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Conjugates and Rationalization

Conjugates and Rationalization

Introduction:

The term conjugate means a pair of things joined together. Let's look at these smileys:

These two smileys are exactly the same except for one pair of features that are actually opposite of each other. If you look at these smileys, you will notice that they are the same except that they have opposite facial expressions: one has a smile and the other has a frown.

What are the math Conjugates?

math conjugate is formed by changing the sign between two terms in a binomial. For instance, the conjugate of $$x + y$$ is $$x - y$$. We can also say that $$x + y$$ is a conjugate of $$x - y$$. In other words, the two binomials are conjugates of each other. Instead of a smile and a frown, math conjugates have a positive sign and a negative sign, respectively.

Let's consider a simple example: The conjugate of $$3 + 4x$$ is $$3 - 4x$$.

Now, Consider the surd $$3 + \sqrt 2$$,

If we change the plus sign to minus, we get the conjugate of this surd: $$3 - \sqrt 2$$.

What is special about conjugate of surds?

Rationalization of surds:

The special thing about conjugate of surds is that if you multiply the two (the surd and it's conjugate), you get a rational number. For example,

$\left( {3 + \sqrt 2 } \right)\left( {3 - \sqrt 2 } \right) = {3^2} - 2 = 7$

We note that for every surd of the form $$a + b\sqrt c$$, we can multiply it by its conjugate $$a - b\sqrt c$$  and obtain a rational number:

$\left( {a + b\sqrt c } \right)\left( {a - b\sqrt c } \right) = {a^2} - {b^2}c$

Let’s call this process of multiplying a surd by something to make it rational – the process of rationalization. In the example above, that something with which we multiplied the original surd was its conjugate surd.

How will we rationalize the surd $$\sqrt 2 + \sqrt 3$$? The conjugate surd (in the sense we have defined) in this case will be $$\sqrt 2 - \sqrt 3$$, and we have,

$\left( {\sqrt 2 + \sqrt 3 } \right)\left( {\sqrt 2 - \sqrt 3 } \right) = 2 - 3 = - 1$

How about rationalizing $$2 - \sqrt[3]{7}$$ ? The conjugate surd in this case will be  $$2 + \sqrt[3]{7}$$, but if we multiply the two, we have

$\left( {2 - \sqrt[3]{7}} \right)\left( {2 + \sqrt[3]{7}} \right) = 4 - \sqrt[3]{{{7^2}}} = 4 - \sqrt[3]{{49}}$

which is not a rational number. Thus, the process of rationalization could not be accomplished in this case by multiplying with the conjugate. The rationalizing factor (the something with which we have to multiply to rationalize) in this case will be something else. But what?

✍Note: The process of rationalization of surds by multiplying the two (the surd and it's conjugate) to get a rational number will work only if the surds have square roots.

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# Irrational Numbers

## Introduction of Irrational Numbers

An irrational number is a real number that cannot be expressed as a ratio of integers. example, √ 2. The real numbers which are not rational i.e. which cannot be expressed in the form of p/q, where p and q are integers and q ≠ 0 are known as irrational numbers. The decimal expansion of an irrational number is neither terminating nor recurring.

For Example  √ 2, √ 3, √ pi, etc.

The above numbers are not a rational number as any number which can be represented in the form of p/q, such that, p and q are integers and q ≠ 0 is known as a rational number.

For Example, 2, 5/11, -5.12, 0.31 are all examples of rational numbers because the decimal expansion of an irrational number either terminates or repeats.

From the study of irrational numbers, if p is a prime number, then p is irrational. From this, it can be said that 2, √ 3, √ pi, etc. are all irrational numbers.

## Proof that root 2 is irrational.

Let us assume that 2 is a rational number.

Going by the definition of rational numbers, it can be written that,

2 =p/q    …….(1)

Where p and q are co-prime integers and q ≠ 0 (Co-prime numbers are those numbers whose common factor is 1).

Squaring both the sides of equation (1), we have

2 = p2/q2

⇒ p2 = 2 q 2    ………. (2)

From the theorem stated above, if 2 is a prime factor of p2, then 2 is also a prime factor of p.

So, = 2 × c, where c is any integer.

Substituting this value of p in equation (3), we have

(2c)2 = 2 q 2

⇒ q2 = 2c 2

This implies that 2 is a prime factor of q2 also. Again from the theorem, it can be said that 2 is also a prime factor of q.

Since according to initial assumption, p and q are co-primes but the result obtained above contradicts this assumption as p and q have 2 as a common prime factor other than 1. This contradiction arose due to the incorrect assumption that 2  is rational.

So, root 2 is irrational.

Similarly, we can justify the statement discussed in the beginning that if p is a prime number, then  p  is an irrational number. Similarly, it can be proved that for any prime number p, p is irrational.

### Irrational Number theorem

The following theorem is used to prove the above statement

Theorem: Given p is a prime number and a2 is divisible by p, (where a is any positive integer), then it can be concluded that p also divides a.

Proof: Using the Fundamental Theorem of Arithmetic, the positive integer a can be expressed in the form of the product of its primes as:

a = p1 × p× p3………..  × pn …..(1)

Where, p1, p2p3, ……, pn represent all the prime factors of a.

Squaring both the sides of equation (1),

a2 = ( p1 × p× p3………..  × pn) ( p1 × p2 × p3………..  × pn)

⇒a2 = (p1)2 × (p2)2 × (p3 )2………..× (pn)2

According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique, except for the order of its factors.

The only prime factors of a2 are p1 , p2 , p3……….., pn. If p is a prime number and a factor of a2, then p is one of  p1 , p2 , p3……….., pn . So, p will also be a factor of a. Hence, if a2  is divisible by p, then p also divides a.

Now, using this theorem, we can prove that 2 is irrational.

### Solved examples of Irrational Numbers

Question: Which of the following are Rational Numbers or Irrational Numbers?

2, -.45678…, 6.5,  3,  2

Solution: Rational Numbers – 2, 6.5 as these are terminating numbers.

Irrational Numbers – -.45678…,  3,  2 as these are non-terminating numbers.


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# Understanding Linux CPU Load - when should you be worried?

You might be familiar with Linux load averages already. Load averages are the three numbers shown with the uptime and top commands - they look like this:

load average: 0.09, 0.05, 0.01

Most people have an inkling of what the load averages mean: the three numbers represent averages over progressively longer periods of time (one, five, and fifteen minute averages), and that lower numbers are better. Higher numbers represent a problem or an overloaded machine. But, what's the the threshold? What constitutes "good" and "bad" load average values? When should you be concerned over a load average value, and when should you scramble to fix it ASAP?

First, a little background on what the load average values mean. We'll start out with the simplest case: a machine with one single-core processor.

## The traffic analogy

A single-core CPU is like a single lane of traffic. Imagine you are a bridge operator ... sometimes your bridge is so busy there are cars lined up to cross. You want to let folks know how traffic is moving on your bridge. A decent metric would be how many cars are waiting at a particular time. If no cars are waiting, incoming drivers know they can drive across right away. If cars are backed up, drivers know they're in for delays.

So, Bridge Operator, what numbering system are you going to use? How about:

• 0.00 means there's no traffic on the bridge at all . In fact, between 0.00 and 1.00 means there's no backup, and an arriving car will just go right on.
• 1.00 means the bridge is exactly at capacity. All is still good, but if traffic gets a little heavier, things are going to slow down.
• over 1.00 means there's backup. How much? Well, 2.00 means that there are two lanes worth of cars total -- one lane's worth on the bridge, and one lane's worth waiting. 3.00 means there are three lane's worth total -- one lane's worth on the bridge, and two lanes' worth waiting. Etc.

= load of 1.00

= load of 0.50

= load of 1.70

This is basically what CPU load is. "Cars" are processes using a slice of CPU time ("crossing the bridge") or queued up to use the CPU. Unix refers to this as the run-queue length : the sum of the number of processes that are currently running plus the number that are waiting (queued) to run.

Like the bridge operator, you'd like your cars/processes to never be waiting. So, your CPU load should ideally stay below 1.00. Also like the bridge operator, you are still ok if you get some temporary spikes above 1.00 ... but when you're consistently above 1.00, you need to worry.

## So you're saying the ideal load is 1.00?

Well, not exactly. The problem with a load of 1.00 is that you have no headroom. In practice, many sysadmins will draw a line at 0.70:

• The "Need to Look into it" Rule of Thumb: 0.70 If your load average is staying above > 0.70, it's time to investigate before things get worse.

• The "Fix this now" Rule of Thumb: 1.00 . If your load average stays above 1.00, find the problem and fix it now. Otherwise, you're going to get woken up in the middle of the night, and it's not going to be fun.

• The "Arrgh, it's 3AM WTF?" Rule of Thumb: 5.0 . If your load average is above 5.00, you could be in serious trouble, your box is either hanging or slowing way down, and this will (inexplicably) happen in the worst possible time like in the middle of the night or when you're presenting at a conference. Don't let it get there.

## What about Multi-processors? My load says 3.00, but things are running fine!

Got a quad-processor system? It's still healthy with a load of 3.00.

On multi-processor system, the load is relative to the number of processor cores available. The "100% utilization" mark is 1.00 on a single-core system, 2.00, on a dual-core, 4.00 on a quad-core, etc.

If we go back to the bridge analogy, the "1.00" really means "one lane's worth of traffic". On a one-lane bridge, that means it's filled up. On a two-late bridge, a load of 1.00 means its at 50% capacity -- only one lane is full, so there's another whole lane that can be filled.

= load of 2.00 on two-lane road

Same with CPUs: a load of 1.00 is 100% CPU utilization on single-core box. On a dual-core box, a load of 2.00 is 100% CPU utilization.

## Multicore vs. multiprocessor

While we're on the topic, let's talk about multicore vs. multiprocessor. For performance purposes, is a machine with a single dual-core processor basically equivalent to a machine with two processors with one core each? Yes. Roughly. There are lots of subtleties here concerning amount of cache, frequency of process hand-offs between processors, etc. Despite those finer points, for the purposes of sizing up the CPU load value, the total number of cores is what matters, regardless of how many physical processors those cores are spread across.

Which leads us to a two new Rules of Thumb:

• The "number of cores = max load" Rule of Thumb: on a multicore system, your load should not exceed the number of cores available.

• The "cores is cores" Rule of Thumb: How the cores are spread out over CPUs doesn't matter. Two quad-cores == four dual-cores == eight single-cores. It's all eight cores for these purposes.

## Bringing It Home

Let's take a look at the load averages output from uptime :

~ \$ uptime
23:05 up 14 days, 6:08, 7 users, load averages: 0.65 0.42 0.36

This is on a dual-core CPU, so we've got lots of headroom. I won't even think about it until load gets and stays above 1.7 or so.

Now, what about those three numbers? 0.65 is the average over the last minute, 0.42 is the average over the last five minutes, and 0.36 is the average over the last 15 minutes. Which brings us to the question:

Which average should I be observing? One, five, or 15 minute?

For the numbers we've talked about (1.00 = fix it now, etc), you should be looking at the five or 15-minute averages. Frankly, if your box spikes above 1.0 on the one-minute average, you're still fine. It's when the 15-minute average goes north of 1.0 and stays there that you need to snap to. (obviously, as we've learned, adjust these numbers to the number of processor cores your system has).

So # of cores is important to interpreting load averages ... how do I know how many cores my system has?

cat /proc/cpuinfo to get info on each processor in your system. Note: not available on OSX, Google for alternatives . To get just a count, run it through grep and word count: grep 'model name' /proc/cpuinfo | wc -l

## Monitoring Linux CPU Load with Scout

Scout provides 2 ways to modify the CPU load. Our original server load plugin and Jesse Newland's Load-Per-Processor plugin both report the CPU load and alert you when the load peaks and/or is trending in the wrong direction:

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## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

$v = \sqrt{\frac{m_2~g~r}{m_1}}$
The weight of $m_2$ should be equal to the centripetal force required to keep $m_1$ moving in a circle. We can find the required speed of $m_1$ as; $F_c = \frac{m_1~v^2}{r} = m_2~g$ $v^2 = \frac{m_2~g~r}{m_1}$ $v = \sqrt{\frac{m_2~g~r}{m_1}}$


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# Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

+1 vote
4.3k views

edited
Write one equation each for decomposition reactions where energy is supplied in the form of heat, light or electricity.

by (13.6k points)
selected

Decomposition reactions are:


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# Find the equation of circle touching given lines and a given point. [closed]

$U: 3x+4y-20=0$ and $v:3x+4y+10=0$ are two straight lines. Find the equation of circles touching the given lines and passing through point $P(1,2)$.

## closed as off-topic by user147263, user223391, colormegone, Yiorgos S. Smyrlis, CrostulJun 5 '15 at 10:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Community, colormegone, Yiorgos S. Smyrlis, Crostul
If this question can be reworded to fit the rules in the help center, please edit the question.

• Have you drawn a diagram? – Mark Bennet Jun 4 '15 at 17:47

Since the lines are parallel, the center of circle lays between these two, on the line $3x+4y+\frac{-20+10}{2}=0$. It's radius is the distance form $3x+4y-5=0$ to one of that lines: $$\frac{|3x+4y-20|}{\sqrt{3^2+4^2}}=\frac{|5-20|}{5}=3$$ So we find the center $(a,b)$:\begin{cases}3a+4b-5=0\\(a-1)^2+(b-2)^2=3^3\end{cases} Can you take it from here?

• I think the circle(s) pass through $P$, they are not centred at $P$... – danimal Jun 4 '15 at 13:26
• Hm, of course you're right. We find $(x,y)$ the center of the circle. It lays on $3x+4y-5=0$ and distance from it to $P$ $\sqrt{(x-1)^2+(y-2)^2}$ is $3$. – Alexey Burdin Jun 4 '15 at 13:30
• i was confused for a moment too; it would help if you use $(a, b)$ for the center instead of the $(x,y)$ you are using. nice work(+1). – abel Jun 4 '15 at 13:34

these are two parallel lines. first we will find a circle touching the two lines without the circle going through the point $(1,2).$ we may be able translate the circle to make it go through the point $(1,2).$

the points on contact is a diameter of the circle and goes through the origin. contact points are of the form $(3t, 4t).$ subbing in $3x+4y = 20 \to t = 4/5$ and $3x+4y = -10 \to t = -2/5.$ therefore the center is $(3/5, 4/5)$ and the radius is $3.$

$$(x-3/5)^2 + (y-4/5)^2 = 9$$ touches the line $3x+4y = 20$ at $(12/5, 16/5)$ and the line $3x+4y=-10$ at $(-6/5, -8/5).$

now we will move the center parallel to the lines $3x+4y = 0$ by making the center $(3/5+4k/5, 4/5 -3k/5).$ the equation of the circle is $$(x - (3+4k)/5)^2 +(y-(4-3k)/5)^2 = 9.\tag 1$$ we will fix $k$ by making $(1)$ go through $(1,2).$ that is $$(2-4k)^2+(1+3k)^2=225 \to k = \frac{1 \pm \sqrt{221} }5.$$ so there are two circles $(1)$ corresponding to $k = \frac{1 \pm \sqrt{221} }5.$


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# Top self-intersection of the tautological line bundle

Let $\mathcal E$ be a rank $n$ vector bundle over a curve $Y$ and let $X=\mathbb P(\mathcal E)$ and let $\pi: X \to Y$ be the projection. I would like to compute the value of the top self-intersection of the tautological line bundle $(\mathcal O_X(1))^n$.

-

Let $\xi \in A^1(X)$ be the class of $\mathcal{O}_X(1)$.

Since $\dim Y=1$, by [Hartshorne, Algebraic Geometry, p. 429] we have the equality $$\xi^n=\pi^*c_1(\mathcal{E}) \cdot \xi^{n-1}.$$

But $\mathcal{O}_X(1)$ is a relative hyperplane, so $\pi^*(\textrm{point}) \cdot \xi^{n-1}=1$. Then $$\xi^n= \deg (\mathcal{E}).$$

-

As Francesco showed you this is quite straightforward, therefore I think is a good opportunity to give a more general answer here.

If you have a rank $r$ holomorphic vector bundle $E$ over a compact complex manifold $X$ of dimension $n$, and you call $\pi\colon\mathbb P(E)\to X$ the projectivized bundle of hyperplanes of $E$ and $\mathcal O_E(1)\to\mathbb P(E)$ the tautological line bundle of one dimensional quotients of $E$, then $$\pi_*c_1(\mathcal O_E(1))^{r-1+k}=(-1)^ks_k(E),\quad k=1,\dots,n.$$ Here, $s_k(E)$ is the $k$-th Segre class which is defined for instance by the following relation: $$c_\bullet(E)\cdot s_\bullet(E)=1,$$ where $$c_\bullet(E)=1+c_1(E)+\cdots+c_n(E)$$ is the total Chern class of $E$ and $$s_\bullet(E)=1+s_1(E)+\cdots+s_n(E)$$ is the total Segre class.

Thus, the first terms for $s_\bullet$ are given in terms of the Chern classes by $$s_1(E)=-c_1(E),\quad s_2(E)=c_1(E)^2-c_2(E),\quad\dots$$ The top Segre class $s_n(E)$ has an expansion which always contains the term $(-1)^nc_1(E)^n+$ terms involving higher Chern classes.

In particular, all such terms vanish if $X$ has dimension $1$ (or the bundle has rank one).

The answer to your question is now a particular case of all this and you get that your top self-intersection is just $c_1(\mathcal E)$, that is the degree of $\mathcal E$.

-
Isn't self-intersection on a curve just degree? For instance, because it should be a continuous map to $\mathbb Z$ but the degree-zero line bundles form a connected component? Intersection number is the generalization of the "degree" concept to higher-dimensional varieties? –  Will Sawin Oct 18 '11 at 4:46
In the question of fds (as well as in Francesco's answer and mine) the intersections considered ar not on a curve! They are on the projectivized bundle of a holomorphic vector bundle over a curve! –  diverietti Oct 18 '11 at 7:22


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MathForces: Math Olympiads

# Special square

Author: admin
Problem has been solved: 54 times

Русский язык | English Language

Find the number of natural numbers $n <20202020$ such that for any positive integer $m$ the sum $n ^ m + 1$ is representable as the sum of the squares of two integers.

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# Can Milky Way be seen in South hemisphere only?

I heard something(long time ago) that Milky Way can be seen in South hemisphere only, is it true? If so, what is the reason?

• You might be thinking of the Magellanic Clouds, island galaxies closer to the Milky Way than the Andromeda Galaxy. – user21 Jul 25 '17 at 2:36

That is not true. The band of the Milky Way can be seen from any latitude on the planet. It just requires the right time of year and a dark sky.

The band of milky way is inclined by more than $60\deg$ to the celestial equator. Hence, it can be easily seen from the southern hemisphere.

In fact, the milky way center lies in Sagittarius (a southern hemisphere constellation), which means,

The Milky Way would look better in the Southern Hemisphere.


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# hello¶

Syntax: hello(name)

Shows a greetings message. This command is used for testing things.

This command can be used in the distributed processing mode, but it does not participate in distributed processing.

## Arguments¶

### name [input]¶

Data type: string

Default value: there

Name of caller. Omit to output a generic greeting.


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Question

# A mercury thermometer can be used to accurately infer whether mercury expands linearly with temperature

A
True
B
False

Solution

## The correct option is B FalseNo! We cannot use the thermometer to verify the linear expansion of mercury. This is because the temperature scale itself is defined by making equal markings between 00 C and 1000 C, assuming that mercury expands linearly. In fact, scientists have now established using statistical physics and other concepts that mercury does not expand exactly in a linear fashion. Before all this, the only way to guess if a thermometer fluid is a good choice (i.e, expands linearly) was to observe which of them gave consistent results.  Physics

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# Concepts¶

This block is about how we pull off the trick to turn geography into numbers statistics can understand. At this point, we dive right into the more methodological part of the course; so you can expect a bit of a ramp up in the conceptual sections. Take a deep breath and jump in, it’s well worth the effort! At the same time, the coding side of each block will start looking more and more familiar because we are starting to repeat concepts and we will introduce less new building blocks and instead rely more and more on what we have seen, just adding small bits here and there.

## Space, formally¶

How do you express geographical relations between objects (e.g. areas, points) in a way that can be used in statistical analysis? This is exactly the core of what we get into in here. There are several ways, of course, but one of the most widespread approaches is what is termed spatial weights matrices. We motivate their role and define them in the following clip.

Once you have watched the clip above, here’s a quiz for you!

Imagine a geography of squared regions (ie. a grid) with the following structure:

Each region is assigned an ID; so the most north-west region is 1, while the most south-east is 9. Here’s a question:

What is the dimension of the Spatial Weights Matrix for the region above?

## Types of Weights¶

Once we know what spatial weights are generally, in this clip we dive into some of the specific types we can build for our analyses.

Here is a second question for you once you have watched the clip above:

What does the rook contiguity spatial weights matrix look like for the region abvoe? Can you write it down by hand?

## The Spatial Lag¶

We wrap up the the set of concepts in this block with one of the applications that makes spatial weights matrices so important: the spatial lag. Watch the clip to find out what it is and then jump over the next part to see how all of these ideas translate into delicious, juicy Python code!

## More materials¶

If you want a similar but slightly different take on spatial weights by Luc Anselin, one of the biggest minds in the field of spatial analysis, I strongly recommend you watch the following two clips, part of the course offered by the Spatial Data Center at the University of Chicago:

• Lecture on “Spatial Weights”

• Lecture on “Spatial Lag”, you can ignore the last five minutes as they are a bit more advanced

• Spatial weights chapter on the GDS book (in progress) :cite:reyABwolf.
• For a more advanced and detailed treatment, the chapters on spatial weights in the Anselin & Rey book :cite:anselin2014modern are the best source.


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## Class/Course - Class XI

### Subject - Physics

#### Total Number of Question/s - 6437

Just Exam provide question bank for Class XI standard. Currently number of question's are 6437. We provide this data in all format (word, excel, pdf, sql, latex form with images) to institutes for conducting online test/ examinations. Here we are providing some demo contents. Interested person may contact us at info@justexam.in

• 1. Physical World - Quiz

1. The range of masses we study in Physics is
a) 10-30kg to 1055 kg
b) 10-27 kg to 1055 kg
c) 10-27 kg to 1055 kg
d) 10-27 g to 1060 g

2. Einstein was awarded Nobel Prize in Physics for
a) Theory of relativity
b) Law f gravitation
c) Uncertainty Principle
d) Photo electricity

• 2. Units and Measurements - Quiz

1. A student uses a simple pendulum of exactly 1 m length to determine g, the acceleration due to gravity. He uses a stop watch with the least count of 1 sec for this and records 40 seconds for 20 oscillations. For this observation, which of the following statement (s) (are) true?
a) Error ΔT in measuring T, the time period , is 0.05 seconds
b) Error ΔT in measuring T, the time period, is 1 second
c) Percentage error in the determination of g is 5%
d) Perecentage error in the determination of g is 2.5%

2. If Planck constant (h) and speed of light in vacuum (c ) are taken as two fundamental quantities, which one of the following can, in addition, be taken to express length, mass and time in terms of the three chosen fundamental quantities?
a) Mass of electron (me)
b) Universal gravitational constant ( G)
c) Charge of electron (e )
d) Mass of proton (mp)

• 3. Motion in a Straight Line - Quiz

1. A body is projected upwards with a velocity u. It passes through a certain point above the ground after t1. The time after which the body passes through the same point during the return journey is
a) $\left (\frac{u}{g} - t_{1}^{2} \right )$
b) $2\left (\frac{u}{g} - t_{1} \right )$
c) $3\left (\frac{u^{2}}{g} - t_{1} \right )$
d) $3\left (\frac{u^{2}}{g^{2}} - t_{1} \right )$

2. Two forces 5 kg wt and 10 kg wt are acting with an inclination of 1200 between them. The angle which the resultant makes with 10 kg wt is 300
a) True
b) False

• 4. Motion in a Plane - Quiz

1. The horizontal range is same whether the angle of projection is α or β, where α+β = ______
a) 00
b) 900
c) 1800
d) none of these

2. A ball is projected from the ground at angle θ with the horizontal . After 1s it moving at angle 450 with the horizontal and after 2s it is moving horizontally. What is the velocity of projection of the ball?
a) 10$\sqrt{3}$ ms-1
b) 20$\sqrt{3}$ ms-1
c) 10$\sqrt{5}$ ms-1
d) 20$\sqrt{2}$ ms-1

• 5. Laws of Motion - Quiz

1. An insect crawls up a hemispherical surface very slowly, (figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre centre of the hemispherical surface to the insect makes an angle α with vertical, the max.possible value of α is given by

a) cot α = 3
b) sec α = 3
c) cosec α = 3
d) None

2. A trolley T of mass 5kg on a horizontal smooth surface is pulled by a load of 2kg through a uniform rope ABC of length 2m and mass 1kg. As the load falls from BC = 0 to BC = 2m, its acceleration (in ms-2) changes from

a) $\frac{20}{6} to \frac{30}{6}$
b) $\frac{20}{8} to \frac{30}{8}$
c) $\frac{20}{5} to \frac{30}{6}$
d) None of these

• 6. Work, Energy and Power - Quiz

1. A block of mass 10 kg is moving in x-direction with a constant speed of 10 m/s. It is subjected to a retarding force F = 0.1 x joule/metre during its travel from x = 20 m to x =30 m. Its final K.E. will be
a) 475 J
b) 450 J
c) 275 J
d) 250 J

2. The kinetic energy K of a particle moving along a circle of radius R depends upon the distance s as K = as2. The force acting on the particle is
a) $2a\frac{s^{2}}{R}$
b) $2as\left [1 + \frac{s^{2}}{R} \right ]^{1/2}$
c) 2as
d) 2a

• 7. System of Particles and Rotational Motion - Quiz

1. A grindstone of moment of interia 6 kg m2 is found to have a speed of 150 rpm, 10 sec. after starting from rest. Torque applied is
a) 6 л Ns
b) 3 π Ns
c) 9 π Ns
d) 12 π Ns

2. A boy is pushing a ring of mass 2 kg and radius 0.5 m with a stick as shown in the (figure)

The stick applies a force of 2 N on the ring and rolls it without slipping with an acceleration of 0.3 m/s2.
The coefficient of friction between the ground and the ring is large enough that rolling always occurs and coefficient of friction between the stick and the ring is (P/10). The value of P is
a) 3
b) 4
c) 5
d) 6

• 8. Gravitation - Quiz

1. The density of a newly discovered planet is twice that of earth . The acceleration due to the gravity the surface of the planet is equal to that at the surface of the earth. If the radius of the earth is R. The radius of the planet would be
a) 2R
b) 4R
c) R/4
d) R/2

2. Figure shows a planet in an elliptical orbit around the sun S. Whare is the kinetic energy of the planet maximum?

a) P1
b) P2
c) P3
d) P4

• 9. Mechanical Properties of Solids - Quiz

1. A brass bar of length 0.2 m and cross-sectional are 1 cm2 is compressed with a load of 5 kg increase in elastic of brass is 2.4 x 10-3 J.
a) True
b) False

2. Two wires of the same length and same material but radii in the ratio of 1:2 are stretched by unequal forces to produce equal elongation. The ratio of the two forces is
a) 1:1
b) 1:2
c) 1:3
d) 1:4

• 10. Mechanical Properties of Fluids - Quiz

1. A ball whose density is 0.4 x 103 falls into water from a height of 9 cm. To what depth does the ball sink?
a) 4.5 cm
b) 6.0 cm
c) 7.5 cm
d) 800

2. The velocity of liquid flow, upto which its flow is streamlined and above which its flow becomes turbulent is called __________.
a) Critical velocity
b) velocity
c) Both a and b
d) None of these

• 11. Thermal Properties of Matter - Quiz

1. Heat is required to change 1kg of ice at = 200C into steam Q1 is the heat needed to warm the ice from -200C to 00C, Q2 is the heat needed to melt the iice, Q3 is the heat needed to warm the water from 00C to 1000C and Q1 is the heat needed to vapourize the water. Then
a) $Q_{4} > Q_{3} > Q_{2} > Q_{1}$
b) $Q_{4} > Q_{3} > Q_{1} > Q_{2}$
c) $Q_{4} > Q_{2} > Q_{3} > Q_{1}$
d) $Q_{4} > Q_{2} > Q_{1} > Q_{3}$

2. A uniform metal rod is used as a bar pendulum. If the room temperature rises by 100C, and the coefficient of linear expansion of the metal of the rod is 2 x 10-6per0C. the period of the pendulum will have percentage increase of
a) -2 x 10-3
b) -1 x 10-3
c) 2 x 10-3
d) 1 x 10-3

• 12. Thermodynamics - Quiz

1. Net work done on the gas in cycle ABCDA in fig. is

a) 276 R
b) 1076 R
c) 1904 R
d) zero

2. A thermically insulted vessel contains an ideal gas of molecular mass M and ratio of specific heats γ. It is moving with speed υ and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increased by
a) (γ-1)mυ2K/2γR
b) γMυ2K/2R
c) (γ-1)Mυ2K/2R
d) (gamma;-1)Mupsilon;K/2(gamma;+1)R

• 13. Kinetic Theory - Quiz

1. All curves foe real gases approach the ideal gas behavior at
a) low pressure only
b) high temperature only
c) low pressure and high temperature
d) high pressures and low temperatures

2. Temperature remaining constant, the pressure of gas is decreasedby 20%. The percentage change in volume is
a) increased by 20%
b) decreased by 20%
c) increased by 25%
d) decreased by 25%

• 14. Oscillations - Quiz

1. A particle executes simple harmonic motion between x- -A and x= +A. The time taken for it to go from 0 to A/2 is T1, and to go from A/2 to A is T2 . Then
a) T1 < T2
b) T1 > T2
c) T1 = T2
d) T1 = 2T2

2. A sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges, when this mass of 9kg is suspended from the wire. When this mass is replaced by mass M, the wire resonates with the same tuning fork forming three antinodes for the same position of the bridges. The value of M is
a) 25 kg
b) 5 kg
c) 12.5 kg
d) 1/25 kg

• 15. Waves - Quiz

1. A sonometer wire resonates with a given tuning fork forming 5 antinodes when a mass 9kg is suspended from the wire. When this time is replaced by a mass m, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is
a) 25 kg
b) 5 kg
c) 12.5 kg
d) (1/25)kg

2. A train moving at a speed of 220ms-1 towards a stationary object, emits a sound of frequency 1000Hz. Some of the sound reaching the object gets reflected back to the train as echo. The frequency of the echo as detected by a person in the train is:(speed of sound in air is 330ms-1
a) 3500 Hz
b) 4000Hz
c) 5000Hz
d) 3000Hz

• 16. Scalar and Vector - Quiz

1. If $\vec{b}$ = $3\hat{i}+ 4\hat{j}$ and $\vec{a}$ = $\hat{i}$ - $\hat{j}$, the vector having the same magnitude as that of $\vec{b}$ and parallel to $\vec{a}$ is
a) $\frac{5}{\sqrt{2}}\left (\hat{i} - \hat{j} \right )$
b) $\frac{5}{\sqrt{2}}\left (\hat{i}+ \hat{j} \right )$
c) $5\left (\hat{i} - \hat{j} \right )$
d) $5\left (\hat{i} + \hat{j} \right )$

2. The sum of the magnitudes of two forces acting at a point is 16N. The resultant of these forces is perpendicular to the smaller force and has a magnitude of 8N. If the smaller force is of magnitude x, then the value of x is
a) 2N
b) 4N
c) 6N
d) 7N

• 17. Conservation of momentum and Collision - Quiz

1. Two blocks of masses 6 kg and 4 kg are attached to the ends of a massless string passing over a smooth fixed pulley. If the system is released, the acceleration of the centre of mass of the system will be
a) g, vertically downwards
b) g/5, vertically downwards
c) g/25, vertically downwards
d) zero

2. A ball released from a height h0 above a horizontal surface rebounds to a height h1 after one bounce. The graph that relates h0 to h1 is shown in Figure. If the ball ( of the mass m) was dropped from an initial height h and made three bounces, the kinetic energy of the ball immediately after the third impact with the surface was

a) (0.8)3 mgh
b) (0.8)2 mgh
c) 0.8 mg (h/3)
d) [1-(0.8)3] mgh

• 18. Rotational Dynamics - Quiz

1. A uniform disc of radius R lies in the x-y plane, with its centre at origin. Its moment of inertia about z-axis is equal to its moment of inertia about line y = x+c.The value of c will be
a) $-\frac{R}{2}$
b) $\pm \frac{R}{\sqrt{2}}$
c) $+\frac{R}{4}$
d) -R

2. Two painters are working from a wooden board 5 cm long suspended from the top of a building by two ropes attached to the ends of the plank. Either rope can withstand a maximum tension of 1040 N. Painter A of mass 80 kg is working at a distance of 1 m from one end. Painter B of mass 60 kg is working at a distance of x m from the centre of mass of the board on the other side. Take mass of the board as 20 kg and g = 10 m/s2. The range of x so that both the painters can work safely is
a) $\frac{1}{3}< x < \frac{11}{6}$
b) $0< x < \frac{11}{6}$
c) 0 < x < 2
d) $\frac{1}{3}< x < 2$

• 19. Fluid Dynamics - Quiz

1. Three identical vessels A, B and C contain same quantity of liquid. In each vessel balls of different densities but same masses are placed. In vessel A, the ball is partly submerged; in vessel B, the ball is completely submerged but floating and in vessel C, the ball has sunk to the base. If FA , FB and FC are the total forces acting on the base of vessels A, B and C, respectively, then

a) FA = FB = FC
b) FA < FB < FC
c) FA = FB < FC
d) FA < FB < FC

2. A pressure meter attached to a closed water tap reads 1.5 × 105 Pa. When the tap is opened, the velocity of flow of water is 10 ms-1 and the reading of the pressure meter is
a) 1.5 × 105 Pa
b) 3 × 105 Pa
c) 0.5 × 105 Pa
d) 105 Pa


text:
# Half-Life

In this half-life learning exercise, students are guided through the steps of solving radioactive decay problems. They solve nine problems finding rates of decay, half-life and quantities of substances remaining after given amounts of time.

Concepts

Resource Details


text:
# Incredible Tangential Sum

Geometry Level 4

Amazingly, the value of $$\tan\bigg(\displaystyle\sum_{n=1}^{10} \tan^{-1}(n)\bigg)$$ can be written in the form $$\frac{a}{b}$$ where $$a$$ and $$b$$ are coprime integers. What is $$a+b$$?

×


text:
# Chapter 1: Real Variables: examples II

Examples II.

1) Show that no rational number can have its cube equal to 2.

Proof 1.

Proof by contradiction. Let $x=p/q$. $q \neq 0, p, q \in Z.$ (p, q have no common factors).

Let $x^{3}=2$. Hence, $\frac{p^{3}}{q^{3}}=2$. Hence, $p^{3}=2q^{3}$. Hence, p^{3} is even because we know that even times even is even and even times odd i also even and odd times odd is odd. Hence, p ought to be even. Let $p=2m$. Then, again $q^{3}=4m^{3}$. Hence, $q^{3}$ is even. Hence, q is even. But, this means that p and q have a common factor 2 which contradicits our hypothesis. Hence, the proof.. QED.

Proof 2)

Let given rational fraction be $\frac{p}{q}$, $q \neq 0, p, q \in Z$.

Let $\frac{p}{q}=\frac{m^{3}}{n^{3}}$, $n \neq 0, m,n \in Z$.

Since p and q do not have any common factors, m and n also do not have any common factors.

Case I: p is even, q is odd so clearly, they do not have any common factors.

Case IIL p is odd, q is odd but with no common factors.

Case I: since m and n are without any common factors, and $m^{3}, n^{3}$ are also in its lowest terms, we have $p=m^{3}, q=n^{3}$.

Case II: similar to case I above.

Proof 3.

A more general proposition, due to Gauss, includes those two above problems as special cases. Consider the following algebraic equation;

$x^{n}+p_{1}x^{n-1}+p_{2}x^{n-2}+\ldots +p_{n}=0$.

with integral coefficients,, cannot have a rational root but non integral root.

Proof 3:

For suppose that the equation has a root a/b, where a and b are integers without a common factor, and b is positive. Writing a/b for x, and multiply both the sides of the equation $b^{n-1}$, e obtain

$-\frac{a^{n}}{b}=p_{1}a^{n-1}+p_{2}a^{n-2}b+\ldots +p_{n}b^{n-1}$,

a fraction in the lowest terms equal to an integer, which is absurd. thus, b=1, and the root is a. It is clear that a must be a divisor of $p_{n}$

Proof 4.

Show that if $p_{n}=1$ and neither of

$1+p_{1}+p_{2}+p_{3}+\ldots$,, $1-p_{1}+p_{2}-p_{3}+\ldots$ is zero, then the equation cannot have a rational root.

I will put the proof later.

Problem 5.

Find the rational toots, if any of $x^{4}-4x^{3}-8x^{2}+13x+10=0$.’

Solution.

The roots can only be integral and so $\pm 1, \pm 2, \pm 3, \pm 5 pm 10$ are the only possibilities: whether these are roots can be determined by tiral. it is clear that can in this way determine the rational roots of any equation.

More later,

Nalin Pithwa

# Analysis: Chapter 1: part 10: algebraic operations with real numbers

Algebraic operations with real numbers.

We now proceed to meaning of the elementary algebraic operations such as addition, as applied to real numbers in general.

(i),  Addition. In order to define the sum of two numbers $\alpha$ and $\beta$, we consider the following two classes: (i) the class (c) formed by all sums $c=a+b$, (ii) the class (C) formed by all sums $C=A+B$. Clearly, $c < C$ in all cases.

Again, there cannot be more than one rational number which does not belong either to (c) or to (C). For suppose there were two, say r and s, and let s be the greater. Then, both r and s must be greater than every c and less than every C; and so $C-c$ cannot be less than $s-r$. But,

$C-c=(A-a)+(B-b)$;

and we can choose a, b, A, B so that both $A-a$ and $B-b$ are as small as we like; and this plainly contradicts our hypothesis.

If every rational number belongs to (c) or to (C), the classes (c), (C) form a section of the rational numbers, that is to say, a number $\gamma$. If there is one which does not, we add it to (C). We have now a section or real number $\gamma$, which must clearly be rational, since it corresponds to the least member of (C). In any case we call $\gamma$ the sum of $\alpha$ and $\beta$and write

$\gamma=\alpha + \beta$.

If both $\alpha$ and $\beta$ are rational, they are the least members of the upper classes (A) and (B). In this case it is clear that $\alpha + \beta$ is the least member of (C), so that our definition agrees with our previous ideas of addition.

(ii) Subtraction.

We define $\alpha - \beta$ by the equation $\alpha-\beta=\alpha +(-\beta)$.

The idea of subtraction accordingly presents no fresh difficulties.

More later,

Nalin Pithwa


text:
# How to prove $d\omega=(\nabla_\mu\omega)_\nu dx^\mu\wedge dx^\nu$ without using coordinates

This is exercise 7.8 b) of Nakahara's GTaP: Let $\omega\in\Omega^1(M)$ be a 1-form on a Riemannian manifold with Levi-Civita connection $\nabla$. Prove that

$$\mathrm{d}\omega=(\nabla_\mu\omega)_\nu\, \mathrm dx^\mu\wedge\mathrm dx^\nu$$

I proved it using the fact that $\mathrm dx^\mu\wedge\mathrm dx^\nu=-\mathrm dx^\nu\wedge\mathrm dx^\mu$, so:

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\,\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu+({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \sum_{\mu<\nu}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

I hope this is correct and makes sense.

I don't like my solution because starting at the second line, it "quits" Einstein summation convention and needs an explicit summation symbol.

1. Is there a way to prove this without "quitting" Einstein summation convention?
2. Is there maybe even a way to prove it in a coordinate-free way?

## 2 Answers

Here's an outline of a coordinate-free proof.

(1) For any $1$-form $\omega$ and any vector fields $X,Y$, there is the formula $$d\omega(X,Y) = X(\omega(Y)) - Y(\omega(X)) - \omega([X,Y]).$$ (2) For any affine connection $\nabla$, there is the formula $$X (\omega(Y)) = (\nabla_X\omega)(Y) + \omega(\nabla_XY).$$ By switching the roles of $X, Y$ and subtracting, this gives $$X \omega(Y) - Y \omega(X) = (\nabla_X \omega)(Y) - (\nabla_Y \omega)(X) + \omega(\nabla_XY - \nabla_YX).$$ But $\nabla$ is torsion-free, so.... (left to you)

(3) By plugging in $X = \frac{\partial}{\partial x^\mu}$ and $Y = \frac{\partial}{\partial x^\nu}$, we get..... (left to you)

Remarks: Note that the proof works for any torsion-free affine connection $\nabla$, not just the Levi-Civita connection. This formula is an instance of "Cartan's First Structure Equation." Generalizations exist to $k$-forms for any $k \geq 1$.

There are some wrong expressions in your solution. A possible correct solution is as follows.

\begin{align} (\nabla_\mu\omega)_\nu\,\mathrm dx^\mu\wedge\mathrm dx^\nu & = (\partial_\mu\omega_\nu-{\Gamma^\lambda}_{\mu\nu}\omega_\lambda)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu-({\Gamma^\lambda}_{\mu\nu}-{\Gamma^\lambda}_{\nu\mu})\omega_\lambda\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \frac{1}{2}\left(\partial_\mu\omega_\nu-\partial_\nu\omega_\mu\right)\,\mathrm dx^\mu\wedge\mathrm dx^\nu\\ & = \mathrm d\omega \end{align}

Do you agree?

• Corrected my question, do you still see any errors? Your solution looks good, thanks.
– Bass
Dec 14, 2015 at 21:09


text:
Paul Shaman delves into the history of the Putnam Mathematics Competition, and some of its notable laureates. He writes:

The William Lowell Putnam Mathematics Competition is held annually. Undergraduates at schools in the US and Canada are eligible, and many of those who do well go on to have distinguished careers in the mathematical sciences and other fields. In addition to the competition between individuals, there is a team contest among schools, with three people on a team. Since 1962, the examination has involved a three-hour morning session and a three-hour afternoon session, each with six questions. The top five individual performers are designated Putnam Fellows and receive monetary awards. The top teams are also awarded monetary prizes. Originally three teams received the prizes, and later the awards were expanded to four teams, and then to the current five. Several of the Putnam Fellows went on to become Fields Medalists, and two received the Nobel Prize in Physics.

Over the history of the competition, many entrants who achieved strong performances and were on winning teams went to become prominent statisticians and probabilists. For statisticians and probabilists, two Putnam outcomes are notable. In the February 1958 competition, three of the five Putnam Fellows subsequently became leading researchers in statistics and probability. They were David Brillinger, University of Toronto; Richard Dudley, Harvard; and Lawrence Shepp, Polytechnic Institute of Brooklyn. In addition, Edward L. Kaplan was a Putnam Fellow for three consecutive years, 1939–1941, while a student at Carnegie Institute of Technology.

Currently, Putnam Fellows are awarded \$2,500; lesser amounts to other top finishers. The department of the top team is awarded \$25,000, and each team member receives \$1,000. In the February 1958 competition, these awards were \$50, \$400, and \$40, respectively.

The Putnam Competition is underwritten by a trust established by Elizabeth Lowell Putnam in her will, written in 1927, to honor her late husband, William Lowell Putnam, a member of the Harvard class of 1882 and a prominent lawyer and banker. Mrs. Putnam’s intent was to sponsor intellectual competition among schools as had been advocated by her husband. After Mrs. Putnam died in 1935, her sons consulted with George David Birkhoff and Garrett Birkhoff of the Harvard Mathematics Department, both friends of the Putnam family, as to the use of her bequest. The structure of the Competition was formulated, with the Mathematical Association of America in charge of administration, and the first examination took place in 1938. With several exceptions, the examination has been given annually. There was a three-year hiatus during World War II, and two examinations were held in 1958. Now the Competition is staged on the first Saturday in December. Birkhoff (1965) and Bush (1965) describe the history of events leading to the Competition and the early years.

In the first eight years of the Competition, Toronto fielded the top-ranked team four times and Brooklyn College three times. Over the history of the Competition, Harvard has had the greatest number of first place team finishes, with Caltech second and MIT third. See Gallian (2004, 2014) for a detailed accounting of the Putnam results. Additional information is given in the Wikipedia article.

Table 1, below, lists the Putnam Fellows who became active in statistics and probability, with their undergraduate institutions. Their doctoral degree information is given in Table 2. J. Arthur Greenwood earned the first doctorate awarded by Harvard’s Department of Statistics.
For each Putnam Competition, the results, including the questions and their solutions, are reported subsequently in The American Mathematical Monthly. In addition to the Putnam Fellows, top team performances, and those individuals receiving honorable mention and above are cited. Tables 3 and 4 list such performances by persons who have contributed to research in statistics and probability.

Table 1. Putnam Fellows and their Undergraduate Institutions

 Name Years Institution Edward L. Kaplan 1939, 1940, 1941 Carnegie Institute of Technology Donald A. S. Fraser 1946 University of Toronto J. Arthur Greenwood 1946 Harvard Arthur P. Dempster 1951 University of Toronto David R. Brillinger S1958 University of Toronto Richard M. Dudley S1958 Harvard Lawrence A. Shepp S1958 Polytechnic Institute of Brooklyn Louis Jaeckel 1960 UCLA Theodore C. Chang 1966 MIT Robin Pemantle 1981 University of California Berkeley Martin V. Hildebrand 1985 Williams College John S. Tillinghast 1987 University of California Davis

Table 2. Putnam Fellows and their Doctoral Institutions and Advisers

 Name Year Institution Advisers Edward L. Kaplan 1951 Princeton John W. Tukey Donald A. S. Fraser 1949 Princeton John W. Tukey and Samuel S. Wilks J. Arthur Greenwood 1959 Harvard Frederick Mosteller and John W. Pratt Arthur P. Dempster 1956 Princeton John W. Tukey David R. Brillinger 1961 Princeton John W. Tukey Richard M. Dudley 1962 Princeton Gilbert A. Hunt and J. Edward Nelson Lawrence A. Shepp 1961 Princeton William Feller Louis Jaeckel 1969 UC Berkeley Erich L. Lehmann Theodore C. Chang 1972 UC Berkeley Wu-Yi Hsiang Robin Pemantle 1988 MIT Persi Diaconis Martin V. Hildebrand 1990 Harvard Persi Diaconis

Table 3. Members of Winning Teams

 Name Year Institution Team rank Herman Chernoff 1942 CCNY 4 Donald A. S. Fraser 1946 Toronto 1 Julian Keilson 1946 Brooklyn College 3 Hale F. Trotter 1952 Queen’s University 1 Charles J. Stone 1957 Caltech 4 Lawrence A. Shepp S1958 Brooklyn Poly 1 David R. Brillinger S1958 Toronto 3 Richard M. Dudley F1958 Harvard 1 David R. Brillinger F1958 Toronto 2 Theodore C. Chang 1966 MIT 2 Michael Klass 1968 UCLA 3 Richard Arratia 1971 MIT 5 Robin Pemantle 1980 UC Berkeley 5

Table 4. Honorable Mention and Above

 Name Institution Year and rank range Julian Keilson Brooklyn College 1946, 11–15; 1947, 6–10 Hale F. Trotter Queen’s University 1951, 12–20; 1952, 6–10 Arthur P. Dempster Toronto 1952, 6–10 Gian Carlo Rota Princeton 1953, 11–25 Lawrence A. Shepp Brooklyn Poly 1957, 6–11 Richard M. Dudley Harvard F1958, 6–10 Stanley A. Sawyer Caltech 1959, 11–28 Fred L. Bookstein University of Michigan 1964, 12–35; 1965, 12–35 Theodore C. Chang MIT 1965, 12–35 Richard Arratia MIT 1968, 13–35; 1971, 11–42 James A. Reeds University of Michigan 1968, 13–35 Neal Madras McGill 1977, 11–39; 1978, 11–42 Kenneth S. Alexander University of Washington 1978, 11–42 Robin Pemantle UC Berkeley 1980, 11–43; 1982, 6–11 Robert Shapire Brown 1985, 11–44 John S. Tillinghast UC Davis 1988, 12–49 Tong Zhang Cornell 1992, 6–11; 1993, 29–57

References

Birkhoff, G. (1965). The William Lowell Putnam Mathematical Competition: Early history. Amer. Math. Monthly, Vol. 72, pp. 469–473
Bush, L. E. (1965). The William Lowell Putnam Competition: Later history and summary of Results. Amer. Math. Monthly, Vol. 72, pp. 474–483
Gallian, J. A. (2004). The first sixty-six years of the Putnam Competition. Amer. Math. Monthly, Vol. 111, pp. 691–699
Gallian, J. A. (2014). The Putnam Competition from 1938–2014. www.d.umn.edu/~jgallian/putnam14.pdf
William Lowell Putnam Mathematical Competition. https://en.wikipedia.org/wiki/William_Lowell_Putnam_Mathematical_Competition


text:
# I would appreciate if i could get help with creating Hangman code for C++ in visual...

###### Question:

I would appreciate if i could get help with creating Hangman code for C++ in visual studio.

It must include: vector, operator, Classes,string
option to read random word for guessing in txt file would be plus , also line for hangin for wrong word if possible.

thanks
regards

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12. a) Suppose that Veclr in R" such that U eu = 1 Let P -uut and Q -% 2P. Show that Q? = I...
##### 1. A nurse is caring for a client who was involved in heavy combat and observed...
1. A nurse is caring for a client who was involved in heavy combat and observed war casualties. The nurse should suspect that the client is suffering from posttraumatic stress disorder (PTSD) if the client makes which of the following statements? A "I check any room I enter because the enemy is ...
##### For the matrix below find a nonzero vector in Nul A and a nonzero vector in Col A-1215A =1620 10nonzero vector in Nul A isA nonzero vector in Col A is
For the matrix below find a nonzero vector in Nul A and a nonzero vector in Col A -12 15 A = 16 20 10 nonzero vector in Nul A is A nonzero vector in Col A is...
##### 6.Which statement is not true? a.Disclosures related to receivables are reported in the financial statement notes....
6.Which statement is not true? a.Disclosures related to receivables are reported in the financial statement notes. b.Cash and cash equivalents are the first items reported under Current assets. c.Current assets are normally reported in order of their liquidity. d.All receivables that are expected to...


text:
Search
Question: Could you help me
0
13.4 years ago by
Liao, Guiping10 wrote:
Dear everyone: I am employing the Bioconductor Packages such as Siggenes to analyse Microarray Data. My experiment is the two class paired case, according to arguments of siggenes, I set the cl as follows: > npr4.cl V1 V2 1 1 2 2 -1 2 3 1 5 4 -1 5 5 1 1 6 -1 1 7 1 3 8 -1 3 9 1 4 10 -1 4 I run the find.a0, it gives me a wrong message:unpaired case??? :: a0 <- find.a0(npr4,npr4.cl,B=26282,balanced=FALSE,mat.samp=NULL,delta=0.485, + alpha=(0:9)/10, include.0=TRUE,p0=NA,plot.legend=TRUE,na.rm=FALSE,rand=1234567) EBAM Analysis for the two class unpaired case. Warning: Expected classlabels are 0 and 1. cl will thus be set to 0 and 1. Error in "[<-.data.frame"(*tmp*, which(cl == min(lev)), value = 0) : new columns would leave holes after existing columns So, I dont know what's wrong? Yet that cl will be set to 0 and 1is for the unpaired case. I am sure that you can help me. Tthanks so much. Guiping Liao PBI, NRC, Canada [[alternative HTML version deleted]]
modified 13.4 years ago by Holger Schwender900 • written 13.4 years ago by Liao, Guiping10
0
13.4 years ago by
United States
James W. MacDonald46k wrote:
Liao, Guiping wrote: > Dear everyone: > > I am employing the Bioconductor Packages such as Siggenes to analyse > Microarray Data. My experiment is the two class paired case, according to > arguments of siggenes, I set the cl as follows: > >>npr4.cl > > V1 V2 > 1 1 2 > 2 -1 2 > 3 1 5 > 4 -1 5 > 5 1 1 > 6 -1 1 > 7 1 3 > 8 -1 3 > 9 1 4 > 10 -1 4 From the help page for find.a0: cl: a vector containing the class labels of the samples. In the two class unpaired case, the label of a sample is either 0 (e.g., control group) or 1 (e.g., case group). In the two class paired case, the labels are the integers between 1 and n/2 (e.g., after treatment group) and between -1 and -n/2 (e.g., before treatment group), where n is the length of 'cl' and k is paired with -k. For one group data, the label for each sample should be 1. So find.a0 is expecting a vector rather than the matrix you are using. I believe you want something like: npr4.cl <- npr4.cl[,1]*npr4.cl[,2] You also don't have enough samples to do 26282 permutations, nor (I suspect) do you really want to wait that long. I would suggest something more reasonable like 500 or 1000. Jim > > I run the find.a0, it gives me a wrong message:unpaired case??? :: > > a0 <- > find.a0(npr4,npr4.cl,B=26282,balanced=FALSE,mat.samp=NULL,delta=0.485, > + alpha=(0:9)/10, > include.0=TRUE,p0=NA,plot.legend=TRUE,na.rm=FALSE,rand=1234567) > EBAM Analysis for the two class unpaired case. > > Warning: Expected classlabels are 0 and 1. cl will thus be set to 0 and 1. > > Error in "[<-.data.frame"(*tmp*, which(cl == min(lev)), value = 0) : > new columns would leave holes after existing columns > > So, I dont know what's wrong? Yet that cl will be set to 0 and 1is for the > unpaired case. > > I am sure that you can help me. > > Tthanks so much. > > Guiping Liao > PBI, NRC, Canada > > > [[alternative HTML version deleted]] > > _______________________________________________ > Bioconductor mailing list > Bioconductor@stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor -- James W. MacDonald Affymetrix and cDNA Microarray Core University of Michigan Cancer Center 1500 E. Medical Center Drive 7410 CCGC Ann Arbor MI 48109
0
13.4 years ago by
Holger Schwender900 wrote:
The problem is that the specification of cl currently differs between SAM and EBAM. When performing a SAM analysis you can specify cl in the way that you have specified it. In an EBAM analysis, this way of specification is currently not implemented. I hope that I can implement this soon. This should be available latest in the next Bioconductor release. > Liao, Guiping wrote: > > Dear everyone: > > > > I am employing the Bioconductor Packages such as Siggenes to analyse > > Microarray Data. My experiment is the two class paired case, according > to > > arguments of siggenes, I set the cl as follows: > > > >>npr4.cl > > > > V1 V2 > > 1 1 2 > > 2 -1 2 > > 3 1 5 > > 4 -1 5 > > 5 1 1 > > 6 -1 1 > > 7 1 3 > > 8 -1 3 > > 9 1 4 > > 10 -1 4 > > From the help page for find.a0: > > cl: a vector containing the class labels of the samples. In the > two class unpaired case, the label of a sample is either 0 > (e.g., control group) or 1 (e.g., case group). In the two > class paired case, the labels are the integers between 1 and > n/2 (e.g., after treatment group) and between -1 and -n/2 > (e.g., before treatment group), where n is the length of 'cl' > and k is paired with -k. For one group data, the label for > each sample should be 1. > > So find.a0 is expecting a vector rather than the matrix you are using. I > believe you want something like: > > npr4.cl <- npr4.cl[,1]*npr4.cl[,2] > > You also don't have enough samples to do 26282 permutations, nor (I > suspect) do you really want to wait that long. I would suggest something > more reasonable like 500 or 1000. > > Jim > > > > > > I run the find.a0, it gives me a wrong message:unpaired case??? :: > > > > a0 <- > > find.a0(npr4,npr4.cl,B=26282,balanced=FALSE,mat.samp=NULL,delta=0.485, > > + alpha=(0:9)/10, > > include.0=TRUE,p0=NA,plot.legend=TRUE,na.rm=FALSE,rand=1234567) > > EBAM Analysis for the two class unpaired case. > > > > Warning: Expected classlabels are 0 and 1. cl will thus be set to 0 and > 1. > > > > Error in "[<-.data.frame"(*tmp*, which(cl == min(lev)), value = 0) : > > new columns would leave holes after existing columns > > > > So, I dont know what's wrong? Yet that cl will be set to 0 and 1is for > the > > unpaired case. > > > > I am sure that you can help me. > > > > Tthanks so much. > > > > Guiping Liao > > PBI, NRC, Canada > > > > > > [[alternative HTML version deleted]] > > > > _______________________________________________ > > Bioconductor mailing list > > Bioconductor@stat.math.ethz.ch > > https://stat.ethz.ch/mailman/listinfo/bioconductor > > > -- > James W. MacDonald > Affymetrix and cDNA Microarray Core > University of Michigan Cancer Center > 1500 E. Medical Center Drive > 7410 CCGC > Ann Arbor MI 48109 > > _______________________________________________ > Bioconductor mailing list > Bioconductor@stat.math.ethz.ch > https://stat.ethz.ch/mailman/listinfo/bioconductor > -- DSL Komplett von GMX +++ Supergünstig und stressfrei einsteigen! AKTION "Kein Einrichtungspreis" nutzen: http://www.gmx.net/de/go/dsl


text:
# [texhax] dotfill leaders on first line of paragraph only ?

Reinhard Kotucha reinhard.kotucha at web.de
Sun Dec 11 02:42:57 CET 2011

On 2011-12-10 at 23:48:23 +0000, Philip TAYLOR wrote:

> My sincere thanks to Heiko and to all who gave advice about
> converting my RGB source into CMYK.  I got there in the end
> (two days after the deadline for submitting CRC), but at
> least I know it that is now achievable and have a nice piece of
> code for converting from XeTeX's extended font syntax for
> RGB to XeTeX \specials in CMYK written in pure TeX :-)
>

Ok.  Red could be a little bit darker.  The last step is to choose a
good type of paper and avoid lamination.  Laminated menu cards are
only good for fast food restaurants.  They always look cheap.

Menu cards printed on plain paper look more more elegant.  You need
more copies then, but you also save the money for lamination.

Paper looks more elegant if it's a bit brownish instead of white.
However, I fear that you don't have much freedom here because people
expect that rice is white.

card is machine-washable.  Otherwise lamination is useless and the
money is wasted.

Regards,
Reinhard

--
----------------------------------------------------------------------------
Reinhard Kotucha                                      Phone: +49-511-3373112
Marschnerstr. 25
D-30167 Hannover                              mailto:reinhard.kotucha at web.de
----------------------------------------------------------------------------
Microsoft isn't the answer. Microsoft is the question, and the answer is NO.
----------------------------------------------------------------------------


text:
This data set contains simulated effect sizes of randomized controlled trials evaluating the (relative) effectiveness of cognitive behavioral therapy for depression, using different delivery formats.

## Usage

data("TherapyFormatsGeMTC")

## Format

A list with 3 elements.

data

data.frame. The data frame containing the network meta-analysis data. See the documentation of the TherapyFormats data set for more information.

study.info

data.frame. Contains information on the risk of bias of the included studies. Can be used for network meta-regression.

treat.codes

data.frame. Contains the long-form treatment names associated with each label.

## Source

Cuijpers, P., Noma, H., Karyotaki, E., Cipriani, A., & Furukawa, T. A. (2019). Effectiveness and Acceptability of Cognitive Behavior Therapy Delivery Formats in Adults with Depression: A Network Meta-analysis. JAMA Psychiatry, 76(7), 700-707.

## Details

This simulated data set is based on pooled results of a network meta-analysis examining the effects of different cognitive behavioral therapy (CBT) delivery formats in adults with depression (Cuijpers et al., 2019). Using network meta-analysis techniques on this data set will yield roughly similar, although not identical results to the ones reported in the original study.

For the simulation, pooled effects within designs as reported in Cuijpers et al. along with the estimated between-study heterogeneity variance were assumed as true population parameters. Based on these "true" population parameters, simulations were generated assuming a (random-effects) bayesian hierarchical model, using a non-central chi-squared distribution for the simulation of study-specific sampling errors. The number of effect size draws was chosen to (approximately) mimic the number of comparisons in the original analysis.

This is an alternative version of the data set, adapted for out-of-the-box usage in gemtc.

Mathias Harrer


text:
# I have a rational function f(x)=1/(x^2−4). We know that f(x) is not defined at x=2 and x=−2 and has an infinite discontinuity at these x-values. However, I wanted to know if the function is continuous on the interval (0,2] because we know that it is approaching −infty as x approaches 2 but if we only have the interval (0,2], it is continuously going to negative infinity. So, is this function continuous in this interval or not?

I have a rational function $f\left(x\right)=1/\left({x}^{2}-4\right)$. We know that f(x) is not defined at x=2 and x=−2 and has an infinite discontinuity at these x-values. However, I wanted to know if the function is continuous on the interval (0,2] because we know that it is approaching $-\mathrm{\infty }$ as x approaches 2 but if we only have the interval (0,2], it is continuously going to negative infinity. So, is this function continuous in this interval or not? Thank you so much.
You can still ask an expert for help

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procjenomuj
To be continuous on an interval [a,b], your function must (as a starting point!) be defined for every $x\in \left[a,b\right]$. In your case, your function is not defined at x=2, so it cannot be continuous on (0,2]. It is, however, continuous on the open interval $0.


text:
# Category of Trees as sub-category of Category of Graphs

A tree (like a binary search tree) is a direct graph with some limitations (no cycles, connected). How can I express the category of trees as "sub-category" of a graphs? There is a way? I'm not sure the term "sub-category" is correct.

-
What do you mean by "express"? You can just say that $\mathcal{T}$ is a category with objects being trees and morphisms being homomorphisms of trees (graphs -- it's a full subcategory of the category $\mathcal{G}$ of graphs). – Damian Sobota Feb 12 '12 at 13:27
"define" is better than "express" – tyranitar Feb 12 '12 at 13:41
So I wrote you an answer. What do you need more? – Damian Sobota Feb 12 '12 at 13:55

I think the term you're looking for is "full subcategory". If you have a category $C$, and a set $d$ of objects in $C$, then the full subcategory of $C$ defined by $d$ is the category $D$ whose objects are the elements of $d$, and whose morphisms are all the morphisms of $C$ whose domain and codomain are in $d$. Thus, you can simply say:

"Define the category Tree as the full subcategory of Gph whose objects are trees."

-
That's exactly what I said in a comment. – Damian Sobota Feb 12 '12 at 21:14
@DamianSobota: you could make that a full answer rather than a comment. – Mitch Feb 13 '12 at 2:34

Defining what a tree is is a subtle thing and depends on what you want to use the trees for. Let us assume that you are interested in the notion of an "undirected, rooted, finite non-planar tree". The definition 'a connected undirected graph with no cycles, and with a chosen leaf as root' captures that notion and gives rise to a category of trees as the full subcategory of the category of undirected graphs spanned by these trees. However, the definition 'a contractible 1-dimensional compact connected and simply connected space with a chosen boundary point' also captures the the same notion and thus gives rise to another category of trees, the full subcategory of the category of topological spaces spanned by those trees.

Another possibile definition is 'a finite poset with a smallest element and such that for every element the down-set determined by it is linearly ordered'. It too captures the same notion of "tree" as above and yields yet another category of trees, the full subcategory of the category of posets spanned by those trees. There is also the dendroidal category $\Omega$ whose objects are trees and can be defined in at least three different ways.

More possible definitions can be found in http://ncatlab.org/nlab/show/tree.

The point is that all of these categories are radically different. In particular the notion of subtree is highly sensitive to which of these categories you consider.

All this just goes to show that much care is needed with such formulations, even if you wish your trees form a subcategory of graphs, depending on your choice of axiomatization of the tree concept you might get very different categories.

-


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# Voltage drop across a diode [closed]

For the sake of simplicity, the voltage drop across a diode, $V_d$ made of silicon is assumed to be $0.7\mathrm{V}$ after the input voltage exceeds this voltage; and $V_d$ is assumed to stay constant regardless how large the input voltage, $V_i$ is after the input voltage exceeds the threshold. In practical, there will be an increase in $V_d$, albeit a very small one, as compared to the large increase in $V_i$. There is an ideal diode equation: $$I=I_s\left(e^\frac{qV_i}{kT}-1\right)$$ which relates the current flowing through diode $I$ with the input voltage $V_i$. So a little increase in $V_i$ will result in a large increase in $I$, as $I$ is an exponential function of $V_i$.

I didn't express my question clearly in my previous attempt. Here is my question:

We consider a circuit consists of a diode is connected to a DC power supply. If we were to connect a voltmeter parallel to the diode, and plot the reading of voltmeter(voltage drop across the diode) against the value of the DC input, how will the graph look like? Is there any equation to describe the graph(like the ideal diode equation relating $I$ with $V_i$)?

Should the equation look something like this? $$V=IR_1+V_t\ln{\left(1+\frac{I}{I_s}\right)}$$ based on an article in Wikipedia under the title Shockley diode equation.

• You are kinda mixing two models. Keep also in mind that $V_d$ has no physical equivalent, i.e. it is chosen more or less arbitrarily. Starting from the true diode equation you can't possibly derive $V_d$. – Vladimir Cravero Apr 27 '16 at 14:48
• In this case $V_d = V_i$ – G36 Apr 27 '16 at 14:53
• I think the answer you are looking for is that, unlike resistors, capacitors and inductors, diodes are non-linear devices. You can't use one equation to describe how they work across commonly used voltage / current ranges. – st2000 Apr 27 '16 at 14:56
• @VladimirCravero Agree, I didn't realize I had been mixing two models. Is there any equation concerning the voltage drop across the diode? I have been looking up different semiconductor devices textbooks to no avail. – Dave Clifford Apr 27 '16 at 15:07
• My suggestion would be to measure the diode drop for a number of different currents and plot up your results. (say 100 nA to 1 mA) – George Herold Apr 27 '16 at 16:21

Let me explain a mathematical origin of the magical number 700 mV. This may help you to understand what's wrong with your question.

Let $V_T = \frac{kT}{q} \approx 26\,\text{mV}$ at room temperature. For $V_d > V_T$ we can rewrite the diode equation this way: $$I = I_s \left[\exp\left(\frac{V_d}{V_T}\right) - 1\right] \approx I_s \exp\left(\frac{V_d}{V_T}\right) = 1\,\text{A}\cdot\exp\left(\frac{V_d + V_T\log(I_s)}{V_T}\right)$$

A typical value of $I_s$ for a silicon p-n diode is around $1\cdot10^{-12}\,\text{A}$, so $\log(I_s) \approx -28$.

The numerator becomes positive if $V_d > -V_T\log(I_s) \approx 28\cdot26\,\text{mV} = 728\,\text{mV}$. Nothing abrupt happens at this point; just a border between negative and positive values of $x$ in $\exp(x)$, where $x = \frac{V_d + V_T\log(I_s)}{V_T}$.

Define $V_d = 728\,\text{mV}$ as "the threshold voltage", and the magical number is ready.

Diode is a "current driven" devices. And you should never connect a diode directly across the voltage source without current limiting device in series with the diode. Otherwise the diode will burn. Because the ideal diode act just like a "short circuit" when diode is in forward biased or like a voltage source with constant voltage (0.7V).

As for voltage across the diode versus current we have this $$Vd=\frac{kT}{q}*ln(\frac{Id}{Is})$$

And notice that the diode forward voltage have to be increased by only 60mV in order to increase the diode forward current by the factor of 10. And this is why in most application we assume Vd = 0.7V http://www.ittc.ku.edu/~jstiles/312/handouts/The%20Junction%20Diode%20Forward%20Bias%20Equation.pdf http://www.ittc.ku.edu/~jstiles/312/handouts/The%20Junction%20Diode%20Curve.pdf


text:
×

Seebeck Effect

Top
FAQ

What is Seebeck Effect?

View Notes

The German physicist Thomas Johanan Seebeck first observed the Seebeck effect when a magnetic compass was brought near a loop connecting two conductors, he noticed a variation in the compass. The Seebeck effect mainly focuses on thermoelectric effects. The Seebeck effect is also known as the thermoelectric effect.

While studying the thermoelectric effect in detail, one of the important concepts we come across is the Seebeck effect. The Seebeck effect explains the presence of potential difference when a semiconductor or the conductors are subjected to the temperature difference. The Seebeck effect is used to measure the temperature with accuracy and generates the electric current for some applications.

Understanding Seebeck Effect

• Thomas Johann Seebeck in the year 1821 gave a detailed description of the thermoelectric effect or the Seebeck effect. He discovered that in a circuit consisting of two dissimilar metals (like iron and copper) an EMF is developed when the junctions are maintained at different temperatures.

• Two dissimilar metals connected to form a junction are known as a thermocouple. The EMF developed is known as the thermoelectric EMF. And the current through the circuit is called thermoelectric current.

• The Seebeck effect explained the production of an electromotive force and the electric current in a loop of materials consisting of at least two dissimilar conductors maintained at two different temperatures, known as the thermocouples. It can be termed as the Seebeck effect thermocouple.

• The Seebeck effect is a reversible process. If the hot and cold junctions are interchanged then the direction of the current will also change. Therefore, the thermoelectric effect is a reversible process.

• The magnitude and sign of thermo EMF depend on the materials of the two conductors and the temperature of the hot and cold junction.

• Seebeck after discovering thermal properties of different pair of metals arranged them in series is called thermoelectric series.

• The thermoelectric effect is the conversion of temperature differences into electrical potential differences or vice versa using a thermocouple.

• The Seebeck effect is the best example of an electromotive force. Through the Seebeck effect, we can also calculate the measurable electric currents or voltages in the same way as electromotive forces.

• The local current density can be calculated using the formula,

⇒ J = σ(-ΔV + E$_{emf}$)

Where,

ΔV - The potential difference developed

E$_{emf}$ - Electromotive force

σ - The local conductivity

• The electromotive force created will explain the Seebeck effect and the equation of electromotive force in terms of the Seebeck coefficient is given by,

⇒ E$_{emf}$ = -SΔT

Where,

S - The Seebeck coefficient

• The Seebeck coefficient is a measure of the amount of potential induced per difference in temperature. S is one of the transport properties of the material used.

• The Seebeck coefficients generally vary with the temperature and the Seebeck coefficient depends on the composition of the conductor. Generally, the Seebeck coefficient at room temperature ranges from -100V/K to 1000V/K.

Application of Seebeck Effect:

There are many properties of the materials that vary with temperatures. The Seebeck effect will be a useful tool for monitoring the thermoelectric effects. Following are the few applications of the Seebeck effect:

1. The Seebeck effect is used in thermoelectric generators. Thermoelectric generators are used in power plants where it converts waste heat into electricity.

2. The Seebeck effect is used in automobiles as an automotive generator or an automotive thermoelectric generator that will help in increasing fuel efficiency.

3. The Seebeck effect is used in thermocouples. The thermocouples are electrical devices made of two dissimilar metals or conductors placed at two different temperatures. These are used to measure the potential difference developed between the two conductors.

4. The Seebeck effect is also used in thermophiles. The thermophiles are the number of thermocouples connected in series.

Did You Know?

In thermal plants, we know that heat energy is converted into electrical energy and is later distributed according to the need through powerlines. While converting heat energy into electrical energy we will encounter some energy loss. Then, The Seebeck effect is used in such thermoelectric generators. These thermoelectric generators are used in power plants where it converts waste heat into electricity.


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# Intersection of parabola and vertical line at a particular point

I want to create a parabola that intersects a vertical line at a particular point. Can someone help me with this?

my MWE is

\usepackage{tikz}
\usetikzlibrary{intersections,shapes,arrows,calc}
\usepackage{esvect}
\usepackage{relsize}

\begin{tikzpicture}
% New graph - bottow left
\draw[<->,rounded corners, thick]  (0,2.5) node(yline)[left] {$r$} -- (0,0) -- (3,0)  node(yline)[below] { $Y$};
\draw (1.5,0) -- (1.5,2.5);
\node[label=right:$\overline{S}(Y_{A})$] at (1.5,2.5) {};
\node[ circle,fill=black,minimum size=2pt,inner sep=0pt, outer sep=-1pt, label=left:$r^*_{A}$] at (0,1.5) {};
\draw[dashed]  (0,1.5) -- (1.5,1.5) ;
\draw (0.8,2.5) parabola[bend at end] (1.5,1.5) ;
\draw (1.5,1.5) parabola[bend at end] (2.5,1.2) ;
\end{tikzpicture}

Which produces the following agly picture:

I am very rigid about the intersection point. In my case the parabola has to pass through the (1.5,1.5) point

Any help to make the graph look better ?

• \draw (0.5,2.5) parabola[bend at end] (2.5,3.5/3) ; contains the point (1.5,1.5). – TonioElGringo Jul 3 '17 at 15:49

See, if you liked ...

\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{intersections}

\begin{document}
\begin{tikzpicture}
% axes
\draw[thick,->]  (-0.1,0) coordinate (O)
-- ++ (4.1,0) node [below left] {$X$};
\draw[thick,->]  (0,-0.1) -- ++ (0,4.1) node [below left] {$r$};
\draw[ultra thin, gray] (0,0) grid + (4,4); %  only to show that intersection is at (1.5,1.5)
% curve
\begin{scope}[font=\footnotesize]
\draw[red, very thick,name path=A]
plot[domain=0.32:4, samples=64] (\x,0.8333+1/\x);
\draw[name path=B]
(1.5,-0.1) node[below] {1.5}
-- ++ (0,4.1) node[below right] {$\overline{S}(Y_{A})$};
\draw[dashed, name intersections={of =A and B, by={r}}]
(r) -- (r -| O) node[left]  {$r^*_{A}$};
\end{scope}
\end{tikzpicture}
\end{document}

Addendum: In the first version of answer I assume, that the main problem is redesign LaTeX part of image, e.g. defining of function is not problem. It seems, that I was wrong, so now I correcting this my faulty performance :) but on the way of minimal effort. You can find adequate function yourself, and of course change tick labels to what you like to have. Added grid is only indicative and had to be removed in real use of MWE (as I understand your wish).

Note: shoved curve is not parabola, I only mimic what you shown in your question with inverse function i.e. hyperbole.

• The plot was supposed to pass through (1.5, 1.5) I think, here it passes through (1.5, 0.667). – Torbjørn T. Jul 3 '17 at 16:06
• @TorbjørnT. is now better? – Zarko Jul 3 '17 at 17:37


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Q

# Chlorine is used to purify drinking water Excess of chlorine is harmful The excess of chlorine is removed by treating with sulphur dioxide

8.23  Chlorine is used to purify drinking water. Excess of chlorine is harmful. The excess of chlorine is removed by treating with sulphur dioxide. Present a balanced equation for this redox change taking place in water

Views

Base equation-  $SO_{2}+Cl_{2}\rightarrow Cl^{-}+SO_{4}^{2-}$--------------(have to remember)

Now we have to balance the oxidation half and reduction half.

oxidation half$SO_{2}\rightarrow SO_{4}^{2-}$

balancing-  Oxygen is balanced by adding water molecule, Hydrogen is balanced by $H^{+}$ ion and for charge add $e^{-}$(electron)$SO_{2} +2H_{2}O\rightarrow SO_{4}^{2-}+4H^{+}+2e^{-}$

Reduction hallf$Cl_{2}\rightarrow Cl^{-}$

Balancing-  to balance charge add an electron   $Cl_{2}+2e^{-}\rightarrow 2Cl^{-}$

Now add both balanced oxidation half and reduction half,  we get

$Cl_{2}+SO_{2}+2H_{2}O\rightarrow 2Cl^{-}+SO_{4}^{2-}+4H^{+}$

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text:
# 5.8: Human Stability

When asking what makes a structure more or less stable, we find that a high center of gravity or a small support base makes a structure less stable. In these cases a small displacement is need in order to move the center of gravity outside the area of support. Structures with a low center of gravity compared to the size of the support area are more stable. One way to visualize stability is to imagine displacement of the center of gravity caused by placing the object on a slope. For example, a 10° displacement angle might displace the center of gravity of a toddler beyond the support base formed by its feet, while an adult would still be in equilibrium.

Compared to an adult, a smaller displacement will move a toddlers center of gravity outside the base of support. Image adapted from A man and a toddler take a leisurely walk on a boardwalk by Steve Hillibrand via Wikimedia Commons.

The center of gravity of a person’s body is above the pivots in the hips, which is relatively high compared to the size of the support base formed by the feet, so displacements must be quickly controlled. This control is a nervous system function that is developed when we learn to hold our bodies erect as infants. For increased stability while standing, the feet should be spread apart, giving a larger base of support. Stability is also increased by bending the knees, which lowers the center of gravity toward the base of support. A cane, a crutch, or a walker increases the stability of the user by widening the base of support. Due to their disproportionately large heads, young children have their center of gravity between the shoulders, rather than down near the hips, which decreases their stability and increases the likelihood of reaching a tipping point.[2]

Warning label on a bucket indicating the danger of children falling into a bucket and drowning. This danger is caused by the inherent instability of the toddler body. Image Credit: GodsMoon via Wikimedia Commons.

1. " A man and toddler take a leisurely walk on a boardwalk" by Steve Hillebrand, U.S. Fish and Wildlife Service, Wikimedia Commons, is in the Public Domain
2. OpenStax, College Physics. OpenStax CNX. Aug 3, 2018 [1]http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@12.1.