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# NIELIT 2017 July Scientist B - Section A: 51

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Choose the most appropriate words from the options given below to complete the following sentence ;
Female sparrows and immatures are _____, while the typical adult male sparrow is _____ by its markings; a black bib, a grey cap, and white lines trailing down from the mouth.

1. somewhat nondescript, easily recognized
2. difficult to spot, better camouflaged
3. considered to be endangered, characterized
4. comparatively small, made more conspicuous

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Female sparrows and immatures are somewhat nondescript, while the typical adult male sparrow is easily recognized by its markings; a black bib, a grey cap, and white lines trailing down from the mouth.

This sentence compares and contrasts the observable characteristics of the female and male sparrow. It can be inferred by the contrasting phrases in Option A.

In Option B, “difficult to spot, better camouflaged” both mean the same thing, and there is no contrast.

As Option C states, there is no data to support the fact that female sparrows are “considered to be endangered”, so this is INCORRECT.

Option D would mean that female sparrows are “comparatively small”. There is no evidence, within the given sentence, to support this statement about size. The only characteristics mentioned are markings (a black bib, a grey cap, and white lines trailing down from the mouth).

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A Huge majority - $84$ percent - of the population identifies as Hindu. There are many variations of Hinduism, and four predominant sects - Shiva, Vaishnava, Shakteya and Smarta. About $13$ percent of Indians are Muslim, making it one of the largest ... a multilingual country. India is identified as the birthplace of Hinduism and Buddhism. India is a country where diversity in religion exists.

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# Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

You can Download Samacheer Kalvi 9th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Additional Questions

Exercise 3.1

Question 1.
Classify the following polynomials based on number of terms.
(i) x3 – x2
(ii) 5x
(iii) 4x4 + 2x3 + 1
(iv) 4.x3
(v) x + 2
(vi) 3x2
(vii) y4 + 1
(viii) y20 + y18 + y2
(ix) 6
(x) 2u3 + u2 + 3
(xi) u23 – u4
(xii) y
Solution:
5x, 3x2, 4x3, y and 6 are monomials because they have only one term.
x3 – x2, x + 2, y4 + 1 and u23 – u4 are binomials as they contain only two terms.
4x4 + 2x3 + 1 , y20 + y18 + y2 and 2u3 + u2 + 3 are trinomials as they contain only three terms.

Question 2.
Classify the following polynomials based on their degree.
Solution:
p(x) = 3, p(x) = -7, p(x) = $$\frac{3}{2}$$ are constant polynomials
p(x) = x + 3, p(x) = 4x, p(x) = $$\sqrt{3}$$x + 1 are linear polynomials, since the highest degree of the variable x is one.
p(x) = 5x2 – 3x + 2, p(y) = $$\frac{5}{2}$$ y2 + 1, p(x) = 3x2 are quadratic polynomials, since the highest degree of the variable is two.
p(x) = 2x3 – x2 + 4x + 1, p(x) = x3 + 1, p(y) = y3 + 3y are cubic polynomials, since the highest degree of the variable is three.

Question 3.
Find the product of given polynomials p(x) = 3x3+ 2x – x2 + 8 and q (x) = 7x + 2.

Solution:
(7x +2) (3x3 + 2x – x2 + 8) = 7x(3x3 + 2x – x2 + 8) + 2x
(3x3 + 2x – x2 + 8) = 21x4 + 14x2 – 7x3 + 56x + 6x3 + 4x – 2x2+ 16 = 21x4 – x3 + 12x4 + 60x + 16

Question 4.
Let P(x) = 4x4 – 3x + 2x3 + 5 and q(x) = x2 + 2x + 4 find p (x) – q(x).

Solution:

Exercise 3.2

Question 1.
If p(x) = 5x3 – 3x2 + 7x – 9, find
(i) p(-1)
(ii) p(2).
Solution:
Given that p(x) = 5x3 – 3x2 + 7x – 9
(i) p(-1) = 5(-1)3 – 3(-1)2 + 7(-1) – 9 = -5 – 3 – 7 – 9
= -24
(ii) p(2) = 5(2)3 – 3(2)3 + 7(2) – 9 = 40 – 12 + 14 – 9
∴ p(2) = 33

Question 2.
Find the zeros of the following polynomials.
(i) p(x) = 2x – 3
(ii) p(x) = x – 2
Solution:

Exercise 3.3

Question 1.
Find the remainder using remainder theorem, when
(i) 4x3 – 5x2 + 6x – 2 is divided by x – 1.
(ii) x3 – 7x2 – x + 6 is divided by x + 2.
Solution:
(i) Let p(x) = 4x3 – 5x2 + 6x – 2. The zero of (x – 1) is 1.
When p(x) is divided by (x – 1) the remainder is p( 1). Now,
p(1) = 4(1)3 – 5(1)2 + 6(1) – 2 = 4 – 5 + 6 – 2 = 3
∴ The remainder is 3.

(ii) Let p(x) = x3 – 7x2 – x + 6. The zero of x + 2 is -2.
When p(x) is divided by x + 2, the remainder is p(-2). Now,
p(-2) = (-2)3 – 7(-2)2 – (-2) +6 = – 8 – 7(4) + 2 + 6
= – 8 – 28 + 2 + 6 = -28.
∴ The remainder is -28.

Question 2.
Find the value of a if 2x3 – 6x2 + 5ax – 9 leaves the remainder 13 when it is divided by x – 2.
Solution:
Let p(x) = 2x3 – 6x2 + 5ax – 9
When p(x) is divided by (x – 2) the remainder is p(2).
Given that p(2) = 13
⇒ 2(2)3 – 6(2)2 + 5a(2) – 9 = 13
⇒ 2(8) – 6(4) + 10a – 9 = 13
⇒ 16 – 24 + 10a – 9 = 13
⇒ 10a – 17 = 13
⇒ 10a = 30
⇒ ∴ a = 3

Question 3.
If the polynomials f(x) = ax3+ 4x2 + 3x – 4 and g (x) = x3 – 4x + a leave the same remainder when divided by x – 3. Find the value of a. Also find the remainder.

Solution:
Let f(x) = ax3 + 4x2 + 3x – 4 and g(x) = x3 – 4x + a,When f(x) is divided by (x – 3), the remainder is f(3).
Now f(3) = a(3)34(3)2 + 3(3) – 4 = 27a + 36 + 9 – 4
f (3) = 27a + 41 …(1)
When g(x) is divided by (x – 3), the remainder is g(3).
Now g(3) = 33 – 4(3) + a = 27 – 12 + a = 15 + a …(2)
Since the remainders are same, (1) = (2)
Given that, f(3) = g(3)
That is 27a + 41 = 15 +a
27a – a = 15 – 41
26a = -26

Sustituting a = -1, in f(3), we get
f(3) = 27(-1) + 41 = -27 + 41
f(3) = 14
∴ The remainder is 14.

Question 4.
Show that x + 4 is a factor of x3 + 6x2 – 7x – 60.
Solution:

Question 5.
In (5x + 4) a factor of 5x3 + 14x2 – 32x – 32
Solution:

Question 6.
Find the value of k, if (x – 3) is a factor of polynomial x3 – 9x2 + 26x + k.
Solution:
Let p (x) = x3 – 9x2 + 26x + k
By factor theorem, (x – 3) is a factor of p(x), if p(3) = 0
p(3) = 0
33 – 9(3)2 + 26 (3) + k = 0
27 – 81 + 78 + k = 0
k = -24

Question 7.
Show that (x – 3) is a factor of x3 + 9x2 – x – 105.
Solution:
Let p(x) = x3 + 9x2 – x – 105
By factor theorem, x – 3 is a factor of p(x), if p(3) = 0
p (3) = 33 + 9(3)3 – 3 – 105
= 27 + 81 – 3 – 105
= 108 – 108
p(3) = 0

Therefore, x – 3 is a factor of x3 + 9x2 – x – 105.

Question 8.
Show that (x + 2) is a factor of x3 – 4x2 – 2x + 20.
Solution:
Let p(x) = x3 – 4x2 – 2x + 20
By factor theorem,
(x + 2) is factor of p(x), if p(-2) = 0
p(-2) = (-2)3 – 4(-2)2 – 2(-2) + 20 = -8 – 4(4) + 4 + 20
P(- 2) = 0
Therefore, (x + 2) is a factor of x3 – 4x2 – 2x + 20

Exercise 3.4

Question 1.
Expand the following using identities :
(i) (7x + 2y)2
(ii) (4m – 3m)2
(iii) (4a + 3b) (4a – 3b)
(iv) (k + 2)(k – 3)
Solution:
(i) (7x + 2y)2 = (7x)2 + 2 (7x) (2y) + (2y)2 = 49x2 + 28xy + 4y2
(ii) (4m – 3m)2 = (4m)2 – 2(4m) (3m) + (3m)2 = 16m2 – 24mn + 9n2
(iii) (4a + 3b) (4a – 3b) [We have (a + b ) (a – b) = a2 – b2 ]
Put [a = 4a, b = 3b]
(4a + 3b) (4a – 3b) = (4a)2 – (3b)2 = 16a2 – 9b2
(iv) (k + 2)(k – 3) [We have (x + a) (x – b) = x2 + (a – b)x – ab]
Put [x = k, a = 2, b = 3]
(k + 2) (k – 3) = k2 + (2 – 3)x – 2 × 3 = k2 – x – 6

Question 2.
Expand : (a + b – c)2
Solution:
Replacing ‘c’ by ‘-c’ in the expansion of
(a + b + c)2 = a2 + b2 + c2 + 2 ab+ 2 bc + 2ca
(a + b + (-c))2 = a2 + b2 + (-c)2 + 2ab + 2b(-c) + 2(-c)a
= a2 + b2 + c2 + 2ab – 2bc – 2ca

Question 3.
Expand : (x + 2y + 3z)2
Solution:
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
Substituting, a = x, b = 2y and c = 3z
(x + 2y + 3 z)2 = x2 + (2y)2 + (3z)2 + 2(x)(2y) + 2(2y)(3z) + 2(3z)(x)
= x2 + 4y2 + 9z2 + 4xy + 12y2 + 6zx

Question 4.
Find the area of square whose side length is m + n – q.
Solution:
Area of square = side × side = (m + n – q)2
We know that,
(a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
[m + n + (-q)]2 = m2 + n2 + (-q)2 + 2mn + 2n(-q) + 2(-q)m
= m2 + n2 + q2 + 2mn – 2nq – 2qm
Therefore, Area of square = m2 + n2 + q2 + 2mn – 2nq – 2qm = [m2 + n2 + q2 + 2mn – 2nq – 2qm] sq. units.

EXERCISE 3.5

Question 1.
Factorise the following
(i) 25m-2 – 16n2
(ii) x4 – 9x2
Solution:
(i) 25m2 – 16n2 = (5m)2 – (4n)2
= (5m – 4n) (5m + 4n) [∵ a2 – b2 = (a – b)(a + b)]
(ii) x4 – 9x2 = x2(x2 – 9) = x2(x2 – 32) = x2(x – 3)(x + 3)

Question 2.
Factorise the following.
(i) 64m3 + 27n3
(ii) 729x3 – 343y3
Solution:
(i) 64m3 + 27n3 = (4m)3 + (3n)3
= (4m + 3n) ((4m)2 – (4m) (3n) + (3n)2) [∵ a2 + b2 = (a + b) (a2 – ab + b2)]
= (4m + 3n) (16m2 – 12mn + 9n2)
(ii) 729x3 – 343y3 = (9x)3 – (7y)3 = (9x – 7y) ((9x)2 + (9x) (7y) + (7y)2
= (9x – 7y) (81x2 + 63xy + 49y2)
[∵ a3 – b3 = (a – b) (a2 + ab + b2)]

Question 3.
Factorise 81x2 + 90x + 25
Solution:
81x2 + 90x + 25 = (9x)2 + 2 (9x) (5) + 52
= (9x + 5)2 [ ∵ a2 + 2ab + b2 = (a + b)2]

Question 4.
Find a3 + b2 if a + b = 6, ab = 5.
Solution:
Given, a + b = 6, ab = 5
a2 + b2 = (a + b)2 – 3ab(a + b) = (6)3 – 3(5)(6) = 126

Question 5.
Factorise (a – b)2 + 7(a – b) + 10
Solution:
Let a – b = p, we get p2 + 7p + 10,

p2 + 7p + 10 = p2 + 5p + 2p + 10
= p(p + 5) + 2(p + 5) = (p + 5)(p + 2)
Put p = a – b we get,
(a – b)2 + 7(a – b) + 10 = (a – b + 5)(a – b + 2)

Question 6.
Factorise 6x2 + 17x + 12
Solution:

Exercise 3.6

Question 1.
Factorise x2 + 4x – 96
Solution:

Question 2.
Factorise x2 + 7xy – 12y2
Solution:

Question 3.
Find the quotient and remainder when 5x3 – 9x2 + 10x + 2 is divided by x + 2 using synthetic division.
Solution:

Hence the quotient is 5x2 – 19x + 48 and remainder is -94.

Question 4.
Find the quotient and remainder when -7 + 3x – 2x2 + 5x3 is divided by -1 + 4x using synthetic division.
Solution:

Question 5.
If the quotient on dividing 5x4 + 4x3 + 2x + 1 by x + 3 is 5x3 + 9x2 + bx – 97 then find the values of a, b and also remainder.
Solution:

Quotient 5x3 + 11x2 + 33x – 97 is compared with given quotient 5x3 + ax2 + bx – 97
co-efficient of x2 is – 11 = a and co-efficient of x is 33 = b.
Therefore a = -11, b = 33 and remainder = 292.

Exercise 3.7

Question 1.
Find the quotient and the remainder when 10 – 4x + 3x2 is divided by x – 2.
Solution:
Let us first write the terms of each polynomial in descending order (or ascending
order). Thus the given problem becomes (3x2 – 4x + 10) ÷ (x – 2)

∴ Quotient = 3x + 2 and Remainder = 14, i.e 3x2 – 4x + 10 = (x – 2) (3x + 2) + 14 and is in the form
Dividend = (Divisor × Quotient) + Remainder

Question 2.
If 8x3 – 14x2 – 19x – 8 is divided by 4x + 3 then find the quotient and the remainder.
Solution:

Exercise 3.8

Question 1.
Factorise 2x3 – x2 – 12x – 9 into linear factors.
Solution:
Let p(x) = 2x3 – x2 – 12x – 9
Sum of the co-efficients = 2 – 1 – 12 – 9 = -20 ≠ 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = -1 – 9 = -10
Sum of co-efficients of odd powers = 2 – 12 = -10
Hence x + 1 is a factor of x.
Now we use synthetic division to find the other factors.

Then p (x) = (x + 1)(2x2 – 3x – 9)
Now 2x2 – 3x – 9 = 2x2 – 6x + 3x – 9 = 2x(x – 3) + 3(x – 3)
= (x – 3)(2x + 3)
Hence 2x3 – x2 – 12x – 9 = (x + 1) (x – 3) (2x + 3)

Question 2.
Factorize x3 + 3x2 – 13x – 15.
Solution:
Let p(x) = x3 + 3x2 – 13x – 15
Sum of all the co-efficients = 1 + 3 – 13 – 15 = -24 + 0
Hence x – 1 is not a factor
Sum of co-efficients of even powers with constant = 3 – 15 = – 12
Sum of coefficients of odd powers = 1 – 13 = – 12
Hence x + 1 is a factor of p (x)
Now use synthetic division to find the other factors.

Hence p(x) = (x + 1) (x2 + 2x – 15)
Now x3 + 3x2 – 13x – 15 = (x + 1)(x2 + 2x – 15)
Now x2 + 2x – 15 = x2 + 5x – 3x – 15 = x(x + 5) – 3 (x + 5) = (x + 5)(x – 3)
Hence x3 + 3x2 – 13x – 15 = (x + 1)(x + 5)(x – 3)

Question 3.
Factorize x3 + 13x2 + 32x + 20 into linear factors.

Solution:
Let, p(x) = x3 + 13x2 + 32x + 20
Sum of all the coefficients = 1 + 13 + 32 + 20 = 66 ≠ 0
Hence, (x – 1) is not a factor.
Sum of coefficients of even powers with constant = 13 + 20 = 33
Sum of coefficients of odd powers = 1 + 32 = 33
Hence, (x + 1) is a factor of p(x)
Now we use synthetic division to find the other factors

Exercise 3.9

Question 1.
Find GCD of 25x3y2 z, 45x2y4z3b
Solution:

Question 2.
Find the GCD of (y3 – 1) and (y – 1).
Solution:
y3 – 1 = (y – 1)(y3 + y + 1)
y – 1 = y – 1
Therefore, GCD = y – 1

Question 3.
Find the GCD of 3x2 – 48 and x2 – 7x + 12.
Solution:
3x2 – 48 = 3(x2 – 16) = 3(x2 – 43) = 3(x + 4)(x – 4)
x2 – 7x + 12 = x2 – 3x – 4x + 12 = x(x – 3) – 4 (x – 3) = (x – 3) (x – 4)
Therefore, GCD = x – 4

Question 4.
Find the GCD of ax , ax + y, ax + y + z.
Solution:

Exercise 3.10

Question 5.
Solve graphically. x – y = 3 ; 2x – y = 11
Solution:

Exercise 3.11

Question 1.
Solve using the method of substitution. 5x – y = 5, 3x + y = 11
Solution:

Exercise 3.12

Question 2.
Solve by the method of elimination. 2x + y = 10; 5x – y = 11
Solution:

Exercise 3.13

Question 3.
Solve 5x – 2y = 10 ; 3x + y = 17 by the method of cross multiplication.
Solution:

Exercise 3.14

Question 1.
The age of Arjun is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Solution:
Let the age of the father be x
Let the sum of the age of his sons be y
At present x = 2y ⇒ x – 2y = 0 …. (1)
After 20 years x + 20 = y + 40
x – y = 40 – 20

Exercise 3.15

Multiple Choice Questions :

Question 1.
The polynomial 3x – 2 is a
(1) linear polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is linear if its degree is 1
Solution:
(1) linear polynomial

Question 2.
The polynomial 4x2 + 2x – 2 is a
(1) linear polynomial
(3) cubic polynomial
(4) constant polynomial
Hint: A polynomial is quadratic of its highest power of x is 2
Solution:

Question 3.
The zero of the polynomial 2x – 5 is

Hint:

Solution:
(1) $$\frac{5}{2}$$

Question 4.
The root of the polynomial equation 3x – 1 = 0 is

Hint:

Solution:
(2) x = $$\frac{1}{3}$$

Question 5.
Zero of (7 + 4x) is ___

Solution:
(2) $$\frac{-7}{4}$$

Question 6.
Which of the following has as a factor?
(1) x2 + 2x
(2) (x – 1)2
(3) (x + 1)2
(4) (x2 – 22)
Solution:
(1) x2 + 2x

Question 7.
If x – 2 is a factor of q (x), then the remainder is _____
(1) q (-2)
(2) x – 2
(3) 0
(4) -2
Solution:
(3) 0

Question 8.
(a – b) (a2 + ab + b2) =
Solution:
(1) a3 + b3 + c3 – 3abc
(2) a2 – b2
(3) a3 + b3
(4) a3 – b3
Solution:
(4) a3 – b3

Question 9.
The polynomial whose factors are (x + 2) (x + 3) is
(1) x2 + 5x + 6
(2) x2 – 4
(3) x2 – 9
(4) x2 + 6x + 5
Solution:
(1) x2 + 5x + 6

Question 10.
(-a – b – c)2 is equal to
(1) (a – b + c)2
(2) (a + b – c)2
(3) (-a + b + c)2
(4) (a + b + c)2
Solution:
(4) (a + b + c)2

Text Book Activities

Activity – 1
Write the Variable, Coefficient and Constant in the given algebraic expression,
Solution:

Activity – 2
Write the following polynomials in standard form.

Activity – 3

Question 1.
Find the value of k for the given system of linear equations satisfying the condition below:
(i) 2x + ky = 1; 3x – 5y = 7 has a unique solution
(ii) kx + 3y = 3; 12x + ky = 6 has no solution
(iii) (k – 3)x + 3y = k; kx + ky = 12 has infinite number of solution
Solution:

Question 2.
Find the value of a and b for which the given system of linear equation has infinite number of solutions 3x – (a + 1)y = 2b – 1, 5x + (1 – 2a)y = 3b
Solution:

Activity – 4
For the given linear equations, find another linear equation satisfying each of the given condition
Solution:


text:
Note: This is a working paper which will be expanded/updated frequently. One way to think about this paper is as an update of De Leeuw (2014), using simpler and more direct majorization methods. The directory deleeuwpdx.net/pubfolders/rstress has a pdf copy of this article, the complete Rmd file that includes all code chunks, and R files with the code.

# 1 Problem

Define the multidimensional scaling (MDS) loss function $$$\sigma_r(x)=\sum_{i=1}^nw_i(\delta_i-(x'A_ix)^r)^2,\label{E:rstress}$$$

with $$r>0$$ and the $$A_i$$ positive semi-definite. The $$w_i$$ are positive weights, the $$\delta_i$$ are non-negative dissimilarities. We call this rStress. Special cases are stress (Kruskal 1964) for $$r=\frac12$$, sstress (Takane, Young, and De Leeuw 1977) for $$r=1$$, and the loss function used in MULTISCALE (Ramsay 1977) for $$r\rightarrow 0$$.

Usually MDS is formulated uses Euclidean distances $$d_{j\ell}(X)$$ between points $$j$$ and $$\ell$$, which are rows of an $$n\times p$$ configuration matrix $$X$$. This fits into our formulation by setting $$x=\mathbf{vec}(X)$$ and by setting the $$A_i$$ to $$np\times np$$ matrices of the form $$I_p\otimes E_{j\ell}$$, where the matrix $$E_{j\ell}$$ has elements $$(j,j)$$ and $$(\ell,\ell)$$ equal to $$+1$$ and elements $$(j,\ell)$$ and $$(\ell,j)$$ equal to $$-1$$. Thus the $$A_i$$ are direct sums of $$p$$ copies of the $$E_{j\ell}$$. Now $$x'A_ix=d_{j\ell}^2(X)$$.

The problem we are trying to solve is to find an algorithm to minimize $$\sigma_r$$ over $$x$$ in $$\mathbb{R}^m$$ for all values of $$r>0$$. It is understood that parts of the algorithm may be different for different values of $$r$$.

# 2 Minorization and Majorization

We will design a convergent iterative majorization algorithm. Briefly this means constructing for each $$r>0$$ a majorization function $$\gamma_r$$ on $$\mathbb{R}^m\otimes\mathbb{R}^m$$ such that $$\sigma_r(x)\leq\gamma_r(x,y)$$ for all $$x$$ and $$y$$ and such that $$\sigma_r(x)=\gamma_r(x,x)$$ for all $$x$$. One step of the iterative algorithm is $x^{(k+1)}=\text{arg}\min_x\gamma_r(x,x^{(k)}),$ unless $$x^{(k)}$$ already minimizes $$\gamma_r(x,x^{(k)})$$, in which case we stop. If we do not stop we have the sandwich inequality $\sigma_r(x^{(k+1)})\leq\gamma_r(x^{(k+1)},x^{(k)})<\gamma_r(x^{(k)},x^{(k)})=\sigma_r(x^{(k)}).$ Thus majorization algorithms exhibit monotone convergence of loss function values.

The history of majorization algorithms is complicated. They were first developed in the specific contexts of step-size estimation in descent algorothms (Ortega and Rheinboldt 1970), maximum likelihood estimation with missing data (Dempster, Laird, and Rubin 1977), and multidimensional scaling (De Leeuw 1977). Subsequently they were presented as a general class of optimization methods, and as a special case of block relaxation and augmentation methods, by De Leeuw (1994), see also Heiser (1995). The material in De Leeuw (1994) is (slowly, slowly) expanded into an e-book, with one part on block relaxation, augmentation and alternating least squares (“Homogeneity Analysis” 2015), one part on majorization (De Leeuw 2015a), and one part on mathematical background (De Leeuw 2015b). There have been many applicatons of the general majorization principle by The Dutch School in psychometrics, which includes Ten Berge, Kiers, Heiser, Meulman, Groenen, and Vermunt.

Systematic use of majorization in statistics started with Lange, Hunter, and Yang (2000). There have been many further reviews, developments, and applications by Ken Lange and his co-workers. Theyo use the name MM Algorithms, where MM stands for either majorization-minimization or minorization-maximization. Lange (2013) has a very nice chapter on the MM algorithm, and he is currently working on an MM methods book.

# 3 Use of Homogeneity

We start by exploring the obvious fact that $\min_x\sigma_r(x)=\min_{\theta\geq 0}\min_{x'x=1}\sigma(\theta x).$ Let us introduce some notation to simplify the discussion. Without loss of generality we assume the dissimilarities are scaled by $\sum_{i=1}^nw_i\delta_i^2=1.$ Next, it is convenient to define $\rho_r(x):=\sum_{i=1}^nw_i\delta_i(x'A_ix)^r,$ and $\eta_r(x):=\sum_{i=1}^nw_i(x'A_ix)^{2r},$ so that $$$\sigma_r(x)=1-2\rho_r(x)+\eta_r(x).\label{E:expand}$$$ Now $$$\sigma_r(\theta x)=1-2\theta^{2r}\rho_r(x)+\theta^{4r}\eta_r(x).\label{E:homo}$$$ Thus, as in De Leeuw (1977), thinking of rStress as a function of both $$x$$ and $$\theta$$, $$$\min_\theta\sigma_r(\theta, x)=1-\frac{\rho_r^2(x)}{\eta_r(x)},\label{E:theta}$$$ where the minimium is attained at $$$\theta^{2r}=\frac{\rho_r(x)}{\eta_r(x)}.\label{E:optt}$$$

Thus minimizing $$\sigma_r$$ can be done by maximizing the ratio in $$\eqref{E:theta}$$ over $$x'x=1$$. This is used in the majorization method proposed by De Leeuw (2014), by combining Dinkelbach (1967) and quadratic majorization.

The approach of De Leeuw (2014) is somewhat cumbersome, because we first use homogeneity to reduce the MDS problem to a fractional programming problem, and then we use Dinkelbach to get rid of the fractional objective funcion again. Moreover one of the special cases requires us to solve a modified eigenvalue problem of the type discussed, for example, by Hager (2001) in each iteration.

In this paper we combine majorization with alternating least squares, working directly with (3). Thus we think of rStress as a function of both $$\theta$$ and $$x$$, and we alternate minimization over $$\theta$$ for fixed $$x$$ and over $$x$$ for fixed $$\theta$$. Thus $\begin{split} \theta^{(k)}&=\text{arg}\min_\theta\sigma_r(\theta, x^{(k)}),\\ x^{(k+1)}&=\text{arg}\min_{x'x=1}\sigma_r(\theta^{(k)},x). \end{split}$ The first step is trivially solved using $$\eqref{E:optt}$$. Updating $$x$$, however, is far from trivial and needs the majorization machinery we will develop in this paper. Since updating $$x$$ is iterative, we have to choose how many majorization steps to take for updating $$x$$, before going to the next update for $$\theta$$.

Thus the main focus of the paper will be a majorization method for minimizing $$$\sigma_r(x,\alpha):=1-2\alpha\rho_r(x)+\alpha^2\eta_r(x)\label{E:lambda}$$$

over $$x'x=1$$ with $$\alpha:=\theta^{2r}$$ fixed at its current value. It is clear from $$\eqref{E:lambda}$$ that we can find a majorization of $$\sigma_r$$ by combining a minorization of $$\rho_r$$ and a majorization of $$\eta_r$$.

# 4 Powers of Quadratic Forms

We start with some lemmas we will need to construct our minorizations and majorizations.

Lemma 4.1: $$f_r(x):=(x'Ax)^r$$ is convex on $$x'Ax>0$$ if and only if $$r\geq\frac12$$.

Proof: The first and second derivative are $\mathcal{D}f_r(x)=2r(x'Ax)^{r-1}Ax,$ and $\mathcal{D}^2f_r(x)=2r(x'Ax)^{r-1}\left(A+2(r-1)\frac{Axx'A}{x'Ax}\right).$ The matrix $$H_r(x):= A+ 2(r-1)\frac{Axx'A}{x'Ax}$$ is psd for $$r=\frac12$$, and its eigenvalues increase with $$r$$. Thus it is psd for all $$r\geq\frac12$$.

Also, if $$0<r<\frac12$$ then, by Sylvester’s Law of Inertia, $$\mathcal{D}^2f_r(x)$$ has precisely one negative eigenvalue, as well as $$\mathbf{rank}(A)-1$$ positive eigenvalues, and $$n-\mathbf{rank}(A)$$ zero eigenvalues. Thus in this case $$f_r$$ is not convex (and not concave either). QED

By the way, we can use the definition of convexity to show lemma 4.1 is true for all $$x$$, not just for $$x$$ with $$x'Ax>0$$.

Lemma 4.2: If $$r\geq\frac12$$ then $f_r(x)\geq(1-2r)f_r(y)+2rf_{r-1}(y)y'Ax.$

Proof: Follows from the fact that $$f_r$$ is convex, i.e. above each of its tangents. QED

Now write $$\overline{\lambda}(X)$$ or $$\overline{\lambda}_X$$ for the largest eigenvalue of a matrix $$X$$, and $$\underline{\lambda}(X)$$ or $$\underline{\lambda}_X$$ for the smallest eigenvalue. Note that if $$A=I\otimes E_{j\ell}$$ then $$\overline{\lambda}_A=2$$.

Lemma 4.3: If $$r\geq 1$$ then $\overline{\lambda}(\mathcal{D}^2f_r(x))\leq 2r(2r-1)\overline{\lambda}^r_A\ (x'x)^{r-1}.$ If $$x'x\leq 1$$ then $\overline{\lambda}(\mathcal{D}^2f_r(x))\leq 2r(2r-1)\overline{\lambda}^r_A.$ Proof: If $$r\geq 1$$, then $u'H_r(x)u=u'Au+2(r-1)\frac{(u'Ax)^2}{x'Ax}\leq (2r-1)u'Au.$ Thus $\overline{\lambda}(H_r(x))\leq (2r-1)\overline{\lambda}_A,$ and $\overline{\lambda}(\mathcal{D}^2f_r(x))\leq 2r(2r-1)\overline{\lambda}_A(x'Ax)^{r-1} \leq 2r(2r-1)\overline{\lambda}^r_A\ (x'x)^{r-1}.$ Finally, if $$x'x\leq 1$$ then $$(x'x)^{r-1}\leq 1$$. QED

Lemma 4.4: If $$0<r\leq 1$$ then $f_r(x)\leq (1-r)f_r(y)+rf_{r-1}(y)x'Ax.$

Proof: If $$r\leq 1$$ then $$(x'Ax)^r$$ is concave in $$x'Ax$$, although not in $$x$$. Thus $$f_r$$ is below its tangents, and $f_r(x)\leq f_r(y)+r(y'Ay)^{r-1}(x'Ax-y'Ay),$ which simplifies to the required result. QED

Lemma 4.5: If $$0<r\leq 1$$ and $$x'x=y'y=1$$ then $f_r(x)\leq(1-2r)f_r(y)+2rf_{r-1}(y)(x'(A-\overline{\lambda}_A I)y+\overline{\lambda}_A).$

Proof: From $x'Ax=y'Ay+2y'A(x-y)+(x-y)'A(x-y)\leq -y'Ay+2y'Ax+\overline{\lambda}_A(x-y)'(x-y)$ we see that if $$x'x=y'y=1$$ that $x'Ax\leq 2x'(A-\overline{\lambda}_A I)y+2\overline{\lambda}_A-y'Ay.$ Substitute this in lemma 4.4 and simplify. QED

Lemma 4.6: If $$0<r<\frac12$$ then $\underline{\lambda}(\mathcal{D}^2f_r(x))\geq 2r(2r-1)\overline{\lambda}^r_A(x'x)^{r-1}.$ If $$x'x\leq 1$$ then $\underline{\lambda}(\mathcal{D}^2f_r(x))\geq 2r(2r-1)\overline{\lambda}^r_A.$

Proof: We have $\frac{(u'Ax)^2}{x'Ax}\leq u'Au$ as before. Thus $u'H(x)u\geq(2r-1)u'Au\geq(2r-1)\overline{\lambda}_Au'u.$ The result follows because in addition $$x'Ax\leq\overline{\lambda}_A x'x$$, and consequently $$(x'Ax)^{r-1}\geq\overline{\lambda}_A^{r-1}(x'x)^{r-1}$$. Moreover if $$x'x\leq 1$$ then $$(x'x)^{r-1}\geq 1$$. QED

The following lemma, defining a type of uniform quadratic majorization (De Leeuw and Lange 2009), is an additional useful tool. Because we work on the unit sphere, the quadratic majorization actually becomes linear. This relies heavily on the fact that if $$x'x=y'y=1$$ and $$z$$ is on the segment $$[x,y]$$ connecting $$x$$ and $$y$$, then $$z'z\leq 1$$.

Lemma 4.7: Suppose $$\phi$$ is any homogeneous function of degree $$s$$, $$x'x=y'y=1$$ and $$\overline{\lambda}(\mathcal{D}^2\phi(z))\leq\kappa$$ for all $$z\in[x,y]$$. Then $\phi(x)\leq(1-s)\phi(y)+\kappa+x'(\mathcal{D}\phi(y)-\kappa y).$ In the same way, if $$\underline{\lambda}(\mathcal{D}^2\phi(z))\geq\kappa$$ for all $$z\in[x,y]$$ we have $\phi(x)\geq(1-s)\phi(y)+\kappa+x'(\mathcal{D}\phi(y)-\kappa y).$

Proof: We only prove the first part. The proof of the second part goes the same. By Taylor’s theorem we have for all $$x$$ and $$y$$ $\phi(x)\leq\phi(y)+(x-y)'\mathcal{D}\phi(y)+\frac12\kappa(x-y)'(x-y),$ which simplifies to the stated result if $$x'x=y'y=1$$ and $$\phi$$ is homogeneous of degree $$s$$. QED

# 5 Majorizing rStress

We will now use the inequalities from the lemmas in the previous section to derive linear minorizations of $$\sigma_r$$, by combining majorization of $$\eta_r$$ and minorization of $$\rho_r$$. Our majorizations are in the form $$f(x)\leq x'g(y)+\epsilon(y)$$, where both $$g$$ and $$\epsilon$$ are functions which do not depend on $$x$$, only on $$y$$. Since the precise form of $$\epsilon$$ has no influence on the iterative algorithm we usually leave it unspecified, so in different formulas $$\epsilon$$ can refer to different funcions. Clearly the majorization algorithm will have the form $$$x^{(k+1)}=-\frac{g(x^{(k)})}{\|g(x^{(k)})\|}.\label{E:major}$$$ For the two key results of this paper we need to define matrices $$$B_r(y):=\sum_{i=1}^n w_i\delta_i(y'A_iy)^{r-1}A_i,\label{E:bdef}$$$ and $$$C_r(y):=\sum_{i=1}^n w_i(y'A_iy)^{2r-1}A_i,\label{E:cdef}$$$ and scalars $$$\beta_r:=(2r-1)\sum_{i=1}^n w_i\delta_i\overline{\lambda}^r(A_i),\label{E:bedef}$$$ and $$$\gamma_r:=\sum_{i=1}^n w_i(y'A_iy)^{2r-1}\overline{\lambda}(A_i),\label{E:gadef}$$$ and $$$\delta_r:=(4r-1)\sum_{i=1}^n w_i\overline{\lambda}^{2r}(A_i).\label{E:dedef}$$$

Note that $$\delta_r=\beta_{2r}$$. Also $$y'B_r(y)y=\rho_r(y)$$ and $$y'C_r(y)y=\eta_r(y)$$.

Theorem 5.1: If $$0<r\leq\frac12$$ then $$\eqref{E:major}$$ with $$$g(y)=-\{(B_r(y)-\beta_rI)-\alpha(C_r(y)-\gamma_rI)\}y\label{E:gsmall}$$$

is a convergent majorization algorithm.

Proof: To majorize $$\eta_r$$ we use lemma 4.5. After some calculation we find the linear majorization $\eta_r(x)\leq 4rx'(C_r(y)-\gamma_rI)y+\epsilon(y),$ where $$\epsilon$$ is some function which only depends on $$y$$ and not on $$x$$.

To minorize $$\rho_r$$ we use lemma 4.7 with $$\phi$$ equal to $$f_r$$. Note that $$f_r$$ is homogeneous of order $$2r$$. Thus $(x'A_ix)^r\geq(1-2r)(y'A_iy)^r+\kappa_i+x'(2r(y'A_iy)^{r-1}A_i-\kappa_i I)y$ By lemma 4.6 we have $$\kappa_i=2r(2r-1)\overline{\lambda}^r(A_i)$$. Thus $\rho_r(x)\geq 2rx'(B_r(y)-\beta_r I)y + \epsilon(y).$

By combining the results for $$\rho_r$$ and $$\eta_r$$ we see that $\sigma_r(x)\leq 4\alpha rx'g(y)+\epsilon(y)$ with $$g(y)$$ defined by $$\eqref{E:gsmall}$$. QED

Theorem 5.2: If $$r\geq\frac12$$ then $$\eqref{E:major}$$ with $$$g(y):=-\{(B_r(y)-\alpha(C_r(y)-\delta_r I)\}y\label{E:glarge}$$$

is a convergent majorization algorithm.

Proof: We can use lemma 4.2 to get $\rho_r(x)\geq 2rx'B_r(y)y+\epsilon(y).$

Because $$\eta_r$$ is homogeneous of order $$4r$$ we see from lemma 4.7 that $\eta_r(x)\leq 4rx'(C_r(y)-\delta_r I)y+\epsilon(y).$

By combining the results for $$\rho_r$$ and $$\eta_r$$ we see that $\sigma_r(x)\leq 4r\alpha x'g(y)+\epsilon(y)$ with $$g(y)$$ defined by $$\eqref{E:glarge}$$. QED

# 6 Special Considerations

If $$r=\frac12$$ we have $$\beta_r=0$$, and $B_r(y)=\sum_{i=1}^n w_i\frac{\delta_i}{\sqrt{y'A_iy}}A_i,$ and $C_r(y)=\sum_{i=1}^n w_iA_i.$ These are precisely the matrices used in the SMACOF algorithm of De Leeuw (1977), implemented in De Leeuw and Mair (2009), which is $x^{(k+1)}=C_r^+(x^{(k)})B_r(x^{(k)})x^{(k)},$ where $$C_r^+$$ is the Moore-Penrose inverse. Also note that if $$r=\frac12$$ then $$\gamma_r=\delta_r$$ and the algorithms of theorems 5.1 and 5.2 are the same.

The case $$r\rightarrow 0$$ also requires some special attention. Define $\sigma_0(x):=\sum_{i=1}^nw_i(\log\delta_i-\log x'A_ix)^2.$ Note that $\sum_{i=1}^nw_i(\log\delta_i^r-\log (x'A_ix)^r)^2=r^2\sigma_0(x),$ which means that minimizing $$\sum_{i=1}^nw_i(\log\delta_i^r-\log (x'A_ix)^r)^2$$ for any $$r>0$$ can be done by minimizing $$\sigma_0$$.

Now use $\log x=\lim_{r\rightarrow 0}\frac{x^r-1}{r}.$ Thus, for small $$r$$, $\log\delta_{i}-\log x'A_ix\approx\frac{\delta_i^r-(x'A_ix)^r}{r},$ and $\sigma_0(x)\approx \frac{1}{r^2}\sum_{i=1}^n w_i(\delta_i^r-(x'A_ix)^r)^2.$ On the other hand, if $$\delta_i\approx x'A_ix$$ then for any $$r$$ we have the approximation $\log (x'A_ix)^r\approx\log\delta_i^r+\frac{1}{\delta_i^r}((x'A_ix)^r-\delta_i^r),$ so that $\sigma_0(x)\approx \sum_{i=1}^n \frac{w_i}{\delta_i^{2r}}(\delta_i^r-(x'A_ix)^r)^2.$ Both approximatations to $$\sigma_0$$ can be minimized by our iterative majorization algorithm for $$r<\frac12$$.

Note that our definition of rStress compares dissimilarities with powers of Euclidean distances. Alternatively, we can compare powers of dissimilarities with corresponding powers of Euclidean distances, i.e. we can define another rStress as $$$\tilde\sigma_r(x):=\sum_{i=1}^n w_i((\delta_i^2)^r-(x'A_ix)^r)^2.\label{E:tilde}$$$

This is similar to our approach in the limiting case $$r\rightarrow 0$$. We do not need an additional algorithm to minimize $$\tilde\sigma_r$$, because we can just input the $$(\delta_i^2)^r$$ instead of $$\delta_i$$ in our previous majorization method.

The difference between $$\sigma_r$$ and $$\tilde\sigma_r$$ disappears, of course, if we make the algorithm nonmetric. In that case we alternate minimization over $$x$$ with minimization over monotone transformations of $$\delta$$ (or of $$\delta^r$$, which is of course the same thing). We could easily add a monotone regression step to our alternating least squares algorithm.

# 7 Examples

## 7.1 Artificial

Let’s start with a small example that shows the flow of the algorithm. We perform one iteration and write out the intermediary results in detail.

Suppose $$n=3$$, the weights $$w$$ are all one, and the dissimilarities are 1, 2, 3. The matrices $$A$$ are

##      [,1] [,2] [,3] [,4]
## [1,]    1    1   -1   -1
## [2,]    1    1   -1   -1
## [3,]   -1   -1    1    1
## [4,]   -1   -1    1    1
##      [,1] [,2] [,3] [,4]
## [1,]    1   -1   -1    1
## [2,]   -1    1    1   -1
## [3,]   -1    1    1   -1
## [4,]    1   -1   -1    1
##      [,1] [,2] [,3] [,4]
## [1,]    1   -1    1   -1
## [2,]   -1    1   -1    1
## [3,]    1   -1    1   -1
## [4,]   -1    1   -1    1

The largest eigenvalues of all three $$A$$ matrices are equal to 4. Thus $$\delta_r$$ of $$\eqref{}$$ is 144.

Let’s start with $$x$$ four random normals. We want to minimize rStress for $$r=1$$, i.e. sStress.

## [1] 3.4514035 0.2191467 0.0485130

At the start sStress is 17.892093. Also $$\rho$$ is 4.0352358 and $$\eta$$ is 11.9625647. This means $$\alpha$$ is 0.337322 and the minimum of rStress over $$\alpha$$ is 12.6388263.

We now compute the matrices $$B_r$$ and $$C_r$$ at $$x$$.

For $$B_r$$ we have

##      [,1] [,2] [,3] [,4]
## [1,]    6   -4    0   -2
## [2,]   -4    6   -2    0
## [3,]    0   -2    6   -4
## [4,]   -2    0   -4    6

and for $$C_r$$

##           [,1]      [,2]      [,3]      [,4]
## [1,]  3.719063  3.183744 -3.622037 -3.280770
## [2,]  3.183744  3.719063 -3.280770 -3.622037
## [3,] -3.622037 -3.280770  3.719063  3.183744
## [4,] -3.280770 -3.622037  3.183744  3.719063

Also, $$\delta_r$$ is equal to 144 and thus

Now $$g$$ is

## [1]  9.408613 11.604293 -1.162210 -7.853545

and the new $$x$$ is

## [1] -0.55613830 -0.68592383  0.06869765  0.46421904

With this $$x$$ sStress is 12.4979159. We now go back, compute a new $$\alpha$$, and so on.

## 7.2 Dutch Political Parties

De Gruijter (1967) collected dissimilarity measures for nine Dutch political parties.

##       KVP PvdA  VVD  ARP  CHU  CPN  PSP   BP
## PvdA 5.63
## VVD  5.27 6.72
## ARP  4.60 5.64 5.46
## CHU  4.80 6.22 4.97 3.20
## CPN  7.54 5.12 8.13 7.84 7.80
## PSP  6.73 4.59 7.55 6.73 7.08 4.08
## BP   7.18 7.22 6.90 7.28 6.96 6.34 6.88
## D66  6.17 5.47 4.67 6.13 6.04 7.42 6.36 7.36
The solutions for 0.1, 0.25, 0.5, 0.75, 1, 2 are given in figure 1.
Figure 1: De Gruijter Data: Separate Configurations

The basic structure for all solutions is the same, we see the familiar horseshoe, in which the left-right scale is bended so that extreme left (CPN) and extreme right (BP) are close. This is perhaps most clear for r = .75. If r becomes large, then we see increasing clustering.

We also give all six solutions plotted on top of each other (after Procrustus matching) in figure 9. It shows that the two extreme parties are stable over solutions, but the middle is more variable.
Figure 2: De Gruijter Data: Joint Configuration

Plotting rStress in figure 3 as a function of r shows, somewhat surprisingly perhaps, an increasing function. Because the data are average dissimilarities, it is likely there is a fairly large additive constant, and additive constants correspond with lower values of r.
Figure 3: De Gruijter Data: rStress

We put the nuber of iterations and the rStress values in a small table.

## r:  0.10  iterations:    29103  rStress:  0.005464
## r:  0.25  iterations:     3605  rStress:  0.006310
## r:  0.50  iterations:     3566  rStress:  0.044603
## r:  0.75  iterations:     3440  rStress:  0.107113
## r:  1.00  iterations:   100000  rStress:  0.155392
## r:  2.00  iterations:   100000  rStress:  0.234877
Figure 4 shows the six Shepard diagrams. We see the clustering for r = 2. For r = .1 the Shepard diagram becomes concave, indicating that the larger dissimilarities are underestimated and reflecting the fact that for small powers the powered distances will all be close to one.
Figure 4: De Gruijter Data: Shepard Diagrams

## 7.3 Ekman Color Data

We use the color data from Ekman (1954), with two dimensions and unit weights. As we know from previous analyses, MDS of the Ekman data gives a very good fit and a very clear representation of the color circle.

##      434  445  465  472  490  504  537  555  584  600  610  628  651
## 445 0.14
## 465 0.58 0.50
## 472 0.58 0.56 0.19
## 490 0.82 0.78 0.53 0.46
## 504 0.94 0.91 0.83 0.75 0.39
## 537 0.93 0.93 0.90 0.90 0.69 0.38
## 555 0.96 0.93 0.92 0.91 0.74 0.55 0.27
## 584 0.98 0.98 0.98 0.98 0.93 0.86 0.78 0.67
## 600 0.93 0.96 0.99 0.99 0.98 0.92 0.86 0.81 0.42
## 610 0.91 0.93 0.98 1.00 0.98 0.98 0.95 0.96 0.63 0.26
## 628 0.88 0.89 0.99 0.99 0.99 0.98 0.98 0.97 0.73 0.50 0.24
## 651 0.87 0.87 0.95 0.98 0.98 0.98 0.98 0.98 0.80 0.59 0.38 0.15
## 674 0.84 0.86 0.97 0.96 1.00 0.99 1.00 0.98 0.77 0.72 0.45 0.32 0.24
In our algorithm we always start with a classical metric scaling solution (Torgerson 1958). We minimize rStress for r equal to 0.1, 0.25, 0.5, 0.75, 1, 2. The six solutions are plotted jointly in Figure 5. Because of the very good fit, solutions for different values of r are similar and very much on the color circle.
Figure 5: Ekman Color Data: Configuration

The Shepard diagrams in figure 6 are interesting, because, unlike the configurations, they are quite different, becoming more and more convex as r increases. The plot for r = .25 looks best.


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Isotones : Those elements having the different atomic number and atomic weight but the same number of neutrons in the nucleus are called Isotones.

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## Bulletin of the Belgian Mathematical Society - Simon Stevin

### On the approximation of functions by Fourier Stieltjes Series

#### Abstract

Recently, Leindler [2] introduced the sequences of Head Bounded Variation ($HBVS$) and the sequences of Rest Bounded Variation ($RBVS$), which are nontrivial generalizations of nondecreasing sequences and nonincreasing sequences respectively. In the present paper, we generalize a classical result of Mazhar([1]) on the approximation by means of Fourier Stieltjes series by using the $HBVS$ and $RBVS.$

#### Article information

Source
Bull. Belg. Math. Soc. Simon Stevin, Volume 20, Number 4 (2013), 667-674.

Dates
First available in Project Euclid: 22 October 2013

https://projecteuclid.org/euclid.bbms/1382448187

Digital Object Identifier
doi:10.36045/bbms/1382448187

Mathematical Reviews number (MathSciNet)
MR3129066

Zentralblatt MATH identifier
1280.42002

#### Citation

Lu, Chengbo; Yu, Dansheng. On the approximation of functions by Fourier Stieltjes Series. Bull. Belg. Math. Soc. Simon Stevin 20 (2013), no. 4, 667--674. doi:10.36045/bbms/1382448187. https://projecteuclid.org/euclid.bbms/1382448187


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# Number of points of accumulation of a sequence

Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.

Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.

Every real between $0$ and $1$ is an accumulation point of our sequence.

Yes, this is possible. For example consider the sequence $a_n$ for $n \ge 2$ defined as the smallest divisor of $n$ greater than $1$.

The accumulation points are all the prime numbers. Subsequences witnessing them are for instance the $p$-th powers for each prime $p$.

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# Number of points of accumulation of a sequence

Can a sequence have infinitely many points of accumulation i.e. we can extract infinitely many subsequences from it s.t. they all converge to their respective point of accumulation? I have the feeling it would mean that the period of repetition of something could be infinitely big.

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Start with $0,1$. Then travel backwards to $0$ in steps of $1/2$, so $1/2,0$. Then travel forwards to $1$ in steps of $1/4$, so $1/4,2/4,3/4,1$. Then travel backwards to $0$ in steps of $1/8$, so $7/8$, $6/8$, $5/8$, and so on. Continue.

Every real between $0$ and $1$ is an accumulation point of our sequence.

@MehdiSlimani: You are welcome. If you prefer we could use $0,1,1/2,0,1/3,2/3,1,3/4,2/4,1/4,0,1/5,2/5,\dots$ and then we can quote the result that the rationals in $[0,1]$ are dense in $[0,1]$. - André Nicolas Apr 29 at 16:29

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text:
# WikiBuddy certification guidance week

Dear Balqis,

Thank you for your kind words! You are one of the very active members in our WE Community. Would you like to add your photo in the Infobox_personal on your user page?
Warm regards
Anil Prasad 01:46, 23 April 2012 (UTC)

14:46, 23 April 2012

Hi Anil,

01:39, 24 April 2012

Hi Balqis,

Uploading one's photo on open web can never be a compulsory item :) You have attained the required level of wiki skills and made content contributions and WE community service necessary for a WikiBuddy certification. You will receive the WikiBuddy badge soon.

Regards
Anil

03:22, 24 April 2012
  Balqis ji
congratulation  for your good work  for which you are going to receive WikiBuddy  Badge soon  as stated above.
regards

harbans

05:01, 24 April 2012

Hi Harbanji

Thank you for the encouraging words .I wish you too a wonderful week of WikiBuddy Skill development.

Best regards
Balqis

02:00, 25 April 2012

Hi Balqis

Your  enthusiasm  to learn is attracting to me. To  learn  animation  as have well expressed  in your userpage needs some training? To learn more I  congratulate  you for  joining  l4content 453  as facilitator?  Will you please guide me to join  as you have joined in  l4content 453  just to learn more? thanks

harbans

17:44, 29 April 2012

Hi Harban ji ,

I appreciate your eagerness and thank you for your nice words.I just joined the upcoming eL4C53 and was invited to do facilitation .So I joined it .I have not yet done any courses on animation but my students are doing some of this stuff.So they have helped me in this field :)
Best Regards

Balqis

00:37, 30 April 2012

Hi Anil ,

I am excited to know this !!!!!!!!!!!!!!! I am indebted to you for conducting this event on Wiki Buddy guidance week .This really helped me to get to know what I had to do systematically to attain each level.Your guidance is the key factor here.

Thanks in tonnes and tonnes again :)

Your humble student and Wiki Neighbour

Balqis

11:51, 24 April 2012

CONGRATS BALQIS! YOU ARE NOW A WIKIBUDDY!

Warm Regards
Anil Prasad 06:40, 24 April 2012 (UTC)

19:40, 24 April 2012

You made my day :D I don't know how to you now !!!!! I am sooooooooooo happy to read this and see my Wiki Buddy badge on my page .I'm REALLY THRILLED to do more on WE now .This magical platform is always enticing .

Sincere Thanks again

Balqis

01:58, 25 April 2012


text:
# group cohomology

In this post we will talk about cohomology(from homological algebra), group cohomology and its applications.

Why should we study (co)homology? Of course there are historical motivations. Yet a modern point of view is to compensate for the non-exactness of a functor from a category(with good properties, for example, an abelian category) to another. So, one more fundamental question arises: why should we concern ourselves with exactness of a sequence? Perhaps the motivation for this comes from the fact that a splitting sequence of abelian groups implies that this sequence is exact. This is the most common necessary condition for a sequence to split.

Now let’s get down to some concrete things, we suppose that $\mathfrak{C}$ is an abelian category, one prototype example is the category of modules over a commutative ring with unit. We consider the associated ($\mathbb{Z}$-)graded category $\mathfrak{C}'$, in which objects are $(C_i,\delta_i)_{i\in\mathbb{Z}}$ where $\delta_i:C_i\rightarrow C_{i-1}$ is a morphism in $\mathfrak{C}$ such that $\delta_i\circ \delta_{i-1}=0$. So the first motivational question comes: why de we consider this kind of things? One motivation comes from differential manifolds. In the theory of differential manifolds, we have an important operator, the exterior differential operator, $d$, which acts on differential forms. Differential forms give a $\mathbb{Z}$(or $\mathbb{N}$)-graded real vector space, and $d\circ d=0$. This is an important property of $d$. And the category $\mathfrak{C}'$ can be seen as a generalization of this object.

Now a morphism from $(C_i,\delta_i),(D_i,\partial_i)$ is a sequence of maps $f_i:C_i\rightarrow D_i$ such that $\partial_i\circ f_i=f_{i-1}\circ \delta_i$. This definition is very reasonable.

For each such object $C=(C_i,\delta_i)$, we consider $H_n(C)=Ker(\delta_i)/Im(\delta_{i+1})$. This is of course a most fundamental concept in the theory of (co)homology. One straightforward motivation for these $H_n(C)$ is to compensate for the non-exactness of these maps $\delta_i$. If $f:C=(C_i,\delta_i)\rightarrow D=(D_i,\partial_i)$ is a morphism, then it is easy to verify that $f$ induces maps $H_n(f):H_n(C)\rightarrow H_n(D)$.

Now let’s consider another different question: given two maps $f,g:C=(C_i,\delta_i)\rightarrow D=(D_i,\partial_i)$, when will $H(f)=H(g)$? This is a rather interesting question. One sufficient condition comes from homotopy theory, which is, at first sight, completely different from homology theory. We say that two such maps $f,g$ are homotopy equivalent if there is a sequence of maps $h_n:C_n\rightarrow D_{n+1}$ such that $f_n-g_n=\partial_{n+1}\circ h_n+h_{n-1}\circ \delta_n$. It is very easy to show that if $f, g$ are homotopy equivalent, then they induce the same maps $H_n(f)=H_n(g),\forall n$. This is really a fundamental verification in homology theory. Recall the domains of $H_(f),H(g)$: they are $H_n(C)=Ker(\delta_n)/Im(\delta_{n+1})$. Thus for any $x\in H_n(C)$, we have that $f_n(x)-g_n(x)=\partial_{n+1}\circ h_n(x)+h_{n-1}\circ \delta_n(x)$. Note that $\delta_n(x)=0$, besides $\partial_{n+1}\circ h_n(x)\in Im(\partial_{n+1})$, thus $f_n(x)=g_n(x),\forall x\in H_n(C)$. It is also easy to show that homotopy equivalent is an equivalent relation. Note here why should we need the first term $\partial_{n+1}\circ h_n$? In fact these two terms can be seen from homotopy theory. That is what we are going to say in the next post.


text:
# Can we generalize it?

Solve :$$\displaystyle \frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}=1$$ for positive integers $$(a_1,a_2,a_3,\cdots,a_n)$$ .

There could be a remarkable proof/method.

1 year, 11 months ago

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Sylvester's sequence!

- 1 year, 11 months ago

What about 1/(n)(n+1)? Or for generalization, Leibniz's triangle?

- 1 year, 10 months ago

Yes I've tried that and thanks for mentioning too , but I think those are not only solutions. Indeed suppose we are to generalize it for 5 variables & Leibniz's triangle gives us 5 different values only. But there exists many more pairs than that

- 1 year, 10 months ago

How should we ? It's Its ...... How ?

- 1 year, 11 months ago


text:
# 14 Resampling and simulation

The use of computer simulations has become an essential aspect of modern statistics. For example, one of the most important books in practical computer science, called Numerical Recipes, says the following:

“Offered the choice between mastery of a five-foot shelf of analytical statistics books and middling ability at performing statistical Monte Carlo simulations, we would surely choose to have the latter skill.”

In this chapter we will introduce the concept of a Monte Carlo simulation and discuss how it can be used to perform statistical analyses.

## 14.1 Monte Carlo simulation

The concept of Monte Carlo simulation was devised by the mathematicians Stan Ulam and Nicholas Metropolis, who were working to develop an atomic weapon for the US as part of the Manhattan Project. They needed to compute the average distance that a neutron would travel in a substance before it collided with an atomic nucleus, but they could not compute this using standard mathematics. Ulam realized that these computations could be simulated using random numbers, just like a casino game. In a casino game such as a roulette wheel, numbers are generated at random; to estimate the probability of a specific outcome, you could play the game hundreds of times. Ulam’s uncle had gambled at the Monte Carlo casino in Monaco, which is apparently where the name came from for this new technique.

There are four steps to performing a Monte Carlo simulation:

1. Define a domain of possible values
2. Generate random numbers within that domain from a probability distribution
3. Perform a computation using the random numbers
4. Combine the results across many repetitions

As an example, let’s say that I want to figure out how much time to allow for an in-class quiz. Say that we know that the distribution of quiz completion times is normal, with mean of 5 minutes and standard deviation of 1 minute. Given this, how long does the test period need to be so that we expect all students to finish the exam 99% of the time? There are two ways to solve this problem. The first is to calculate the answer using a mathematical theory known as the statistics of extreme values. However, this involves complicated mathematics. Alternatively, we could use Monte Carlo simulation. To do this, we need to generate random samples from a normal distribution.

## 14.2 Randomness in statistics

The term “random” is often used colloquially to refer to things that are bizarre or unexpected, but in statistics the term has a very specific meaning: A process is random if it is unpredictable. For example, if I flip a fair coin 10 times, the value of the outcome on one flip does not provide me with any information that lets me predict the outcome on the next flip. It’s important to note that the fact that something is unpredictable doesn’t necessarily mean that it is not deterministic. For example, when we flip a coin, the outcome of the flip is determined by the laws of physics; if we knew all of the conditions in enough detail, we should be able to predict the outcome of the flip. However, many factors combine to make the outcome of the coin flip unpredictable in practice.

Psychologists have shown that humans actually have a fairly bad sense of randomness. First, we tend to see patterns when they don’t exist. In the extreme, this leads to the phenomenon of pareidolia, in which people will perceive familiar objects within random patterns (such as perceiving a cloud as a human face or seeing the Virgin Mary in a piece of toast). Second, humans tend to think of random processes as self-correcting, which leads us to expect that we are “due for a win” after losing many rounds in a game of chance, a phenonenon known as the “gambler’s fallacy”.

## 14.3 Generating random numbers

Running a Monte Carlo simulation requires that we generate random numbers. Generating truly random numbers (i.e. numbers that are completely unpredictable) is only possible through physical processes, such as the decay of atoms or the rolling of dice, which are difficult to obtain and/or too slow to be useful for computer simulation (though they can be obtained from the NIST Randomness Beacon).

In general, instead of truly random numbers we use pseudo-random numbers generated using a computer algorithm; these numbers will seem random in the sense that they are difficult to predict, but the series of numbers will actually repeat at some point. For example, the random number generator used in R will repeate after $2^{19937} - 1$ numbers. That’s far more than the number of seconds in the history of the universe, and we generally think that this is fine for most purposes in statistical analysis.

In R, there is a function to generate random numbers for each of the major probability distributions, such as:

• runif() - uniform distribution (all values between 0 and 1 equally)
• rnorm() - normal distribution
• rbinom() - binomial distribution (e.g. rolling the dice, coin flips)

Figure 14.1 shows examples of numbers generated using the runif() and rnorm() functions, generated using the following code:

You can also generate random numbers for any distribution if you have a quantile function for the distribution. This is the inverse of the cumulative distribution function; instead of identifying the cumulative probabilities for a set of values, the quantile function identifies the values for a set of cumulative probabilities. Using the quantile function, we can generate random numbers from a uniform distribution, and then map those into the distribution of interest via its quantile function.

By default, R will generate a different set of random numbers every time you run one of the random number generator functions described above. However, it is also possible to generate exactly the same set of random numbers, by setting what is called the random seed to a specific value. We will do this in many of the examples in this book, in order to make sure that the examples are reproducible.

If we run the rnorm() function twice, it will give us different sets of pseudorandom numbers each time:

print(rnorm(n = 5))
## [1]  1.48  0.18  0.21 -0.15 -1.72
print(rnorm(n = 5))
## [1] -0.691 -2.231  0.391  0.029 -0.647

However, if we set the random seed to the same value each time using the set.seed() function, then it will give us the same series of pseudorandom numbers each time:

set.seed(12345)
print(rnorm(n = 5))
## [1]  0.59  0.71 -0.11 -0.45  0.61
set.seed(12345)
print(rnorm(n = 5))
## [1]  0.59  0.71 -0.11 -0.45  0.61

## 14.4 Using Monte Carlo simulation

Let’s go back to our example of exam finishing times. Let’s say that I administer three quizzes and record the finishing times for each student for each exam, which might look like the distributions presented in Figure 14.2.

However, what we really want to know is not what the distribution of finishing times looks like, but rather what the distribution of the longest finishing time for each quiz looks like. To do this, we can simulate the finishing time for a quiz, using the assumption that the finishing times are distributed normally, as stated above; for each of these simulated quizzes, we then record the longest finishing time. We repeat this simulation a large number of times (5000 should be enough) and record the distribution of finishing times, which is shown in Figure 14.3.

This shows that the 99th percentile of the finishing time distribution falls at 8.81, meaning that if we were to give that much time for the quiz, then everyone should finish 99% of the time. It’s always important to remember that our assumptions matter – if they are wrong, then the results of the simulation are useless. In this case, we assumed that the finishing time distribution was normally distributed with a particular mean and standard deviation; if these assumptions are incorrect (and they almost certainly are), then the true answer could be very different.

## 14.5 Using simulation for statistics: The bootstrap

So far we have used simulation to demonstrate statistical principles, but we can also use simulation to answer real statistical questions. In this section we will introduce a concept known as the bootstrap that lets us use simulation to quantify our uncertainty about statistical estimates. Later in the course, we will see other examples of how simulation can often be used to answer statistical questions, especially when theoretical statistical methods are not available or when their assumptions are too difficult to meet.

### 14.5.1 Computing the bootstrap

In the section above, we used our knowledge of the sampling distribution of the mean to compute the standard error of the mean and confidence intervals. But what if we can’t assume that the estimates are normally distributed, or we don’t know their distribution? The idea of the bootstrap is to use the data themselves to estimate an answer. The name comes from the idea of pulling one’s self up by one’s own bootstraps, expressing the idea that we don’t have any external source of leverage so we have to rely upon the data themselves. The bootstrap method was conceived by Bradley Efron of the Stanford Department of Statistics, who is one of the world’s most influential statisticians.

The idea behind the bootstrap is that we repeatedly sample from the actual dataset; importantly, we sample with replacement, such that the same data point will often end up being represented multiple times within one of the samples. We then compute our statistic of interest on each of the bootstrap samples, and use the distribution of those estimates.

Let’s start by using the bootstrap to estimate the sampling distribution of the mean, so that we can compare the result to the standard error of the mean (SEM) that we discussed earlier.

Figure 14.4 shows that the distribution of means across bootstrap samples is fairly close to the theoretical estimate based on the assumption of normality. We can also use the bootstrap samples to compute a confidence interval for the mean, simply by computing the quantiles of interest from the distribution of bootstrap samples.

Table 14.1: Confidence limits for normal distribution and bootstrap methods
type 2.5% 97.5%
Normal 165 172
Bootstrap 165 172

We would not usually employ the bootstrap to compute confidence intervals for the mean (since we can generally assume that the normal distribution is appropriate for the sampling distribution of the mean, as long as our sample is large enough), but this example shows how the method gives us roughly the same result as the standard method based on the normal distribution. The bootstrap would more often be used to generate standard errors for estimates of other statistics where we know or suspect that the normal distribution is not appropriate.

## 14.6 Learning objectives

After reading this chapter, you should be able to:

• Describe the concept of a Monte Carlo simulation.
• Describe the meaning of randomness in statistics
• Obtain random numbers from the uniform and normal distributions
• Describe the concept of the bootstrap


text:
## Plot the Impulse Response Function

### Moving Average Model

This example shows how to calculate and plot the impulse response function for a moving average (MA) model. The MA(q) model is given by

`${y}_{t}=\mu +\theta \left(L\right){\epsilon }_{t},$`

where $\theta \left(L\right)$ is a q-degree MA operator polynomial, $\left(1+{\theta }_{1}L+\dots +{\theta }_{q}{L}^{q}\right).$

The impulse response function for an MA model is the sequence of MA coefficients, $1,{\theta }_{1},\dots ,{\theta }_{q}.$

Create MA Model

Create a zero-mean MA(3) model with coefficients ${\theta }_{1}=0.8$, ${\theta }_{2}=0.5$, and ${\theta }_{3}=-0.1.$

`Mdl = arima('Constant',0,'MA',{0.8,0.5,-0.1});`

Step 2. Plot the impulse response function.

`impulse(Mdl)`

For an MA model, the impulse response function cuts off after q periods. For this example, the last nonzero coefficient is at lag q = 3.

### Autoregressive Model

This example shows how to compute and plot the impulse response function for an autoregressive (AR) model. The AR(p) model is given by

`${y}_{t}=\mu +\varphi \left(L{\right)}^{-1}{\epsilon }_{t},$`

where $\varphi \left(L\right)$ is a $p$-degree AR operator polynomial, $\left(1-{\varphi }_{1}L-\dots -{\varphi }_{p}{L}^{p}\right)$.

An AR process is stationary provided that the AR operator polynomial is stable, meaning all its roots lie outside the unit circle. In this case, the infinite-degree inverse polynomial, $\psi \left(L\right)=\varphi \left(L{\right)}^{-1}$, has absolutely summable coefficients, and the impulse response function decays to zero.

Step 1. Specify the AR model.

Specify an AR(2) model with coefficients ${\varphi }_{1}=0.5$ and ${\varphi }_{2}=-0.75$.

`modelAR = arima('AR',{0.5,-0.75});`

Step 2. Plot the impulse response function.

Plot the impulse response function for 30 periods.

`impulse(modelAR,30)`

The impulse function decays in a sinusoidal pattern.

### ARMA Model

This example shows how to plot the impulse response function for an autoregressive moving average (ARMA) model. The ARMA(p, q) model is given by

`${y}_{t}=\mu +\frac{\theta \left(L\right)}{\varphi \left(L\right)}{\epsilon }_{t},$`

where $\theta \left(L\right)$ is a q-degree MA operator polynomial, $\left(1+{\theta }_{1}L+\dots +{\theta }_{q}{L}^{q}\right)$, and $\varphi \left(L\right)$ is a p-degree AR operator polynomial, $\left(1-{\varphi }_{1}L-\dots -{\varphi }_{p}{L}^{p}\right)$.

An ARMA process is stationary provided that the AR operator polynomial is stable, meaning all its roots lie outside the unit circle. In this case, the infinite-degree inverse polynomial, $\psi \left(L\right)=\theta \left(L\right)/\varphi \left(L\right)$ , has absolutely summable coefficients, and the impulse response function decays to zero.

Specify ARMA Model

Specify an ARMA(2,1) model with coefficients ${\varphi }_{1}$ = 0.6, ${\varphi }_{2}=-0.3$, and ${\theta }_{1}=0.4$.

`Mdl = arima('AR',{0.6,-0.3},'MA',0.4);`

Plot Impulse Response Function

Plot the impulse response function for 10 periods.

`impulse(Mdl,10)`


text:
# Find if a point lies within the angle formed by three other points

I have three points (A, B, C) in a 3d space that define an angle like this:

I need a fast way to know if another point D (that I already know is coplanar with the plane ABC) lies within the angle formed by <ABC.

You can express vector AD as a linear combination of AB and AC. If the coefficients of the linear combination are both positive, then D is inside the angle.

So basically, you are solving the system of equations (3 equations, 2 unknowns):

$\left[\begin{array}{rr|r} b_x-a_x & c_x-a_x& d_x-a_x \\ b_y-a_y & c_y-a_y& d_y-a_y \\ b_z-a_z & c_z-a_z& d_z-a_z \\ \end{array}\right]$

which represent the following equations:

$(b_x-a_x) \times I + (c_x-a_x) \times J = d_x-a_x \\ (b_y-a_y) \times I + (c_y-a_y) \times J = d_y-a_y \\ (b_z-a_z) \times I + (c_z-a_z) \times J = d_z-a_z \\$

and if your solution vector has positive components ($I>0, J>0$), then D is inside that angle. If there is no solution, then D is not in the ABC plane.

• Shouldn't that be $d_y-a_y$ and $d_z-a_z$ on the right side? – almosnow Oct 8 '15 at 18:57
• Yeah, my bad. I've fixed the typo, thanks! – jh314 Oct 8 '15 at 19:00
• I really don't know how to solve that... shame... I know D is in the plane, so there exists a solution, but how would I arrive to that, IDK... I'm not used to that notation. – almosnow Oct 8 '15 at 19:01
• I've added the 3 basic equations, now you just have to solve for I and J. – jh314 Oct 8 '15 at 19:23
• Oh ok, now I see, thanks! – almosnow Oct 8 '15 at 19:25

Precompute $v = C - A$, and let $w = (-v_y, v_x)$. Then $D$ is on the right side of $AC$ if and only iff $(D - A) \cdot w > 0$. A similar construction tells you whether $D$ is on the upper side of $AB$. So in that case, you need to check $(D-A) \cdot w' < 0$. If you pass both those tests, then $D$ is in the wedge.

(This assumes that $B$ is on the 'right side" of $AC$, i.e., $(B- A) \cdot w > 0$. If that's not the case, you need to swap $B$ and $C$.).

• How would you do that in 3D? I'm curious because of $w = (-v_y, v_x)$ – almosnow Oct 8 '15 at 18:47
• Ah...we're in 3-space. Then let $n = (C - A) \times (B - A)$, and let $w = n \times (C - A)$ and $w' = n \times (B - A)$. Then, if you're applying this to many points $D$, you have only two dot products to compute, rather than trying to solve a degenerate $3 \times 3$ system of equations as you need to do in @jh314's solution. – John Hughes Oct 9 '15 at 0:06
• Thank you @John Hughes. I will try out this method as it would be faster and easier to implement. – almosnow Oct 11 '15 at 3:32

Project the points onto one of the coordinate planes and test $$AC \times AD > 0 \text{ and } AD \times AB > 0$$ (use the projected points here; $\times$ is the 2D cross product, a number)


text: Although composed of only four tests, the Gf-Gc Composite yields highly stable scores, with reliability values comparable to those of the GIA cluster score. Good to know. V(δi) = variance of the error term for the ith indicator, 3. p = number of indicators @chl: nice citation. Composite reliability is based on the factor loadings in a confirmatory factor analysis (CFA). From the results below, we could see all the composite reliability for the constructs ranges from 0.758 to 0.898, which exceeds 0.70. mRNA-1273 vaccine: How do you say the “1273” part aloud? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. Reporting and Interpreting Scores Derived from Likert-type Scales J. Robert Warmbrod1 ... with at least one variable measured by a Likert-type scale. Ideally, we want to have fewer than 1.5% defective widgets.So, we set up a simple lot sampling plan. for factor analysis or reliability studies. I came across his website because some of my colleagues use SPSS for psychometrics, esp. Why aren't "fuel polishing" systems removing water & ice from fuel in aircraft, like in cruising yachts? Researchers commonly calculate the Cronbach’s alpha to evaluate the reliability of the items comprising a composite score. How to calculate composite score with a 2-factor scale. Thanks for contributing an answer to Cross Validated! Reliability analysis allows you to study the properties of measurement scales and the items that compose the scales. Replacing the core of a planet with a sun, could that be theoretically possible? Composite reliability does not assume this but take into consideration the varying factor loadings of the items. If none fail to accept the lot, if on… Centering remains useful when one wants to avoid spurious collinearity effects or to interpret the effect of a predictor as deviation from its means (e.g., age). How can I quickly grab items from a chest to my inventory? In this case, the convergent validity of the construct is questionable. It is very shortly discussed on Garson's SPSS website. Sorry for the lack of a definitive answer, but I felt that some response was definitely better than none. Quite likely, people will guess differently, the different measures will be inconsistent, and therefore, the “guessing” technique of measurement is unreliable. Academic theme and In interpreting correlation coefficients, two properties are important. Definition of Composite Reliability: Composite reliability (CR), ? Construction of composite measures by the canonical-factor-regression method . Al). Measures of Reliability Reliability: the fact that a scale should consistently reflect the construct it is measuring. In other words, if we use this scale to measure the same construct multiple times, do we get pretty much the same result every time, assuming the underlying phenomenon is not changing? Dog likes walks, but is terrified of walk preparation. Blogdown, Apr 19, 2020. AVE can be calculated by using auto design by James Gaskin by visiting this website, and click on Excel StatTools on the left hand menu, it is an excel file with calculator for calculating AVE, reliability and validity test. Let’s face it; sometimes understanding bull proofs can be like reading a document in a foreign language. The results suggest that composite reliability may be used as an assessment tool, but should not be used as an item selection tool in structural equations modeling. Getting the same or very similar results from slight variations on the … There is a popular technique called the split half reliability. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. The composite reliability for the extraversion factor is .90. Allen, M. P. ( 1974 ). Grounded in the classical test theory definition of reliability and the tenets basic to Likert- ... measurement methodology is that scores yielded by a Likert scale are composite (summated) Asking for help, clarification, or responding to other answers. The formula (Netemeyer, 2003) is: Where: 1. λi = completely standardized loading for the ith indicator, 2. Estimation of composite reliability for congeneric measures. AVEmeasures the level of variance captured by a construct versus the level due to measurement error, values above 0.7 are considered very good, whereas, the level of 0.5 is acceptable. The results suggest that composite reliability may be used as an assessment tool, but should not be used as an item selection tool in structural equations modeling. This method splits the data into two parts. What does it mean when an aircraft is statically stable but dynamically unstable? What authority does the Vice President have to mobilize the National Guard? This statistic allows you to make a statement regarding the acceptability of the combination of items to represent your variable. Hugo. Why Business Credit Reports are Important. One way to think of reliability is that other things being equal, a person should get the same score on a questionnaire if they complete it at two different points in time (test-retest reliability. The Reliability Analysis procedure calculates a number of commonly used measures of scale reliability and also provides information about the relationships between individual items in the scale. There are 1,000 said widgets per shipment. A … reliability" you mean internal AVE- average variance extracted (AVE) should not be less than .05, this is to show that more than half of the variances is observed (Janssens, et. The median Gf-Gc Composite reliability coefficient is .96 compared to the median GIA reliability coefficient of .97, as reported in Table 2. I haven't used SPSS in some time, and I don't remember seeing an option to perform these calculations, but you can certainly do it using the syntax. Parallel Forms Reliability. Garson sure has a lot to offer, although [side comment] I believe I have found a chink in his armor when it comes to centering in regression (he advocates it unnecessarily). $\endgroup$ – chl Mar 30 '11 at 21:02 What I would do from here is examine the syntax SPSS gives you when you perform an FA, and use the variables named to compute the average variance explained. It is most commonly used when you have multiple Likert questions in a survey/questionnaire that form a scale and you wish to determine if the scale is reliable. Correlation coefficients index the extent to which two scores are related, and the direction of that relationship. The line graph in the top area, highlighted in Figure B, displays a composite stability index for your PC on a scale from 1 to 10. Applied Psychological Measurement, 21(2), 173-184. Tutorial for Reliability Analysis. consistency reliability. Can I print plastic blank space fillers for my service panel? If the average variance extracted is less than .50, then the variance due to measurement error is greater than the variance due to the construct. 1. Therefore, it can be concluded that the measurement model and data have passed validity and reliability tests. Can I hang this heavy and deep cabinet on this wall safely? The average variance extracted (AVE) calculated as follows: In statistical models applied to psychometrics, congeneric reliability ("rho C") a single-administration test score reliability (i.e., the reliability of persons over items holding occasion fixed[2]) coefficient, commonly referred to as composite reliability, construct reliability, and coefficient omega. Should the stipend be paid if working remotely? Draw 50 random widgets from the shipment of 1,000 and inspect them. Very different results of principal component analysis in SPSS and Stata after rotation. Why is an early e5 against a Yugoslav setup evaluated at +2.6 according to Stockfish? High composite reliability is a very good indication that all your items constantly measure the same construct. Confirmatory Factor Analysis is one way to measure composite reliability, and it is widely available in many different statistical software packages. In the case of a unidimensional scale (like extraversion here), we define a one-factor CFA, and then use the factor loadings to compute our internal consistency estimate. 3 Split-Half Reliability KR-20 • NOTE: Only use the KR-20 if each item has a right answer. Raykov, Tenko. Coach to clarify the contents of the Variance To learn more, see our tips on writing great answers. There is a certain level of comfort and affection with these aspects that help strengthen a bond between two people. rev 2021.1.7.38271, Sorry, we no longer support Internet Explorer, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. Use MathJax to format equations. The reliability of the sum score of the observed variables is estimated by the quotient between the estimate of the true composite variance (F4) and the variance of the composite … coefficient, is obtained by combining all of the true score variances and covariances in the composite of indicator variables related to constructs, and by dividing this sum by the total variance in the composite. Printing message when class variable is called. The larger the absolute value of the coefficient (the size of the number without regard to the sign) the greate… statmodel.com/discussion/messages/9/258.html?1237657036, faculty.chass.ncsu.edu/garson/PA765/reliab.htm, Interpreting discrepancies between R and SPSS with exploratory factor analysis. It is recommended that the reliability of a construct is at least 0.70. The CR (composite reliability) values are above.95 which indicates redundant items. What causes that "organic fade to black" effect in classic video games? Composite reliability is based on the factor loadings in a confirmatory factor analysis (CFA). An aspect in which the researcher can use reliability analysis is when two observations under study that are equivalent to each other in terms of the construct being measured also have the equivalent outcome. One appealing aspect of composite reliability is that we can calculate it for multiple factors in the same model. What is the relationship between scale reliability measures (Cronbach's alpha etc.) The next output of confirmatory factory analysis is Goodness-of-fit indices. , which has been also referred to as McDonald’s ? Explained table, searching the Help files or To subscribe to this RSS feed, copy and paste this URL into your RSS reader. I have some issues with an exploratory factor analysis. The following is shamelessly extracted from the following link. When it comes to business credit, making the right decision depends on having the right information. Let’s say we will periodically receive a shipment of widgets. the total of the squared multiple correlations plus the total sum of each variable, then divides it by the number of factors in that variable. If not with SPSS, Stata might help too. Making statements based on opinion; back them up with references or personal experience. CRL over HTTPS: is it really a bad practice? By hand, the calculations are a little cumbersome. It only takes a minute to sign up. Centering remains useful when one wants to avoid spurious collinearity effects or to interpret the effect of a predictor as deviation from its means (e.g., age). These two links give you an introduction to SPSS syntax. They reflect the tendency of the variables to “co-vary”; that is, for changes in the value of one variable to be associated with changes in the value of the other. According to this criterion, theconvergent validityof the measurement model can be assessed by the Average Variance Extracted (AVE) and Composite Reliability (CR). the Gf-Gc Composite. Any shortcuts to understanding the properties of the Riemannian manifolds which are used in the books on algebraic topology, Swap the two colours around in an image in Photoshop CS6, right-clicking your SPSS factor Correlations range in magnitude from -1.00 to 1.00. Allen, M. P. ( 1974 ). We know the supplier’s manufacturing process is not in control and they ship a lot with more than 5% defective items on occasion. I came across his website because some of my colleagues use SPSS for psychometrics, esp. In consequence to this I nneed to reduce the CR because it is indicating that there are redundant items. research, values of composite reliability/Cronbach alpha between 0.60 to 0.70 are acceptable, while in more advanced stage the value have to be higher than 0.70. What is Composite Reliability? (1997). Estimating the Reliability of Composite Scores. Can anybody please tell me how to calculate the Average Variance Extracted (AVE) and the Composite Reliability from two factors, each with three items using SPSS? With the Bullvine has developed this cheat sheet to help you understand North American Genetic Evaluations easier. With all the letters, numbers and acronyms on a proof sheet, it is enough to confuse even the most passionate dairy breeder. Construction of composite measures by the canonical-factor-regression method . MathJax reference. What drives the test reliability of a composite based on three components? Altering table row colors trought multirow. Meanwhile, all composite reliability scores and Cronbach’s Alpha values of each variable exceed 0.7 which indicates that the data has passed the reliability test according to Hair et al. Should EFA be repeated after deletion of items with low factor loadings? How do you take into account order in linear programming? Composite Sun trine or sextile Moon helps a relationship significantly. for factor analysis or reliability studies. Finally, the "average variance extracted" measures the amount of variance that is captured by the construct in relation to the amount of variance due to measurement error and can be calculated using the following formula: (summation of squared factor loadings)/(summation of squared factor loadings) (summation of error variances) (Fornell & Larcker). In the case of a unidimensional scale, we define a one-factor CFA, and then use the factor loadings to compute our internal consistency estimate. Magnitude. # Compute residual variance of each item re - 1 - sl^2 # Compute composite reliability sum(sl)^2 / (sum(sl)^2 + sum(re)) #> [1] 0.9029523 There you have it. An example of an unreliable measurement is people guessing your weight. Do NOT use with a Likert scale. If the one-parameter logistic model (1PL) or the Rasch model are used, the result is equivalent to adding raw … (2010). Last updated on Composite Venus trine or sextile Mars offers a couple a certain level of comfort and lazy pleasure. analysis output and choosing Results Is it possible to assign value to set (not setx) value %path% on Windows 10? However, the value that is more than 0.90 is not desirable and the value that is 0.95 or above is definitely undesirable. Consulting a business credit report is an important first step in evaluating partners to avoid payment defaults, reduce exposure to bad … Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The closer the line is to 10 during any given period, the better. and component/factor loadings? My question is: Is there a solid way to solve this problem? You can also view the video on YouTube through this link. I seem to remember that Cronbach's alpha will only equal composite score reliability for essentially tau-equivalent tests. model used. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. Cronbach's alpha is the most common measure of internal consistency ("reliability"). In contrast, with composite reliability, the items are weighted based on the construct indicators’ individual loadings and, hence, this reliability is higher than Cronbach’s alpha. In order to get square multiple correlation of each item, you need to find square of each item Standardized Regression Weight / Estimate. Made with I'm thinking that by "composite Reliability is the degree to which the measure of a construct is consistent or dependable. • Formula: – rKR20 is the Kuder-Richardson formula 20 – k is the total number of test items I want to identify which items are redudant. : how do you take into account order in linear programming weight / Estimate by Likert-type. 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In a confirmatory factor analysis ( CFA ) the median Gf-Gc composite for... Cheat sheet to help you understand North American Genetic Evaluations easier to find square of each item a. To calculate composite score with a 2-factor scale reduce the CR because it is enough confuse... Relationship between scale reliability measures ( Cronbach 's alpha is the relationship between scale reliability measures Cronbach. Vice President have to mobilize the National Guard Psychological measurement, 21 ( 2 ), 173-184 your constantly! Items that compose the scales you agree to our terms of service, privacy policy cookie! Youtube through this link you to study the properties of measurement scales and the of! Value that is more than 0.90 is not desirable and the value is. Mobilize the National Guard it possible to assign value to set ( not setx value. The ith indicator, 2 square multiple correlation of each item, you agree to our of... 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# Are Cumulative Distribution Functions measurable?

It is well-known that CDFs (Cumulative Distribution Functions) of one dimensional random variables are Borel measurable. But does the same apply to CDFs of multi-dimensional random variables (rvecs)? It suffices, for my purposes, to consider finite dimensional rvecs.

Relevant definitions

Let $n$ be an integer $>= 2$.

Call a probability measure over the Borel field on $\mathbb R^n$ an $n$-dimensional probability measure.

To every $n$-dimensional probability measure, $m$, define the following function $F:\mathbb R^n\rightarrow\mathbb R$, called $m$'s CDF: $F(x)=m\left((-\infty, x]\right)$ where $(-\infty, x]$ is the set of all $y$ in $\mathbb R^n$ such that $y\leq x$ component-wise.

An $n$-dimensional CDF is a function $F:\mathbb R^n\rightarrow\mathbb R$ that can be obtained as some $n$-dimensional probability measure's CDF.

$n$-dimensional CDFs are known to be characterized by certain properties.

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For any $x \in \mathbb R^n$, consider the measurable box-like function $M_x(y) = m((-\infty,x]) \cdot \mathbb1_{[x,\infty)}(y)$. Here $\mathbb1_{[x,\infty)}(y)$ is the indicator function of "$y \ge x$ component-wise".
It should be easy to show that $F(y) = \sup_{x \in \mathbb Q^n} M_x(y)$ for any $y$ (take an increasing sequence of $x$ converging to $y$), and this makes $F$ the countable supremum of measurable functions.
Thank you. What a great idea, though i suspect your $F(y)$ fails to work, since it's left-continuous, whereas a CDF should be right-continuous. However, a slight modification will fix it: define $M_x(y) = m((-\infty,x])\cdot\mathbb1_{(-\infty,x]}(y) + (1+\epsilon)\cdot\mathbb1_{\mathbb R^n \setminus (-\infty,x]}(y)$ and $F(y) = \inf_{x\in\mathbb Q^n}M_x(y)$. –  Evan Aad Aug 3 '12 at 15:01
@EvanAad Oops, of course you are correct! Thanks for the fix. How silly of me to try to write $(-\infty,x]$ as a countable union rather than a countable intersection. –  Erick Wong Aug 3 '12 at 22:21
Actually the $\epsilon$ is redundant. –  Evan Aad Aug 4 '12 at 5:38


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# For the function f(x) = \frac{1}{\sqrt x} , find an equation of the tangent line the graph of...

## Question:

For the function {eq}f(x) = \frac{1}{\sqrt x} {/eq}, find an equation of the tangent line the graph of the above function at {eq}(4, \frac{1}{2}) {/eq}

## Tangent Line:

The tangent line determines a first degree approximation of the function, for differentiable functions, we have the general expression of the tangent line as: {eq}y = f(x),x = a \to y = f(a) + f'(a)\left( {x - a} \right). {/eq}

Taking into account the function and the given value:

{eq}f(x) = \frac{1}{{\sqrt x }}\quad ,\left( {4,\frac{1}{2}} \right) \to f(4) = \frac{1}{2} {/eq}

With the first derivative, at the point:

{eq}f(x) = \frac{1}{{\sqrt x }} = {x^{ - 1/2}}\\ f'(x) = - \frac{1}{2}{x^{ - 3/2}}\\ f'(4) = - \frac{1}{2}{4^{ - 3/2}} = - \frac{1}{{16}} {/eq}

We can write the tangent line as:

{eq}y = f(4) + f'(4)\left( {x - 4} \right)\\ y = \frac{1}{2} - \frac{1}{{16}}\left( {x - 4} \right) {/eq}