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# 1st PUC Maths Question Bank Chapter 2 Relations and Functions

Students can Download Maths Chapter 2 Relations and Functions Questions and Answers, Notes Pdf, 1st PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka 1st PUC Maths Question Bank Chapter 2 Relations and Functions

Question 1.
Define ordered pair.
Two numbers a and b listed in specific order and enclosed in parentheses form is called an ordered pair (a, b).
Keen Eye: Equality of two ordered pairs:
We have {a, b)-(c,d)⇔a-c and b – d.

Question 2.
Define a Cartesian product of two sets.
Let A and B two non-empty sets. Then, the Cartesian product of A and B is the set denoted by Ax B, consisting of all ordered pairs (a, b) such that a e A and be B.
A x B= {(a, b): a ∈ A, b∈B}
Keen Eye:

• If n(A) = p and n(B) = q, then n (A x B) = pq and n (B x A) = pq
• If at least one of A and B is infinite then AxB is infinite and B x A is infinite,
• In general, A x B ≠ B x A
• A x A x A = {(a, b, c) : a, b, c ∈A}. Here (a, b, c) is called an ordered triplet.

Question 3.
If (x + 1, y – 2) = (3,1), find the values of x
Given (x + 1, y – 2) = (3,1)
⇒ x+1=3  ∴x=2
y-2=1  ∴ y=3

Question 4.
If $$\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$$
Given $$\left(\frac{x}{3}+1, y-\frac{2}{3}\right)=\left(\frac{5}{3}, \frac{1}{3}\right)$$

Question 5.
If P={a,b,c}and Q={r},from P×Q and Q x P.Are these two products equal?
PxQ = {(a,r),(b,r)(c,r)}
QxP = {(r, a), (r, b), (r, c)}
Clearly PxQ≠QxP

Question 6.
If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in
Given n(A) = 3; n(B) = 3.
∴ n(AxB) = 3×3 = 9

Question 7.
If G=(7, 8} and H={5, 4, 2), find G x II and
GxH = {(7,5),(7,4),(7,2),(8,5),(8,4),(8,2)}
HxG = {(5,7),(5,8),(4,7),(4,8),(2,7),(2,8)}

Question 8.
State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P={m, n} and Q = {n, m}, then P x Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A x B is a non-empty set of ordered pairs (at, y) such that x∈B and y∈A
(iii) If A = {1,2}, B = {3,4}, then A x {B∩φ ) = φ
(i) Given statement is false:
Correct statement:
PxQ={(m, n), (m, m), (n, n), (n, m)}.

(ii) Given statement is false:
Correct statement:
AxB = {(x, y) :x∈A, y ∈B}.                                 ‘

(iii) True statement,

Question 9.
If A x B = {(a, x), (a, y), (b, x), (b, y)}. Find A and
A = {a, b} and B – {x, y}

Question 10.
If A x B = {(p, q), (p, r), (m, q), (m, r)}, find A and
A = set of first elements = {p, m}
B = set of second elements = {q, r}

Question 11.
Let A = (1, 2}, B = [1, 2, 3, 4}, C = { 5, 6} and D = (5,6,7,8}. Verify that
(i) A x (B∩C) = (A x B)∩(A x C).
(ii) A x C is a subset
(i) B∩C = { }
∴ Ax(B∩C)=φ ………….. (1)
A x B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2,4)}
A x C = {(1, 5), (1,6), (2, 5) (2,6)}
∴ (A x B) ∩ (A x C) = φ ………………. (2)
From (1) and (2), we get
A x (B∩C) = (A x B)∩(A x C)

(ii) A x C = {(1, 5), (1,6), (2, 5), (2, 6)}
B x D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}.
Clearly every elements of A x C is an element of B x D.
A x C ⊂B x D.

Question 12.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
(i) A x (B ∩ C)
(ii) (A x B) ∩ (A x C)
(iii) A x (B∪C)
(iv) (A x B)∪(A x C)
(i) B∩C={4}
A x (B∩C) = (1,4), (2, 4), (3,4)}

(ii) A x B = {(1, 3), (1,4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5) (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∩(A x C)= {(1, 4), (2, 4), (3, 4)}

(iii) B ∪ C={3,4, 5, 6}
∴ Ax(B∪C)  = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)}

(iv) A x B = {(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)}
A x C = {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
(A x B)∪(A x C) = {(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6) (3, 3), (3, 4), (3, 5), (3,6)}.

Question 13.
Let A = {1, 2} and B = {3, 4}. Write A x B. How many subsets will A x B have? List them.
Given A = {1, 2} and B = {3,4}
A x B= {(1, 3), (1,4), (2, 3), (2, 4)}
∴n (A x B) = 4
Number of subsets of A x B = 24=16
Subsets of A x B are: A x B, φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2,4)}, {(1,4), (2, 3)}, {(1, 4), (2, 4)} {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1,4), (2, 3), (2, 4)}, {(2, 3), (2,4), (1, 3)}.

Question 14.
Let A and B be two sets such that n(A) = 3 and n(B) = 2. If (x, 1), (y, 2), (z, 1) are in A x B, find A and B, where x, y, z are distinct elements.
A = {x, y, z} and B = {1, 2}.

Question 15.
The Cartesian product A x A has 9 elements along which are found (-1, 0) and (0,1). Find the set A and the remaining elements Ax A.
Given n(A x A) = 9 = 32
⇒n(A) = 3
But (-1, 0) and (0, 1) are in A x A
∴ A= {-1,0,1}.
Remaining elements of A x A: (-1, -1), (-1, 1),
(0,-1), (0,0), (1,-1), (1,0), (1,1).

Question 16.
If P = {1,2}, form the set
P x P x P = {(1, 1, 1), (1, 1, 2), (1, 2, 1),
(1, 2,2), (2, 1,1), (2, 1, 2), (2, 2,1), (2, 2, 2)}

Question 17.
If A = {-1,1}, find A x A x A.
A x A x A = {(-1, -1, -1), (-1, -I, 1), (-1, 1, -1), (-1, 1, 1), (1, -1, -1), (1, -1, 1), (1,1,-1), (1,1,1)}

Question 18.
If R is the set of all real numbers, what do the Cartesian products R x R and R x R x R represent?
We have R x R = {(x, y) : x, y ∈ R } which represents the coordinates of all the points in two dimensional space and R x R x R = {(x, y, z)  x,y,z ∈ R } which represents the coordinates of all the points in three-dimensional space.

Question 19.
Define a relation.
A relation R from a non-empty set A to non empty set B is a subset of the Cartesian product A x B.

Question 20.
Define domain of a relation.
The set of all first elements of the ordered pairs in a relation R from a set A to a set B is called the domain of the relation R.

Question 21.
Define range of a relation.
The set of all second elements in a relation R from a set A to a set B is called the range of the relation R. The whole set B is called the co-domain of the relation R.

Question 22.
Let A = {1, 2, 3, 4, 5, 6}. Define a relation R from A to A by R = {(x,y): y = x + 1}
(i) Depict this relation using an arrow diagram
(ii) Write down the domain, condomain and range of
Given R = {(x, y): y = x + 1}
= {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

Domain = {1, 2, 3,4, 5,}; Co-domain = A
Range = {2, 3,4, 5, 6}.

Question 23.
Let A = {1, 2, 3, ………….14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, x,
y ∈ A}. Write down its domain, co-domain and range.
Given R = {(x, y): 3x -y = 0, x, y ∈ A}
= {(1, 3), (2, 6), (3, 9), (4,12)}
Domain = {1, 2, 3,4}
Co-domain = A Range = {3, 6, 9,12}

Question 24.
Define a relation R on the set N of natural numbers by R – {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ . N}. Depict this relationship using roster form. Write down the domain and the range.
Given R = {(x, y): x, y ∈ N and y = x + 5, x < 4}
= {(1,6), (2,7), (3, 8)}
Domain = {1, 2, 3}
Range = {6, 7, 8}

Question 25.
A = {1, 2, 3, 5} and B = (4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y∈ B}. Write R in roster form.
Given A = {1, 2, 3, 5} and B = {4, 6, 9}
R = {(x, y): the difference between x and y is odd; x∈ A,y∈B}
= {(1,4), (1, 6), (2, 9), (3, 4), (3, 6), (5,4), (5, 6)}

Question 26.
The figure shows a relationship between the sets P and Write this relation
(i) in set- builder-form
(ii) roster form. What is its domain and range?

Given P={5,6,7} and Q={3,4,5}
(i) Set builder form
R= {(x,y):x-y = 2; x∈P,y ∈Q)

(ii) Roster form
R = {(5, 3), (6,4), (7, 5)}
Domain of R = P
Range of R = Q.

Question 27.
Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by {(a, b): a, b ∈A, b is exactly divisible by a},
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R
Given A = { 1, 2, 3,4, 6}
R- {(a, b),a,b∈A,bis exactly divisible by a}

(i) Roster form:
R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4,4), (6, 6)}

(ii) Domain of R = {1, 2, 3,4, 6} = A

(iii) Range of R = {1, 2, 3,4, 6} = A

Question 28.
Determine the domain and range of the relation R .defined by R = {(x, x + 5): x e {0,1, 2,3,4,5}}.
Given R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}.
Domain of R = {0, 1, 2, 3,4, 5}
Range of R = {5, 6, 7, 8, 9, 10}

Question 29.
Write the relation R = {(x, x3) : x is a prime number less than 10} in roster form.
Given R = {(x, x3): x g {2, 3, 5, 7}}
= {(2, 8), (3, 27), (5, 125), (7, 343)}

Question 30.
Let A – {x, y, z} and B = (1, 2}. Find the number of relations from A to
Given n(A) = 3 and n(B) = 2
∴ n (A x B) = 3 x 2 = 6
Number of relations from A to B = 2n (A x B) = 26 = 64

Question 31.
Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of
Given: R = {(a, b): a, b ∈Z,a-b is an integer} Domain of R-Z Range of R = Z

Question 32.
Let R be a relation from Q to Q defined by R = {(a, b)\ a,b ∈Q and a – b ∈ Z}. Show that (a, a) g R, for all a ∈Q
(ii) (a, b) ∈ R implies that (b, a) ∈R
(iii) (a, b) ∈ R and (b, c) ∈R implies that (a, c) ∈ R.
Given R = {{a, b): a,b∈Q and a-b ∈ Z)
(i) ∀ a ∈ Q, a – a = 0 ∈ Z ⇒ (a, a)∈R

(ii) Let (a, b) ∈ R⇒ a- b ∈ Z
b – a ∈ Z ⇒ (b, a) ∈ R

(iii) Let (a, b) and (b, c) g R ⇒ a – b ∈Z and
b – c ∈ Z
a- c = (a-b) + (b – c) ∈Z
∴ (a, c)∈ R

Question 33.
Let R be a relation from A to A defined by R = {(a, b): a,b ∈ N and a = b2} Are the following true?
(i) (a, a) ∈ R for all a ∈ A
(ii) (a, b) ∈ R, implies (b, a) ∈ R
(iii) (a, b) ∈ R, (b, c) ∈ R, implies (a, c) ∈ R.
Given R= {(a, b): a,b∈N and a = b2}
= {(1,1), (2,4), (3, 9), (4, 16),…}
(i) (a, a)∈ R for all a ∈ N is not true because (2, 2) ∉ R.
(ii) (a, b) ∈ R implies (b, a) ∈ R is not true, because  (2,4) ∈ R but (4,2) ∉ R
(iii) (a, b) ∈ R, (b, c) ∈ R implies (a,c) ∈ R is not true because (2,4) and (4,16) ∈ R but (2,16) ∉ R.

Question 34.
Define a function.
A relation/from a set A to a set B is said to be a function if every element of set A has one and only one image in set B, and
we write f : A → B

Question 35.
Define
(i) a real valued function
(ii) a real function
A function which has either R or one of its subsets as its range is called a real valued function. A function f: A → B is said to be a real function if both A and B are subsets of R.

Question 36.
Let N be the set of natural numbers and the relation R be defined on N such that
R = {(x,y):y = 2x,x,y ∈ N} What is the domain, co-domain and range of R? Is this relation a function?
Given R = {x, y): y = 2x; x, y ∈ N}
Domain of R = set of natural numbers Co-domain of
R = set of natural number
Range of R = set of even natural numbers
Clearly, every natural number is related to unique image, so this relation is a function.

Question 37.
Examine each of the following relations given below and state in each case, giving reasons whether it is a function or not?
(i) R = {(2,1), (3,1), (4,2)}
(ii) R = {(2,2), (2,4), (3,3), (4,4)}
(iii) R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6,7)}
(i) Given R= {(2,1), (3,1), (4, 2)}
Here every element of domain is related to unique element of co-domain, so it is a function.

(ii) Given R = {(2, 2), (2,4), (3, 3), (4, 4)}
Since the element 2 has two images namely 2 and 4, so it is not a function.

(iii) Given R = {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7)}
Since every element of domain is related to unique

Question 38.
Let N be the set of natural numbers. Define a real valued function f: N → N by
f(x) = 2x + 1. Using this definition, complete the table given below:

 X 1 2 3 4 5 6 7 y F : (1) = F : ( 2) = F : (3) = F : (4) = F : (5) = F : (6) = F : (7) =

Given function is f(x) = 2x + 1.

 X 1 2 3 4 5 6 7 y F:(1)=3 F:(2)=5 F:(3)=7 F:(4)=9 F:(5)=11 F:(6)=13 F:(7)=15

Question 39.
Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2,1), (5,1), (8,1), (11,1), (14,1), (17,1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14,7)}
(iii) {(1,3), (1,5), (2,5)}.
(i) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2, 5, 8, 11, 14, 17}
Range = {1}

(ii) Clearly, every element of domain is related to unique element of co-domain, so it is a function.
Domain = {2,4, 6, 8,10,12,14}
Range = {1,2, 3,4, 5, 6,7}

(iii) 1 is related to two elements of co-domain, namely 3 and 5, so it is not a function.

Question 40.
Let A={1,2,3,4), B={1,5,9,11,15,16} and f = {(1,5),(2,9), (3,1), (4,5),(2, 11)). Are the
following true?
(i) f is a relation from A to B
(ii) f is a function from A to B. Justify your answer in each case.
(i) Every element off is an element of A x B, so f is a relation.
(ii) ‘f’ is not a function the element 2 has two images.

Question 41.
Let f be the subset of Z x Z defined by f ={(ab,a +b):a,b ∈z) Is f a function from Z to Z? Justify your answer.
Given f={(ab,a+b):a,b∈Z)
If a = 1 and b = 4 = ab = 4 and a+b=5
∴ (4,5) ∈ fIf a=2 and b=2⇒ab=4 and a+b=4
∴ (4,4) ∈ f∴ The element 4 has two images, so f is not a function.

Question 42.
The relation f is defined by
$$f(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 3} \\ {3 x,} & {3 \leq x \leq 10}\end{array}\right.$$
The relation g is defined by
$$g(x)=\left\{\begin{array}{ll}{x^{2},} & {0 \leq x \leq 2} \\ {3 x,} & {2 \leq x \leq 10}\end{array}\right.$$ .
Show that/is a function and g is not a function.
Since f(x) is unique for 0 ≤ x ≤ 10.
f(x) is a function. g(2) = 22 =4 and g(2) = 3(2) = 6
∴ z has two images under g.
∴ g is not a function.

Question 43.
Let f= {(1,1),(2,3),(0,-1),(-1,-3)} be a linear function from Z into Z. Find f(x) ).
Since/is a linear function.
∴f{x) = ax + b
But (0,-1) ∈ f . f(0) ∴ a(0) + b ⇒ -1 = b
Similarly, (1,1) ∈ f ∴ f(1) = a(1) +a(1)+b
⇒ 1=a+b ∴ a=2 ∴ f(x) = 2x -1

Question 44.
A function f is defined by f(x) = 2x-5. Write down the values of (i) f(0) (ii) f(17) (iii) f(-3).
Given: f(x) = 2x-5

• f(0) = 2(0) – 5 = -5
• f(17) = 2(17)-5 = 29
• f(-3) = 2(-3) – 5 = -11

Question 45.
The function ‘t’ which maps temperature in degree Celsius into temperature in degree. Fahrenheit is defined by $$t(c)=\frac{9 c}{5}+32$$. Find
(i) t (0)
(ii) t (28)
(iii) t (-10)
(iv) The value of c, when t(c) – 212

Question 46.
$$\text { If } f(x)=x^{2}, \text { find } \frac{f(1 \cdot 1)-f(1)}{1 \cdot 1-1}$$

Question 47.
Find the range of each of the following functions:
(i) f{x) = 2-3x, x∈R,x>0
(ii) f(x) = x2 + 2, x is a real number
(iii) f(x) = x, x is a real number.
(i) Given f(x) = 2-3x, x∈R,x>0
For x > 0,f(x) = 2 – 3x < 2
∴ Range of f= (-∞, 2)

(ii) Given f(x) = x2 + 2, x ∈ R
For x ∈ R, f(x) = x2 + 2 ≥ 2
Range of f = [2, ∞)

(iii) Given f(x) = x, x∈ R For r∈ E, f(x) = x∈R
Range of f = R

Question 48.
Let A = (9, 10, 11, 12, 13} and let f:A→N be defined by f(n) = the highest prime factor of n. Find the range of f.
Given f(n) = the highest prime factor of n.
f(9) = the highest prime factor of 9 = 3
f(10) = the highest prime factor of 10 = 5
f(11) = the highest prime factor of 11 = 11
f(12) = the highest prime factor of 12 = 3
f(13) = the highest prime factor of 13 = 13
Range of f ={3,5,11,13}

Question 49.
$$\text { Let } f=\left\{\left(x, \frac{x^{2}}{1+x^{2}}\right): x \in \mathbb{R}\right\}$$ be a function from R to R .Determine the range of f.

Question 50.
Find the domain of the function $$f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$
Given $$f(x)=\frac{x^{2}+3 x+5}{x^{2}-5 x+4}$$
f(x) is defined for all real numbers except x2 – 5x + 4 = 0
But x2 -5x + 4 = (x-4) (x-1)
Domain of f= R-{1,4}

Question 51.
Find the domain of the function
$$f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$
$$f(x)=\frac{x^{2}+2 x+1}{x^{2}-8 x+12}$$
is not defined when
x2 – 8+12 = 0.
∴ (x-6)(x-2) = 0
∴ Domain of f = R-{2,6}

Question 52.
Find the domain and range of the following real functions:
(i) $$f(x)=-|x|$$
(ii) $$f(x)=\sqrt{9-x^{2}}$$
(iii) $$f(x)=\sqrt{x-1}$$
(iv) $$f(x)=|x-1|$$

Remark: Operations of functions:
(i) Addition of two real functions:
Let f: X → R and g : X → R Then.
(f + g):X → R; (f + g)(x) = f(x) + g(x) for all x ∈ X

(ii) Difference of two real functions:
Let f : X → R and g : X → R Then.
(f – g): X → R; (f- g)(x) = f(x) – g(x) for all x ∈ X

(iii) Scalar multiplication of a function:
Let f: X → R and let ‘a’ he a scalar. Then,
(αf):X → R; (fg)(x) = f(x) for all x ∈ X

(iv) Multiplication of two real functions:
Let f: X → R and g : X → R .Then
(fg): X → R; (fg)(x)= f(x)g(x), for all x ∈ X

(v) Quotient of two real functions:
Let f: X → R and  g : X → R for all x for which g(x) ≠ 0. Then
$$\left(\frac{f}{g}\right): X \rightarrow \mathbb{R} ;\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}$$

Question 53.
Let f(x) = x2 and g(x) = 2x + 1 be two real functions. Find (f+g)(x),(f-g)(x)
$$(f g)(x),\left(\frac{f}{g}\right)(x)$$

Question 54.
Let $$f(x)=\sqrt{x} \text { and } g(x)=x$$ be two functions defined over the set of non-negative real numbers. Find (f+ g)(x), (f – g)(x) $$(f g)(x) \text { and }\left(\frac{f}{g}\right)(x)$$

Question 55.
Let f,g: R→R be defined, respectively by f(x) = x + 1,g(x) = 2x-3- Find f + g,f-g and $$\frac{f}{g}$$
Given: f(x) = x + 1,g(x) = 2x-3
(f + g)(x) = f (x) + g(x) = x + 1 + 2x-3 = 3x-2
(f – g)(x) = fix) – g(x) = x + 1-2x + 3 = -x + 4
$$\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{x+1}{2 x-3}, x \neq \frac{3}{2}$$

Question 56.
Define an identity function and draw its graph also find its domain and range.
The function f: M → R; f(x) = x for a II x∈R is called an identity function on R.

Question 57.
Define a constant function and draw its graph also find its domain and range.
Let c be a fixed real number. Then, the function f : R→R, f(x) = c for all x∈R is called the constant function.

f(x) is defined for all real number,
∴ Domain = M
Range = { c }

Question 58.
Define a polynomial function.
A function f : M → R is said to be polynomial function if for each x in IR,
y = f (x) = a0 + axx + a2x2 + ……………. + anxn
where n is a non-negative integer and a0,a1,a2………………….an ∈ R .

Question 59.
Draw the graph of the function f (x) = x2 and write its domain and range.
Given function: f(x)= x2

Domain = R
Range = set of non-negative reals.

Question 60.
Draw the graph of the function f: R → R defined by f(x) = x3 Find its domain and range.
Let f : R → R: f(x) = x3, ∀ ∈ R
Then, domain of f = R and range of f = R . we have

Question 61.
Define a rational function
The functions of the type $$\frac{f(x)}{g(x)}$$ where f(x) g(x) and g(x) are polynomial functions of x, defined in a domain, where g(x) ≠ 0

Question 62.
Let f : R – {0} → R defined by $$f(x)=\frac{1}{x}, \forall x \in \mathbb{R}-\{0\}$$ . Find its domain and x range. Also, draw its graph.
Given function is f : R – {0} → R defined by $$f(x)=\frac{1}{x}$$
∴ Domain = R-{0} and range =R-{0}

 X -4 -2 -1 -0-5 -0-25 0-25 0-5 1 2 f(x)=1/x -0-25 -0-5 -1 -2 -4 4 2 1 0-5

Question 63.
Define a modulus function. Find its domain and range. Also, draw its graph.
Let f: R → R defined by f(x) =1 x I, for each x ∈ R, is called modulus function.
$$\text { i.e., } f(x)=|x|=\left\{\begin{array}{ll}{x,} & {\text { if } x \geq 0} \\ {-x,} & {\text { if } x<0}\end{array}\right.$$

Domain = R
Range = set of non negative real numbers

Question 64.
Define Signum function. Draw its graph and find its domain and range.
The function f: R → R defined by
$$f(x)=\left\{\begin{array}{lll}{1,} & {\text { if }} & {x>0} \\ {0,} & {\text { if }} & {x=0} \\ {-1,} & {\text { if }} & {x<0}\end{array}\right.$$ is called signum function
We have

Domain = R
Range = {-1,0,1}

Question 65.
Define a greatest integer function. Draw its graph and find its domain and range.

The function f : R → R define by f(x) = [x], x∈ R assumes the value of the greatest integer, less than or equal to x. Such a’ function is called the greatest integer function or step function. We have
[x] = -2 for – 2 ≤ x < -1
[x] = -1 for -1≤x<0
[x] = 0 for 0≤x≤1
[x] = 1 for 1 ≤ x < 2
[x] = 2 for 2 ≤ x < 3.
Hence, domain of f = R and range = Z.

Question 66.
Define a linear function.
The function f : R → R defined by f(x) = mx + c, x ∈ R is called linear function, where m and c are constant.

Question 67.
Let R be the set of real numbers. Define the real function f : R →R by f(x) = x + 10 and sketch the graph of this function.
$$f(x)=\left\{\begin{array}{cl}{1-x,} & {x<0} \\ {1,} & {x=0} \\ {x+1,} & {x>0}\end{array}\right.$$.


text:
HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS

Asset to Sales Ratio

The asset to sales ratio is calculated by dividing total assets by sales revenues. The asset to sales formula can be used to compare how much in assets a company has relative to the amount of revenues the company can generate using their assets

$\frac{a}{s}$

Here,a=total assests,s=sales revenue

ENTER THE VARIABLES TO BE USED IN THE FORMULA

Similar formulas which you may find interesting.

format: plain


text:
Here's the question you clicked on:

## janessia Find x. picture attached! one year ago one year ago

• This Question is Closed
1. janessia

2. janessia

do i just do the pythagorean therom?

3. ChmE

use the Pythagorean theorem. $a ^{2}+b ^{2}=c ^{2}$Where a and b are the sides that make up the right angle, and c is the longest side.

4. janessia

i got 676 when i did the pythag theorem

5. ChmE

So.. $10^{2}+24^{2}=c ^{2}$

6. etemplin

you forgot to take the square root of that.

7. janessia

yea i got 676 is that right?

8. janessia

oh ohk thanks @etemplin

9. janessia

its 26!?

10. janessia

@etemplin

11. ChmE

yes

12. etemplin

yes

13. janessia

thanks you guys!

14. ChmE

$676=c^{2} \rightarrow \sqrt{676}=\sqrt{c ^{2}}=c$

15. ChmE

yup


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# Understanding Sweedler's notation for the structure map of a comodule

I was hoping someone might be able to shed some light on the choice of indices for expressing the coaction using Sweedler notation.

For example, in the paper of Andruskiewitsch About finite-dimensional Hopf algebras, (see page 3) http://cel.archives-ouvertes.fr/docs/00/37/43/83/PDF/2000Andrusk.pdf, $C$ is a coalgebra, $V$ is a left comodule, and the structure map is expressed as

$$\delta: V \longrightarrow C \otimes V$$ $$\delta(x) = x_{(-1)} \otimes x_{(0)}$$

Can someone explain this choice of indices and how it is meaningful for understanding the coaction? I am working on building an intuition for comodules apart from the fact that they are dual to modules, and understanding this notation would be helpful.

The use of Sweedler notation for the comultiplication is familiar and clear to me:

$$\Delta(x)= x_{(1)} \otimes x_{(2)}$$

but I am unsure of the significance of the $(-1)$ and $(0)$.

• Maybe it is helpful: Sweddler's book on Hopf algebras has several exercises about this notation. – Leandro Vendramin Nov 15 '13 at 22:56

I would start with considering the case of a right $C$-comodule $U$, for which the structure map $\delta\colon U\to U\otimes C$ would be expressed as $\delta(u) = u_{(0)}\otimes u_{(1)}$. The double (iterated) coaction map $U\to U\otimes C\otimes C$ would then be denoted by $u\mapsto u_{(0)}\otimes u_{(1)}\otimes u_{(2)}$, with the coassociativity equiation (for the coaction) being $u_{(0)}\otimes u_{(1)(1)}\otimes u_{(1)(2)} = u_{(0)}\otimes u_{(1)}\otimes u_{(2)} = u_{(0)(0)}\otimes u_{(0)(1)}\otimes u_{(1)}$.
Now this looks quite natural, with the index ${}_{(0)}$ distinguishing the factors belonging to $U$, while the factors with positive indices (or in more complicated formulas, with at least one positive index) belonging to $C$.
If this is clear, notice that the indices always go in the order of succession of the integers in our tensor notation, that is one has $j=i+1$ whenever there are two successive factors like $x_{(i)}\otimes x_{(j)}$ in our expressions. Given that, and the desire to reserve the index ${}_{(0)}$ for the factors belonging to comodules (as opposed to coalgebras), the notation $\delta(v)= v_{(-1)}\otimes v_{(0)}$ for a left coaction map $\delta\colon V\to C\otimes V$ comes out pretty naturally, too.


text:
1,046 views

If $r$ is a constant such that $|x^{2} – 4x -13| = r$ has exactly three distinct real roots, then the value of $r$ is

1. $15$
2. $18$
3. $17$
4. $21$

Given that,  $|x^{2}-4x-13| = r\quad \longrightarrow (1)$

We know that, $|x| = \left\{\begin{matrix}x \;;&x \geq 0 \\ -x\;; &x<0 \end{matrix}\right.$

$\textbf{Case 1:}\; x^{2}-4x-13 = r$

$\Rightarrow x^{2}-4x-13-r = 0$

For two distinct real roots, the discriminant will be greater than zero.

$\Rightarrow D_{1}>0$

$\Rightarrow b^{2}-4ac>0 \quad [\because \text{For the quadratic equation}\; ax^{2} + bx +c = 0]$

$\Rightarrow (-4)^{2}-4(-13-r)>0$

$\Rightarrow 16+52+4r>0$

$\Rightarrow 4r+68>0$

$\Rightarrow \boxed{r>-17}$

$\textbf{Case 2:}\; x^{2}-4x-13 = -r$

$\Rightarrow x^{2}-4x-13+r = 0$

For one real root, the discriminant will be zero.

$\Rightarrow D_{2} =0$

$\Rightarrow b^{2}-4ac=0$

$\Rightarrow (-4)^{2}-4(1)(-13+r)=0$

$\Rightarrow 16+52-4r=0$

$\Rightarrow 68-4r=0$

$\Rightarrow 4r=68$

$\Rightarrow \boxed{r=17}$

$\therefore$ The value of $r = 17$

Correct Answer $:\text{C}$

$\textbf{PS:}$

We know the quadratic formula is

$$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

for any quadratic equation written in standard form of $ax^2+bx+c=0.$ The discriminant $D$ for the quadratic equation is

$$D=b^2-4ac,$$

where

$$\begin{cases} b^2-4ac \gt 0: & \text{two distinct real roots} \\ b^2-4ac=0: & \text{equal and real roots} \\ b^2-4ac \lt 0: & \text{imaginary roots}. \end{cases}$$

10.3k points

1
101 views
1 vote


text:
CGAL 5.0.2 - Polynomial
PolynomialTraits_d::TotalDegree Concept Reference

## Definition

This AdaptableUnaryFunction computes the total degree of a PolynomialTraits_d::Polynomial_d.

Given a (multivariate) monomial the sum of all appearing exponents is the total degree of this monomial. The total degree of a polynomial $$p$$ is the maximum of the total degrees of all appearing (multivariate) monomials in $$p$$.

For instance the total degree of $$p = x_0^2x_1^3+x_1^4$$ is $$5$$.

The total degree of the zero polynomial is set to $$0$$. From the mathematical point of view this should be $$-\infty$$, but this would imply an inconvenient return type.

Refines:

AdaptableUnaryFunction

CopyConstructible

DefaultConstructible

Polynomial_d
PolynomialTraits_d
PolynomialTraits_d::Degree
PolynomialTraits_d::DegreeVector

## Types

typedef int result_type

typedef PolynomialTraits_d::Polynomial_d argument_type

## Operations

result_type operator() (argument_type p)
Computes the total degree of $$p$$.


text:
# How do I align enumerate items inside empty description?

I have an enumerate list inside an empty description item.

The minimal file is,

\documentclass[11pt]{article}

\begin{document}
\begin{description}
\item [A Label:]

\begin{enumerate}
\item A1.
\item A2.
\item A3.
\end{enumerate}
\end{description}

\end{document}

The problem is, A1 comes after the colon (:). As you can see here,

But I want A1 in a new line, with 1. aligned with 2.

I tried putting a \relax in the empty description item. Nothing was improved.

Any suggestion will be appreciated.

-


text:
# American Institute of Mathematical Sciences

October  2021, 8(4): 447-494. doi: 10.3934/jcd.2021017

## Tracking the critical points of curves evolving under planar curvature flows

 1 MTA-BME Morphodynamics Research Group 2 Department of Morphology and Geometric Modeling, Budapest University of Technology and Economics, Müegyetem rakpart 1-3., 1111 Budapest, Hungary 3 Department of Mechanics, Materials and Structures, Budapest University of Technology and Economics, Müegyetem rakpart 1-3., 1111 Budapest, Hungary 4 Department of Mathematics, University of Auckland, Private Bag 92019, Auckland 1142, New Zealand

Received  June 2021 Revised  October 2021 Published  October 2021 Early access  November 2021

We are concerned with the evolution of planar, star-like curves and associated shapes under a broad class of curvature-driven geometric flows, which we refer to as the Andrews-Bloore flow. This family of flows has two parameters that control one constant and one curvature-dependent component for the velocity in the direction of the normal to the curve. The Andrews-Bloore flow includes as special cases the well known Eikonal, curve-shortening and affine shortening flows, and for positive parameter values its evolution shrinks the area enclosed by the curve to zero in finite time. A question of key interest has been how various shape descriptors of the evolving shape behave as this limit is approached. Star-like curves (which include convex curves) can be represented by a periodic scalar polar distance function $r(\varphi)$ measured from a reference point, which may or may not be fixed. An important question is how the numbers and the trajectories of critical points of the distance function $r(\varphi)$ and of the curvature $\kappa(\varphi)$ (characterized by $dr/d\varphi = 0$ and $d\kappa /d\varphi = 0$, respectively) evolve under the Andrews-Bloore flows for different choices of the parameters.

We present a numerical method that is specifically designed to meet the challenge of computing accurate trajectories of the critical points of an evolving curve up to the vicinity of a limiting shape. Each curve is represented by a piecewise polynomial periodic radial distance function, as determined by a chosen mesh; different types of meshes and mesh adaptation can be chosen to ensure a good balance between accuracy and computational cost. As we demonstrate with test-case examples and two longer case studies, our method allows one to perform numerical investigations into subtle questions of planar curve evolution. More specifically — in the spirit of experimental mathematics — we provide illustrations of some known results, numerical evidence for two stated conjectures, as well as new insights and observations regarding the limits of shapes and their critical points.

Citation: Eszter Fehér, Gábor Domokos, Bernd Krauskopf. Tracking the critical points of curves evolving under planar curvature flows. Journal of Computational Dynamics, 2021, 8 (4) : 447-494. doi: 10.3934/jcd.2021017
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show all references

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The piecewise polynomial central radial distance function $r$ (a1) and its derivatives $r'$ (a2) and $r''$ (a3), generated by the phase-uniform mesh (black dots) of size $N = 20$ on the standard example curve (b), as defined in (36)
The piecewise polynomial central radial distance function $r$ generated by a uniform mesh of size $N = 20$ on the standard ellipse (38) with $a = 1$ and $b = 0.5$, for a phase-uniform mesh in panels (a1) and (a2), and for an arclength-uniform mesh in panels (b1) and (b2)
">Figure 3.  Illustration of step $i$ in panels (a) and of step $i+1$ in panels (b) of the mesh propagation for a mesh of size $N = 20$. The radial distance function $r_{i-1}$ in panel (a1) generates the parameterization of the curve $\Gamma_{i-1}$ in panel (a2) with normals $\mathbf{n}_{i-1}$ (red arrows) at the mesh points (black dots). Euler steps at each mesh point give the new mesh that defines the next radial distance function $r_{i}$ shown in panel (b1). It generates the parameterization of the next curve $\Gamma_{i}$ in panel (b2) and its normals $\mathbf{n}_{i}$ (red arrows) at the mesh points (black dots). The figure is for the standard example curve from (36); compare with Fig. 1
Evolution of the standard curve (36) under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, as computed for stepsize parameter $S = 10^{-5}$ up to $t = 0.1062$ in 36,599 steps with a phase-uniform mesh of size $N = 20$ at each step. Panels (a1) and (b1) show the radial distance functions of the final curve for the centroid $C_t$ and for $O = (0.05, 0.1)$ as the reference point, respectively. Panels (a2) and (b2) show seven curves at equally distributed time steps from $t = 0$ to $t = 0.1062$, together with trajectories of the critical points, determined from the minima (green) and maxima (red) of the respective radial distance function
Evolution under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, of the trajectories of the critical points (green for minima and red for maxima) of the radial distance function $r_{O}$ for the fixed reference point $O = (0.01,0.05)$ (black dot) for the standard curve (36) in column (a), the curve given by (40) in column (b), and the curve given by (41) in column (c). Each computation is for an arclength-uniform mesh with $N = 80$ points, $S = 10^{-5}$ and remeshing at every 10th time step. The top row shows the trajectories of the critical points of $r_{O}$ with the respective initial curve and mesh in $\mathbb{R}^2$, and the bottom row shows them on the evolution surface in $\mathbb{R}^2 \times \mathbb{R}^+$
Evolution under the curve-shortening flow (1), (5) with $c = 0$ and $\alpha = 1$, of the trajectories of the critical points (green for minima and red for maxima) of the radial distance function $r_{C}$, with reference point at the (moving) centroid $C(t)$, for the standard curve (36) in column (a), the curve given by (40) in column (b), and the curve given by (41) in column (c). Also shown are the trajectories of the critical points (light blue for minima and dark blue for maxima) of the signed curvature function $\kappa$. Each computation is for an arclength-uniform mesh with $N = 80$ points, $S = 10^{-5}$ and remeshing at every 10th time step. The top row shows the trajectories of the critical points of $r_{C}$ and of $\kappa$ with the respective initial curve and mesh in $\mathbb{R}^2$, and the bottom row shows them on the evolution surface in $\mathbb{R}^2 \times \mathbb{R}^+$
Evolution of the standard ellipse (38) with $a = 1$ and $b = 0.5$ under (5) with $c = 0$ for $\alpha = 1, 0.8, 0.6, 0.4, \frac{1}{3}, 0.25, 0.125, 0.1$, computed with an arclength-uniform mesh with $N = 80$, remeshing at every 10th time step and $S = 10^{-5}$. Panel (a) shows the respective time series of the isoperimetric quotient $Q$ and panel (b) presents the rescaled final curves
Evolution of the standard curve (36) under (5) with $c = 0$ for $\alpha = 1, 0.8, 0.6, 0.4, \frac{1}{3}, 0.25, 0.125, 0.1$, computed with an arclength-uniform mesh with $N = 80$, remeshing at every 10th time step and $S = 10^{-5}$. Panel (a) shows the respective time series of the isoperimetric quotient $Q$ and panel (b) presents the rescaled final curves
Evolutions under (5) with $c = 0$ and $\alpha = 0.1$ of the $D_n$-symmetric curves (43) with $n = 3$ and $\delta = 0.05$ (a), and with $n = 4$ and $\delta = 0.03$ (b), computed with an arclength-uniform mesh with $N = 60$ for $n = 3$ and $N = 64$ for $n = 4$, remeshing at every 10th time step and $S = 10^{-4}$. Shown are a selection of curves with their respective meshes and the trajectories of the stationary points (green for minima and red for maxima); the computed final shape of the respective evolution is at the bottom right of each panel
">Figure 10.  Evolutions under (5) with $c = 0$ and $\alpha = 0.05$ of the $D_n$-symmetric curves (43) with $n = 3$ and $\delta = 0.05$ (a), with $n = 4$ and $\delta = 0.03$ (b), with $n = 5$ and $\delta = 0.02$ (c), and with $n = 6$ and $\delta = 0.01$ (d), computed with an arclength-uniform mesh with $N = 64$ for $n = 4$ and $N = 60$ otherwise, remeshing at every 10th time step and $S = 10^{-4}$; illustrated as in Fig.9
">Figure 11.  Evolutions under (5) with $c = 0$ and $\alpha = 0.01$ of the $D_n$-symmetric curves (43) with $n = 3$ and $\delta = 0.01$ (a), with $n = 4$ and $\delta = 0.03$ (b), with $n = 5$ and $\delta = 0.02$ (c), and with $n = 6$ and $\delta = 0.01$ (d), computed with an arclength-uniform mesh with $N = 64$ for $n = 4$ and $N = 60$ otherwise, remeshing at every 10th time step and $S = 10^{-4}$; illustrated as in Fig.9
and Fig. 11, respectively; the horizontal dashed lines are at the values $Q_n$ of the regular $n$-gons">Figure 12.  Evolution under (5) with $c = 0$ and $\alpha = 0.05$ (a) and $\alpha = 0.01$ (b) of the isoperimetric quotient $Q$ for the $D_n$-symmetric curves from Fig. 10 and Fig. 11, respectively; the horizontal dashed lines are at the values $Q_n$ of the regular $n$-gons
">Figure 13.  Evolutions under (5) with $c = 0$ and $\alpha = 0.1$ of the $C_n$-symmetric curves (45) with $n = 3$, $\delta = 0.05$ and $\xi = 0.01$ (a), and with $n = 4$, $\delta = 0.03$ and $\xi = 0.01$ (b), computed with an arclength-uniform mesh with $N = 60$ for $n = 3$ and $N = 64$ for $n = 4$, remeshing at every 10th time step and $S = 10^{-5}$; illustrated as in Fig. 9
Discrete and integral error norms for the discretizations with a phase-uniform mesh of size $N$ with the centroid as reference point of the standard curve (36) in column (a) and of the standard ellipse (38) with $a = 1$ and $b = 0.5$ in column (b)
Evolution of the standard curve (36) under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, for a phase-uniform mesh with remeshing at every time step, for mesh size $N = 10$ (a), $N = 60$ (b), $N = 80$ (c) and $N = 100$ (d). Here $S = 10^{-5}$, the reference point is the centroid, and the stationary points are determined from the minima (green) and maxima (red) of the central radial distance function
Evolution of the standard curve (36) in row (a) and of the ellipse (38) with $a = 1$ and $b = 0.5$ in row (b) under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, computed with a phase-uniform mesh with $N = 80$ points. Panels (a1) and (b1) are computed without remeshing, panels (a2) and (b2) with remeshing at every 10th time step, and panels (a3) and (b3) with remeshing at every time step. Here $S = 10^{-5}$, the reference point is the centroid, and the stationary points are determined from the minima (green) and maxima (red) of the central radial distance function
Evolution of the standard curve (36) in row (a), of the ellipse (38) with $a = 1$ and $b = 0.5$ in row (b) and of the ellipse (38) with $a = 2$ and $b = 0.5$ in row (c) under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, computed with an initial arclength-uniform mesh with $N = 80$ points. Panels (a1), (b1) and (c1) are computed without remeshing, panels (a2), (b2) and (c2) with remeshing at every 10th time step, and panels (a3), (b3) and (c3) with remeshing at every time step; during the remeshing the number $N$ of mesh points is reduced to keep the arclength distance $L/N$ between mesh points approximately constant, down to a minimum of $N_{\rm min} = 40$ mesh points. Here $S = 10^{-5}$, the reference point is the centroid, and the stationary points are determined from the minima (green) and maxima (red) of the central radial distance function
Evolution under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, of three rotations of the standard curve (36) over $\varphi_0$ as indicated, computed with a phase-uniform mesh of $N = 80$ points with remeshing at every 10th time step. The left column shows the mesh on the standard curve and the trajectories of the stationary points, and the right column shows these trajectories on the evolution surface in $\mathbb{R}^2 \times \mathbb{R}^+$. Here $S = 10^{-5}$, the reference point is the centroid, and the stationary points are determined from the minima (green) and maxima (red) of the central radial distance function
Evolution under the curve-shortening flow, (5) with $c = 0$ and $\alpha = 1$, of three rotations of the standard ellipse (38) with $a = 1$ and $b = 0.5$ over $\varphi_0$ as indicated, computed with a phase-uniform mesh of $N = 80$ points with remeshing at every 10th time step. The left column shows the mesh on the standard curve and the trajectories of the stationary points, and the right column shows these trajectories on the evolution surface in $\mathbb{R}^2 \times \mathbb{R}^+$. Here $S = 10^{-5}$, the reference point is the centroid, and the stationary points are determined from the minima (green) and maxima (red) of the central radial distance function
Evolution under the affine shortening flow, (5) with $c = 0$ and $\alpha = \frac{1}{3}$, of the standard ellipse (38), with $a = 1$ and $b = 0.5$ in column (a) and with $a = 2$ and $b = 0.5$ in column (b). Each computation is for an arclength-uniform mesh with $N = 80$ points, $S = 10^{-4}$ and the centroid as the reference point. Panels (a1) and (b1) show selected curves during the computation with the trajectories of the stationary points, determined from the extrema of the central radial distance function. Panels (a2) and (b2) show the distance of the corresponding isoperimetric quotient $Q$ from its value for the respective ellipse as a function of time $t$, and panels (a3) and (b3) show the integral distance of the computed curves $\Gamma_{t_i}$ from the ellipse with same area and axis ratio, determined for the functions $r$, $r'$, $r''$ and $\kappa$
A star-like curve whose centroid $C$ is not a reference point (a), and a non-convex star-like curve whose centroid $C$ can be chosen as a reference point (b). Also shown are the stationary points (red dots), the inflection points (blue dots) and their tangent lines (blue lines), which bound the set of reference points (shaded)
The critical point evolution algorithm with mesh propagation by an Euler step and remeshing
 Require: a star-like closed curve $\Gamma$; parameters $c$ and $\alpha$ of (5); number of mesh points $N$; stepsize parameter $S$; accuracy parameters $h_{\rm min}$ and $A_{\rm min}$; Ensure: construct the piecewise polynomial radial distance function $r_0$ for a uniform mesh $M_0$ of size $N$ with dual mesh $M^\varphi_0$ for a suitable reference point $O_0$; ensure that the associated curve $\Gamma_0$ with parameterization $\gamma_0$ is sufficiently close to $\Gamma$; set the step counter $i$ to 1; While none of criteria (ⅰ)-(ⅳ) from Sec. 4.3 are satisfied do calculate the polynomial derivatives $r_{i-1}'$ and $r_{i-1}''$ symbolically;     determine the stepsize $h_i$ from $\kappa_{i-1}$;     evaluate the associated normals $\mathbf{n}_{i-1}$ and the curvature $\kappa_{i-1}$ in the points of $M^\varphi_{i-1}$;     perform the Euler step (39) at the mesh points to obtain the propagated mesh $M_{i} = E_{h_i}(M_{i-1})$ with dual $M^\varphi_i$;     choose a suitable reference point $O_i$ for the new curve $\Gamma_{i} = E_{h_i}(\Gamma_{i-1})$;     construct the piecewise polynomial radial distance function $r_i$ for $M_i$ and $O_i$;     [when required: remesh the curve with a uniform mesh $M^*$ by computing the associated piecewise polynomial radial distance function $r^*_i$ with respect to $O_i$; set $r_i = r^*_i$];     increase the step counter $i$ by 1; end while
 Require: a star-like closed curve $\Gamma$; parameters $c$ and $\alpha$ of (5); number of mesh points $N$; stepsize parameter $S$; accuracy parameters $h_{\rm min}$ and $A_{\rm min}$; Ensure: construct the piecewise polynomial radial distance function $r_0$ for a uniform mesh $M_0$ of size $N$ with dual mesh $M^\varphi_0$ for a suitable reference point $O_0$; ensure that the associated curve $\Gamma_0$ with parameterization $\gamma_0$ is sufficiently close to $\Gamma$; set the step counter $i$ to 1; While none of criteria (ⅰ)-(ⅳ) from Sec. 4.3 are satisfied do calculate the polynomial derivatives $r_{i-1}'$ and $r_{i-1}''$ symbolically;     determine the stepsize $h_i$ from $\kappa_{i-1}$;     evaluate the associated normals $\mathbf{n}_{i-1}$ and the curvature $\kappa_{i-1}$ in the points of $M^\varphi_{i-1}$;     perform the Euler step (39) at the mesh points to obtain the propagated mesh $M_{i} = E_{h_i}(M_{i-1})$ with dual $M^\varphi_i$;     choose a suitable reference point $O_i$ for the new curve $\Gamma_{i} = E_{h_i}(\Gamma_{i-1})$;     construct the piecewise polynomial radial distance function $r_i$ for $M_i$ and $O_i$;     [when required: remesh the curve with a uniform mesh $M^*$ by computing the associated piecewise polynomial radial distance function $r^*_i$ with respect to $O_i$; set $r_i = r^*_i$];     increase the step counter $i$ by 1; end while
 [1] Leif Arkeryd, Raffaele Esposito, Rossana Marra, Anne Nouri. Ghost effect by curvature in planar Couette flow. Kinetic & Related Models, 2011, 4 (1) : 109-138. doi: 10.3934/krm.2011.4.109 [2] Gioconda Moscariello, Antonia Passarelli di Napoli, Carlo Sbordone. Planar ACL-homeomorphisms : Critical points of their components. Communications on Pure & Applied Analysis, 2010, 9 (5) : 1391-1397. doi: 10.3934/cpaa.2010.9.1391 [3] Jaime Arango, Adriana Gómez. Critical points of solutions to elliptic problems in planar domains. Communications on Pure & Applied Analysis, 2011, 10 (1) : 327-338. doi: 10.3934/cpaa.2011.10.327 [4] Massimo Grossi. On the number of critical points of solutions of semilinear elliptic equations. Electronic Research Archive, 2021, 29 (6) : 4215-4228. doi: 10.3934/era.2021080 [5] Tetsuya Ishiwata. On the motion of polygonal curves with asymptotic lines by crystalline curvature flow with bulk effect. Discrete & Continuous Dynamical Systems - S, 2011, 4 (4) : 865-873. doi: 10.3934/dcdss.2011.4.865 [6] Petr Pauš, Shigetoshi Yazaki. Segmentation of color images using mean curvature flow and parametric curves. Discrete & Continuous Dynamical Systems - S, 2021, 14 (3) : 1123-1132. doi: 10.3934/dcdss.2020389 [7] Miroslav KolÁŘ, Michal BeneŠ, Daniel ŠevČoviČ. Area preserving geodesic curvature driven flow of closed curves on a surface. Discrete & Continuous Dynamical Systems - B, 2017, 22 (10) : 3671-3689. doi: 10.3934/dcdsb.2017148 [8] Kensuke Yoshizawa. The critical points of the elastic energy among curves pinned at endpoints. Discrete & Continuous Dynamical Systems, 2022, 42 (1) : 403-423. doi: 10.3934/dcds.2021122 [9] Leif Arkeryd, Raffaele Esposito, Rossana Marra, Anne Nouri. Erratum to: Ghost effect by curvature in planar Couette flow [1]. Kinetic & Related Models, 2012, 5 (3) : 669-672. doi: 10.3934/krm.2012.5.669 [10] Nicolas Dirr, Federica Dragoni, Max von Renesse. Evolution by mean curvature flow in sub-Riemannian geometries: A stochastic approach. Communications on Pure & Applied Analysis, 2010, 9 (2) : 307-326. doi: 10.3934/cpaa.2010.9.307 [11] Jaume Llibre, Jesús S. Pérez del Río, J. Angel Rodríguez. Structural stability of planar semi-homogeneous polynomial vector fields applications to critical points and to infinity. Discrete & Continuous Dynamical Systems, 2000, 6 (4) : 809-828. doi: 10.3934/dcds.2000.6.809 [12] Tetsuya Ishiwata, Takeshi Ohtsuka. Evolution of a spiral-shaped polygonal curve by the crystalline curvature flow with a pinned tip. Discrete & Continuous Dynamical Systems - B, 2019, 24 (10) : 5261-5295. doi: 10.3934/dcdsb.2019058 [13] Sujay Jayakar, Robert S. Strichartz. Average number of lattice points in a disk. Communications on Pure & Applied Analysis, 2016, 15 (1) : 1-8. doi: 10.3934/cpaa.2016.15.1 [14] Georgi I. Kamberov. Recovering the shape of a surface from the mean curvature. Conference Publications, 1998, 1998 (Special) : 353-359. doi: 10.3934/proc.1998.1998.353 [15] Enrique R. Pujals, Federico Rodriguez Hertz. Critical points for surface diffeomorphisms. Journal of Modern Dynamics, 2007, 1 (4) : 615-648. doi: 10.3934/jmd.2007.1.615 [16] Keith Promislow, Hang Zhang. Critical points of functionalized Lagrangians. Discrete & Continuous Dynamical Systems, 2013, 33 (4) : 1231-1246. doi: 10.3934/dcds.2013.33.1231 [17] Matteo Novaga, Enrico Valdinoci. Closed curves of prescribed curvature and a pinning effect. Networks & Heterogeneous Media, 2011, 6 (1) : 77-88. doi: 10.3934/nhm.2011.6.77 [18] Françoise Pène. Asymptotic of the number of obstacles visited by the planar Lorentz process. Discrete & Continuous Dynamical Systems, 2009, 24 (2) : 567-587. doi: 10.3934/dcds.2009.24.567 [19] Lekbir Afraites, Chorouk Masnaoui, Mourad Nachaoui. Shape optimization method for an inverse geometric source problem and stability at critical shape. Discrete & Continuous Dynamical Systems - S, 2022, 15 (1) : 1-21. doi: 10.3934/dcdss.2021006 [20] Suzana Antunović, Tonči Kokan, Tanja Vojković, Damir Vukičević. Exponential generalised network descriptors. Advances in Mathematics of Communications, 2019, 13 (3) : 405-420. doi: 10.3934/amc.2019026

Impact Factor:


text:
# Revision history [back]

### remove_vertical_border_strip

Hello, I'm interested to know if there is a way to define the analoge of "remove_horizontal_border_strip(k)" for integer partitions, but for vertical border strip. I think that maybe with IntegerListLex.

### remove_vertical_border_strip

Hello, I'm interested to know if there is a way to define the analoge of "remove_horizontal_border_strip(k)" for integer partitions, remove_horizontal_border_strip(k) which returns the partitions obtained from self by removing an horizontal border strip of length k.

For example:

Partition([5,3,1]).remove_horizontal_border_strip(1).list()
[[5, 3], [5, 2, 1], [4, 3, 1]]

but for vertical border strip. I think that maybe with IntegerListLex.strip.


text:
# Senkou Span A

## DEFINITION of 'Senkou Span A'

A component of the Ichimoku Kinko Hyo indicator that is used to measure momentum and future areas of support and resistance. Senkou span A is always plotted alongside Senkou span B and the area between the two lines is filled with shaded indicator lines, also known as the cloud, which is used by traders to predict levels of future support/resistance. Senkou span A is calculated by using the following formula:

## BREAKING DOWN 'Senkou Span A'

The trend is deemed to be downward when Senkou span A is located below senkou span B. In practice, the indicator is most commonly used to predict the reversal of a current trend when the senkou spans cross over each other.


text:
# Algorithmic problem classification

I'm trying to classify the following problem:

I have $N$ empty boxes ($n_i$ is the volume of the $i$-th box, $1 \leq i \leq N$) and $M$ divisible items ($m_j$ is the volume of $j$-th item, $1 \leq j \leq M$). The total volume of all boxes is exactly equal to the total volume of all items. I need to find a distribution of items among boxes which minimizes the number of item divisions.

I suppose this problem is NP-complete, and is some kind of set coverage problem, but I can't find appropriate variation of it.


text:
# 7.1 Existential Types and Eliminators

## Exercise 7.1 (i)

I think that the only functions of type forall a. a -> r are the constant functions, since one of the things an a can be is () and the only thing we can do from unit is return a constant. That is, we can’t use any information from a to determine r, because a might not give us any information at all.

## Exercise 7.1 (ii)

Is this question just asking me to relay GHC’s error message? If so:

07/Existential.hs:16:36: error:
• No instance for (Show t) arising from a use of ‘show’
Possible fix:
add (Show t) to the context of the data constructor ‘HasShow’
• In the second argument of ‘(++)’, namely ‘show s’
In the expression: "HasShow " ++ show s
In an equation for ‘show’: show (HasShow s) = "HasShow " ++ show s
|
16 |   show (HasShow s) = "HasShow " ++ show s
|

see 07/Existential.hs

## Exercise 7.1 (iii)

Since the only information elimHasShow has about the a in

forall a. Show a => a -> r

is that a has a Show instance, the only functions elimHasShow can take are either constant with respect to a, or use a’s Show instance.

## 7.1.1 Dynamic Types

It’s cool how existential types, and a Typeable constraint to allow you to recover the type turns out to be equivalent to dynamic typing. Yes, it’s unityping, as per the common saying “dynamic typing is just static typing with a single type”, but I think it’s actually a little more interesting than that here: We’re bijecting the universe of Haskell types (constrained by Typeable) into the Dynamic unitype. There no actual loss of information, we’re just turning static checking into runtime checking.

I’m not really a static typing ideologue; I think dynamic typing is an important (though often abused) technique, because sometimes being able to selectively turn off static checks gives you extraordinary power (“with great power comes great responsibility”).

Interestingly, the constraint synonym

class (Monoid a, Eq a) => MonoidEq a
instance (Monoid a, Eq a) => MonoidEq a

required turning on UndecidableInstances. Not sure why.

## Scoping Information with Existentials

All the seqs in the typeclass instances are to make sure the a in ST s a is strict.

This ST trick is such a hack. Like, it’s clever to use forall scoping to teach the type-system that we don’t want to leak references, but the fact we have to add an otherwise useless parameter s to ST just feels clunky.

I wonder how this trick relates to linear types?


text:
0
RESEARCH PAPERS: Forced Convection

# Rheological Characteristics and Turbulent Friction Drag and Heat Transfer Reductions of a Very Dilute Cationic Surfactant Solution

[+] Author and Article Information
Jinjia Wei

State Key Laboratory of Multiphase Flow in Power Engineering, Xi’an Jiaotong University, Xi’an, Shaanxi 710049, China

Yasuo Kawaguchi1

Department of Mechanical Engineering, Faculty of Science and Technology, Tokyo University of Science, Noda, Chiba 278-8510, Japanyasuo@rs.noda.tus.ac.jp

Bo Yu

Oil and Gas Storage and Transportation Engineering, China University of Petroleum, Beijing, 102249, People’s Republic of China

Ziping Feng

Daikin Air Conditioning Engineering Laboratory Ltd., Sakai, 591-8511, Japan

1

Corresponding author.

J. Heat Transfer 128(10), 977-983 (Feb 24, 2006) (7 pages) doi:10.1115/1.2345422 History: Received August 31, 2004; Revised February 24, 2006

## Abstract

Turbulent friction drag and heat transfer reductions and rheological characteristics of a very dilute cationic surfactant solution, cetyltrimethyl ammonium chloride (CTAC)/sodium salicylate $(NaSal)$ aqueous solution, were experimentally investigated at various temperatures. It was found that there existed a critical temperature above which drag and heat transfer reductions disappeared and shear viscosities rapidly dropped to that of water. It was surmised that drag and heat transfer reductions had a certain relationship with rheological characteristics and a rheological characterization of $CTAC∕NaSal$ surfactant solutions was performed to clarify this relationship. The effects of Reynolds number and fluid temperature and concentration on drag and heat transfer reductions were qualitatively explained by analyzing the measured shear viscosity data at different shear rates and solution temperatures and concentrations. The Giesekus model was found to fit the measured shear viscosities reasonably well for different temperatures and concentrations of the surfactant solution and the model parameter values obtained by fitting were correlated with temperature at certain solution concentrations. From the correlation results, the temperature effect on viscoelasticity of surfactant solutions was analyzed to relate the rheological characteristics with drag and heat transfer reduction phenomena.

###### FIGURES IN THIS ARTICLE
<>
Copyright © 2006 by American Society of Mechanical Engineers
Your Session has timed out. Please sign back in to continue.

## Figures

Figure 1

Schematic diagram of two-dimensional water channel: (1) reservoir tank; (2) pump; (3) filter; (4) contraction; (5) honeycomb; (6) 2D channel; (7) heater plate; (8) pressure tap; (9) pressure transducer; (10) diffuser; (11) flow meter; (12) stirrer; (13) cooling coil; (14) heater; (15) heater regulator; (16) thermocouple

Figure 2

Schematics of flow geometries of ARES rheometers

Figure 3

Friction factors versus Reynolds number in a two-dimensional channel

Figure 4

Colburn j factors versus Reynolds number in a two-dimensional channel

Figure 8

Least squares fitting result of the measured shear viscosities with the Giesekus model

Figure 9

Zero-shear viscosity η0 versus temperature T

Figure 10

Relaxation time λ versus temperature T

Figure 11

Comparison of the measured shear viscosities with the prediction of the Giesekus model

Figure 7

Solution temperature versus shear rate at the same shear viscosity

Figure 6

Shear viscosities versus solution temperature (a)Cm=50ppm, (b)Cm=75ppm, (c)Cm=100ppm, and (d)Cm=200ppm

Figure 5

Inlet fluid temperature dependence of friction factors and Colburn j factors

## Errata

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text:
# Vertical Visibility among Parallel Polygons in Three Dimensions

Motivated by a paper of Babilon et al. entitled “Visibility Representations of Complete Graphs” we study the following problem. Let $C$ be a collection of homothetic copies of a convex k-gon P in the plane with the given stacking order. The collection $C$ forms a visibility clique if, for every pair of copies $P_1$ and $P_2$ in $C$, the intersection of $P_1$ and $P_2$ is not covered by the union of the copies in $C$ appearing between $P_1$ and $P_2$ in the stacking order.
We obtain an upper bound of $2^{2{k \choose 2}+2}$ on the size of a visibility clique of homothetic copies of a convex $k$-gon. In the case of translates of a regular convex $k$-gon we improve the bound to $O(k^4)$ thereby improving the upper bound of $2^{2^k}$ from the aforementioned paper.
Furthermore, we show that every visibility clique of $n$ translates of a regular k-gon contains a subset of size $cn^{1/6}$, for a fixed $c>0$, that is combinatorially equivalent to the best known lower bound construction, giving the bound of roughly $k/2$, for the visibility clique of homothetic copies of a convex $k$-gon. The last result holds in more general setting, where we consider pseudo-disc arrangements satisfying certain conditions instead of translates of a polygon.


text:
# Determine the depth of an array

A simple challenge for your Monday evening (well, or Tuesday morning in the other half of the world...)

You're given as input a nested, potentially ragged array of positive integers:

[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]

Your task is to determine its depth, which is the greatest nesting-depth of any integer in the list. In this case, the depth of 11 is 6, which is largest.

You may assume that none of the arrays will be empty.

You may write a program or function, taking input via STDIN (or closest alternative), command-line argument or function argument and outputting the result via STDOUT (or closest alternative), function return value or function (out) parameter.

Input may be taken in any convenient list or string format that supports non-rectangular arrays (with nested arrays of different depths), as long as the actual information isn't preprocessed.

You must not use any built-ins related to the shape of arrays (including built-ins that solve this challenge, that get you the dimensions of a nested array). The only exception to this is getting the length of an array.

Standard rules apply.

### Test Cases

[1]                                                               -> 1
[1, 2, 3]                                                         -> 1
[[1, 2, 3]]                                                       -> 2
[3, [3, [3], 3], 3]                                               -> 3
[[[[1], 2], [3, [4]]]]                                            -> 4
[1, [[3]], [5, 6], [[[[8]]]], 1]                                  -> 5
[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14] -> 6
[[[[[[[3]]]]]]]                                                   -> 7

• After discussion in chat I've decided to allow length built-ins, because some languages require them to iterate over an array cleanly. – Martin Ender Feb 8 '16 at 23:09
• Just for general education: ≡ is APL's built-in primitive for exactly this. – Adám Feb 9 '16 at 16:09
• @MartinBüttner I've run into a tiny problem. I started doing this in java, unfourtunetly when testing inputs the commas are causing it to split the inputs into multiple command line arguments rather then one. Can I use the escape character \  in the inputs? EDIT: nevermind just tried it like that. That doesn't even work either. Darn can I not use CMD args? – Ashwin Gupta Feb 10 '16 at 5:21
• @AshwinGupta can't you wrap the command line argument in quotes? You can also read input from STDIN or submit a function that takes an actual array object as a parameter. – Martin Ender Feb 10 '16 at 10:39
• @MartinBüttner oh I didn't know that quotes thing I'll try it out. Currently just using Scanner. (System.in). I believe that is a form of STDIN? – Ashwin Gupta Feb 10 '16 at 22:02

# Ruby, 42 characters

f=->a{a==a.flatten(1)?1:f[a.flatten(1)]+1}

Example:

irb(main):001:0> f.call([[[1,2],[2,3]],[[3,4],[5]]])
=> 3

• Welcome to the site! – DJMcMayhem Sep 7 '17 at 17:02

## Perl 6, 70 bytes

Whole program that reads the given examples terminated with newline or EOF from STDIN.

# perl6 % <<< '[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]'

my $i;dd [max] lines.comb.map:{$i++when '[';$i--when ']'}  lines takes text line by line from STDIN and returns a list of Str. Calling .comb on a list will convert that into a Str and then split it into graphenes. Any block in Perl 6 is born with one positional argument that ends up in $_. map called with any callable expect it to have one positional argument. What is nice because when ']' is short for when \$_ ~~ ']'. The returned list is fed to [max] and reduced to it's biggest value.

## C#, 99 Bytes

int f(string s){int m=0,w=0;foreach(char c in s){if(c=='[')w++;if(c==']')w--;if(w>m)m=w;}return m;}

• Does int m=w=0 work in C# ? – Cyoce Oct 7 '16 at 0:40

# Axiom 106 bytes

RL(a:Union(List(Any),Any)):INT==(a case List(Any)=>(g:List(Any):=a;r:=1;for i in g repeat r:=r+RL(i);r);0)

ungolf

RL(a:Union(List(Any),Any)):INT==
a case List(Any)=>
g:List(Any):=a
r:=1
for i in g repeat r:=r+RL(i)
r
0

results

(3) -> for i in[[1],[1,2,3], [[1,2,3]], [3,[3,[3],3],3]]repeat output[i, RL(i)]
[[1],1]
[[1,2,3],1]
[[[1,2,3]],2]
[[3,[3,[3],3],3],3]

# C++14, 7270 67 bytes

-2 bytes for removing the curly brackets at for and -3 bytes for replacing m=l>m?l:m with m+=l>m.

As unnamed lambda returning via reference parameter:

[](auto&s,int&m){int l=m=0;for(auto c:s)l+=(c==91)-(c==93),m+=l>m;}

s can be std::string or char[]

Ungolfed and usage:

#include<string>
#include<iostream>

auto f=
[](auto&s,int&m){
int l=m=0;
for(auto c:s)
l+=(c==91)-(c==93),
m+=l>m;
}
;

int main(){
std::string S[] = {
"[1]",
"[1, 2, 3]",
"[[1, 2, 3]]",
"[3, [3, [3], 3], 3]",
"[[[[1], 2], [3, [4]]]]",
"[1, [[3]], [5, 6], [[[[8]]]], 1]",
"[1, [[2, 3, [[4], 5], 6, [7, 8]], 9, [10, [[[11]]]], 12, 13], 14]",
"[[[[[[[3]]]]]]]"
};

for (auto s:S){
int m;
g(s,m);
std::cout << m << std::endl;
}

}

Clojure, 57 bytes

(fn[s](apply max(reductions +(map #({$1$ -1}% 0)s))))

Also based on parsing a string:

(f "[3, [3, [3], 3], 3]")

R: 53 bytes.

d=function(L) ifelse(is.list(L),max(sapply(L,d))+1,0)

Recursive function, it adds one point everytime it finds an inner list. It chooses only the element with the maximum depth...


text:
# How to cut a horizontal line by a perpendicular line?

I'd like to cut the line AC at the point p by a perpendicular line.

\documentclass{article}
\usepackage{xypic}
\begin{document}
\xymatrix{
& B\ar @{-}[d] &\\
A\ar @{-}[rr]_{P} &  & C
}
\end{document}

-
@MdKutubuddinSardar: Why don't you react to any of the comments? –  Jake Mar 22 '13 at 14:24
Please, at least have the decency to upvote and accept answers to your questions. –  Jake Mar 22 '13 at 14:30

To extend the vertical line down to the line AC, use the arrow adjustment facility placing +0{} after the arrow. By default the arrow stops short of the center of the node, this adjustment makes it stop distance 0 from the center instead.

\documentclass{article}

\usepackage{xypic}

\begin{document}
\xymatrix{
& B\ar @{-}[d]+0{} &\\
A\ar @{-}[rr]_{P} &  & C
}