text:
# 10 Non-linear least squares. Least absolute deviation. Principal component analysis#

# import all python add-ons etc that will be needed later on
%matplotlib inline
import numpy as np
from numpy import linalg as lina
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from sympy import *
init_printing()                      # allows printing of SymPy results in typeset maths format
plt.rcParams.update({'font.size': 14})  # set font size for plots

## 10.1 Non-linear least squares#

In the least squares method, the derivative of the $$\chi^2$$ is taken with respect to each parameter $$\alpha_i$$ and minimized. The normal equations of linear least squares analysis (equations 28,44), have terms only in $$a_ix$$ or $$a_ix^2$$ etc., and can be solved exactly. In non-linear least squares, the normal equations cannot be solved exactly. The function

$Y = \alpha_1e^{-\alpha_2x} + \alpha_3e^{-\alpha_4x} + \alpha_5$

has a form that is common to several schemes in chemical kinetics and when analysing the decay of excited states, but in this function the normal equations still contain exponentials in $$\alpha_i$$ and $$x$$, for example $$e^{-\alpha_2x}$$ and cannot be solved exactly. An approximate numerical solution has to be sought, usually by an iterative method. Any iterative method follows an algorithm that tries to approach the minimum $$\chi^2$$ without searching through the whole range of possible values of the parameters ($$\alpha$$’s). There are several algorithms to choose from, some, such as the simplex, simulated annealing (Prest et al. 1986), and evolutionary (natural selection) algorithms are not least squares methods; however, the best algorithm for non-linear least squares is the Levenberg - Marquardt algorithm (Bevington 1969; Prest et al. 1986; Bevington & Robinson 2003). Importantly, it is easy to implement and operates stably, i.e. it hardly ever fails.

Using a numerical method implies

(i)$$\quad$$ that starting values have to be defined for each of the parameters sought,

(ii)$$\quad$$ increments in these have to be determined and are changed as the calculation proceeds,

(iii)$$\quad$$ some ending condition has been specified for an acceptable fit to the data.

These requirements mean that this problem is far harder to solve than is linear least-squares, partly for the reasons given, but also because there is no guarantee that the true minimum will be found even if the calculation appears to end satisfactorily. This last effect has two forms. The first is that close to the minimum solution several related sets of parameters can satisfy the same goodness of fit criterion, so that a unique set is not found, only a range of them. The second effect is that the ‘surface’ through and about which the minimum $$\chi^2$$ is sought may be rough, with local minima in which the calculation can become trapped.

The $$\chi^2$$ is defined as

$\displaystyle \chi^2=\sum_{i=1}^n w_i(y_i-Y_i(\alpha_j))^2$

where the subscript $$i$$ identifies the data points and $$j$$ the parameters, $$\alpha_1, \alpha_2\cdots$$. The $$\chi^2$$ surface would ideally be shaped rather like a bowl so that it can be imagined how starting at one point the minimum can be reached. But it is instead highly complex, and can be represented as a three-dimensional object only if two parameters are used and if there are more, and there usually are, it is almost impossible to have an idea what this object may look like. In the best circumstance, the object is smooth and a minimum is found, but generally, it might be supposed that the objects ‘landscape’ is ‘corrugated’ with many local minima separated by ‘hills’ and ‘ridges’. However, not withstanding the complexity of the $$\chi^2$$ ‘surface’, most equations describing physical and chemical phenomena can be success- fully minimized.

Quite often, finding the true minimum starts with initial guesses for the parameters close to the final ones that the calculation will produce. Often, approximate parameters can be guessed by looking at the data; they do not usually have to be that good for the calculation to converge perhaps within a factor of 5 or 10. Alternatively, the literature can be consulted to find values for similar experiments. If this is not possible, as there may be too many variables to find them all, then ranges must be placed on these and perhaps hundreds of sets of parameters can be guessed at random and those with the smallest $$\chi^2$$ chosen as starting points. Once the calculation is run, if the results are not particularly good it may be because (a) the function is sensitive to the starting values; (b) it ended up in a local not the global minimum; (c) the calculation did not run long enough to converge; or, if it did converge, (d) the model did not describe the data. Starting the calculation with different initial conditions will usually help to sort this out.

The Levenberg - Marquardt Method combines two different approaches to finding the minimum $$\chi^2$$; Bevington (1969) and Prest et al. (1986) give detailed derivations and equations. The first approach is a gradient-search method that tries to find the steepest descent from any point towards the minimum of the surface that represents the $$\chi^2$$. This method is good when the calculation is a long way from the minimum, as it allows it to be approached rapidly because the gradient is large. When close to the minimum, this method is poor because the gradient may be almost zero and the calculation will creep along, hardly making any progress.

The gradient in the $$\chi^2$$ is calculated as $$\displaystyle \sum_{j=1}^n\frac{\partial \chi^2}{\partial \alpha_j}$$. The second aspect to the method is to change the equations close to the minimum where the model function $$Y$$ can be accurately expanded as a Taylor series in the parameters $$\alpha_i$$ and the $$\chi^2$$ is minimized, and normal equations calculated. It turns out, fortunately, that the two sets of equations, steepest descent and expansion, can be changed into one another by varying only one parameter conventionally called $$\lambda$$.

It has been shown that the normal equations 44 can be written in matrix form (equation 48)

$\displaystyle \boldsymbol C=\boldsymbol{BA}^{-1}$

In the linear case, these equations can be solved by inverting matrix $$A$$ and left multiplying by matrix $$B$$ as is done in Algorithm 3. The equations in the non-linear case are formally the same as those of equation 48 but now the solution must be obtained iteratively. The iterative part has nothing to do with the Marquardt method; this method changes only the diagonals of matrix $$A$$ and so moves from a gradient search to a linear expansion of the function. Matrix $$A$$ is changed so that

$\displaystyle A_{jj}\to A_{jj}(1+\lambda)$

and $$\lambda$$ is changed in the algorithm to go between gradient expansion where $$\lambda$$ large, to function linearization where $$\lambda$$ is small.

The algorithm (Bevington (1969, Curfit p. 237) and Bevington & Robinson (2003)) is

(i)$$\quad$$ calculate $$\chi^2(\alpha)$$

(ii)$$\quad$$ set $$\lambda = 0.001$$

(iii)$$\quad$$ calculate $$\chi^2(\alpha +\delta\alpha)$$ where $$\delta\alpha$$ are the increments in parameters $$\alpha$$.

(iv)$$\quad$$ if $$\chi^2(\alpha+\delta\alpha)>\chi^2(\alpha)$$ then $$\lambda\to 10\lambda$$ and go to (iii)

(v)$$\quad$$ if $$\chi^2(\alpha+\delta\alpha)<\chi^2(\alpha)$$ then $$\lambda \to \lambda/10$$, make $$\alpha \to \alpha+\delta\alpha$$ and go to (iii).

(vi)$$\quad$$ check at (iii) whether a preset minimum difference in consecutive $$\chi^2$$ is produced or maximum number of iterations reached.

The algorithm is shown below fitted to Poisson distributed data. The data represents a single exponential decay with a constant background and the model function used is $$\displaystyle Y = c_1e^{-c_2x} + c_3$$. The data was simulated, as described in chapter 12.3.4, and the data is

$\begin{split}\displaystyle \begin{array}\\ \hline x &1 & 5 &9 &13 &17 & 21 &25 &29 & 33 &37 &41 &45 &49\\ y &1927 & 1329 &812 &568 &390 &290 &171 &112& 87 &48 &43 &26 &26\\ \hline\\ \hline x &53 & 57 &61 &65 &69 &73 &77 &81 & 85 &89 &93 &97\\ y &14 & 13 &10 &6 &9 &8 &7 &3 & 9 &5 &4 &8\\ \hline \end{array}\end{split}$

The true values used to produce the data are $$c_2 = 0.1, c_3 = 5$$, but the initial value of $$c_1$$ is unknown as the data has been simulated with $$20000$$ events and Poisson distributed noise has been added with a mean value of $$5$$. The total number of counts in the data is $$5925$$, which is rather a small number for an experiment, although single molecule fluorescence measured through a confocal microscope may have far fewer than this. The implementation is based on Bevington (1969, algorithm Curfit).

# Algorithm. Levenburg - Marquardt non-linear least squares.

def chisqrd(n):
s = 0.0
for i in range(n):
s = s + w[i]*(yval[i] - model(xval[i],C) )**2
return s
#------------------------------

def getdata(filename):
xv = []
yv = []
wv = []
with open(filename) as f:               # length not known so read in all data and make list of each
i=0
for line in f:
new_str = ' '.join(line.split())
vals = new_str.split(' ')
xv.append(vals[0])
yv.append(vals[1])
wv.append(vals[2])
f.close()
n = len(xv)                             # we do not know length of data before hand
w = np.zeros(n,dtype=float)
xval = np.zeros(n,dtype=float)
yval = np.zeros(n,dtype=float)
for i in range(n):
w[i] = 1/float(wv[i])               # counting weighting
xval[i] = float(xv[i])
yval[i] = float(yv[i])
return xval,yval,w
#--------------------------------

# fitting model   y = c1.exp(-c2.x) + c3

filename='exponential-data.txt'  # in rows x , y weighting,
# data is at end of book in 'Appendix, some basic Python instructions'

xval,yval,w = getdata(filename)

n = len(xval)
lambd= 0.001            # initial value
reps = 40               # typoical value can be altered
C = [2*yval[0],0.55,5]  # initial guessed values
m = len(C)
model = lambda x,C: C[0]*np.exp( -C[1]*x ) + C[2]  # function to be fitted

nf   = n - m - 1                                   # degrees of freedom
chiB = 1e20                                        # choose initial value to be huge

B = np.zeros(m,dtype=float)
A = np.zeros((m,m),dtype=float)
deriv = np.zeros((3,n),dtype=float)

for L in range(reps):
beta= np.zeros(m,dtype=float)
AA  = np.zeros((m,m),dtype=float)
for i in range(n):
deriv[0,i] = np.exp(-C[1]*xval[i])               # dy/dC[0]
deriv[1,i] = -C[0]*xval[i]*np.exp(-C[1]*xval[i]) # dy/dC[1]
deriv[2,i] = 1.0                                 # dy/dC[2]
pass
chiA = chisqrd(n)/nf

for j in range(m):
s = 0.0
for i in range(n):
s = s + w[i]*(yval[i] - model(xval[i],C) )*deriv[j,i]
beta[j] = s
for k in range(1+j):
s = 0.0
for i in range(n):
s = s + w[i]*deriv[j,i]*deriv[k,i]
AA[j,k] = s
pass

if np.abs(chiA - chiB) <= 0.001:
break
AA = ( np.transpose(AA) + AA )/2.0                   # symmetrise
while chiB > chiA:                                   # step iii
for j in range(m):
for k in range(m):
A[j,k] = AA[j,k]/np.sqrt( AA[j,j]*AA[k,k] )
A[j,j] = 1.0 + lambd
pass
A = np.linalg.inv(A)
for j in range(m):
s = 0.0
for ii in range(m):
s = s + beta[ii]*A[j,ii]/np.sqrt( AA[j,j]*AA[ii,ii] )
B[j] = C[j]+ s
pass
chiB = chisqrd(n)/nf
lambd = lambd*10          # step iv
pass                      # end while step iii  ChiB > chiA
C = B                         # replace coeffs
chiB  = chiA                  # replace chi^2
lambd = lambd/10              # step iv
pass                          # end loop on L

sig = np.zeros( m, dtype = float)
for j in range(m):
sig[j] = np.sqrt( A[j,j]/( AA[j,j] ) )

for i in range(m):
print('{:s}{:d}{:s}{:8.4g}{:s}{:6.2g}'.format('C',i+1,' = ', C[i],' +/- ',sig[i]) )
print('{:s}{:6.3f}'.format('reduced chi sqrd = ',chiA))
print('{:s}{:8.4g}'.format('total calculated counts   =',np.sum(model(xval[:],C[:]))) )
print('{:s}{:8.4g}'.format('total experimental counts =',np.sum(yval[:] ) ) )

C1 =     2128 +/-     30
C2 =   0.1011 +/- 0.0011
C3 =    5.045 +/-   0.72
reduced chi sqrd =  0.883
total calculated counts   =    5910
total experimental counts =    5925

fchi = lambda x: x**(n/2-1) * exp(-x/2)/(2**(n/2)*gamma(n/2)) # chi sqrd function
Q,err = quad(fchi , chiA*n, np.inf )    # integrate from chisqrd to infinity
print('{:s} {:6.3f} {:s} {:6.3f}'.format('probability of getting chi-sqrd > ',chiA*n,' is', Q))

probability of getting chi-sqrd >  22.066  is  0.632

Figure 14. Non-linear least squares fit to $$\displaystyle Y = c_1e^{-c_2x} + c_3$$ with $$c_1= 2128, c_2 = 0.1$$, and $$c_3 = 5.05$$.

The calculation produces a $$\chi^2= 0.9$$ and that corresponds to a probability of $$63$$% which is near perfect; a perfect value of $$1$$ produces a $$50$$% chance. The model curve clearly fits the data as shown in Figure 14. The total counts predicted by the model are almost the same as in the data itself, also indicating a close fit. The normalized residuals are calculated as $$(y_i - Y_i)/y_i$$.

## 10.2 Least absolute deviation#

Any least squares method is very sensitive to outliers in the data; see figure 15. You can ‘eyeball’ data and see roughly where a straight line should go through the majority of the data points; however, the least squares line will generally not follow this trend because it is pulled off by the outliers. In such cases, the function to minimize is the absolute value of the deviation rather than the $$\chi^2$$ and for an unweighted straight-line fit is the sum

$\displaystyle \sum_i |y_i-a-bx_i|$

Using this function now produces a far better fit to the data, but there is no $$\chi^2$$ with which to estimate how good this fit is. Taking the absolute value of the distance from the best fit line, rather than the square of the distance, means that the effect of large deviations is reduced. Prest et al. (1986) give an algorithm with which to calculate the minimum deviation line and this is shown in figure 15.

Figure 15. Outliers produce a least squares fit that does not reflect the trend exhibited by most of the data points. The line passing mainly through the data is the least absolute deviation LAD line.

## 11 Principal component analysis ( PCA )#

In the life sciences, the experimental techniques of chemical physics, for instance, NMR, FTIR, mass spectrometry, and various chromatographic methods are widely used and they produce large quantities of data. The data may consist of hundreds of data sets each containing data from thousands of proteins and metabolites and from which the presence of a few target molecules may be sought. For example, NMR spectra taken on blood plasma will contain signals from thousands of compounds. If samples are taken from a group of patients with a particular disease and a similar group without the disease, the two sets of NMRs can be compared in an attempt to identify markers that may be used to target the disease, either in an attempt either to cure it, or act as an assay to identify its early stages. However, these spectra will be different due to normal variability, will be highly congested and contain numerous overlapping signals from proteins, lipids, sugars, and DNA not involved in the disease. The task is to find the species that are either present or are missing in the diseased group. The problem is that there is an abundance of data but no results.

Data from any instrument is not always in the form that most easily allows the results required to be extracted from it. PCA attempts to rotate the axes with which the data is presented in such a way as to isolate each of its features independently of all of the others. This method produces new variables that are linear combinations of the original data and these are the principal components; mathematically a new orthogonal basis set is produced with which to represent the data (Jolliffe 2002; Miller & Miller 2005). Recall that in changing a basis, the data (a vector) is unaffected only the axes are changed. Some of these new components may contain only background or noise while others contain the data. Sorting out which ones to keep requires some judgement; the process is not automatic. The PCA method is therefore a way of reducing the size of a data set, the penalty is that some data is lost; however, this may not be significant, simply necessary to expose the pertinent data. This type of analysis is therefore always a balance between simplifying the problem to make it understandable and loosing information. Note that principal component analysis is a non-parametric method of analysing data; correlations are sought between data sets without using any model to describe the data. In contrast, the least squares method is a parametric method because a model (equation) is used to test the data.

A set of data can be represented as an expansion of coefficients $$a$$ in a basis set $$w$$ of length, or dimension, $$n$$,

$\displaystyle x=a_1w_1+a_2w_2+\cdots +a_nw_n$

The same data can also be expressed in another basis set $$u$$ as

$\displaystyle x=b_1u_1 +b_2u_2 +\cdots + b_nu_n$

but with a different set of coefficients, $$b$$. This distinction is illustrated in figure 16 where any data point has approximately equal values of coefficients $$a$$, because it is approximately equally positioned between the three experimental x-axes whereas those on the new axes $$p$$, have a large initial value $$b_1$$, along $$p_1$$ and smaller ones along $$p_2$$ and $$p_3$$.

PCA changes the basis set to optimize the variance placing the principal axis along the direction of maximum spread or variance of the data, and other principal components orthogonally to this. Figure 16 also shows the new axes $$p$$ as the principal component axes, two of which are shown on the right and where two regions of data are identified. The missing data in this graph compared to the one on the left is in the direction $$p_3$$ and not $$p_2$$, and would be seen on a plot of $$p_1$$ vs $$p_3$$ and in this way, the change of axes (basis set) can identify different groups of data. The next step in PCA is to reduce the size of the basis set used; this means describing the data in the basis set $$u$$ with a smaller set of terms in the summation; $$s$$ instead of $$n$$. Because the change of axes puts more contributions of each data point along each principal axis, fewer terms are now needed to form an acceptable description of the data, although all the basis set is needed to reproduce the data exactly. A reduced data set is often the rationale for using PCA, with the aim of identifying new signals, removing noise and/or isolating known interfering signals.

Figure 16. Data on the experimental axes (left), and on the axes of principal components (right)

Matrix methods are used to find the principal components (PC). The data comprises m sets where each set has n data points. The whole data forms an n ×m matrix, D, with each data point in a new row of a given column and these vectors form m columns. The steps to find the principal components are:

(i)$$\quad$$ Remove the mean value from each of the individual datasets so that the centroid of the data becomes the origin of the new axes. In matrix form subtracting the mean from each data point is $$$\displaystyle \boldsymbol X= \boldsymbol D -\boldsymbol \mu$$$

where $$\boldsymbol D$$ is the $$n\times m$$ matrix of data and $$\boldsymbol \mu$$ an $$n\times m$$ matrix with the appropriate mean value in each element of its columns. The PC.s are also sensitive to the scale of the data, so it may be necessary to divide the data by the variance giving each set a mean of zero and unit standard deviation during the calculation.
(ii)$$\quad$$ The covariance of the data is then calculated, this is an $$m\times m$$ symmetrical matrix

$\boldsymbol C=\frac{1}{n-1}\boldsymbol{XX^T}\qquad\tag{52}$

The superscript $$T$$ is the transpose, and the matrix multiplication has dimensions $$(m \times n)\cdot(n \times m) = m \times m)$$. (The $$n-1$$ makes an unbiased estimate). The entries in this matrix would be diagonal if the data were random, because each value is then independent of every other. However, this is very unlikely to be the case unless the data contains only random noise; therefore, off-diagonal terms are not zero. The covariance describes how much the data is spread among the axes. If the data were to lie only along each axis the covariance would be diagonal, the rotation of axes accompanying the principal components reduces the off-diagonal component in the covariance.

(iii)$$\quad$$ The m eigenvalues $$\lambda$$ and (column) eigenvectors $$V$$ ($$m\times m$$ matrix) of the covariance matrix $$C$$ are calculated and the eigenvalues are sorted from largest to smallest and the eigenvectors are placed in the same order see chapter 7.12.3. The eigenvalue-eigenvector equation is $$\boldsymbol{CV} = \boldsymbol{\Lambda V}$$ where $$\Lambda$$ is the diagonal matrix of the eigenvalues $$\lambda$$, i.e. a vector. It is assumed that the calculation produces eigenvectors that are normalized, if not they should be normalized by dividing each column by its vector length $$\vec V_i\to \vec V_i/\sqrt{\vec V_i\cdot \vec V_i}$$ where $$\vec V_i$$ is one column vector. The eigenvectors form the new principal component basis set because they are orthogonal one to another.

The eigenvector of the largest eigenvalue forms the principal component of the data, smaller values the other principal components. The largest contains most of the variance or spread of the data and best describes any trend in the data. The smaller eigenvalues do so to lesser extents depending on their size. The largest s eigenvalues are then chosen to approximate the data. The eigenvalues correspond to variances along the principal axes, thus the fractional value of each eigenvalue to the sum of them all is

$\displaystyle \lambda_i/ \sum_i^m\lambda_i$

and this is the fraction that eigenvalue $$i$$ contributes to the data. A measure of the error made by approximating the data by s principal components instead of the total number $$m$$, is the residual variance,

$\displaystyle \sigma^2=1-\frac{\sum_i^s \lambda_i}{\sum)i^m \lambda_i}$

For example, if the first two eigenvalues describe $$85$$% of the data, it may be appropriate to ignore all the others. As a rule of thumb, when selecting data to eliminate, eigenvalues less than $$\approx 1$$ can be ignored.

(iv)$$\quad$$ To project the data points onto the new axes, the principal components, the dot product of the data $$\boldsymbol X$$ is made with each eigenvector $$\boldsymbol V$$,

$\displaystyle \boldsymbol Y = \boldsymbol V^T\boldsymbol X^T\qquad\tag{53}$

The matrix dimensions are $$(m \times m)\cdot(m \times n) = (m \times n)$$. Plotting one row vs another, the data in matrix $$Y$$ produces the data along the various principal component axes as shown on the right of figure 16. The first column is $$p_1$$, the second, $$p_2$$, etc. Because the eigenvector (modal) matrix $$V$$ is square and orthogonal, the relationship $$\boldsymbol V^{-1} = \boldsymbol V^T$$ can be used to reform the original data. The transformation steps are to left multiply both sides of the equation with $$\boldsymbol V$$ giving

$\displaystyle \boldsymbol{VV}^T\boldsymbol X^T = \boldsymbol{VY}$

Simplifying the left-hand side produces

$\displaystyle \boldsymbol{VV}^T\boldsymbol X^T = \boldsymbol{VV}^{-1}\boldsymbol X^T = \boldsymbol X^T = \boldsymbol{VY}$

and then $$\boldsymbol{VY}$$ is transposed again to form $$\boldsymbol X$$ or,

$\displaystyle \boldsymbol X=(\boldsymbol{VY})^T, \qquad \boldsymbol D = \boldsymbol X + \boldsymbol\mu\qquad\tag{54}$

and $$\boldsymbol\mu$$ restores the mean to the data $$\boldsymbol D$$.

(v)$$\quad$$ When only $$s$$ of the eigenvalues and eigenvectors are being used, the $$V$$ matrix is reduced by ignoring all but the first $$s$$ columns and equation 54 is used to reconstruct the data.

A python PCA algorithm is detailed below

# Algorithm Pricipal component analysis

filename='PCA data.txt'                    # 3 sets of data 10 values each
# data is at end of book in 'Appendix, some basic Python instructions'

xv = []
yv = []
zv = []
with open(filename) as ff:                 # length not known so read in all data and make list of each
for line in ff:
new_str = ' '.join(line.split())
vals = new_str.split(' ')
xv.append(vals[0])
yv.append(vals[1])
zv.append(vals[2])
ff.close()

n = len(xv)                               # get length of data
m = 3                                     # 3 axes of data
data = np.zeros((n,m),dtype=float)
for i in range(n):
data[i,0] = float(xv[i])              # make lists into arrays, data set 0,1,2 with n values each
data[i,1] = float(yv[i])
data[i,2] = float(zv[i])

S = 1                                     # select first S components eigenvals
means = np.zeros((n,m),dtype=float)
X     = np.zeros((n,m),dtype=float)
VS    = np.zeros((m,m),dtype=float )
Sev   = np.zeros(m,dtype=float)

for k in range(m):                         # calculate mean value
s = 0
for i in range(n):
s = s + data[i,k]
means[:,k] = s/n
pass
X = data - means                           # subtract mean
C = (np.transpose(X) @ X)/(n-1)            # make covariance matrix ( @ is matrix multiply)
(eigval,eigvec) = lina.eig(C)              # eigvals, eigvects of covariance. eigvects normalised

indx = np.argsort(eigval)

Sev[:] = eigval[  indx[::-1]]              # reverse indx array to get largest first
VS[:,:]= eigvec[:,indx[::-1]]

print('eigenvalues and eigenvectors')
for i in range(m):
print(i+1,Sev[i], VS[:,i])

V = np.zeros((m,S))                        # do PCA here DR is result
V[:,:] = VS[:,0:S]                         # select S sorted eigenvectors

Y  = np.transpose(V) @ np.transpose(X)     # project onto S new axes

DR = np.transpose(V @ Y) + means           # DR is data on new axes

eigenvalues and eigenvectors
1 4.886993943985875 [ 0.28976365  0.9562666  -0.03989011]
2 0.11490529483430316 [ 0.94367801 -0.2784974   0.17863651]
3 0.08368965006871082 [-0.15971484  0.08940578  0.98310619]

Figure 17. Left: Data plotted with mean values subtracted together with the principal axes. The major principal axis is (1). Right: Data partly reconstructed (also with mean values subtracted for comparison) with the first eigenvector (red) and then the first two eigenvectors combined (light blue).

The figure shows the data (centred at zero by subtracting the mean values), and the three principal axes. The first eigenvector, (1) the major axis, is drawn in a red line. The lengths are arbitrary. The eigenvalues are $$4.9, 0.11, 0.084$$. On the right of the figure, two plots are superimposed. The light blue symbols are the data reconstructed from the first two eigenvectors using equations 53 and 54. The original data is flattened onto the plane of the first two principal axes because the third coordinate is ignored. The second set of data (red symbols) is reconstructed using only the first eigenvalue and is a set of points lying along the principal axis at their relative positions when projected from the other axes. If all three eigenvectors are used the original data is reformed exactly.

Related to PCA is the method of singular value decomposition (SVD). Formally, it is a technique in which one matrix is split into three, one of them is diagonal and the other two are row and column orthogonal as are eigenvector modal matrices (Prest et al. 1986). The reason for expanding a matrix in this way is that it can now be inverted even though it is close to being singular, i.e. inverting it would normally produce infinity or a number that approximates this, and normal inversion methods fail. This method could be used to invert the $$A$$ matrix in non-linear least squares, in the example given. The interest in SVD is however not primarily to do with the mathematics but in un-ravelling complex information just as in PCA. SVD can be used either directly on raw data, not the covariance as in PCA, or as a mathematical method to perform PCA (Jolliffe 2002).


text:
# Tag Info

10

Possibly the smallest group where this occurs is the alternating group $A_{4},$ which has order $12$ and yet has no subgroup of order $6$. Note that this is a solvable group.

5

It is not possible for $G$ to be finite and abelian for this to be true. Many examples are given with $n=2m$ by simple groups: any subgroup of index 2 is normal, hence no simple group can have a subgroup of index 2. Furthermore, any nonabelian simple group is of even order, so any nonabelian finite simple group gives an example. As Hagen von Eitzen points ...

5

Prove the following (easy): Lemma: For any group $\;G\;$, the quotient group $\;G/Z(G)\;$ cannot be cyclic non-trivial. From the above it follows that $\;G/Z(G)\;$ cyclic $\;\implies G\;$ is abelian.

3

In the finitely generated case the answer is NO as pointed out by @Timbuc in the comments. In the infinitely generated case I believe that the answer is YES by the following construction. Let $K \subseteq \mathbb{N}$ be a recursively enumerable non-recursive set and consider a countable abelian group given by the following presentation (commutators ...

3

Hints: Fix decompositions for $A,B,C$ and note that they "stack" to become decompositions of $A\oplus B$ and $A\oplus C$. The components of the decompositions must line up exactly, and without loss of generality, we can line up the components of $A$ on both sides because they're identical. What does this say about the remaining components?

3

For your first question: If the proposition just said $n$ coprime integers, that would (or could) be taken to mean that each pair of integers $\{v_i,v_j\}|i \neq j$ is coprime ($\gcd(v_i,v_j) = 1$). This is a much stronger condition than is required for the proposition (which is that there is no "global" divisor $d>1|d\in\Bbb{Z}$ dividing every $v_i$. ...

3

Indeed it is not cyclic: since $\sqrt{2}$ is irrational, the only way the sum is an integer is if $b=0$, so this means there are no additive relations between $a$ and $b$ and the group is isomorphic to it is isomorphic to $\Bbb Z\oplus\Bbb Z$.

3

$$\mathbb{Z}_m\oplus \mathbb{Z}_n=\mathbb{Z}_{mn}.$$ And from here it is easy to show that all the subgroups are $$\mathbb{Z}_k,$$ where $k$ is divisor of $mn$.

2

If you stick to the "$QAP^{-1}$ convention" and do things horizontally, then $(Q^{-1})^{\top}$ is what gives you the basis and $P^{-1})^{\top}$ is what gives you the generators. To avoid transposing, take $\begin{pmatrix}1&1&4\\2&-1&-2\end{pmatrix}$. Using elementary row and column operations and recording them in matrix form, I get that ...

2

We could probably get this information more directly from Gap, but here is at least some of what you are looking for. sage: F.<x, y> = FreeGroup() sage: G=F / [x^3,y^3,x*y*x^-1*y^-1] sage: G.order() 9 sage: H = G.as_permutation_group() sage: H.subgroups() [Subgroup of (Permutation Group with generators [(1,2,3)(4,6,8)(5,7,9), (1,4,5)(2,6,7)(3,8,9)]) ...

2

$n\mathbb{Z}_p\subseteq G$ is just Lagrange's theorem applied to the quotient group $\mathbb{Z}_p/G$ in the form: $|\mathcal G|=n, x\in \mathcal G \implies x^n =1$.

2

This is never true for non-abelian groups if $n=2$, but otherwise, for $n\ge 3$ this can happen for certain non-abelian groups. For more details see here. Example: The quaternion group satisfies $(ab)^n=a^nb^n$ for all $n=2^k$ with $k>1$, but is not abelian.

2

Let $\{P\}=Syl_3(G)$. Since $|P|=9=3^2$ (the square of a prime), $P$ must be abelian. Hints: (1) Study the N/C Theorem. Observe that $P \subseteq C_G(P) \subseteq G=N_G(P)$. Hence $G/C_G(P)$ embeds homomorphically into $Aut(P)$. (2) Note that $Aut(P) \cong C_6$ or $\cong GL(2,3)$. Hence $|Aut(P)|$ divides $6$ or $48$, so $|G/C_G(P)|$ divides these numbers, ...

2

OK, without automorphisms: put $\{P\}=Syl_3(G)$, $Q \in Syl_5(G)$ and $R \in Syl_7(G)$. Observe that $G/P \cong C_{35}$ (I leave it to you to prove this with again Sylow Theory). Hence $QP \unlhd G$, since $QP/P \in Syl_5(G/P)$. Further it is easy to see that $Q$ is the only Sylow $5$-subgroup of $QP$. Hence $Q \text{ char } QP \unlhd G$, so $Q$ is normal in ...

2

I will assume you know how to check that $G'$ is normal subgroup, and thus that quotient $G/G'$ is well defined. I will also assume that you know how to check that canonical map $f= (x\mapsto xG')\colon G\to G/G'$ is indeed homomorphism, and not only that, it is epimorphism. I will also assume that you know Fundamental theorem on homomorphisms (and if you ...

2

$(\Bbb F_3[X],+){}{}{}{}{}{}$.

2

$K_4=\mathbf{Z}/2\mathbf{Z}\times\mathbf{Z}/2\mathbf{Z}$ (product of rings has naturally a ring structure, defined componentwise).

2

The question is unclear, since as Derek Holt mentioned, "solvable word problem" has not been defined. If you have a presentation with a sequence of generators, you can ask two questions: 1) whether this group endowed with this generating family has a solvable word problem 2) whether this group is isomorphic to a computable group (i.e. is finite or ...

2

Hint: (1) $Q_8, S_3.$ (2) $\mathbb Z.$

2

I don't know how 'real' this is, but here's one way to visualise $\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$. There are two copies of $\mathbb{Z}$, call them $\mathbb{Z}\times\{0\}$ and $\mathbb{Z}\times\{1\}$. $$\mathbb{Z}\times\{0\}\qquad :\qquad \dots\qquad -3\qquad -2\qquad -1\qquad 0\qquad 1\qquad 2\qquad 3\qquad \dots$$ $$\mathbb{Z}\times\{1\}\qquad ... 2 Here's a simple geometric interpretation. Consider the figure below, with marked fenceposts (or copies of the letter C) at every integer point on a horizontal line. Its group of symmetries is \Bbb Z\times \Bbb Z/2\Bbb Z, generated by the horizontal translation and reflection across the horizontal. 2 Maybe good to know if you are interested: finite groups for which the property does hold (that is, for every divisor d of the order |G|, G has a subgroup of order d) have been studied extensively. They are called CLT (Converse Lagrange Theorem) groups. It is known that a CLT group must be solvable and that every supersolvable group is a CLT group: ... 1 Unfortunately, your attempt at a proof isn't valid. You define n to be the order of a, then prove that a^n=1 (which doesn't need proof, actually, because that's what n does), and then derive a contradiction because you had claimed n<m. Indeed, I don't think you ever used the hypothesis that m\nmid k, other than to (correctly) derive a^k\ne1 ... 1 I'm assuming that you are asking about group automorphisms because of the notations you used [(\mathbb Z/n\mathbb Z, +), (\mathbb{Q}, +), (\mathbb R, +)]. (1). (\mathbb Z/n\mathbb Z, +): Let G be a cyclic group of order n and let x \in G be a geneator. Let \phi : G \to G be a group homomorphism. First note that, \phi is uniquely determined by ... 1 o(Z(G))=p then o(G/Z(G))=p and every group of prime order is cyclic 1 Subgroup lattices of finite abelian groups have been studied from various points of view, see the following reference: Vogt, Frank : Subgroup lattices of finite Abelian groups: Structure and cardinality. In: Lattice theory and its applications. Hrsg.: K.A. Baker, R. Wille. S. 241-259. Heldermann , Berlin . [Buchkapitel], (1995) 1 Let A denote the set of positive rational numbers, an abelian group with respect to multiplication. Define f\colon A\to A by f(x)=x^3. This is a homomorphism, is injective, but not an automorphism. For example 4 is not in the image. SO the image subgroup f(A) is a proper subgroup. The quotient group A/f(A) meets your requirement. All the cosets ... 1 Suppose that \mathsf{hom}(A, -) has a left adjoint T_A. In other words you have a natural (in B, C) isomorphism$$\mathsf{Ab}(A, \mathsf{hom}(B,C)) \cong \mathsf{Ab}(T_A(B), C).$$Since \mathsf{hom}(\mathbb{Z}, C) \cong C, it follows that T_A(\mathbb{Z}) \cong A. As all left adjoint, T_A preserves colimits, so for any family \{\alpha\}, ... 1 Just a remark - this does hold in much greater generality. According to a theorem of Nikolay Nikolov and Dan Segal, in any topologically finitely-generated profinite group (that is, a profinite group that has a dense finitely-generated subgroup) the subgroups of finite index are open. In particular, this holds for the p-adic integers. 1 No need to refer to \mathbb Q: we have the following short exact sequence of free \mathbb Z-modules (or abelian groups, if you like it)$$0\to\ker\phi\to\mathbb Z^n\to\operatorname{im}\phi\to0 which is split (helpful, but not necessary to conclude).

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# Why haven't we gone back to the moon?

by DruidArmy
Tags: moon
Mentor
P: 15,202
 Quote by vociferous I did read the entire article.
Did you read the part that said "Nobody knows how to do it"?

That is the key sentence in the entire article.

 For instance, I do not interpret the article to imply that he felt it was impossible to use thrusters at all to slow the descent, which would be absurd since the Viking landers used thrusters on Mars.
That's a non sequitur. The Viking landers could use thrusters to land precisely because they were small. One of the myriad of nasty facts that confronts putting humans into space is the square-cube law. If you scaled a mouse up to the size of an elephant it would die. The square-cube law dictates that elephant physiology is very, very different from mouse physiology. The same applies to those little Viking landers. Space vehicles don't scale up because of the square-cube law. More mass needs to be added because of structural integrity issues. More thrusters needed to be added because thrust is proportional to throat area. That means even more mass. The end result is that a large vehicle is elephantine compared to the mouselike characteristics of a small vehicle.

 Furthermore, I am not convinced that the turbulence is an insurmountable engineering program.
It is a huge issue, one we do not know how to solve.

 We've been landing VTOL craft on Earth using nothing but thrust since before we landed men on the moon.
That's another non sequitur. Aircraft don't carry their own oxidizer, and they don't go anywhere close to the speed of a reentering spacecraft. It's more like 1/20th of the speed in the case of a VTOL aircraft vs a reentering spacecraft, and that means 1/400th the energy needs to be dissipated.

 Even if, for the sake of argument, we just agree that it is impossible to land such a single mass on Mars, then we simply use a different engineering paradigm.
It's so simple if you simply hand-wave away all the problems!

There's only so much you can shrink a spacecraft that will safely land humans. The Apollo command modules were 6.5 times as massive as is the Mars Science Laboratory rover. Landing a vehicle that is 6.5 times heavier not just 6.5 times harder. It's harder than we can imagine. We don't know how to do it. The Apollo command modules carried the barest minimum of life support needed to bring humans from separation from the service module down to the Earth, where they would be whisked away via helicopter shortly after splashdown. A Mars lander would need to carry a significant amount of life support (air, water, food, life support equipment, etc.). And a hefty rover. And a launch vehicle.

Why a launch vehicle? Not bringing the astronauts back home is a death sentence. We haven't the foggiest idea of how to grow food away from home. For a little while, and for a few bites of lettuce to augment otherwise boring freeze dried food, yes, we can do that. A fully sustaining farm? We don't know how to do that.

Why a rover? Unless we can land several months worth of life support with that one vehicle, we need to send multiple landing vehicles. Those other landers are likely to land tens of kilometers apart. The Mars Science Laboratory represents the state of the art in precision landing on Mars. It's one sigma landing ellipse was 20 km × 7 km. The probability of landing within that one sigma ellipse is about 39%. Tripling each dimension (60 km × 21 km) yields a probability of 98.9%. Those aren't great odds, but close to the target for human spaceflight activities. We'll need that rover to go pick up the water, food, and equipment that landed tens of kilometers away.

 I think the crux of the difference is between understanding the distinction between us not having a fabrication-ready design for a Mars mission today and a Mars mission simply being beyond our current level of technology.
I think the crux of the problem is that you don't know how impossibly hard it is, even using technology that doesn't exist yet. We don't know how to do it. I've mentioned but three of the things we don't know how to do. There are a number more.
PF Gold
P: 6,506
 Quote by vociferous Fair enough; I am just stating my opinion as you are stating yours .
Well, it does seem that your "opinion" is based on wishful thinking and a very selective reading and interpretation of the information available whereas Drakkith's is based on a pretty straight forward reading of the same information. I don't think anyone here is trying to give you a hard time but it IS frustrating that you don't seem to understand or appreciate the difficulties that have been mentioned to you.
P: 304
 Quote by D H Did you read the part that said "Nobody knows how to do it"? That is the key sentence in the entire article. That's a non sequitur. The Viking landers could use thrusters to land precisely because they were small. One of the myriad of nasty facts that confronts putting humans into space is the square-cube law. If you scaled a mouse up to the size of an elephant it would die. The square-cube law dictates that elephant physiology is very, very different from mouse physiology. The same applies to those little Viking landers. Space vehicles don't scale up because of the square-cube law. More mass needs to be added because of structural integrity issues. More thrusters needed to be added because thrust is proportional to throat area. That means even more mass. The end result is that a large vehicle is elephantine compared to the mouselike characteristics of a small vehicle. It is a huge issue, one we do not know how to solve. That's another non sequitur. Aircraft don't carry their own oxidizer, and they don't go anywhere close to the speed of a reentering spacecraft. It's more like 1/20th of the speed in the case of a VTOL aircraft vs a reentering spacecraft, and that means 1/400th the energy needs to be dissipated. It's so simple if you simply hand-wave away all the problems! There's only so much you can shrink a spacecraft that will safely land humans. The Apollo command modules were 6.5 times as massive as is the Mars Science Laboratory rover. Landing a vehicle that is 6.5 times heavier not just 6.5 times harder. It's harder than we can imagine. We don't know how to do it. The Apollo command modules carried the barest minimum of life support needed to bring humans from separation from the service module down to the Earth, where they would be whisked away via helicopter shortly after splashdown. A Mars lander would need to carry a significant amount of life support (air, water, food, life support equipment, etc.). And a hefty rover. And a launch vehicle. Why a launch vehicle? Not bringing the astronauts back home is a death sentence. We haven't the foggiest idea of how to grow food away from home. For a little while, and for a few bites of lettuce to augment otherwise boring freeze dried food, yes, we can do that. A fully sustaining farm? We don't know how to do that. Why a rover? Unless we can land several months worth of life support with that one vehicle, we need to send multiple landing vehicles. Those other landers are likely to land tens of kilometers apart. The Mars Science Laboratory represents the state of the art in precision landing on Mars. It's one sigma landing ellipse was 20 km × 7 km. The probability of landing within that one sigma ellipse is about 39%. Tripling each dimension (60 km × 21 km) yields a probability of 98.9%. Those aren't great odds, but close to the target for human spaceflight activities. We'll need that rover to go pick up the water, food, and equipment that landed tens of kilometers away. I think the crux of the problem is that you don't know how impossibly hard it is, even using technology that doesn't exist yet. We don't know how to do it. I've mentioned but three of the things we don't know how to do. There are a number more.
I am not trying to downplay the challenges. However, I believe that the challenge of landing people safely on Mars today is very similar to the challenge of landing people safely on the moon in 1960. We have the basic technological know-how, but it is near the edge (although I would argue that we are at a much better starting position for a Mars mission today than we were for a moon landing in 1960).

The cost of the Apollo program was around $100 billion dollars and it would not be surprising to see a Mars mission cost$1 trillion or more. I'm confident if the budget were there, just like the Apollo program, the results would follow.

The only counterpoint to this is a seven year old article about a decade-old conference that took place around the same time that President Bush committed NASA to a manned Mars mission, before any serious planning about how to effect that had occurred.

By contrast, much more recently, a workshop group of more than 60 individuals representing more than 30 government, industry, academic and other organizations has found that a NASA-led manned mission to Mars is feasible if the space agency's budget is restored to pre-sequestration levels. [1]

I am not an engineer, but I'm pretty confident that our current state of technology is commensurate with engineering such a mission in the next decade or two.

REFERENCES:

[1] http://www.space.com/24268-manned-ma...asibility.html
Mentor
P: 15,202
 Quote by vociferous I am not trying to downplay the challenges.
That is exactly what you are trying to do.

 However, I believe that the challenge of landing people safely on Mars today is very similar to the challenge of landing people safely on the moon in 1960.
Getting to Mars now is much harder than was getting to the Moon then. The breakthroughs needed in the 1960s to land on and return from the Moon were simple compared to those needed to land on and return from Mars.

Another problem is politics and money. The Apollo mission was uniquely situated in history. There was a perceived threat in the US that Russia was getting ahead of the US technologically. Apollo would never have happened without that perceived threat, or without the perpetual duck and cover exercises (youtube video: https://www.youtube.com/watch?v=IKqXu-5jw60) throughout the 1950s and into the early 1960s. The US was very afraid of what Russia might do (and Russia was very afraid of what the US might do). It was a time of MADness.

 By contrast, much more recently, a workshop group of more than 60 individuals representing more than 30 government, industry, academic and other organizations has found that a NASA-led manned mission to Mars is feasible if the space agency's budget is restored to pre-sequestration levels. [1]
Did you read your cited article? It targets a "manned mission to Mars By 2030s" -- in other words, 25 years from now. That is perhaps doable with pre-sequestration funding levels, with some international participation to augment that pre-sequestration level funding. It does not require Apollo-style funding. Apollo-style funding is a pipe dream. That kind of funding will not happen until another major perceived threat appears.

Let's say we do start a long term project to send people to land on and return from Mars. Rhetorical question: What happens after that? The answer is to learn from history. What happens after that is what happened to NASA after they landed a few men on the Moon. What politicians from both sides thought was "Mission accomplished, and now we better clamp down on NASA before they bleed us dry." The end result: NASA went through a rather bleak period from the early 1970s to the late 1980s where they received far less money in inflation-adjusted dollars than they receive now, post-sequestration.

I predict the exact same outcome should NASA focus on sending humans to Mars. We might accomplish that mission, but it will be followed by decades of stagnation. IMHO, a much better objective is for NASA to find a way to show that space is profitable. The asteroid recovery plans are the best thing NASA has come up with for a long time. Show that space is a profit center rather than a cost center and we'll have people in space for a long time to come. Some of them will be doing science, a good deal more science than the meager amount done now. Humanity will be stuck here on Earth with little dribbles of funding for space exploration every now and then if space exploration turns out to be not at all profitable.

 I am not an engineer, but I'm pretty confident that our current state of technology is commensurate with engineering such a mission in the next decade or two.
I am an engineer. In fact, I am an aerospace engineer who works with NASA, and I'm pretty confident that this is not the case.
 Sci Advisor PF Gold P: 9,488 Even with our modern engineering savvy, manned space missions are obscenely expensive. The He3 carrot still dangles in front of us, but, who is willing to gamble vast resources without proven technology to utilize it?
PF Gold
P: 6,506
 Quote by vociferous I am not trying to downplay the challenges.
I can believe that you BELIEVE you are not but you have made it abundantly clear that you simply don't understand them and you in fact ARE downplaying them to a ridiculous degree.
P: 190
I am of the considered opinion that it will not only not be mere decades but actually Lifetimes before we land humans on Mars, absent some (and not just a few) fairly extreme technological breakthroughs, not the least of which is propulsion and radiation shielding. I utterly hate concluding this, but "if wishes were horses....".

It seems to me that in a fairly short time, perhaps that decade or two, we could have a Moon Base IF we had the will, which we apparently don't. It is my understanding that despite the apparent fact that such mass conflicts as WWII are a thing of the past, and a more Guerilla/Covert type of conflicts are more likely, the US spends more in a day on Conventional Defense (even though there is no single perceived enemy as during the Cold War) than NASA receives in an entire year.

It is also my understanding that during the height of the Apollo Mission the expenditures reached more than 1% of the National Budget, still a mere fraction of the Defense Budget at that time and even less now.

 Quote by Wikipedia_Apollo_Project In 2009, NASA held a symposium on project costs which presented an estimate of the Apollo program costs in 2005 dollars as roughly $170 billion. This included all research and development costs; the procurement of 15 Saturn V rockets, 16 Command/Service Modules, 12 Lunar Modules, plus program support and management costs; construction expenses for facilities and their upgrading, and costs for flight operations. This was based on a Congressional Budget Office report, A Budgetary Analysis of NASA’s New Vision for Space, September 2004.[71] The Space Review estimated in 2010 the cost of Apollo from 1959 to 1973 as$20.4 billion, or $109 billion in 2010 dollars, averaged over the six landings as$18 billion each
Very few people understand the payoff, considering all the advances in so many areas, of those expenditures. Even such mundane things as textiles and fasteners (affecting many kinds or work wear and hazard wear as well as specialized medical condition protective attire, just to name a few) let alone the huge effect on micro-electronics, without which we would not be having this conversation not to mention a complete shift in economic base to Information.... none of this and much more would have occurred.

However for the average businessman, politician, man-on-the-street this is all too abstract (read as "nebulous pie-in-the-sky") to actually divert expenditures from the closer to home, sexy new bomber (who cares if they rust away in some hangar? ...just in case, right?)

Many, even those "in the business" thought JFK's challenge was impossible to achieve at even twice the timeframe, but somehow a flame was ignited AND FUNDED and the rest is now History. I think it foolish to dismiss this as "just lucky" and impossible to reproduce.

Maybe it would take 20 years instead of 10. That is uncertain. What is certain is that we will never do it unless we begin to take it seriously and make it common knowledge just how beneficial Apollo really was on a global scale to an inestimable number of people, some of whom are alive today because of it.
P: 98
 Quote by enorbet Very few people understand the payoff, considering all the advances in so many areas, of those expenditures. Even such mundane things as textiles and fasteners (affecting many kinds or work wear and hazard wear as well as specialized medical condition protective attire, just to name a few) let alone the huge effect on micro-electronics, without which we would not be having this conversation not to mention a complete shift in economic base to Information.... none of this and much more would have occurred. However for the average businessman, politician, man-on-the-street this is all too abstract (read as "nebulous pie-in-the-sky") to actually divert expenditures from the closer to home, sexy new bomber (who cares if they rust away in some hangar? ...just in case, right?) Maybe it would take 20 years instead of 10. That is uncertain. What is certain is that we will never do it unless we begin to take it seriously and make it common knowledge just how beneficial Apollo really was on a global scale to an inestimable number of people, some of whom are alive today because of it.

Enorbit, they are words plucked straight from my thoughts!
I believe that this is the fundamental problem that the western world faces, and unfortunately no amount of education, reasoning or even scare tactics will convince the 95% of the population (which enjoy western freedoms) that spending in science of ANY disipline is worth the payoff. Give them the internet with every educational book ever written, every science theory ever postulated, every forum like this, and what do they do, search for boobs!

"I stupid people, they are everywhere, they don't even know they are STUPID!"

Damo
 Mentor P: 12,006 Just because someone doesn't know much about science and technology, or has different priorities than you, does not make them stupid. Most people are not stupid.
 P: 190 Given the technological difficulties as well as the socio-political environment, I am betting..... let's say $10,000 that I will not only be dead but long dead before another single human sets foot on the Moon, forget about Mars. If I'm wrong and you wish to collect I will instruct my lawyer to place the 10K on my headstone in a ziploc baggie (he assures me he is trustworthy). If you think my estimate too long and think you can win this bet and this bet interests you, I also deal in bridges. :P  P: 98 Fair point Drakkith, that was a very generalized statement. It is just so frustrating that our society values such trivial things/people over the genuine contributors to the advancement of our species in becoming more than just consumers and another transient lifeform on this planet. What happened to the days when 'our' celebrities were people like Einstein? Damo PF Gold P: 6,506  Quote by enorbet Given the technological difficulties as well as the socio-political environment, I am betting..... let's say$10,000 that I will not only be dead but long dead before another single human sets foot on the Moon, forget about Mars. If I'm wrong and you wish to collect I will instruct my lawyer to place the 10K on my headstone in a ziploc baggie (he assures me he is trustworthy). If you think my estimate too long and think you can win this bet and this bet interests you, I also deal in bridges. :P
Mentor
P: 15,202
 Quote by enorbet Given the technological difficulties as well as the socio-political environment, I am betting..... let's say \$10,000 that I will not only be dead but long dead before another single human sets foot on the Moon, forget about Mars.
That's a fair bet if you are seventy years old. If you are sixty or less, it's not such a good bet.

NASA is working on plans to send humans to Mars in the 2030s (e.g., 2039 counts, so about 25 years from now) and do so within the constraints of a pre-sequestration budget with only small, inflation-level increases over the course of that 25 years. That 25 year period gives more than enough time to develop solutions to the hard engineering problems outlined earlier in this thread, and it gives time to do so incrementally.

Funding at a pre-sequestration budget level means a bit more than half a percent of the federal budget. That's tiny compared to the unsustainable Apollo-level funding which peaked at 4.4% of the federal budget. This pre-sequestration level funding is arguably sustainable for the long term. An Apollo-style program is not, but this is not an Apollo-style program.
PF Gold
P: 6,506
 Quote by D H NASA is working on plans ...
Have you ever really looked at the outcome vs the plans for ANY major NASA mission? I point you specifically to the costs and timelines for both the Space Shuttle and the International Space Station. The plans are a joke compared to what really happens.
 P: 18 Because we have not only the moon....there are many other planets to explore...like mars, for example, and so on..
Mentor
P: 12,006
 Quote by Damo ET What happened to the days when 'our' celebrities were people like Einstein?
Remember that back in those days we didn't have movies, TV, radio, the internet, and other ways of getting information out. The only people who became famous were those who did something that made the newspapers, such as scientists, politicians, and sports stars.

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# Vector analysis, calculate path

Tags:
1. Jan 17, 2016

### Hannibal123

1. The problem statement, all variables and given/known data
I have these vectorfields defined on the specified set. All of them are conservative on their set.

Also i have three unit circles C1, C2 and C3 centered respectively on (0,0) , (-2,0) and (-1,0)

I need to find the line integrals over all of them on H

(beregn = calculate, og = and. (I'm Danish))

2. The attempt at a solution
Since H is conservative i think both integrals in (b) equals zero because you can just "repair" the hole in the middle. Not sure about (c)

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2. Jan 22, 2016

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Jan 23, 2016

### Ray Vickson

You can draw small circles S2 of small radius r around (-2,0) and S0 of small radius r around (0,0). Suppose S2 intersects C3 at A and B (where A is nearly (-2,r) and B is nearly (-2,-r)),and S0 intersects C3 at C and D (where C is nearly (0,-r) and D is nearly (0,r)). Then your integral in (c) equals
$$I_c = I_1 + I_2 + I_3 + I_4,$$
where
$$\begin{array}{l}I_1 = \int_D^A \vec{H} \cdot d\vec{r} \\ I_2 = \int_B^C \vec{H} \cdot d\vec{r} \\ I_3 = \int_C^D \vec{H} \cdot d\vec{r} \; \text{along C3}\\ I_4 = \int_A^B \vec{H} \cdot d\vec{r} \; \text{along C3} \end{array}$$
Also, let $I_3'$ be the integral from C to D along the small circle S0 (clockwise, inside the big circle C3) and let $I_4'$ be the integral from A to B along the small circle S2 (clockwise, inside C3). We have $I_1 + I_2 + I_3' + I_4' = 0$ because there are no singularities inside the corresponding closed path. Thus, the answer $I_c = I_3+I_4-I_3' - I_4'$. We can try to evaluate $I_3 - I_3'$ directly, since inside the small circle S0 the direct path (along C3) from C to D is almost a straight line from (0,-r) to (0,r), so $I_3 - I_3'$ is an integral along a closed semicircle, where the singularity in H at (0,0) does not appear because x = 0 along the vertical line through (0,0); that is, the term $y dx/(x^2+y^2) = 0$ because dx = 0 on the vertical line, and $x dy/(x^2+y^2) = 0$ because $x=0$ on the vertical line.

Last edited: Jan 23, 2016


text:
## Minimum Residual Energy

In the notes on signal spaces we commented on interpreting the integral $\int_{-\pi}^\pi|f(t)|^2dt$ as the energy in a signal. The question that we want to answer here is what "trig polynomial" $T_N(t)=A_0+\sum_{k=1}^N A_k \cos(kt)+B_k\sin(kt)$ will give the best approximation to $f$, in the sense of minimizing the residual energy, $(\dagger) \quad \int_{-\pi}^\pi(f(t)-T_N(t))^2dt.$ We will begin by showing that functions for that $(\ast) \quad \int_{-\pi}^\pi f(t)T_N(t)dt = 2\pi a_0A_0 + \pi \sum_{k=1}^N a_k A_k+b_k B_k.$ To see this, note that \begin{align} \int_{-\pi}^\pi f(t)T_N(t)dt =& \int_{-\pi}^\pi f(t) \bigg(A_0+\sum_{k=1}^N A_k \cos(kt)+B_k\sin(kt)\bigg)dt \\ =& A_0\int_{-\pi}^\pi f(t)dt + \sum_{k=1}^N A_k \int_{-\pi}^\pi f(t)\cos(kt)dt + B_k \int_{-\pi}^\pi f(t)\sin(kt)dt. \end{align} Recall that $a_0=\frac{1}{2\pi}\int_{-\pi}^\pi f(t)dt$, and for $k\ge 1$, $a_k = \frac{1}{\pi}\int_{-\pi}^\pi f(t)\cos(kt)dt$. Similarly, $b_k = \frac{1}{\pi}\int_{-\pi}^\pi f(t)\sin(kt)dt$. Using these in the previous equation results in $(\ast)$. Since $f$ is arbitrary, we may replace it by $T_N$. Doing so in $(\ast)$ results in the formulas below: $(\ast \ast)\quad \int_{-\pi}^\pi T_N(t)^2dt =2\pi A_0^2+ \pi \sum_{k=1}^N A_k^2+B_k^2.$ We can now compute the residual energy defined in $(\dagger)$. \begin{align} \int_{-\pi}^\pi(f(t)-T_N(t))^2dt =& \int_{-\pi}^\pi f(t)^2dt - 2\int_{-\pi}^\pi f(t)T_N(t)dt + \int_{-\pi}^\pi T_N(t))^2dt \\ =& \int_{-\pi}^\pi f(t)^2dt -4\pi a_0A_0 - 2\pi \sum_{k=1}^N \big(a_k A_k+b_k B_k\big) + 2\pi A_0^2+ \pi \sum_{k=1}^N\big( A_k^2+B_k^2)\\. \end{align} Next, complete the square for the terms on the right in the last line above: $2\pi(A_0^2 -2a_0 A_0) = 2\pi (A_0-a_0)^2 - 2\pi a_0^2$, $\pi (A_k^2 -A_k) = \pi (A_k-a_k)^2 - \pi a_k^2$, and finally $\pi (B_k^2 -b_k) = \pi (B_k-b_k)^2 - \pi b_k^2$. The result is $\int_{-\pi}^\pi(f(t)-T_N(t))^2dt = \int_{-\pi}^\pi f(t)^2dt + 2\pi (A_0^2-a_0)^2+ \pi \sum_{k=1}^N \big (A_k-a_k)^2 + \pi (B_k-b_k)^2\big) - 2\pi a_0^2 -\pi \sum_{k=1}^N (a_k^2+b_k^2)$ The middle term will be strictly positive if $A_k\ne a_k$ and $B_k\ne b_k$, even for one $k$; consequently, if $T_N(t)\ne S_N(t)$, $\int_{-\pi}^\pi(f(t)-T_N(t))^2dt > \int_{-\pi}^\pi f(t)^2dt -2\pi a_0^2 -\pi \sum_{k=1}^N (a_k^2+b_k^2)= \int_{-\pi}^\pi (f(t)-S_N(t))^2dt$ Thus the minimum occurs exactly when $T_N(t)=S_N(t)$. Moreover, the minimizer is unique; that is,only the partial sum $S_N$ is the minimizer. Using the two terms on the right above, we see that the minimum residual energy is given by $(\ddagger)\quad \int_{-\pi}^\pi(f(t)-S_N(t))^2dt = \int_{-\pi}^\pi f(t)^2dt -2\pi a_0^2 -\pi \sum_{k=1}^N (a_k^2+b_k^2) = \int_{-\pi}^\pi f(t)^2dt - \int_{-\pi}^\pi S_N(t)^2dt$

## Convergence in the mean

We say that the partial sum $S_N$ converges to $f$ in the mean if and only if $\lim_{N\to \infty}\int_{-\pi}^\pi(F(t)- S_N(t)^2dt =0$. Using the expression for the minimal energy in $(\ddagger)$ we have the following proposition.

Theorem (Parseval) The partial sum $S_N$ converges in the mean to $f$ if and only if Parseval's equation holds: $(\S) \quad \int_{-\pi}^\pi f(t)^2dt = 2\pi a_0^2 +\pi \sum_{k=1}^\infty (a_k^2+b_k^2)$

Proof By $(\ddagger)$, if $S_N$ converges in the mean to $f$, then $0=\lim_{N\to \infty} \int_{-\pi}^\pi(f(t)-S_N(t))^2dt =\int_{-\pi}^\pi f(t)^2dt -2\pi a_0^2 -\pi \lim_{N\to \infty}\sum_{k=1}^N (a_k^2+b_k^2),$ so Parseval's equation holds. Conversely, if $2\pi a_0^2 +\pi \sum_{k=1}^\infty (a_k^2+b_k^2) = \int_{-\pi}^\pi f(t)^2dt$, then, again by $(\ddagger)$, $S_N$ converges to $f$ in the mean.

Theorem (Riesz-Fischer) Suppose that $f$ is in $L^2[-\pi,\pi]$. The $S_N$ converges in the mean to $f$. Conversely, suppose that the sequence $\{a_0, a_1, b_1, a_2,b_2, \ldots a_k,b_k \dots\}$ satisfies $2\pi a_0^2 +\pi \sum_{k=1}^\infty (a_k^2+b_k^2)<\infty$, then there exists a unique $f$ in $L^2[-\pi,\pi]$ such that $S_N(t)= a_0+\sum_{k=1}^N a_k \cos(kt)+b_k\sin(kt)$ converges to $f$ in the mean.

The proof of this theorem is beyond the scope of this course. To describe what it says we need to know the meaning of Parseval's equation $(\S)$. Remember that the left side is interpreted as the energy in the signal $f$. But what about the right side? If consider a single term of the form $a_k \cos(kt)$, then the energy in this term $2\pi a_0^2$, for $k=0$, and $\pi a_k^2$ for $k\ge 1$. The same is true for the $b_k$'s. The energy in $b_k\sin(kt)$ is $\pi b_k^2$. The point is that Parseval's equation $(\S)$ represents conservation of energy. The energy in the signal is the sum of the energies in the individual frequencies.

The Riesz-Fischer Theorem has an interesting interpretation. The first is that for any finite energy signal, the energy in the sum of its frequencies components equals the energy in the signal, as mentioned above. But the reverse is also true. Given an sequence of $a$'s and $b$'s such that the sum of the energies in the individual frequencies is finite, then there is a unique signal $f$, with finite energy, having the $a$'s and $b$'s from the sequence as its Fourier series.

Let's think of this in another way. Each of the terms $a_k\cos(kt)$ or $b_k\sin(kt)$ represents a signal with pure (angular) frequency $k$. The amplitude of each of these, $|a_k|$ or $|b_k|$ represents the intensity of a frequency — for example, how loud the sound corresponding to the frequency component is. What the Riesz-Fischer Theorem implies is that for any sequence of $a$'s and $b$'s, as long as the sum $2\pi a_0^2 +\pi \sum_{k=1}^\infty (a_k^2+b_k^2)<\infty$, there will be a corresponding finite energy signal. Put another way, you can synthesize a signal with arbitrary frequencies in it, provided the total energy in the components is finite.


text:
My Math Forum Physics Myth or Reality?

 Physics Physics Forum

 May 18th, 2018, 05:50 AM #1 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 Physics Myth or Reality? Hi everyone, if you would like to take a look here is my take of a still unsettled Physics question, has anyone successfully performed the experiment?
 May 18th, 2018, 06:37 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Is this what you mean? Thanks from rudimt
 May 18th, 2018, 02:16 PM #3 Senior Member   Joined: Oct 2009 Posts: 406 Thanks: 140 Seriously, drop the letters appearing one by one. It is extremely annoying.
May 18th, 2018, 02:59 PM   #4
Senior Member

Joined: Aug 2012

Posts: 1,919
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Quote:
 Originally Posted by rudimt Hi everyone, if you would like to take a look here is my take of a still unsettled Physics question ...
Didn't look at your vid and I'm about to go off topic ... but the title Physics, Myth or Reality is a wonderful question of philsophy.

We casually called the laws of physics "The laws of the universe." But that actually can't be true, since the laws of physics are historically contingent. Aristotle, Newton, and Einstein all had theories of gravity. The former two are already falsified, and there's no reason to believe relativity is the last word on the subject; especially since we still don't know, after a century of trying, how to reconcile gravity with electromagnetism and quantum theory.

As Newton famously said, "I frame no hypthotheses." That quote (which I won't discuss in detail today) is Newton's admission that he does not know what gravity is; only how it behaves. Newton understand that science is descriptive and not explanatory. Too many contemporary thinkers fail to appreciate this point.

Quote:
 Originally Posted by Newton I have not as yet been able to discover the reason for these properties of gravity from phenomena, and I do not feign hypotheses. For whatever is not deduced from the phenomena must be called a hypothesis; and hypotheses, whether metaphysical or physical, or based on occult qualities, or mechanical, have no place in experimental philosophy. In this philosophy particular propositions are inferred from the phenomena, and afterwards rendered general by induction.
https://en.wikipedia.org/wiki/Hypotheses_non_fingo

And here's Feynman, right from the first volume of his famous Lectures on Physics.

Quote:
 Originally Posted by Feynman Each piece, or part, of the whole of nature is always merely an approximation to the complete truth, or the complete truth so far as we know it. In fact, everything we know is only some kind of approximation, because we know that we do not know all the laws as yet. Therefore, things must be learned only to be unlearned again or, more likely, to be corrected.
[Italics as in the original]

The Feynman Lectures on Physics Vol. I Ch. 1: Atoms in Motion

So when I see the phrase, "Physics, Myth or Reality," this is what pops into my head. And every time I hear some really smart and trendy scientist saying, the world IS a computer, or the world IS quantum, or the world IS any way at all; I know they are not doing science, but rather metaphysics. And credulous reporters tell a credulous public that scientists say the world IS this or IS that; when in fact science by definition is incapable of making such a statement. No wonder there's a postmodern reaction against rationality. The scientists have oversold their case and written checks they can't cash.

Let's get back to the genius and wisdom and deep understanding of the limitations of science as understood by minds like Newton and Feynman.

Well thanks for reading. It was either this or trying to explain quotients to Zylo, so I went with a little philosophy today instead.

Last edited by Maschke; May 18th, 2018 at 03:09 PM.

 May 18th, 2018, 11:18 PM #5 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 That's the effect indeed.
 May 21st, 2018, 05:44 AM #6 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,111 Thanks: 706 Math Focus: Physics, mathematical modelling, numerical and computational solutions The Coriolis effect is much, much too weak to affect things like vortices of water in a sink compared to, say the local hydrodynamics. It's a well known con at the equator to get people to bet that the direction of the swirling water will change by stepping past the equator.
 May 23rd, 2018, 05:37 AM #7 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 There are different opinions on this topic, the main problem is that the size of the effect is extremely small, yet, if you do a properly design experiment you could detect it in a properly chosen sink. The video shows you how small the effect really is, a tiny fraction of g. Any small perturbation to the system would easily mask the effect.
May 24th, 2018, 01:23 AM   #8
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From: Glasgow

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Quote:
 Originally Posted by rudimt There are different opinions on this topic,
There are always different opinions on a topic. However, if you look at the opinions that matter, (i.e. the literature), the Coriolis effect is well understood and is not an open problem.

Quote:
 the main problem is that the size of the effect is extremely small, yet, if you do a properly design experiment you could detect it in a properly chosen sink. The video shows you how small the effect really is, a tiny fraction of g. Any small perturbation to the system would easily mask the effect.
I'm sure that a sensitive enough experiment could detect it at those scales, but that's besides the point. The Rossby numbers are very large at those scales and therefore a Coriolis correction term is unimportant. Wikipedia gives a good explanation and two good examples:

---------------------------------------------------------------------------------
Length scales and the Rossby number
---------------------------------------------------------------------------------

The time, space and velocity scales are important in determining the importance of the Coriolis force. Whether rotation is important in a system can be determined by its Rossby number, which is the ratio of the velocity, U, of a system to the product of the Coriolis parameter, $\displaystyle f=2\omega \sin \varphi$ and the length scale, L, of the motion:

$\displaystyle Ro=\frac{U}{fL}$
The Rossby number is the ratio of inertial to Coriolis forces. A small Rossby number indicates a system is strongly affected by Coriolis forces, and a large Rossby number indicates a system in which inertial forces dominate. For example, in tornadoes, the Rossby number is large, in low-pressure systems it is low, and in oceanic systems it is around 1. As a result, in tornadoes the Coriolis force is negligible, and balance is between pressure and centrifugal forces. In low-pressure systems, centrifugal force is negligible and balance is between Coriolis and pressure forces. In the oceans all three forces are comparable.[26]

An atmospheric system moving at U = 10 m/s (22 mph) occupying a spatial distance of L = 1,000 km (621 mi), has a Rossby number of approximately 0.1.

A baseball pitcher may throw the ball at U = 45 m/s (100 mph) for a distance of L = 18.3 m (60 ft). The Rossby number in this case would be 32,000.

Baseball players don't care about which hemisphere they're playing in. However, an unguided missile obeys exactly the same physics as a baseball, but can travel far enough and be in the air long enough to experience the effect of Coriolis force. Long-range shells in the Northern Hemisphere landed close to, but to the right of, where they were aimed until this was noted. (Those fired in the Southern Hemisphere landed to the left.) In fact, it was this effect that first got the attention of Coriolis himself.

Last edited by Benit13; May 24th, 2018 at 01:31 AM. Reason: errors XD

 May 26th, 2018, 09:28 PM #9 Member   Joined: Oct 2017 From: Japan Posts: 48 Thanks: 2 The effect is well established as far as the equations but is still often misinterpreted. There are still Physics textbooks with a completely misleading explanation, the effect has also been wrongly featured in several tv shows. You are raising interesting points, I will link an article later on.
May 29th, 2018, 01:18 AM   #10
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Joined: Apr 2014
From: Glasgow

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Quote:
 Originally Posted by rudimt The effect is well established as far as the equations but is still often misinterpreted. There are still Physics textbooks with a completely misleading explanation, the effect has also been wrongly featured in several tv shows.
Unfortunately, this is all too common, not just with physics but with a lot of subjects. I wonder how much stuff on TV is actually true these days.

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