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# Fixed point results for generalized contractive mappings involving altering distance functions on complete quasi-metric spaces and applications

Volume 10, Issue 4, pp 1377--1398

Publication Date: 2017-04-20

http://dx.doi.org/10.22436/jnsa.010.04.09

### Authors

Yanbin Sang - School of Science, North University of China, Taiyuan, 030051, China.

### Abstract

In this paper, we introduce $\alpha-\psi-\phi$-Jachymski contractive mappings with generalized altering distance functions in the setting of quasi-metric spaces. Some theorems on the existence and uniqueness of fixed points for such mappings via admissible mappings are established. Utilizing above abstract results, we derive common fixed point theorem for two operators and multidimensional fixed point results for nonlinear mappings satisfying different kinds of contractive conditions on partially ordered metric spaces. Moreover, we present some examples and applications in a Fredholm integral equation and an initial value problem for partial differential equation of parabolic type.

### Keywords

Admissible mapping, altering distance, multidimensional, w-distance, partial order.

### References

[1] C. Alegre, J. Marín, S. Romaguera,/ A fixed point theorem for generalized contractions involving w-distances on complete quasi-metric spaces,/ Fixed Point Theory Appl.,/ 2014 (2014), 8 pages.
[2] S. Al-Homidan, Q. H. Ansari, J.-C. Yao,/ Some generalizations of Ekeland-type variational principle with applications to equilibrium problems and fixed point theory,/ Nonlinear Anal.,/ 69 (2008), 126–139.
[3] H. H. Alsulami, S. Gülyaz I. M. Erhan,/ Fixed points of $\alpha$-admissible Meir-Keeler contraction mappings on quasi-metric spaces,/ J. Inequal. Appl.,/ 2015 (2015), 15 pages.
[4] H. H. Alsulami, S. Gülyaz , E. Karapınar, ˙I. M. Erhan,/ Fixed point theorems for a class of $\alpha$-admissible contractions and applications to boundary value problem,/ Abstr. Appl. Anal.,/ 2014 (2014), 10 pages.
[5] V. Berinde,/ Coupled fixed point theorems for $\phi$-contractive mixed monotone mappings in partially ordered metric spaces,/ Nonlinear Anal.,/ 75 (2012), 3218–3228.
[6] M. Borcut, V. Berinde,/ Tripled coincidence theorems for contractive type mappings in partially ordered metric spaces,/ Appl. Math. Comput.,/ 218 (2012), 5929–5936.
[7] P. Chaipunya, Y. J. Cho, P. Kumam,/ Geraghty-type theorems in modular metric spaces with an application to partial differential equation,/ Adv. Difference Equ.,/ 2012 (2012), 12 pages.
[8] R. H. Haghi, S. Rezapour, N. Shahzad,/ Some fixed point generalizations are not real generalizations,/ Nonlinear Anal.,/ 74 (2011), 1799–1803.
[9] J. Jachymski,/ Equivalent conditions and the Meir-Keeler type theorems,/ J. Math. Anal. Appl.,/ 194 (1995), 293–303.
[10] G. Jungck, B. E. Rhoades,/ Fixed points for set valued functions without continuity,/ Indian J. Pure Appl. Math.,/ 29 (1998), 227–238.
[11] E. Karapınar, I . M. Erhan, A. Öztürk,/ Fixed point theorems on quasi-partial metric spaces,/ Math. Comput. Modelling,/ 57 (2013), 2442–2448.
[12] E. Karapınar, P. Kumam, P. Salimi,/ On $\alpha-\psi$-Meir-Keeler contractive mappings,/ Fixed Point Theory Appl.,/ 2013 (2013), 12 pages.
[13] H. Lakzian, D. Gopal, W. Sintunavarat,/ New fixed point results for mappings of contractive type with an application to nonlinear fractional differential equations,/ J. Fixed Point Theory Appl.,/ 18 (2016), 251–266.
[14] V. La Rosa, P. Vetro,/ Common fixed points for $\alpha-\psi-\phi$-contractions in generalized metric spaces,/ Nonlinear Anal. Model. Control,/ 19 (2014), 43–54.
[15] X.-L. Liu, A. H. Ansari, S. Chandok, C.-K. Park,/ Some new fixed point results in partial ordered metric spaces via admissible mappings and two new functions,/ J. Nonlinear Sci. Appl.,/ 9 (2016), 1564–1580.
[16] N. V. Luong, N. X. Thuan,/ Coupled fixed points in partially ordered metric spaces and application,/ Nonlinear Anal.,/ 74 (2011), 983–992.
[17] J. Marín, S. Romaguera, P. Tirado,/ Generalized contractive set-valued maps on complete preordered quasi-metric spaces,/ J. Funct. Spaces Appl.,/ 2013 (2013), 6 pages.
[18] A. Meir, E. Keeler,/ A theorem on contraction mappings,/ J. Math. Anal. Appl., 28 (1969), 326–329.
[19] S. Park,/ On generalizations of the Ekeland-type variational principles,/ Nonlinear Anal.,/ 39 (2000), 881–889.
[20] P. D. Proinov,/ Fixed point theorems in metric spaces,/ Nonlinear Anal.,/ 64 (2006), 546–557.
[21] A. Roldán, J. Martínez-Moreno, C. Roldán,/ Multidimensional fixed point theorems in partially ordered complete metric spaces,/ J. Math. Anal. Appl.,/ 396 (2012), 536–545.
[22] A. Roldán, J. Martínez-Moreno, C. Roldán, Y. J. Cho,/ Multidimensional fixed point theorems under ( $\psi,\phi$)-contractive conditions in partially ordered complete metric spaces,/ J. Comput. Appl. Math.,/ 273 (2015), 76–87.
[23] A. Roldán, J. Martínez-Moreno, C. Roldán, E. Karapınar,/ Some remarks on multidimensional fixed point theorems,/ Fixed Point Theory,/ 15 (2014), 545–558.
[24] B. Samet, C. Vetro, P. Vetro,/ Fixed point theorems for $\alpha-\psi$-contractive type mappings,/ Nonlinear Anal.,/ 75 (2012), 2154–2165.
[25] Y.-B. Sang, Q. Meng,/ Fixed point theorems with generalized altering distance functions in partially ordered metric spaces via w-distances and applications,/ Fixed Point Theory Appl.,/ 2015 (2015), 25 pages.
[26] T. Suzuki,/ Meir-Keeler contractions of integral type are still Meir-Keeler contractions,/ Int. J. Math. Math. Sci.,/ 2007 (2007), 6 pages.


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# Calculus/Algebra

 ← Precalculus Calculus Functions → Algebra

This section is intended to review algebraic manipulation. It is important to understand algebra in order to do calculus. If you have a good knowledge of algebra, you should probably just skim this section to be sure you are familiar with the ideas.

## Rules of arithmetic and algebra

The following laws are true for all a, b, and c, whether a, b, and c are numbers, variables, functions, or more complex expressions involving numbers, variable and/or functions.

• Commutative Law: $a+b=b+a \,$.
• Associative Law: $(a+b)+c=a+(b+c)\,$.
• Additive Identity: $a+0=a\,$.
• Additive Inverse: $a+(-a)=0\,$.

### Subtraction

• Definition: $a-b = a+(-b)\,$.

### Multiplication

• Commutative law: $a\times b=b\times a$.
• Associative law: $(a\times b)\times c=a\times (b\times c)\,$.
• Multiplicative identity: $a\times 1=a\,$.
• Multiplicative inverse: $a\times \frac{1}{a}=1$, whenever $a \neq 0\,$
• Distributive law: $a\times (b+c)=(a\times b)+(a\times c)\,$.

### Division

• Definition: $\frac{a}{b}=a\times \frac{1}{b}$, whenever $b \neq 0\,$.

Let's look at an example to see how these rules are used in practice.

 $\frac{(x+2)(x+3)}{x+3}$ = $\left[(x+2)\times (x+3)\right]\times \left( \frac{1}{x+3}\right)$ (from the definition of division) = $(x+2)\times \left[(x+3)\times \left(\frac{1}{x+3} \right) \right]$ (from the associative law of multiplication) = $((x+2)\times (1)),\qquad x \neq -3 \,$ (from multiplicative inverse) = $x+2, \qquad x \neq -3.$ (from multiplicative identity)

Of course, the above is much longer than simply cancelling $x+3$ out in both the numerator and denominator. But, when you are cancelling, you are really just doing the above steps, so it is important to know what the rules are so as to know when you are allowed to cancel. Occasionally people do the following, for instance, which is incorrect:

$\frac{2\times (x + 2)}{2}= \frac{2}{2} \times \frac{x+2}{2}=1 \times \frac{x+2}{2}= \frac{x+2}{2}$.

The correct simplification is

$\frac{2\times (x + 2)}{2}= \left( 2 \times \frac{1}{2} \right) \times (x+2)=1 \times (x+2)=x+2$,

where the number $2$ cancels out in both the numerator and the denominator.

## Interval notation

There are a few different ways that one can express with symbols a specific interval (all the numbers between two numbers). One way is with inequalities. If we wanted to denote the set of all numbers between, say, 2 and 4, we could write "all x satisfying 2<x<4." This excludes the endpoints 2 and 4 because we use < instead of $\leq$. If we wanted to include the endpoints, we would write "all x satisfying $2 \leq x \leq 4$." This includes the endpoints.

Another way to write these intervals would be with interval notation. If we wished to convey "all x satisfying 2<x<4" we would write (2,4). This does not include the endpoints 2 and 4. If we wanted to include the endpoints we would write [2,4]. If we wanted to include 2 and not 4 we would write [2,4); if we wanted to exclude 2 and include 4, we would write (2,4].

Thus, we have the following table:

Endpoint conditions Inequality notation Interval notation
Including both 2 and 4 all x satisfying $2 \leq x \leq 4$
$[2,4] \,\!$
Not including 2 nor 4 all x satisfying $2
$(2,4) \,\!$
Including 2 not 4 all x satisfying $2 \leq x < 4$
$[2,4) \,\!$
Including 4 not 2 all x satisfying $2 < x \leq 4$
$(2,4] \,\!$

In general, we have the following table:

Meaning Interval Notation Set Notation
All values greater than or equal to $a$ and less than or equal to $b$ $\left[a,b\right]$ $\left\{x:a\le x\le b\right\}$
All values greater than $a$ and less than $b$ $\left(a,b\right)$ $\left\{x:a < x < b\right\}$
All values greater than or equal to $a$ and less than $b$ $\left[a,b\right)$ $\left\{x:a\le x < b\right\}$
All values greater than $a$ and less than or equal to $b$ $\left(a,b\right]$ $\left\{x:a < x\le b\right\}$
All values greater than or equal to $a$. $\left[a,\infty\right)$ $\left\{x:x\ge a\right\}$
All values greater than $a$. $\left(a,\infty\right)$ $\left\{x:x > a\right\}$
All values less than or equal to $a$. $\left(-\infty,a\right]$ $\left\{x:x\le a\right\}$
All values less than $a$. $\left(-\infty,a\right)$ $\left\{x:x < a\right\}$
All values. $\left(-\infty,\infty\right)$ $\left\{x: x\in\mathbb{R}\right\}$

Note that $\infty$ and $-\infty$ must always have an exclusive parenthesis rather than an inclusive bracket. This is because $\infty$ is not a number, and therefore cannot be in our set. $\infty$ is really just a symbol that makes things easier to write, like the intervals above.

The interval (a,b) is called an open interval, and the interval [a,b] is called a closed interval.

Intervals are sets and we can use set notation to show relations between values and intervals. If we want to say that a certain value is contained in an interval, we can use the symbol $\in$ to denote this. For example, $2\in[1,3]$. Likewise, the symbol $\notin$ denotes that a certain element is not in an interval. For example $0\notin(0,1)$.

There are a few rules and properties involving exponents and radicals that you'd do well to remember. As a definition we have that if n is a positive integer then $a^n$ denotes n factors of a. That is,

$a^n = a\cdot a \cdot a \cdots a \qquad (n~ \mbox{times}).$

If $a \not= 0$ then we say that $a^0 =1 \,$.

If n is a negative integer then we say that $a^{-n} = \frac{1}{a^n} .$

If we have an exponent that is a fraction then we say that $a^{m/n} = \sqrt[n]{a^m} = (\sqrt[n]{a})^m .$

In addition to the previous definitions, the following rules apply:

Rule Example
$a^n \cdot a^m = a^{n+m}$ $3^6 \cdot 3^9 = 3^{15}$
$\frac{a^n}{a^m} = a^{n-m}$ $\frac{x^3}{x^2} = x^{1} = x$
$(a^n)^m = a^{n\cdot m}$ $(x^4)^5 = x^{20} \,\!$
$(ab)^n = a^n b^n \,\!$ $(3x)^5 = 3^5 x^5 \,\!$
$\bigg(\frac{a}{b}\bigg)^n = \frac{a^n}{b^n}$ $\bigg(\frac{7}{3}\bigg)^3 = \frac{7^3}{3^3}.$

## Factoring and roots

Given the expression $x^2 + 3x + 2$, one may ask "what are the values of x that make this expression 0?" If we factor we obtain

$x^2 + 3x + 2 = (x + 2)(x + 1). \,\!$

If x=-1 or -2, then one of the factors on the right becomes zero. Therefore, the whole must be zero. So, by factoring we have discovered the values of x that render the expression zero. These values are termed "roots." In general, given a quadratic polynomial $px^2 + qx + r$ that factors as

$px^2 + qx + r = (ax + c)(bx + d) \,\!$

then we have that x = -c/a and x = -d/b are roots of the original polynomial.

A special case to be on the look out for is the difference of two squares, $a^2 - b^2$. In this case, we are always able to factor as

$a^2 - b^2 = (a+b)(a-b). \,\!$

For example, consider $4x^2 - 9$. On initial inspection we would see that both $4x^2$ and $9$ are squares ($(2x)^2 = 4x^2$ and $3^2 = 9$). Applying the previous rule we have

$4x^2 - 9 = (2x+3)(2x-3). \,\!$

The following is a general result of great utility.

Given any quadratic equation $ax^2+bx+c=0, a\neq0$, all solutions of the equation are given by the quadratic formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.
 Example: Find all the roots of $4x^2+7x-2$ Finding the roots is equivalent to solving the equation $4x^2+7x-2=0$. Applying the quadratic formula with $a=4, b=7, c=-2$, we have: $x=\frac{-7\pm\sqrt{7^2-4(4)(-2)}}{2(4)}$ $x=\frac{-7\pm\sqrt{49+32}}{8}$ $x=\frac{-7\pm\sqrt{81}}{8}$ $x=\frac{-7\pm9}{8}$ $x=\frac{2}{8}, x=\frac{-16}{8}$ $x=\frac{1}{4}, x=-2$

The quadratic formula can also help with factoring, as the next example demonstrates.

 Example: Factor the polynomial $4x^2+7x-2$ We already know from the previous example that the polynomial has roots $x=\frac{1}{4}$ and $x=-2$. Our factorization will take the form $C(x+2)(x-\frac{1}{4})$ All we have to do is set this expression equal to our polynomial and solve for the unknown constant C: $C(x+2)(x-\frac{1}{4})=4x^2+7x-2$ $C(x^2+(-\frac{1}{4}+2)x-\frac{2}{4})=4x^2+7x-2$ $C(x^2+\frac{7}{4}x-\frac{1}{2})=4x^2+7x-2$ You can see that $C=4$ solves the equation. So the factorization is $4x^2+7x-2=4(x+2)(x-\frac{1}{4})=(x+2)(4x-1)$

Note that if $4ac>b^2$ then the roots will not be real numbers.

## Simplifying rational expressions

Consider the two polynomials

$p(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$

and

$q(x) = b_m x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0.$

When we take the quotient of the two we obtain

$\frac{p(x)}{q(x)} = \frac{a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0}{b_m x^m + b_{m-1}x^{m-1} + \cdots + b_1x + b_0}.$

The ratio of two polynomials is called a rational expression. Many times we would like to simplify such a beast. For example, say we are given $\frac{x^2-1}{x+1}.$ We may simplify this in the following way:

$\frac{x^2-1}{x+1} = \frac{(x+1)(x-1)}{x+1} = x-1, \qquad x \neq -1 \,\!$

This is nice because we have obtained something we understand quite well, $x-1$, from something we didn't.

## Formulas of multiplication of polynomials

Here are some formulas that can be quite useful for solving polynomial problems:

$(a+b)^2=a^2+2ab+b^2$
$(a-b)^2=a^2-2ab+b^2$
$(a-b)(a+b)=a^2-b^2$
$(a\pm b)^3=a^3\pm 3a^2b+3ab^2\pm b^3$
$a^3\pm b^3=(a\pm b)(a^2\mp ab+b^2)$

## Polynomial Long Division

Suppose we would like to divide one polynomial by another. The procedure is similar to long division of numbers and is illustrated in the following example:

### Example

 Divide $x^2-2x-15$ (the dividend or numerator) by $x+3$ (the divisor or denominator) Similar to long division of numbers, we set up our problem as follows: $\begin{array}{rl}\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\end{array}$ First we have to answer the question, how many times does $x+3$ go into $x^2$? To find out, divide the leading term of the dividend by leading term of the divisor. So it goes in $x$ times. We record this above the leading term of the dividend: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ \end{array}$ , and we multiply $x+3$ by $x$ and write this below the dividend as follows: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ \end{array}$ Now we perform the subtraction, bringing down any terms in the dividend that aren't matched in our subtrahend: $\begin{array}{rl}&~~\,x\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ &\!\!\!\!~~~~~~-5x-15~~~\\ \end{array}$ Now we repeat, treating the bottom line as our new dividend: $\begin{array}{rl}&~~\,x-5\\ x+3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x^2-2x-15 \end{array}\\ &\!\!\!\!-\underline{(x^2+3x)~~~}\\ &\!\!\!\!~~~~~~-5x-15~~~\\ &\!\!\!\!~~~-\underline{(-5x-15)~~~}\\ &\!\!\!\!~~~~~~~~~~~~~~~~~~~0~~~\\ \end{array}$ In this case we have no remainder.

### Application: Factoring Polynomials

We can use polynomial long division to factor a polynomial if we know one of the factors in advance. For example, suppose we have a polynomial $P(x)$ and we know that $r$ is a root of $P$. If we perform polynomial long division using P(x) as the dividend and $(x-r)$ as the divisor, we will obtain a polynomial $Q(x)$ such that $P(x)=(x-r)Q(x)$, where the degree of $Q$ is one less than the degree of $P$.

### Exercise

1. Factor $x-1$ out of $6x^3-4x^2+3x-5$.

$(x-1)(6x^2+2x+5)$

Solution

### Application: Breaking up a rational function

Similar to the way one can convert an improper fraction into an integer plus a proper fraction, one can convert a rational function $P(x)$ whose numerator $N(x)$ has degree $n$ and whose denominator $D(x)$ has degree $d$ with $n\geq d$ into a polynomial plus a rational function whose numerator has degree $\nu$ and denominator has degree $\delta$ with $\nu<\delta$.

Suppose that $N(x)$ divided by $D(x)$ has quotient $Q(x)$ and remainder $R(x)$. That is

$N(x)=D(x)Q(x)+R(x)$

Dividing both sides by $D(x)$ gives

$\frac{N(x)}{D(x)}=Q(x)+\frac{R(x)}{D(x)}$

$R(x)$ will have degree less than $D(x)$.

#### Example

 Write $\frac{x-1}{x-3}$ as a polynomial plus a rational function with numerator having degree less than the denominator. $\begin{array}{rl}&~~\,1\\ x-3\!\!\!\!&\big)\!\!\!\begin{array}{lll} \hline \,x-1 \end{array}\\ &\!\!\!\!-\underline{(x-3)~~~}\\ &\!\!\!\!~~~~~~~~~2~~~\\ \end{array}$ so $\frac{x-1}{x-3}=1+\frac{2}{x-3}$
 ← Precalculus Calculus Functions → Algebra

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# How to Find the Range of a Function Algebraically [15 Ways]

There are different ways to Find the Range of a Function Algebraically. But before that, we take a short overview of the Range of a Function.

In the first chapter What is a Function? we have learned that a function is expressed as

$y=f(x)$,

where x is the input and y is the output.

For every input x (where the function f(x) is defined) there is a unique output.

The set of all outputs of a function is the Range of a Function.

The Range of a Function is the set of all y values or outputs i.e., the set of all $f(x)$ when it is defined.

We suggest you read this article “9 Ways to Find the Domain of a Function Algebraically” first. This will help you to understand the concepts of finding the Range of a Function better.

1. 5 Steps to Find the Range of a Function,

and in the end you will be able to

1. Find the Range of 10 different types of functions

## Steps to Find the Range of a Function

Suppose we have to find the range of the function $f(x)=x+2$.

We can find the range of a function by using the following steps:

### #1. First label the function as y=f(x)

$y=x+2$

### #2. Express x as a function of y

Here $x=y-2$

### #3. Find all possible values of y for which f(y) is defined

See that $x=y-2$ is defined for all real values of $y$.

### #4. Element values of y by looking at the initial function f(x)

Our initial function $y=x+2$ is defined for all real values of $x$ i.e., $x\epsilon \mathbb{R}$.

So here we do not need to eliminate any value of $y$ i.e., $y\epsilon \mathbb{R}$.

### #5. Write the Range of the function f(x)

Therefore the Range of the function $y=x+2$ is {$y\epsilon \mathbb{R}$}.

Maybe you are getting confused and don’t understand all the steps now.

But believe me, you will get a clear concept in the next examples.

## How to Find the Range of a Function Algebraically

There are different types of functions. Here you will learn 10 ways to find the range for each type of function.

### #1. Find the range of a Rational function

Example 1: Find the range

$f(x)=\frac{x-2}{3-x},x\neq3$

Solution:

Step 1: First we equate the function with $y$

$y=\frac{x-2}{3-x}$

Step 2: Then express $x$ as a function of $y$

$y=\frac{x-2}{3-x}$

or, $y(3-x)=x-2$

or, $3y-xy=x-2$

or, $x+xy=3y+2$

or, $x(1+y)=3y+2$

or, $x=\frac{3y+2}{y+1}$

Step 3: Find possible values of $y$ for which $x=f(y)$ is defined

$x=\frac{3y+2}{y+1}$ is defined when $y+1$ can not be equal to $0$,

i.e., $y+1\neq0$

i.e., $y\neq-1$

i.e., $y\epsilon \mathbb{R}-${$-1$}

Step 4: Eliminate the values of $y$

See that $f(x)=\frac{x-2}{3-x}$ is defined on $\mathbb{R}-${$3$} and we do not need to eliminate any value of y from $y\epsilon \mathbb{R}-${$-1$}.

Step 5: Write the Range

$\therefore$ the range of $f(x)=\frac{x-2}{3-x}$ is {$x\epsilon \mathbb{R}:x\neq-1$}.

Example 2: Find the range

$f(x)=\frac{3}{2-x^{2}}$

Solution:

Step 1:

$y=\frac{3}{2-x^{2}}$

Step 2:

$y=\frac{3}{2-x^{2}}$

or, $2y-xy^{2}=3$

or, $2y-3=x^{2y}$

or, $x^{2}=\frac{2y-3}{y}$

Step 3:

The function $x^{2}=\frac{2y-3}{y}$ is defined when $y\neq 0$ …(1)

Also since $x^{2}\geq 0$,

therefore

$\frac{2y-3}{y}\geq 0$

or, $\frac{2y-3}{y}\times {\color{Magenta} \frac{y}{y}}\geq 0$

or, $\frac{y(2y-3)}{y^{2}}\geq 0$

or, $y(2y-3)\geq 0$ $(\because y^{2}\geq 0)$

or, $(y-0){\color{Magenta} 2}(y-\frac{3}{{\color{Magenta} 2}})$

or, $(y-0)(y-\frac{3}{2})\geq 0$

Next we find the values of $y$ for which $(y-0)(y-\frac{3}{2})\geq 0$ i.e., $y(2y-3)\geq 0$ is satisfied.

Now see the table:

Therefore from the above table and using (1) we get,

$y\epsilon (-\infty,0)\cup [\frac{3}{2},\infty)$ $(\because y\neq 0)$

Step 4:

$y=\frac{3}{2-x^{2}}$ is not a square function,

$\therefore$ we do not need to eliminate any value of y except 0 because if y be zero then the function $y=\frac{3}{2-x^{2}}$ will be undefined.

Step 5:

Therefore the range of the function $f(x)=\frac{3}{2-x^{2}}$ is

$(-\infty,0)\cup [\frac{3}{2},\infty)$.

Example 3: Find the range of a rational equation using inverse

$f(x)=\frac{2x-1}{x+4}$

Solution:

### #2. Find the range of a function with square root

Example 4: Find the range

$f(x)=\sqrt{4-x^{2}}$

Solution:

Step 1: First we equate the function with $y$

$y=\sqrt{4-x^{2}}$

Step 2: Then express $x$ as a function of $y$

$y=\sqrt{4-x^{2}}$

or, $y^{2}=4-x^{2}$

or, $x^{2}=4-y^{2}$

Step 3: Find possible values of $y$ for which $x=f(y)$ is defined

Since $x^{2}\geq 0$,

$\therefore 4-y^{2}\geq 0$

or, $(2-y)(2+y)\geq 0$

or, $(y-2)(y+2)\leq 0$

Now we find possible values for which $(y-2)(y+2)\leq 0$

i.e., $y=-2, y\epsilon (-2,2)$ and $y=2$

i.e., $y\epsilon [-2,2]$

Step 4: Eliminate the values of $y$

As $y=\sqrt{4-x^{2}}$, a square root function,

so $y$ can not take any negative value i.e., $y\geq 0$

Therefore $y\epsilon [0,2]$.

Step 5: Write the range

The range of the function $f(x)=\sqrt{4-x^{2}}$ is $[0,2]$ in interval notation.

We can also write the range of the function $f(x)=\sqrt{4-x^{2}}$ as R(f)={$x\epsilon \mathbb{R}:0\leq y \leq 2$}

Example 5: Find the range of a function $f(x) =\sqrt{x^{2}-4}$.

Solution:

Step 1: First we equate the function with $y$

$y=\sqrt{x^{2}-4}$

Step 2: Then express $x$ as a function of $y$

$y=\sqrt{x^{2}-4}$

or, $y^{2}=x^{2}-4$

or, $x^{2}=y^{2}+4$

Step 3: Find possible values of $y$ for which $x=f(y)$ is defined

Since $x^{2}\geq 0$,

therefore $y^{2}+4\geq 0$

i.e., $y^{2}\geq -4$

i.e., $y\geq \sqrt{-4}$

i.e, $y\geq i\sqrt{2}$, a complex number

$\therefore$ $y^{2}+4\geq 0$ for all $y\epsilon \mathbb{R}$

Step 4: Eliminate the values of $y$

Since $y=\sqrt{x^{2}-4}$ is a square root function,

therefore y can not take any negative value i.e., $y\geq 0$

Step 5: Write the Range

The range of $f(x) =\sqrt{x^{2}-4}$ is $(0,\infty)$.

Example 6: Find the range for the square root function

$f(x)=3-\sqrt{x}$

Solution:

### #3. Find the range of a function with a square root in the denominator

Example 7: Find the range

$f(x)=\frac{1}{\sqrt{x-3}}$

Solution:

Step 1:

$y=\frac{1}{\sqrt{x-3}}$

Step 2:

$y=\frac{1}{\sqrt{x-3}}$

or, $y=\frac{1}{\sqrt{x-3}}$

or, $y^{2}=\frac{1}{x-3}$

or, $y^{2}(x-3)=1$

or, $xy^{2}-3y^{2}=1$

or, $xy^{2}=1+3y^{2}$

or, $x=\frac{1+3y^{2}}{y^{2}}$

Step 3:

For $x=\frac{1+3y^{2}}{y^{2}}$ to be defined,

$y^{2}\neq 0$

i.e., $y\neq 0$

Step 4:

As $f(x)=\frac{1}{\sqrt{x-3}}$, so $y$ can not be negative (-ve).

Step 5:

The range of $f(x)=\frac{1}{\sqrt{x-3}}$ is $(0,\infty)$.

Example 8: Find the range

$f(x)=\frac{1}{\sqrt{4-x^{2}}}$

Solution:

Step 1:

$y=\frac{1}{\sqrt{4-x^{2}}}$

Step 2:

$y=\frac{1}{\sqrt{4-x^{2}}}$

or, $y=\frac{1}{\sqrt{4-x^{2}}}$

or, $y^{2}=\frac{1}{4-x^{2}}$

or, $4y^{2}-x^{2}y^{2}=1$

or, $x^{2}y^{2}=4y^{2}-1$

or, $x^{2}=\frac{4y^{2}-1}{y^{2}}$

Step 3:

For $x^{2}=\frac{4y^{2}-1}{y^{2}}$ to be defined, $y$ can not be equal to zero

i.e., $y\neq 0$

Also since $x^{2}\geq 0$,

$\therefore \frac{4y^{2}-1}{y^{2}}\geq 0$

or, $4y^{2}-1\geq 0 (\because y^{2}\geq 0)$

or, $(2y-1)(2y+1)\geq 0$

or, $4(y-\frac{1}{2})(y+\frac{1}{2})\geq 0$

The above table implies that

$y\epsilon \left ( -\infty,-\frac{1}{2} \right )\cup \left (\frac{1}{2},\infty \right )$ …..(1)

Step 4:

Since $y=\frac{1}{\sqrt{4-x^{2}}}$ is a square root function,

therefore y can not be negative (-ve).

i.e., $y\geq 0$ …..(2)

Now from (1) and (2), we get

$y\epsilon ( \frac{1}{2},\infty )$

Step 5:

Therefore the range of the function $f(x)=\frac{1}{\sqrt{4-x^{2}}}$ is $[ \frac{1}{2},\infty )$

Example 9: Find the range of the function

$f(x)=\sqrt{\frac{(x-3)(x+2)}{x-1}}$

Solution:

### #4. Find the range of modulus function or absolute value function

Example 10: Find the range of the absolute value function

$f(x)=\left | x \right |$

Solution:

We can find the range of the absolute value function $f(x)=\left | x \right |$ on a graph.

If we draw the graph then we get

Here you can see that the y value starts at y=0 and extended to infinity.

$\therefore$ the range of the absolute value function $f(x)=\left | x \right |$ is $[0,\infty)$.

Example 11: Find the range of the absolute value function

$f(x)=-\left | x-1 \right |$

Solution:

The graph of $f(x)=-\left | x-1 \right |$ is

From the graph, it is clear that the y value starts from y=0 and extended to $-\infty$.

Therefore the range of $f(x)=-\left | x-1 \right |$ is $(-\infty,0]$.

Shortcut Trick:

1. If the sign before modulus is positive (+ve) i.e., of the form $+\left | x-a \right |$, then the range will be $[a,\infty)$,
2. If the sign before modulus is negative (-ve) i.e., of the form $-\left | x-a \right |$, then the range will be $(-\infty,a]$.

We can also find the range of the absolute value functions $f(x)=\left | x \right |$ and $f(x)=-\left | x-1 \right |$ using the above short cut trick:

The function $f(x)=\left | x \right |$ can be written as $f(x)=+\left | x-0 \right |$

Now using trick 1 we can say, the range of $f(x)=\left | x \right |$ is $[0,\infty)$

Also using trick 2 we can say, the range of $f(x)=-\left | x-1 \right |$ is $(-\infty,0]$.

Example 12: Find the range of the following absolute value functions

1. $f(x)=\left | x \right |+6$,
2. $f(x)=\left | x+4 \right |$

Solution:

### #5. Find the range of a Step function

Example 13: Find the range of the step function $f(x)=[x],x\epsilon \mathbb{R}$.

Solution:

The step function $f(x)=[x],x\epsilon \mathbb{R}$ is expressed as

$f(x)=0, 0\leq x<1$

=$1, 1\leq x<2$

=$2,2\leq x<3$

$………$

=$-1,-1\leq x<0$

=$-2,-2\leq x<-1$

$………$

You can verify this result from the graph of $f(x)=[x],x\epsilon \mathbb{R}$

i.e., $y\epsilon$ {$…,-2,-1,0,1,2,…$}

i.e., $y\epsilon \mathbb{Z}$, the set of all integers.

$\therefore$ the range of the step function $f(x)=[x],x\epsilon \mathbb{R}$ is $\mathbb{Z}$, the set of all integers.

Example 14: Find the range of the step function $f(x)=[x-3],x\epsilon \mathbb{R}$.

Solution:

By using the definition of step function, we can express $f(x)=[x-3],x\epsilon \mathbb{R}$ as

$f(x)=1,3\leq x<4$

=$2,4\leq x<5$

=$3,5\leq x<6$

$………$

=$0,2\leq x<3$

=$-1,1\leq x<2$

=$-2, 0\leq x<1$

=$-3, -1\leq x<0$

$………$

You can verify this result from the graph of $f(x)=[x-3],x\epsilon \mathbb{R}$

i.e., $y\epsilon$ {$…,-3,-2,-1,0,1,2,3,…$}

i.e., $y\epsilon \mathbb{Z}$, the set of all integers.

$\therefore$ the range of the step function $f(x)=[x-3],x\epsilon \mathbb{R}$ is $\mathbb{Z}$, the set of all integers.

Example 15: Find the range of the step function $f(x)=\left [ \frac{1}{4x} \right ],x\epsilon \mathbb{R}$.

Solution:

### #6. Find the range of an Exponential function

Example 16: Find the range of the exponential function $f(x)=2^{x}$.

Solution:

The graph of the function $f(x)=2^{x}$ is

Here y=0 is an asymptote of $f(x)=2^{x}$ i.e., the graph is going very close and close to the y=0 straight line but it will never touch y=0.

Also, you can see on the graph that the function is extended to $+\infty$.

So we can say $y>0$.

$\therefore$ the range of the exponential function $f(x)=2^{x}$ is $(0,\infty)$.

Example 17: Find the range of the exponential function

$f(x)=-3^{x+1}+2$.

Solution:

The graph of the exponential function $f(x)=-3^{x+1}+2$ is

From the graph of $f(x)=-3^{x+1}+2$ you can see that y=2 is an asymptote of $f(x)=-3^{x+1}+2$ i.e., on the graph $f(x)=-3^{x+1}+2$ is going very close and close to y=2 towards -ve x-axis but it will never touch the straight line y=2 and extended to $-\infty$ towards +ve x-axis.

i.e., $y<2$

$\therefore$ the range of the exponential function $f(x)=-3^{x+1}+2$ is $(-\infty,2)$.

There is a shortcut trick to find the range of any exponential function. This trick will help you find the range of any exponential function in just 2 seconds.

Shortcut trick:

Let $f(x)=a\times b^{x-h}+k$ be an exponential function.

Then

1. If $a>0$, then $R(f)=(k,\infty)$,
2. If $a<0$, then $R(f)=(-\infty,k)$.

Now we try to find the range of the exponential functions $f(x)=2^{x}$ and $f(x)=-3^{x+1}+2$ with the above shortcut trick:

We can write $f(x)=2^{x}$ as $f(x)=1\times 2^{x}+0$, $1>0$ and comparing this result with trick 1 we directly say

The range of $f(x)=2^{x}$ is $(0,\infty)$.

Also $f(x)=-3^{x+1}+2$ can be written as $f(x)=-1\times 3^{x+1}+2$, $-1<0$ and comparing with trick 2 we get

The range of $f(x)=-3^{x+1}+2$ is $(-\infty,2)$.

Example 18: Find the range of the exponential functions given below

$f(x)=-2^{x+1}+3$

Solution:

### #7. Find the range of a Logarithmic function

The range of any logarithmic function is $(-\infty,\infty)$.

We can verify this fact from the graph.

$f(x)=\log_{2}x^{3}$ is a logarithmic function and the graph of this function is

Here you can see that the y value starts from $-\infty$ and extended to $+\infty$,

i.e., the range of $f(x)=\log_{2}x^{3}$ is $(-\infty,\infty)$.

Example 19: Find the range of the logarithmic function

$f(x)=\log_{2}(x+4)+3$

Solution:

### #8. Find the range of a function relation of ordered pairs

A relation is the set of ordered pairs i.e., the set of (x,y) where the set of all x values is called the domain and the set of all y values is called the range of the relation.

In the previous chapter, we have learned how to find the domain of a function using relation.

Now we learn how to find the range of a function using relation.

For that we have to remember 2 rules which are given below:

Rules:

1. Before finding the range of a function first we check the given relation (i.e., the set of ordered pairs) is a function or not
2. Find all the y values and form a set. This set is the range of the relation.

Now see the examples given below to understand this concept:

Example 20: Find the range of the relation

{(1,3), (5,9), (8,23), (12,14)}

Solution:

In the relation {(1,3), (5,9), (8,23), (12,14)}, the set of x coordinates is {1, 5, 8, 12} and the set of y coordinates is {3, 9, 14, 23}.

If we draw the diagram of the given relation it will look like this

Here we can clearly see that each element of the set {1, 5, 8, 12} is related to a unique element of the set {3, 9, 14, 23}.

Therefore the given relation is a Function.

Also, we know that the range of a function relation is the set of y coordinates.

Therefore the range of the relation {(1,3), (5,9), (8,23), (12,14)} is the set {3, 9, 14, 23}.

Example 21: Find the range of the set of ordered pairs

{(5,2), (7,6), (9,4), (9,13), (12,19)}.

Solution:

The diagram of the given relation is

Here we can see that element 9 is related to two different elements and they are 4 and 13 i.e., 9 is not related to a unique element and this goes against the definition of the function.

Therefore the relation {(5,2), (7,6), (9,4), (9,13), (12,19)} is not a Function.

Example 22: Determine the range of the relation described by the table

Solution:

### #9. Find the range of a Discrete function

A Discrete Function is a collection of some points on the Cartesian plane and the range of a discrete function is the set of y-coordinates of the points.

Example 23: How do you find the range of the discrete function from the graph

Solution:

From the graph, we can see that there are five points on the discrete function and they are A (2,2), B (4,4), C (6,6), D (8,8), and E (10,10).

The set of the y-coordinates of the points A, B, C, D, and E is {2,4,6, 8, 10}.

$\therefore$ the range of the discrete function is {2,4,6,8,10}.

Example 24: Find the range of the discrete function from the graph

Solution:

The discrete function is made of the five points A (-3,2), B (-2,4), C (2,3), D (3,1), and E (5,5).

The set of the y coordinates of the discrete function is {2,4,3,1,5} = {1,2,3,4,5}.

$\therefore$ the range of the discrete function is {1,2,3,4,5}.

### #14. Find the range of a piecewise function

Example 25: Find the range of the piecewise function

Solution:

The piecewise function consists of two function:

1. $f(x)=x-3$ when $x\leq -1$,
2. $f(x)=x+1$ when $x>1$.

If we plot these two functions on the graph then we get,

This is the graph of the piecewise function.

From the graph, we can see that

1. the range of the function $f(x)=x-3$ is $(-\infty,-2]$ when $x\leq -1$,
2. the range of the function $f(x)=x+1$ is $(2,\infty)$ when $x>1$,

Therefore from the above results we can say that

The range of the piecewise function $f(x)$ is

$(-\infty,-2]\cup (2,\infty)$.

Example 26: Find the range of a piecewise function given below

Solution:

If you notice the piecewise function then you can see there are functions:

1. $f(x)=x$ defined when $x\leq -1$,
2. $f(x)=2$ defined when $-1,
3. $f(x)=\sqrt{x}$ defined when $x\geq 1$.

Now if we draw the graph of these three functions we get,

This is the graph of the piecewise function.

Here you can see that

The function $f(x)=x$ starts $y=-1$ and extended to $-\infty$ when $x\leq -1$.

So the range of the function $f(x)=x,x\leq -1$ is $(-\infty,-1]$……..(1)

The functional value of the function $f(x)=2, -1 is 2.

The range of the function $f(x)=x$ is {$2$}……..(2)

The function $f(x)=\sqrt{x}$ starts at $y=1$ and extended to $\infty$ when $x\geq 1$.

The range of the function $f(x)=\sqrt{x}$ is $[1,\infty)$ when $x\geq 1$……..(3)

From (1), (2), and (3), we get,

the range of the piecewise function is

$(-\infty,-1]\cup$ {$2$}$\cup [1,\infty)$

= $(-\infty,-1]\cup [1,\infty)$

### #15. Find the range of a composite function

Example 27: Let $f(x)=2x-6$ and $g(x)=\sqrt{x}$ be two functions.

Find the range of the following composite functions:

(a) $f\circ g(x)$

(b) $g\circ f(x)$

Solution of (a)

First we need to find the function $g\circ f(x)$.

We know that,

$f\circ g(x)$

=$f(g(x))$

=$f(\sqrt{x})$ ($\because g(x)=\sqrt{x}$)

=$2\sqrt{x}-6$

Now see that $2\sqrt{x}-6$ is a function with a square root and at the beginning of this article, we already learned how to find the range of a function with a square root.

Following these steps, we can get,

the range of the composite function f of g is

$R(f\circ g)=[-6,\infty)$.

Solution of (b):

$g\circ f(x)$

=$g(f(x))$

=$g(2x-6)$ ($\because f(x)=2x-6$)

=$\sqrt{2x-6}$, a function with a square root

Using the previous method we get,

the range of the composite function $g\circ f(x)$ is

$R(g\circ f(x))=[0,\infty)$

Example 28: Let $f(x)=3x-12$ and $g(x)=\sqrt{x}$ be two functions.

Find the range of the following composite functions

1. $f\circ g(x)$,
2. $g\circ f(x)$

Solution:


text:
### -is-this-fft-'s blog

By -is-this-fft-, history, 21 month(s) ago,

Can someone else staying at this hotel please tell me how to turn off that light:

This is urgent, I need some sleep before the finals.

• +131

 » 21 month(s) ago, # |   +84 It has a small button in the right part of the frame. Near the right side of the TV. The bottom is dark brown, so it can be confused with the frame color.
 » 21 month(s) ago, # |   +62 Use Fast Fourier Transform to find the coordinates of the off switch.
 » 21 month(s) ago, # |   +3 switch off the main
 » 10 months ago, # |   -14 pull down hotel's MCB
 » 10 months ago, # | ← Rev. 4 →   -16 you can turn off the light like Mr. beanshoot the light before sleep...
•  » » 10 months ago, # ^ |   -18 exactly.
 » 6 weeks ago, # |   +107 Though I am quite late. But next time, if you are stuck in a hotel and struggling to switch off lights, use this technique XD.
•  » » 6 weeks ago, # ^ |   +50 Interesting! rubai_99 made this joke 11 months too late and got -16. You made the same joke 20 months later and have a positive score. Is it the way you're telling it?
•  » » » 6 weeks ago, # ^ |   0 Okay,so I didn't see his previous revision(Rev 1) and saw the latest I think call the house keeper. Maybe he didn't use the picture, so got downvotes. Now, I think I will also receive downvotes lol, it already decremented from +7 to +4 XD
•  » » » 6 weeks ago, # ^ |   +29 Yeah the way he told it actually made it funny, like because of the picture and all.Moreover, as far as upvotes and downvotes are concerned, there is also some luck factor involved because of the inductive effect. If you get initially downvoted, other people are more likely to downvote you even further.
•  » » » » 6 weeks ago, # ^ |   +1 Yes if you get negative initial then in 30 minutes it turned into -30. Check the comment before downvote.
•  » » » » » 6 weeks ago, # ^ |   0 If you want upvotes , just write a comment and upvote that from 3-4 alt accounts , usually people follow like sheeps , you will get overall much more upvotes.
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   +11 I think my luck and people are also don't have with me.the result in comments box. Also you can see In my latest comments a red says 'good morning' he get +35 and I reply his comments as "good "morning I get -34. Is it fair?
•  » » » » 6 weeks ago, # ^ |   0 I got 12 downvotes in 3 minutes only for being against cheating here. Till now, I don't know why I got the huge amounts of downvotes.
•  » » » » » 6 weeks ago, # ^ |   +3 I think it's depends on first few people's votes. (Please change the mind if the comment has 1 downvote it doesn't mean that the comment is useless.it useless or impractical according to his/her thinking.) You observe if your comments have -2 at first then it's trun into -20 most of the cases.
•  » » » » » » 6 weeks ago, # ^ |   +3 Yes, people are influenced by that. That is why any downvotes less than -5 are not shown and Mike even stated that in an old blog.
•  » » » » » » 6 weeks ago, # ^ |   0 First of all, you stop cheating in every contest man.
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   +136
 » 6 weeks ago, # |   -16 What competition was that? I guess ICPC?


text:
5

2.A 60-kg student does 60 push-ups in 40 s. With each push-up,the student must lift an average of70% of the body mass a height of 40 cm off the floor. Assuming twos...

Question

2.A 60-kg student does 60 push-ups in 40 s. With each push-up,the student must lift an average of70% of the body mass a height of 40 cm off the floor. Assuming twosignificant digits, calculate thefollowing:(a) the work the student does against the force of gravity for eachpush-up, assuming work is doneonly when the student pushes up(b) the total work done against the force of gravity in 40 s(c) the power achieved for this period.

2.A 60-kg student does 60 push-ups in 40 s. With each push-up, the student must lift an average of 70% of the body mass a height of 40 cm off the floor. Assuming two significant digits, calculate the following: (a) the work the student does against the force of gravity for each push-up, assuming work is done only when the student pushes up (b) the total work done against the force of gravity in 40 s (c) the power achieved for this period.

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# A Borel algebra containing $\infty$

If I adjoint $\infty$ to the real numbers ($\overline{\mathbb{R}}=\mathbb{R}\cup\infty$) is there a reasonable way to define a $\sigma$-algebra "$\mathcal{B}_{\overline{\mathbb{R}}}$" such that $\mathcal{B}_{\overline{\mathbb{R}}}$ restricted to sets not containing $\infty$ is equal to the Borel algebra $\mathcal{B}_{\mathbb{R}}$ ? If so, what topology on $\overline{\mathbb{R}}$ would induce $\mathcal{B}_{\overline{\mathbb{R}}}$ ?

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By the way, I believe that we are more interested in $R \cup \{ -\infty, +\infty\}$ than in $R \cup \{\infty\}$. –  user52350 Dec 9 '12 at 12:02

You can do it by considering the Borel sets of the unit circle in the complex plane. This is homeomorphic to the one-point compactification of the real numbers, which is how one should look at $\overline{\mathbb R}$.

It can be shown that the Borel subsets of the circle are simply Borel sets of the plane intersected with the circle. It also turns out that the sigma algebra is not different than that of the real numbers.

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Equivalently, topologize $\overline{\mathbb R}$ as the one-point compactification of the line. It is a Polish space, and the theory of Borel sets in that setting is very nice. –  GEdgar Dec 7 '12 at 17:03

First note that to any topological space you can define the $\sigma$-algebra of its Borel sets.

In general, given any topological space $X$ and a subspace $Y$ of $X$, if $Y$ is a Borel subset of $X$, the the Borel subsets of $Y$ (taken as a topological space in its own right) are exactly the Borel subsets of $X$ which happen to be subsets of $Y$.

Usually, one takes $\overline{\mathbb{R}}$ to be the one-point compactifaction of $\mathbb{R}$ (as in Asaf's answer and GEdgar's comment to it). The open subsets of $\overline{\mathbb{R}}$ would then be the open subsets of $\mathbb{R}$, together with all sets of the form $U \cup \{ \infty \}$ where $\mathbb{R} \setminus U$ is compact (in $\mathbb{R}$). Then $\mathbb{R}$ is an open (hence Borel) subspace of $\overline{\mathbb{R}}$, and therefore the Borel subsets of $\overline{\mathbb{R}}$ which are subsets of $\mathbb{R}$ coincide with the Borel subsets of $\mathbb{R}$.

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SCIENCE CHINA Information Sciences, Volume 62, Issue 8: 082304(2019) https://doi.org/10.1007/s11432-019-9856-y

## Variation of a signal in Schwarzschild spacetime

• AcceptedMar 28, 2019
• PublishedJul 12, 2019
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### Abstract

In this paper, the variation of a signal in Schwarzschild spacetime is studied and a general equation for frequency shift parameter (FSP) is presented. The FSP is found to depend on the gravitationally modified Doppler effects and the gravitational effects of observers. In addition, the time rates of a transmitter and receiver may differ. When the FSP is a function of the receiver time, the FSP contributed through the gravitational effect (GFSP) or the gravitationally modified Doppler effect (GMDFSP) may convert a bandlimited signal into a non-bandlimited signal. Using the general equation, the FSP as a function of receiver time is calculated in three scenarios: (a) a spaceship leaving a star at constant velocity communicating with a transmitter at a fixed position; (b) a spaceship moving around a star with different conic trajectories communicating with a transmitter at a fixed position; and (c) a signal transmitted from a fixed position in a star system to a receiver following an elliptic trajectory in another star system. The studied stars are a Sun-like star, a white dwarf, and a neutron star. The theory is illustrated with numerical examples.

### Acknowledgment

This work was supported in part by National Natural Science Foundation of China (Grant Nos. 61421001, 61331021, U1833203).

### References

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• Figure 1

(Color online) World lines of (a) a static observer and (b) a moving observer in Schwarzschild spacetime.

• Figure 2

(Color online) A spaceship leaving a star and communicating with a transmitter at a fixed position.

• Figure 3

(Color online) A spaceship moving around a star with (a) an elliptic, (b) a parabolic, and (c) a hyperbolic trajectory while communicating with a transmitter at a fixed position.

• Figure 4

(Color online) A signal propagating from star system A to another faraway star system B.

• Figure 5

(Color online) FSPs from gravitational effect (GFSP; top panels) and gravitationally modified Doppler effect (GMDFSP; bottom panels) vs. time for a spaceship moving away from (a), (d) the Sun, (b), (e) a white dwarf, and (c), (f) a neutron star. Results are plotted for different initial positions of the receiver and different transmitter positions.

• Figure 6

(Color online) (a) Difference between the Doppler effect $\beta'_1(\tau)$ (special relativity version) and the gravitationally modified Doppler effect $\beta_1(\tau)$ as a spaceship moves away from a neutron star at a fixed Newtonian velocity; (b) time $\tau$ of the moving spaceship vs. Newtonian time $T$; (c) difference between $\tau$ and $T$ vs. $T$. The receiver is initially located at $5R_n$.

• Figure 7

(Color online) FFTs of a $10^4$ Hz signal transmitted in (a), (d), (g) the solar system, (b), (e), (h) a white dwarf system, and (c), (f), (i) a neutron star system with or without gravitational effects at $\tau=50$ s (top panels), 150 s (center panels), and 250 s (bottom panels). The transmitters in the solar, white dwarf, and neutron star systems are located at $2R_{\odot}$, $2R_w$, and $2R_n$, respectively. The initial positions of the receivers are $4R_{\odot}$, $4R_w$, and $4R_n$, respectively. The sampling rate is 2.5 times the signal frequency.

• Figure 8

(Color online) FSPs of the gravitationally modified Doppler effect ($\beta_1(\tau)$), gravitational effect ($\beta_2(\tau)$), and the whole frequency shift parameter ($\beta(\tau)$), as functions of time. The spaceship moves around (a), (d), (g) the Sun, (b), (e), (h) a white dwarf, and (c), (f), (i) a neutron star with different conic trajectories: an ellipse ($e=0.2$, top panels), a parabola ($e~=1$, center panels) and a hyperbola ($e~=2$, bottom panels) with $a=5R_\odot$, $5R_w$ and $5R_n$, respectively.

• Figure 9

(Color online) FFTs of a $10^5$ Hz signal under (a) gravitational effects only $(\beta_2(\tau))$, (b) gravitationally modified Doppler effect only $(\beta_2(\tau))$, and (c) the whole effect $(\beta(\tau))$. The signals are transmitted in the neutron star system from $2R_n$ to the spaceship following an elliptic trajectory with $e=0.7$ and semi-major axis $a=5R_n$. The sampling rate is 2.5 times the signal frequency.

• Figure 10

(Color online) FSPs of the gravitationally modified Doppler effect ($\beta_1(\tau)$), gravitational effect ($\beta_2(\tau)$), and the whole frequency shift parameter ($\beta(\tau)$), as functions of time. The signal is received at a position with an elliptic trajectory ($e=0.0167$) in (a), (d), (g) a Sun-like star system, (b), (e), (h) a white dwarf system, and (c), (f), (i) a neutron star system, and is transmitted from a fixed position in another Sun-like star, white dwarf, and neutron star systems.

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text:
# Which categorical (coproduct-like) operation captures integration of measures?

Suppose we have a measure space $(X,a)$, a measurable space $Y$ and for every $x\in X$ we have a measure $b_x$ on $Y$. Suppose that $(Z,c)$ is a measure space such that as a measurable space $Z=X\times Y$ and the measure $c$ is the integral of measures $b_x$ with respect to the measure $a$.

I have a vague intuition that $(Z,c)$ is a coproduct of the family $(Y,b_x)$ along $(X, a)$

Question: Is there a category of measure spaces and a categorical construction in it which captures this intuition?

One could ask a similar question about a category of Hilbert spaces and integrals of Hilbert spaces over measure spaces, and preferably the "categorical construction" in question should also answer this problem.

The categorical construction in question should preferably be similar to coproduct, and it would be very nice if it actually was a coproduct in some category.

On the other hand, maybe my vague intuition is wrong, and comments on that would also be appreciated.

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It looks like the measure on $Z$ is specified in a very asymmetric way, so I can't see how it could be a coproduct. – S. Carnahan Jan 8 '11 at 18:44
@Scott: if (X,a) is a finite set with the counting measure then (Z,c) is a coproduct of (Y,b_x) in the category of measure spaces with measure preserving maps. If the measure is other than counting - more "assymetric", as you could maybe put it - then the answer should be more complicated and involve some additional structure, but imho there should be a satisfactory answer. – Łukasz Grabowski Jan 8 '11 at 20:43
Dependent sum? See ncatlab.org/nlab/show/dependent+product – David Roberts Jan 8 '11 at 23:44
I've heard words along the lines of "direct integral" --- something that generalizes direct sums. I think the context is slightly different, but maybe it's close enough that you can adapt the definition? – Theo Johnson-Freyd Jan 9 '11 at 5:02


text:
# Recover the mutated source code

In a very unusual accident involving a small sample of radium, an electrocuted whale, and three gummy bears, some of The Management™'s source code has been mutated. Little does The Management™'s boss know, it was actually the Cops© who had been responsible, in an attempt to thwart The Management™'s "evil" plans. So the Robbers® have been hired in an attempt to retrieve the original code, because who doesn't like being evil sometimes?

note: This challenge was heavily inspired by Unscramble the Source Code.

# Description

This is a challenge.

• The cops will write a program (the mutated code) that performs Task #1 (and also write a program that performs Task #2, but is kept secret).
• The robbers will attempt to reverse the "mutation" and change this original code into code that performs Task #2.

In this challenge, Task #1 will be to output the nth prime number, and Task #2 will be to output the nth Fibonacci number (which is somehow evil, according to the Cops© anyway). The Fibonacci sequence is defined as (n=11; n=21; n=32; ...), and the prime numbers are defined as (n=12; n=23; n=35; ...).

The cops' goal is to minimize the difference between the programs that complete Task #1 and Task #2, while preventing the robbers from recreating the code that completes Task #2.

# Cop Rules

The cops will write two programs (one that completes Task #1, and one that completes Task #2), and make the following information public:

• The first program (that outputs the nth prime number)
• The Levenshtein edit distance between the first program and the second program
• The programming language that both programs are written in (must be the same language for both programs)

The following restrictions apply to both programs:

• They must be 128 characters in length or less.
• They must only use printable ASCII (plus newlines, which are also allowed).
• They must take less than 10 seconds to run for n=45, and they are not required to produce the correct output for any n>45.
• They must not use any hashing or cryptographic functions.

# Robber Rules

The robber will attempt to change the cop's program (which completes Task #1) into a program that completes Task #2 (not necessarily the original program written by the cop) in the edit distance specified by the cop.

An already-cracked submission cannot be cracked again (only the first robber who cracks a submission gets credit).

After cracking a submission, please do the following:

• Edit the cop's answer if you have edit privileges (if you do not, either wait until someone else with the required privileges does so for you or suggest an edit).

# Scoring

If a cop's program remains uncracked for 1 week, the cop can post the original code that completes Task #2 (in the specified edit distance), and the submission is from then on considered "safe." The safe submission that has the smallest edit distance will win. In the event of a tie, the shortest program (the original that completes Task #1) wins. If two submissions are still tied, the one posted earlier wins.

If a robber successfully cracks a cop's submission, the robber's score goes up by the edit distance of that submission. For example, a robber that cracks a submission with an edit distance of 3 and one with a distance of 5 earns 8 points. The robber with the highest score wins. In the event of a tie, the robber who earned the score first wins.

# A small tool to compute the Levenshtein distance

var f=document.getElementById("f"),g=document.getElementById("s"); function h(){var a=f.value,e=g.value;if(128<a.length)a="First program is too long!";else if(128<e.length)a="Second program is too long!";else{if(0===a.length)a=e.length;else if(0===e.length)a=a.length;else{var d=[],b;for(b=0;b<=e.length;b++)d[b]=[b];var c;for(c=0;c<=a.length;c++)d[0][c]=c;for(b=1;b<=e.length;b++)for(c=1;c<=a.length;c++)d[b][c]=e.charAt(b-1)===a.charAt(c-1)?d[b-1][c-1]:Math.min(d[b-1][c-1]+1,Math.min(d[b][c-1]+1,d[b-1][c]+ 1));a=d[e.length][a.length]}a="Distance = "+a+""}document.getElementById("d").innerHTML=a}f.onkeyup=h;g.onkeyup=h;

First program

<textarea id=f rows=7 cols=80 style="width:100%"></textarea>

Second program

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• What is the 1st Fibonacci number? 0 or 1? Or does it not matter – kukac67 Dec 29 '14 at 3:06
• @kukac67 It's 1; I've edited the post. – Doorknob Dec 29 '14 at 3:13
• What should the output of the programs be, in the case of overflow? – es1024 Dec 29 '14 at 5:44
• Must it be a full program or can it be a function? What about an anonymous function? – Tyilo Dec 29 '14 at 8:11
• What counts as a "hashing or cryptographic function"? Can I base-convert stuff? Can I take large exponentials modulo large primes? – Martin Ender Dec 29 '14 at 22:53

# Python 2, distance = 8 [cracked]

from fractions import*
n=input()
k,=P=[2]
while n>len(P):k+=1;z=reduce(lambda x,y:x*y,P,1);P+=[k]*(gcd(z,k)<2)
print P[-1]

Finally got this one under the char limit. Shouldn't be too hard, but I thought the idea was interesting.

Intended solution:

from fractions import* n=input() k,=P=[1] while n>len(P):k+=1;z=reduce(lambda x,y:x+y,P[:-1],1);P+=[z]*(gcd(z,k)<2) print P[-1]

The idea was to use that F(n+2) = 1 + (sum over F(k) from k = 1 to n), and the fact that consecutive Fibonacci numbers are coprime. The 1 in the reduce argument was supposed to provide the +1.

Looks like feersum found a different line of attack!

• Cracked – feersum Dec 29 '14 at 15:18

## J, distance = 5 [Cracked]

f=:3 :'{.p:|.i.y'

Usage:

> f 45
197

• Cracked – randomra Dec 29 '14 at 21:28

# Ruby, distance 6 [safe]

require'prime';p Prime.take(n=gets.to_i)[-1]
#p (((807462154311276410)**n/(5**0.5)).round)

Coming up with pairs of formulas with short edit distances is fun, but it seems like this approach might be more effective/annoying. You may understand exactly what I did, but that doesn't mean you can reverse it.

Solution:

require'prime';p=Prime.take(n=gets.to_i)[-1] p (((0742154311276/4e10)**n/(5**0.5)).round)

Explanation:

The code generates the Golden Ratio to 11 decimal places and uses it to directly calculate the Fibbonaci sequence. It's just enough precision to get the required number of terms right. That part wasn't obfuscated at all, if you happen to know the formula. To make it harder to brute-force reverse my mutations and recover the constant, I used octal notation (the leading 0) and scientific notation (4e10). Dividing by 4e10 rather than 1e11 makes it look more like I'm dividing by something .0 to force float division, when actually anything in scientific notation is for some reason always a Float in Ruby, even when a Bignum might seem to make more sense. I thought I was being clever with the p= stuff, but the way I wrote it you can just delete the p. I could've reduced the distance by 1 and forced the p= solution by using p&& instead of # on the second line, but I didn't think of it.

• Didn't think of trying inserting an e down there when doing brute force. Really sneaky solution. :) – Vectorized Jan 15 '15 at 23:02

## Python 2 - LD = 13 Cracked

n=1;j=input();
while j>0:
n+=1;j-=1;
while~-all(n%i for i in range(2,n)):n+=1;
print n

A nice, easy (hopefully not too easy) one to start things off :)

Looks like it was too easy ;) I feel rather silly that I forgot you could use comments :/

• Cracked. – Sp3000 Dec 29 '14 at 4:21

import Data.List
x=id=<<snd(mapAccumL(\m n->(,)=<<(++m)$[n|and[nmodm1/=0|m1<-m]])[1+1][3..]) main=print.(!!)(0:2:x)=<<readLn  This could be more readable, but the import ate too many bytes, so I had to golf it a bit. # Ruby, distance 14 (Cracked) p [x=2,y=1,*(1..200).map{|i|y==(y*x**(i-1)+x%2).divmod(i)[x-1]?i:1}].-([x-1])[gets.to_i-1]  • Cracked? – Vectorized Dec 31 '14 at 0:39 • Hm, your Fibbonaci sequence starts with 0, where the rules say to start with 1. Otherwise checks out (although very different from my intended solution). – histocrat Dec 31 '14 at 15:14 • Ok, fixed. Nice use of Fermat's btw. – Vectorized Dec 31 '14 at 15:25 # CJam, distance 10 (Cracked) 1l~{{)_mp!}g}*  Just put n on STDIN. Test it here. For reference, the original solution used the rare j. Original: T1]l~\{(_j\(j+}j # J, distance = 4 [safe] f =: 3 : '{. (p:) (+) / 0 , y - 1x'  Solution: f =: 3 : '{. 2(x:) (+%) / 0 , y$ 1x'

Method:

Denominator {. 2(x:) of the continued fraction (+%) 1+1/(...(1+1/(1+1/(1+1/(1))))).

# Python 3, distance = 14 [cracked]

n = int(input())
P = [2]
k = 2
while n > len(P):
k += 1
for x in P:
if k%x == 0: break
else: P += [k]
print(k)

I had some spare chars so I put some whitespace in for clarity :)

• Cracked – feersum Dec 29 '14 at 6:31

# JAGL Alpha 1.2 - Distance = 16 [Cracked]

T~2S]{]S1{D[dmn}wDS}wSP

Shouldn't be too hard, We will see what happens...

• – matsjoyce Dec 29 '14 at 20:16

# TI-BASIC, distance 38

Input N:{2>L1:For(A,3,E99:If min(1=gcd(A,seq(A,A,2,$(A:A>L1(1+dim(L1:End:L1(N  > represents the STO→ key and $ represents the square root symbol.

• Some of these characters don't seem to be printable ASCII. – feersum Dec 30 '14 at 5:58
• @feersum Thanks, corrected. Distance is still 38. – Timtech Dec 30 '14 at 14:04

# Python 2 – distance = 12 [Cracked]

I'm kinda happy with how this turned out.

f=lambda p,i:p if p[45:]else f(p+[i]if all(i%q for q in p[1:])else p,-~i)
print f([1,2,3],2)[input()]

Let's see how long it takes... I assume it'll still be cracked.

Edit: shortened code a tiny bit, no effect on operation/distance.

## Intended solution

I tried to go for no comments or newline changes.

f=lambda p,i:p if p[45:]else f(p+[p[i]+p[-2]]if all(i|q for q in p[1:])else p,-~i) print f([1,1,1],2)[input()]

• Cracked – feersum Dec 29 '14 at 22:55

# Python 3 - Distance = 14 [Cracked]

a,c,n=1,2,int(input())
while n-1:
c+=1
while 1!=list(map(c.__mod__,range(2,46))).count(0):
c,a=a+c,a
n-=1
print(c)

We'll see how long this lasts...

• Cracked – feersum Dec 29 '14 at 19:02


text:
# How to get \baselineskip instead of \lineskip when using a stored box?

Consider this code:

\documentclass{article}
\begin{document}
\pagestyle{empty}

\def\contents{%
\parindent = 0pt
\LARGE
\strut
\raise\baselineskip
\vtop
{%
{\Large\smash{Chapter}\par}%
title\par
}%
\par
}
\newbox\tempbox

\loggingoutput

\vspace*{20pt}%
\begingroup
\contents
\endgroup

\clearpage

\vspace*{20pt}%
\setbox\tempbox\vbox
\bgroup
\contents
\egroup
\noindent \box \tempbox

\end{document}

The \contents are supposed to be the a chapter heading, with Chapter written in one line and the chapter title in the line below.

I use that \raise\baselineskip\vtop{...} contraption to allow me to remove the Chapter text from starred chapters but still have the chapter titles in the same vertical position of the page.

All was fine until I needed to set this chapter head in a box before putting it on the output page. I thought that simply enclosing the whole thing in a \vbox to use later with \box would work, but I wouldn't be asking here if it did :)

The log shows this for the first page:

...\glue 20.0
...\glue 0.0
...\glue(\parskip) 0.0 plus 1.0
...\glue(\baselineskip) 0.0
...\hbox(22.0+6.60007)x345.0

and this for the second:

...\glue 20.0
...\glue 0.0
...\glue(\parskip) 0.0 plus 1.0
...\glue(\lineskip) 1.0
...\hbox(22.0+6.60007)x345.0

the only difference being that in the first one TeX inserts \baselineskip before the contents and in the second it inserts a \lineskip.

Why does it do that? And how to do otherwise?

• Try \unvbox instead of \box. A box only has one baseline. Mar 19 '19 at 21:57
• @JohnKormylo Hm... Seems to work as well. I swear I had a good reason to use \box instead of \unvbox, but I can't remember it right now :P Thanks for the suggestion :-) Mar 19 '19 at 22:15

You have used a \vbox so the reference point is at the bottom line so the box has height 22pt, as this is already more than \baselineskip tex can not ensure \baselineskip space between the reference point of the previous node and that of the box being added so it inserts \lineskip instead. You could use a \vtop instead, which makes it easier to get the correct space at the start (and harder to get the correct space after)
Hmm actually I suppose the main difference here is that in the first case \baselineskip is coming from the large font so it just fits (hence gets added as 0pt) but in the second case the large font is inside the box so you are using a small baselineskip.
you get \baselineskip (0pt) with
{\LARGE\noindent \box \tempbox\par}

• Hm... Setting the font to \LARGE seems to to the job. Thanks :-) One more thing (about your edited out suggestion): If I were to use \vtop instead of \vbox, is there an easy way to force TeX into using either \baselineskip or \lineskip every time? Mar 19 '19 at 22:10
• @PhelypeOleinik the two things are different \baselineskip is a target reference to reference space so the amount added depends on the depth of the previous line and the height of this, \lineskip is just an "emergency padding that tex adds to stop lines touching if baseline spacing can not be maintained. you can use \nointerlineskip before the line then neither is added and you can add an explicit vkip. Mar 19 '19 at 22:18
• Yes, but in my imagination there would be a to coerce TeX into using one or another to compute the vertical glue. But for now \nointerlineskip will do (If I knew about it I forgot). Thanks :-) Mar 19 '19 at 22:25
• @PhelypeOleinik you can set \lineskiplimit to be -\maxdimen but then it will always use \baselineskip so if the height of the row being added is more then that it will add negative space making the lines overprint, forcing the reference point of the new line to be \baselineskip below the previous. So you can force \baselineskip` if you really want that but it is very rarely useful. Mar 19 '19 at 22:30


text:
# W2: Scattering and Refraction of Low-Mode Internal Tides by Interaction With Mesoscale Eddies

Principal investigators: Dr. Janna Köhler (MARUM/University of Bremen), Prof. Dirk Olbers (MARUM/University of Bremen), Prof. Monika Rhein (MARUM/University of Bremen), Prof. Jin-Song von Storch (Max Planck Institute for Meteorology)

Internal gravity waves in the ocean are generated by tides, wind, and interaction of currents with rough seafloor topography. Models predict a global energy supply for the internal wave field of about 0.7–1.3 TW by the conversion of barotropic tides at mid-ocean ridges and abrupt topographic features. Winds acting on the oceanic mixed layer contribute 0.3–1.5 TW and mesoscale flow over rough topography adds an additional amount of 0.2 TW. Globally, 1–2 TW are needed to maintain the observed stratification of the deep ocean by diapycnal mixing that results from the breaking of internal waves. Ocean circulation models show significant impact of the spatial distribution of internal wave dissipation and mixing on the ocean state, e.g. thermal structure, stratification, and meridional overturning circulation.

The scattering and refraction of low-mode internal tides by interactions with mesoscale eddies provides a direct and important energetic link between mesoscale processes and the internal wavefield. Despite their relevance for the geographical distribution and intensity of mixing in the ocean, these processes have not been well studied or understood. In the second phase of the TRR181 W2 aims to improve our understanding of the involved processes and to quantify the eddy-induced changes in the magnitude, direction, and vertical structure of internal tides in the mid-latitude ocean.

For this we will combine
(a) In-situ observations to infer eddy-induced refraction and changes in the vertical structure and magnitude of the internal tides, dissipation rates from finestructure, and spectral energy transfers.
(b) An ICON-o telescope simulation with a resolution of up to O(500m) that will include tides, resolve higher vertical modes and simulate interactions between eddies and internal tides.
(c) Kinetic equations in a modal framework, evaluated for different parameters of the eddy field to estimate the energy transfer between tide and eddies.

The Walvis Ridge region in the southeast Atlantic (Fig. 1) is chosen as a representative for the mid- latitude open ocean as here energetic low-mode $$M_{2}$$ internal tides propagate away from the ridge and cross the path of eddies in the form of both Agulhas rings and other mesoscale features.

Low-mode internal waves in the ocean possess a simple vertical structure, and hold a large part of the total energy of the internal wave field. They are able to travel basin-wide before they are dissipated. The spatial distribution of mixing related to internal wave dissipation affects the global overturning circulation in numerical models, but the observation as well as the representation of internal waves in ocean general circulation models (OGCMs) is still challenging.

To improve our understanding of the life cycle of internal waves, we studied the radiated low mode internal waves including processes operating along their paths. To identify sources and sinks, as well as to quantify the contribution to local dissipation, in-situ hydrographic measurements were combined with observations of internal wave energy fluxes based on satellite altimetry and STORMTIDE. STORMTIDE is a 1/10°-simulation from the Max-Planck Institute Ocean Model (MPIOM) forced by the full lunisolar tidal potential that resolves low-mode internal waves. A realistic surface forcing with 6-hourly wind stress was included in STORMTIDE2, currently the only global OGCM that is driven by tidal and wind forcing. The region of in-situ observations was south of the Azores in the NE Atlantic, where the satellite altimetry and STORMTIDE show beams of converging low mode internal waves generated at a seamount chain.

The modal structure of the internal tide energy flux is important for the ratio of local (close to the generation site) to remote energy dissipation, as low-mode internal waves are less prone to breaking and more likely to propagate over large distances compared to waves with a complex vertical structure. Our in-situ observations show that even close to the generation site, the energy flux is primarily contained in the first mode and thus likely to propagate over long distances before dissipating. Superposition of modes 1 and 2 captures over 84% of the total energy flux, indicating that OGCMs that resolve the first two modes are potentially able to capture the main characteristics of internal tide energy fluxes related to topography on scales typical for seamounts.

The direct comparison of the energy fluxes from the in-situ measurements, satellite altimetry, and OGCM shows that in-situ fluxes are generally higher than the corresponding fluxes in STORMTIDE and those derived from altimetry (Fig 1). Differences between in-situ data and altimetry are expected since the spatial resolution of satellite data limits the resolution of the inferred energy fluxes. The long-range propagation of the internal tides observed in altimetry generally agrees with the results from the in-situ measurements, although the in-situ energy fluxes are more variable and show a less monotonic decrease away from the generation sites. STORMTIDE resolves small scale structures in the energy flux distribution, but high energy fluxes are restricted to the vicinity of the generation sites.

The simulated internal-tide generation was further quantified in a more detailed way, requiring 3-dimensional high-resolution data, not easily achievable from available observations. We showed, using STORMTIDE2, that the $$M_{2}$$internal tide generation is characterized by a systematic pressure drop from the upstream to the downstream side of an obstacle (seamount, ridge), consistent with theoretical considerations. Generation is affected by both the tidal velocity and the topographic slope, although the former is dominant. This dominance arises not only because the internal-tide generation varies linearly with topographic slope but essentially quadratically with tidal velocity, but also because the tidal velocity can change up to one order of magnitude from the top to the foot of a high ridge. STORMTIDE2 shows that the intense generation occurs near the summits of the major high underwater ridges and rises, leading to an $$M_{2}$$internal-tide generation being strongest above 1200 m and weakening drastically below 3000m.

While the in-situ study region south of the Azores was chosen because of the unimpaired spreading of internal tides after generation, a mooring time series reveals the presence of eddies during two time periods (Fig. 2): one surface intensified eddy with a maximum horizontal velocity of approx. 20 cm s-1, and a weaker subsurface one. The temporal variability in the time series of energy flux is dominated by two factors: A strong coupling of the flux magnitude to the spring-neap variability in the barotropic tidal forcing (Fig. 2), and the decrease of energy flux during phases of higher eddy kinetic energy. Especially the surface-intensified eddy is correlated with a significant weakening of the energy flux compared to the time period with no eddy activity (Fig. 2). A potential transfer of internal tide energy from modes 1 and 2 into higher mode internal waves would result in a changed ratio of local to remote energy dissipation and hence be important for the global distribution of internal wave energy.

Whether the decrease in the low-mode internal tide energy flux observed in the mooring time series is due to increased dissipation, refraction or scattering of energy into higher modes is the subject of the 2nd phase.

## Investigating internal wave energy fluxes

In my current work, I also look into the impact of mesoscale motion on the energy flux in this dataset.

Jonas Löb, PhD W2

My name is Jonas and I am a PhD Student in the subproject W2 “Low mode waves” in the working group Oceanography at the University Bremen. In this project I calculate low mode internal wave energy fluxes from mooring measurements and compare the results with measurements from satellite altimetry and a 1/10° ocean model (STORMTIDE2). Energy flux is an important quantity for these models because its divergence identifies sources and sinks.

Internal gravity waves occur all over the stratified ocean and can be grouped in different categories varying on their generation mechanism. I focus mainly on internal tides in the semidiurnal frequency M2 generated by the barotropic tides over rough topography. Internal tides are a response of the astronomical gravitational forces of the ocean via oscillations in the sea surface elevation with horizontal tidal currents through the entire water column. These waves in the stratified ocean take the form of standing vertical oscillations of horizontal currents, called modes. The “zeroth” (barotropic) mode of horizontal velocity corresponds to horizontal ocean currents that are uniform from top to bottom. The first depth dependent (baroclinic) mode is characterized by flow in one direction at the top and in the opposite direct at the bottom. Higher modes have a more complicated vertical structure and their phase speed decreases with increasing mode number. The vertical structure of a mode can be calculated by the stratification, and velocity profiles can be fitted onto a linear combination of these modes. Low mode motions contain appreciable energy but quickly propagate away laterally. To study these low mode internal waves, we deployed a mooring inside a tidal beam in the eastern North Atlantic, south of the Azores, where a seamount chain stands out as a generation site for internal tides. In our study region the energy flux correlates reasonably well in direction, coherent – uncoherent portioning and mode ratio between mooring and model time series and satellite data. With regard to the total energy flux, the model and satellite observations underestimate the flux compared to the in situ data.

In my current work, I also look into the impact of mesoscale motion on the energy flux in this dataset. A surface eddy was crossing the mooring, and in the process dampening the energy flux in the first two modes by about one third, while a passing subsurface eddy dampened the energy mainly in the second mode. These observations support the idea that eddy interactions transfer energy from low modes into higher modes that can lead to increased dissipation. An open question is how much of the energy converted from lower to higher modes result in local dissipation, which is a crucial information in creating energy consistent ocean-climate models.

## The fate of low mode internal tides

The major goal of our subproject is to study the fate of low mode internal tides and the processes that operate along their pathways.

Janna Köhler, Postdoc in W2

I am an observational physical oceanographer working as a Postdoc at the University Bremen. Prior to joining the TRR I mainly studied temporal variability of internal waves using time series of moored instruments, and related the observed changes in the internal wave energy to generation processes such as topography-current interaction or energy input into the internal wave field by the wind.

In the TRR I am part of the subproject W2 “Energy transfer through low mode internal waves”. Low mode internal waves possess a major part of the entire energy of the internal wave field, which makes them an important component of the oceanic energy pathways.

The major goal of our subproject is to study the fate of low mode internal tides and the processes that operate along their pathways. A seamount south of the Azores provides a good study case for these processes, as it is one of the main generation sites of internal tides in the Atlantic. Here we will conduct shipboard measurements and deploy a mooring for about one year to study spatial and temporal variability of internal tide energy. We will then compare the observed spatial and temporal variability along the tidal beam with output of dedicated runs of the STORMTIDE model carried out by our project partners at the MPI in Hamburg.