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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def errors(self):
 """ Computes and returns the error and standard deviation of the filter at this time step. Returns ------- error : np.array size 1xorder+1 std : np.array size 1xorder+1 """


        n = self.n
        dt = self.dt
        order = self._order
        sigma = self.sigma

        error = np.zeros(order + 1)
        std = np.zeros(order + 1)

        if n == 0:
            return (error, std)

        if order == 0:
            error[0] = sigma/sqrt(n)
            std[0] = sigma/sqrt(n)

        elif order == 1:
            if n > 1:
                error[0] = sigma * sqrt(2*(2*n-1) / (n*(n+1)))
                error[1] = sigma * sqrt(12. / (n*(n*n-1)*dt*dt))
            std[0] = sigma * sqrt((2*(2*n-1)) / (n*(n+1)))
            std[1] = (sigma/dt) * sqrt(12. / (n*(n*n-1)))

        elif order == 2:
            dt2 = dt * dt

            if n >= 3:
                error[0] = sigma * sqrt(3*(3*n*n-3*n+2) / (n*(n+1)*(n+2)))
                error[1] = sigma * sqrt(12*(16*n*n-30*n+11) /
                                        (n*(n*n-1)*(n*n-4)*dt2))
                error[2] = sigma * sqrt(720/(n*(n*n-1)*(n*n-4)*dt2*dt2))

            std[0] = sigma * sqrt((3*(3*n*n - 3*n + 2)) / (n*(n+1)*(n+2)))
            std[1] = (sigma/dt) * sqrt((12*(16*n*n - 30*n + 11)) /
                                       (n*(n*n - 1)*(n*n - 4)))
            std[2] = (sigma/dt2) * sqrt(720 / (n*(n*n-1)*(n*n-4)))

        return error, std