mathbb{M}_{41}^{0}&0&0&\mathbb{M}_{44}^{0}\end{array}\right)=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&0&0&0\\\ 0&-1&0&0\\\ 1&0&0&0\end{array}\right).$ (3.35)

This constant matrix plays a crucial role in determining the non-perturbative
solution. Considering the EOM (3.30) at the first order of $\alpha^{\prime}$,
we have:

$\displaystyle\ddot{\Omega}_{1}+\frac{\Omega_{0}\left(\dot{b}_{0}^{2}-4a_{0}^{2}\dot{a}_{0}^{2}\right){}^{2}}{16a_{0}^{8}}$
$\displaystyle=$ $\displaystyle 0,$ $\displaystyle
2\dot{\Omega}_{0}\dot{\Omega}_{1}+\frac{\Omega_{0}\Omega_{1}\left(\dot{b}_{0}^{2}-4a_{0}^{2}\dot{a}_{0}^{2}\right)}{a_{0}^{4}}$
$\displaystyle+\Omega_{0}^{2}\left(\frac{3\left(\dot{b}_{0}^{2}-4a_{0}^{2}\dot{a}_{0}^{2}\right){}^{2}}{16a_{0}^{8}}+\frac{a_{0}\dot{b}_{0}\dot{b}_{1}-2a_{1}\dot{b}_{0}^{2}-4a_{0}^{3}\dot{a}_{0}\dot{a}_{1}+4a_{1}a_{0}^{2}\dot{a}_{0}^{2}}{a_{0}^{5}}\right)$
$\displaystyle=$ $\displaystyle 0,$

and

$\frac{d}{dy}\left(\begin{array}[]{cccc}\mathbb{M}_{11}^{1}&0&0&\mathbb{M}_{14}^{1}\\\
0&\mathbb{M}_{11}^{1}&-\mathbb{M}_{14}^{1}&0\\\
0&-\mathbb{M}_{41}^{1}&\mathbb{M}_{44}^{1}&0\\\
\mathbb{M}_{41}^{1}&0&0&\mathbb{M}_{44}^{1}\end{array}\right)=0,$ (3.36)

where

$\displaystyle\mathbb{M}_{11}^{1}$ $\displaystyle=$
$\displaystyle-\frac{1}{8a_{0}^{8}}\left[4a_{0}^{4}\left(\left(b_{1}\Omega_{0}+b_{0}\Omega_{1}\right)\dot{b}_{0}+\Omega_{0}b_{0}\dot{b}_{1}\right)-2a_{0}^{3}\Omega_{0}\dot{b}_{0}\left(\dot{a}_{0}\dot{b}_{0}+8a_{1}b_{0}\right)-4\Omega_{0}a_{0}^{2}\dot{a}_{0}^{2}b_{0}\dot{b}_{0}\right.$
$\displaystyle\left.-8a_{0}^{7}\left(\Omega_{1}\dot{a}_{0}+\Omega_{0}\dot{a}_{1}\right)+8\Omega_{0}a_{0}^{6}\dot{a}_{0}a_{1}+8\Omega_{0}a_{0}^{5}\dot{a}_{0}^{3}+\Omega_{0}b_{0}\dot{b}_{0}^{3}\right],$
$\displaystyle\mathbb{M}_{41}^{1}$ $\displaystyle=$
$\displaystyle\frac{1}{8a_{0}^{8}}\left[4\Omega_{0}a_{0}^{4}\dot{b}_{1}+4a_{0}^{2}\dot{b}_{0}\left(-\Omega_{0}\dot{a}_{0}^{2}+a_{0}^{2}\Omega_{1}-4\Omega_{0}a_{1}a_{0}\right)+\Omega_{0}(y)\dot{b}_{0}^{3}\right],$
$\displaystyle\mathbb{M}_{14}^{1}$ $\displaystyle=$
$\displaystyle-\frac{1}{8a_{0}^{8}}\left[4a_{0}^{8}\left(\Omega_{1}\dot{b}_{0}+\Omega_{0}\dot{b}_{1}\right)-16a_{0}^{7}\left(\dot{a}_{0}\left(\Omega_{0}b_{1}+\Omega_{1}b_{0}\right)+\Omega_{0}\dot{a}_{1}b_{0}\right)-4a_{0}^{6}\Omega_{0}\dot{a}_{0}\left(\dot{a}_{0}\dot{b}_{0}-4a_{1}b_{0}\right)\right.$
$\displaystyle+16\Omega_{0}a_{0}^{5}\dot{a}_{0}^{3}b_{0}+a_{0}^{4}\left(\Omega_{0}\dot{b}_{0}^{3}+4b_{0}\left(2\Omega_{0}b_{1}+\Omega_{1}b_{0}\right)\dot{b}_{0}+4\Omega_{0}b_{1}^{2}\dot{b}_{1}\right)$
$\displaystyle\left.-4\Omega_{0}a_{0}^{3}b_{0}\dot{b}_{0}\left(\dot{a}_{0}\dot{b}_{0}+4a_{1}b_{0}\right)-4\Omega_{0}a_{0}^{2}\dot{a}_{0}^{2}b_{0}^{2}\dot{b}_{0}+\Omega_{0}b_{0}^{2}\dot{b}_{0}^{3}\right],$
$\displaystyle\mathbb{M}_{44}^{1}$ $\displaystyle=$
$\displaystyle\frac{1}{8a_{0}^{8}}\left[4a_{0}^{4}\left(\left(\Omega_{0}b_{1}+\Omega_{1}b_{0}\right)\dot{b}_{0}+\Omega_{0}b_{0}\dot{b}_{1}\right)-2\Omega_{0}a_{0}^{3}\dot{b}_{0}\left(\dot{a}_{0}\dot{b}_{0}+8a_{1}b_{0}\right)-4\Omega_{0}a_{0}^{2}\dot{a}_{0}^{2}b_{0}\dot{b}_{0}\right.$
(3.37)
$\displaystyle\left.-8a_{0}^{7}\left(\Omega_{1}\dot{a}_{0}+\Omega_{0}\dot{a}_{1}\right)+8\Omega_{0}a_{0}^{6}\dot{a}_{0}a_{1}+8\Omega_{0}a_{0}^{5}\dot{a}_{0}^{3}+\Omega_{0}b_{0}\dot{b}_{0}^{3}\right].$

The solution is

$\Omega_{1}=0,\qquad a_{0}=-\frac{C}{y^{3/2}},\qquad b_{0}=-\frac{2C}{y^{2}}.$
(3.38)

where $C$ is a non-zero constant and

$\left(\begin{array}[]{cccc}\mathbb{M}_{11}^{1}&0&0&\mathbb{M}_{14}^{1}\\\
0&\mathbb{M}_{11}^{1}&-\mathbb{M}_{14}^{1}&0\\\
0&-\mathbb{M}_{41}^{1}&\mathbb{M}_{44}^{1}&0\\\
\mathbb{M}_{41}^{1}&0&0&\mathbb{M}_{44}^{1}\end{array}\right)=\left(\begin{array}[]{cccc}0&0&0&0\\\
0&0&0&0\\\ 0&0&0&0\\\ 0&0&0&0\end{array}\right).$ (3.39)

Therefore, the perturbative solution, including the first two orders of
$\alpha^{\prime}$, can be expressed as:

$\displaystyle a\left(y\right)$ $\displaystyle=$
$\displaystyle\frac{1}{\sqrt{y}}-\frac{C}{y^{3/2}}\alpha^{\prime}+\cdots,$
$\displaystyle b\left(y\right)$ $\displaystyle=$
$\displaystyle\frac{1}{y}-\frac{2C}{y^{2}}\alpha^{\prime}+\cdots,$
$\displaystyle\Omega\left(y\right)$ $\displaystyle=$ $\displaystyle
1+0\cdot\alpha^{\prime}+\cdots.$ (3.40)

Next, let us proceed with the calculation of the non-singular and non-
perturbative solution. To do it, we can rewrite the EOM (3.29) as follows:

$\displaystyle\ddot{\Phi}-y\left(\mathcal{S}\right)$ $\displaystyle=$
$\displaystyle 0,$
$\displaystyle\frac{d}{dy}\left(e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4kc_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}\right)$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\dot{\Phi}^{2}+g\left(\mathcal{S}\right)$ $\displaystyle=$
$\displaystyle 0,$ (3.41)

where

$\displaystyle y\left(\mathcal{S}\right)$ $\displaystyle=$
$\displaystyle\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}kc_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k},$
$\displaystyle g\left(\mathcal{S}\right)$ $\displaystyle=$
$\displaystyle-\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}\left(2k-1\right)c_{k}\mathrm{Tr}\left(\dot{\mathcal{S}}\right)^{2k}.$
(3.42)

Based on the perturbative solution (3.40), we can determine the non-
perturbative solution for the $O\left(d,d\right)$ dilaton:

$\Phi=0,$ (3.43)

which covers the perturbative solution (3.40) as $\alpha^{\prime}\rightarrow
0$. From the first and second equations of (3.41), we obtain:

$y\left(\mathcal{S}\right)=g\left(\mathcal{S}\right)=0.$ (3.44)

Hence, the crucial equation is the second one in (3.41). However, this
equation represents a differential equation involving an infinite summation of
matrices. Finding the solution for $\mathcal{S}$ seems impossible in this
case. To make progress, we use a trick here. Let us begin with the second
equation in (3.41)

$\frac{d}{dy}\left(e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4kc_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}\right)=0.$
(3.45)

After integrating out $d/dy$, we obtain:

$e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4kc_{k}\mathcal{S}\left(\dot{\mathcal{S}}\right)^{2k-1}=\mathbb{C},$
(3.46)

where $\mathbb{C}$ is a $4\times 4$ constant matrix. Multiplying both sides by
$\mathcal{S}^{-1}$ and then $\dot{\mathcal{S}}$, we have:

$e^{-\Phi}\stackrel{{\scriptstyle[}}{{k}}=1]{\infty}{\sum}\left(\alpha^{\prime}\right)^{k-1}4kc_{k}\left(\dot{\mathcal{S}}\right)^{2k}=\mathbb{C}\mathcal{S}^{-1}\dot{\mathcal{S}}.$
(3.47)

Finally, by taking the trace on both sides of the equation, we obtain:

$4e^{-\Phi}y\left(\mathcal{S}\right)=\mathrm{Tr}\left(\mathbb{C}\mathcal{S}^{-1}\dot{\mathcal{S}}\right).$
(3.48)

Therefore, the infinite summation of matrices is reduced to a finite-term
differential equation. To solve this equation, we follow two steps. The first
step involves guessing the regular solution for the $b$ field, which is

$b\left(y\right)=\frac{1}{y+2\alpha^{\prime}C}.$ (3.49)

It covers the perturbative solution (3.40) as $\alpha^{\prime}\rightarrow 0$.
The next step is to determine the constant matrix $\mathbb{C}$. Based on the
previous results (3.35) and (3.39), we can assign the constant matrix
$\mathbb{C}$ as

$\mathbb{C}=\left(\begin{array}[]{cccc}0&0&0&0\\\ 0&0&0&0\\\ 0&-1&0&0\\\
1&0&0&0\end{array}\right).$ (3.50)

Consequently, the equation (3.48) simplifies to

$2+2a^{3}\left(y+2\alpha^{\prime}C\right)^{2}\left(a+4\dot{a}\left(y+2\alpha^{\prime}C\right)\right)=0.$
(3.51)

The solution of this equation is

$a\left(y\right)=\frac{1}{\sqrt{y+2\alpha^{\prime}C}}.$ (3.52)

In summary, we have successfully derived non-perturbative solutions that
satisfy the EOM (3.28):

$ds^{2}=a\left(y\right)^{2}\left(-dt^{2}+dx^{2}\right)+dy^{2},$ (3.53)

with

$\Phi=0,\qquad a\left(y\right)=\frac{1}{\sqrt{y+2\alpha^{\prime}C}},\qquad
b\left(y\right)=\frac{1}{y+2\alpha^{\prime}C},$ (3.54)

It is straightforward to verify that this solution matches with the
perturbative solution as $\alpha^{\prime}\rightarrow 0$. Although our regular
solution matches only the first two orders of the perturbative solution, it
can be easily generalized to arbitrary orders using our previous method [10].
Using this solution, we can calculate the Kretschmann scalar:

$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}=\frac{4\left(2a^{2}\ddot{a}^{2}+\dot{a}^{4}\right)}{a^{4}}=\frac{19}{4\left(y+2\alpha^{\prime}C\right){}^{4}}.$
(3.55)

which is regular for arbitrary $y\geq 0$. To illustrate how $\alpha^{\prime}$
corrections affects the naked singularity, we provide a representation in
Figure (1). It is evident that once the $\alpha^{\prime}$ corrections are
introduced, the singularities in the metric and Kalb-Ramond field disappear.

$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\left(y,\alpha^{\prime}\right)$
$b\left(y,\alpha^{\prime}\right)$

Figure 1:  The figures display
$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\left(y,\alpha^{\prime}\right)$ and
$b\left(y,\alpha^{\prime}\right)$, with the assumption $C=1$. The red lines
represent the singular solutions
$R_{\mu\nu\rho\sigma}R^{\mu\nu\rho\sigma}\left(y,0\right)$ and
$b\left(y,0\right)$ obtained from (2.20). The singularities are located at
$y=0$ and $\alpha^{\prime}=0$.

##  4 Conclusion

In this paper, we have successfully calculated the non-perturbative and non-
singular solution for the Hohm-Zwiebach action, considered the presence of a
non-trivial Kalb-Ramond field. In order to calculate the matrix differential
equations, we transformed the infinite summation of matrices into a simple
trace of a matrix, enabled us to solve the matrix differential equations. As a
result, we obtained regular solutions for all the fields, including the
spacetime metric. This implies that the naked singularity of the extremal
black string can be eliminated by the $\alpha^{\prime}$ corrections. In our
future work, we plan to study the following aspects:

  * •

Extension to anisotropic backgrounds: In this study, we focused on the
isotropic background. However, it would be interesting to investigate the
anisotropic Hohm-Zwiebach action. It implies that we need to consider the
multitrace terms in the Hohm-Zwiebach action and refine our method
accordingly.

  * •

Application to BTZ black hole: The BTZ black hole solution in string theory
also requires a non-trivial Kalb-Ramond field. Therefore, if we aim to address
the curvature singularity in the BTZ black hole, it is crucial to incorporate
the Kalb-Ramond field in the Hohm-Zwiebach action.

By addressing these directions, we can further broaden our understanding of
non-perturbative solutions and their implications for resolving singularities
in black hole spacetimes.

Acknowledgements We are very grateful to Xin Li, Peng Wang, Houwen Wu and
Haitang Yang for many illuminating discussions and suggestions. This work is
supported in part by NSFC (Grant No. 12105031), and the Postdoctoral Science
Foundation of Chongqing (Grant No. cstc2021jcyj-bshX0227).

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# Application of a Shallow Neural Network to Short-Term Stock Trading

Abhinav Madahar  Rutgers University<EMAIL_ADDRESS>Yuze Ma
Rensselaer Polytechnic Institute<EMAIL_ADDRESS>Kunal Patel  Cherokee High
School<EMAIL_ADDRESS>
(2017 March 6)

###### Abstract

Machine learning is increasingly prevalent in stock market trading. Though
neural networks have seen success in computer vision and natural language
processing, they have not been as useful in stock market trading. To
demonstrate the applicability of a neural network in stock trading, we made a
single-layer neural network that recommends buying or selling shares of a
stock by comparing the highest high of 10 consecutive days with that of the
next 10 days, a process repeated for the stock’s year-long historical data. A
$\chi^{2}$ analysis found that the neural network can accurately and
appropriately decide whether to buy or sell shares for a given stock, showing
that a neural network can make simple decisions about the stock market.

##  1 Intro

Large companies (e.g. Google, Exxon Mobil, etc.) are not owned by a single
person or a private group of individuals. Rather, these companies are split up
into small pieces (shares) which are then sold to any individual who can
afford them. The value of a single share depends on many factors, primarily
its current and expected future profits. For example, a share in Apple (the
largest company by total share value) is sold at approximately $140 [1]. As a
company grows, its shares increase in value; Amazon shares sold for $18 each
in 1997 [2], but are now worth over $800 because of Amazon’s massive growth
since [3]. A company’s shares are refered to as its ”stock”. Stock investors
try to buy shares in companies that will grow dramatically; an investor who
bought shares of Amazon in 1997 would now be very rich.

Many artificial intelligence (AI) techniques are applied to stock trading,
such as genetic algorithms and probabilistic logic network [4]. Neural
networks, an AI technique that simulates the human brain, have been applied to
stock trading, but have seen limited success because they work best for
systems with clear patterns, which stock markets do not have [5].

In a neural network, neurons are grouped into layers, which are combined to
form a network that simulates a brain. A single neuron takes in many inputs
from the neurons of the previous layer, multiples each by a weight, sums the
products, adds a constant bias, and finally runs that sum through an
activation function, often the sigmoid or hyperbolic tangent function. If we
represent the inputs as a vector $x$, the weights as a vector $w$, the bias as
a scalar $b$, the activation function as $f$, and the final output as a scalar
$a$, then

$a=f(x\cdot w+b)$

A neuron accepts a vector of the previous layer’s values and outputs a scalar
value, which is then sent to the next layer. By tuning the weights and biases,
the neural network can learn how to interpret given data, a process called
”training.” A deep neural network has many layers, allowing it to find deeper
patterns, leading to a dramatic rise in usage in the past few years,
especially for tasks like image recognition and natural language processing
[4]. A shallow neural network is a neural network with only 1 or a handful of
layers, making it simpler but less powerful.

This study explores the utility of a shallow neural network on stock trading,
specifically on deciding whether to buy or sell shares of a given company when
given only stock information on said company.

##  2 Methods & Procedure

###  2.1 Experimental Trial

The neural network was created in Python 2.7 using Numpy, but a similar design
in another language or with other tools is expected to yield similar results.

From Yahoo Finance, the daily highs over the past year from for stocks in the
Standards & Poors 500 (S&P 500) were downloaded. These highs were then grouped
into $n$ chunks, each 10 days long.

The neural network takes a chunk as a 10-vector input. It has only 1 hidden
layer, which has 20 neurons. It outputs a 2-vector of the form $(a,b)$. $a$
and $b$ are both recommendations in the interval $[0,1]$ with 1 being a strong
recommendation. $a$ recommends selling shares and $b$ recommends buying
shares. The recommendation with the larger magnitude is used. For simplicity,
the network’s output $(a,b)$ is used to generate a Boolean $c$ equal to the
index of the larger component. If the two recommendations are equally strong,
then buying more shares is suggested.

$a>b\implies c=0$ $a\ngtr b\implies c=1$ $\therefore$ $c=a\ngtr b$

It is trivially shown that $c$ represents the recommendation to buy shares. If
the network recommends buying, $c=1$. If the network recommends selling (i.e.
not buying), then $c=0$.

To generate the training data, all but the most recent chunk were considered.
To generate the $k^{th}$ training datum $(x_{k},y_{k})$, let $x_{k}$ be the
$k^{th}$ 10-vector chunk and $y_{k}$ be the correct value of $c$ for said
chunk. To calculate $y_{k}$, compare the highest high $H_{k}$ of the $k^{th}$
chunk with $H_{k+1}$, that of the $(k+1)^{th}$ chunk. If the price rises (i.e.
$H_{k+1}>H_{k}$), then shares shought be bought (i.e. $y_{k}=1$). If the price
does not rise, then shares should be sold (i.e. $y_{k}=0$). To summarize,

$H_{k+1}>H_{k}\implies y_{k}=1$ $H^{\prime}\ngtr H_{k}\implies y_{k}=0$
$\therefore$ $y_{k}=H_{k+1}>H_{k}$

It is trivially shown that

$y_{k}=max(x_{k+1})>max(x_{k})$

The training data

$\\{(x_{1},y_{1}),(x_{2},y_{2}),(x_{3},y_{3}),...,(x_{n-1},y_{n-1})\\}$

ignores the most recent chunk $(x_{n},y_{n})$ so that it can be used to test
the neural network. red The neural network then trains on the training data
using standard backpropagation and gradient descent. To test it, the network
makes its decision $c_{n-1}$ on the second-most recent chunk $x_{n-1}$. The
decision’s accuracy (i.e. $c_{n-1}=y_{n-1}$) was then recorded.

Of the 500 stocks in the S&P 500 available, 385 stocks were given to the
neural network, which trained and tested itself over each.

###  2.2 Control Trial

To know if the neural network is useful, a control that randomly recommends
buying or selling was applied to the S&P 500 data. It generated a $c_{n-1}$
that was equally likely to be $0$ or $1$. $y_{n-1}=max(x_{n})>max(n_{n-1})$ as
in the experimental group. This decision’s accuracy (i.e. $c_{n-1}=y_{n-1}$)
was then recorded.

##  3 Results

The null hypothesis argues that the neural network knows nothing of the stocks
on which it makes decisions and thus is no better than a random suggestion of
either buying or selling.

Of the 500 stocks given to the control, 278 were decided correctly and 222
were decided incorrectly. Assume that the control’s results are not
statistically significant. If so, then a chi-squared analysis should show that
the control is not statistically significant. Applying the standard formula,

$\chi^{2}\equiv\sum_{i}\frac{(O_{i}-E_{i})^{2}}{E_{i}}$
$\chi^{2}=\frac{(278-250)^{2}}{250}+\frac{(222-250)^{2}}{250}$
$\chi^{2}=6.272$

Because there are two possible classifications of a result, correct or
incorrect, there are 2 degrees of freedom. Using a significance level
$\alpha=0.01$ with a single degree of freedom would require $\chi^{2}>6.63$,
which is not the case for the control group.

In the experimental trial, the neural network was given 385 stocks. If the
null hypothesis is true, then 192.5 decisions would be correct and 192.5 would
be incorrect. The neural network made an incorrect decision for 57 stocks and
a correct decision for the remaining 328. Applying chi-squared,

$\chi^{2}\equiv\sum_{i}\frac{(O_{i}-E_{i})^{2}}{E_{i}}$
$\chi^{2}=\frac{(328-192.5)^{2}}{192.5}+\frac{(57-192.5)^{2}}{192.5}$
$\chi^{2}=190.755844$

190.755844 is significantly higher than the required 6.63, so the null
hypothesis is rejected, showing that the neural network has learned from the
stock data and can make an informed decision based on it.

##  4 Conclusions & Discussion

Neural networks have seen little success in stock-trading due to the lack of
an obvious pattern in the stock market. This particular experiment was
designed with simplicity in mind. The neural network has been shown to
accurately predict whether the highest value of a stock for 10 consecutive
days is higher than the highest value of said stock for the next 10 days.
However, this is not very applicable to real-world stock trading because the
network does not determine what the next high is, so the stock trader does not
know when exactly to buy or sell shares in the stock, only that he or she
should at some point.

Changes to the configurations of the neural network, such as its dimensions
and learning rate, may improve the neural network’s accuracy. Improvements may
also be found by an alternative design, such as a convolution neural network,
or a significant alteration to the existing network, such as a deep neural
network.

##  5 Acknowledgements

The authors thank hackNYU for providing food, drink, lodging, internet access,
and countless other utilities during the development of the neural network. We
also thank Yahoo for freely distributing stock data. We also thank Rufus
Pollock for distributing a list of the tickers of the S&P 500 companies.

## References

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  * Yahoo [2017b] Yahoo.  Amazon.com Inc. (AMZN).  February 2017b.
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