# How to Find an Inverse Function: Conflicting Approaches

While pondering issues that often give students trouble in algebra, I decided to check what we have said in Ask Dr. Math about inverse functions. I discovered four answers (all, as it happens, written by me – I tend to be attracted to certain topics!) to essentially the same question, spread over 13 years. It is one that I have pondered in teaching: having taught the subject from books that use two different approaches to the task, I find that I prefer the one that seems to be used less commonly, but each of them has its benefits. Let’s take a look at one of these four discussions, then quickly look at the others to supplement it. In 2010, Maureen wrote this:

To Invert Functions, First Subvert Routine

The inverse of a function is found by interchanging x's and y's, right? However, on Wikipedia they determine the inverse in a way that I find confusing. Specifically, I am writing what they do on the left and my confusion on the right.

f(x) = 3x + 7

Normally, I would now switch
y = 3x + 7     the x's and y's and then solve
for y -- but Wikipedia doesn't

From my point of view this is
(y - 7)/3 = x          NOT the inverse -- it is the
original function

f(inv) of y = (y - 7)/3  This is the inverse using y as
the variable

Most books do not do it this way; and although I agree with the final answer, I find it somewhat meaningless. Would you agree?

I am including the usual way of finding the inverse.

f(x) = 3x + 7
y = 3x + 7     given the original function
x = 3y + 7     switch x and y
y = (x - 7)/3  solve for y to get inverse function

My reply was that, in fact, I prefer the method used by Wikipedia here, which I imagine is less often used in current textbooks (at least at an introductory level).  There are a couple reasons to prefer it. One is that only this way is appropriate to applications in which the variables, unlike the typical generic x and y, have specific meanings, so that they can’t be interchanged (see below for an example); another is that this method forces us to get out of the rut of assuming that x is always the independent variable, and y is always dependent.

What they're doing is correct, and in fact is what I prefer. The confusion is probably because you are used to always thinking of y as a function of x.

(It troubles me that texts often ask questions like "Does the equation x + y^2 = 1 represent a function?" when they really mean to ask if it represents y as a function of x. In that example, x is a function of y, though y is not a function of x. A function is about the relationship between two variables, not what they are called.)

What Wikipedia has done is not to exchange the NAMES of the variables in the function, as usual, but just to change their ROLES. By solving for x, they are determining how x (the "input" of f) can be found given y (the "output" of f). That is exactly what it means to find an inverse.

If you recognize that x and y are just dummy variables, you see that Wikipedia’s form,  $$f^{-1}(y) = \frac{y \, – \, 7}{3}$$, and Maureen’s form, $$f^{-1}(x) = \frac{x \, – \, 7}{3}$$, are really the same function, just with different names for the variable. By not changing the variable names at the start, Wikipedia ends up with y as the independent variable – not what most students are used to, but perfectly  legal. And in fact, the main idea of an inverse function is precisely that we are changing the roles of the variables: what was the input of the original function becomes the output of the inverse function, and vice versa. What they are called is far less important that what they mean.

But Maureen needed more help:

Maybe my difficulty is from a graphing point of view.

If you graph ...

y = 3x + 1

... and you create a table of values for x and y, and then you graph ...

x = (y - 1)/3

... you get the same graph, since the tables are the same. When graphing an inverse from a table of values, you specifically interchange the x's and y's, because if you do not, you have the same table.

I feel students (or possibly just me) confuse a different way of writing a function with the inverse function. Specifically y = ln x is the same function as e^y = x. One is NOT the inverse of the other. The inverse of y = ln x is x = ln y, or e^x = y.

The two equations she graphed are actually  different functions (y as a function of x, and x as a function of y); but they are equivalent equations (relating the same pairs of x and y), which is why they are represented by the same graph. Maureen has confused functions with equations. In reality, the second function, $$f^{-1}(y) = \frac{y \, – \, 1}{3}$$, is the inverse of the first, $$f(x) = 3x + 1$$, where x and y have kept the same name but changed roles.

Now consider the logarithm. This is written explicitly as a function the name of which is "ln" rather than "f" or "g," but it is the same idea. If I write ...

y = ln(x)

... I am using the ln function to express y as a function of x. This equation also expresses (implicitly) x as a function of y, since ln is one-to-one. When you solve for x to make the latter function explicit, you have

x = e^y

This explicitly states a different function than ...

y = ln(x)

... although the relation between the variables is the same.

If we were to name the new function, calling it exp, so that ...

x = exp(y)
= e^y

... the named function exp clearly is not the same function as ln. But nothing has changed except for giving it a name. The equation expresses x as a function of y, named or not.

So when we interchange the variables and change y = ln(x) to x = ln(y), we are, as you say, inverting the function -- in the sense of what function y is of x. We have changed the relationship between x and y. But in another sense, it is still the same function (as explicitly written), just expressed with different placeholders.

So there are two perspectives on what constitutes an inverse. In one sense, just swapping the variables implicitly inverts the relationship, by swapping implied roles (x becomes y); but the inverse is shown explicitly by solving for the other variable, regardless of what you call it (input becomes output).

My answer to another of the four questions I referred to above includes an example of the situation where swapping of variables is meaningless, and you must use the other approach:

Inverting, Subverted

Both approaches are valid. The first is best when the variable names mean something, so that changing their names would not make sense. For example, you wouldn't say that the inverse of C = f(r) = 2 pi r is f^-1(r) = r/(2 pi), where r has now become the circumference. Rather, you would say that r = f^-1(C) = C/(2 pi).

Your trouble, I believe, is that you are thinking of the variable names as if they had a fixed meaning somewhat like that, but using the method of inverting that requires you to swap names. You ask about "the inverse ofy" rather than of the FUNCTION, making y a name for the function itself. You can't invert a variable, only a function. And if you do think of y as the same thing as f(x), you can't swap variable names and still say that.

In fact, in my circumference example above, I was initially going to call the function C(r), which is done commonly, naming the function for its output; but it would make no sense to call the inverse C^-1. If anything, we would call the inverse r(C), since it gives the value of r that yields a given circumference C. I didn't do that because it would not fit the inverse notation you are using, and would just add confusion. (If it does, ignore this paragraph!)

I prefer the first method of inverting I showed, because it helps to free students from fixating on x as the independent variable, and avoids the change of meaning that is confusing to you and others. Unfortunately, I have to teach the second in algebra classes, because it is what most students seem to see elsewhere.

For the record, the other two questions I referred to are

Graphs of Inverse Functions
Inverting Functions

The final thanks in the latter, a long, rambling discussion, are worth ending with:

The terminology and explanation has solidified my understanding. I believe my confusion was, as you pointed out, that "An equation does not define a function!" This is very important and I have never thought of it that way. It could be because I never read that somewhere or was never taught that or most importantly never had to think of them that way until now.

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# CO-Pipelining-Q1

A pipeline of 2 stages with a delay of x units each is split to n stages. In the new design, each stage has a delay of x/n units. To get the throughput increase of 1700% what would be the n value?

reshown Jun 13

1700/100=17

17+1=18

n=18
answered Jun 13 by (19,690 points)
answered Jun 13 by (160 points)
+1 vote

let initial throughput be 100. required throughput is 1800.

speedup = 1800/100 = 18

In the first pipeline, each instruction will get completed after x units, and in the second pipeline, each instruction will get completed after x/n units

speedup = x / (x/n)

= n

therefore n = 18

answered Jun 14 by (4,100 points)

Comment if you find any issues.

Pipeline 1

One instruction <----- x units
? <-------------------------1 unit

$= \frac{1}{x} units$

Pipeline 2

One instruction ----> x/n units

? -----------------------> 1 units
$= \frac{n}{x} units$

$percentage \ of \ gain\ in \ throughput = \frac{old value- new value}{old value} * 100$

1700 = (n-1)* 100

17 = n-1

18=n

answered Jun 14 by (31,090 points)
edited Jun 15
In pipeline 1, stage delay= x units. If we consider the CPI(cycles per instruction as 1), which is considered until no stalls are given.

Thus each instruction takes x time units.

Throughput in Pipeline 1 = 1/x

Similarly, every instruction in pipeline 2 takes x/n time units.

Thus, throughput in pipeline 2 = 1/(x/n) = n/x

% increase in throughput = ((new-old)/old)*100

1700       =(((n/x)-(1/x))/(1/x)*100

17           =n-1

18           = n
Yes you are correct. corrected the answer.
In this question (http://thegatebook.in/qa/4507) the throughput is (no. of stages / max stage time) so why in the above question. its only 1/x?


# Open subset containing all terms of a sequence but a finite number

Does an open subset containing all points of a concergent sequence but a finite number of them necessarily contain the sequence's limit ?

Ideally, I would like an answer for each of the following :

-general topology

-Haussdorf (T2)

-metric

My intuition is that it doesn't, because even in a metric space, the radius of a ball centered at each point of the sequence and contained in the open subset may be forced to vary over the sequence.

No. Take, in $\mathbb R$ with the usual topology, the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ and the open set $(0,+\infty)$.
The answer is no for all three. Consider the sequence $\frac{1}{n} \in \mathbb{R}$, and consider the open set $(0, 2)$. This open set contains all but finitely many of the sequences points (in fact, it contains all the sequences points), but it does not contain the sets limit, 0.


# Thread: Equations with the exponential function.

1. ## Equations with the exponential function.

Hello

I know this is elementary, but I don't know how solve this:

The question is, find the exact value(s) of x which satisfy the equation.

$e^{2x} = e^x +12$

I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

I know if the question was:

$e^{2x} = 12$

$x = \dfrac{1}{2}\ln12$

Some help would be very much appreciated. Thank you.

2. ## Re: Equations with the exponential function.

Originally Posted by Furyan
Hello

I know this is elementary, but I don't know how solve this:

The question is, find the exact value(s) of x which satisfy the equation.

$e^{2x} = e^x +12$

I am not confident working with logarithms, although I do know the laws for the sum and difference of logarithms. I can do the other questions in this exercise, but not this one.

I know if the question was:

$e^{2x} = 12$

$x = \dfrac{1}{2}\ln12$

Some help would be very much appreciated. Thank you.
$e^{2x} - e^x - 12 = 0$

$(e^x - 4)(e^x + 3) = 0$

finish it

3. ## Re: Equations with the exponential function.

Dear Skeeter

Originally Posted by skeeter
$e^{2x} - e^x - 12 = 0$

$(e^x - 4)(e^x + 3) = 0$

finish it
A quadratic in $e$, that was left field and I wasn't looking for it. Should have seen it though.

$x = \ln4 = 2\ln2$

$e^x = -3$ has no solutions.

Thank you very much indeed.

4. ## Re: Equations with the exponential function.

Hopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.

5. ## Re: Equations with the exponential function.

Hello HallsofIvy

Originally Posted by HallsofIvy
Hopefully, you know that $e^{2x}= (e^x)^2$. If you let $y= e^x$, the equation becomes $y^2= y+ 12$ which is the same as $y^2- y- 12= 0$, a quadratic equation in y. If you did not notice that this could be factored as $(y- 4)(y+ 3)$, you could solve it by completing the square or using the quadratic formula.
Thank you. I do find letting $y= e^x$ and solving the quadratic in $y$ easier to get my head round. I was able to factor that one.

6. ## Re: Equations with the exponential function.

Hello, Furyan!

Find the exact value(s) of $x$ which satisfy the equation."

. . $e^{2x} - e^x -12\:=\:0$

As skeeter pointed out, this is a quadratic equation.

There is a way to recognize this phenomenon.

There are usually three terms, written in standard order.

If the first term has twice the exponent of the second term,
. . we may have a quadratic.

In your problem: . $e^{2(x)} - e^x - 12 \:=\:0$

Another example: . $x^6 - 7x^3 - 8 \:=\:0$

Note that we have: . $\left(x^3\right)^2 - 7(x^3) - 8 \:=\:0$

Let $u \,=\,x^3$

Then we have: . $u^2 - 7u - 8 \:=\:0$
. . $(u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1$

Back-substitute: . $\begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}$

And there are four complex roots as well.

7. ## Re: Equations with the exponential function.

Hello Soroban

Originally Posted by Soroban
Hello, Furyan!

As skeeter pointed out, this is a quadratic equation.

There is a way to recognize this phenomenon.

There are usually three terms, written in standard order.

If the first term has twice the exponent of the second term,
. . we may have a quadratic.

In your problem: . $e^{2(x)} - e^x - 12 \:=\:0$

Another example: . $x^6 - 7x^3 - 8 \:=\:0$

Note that we have: . $\left(x^3\right)^2 - 7(x^3) - 8 \:=\:0$

Let $u \,=\,x^3$

Then we have: . $u^2 - 7u - 8 \:=\:0$
. . $(u-8)(u+1) \:=\:0 \quad\Rightarrow\quad u \:=\:8,\text{-}1$

Back-substitute: . $\begin{Bmatrix}x^3 \:=\:8 & \Rightarrow & x \:=\:2 \\ x^3 \:=\:\text{-}1 & \Rightarrow & x \:=\:\text{-}1 \end{Bmatrix}$

And there are four complex roots as well.

Thank you for that. I will be sure to look out for this sort of equation in the future.


If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4,$ then $(a-b)^2=\;$?

If $ab=3$ and $\frac1{a^2}+\frac1{b^2}=4$, what is the value of $(a-b)^2$? I think $a^2+b^2=36$, please confirm and is it possible to to figure out one of the variables?

-
You are right and going well, and yes we can figure the values of each variable. But it is not necessary to do so :) – chubakueno Jun 3 '14 at 18:03
What happens if you expand $(a - b)^2$? – David K Jun 3 '14 at 18:05

$$\frac1{a^2}+\frac1{b^2}=4$$ $$a^2b^2\left(\frac1{a^2}+\frac1{b^2} \right)=4a^2b^2$$ $$b^2+a^2=(2ab)^2$$ $$a^2-2ab+b^2=(2ab)^2-2ab$$ $$(a-b)^2=(2ab)^2-2ab=(2\cdot3)^2-2\cdot3=30$$

-
Your answer is good but try to add some words for the explanation. – Tunk-Fey Jun 3 '14 at 18:24
@Tunk-Fey - Why? each step flows well from the prior one. – JoeTaxpayer Jun 3 '14 at 19:17
@JoeTaxpayer Just in case the OP doesn't understand. You may take a look Andre Nicolas' answer. – Tunk-Fey Jun 3 '14 at 19:27
He states 'you know that $(a-b)^2=30$ ' - but it's not clear that we do. That's the question OP is asking. – JoeTaxpayer Jun 3 '14 at 19:39

Hints:

$\frac{1}{a^2}+\frac{1}{b^2} = 4 \rightarrow b^2 + a^2 = 4(ab)^2$

$(a-b)^2 = (a^2+b^2) -2(ab)$

edit:

To solve for a particular variable, you can use $ab=3 \rightarrow a = \frac{3}{b}$ to eliminate a variable. For example $a^2+b^2 = \frac{9}{b^2}+b^2$

-
I already know that (a-b)² =30 so how would you figure out one of the variables – user154989 Jun 3 '14 at 18:10

You know that $(a-b)^2=30$. The same strategy tells you that $(a+b)^2=42$. Thus $$a+b=\pm \sqrt{42}\quad\text{and}\quad a-b=\pm\sqrt{30}.$$ Now by adding and subtracting, we can find $2a$ and $2b$. and hence $a$ and $b$. Note that there are $4$ combinations, though if we have found one solution $a=p$, $b=q$, the other three are $a=-p$, $b=-q$, and $a=q$, $b=p$, and $a=-q$, $b=-p$.

One of the solutions is $a=\frac{\sqrt{42}+\sqrt{30}}{2}$, $b=\frac{\sqrt{42}-\sqrt{30}}{2}$.

-
This is right, of course, but it misses the point: this question is all about the symmetry in the equations, and avoiding having to solve for the values of $a$ and $b$ separately. I wish any of my high school teachers would have explained the idea that you sometimes don't care about the values of each part of an expression. – symplectomorphic Jun 3 '14 at 19:30
I was answering the question that OP had, in a comment, about solving for $a$ and $b$, given that OP already knew the value of $(a-b)^2$. The point about the solution is that it "breaks symmetry" very late in the game. Completely agree about the importance of exposure to symmetric functions. – André Nicolas Jun 3 '14 at 19:34
In fact, since I trust your judgment, Andre, do you know of a textbook that looks closely at these kinds of algebraic tricks? I think heavily symmetric equations often arise in geometry (and math competitions). But I can't think of a reference that systematically discusses problems like "if $a+b=5$ and $a^2+b^2=10$, what's the value of $ab$?" where the point isn't to find $a$ and $b$ separately. – symplectomorphic Jun 3 '14 at 19:35
I cannot think of a good source. Problem collections aimed below the Olympiad level often have symmetric examples. Algebra books used to cover this kind of material. Alas, no more. – André Nicolas Jun 3 '14 at 19:39
Alas indeed. I've written my own notes on this material and was always surprised I could never ever find a thorough, exhaustive discussion of this sort of technique. The competition books are always terse: they present a few examples but leave out the really interesting discussion of what's going on. Thanks though. – symplectomorphic Jun 3 '14 at 19:42


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Jerry purchased a 1-year $5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity? A.$5102
B. $408 C.$216
D. $202 E.$200
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Re: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 13 Sep 2017, 11:37 2 Top Contributor 1 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?

A. $5102 B.$408
C. $216 D.$202
E. $200 The ANNUAL interest rate = 4% So, every 6 MONTHS, we get 2% interest. Since there are only 2 "compoundings," we can forgo the formula and do some quick mental calculations... INITIAL VALUE:$5000

AFTER 6 MONTHS: 2% of $5000 =$100 in interest
So, the value after 6 months = $5000 +$100 = $5100 AFTER 12 MONTHS: 2% of$5100 = $102 in interest So, the value after 12 months =$5100 + $102 =$5202

How much interest had this bond accrued at maturity?

### Show Tags

12 Sep 2017, 03:09
Here P = 5000
r=4%
time = 1 year
As the amount is compounded annually n = 2

Compound interest formula = $$A = p* (1+\frac{r}{n*100}) ^n^t$$
$$A = 5000* (1+\frac{4}{2*100}) ^(2*1)$$
A= 5202

So Interest earned = 5202 - 5000 = 202

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Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 13 Sep 2017, 10:14 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?

A. $5102 B.$408
C. $216 D.$202
E. $200 Estimate Without the compound interest formula (I use it, but with short periods and low interest rate, often you can estimate): Simple interest would yield .04 * 5,000 =$200 in one year

Compound interest at this low rate (halved and paid twice), after only one year, will be barely above that.

Answer D, $202 is barely above$200.

Stages

If unsure about Answer C ($216), run a quick "in stages" calculation. If interest is paid every 6 months, the 4 percent is split in half (two periods of six months in one year). After first six months, interest payment is$5,000 * 02 = $100 (add$100 to principal for next stage)

After next six months,
.02 * $5,100 =$102

Total interest paid: $202 ANSWER D _________________ Visit SC Butler, here! Get two SC questions to practice, whose links you can find by date. Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has. -- Margaret Mead Director Joined: 02 Sep 2016 Posts: 629 Re: Jerry purchased a 1-year$5,000 bond that paid an annual  [#permalink]

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14 Sep 2017, 04:57
Simple interest= 5000*4*1/100= 200

So the answer would be a number that is slightly more than 200 i.e. 202.
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Re: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 09 Oct 2018, 07:18 guygmat wrote: Jerry purchased a 1-year$5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity?

A. $5102 B.$408
C. $216 D.$202
E. $200 $$5000( 1 + \frac{4}{200})^2 - 5000$$ $$= 5202 - 5000$$ $$= 202$$, Answer must be (D) _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 9952 Location: United States (CA) Re: Jerry purchased a 1-year$5,000 bond that paid an annual  [#permalink]

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12 Oct 2018, 06:52
guygmat wrote:
Jerry purchased a 1-year $5,000 bond that paid an annual interest rate of 4% compounded every six months. How much interest had this bond accrued at maturity? A.$5102
B. $408 C.$216
D. $202 E.$200

Since the annual interest rate is 4%, we see that the bond earns 2% every half-year, or 6 months. Thus, for the first 6 months, the amount of interest earned was 0.02 x 5,000 = 100 dollars. Thus, the new principal at the end of the first 6 months was 5,000 + 100 = 5,100 dollars.

For the next 6 months, the amount of interest earned was 0.02 x 5100 = 102 dollars.

So total interest earned is 202 dollars.

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Re: Jerry purchased a 1-year $5,000 bond that paid an annual [#permalink] ### Show Tags 06 Jan 2020, 17:49 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: Jerry purchased a 1-year$5,000 bond that paid an annual   [#permalink] 06 Jan 2020, 17:49
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My Math Forum One number is four more than five times another. If their sum is decreased by six....

 Algebra Pre-Algebra and Basic Algebra Math Forum

 February 29th, 2016, 08:36 AM #1 Senior Member   Joined: Feb 2016 From: seattle Posts: 377 Thanks: 10 One number is four more than five times another. If their sum is decreased by six.... One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? x+4+5x-6=10 Thanks. Last edited by skipjack; February 29th, 2016 at 10:07 PM.
February 29th, 2016, 08:49 AM   #2
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Quote:
 Originally Posted by GIjoefan1976 One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? x+4+5x-6=10
Solve the equation for x.

Then test it yourself by substituting...

February 29th, 2016, 09:13 AM   #3
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Joined: Feb 2016
From: seattle

Posts: 377
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Quote:
 Originally Posted by Denis Solve the equation for x. Then test it yourself by substituting...
Okay, I don't understand what you mean by "substituting"

and both this time, and last time I can get the first number being 2.

yet then I go 2 plus 4 =6 * 5 is 30 -6 =24

so not yet figuring out why it is not working for me.

 February 29th, 2016, 09:18 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra First of all, you have two numbers. Call them $x$ and $y$. The information gives you two equations relating $x$, $y$ and numbers you are given. See if you can build those equations from the text of the question. Thanks from GIjoefan1976
February 29th, 2016, 09:31 AM   #5
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Quote:
 Originally Posted by GIjoefan1976 Okay, I don't understand what you mean by "substituting" and both this time, and last time I can get the first number being 2. yet then I go 2 plus 4 =6 * 5 is 30 -6 =24
OK, you got x = 2

"One number is four more than five times another."
other number= 5*2 + 4 = 14 ; OK?

"If their sum is decreased by six, the result is ten."
sum = 2 + 14 = 16
16 - 6 = 10 : Bingo! You're correct

February 29th, 2016, 09:50 AM   #6
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Quote:
 Originally Posted by GIjoefan1976 One number is four more than five times another. If their sum is decreased by six, the result is ten. Find the numbers. I tried to write the equation from reading this. Is this anywhere near close? X+4+5x-6=10 Thanks
Let one number be x

Other number is 4+5x

Quote:
 Originally Posted by GIjoefan1976 If their sum is decreased by six, the result is ten
(x + 4+5x)-6 = 10

The only difference is brackets.
Continue.

February 29th, 2016, 10:24 AM   #7
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Quote:
 Originally Posted by Denis OK, you got x = 2 "One number is four more than five times another." other number= 5*2 + 4 = 14 ; OK? "If their sum is decreased by six, the result is ten." sum = 2 + 14 = 16 16 - 6 = 10 : Bingo! You're correct
Thanks i think it was the sum part that really confused me. Was not sure what they meant by that. but now I will try to remember they mean for me to add the 2 numbers I end finding after solving the problem not before solving the problem.

February 29th, 2016, 10:30 AM   #8
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Quote:
 Originally Posted by Prakhar Let one number be x Other number is 4+5x (x + 4+5x)-6 = 10 The only difference is brackets. Continue.
see I was thinking this too, yet was not sure how to write it.

Yet if i did it this way it confused me too as i would have gone and got

-36x-24=10

?

February 29th, 2016, 11:02 AM   #9
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Quote:
 Originally Posted by GIjoefan1976 (x + 4+5x)-6 = 10 see I was thinking this too, yet was not sure how to write it. Yet if i did it this way it confused me too as i would have gone and got -36x-24=10
C'mon GIJoe; quit running at 100 mph!
There's no multiplication there; just remove the brackets:
x + 4 + 5x - 6 = 10
6x = 10 - 4 + 6
6x = 12
x = 2

 February 29th, 2016, 11:05 AM #10 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra If you write that you have two numbers $x$ and $y$, then you get two equations $y = 4+5x$ and $x+y - 6= 10$. If you have learned how to solve simultaneous equations, this is not too difficult to solve (indeed one approach gives you exactly the equation that Denis was working you through. The important thing is to be able to build the equations. To understand what the sentences tell you and be confident that you have written it down correctly. The rest is relatively simple algebra.

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# Position vector

1. Nov 10, 2005

### Ratzinger

Given the typical cartesian xyz- coordinate system, is it correct to speak of a position vector? Isn't (x,y, z) just shothand for the coordinates? Distance vectors, force, velocity are real vectors with magnitude and direction in position space, but what is with a position vector in position space?

I'm confused, help needed

2. Nov 10, 2005

### whozum

The position vector points from the origin to the position of the particle. For your point it is < x , y , z >

Positions that vary with time are expressed as functions of x y and z, and you get parametric functions:

$\vec{r}_t = < x(t) , y(t), z(t) >$

3. Nov 10, 2005

### Ratzinger

yes, but isn't what you describe rather a parameterized curve, not these geometrical objects with head and tail that you can move parallel around?
I mean vectors in position coordinates without any time parameter mentioned.

4. Nov 10, 2005

### FredGarvin

Your question is a bit confusing to me. A specified position vector without any time reference just means that the function that produces the curve for the thing's position in space has been evaluated at some specific time. Ultimately the position has to be a function of time or how can you come up with velocity and acceleration?

5. Nov 10, 2005

### pmb_phy

A displacement vector is a vector which has its tail at one point and its tip at another point. If we refer to the point where the tail is as the "origin" then we refer to the displacement vector as the "position vector" of the point at the tip. We can label that point in any way we please, not only as R = (x, y, z). Quite often you'll see the position vector expressed as

$$R = (r, \theta, \phi)$$

This means that R has the value

$$R = A_r e_r + A_{\theta} e_{\theta} + A_{\phi} e_{\phi}$$

where ek is a unit vector pointing in the direction of the increasing kth coordinate. The coeficients have the value Ak = R*ek ("*" = dot product). The position vector is a bound vector, i.e. attached to the space in which it lies.
Pete

Last edited: Nov 10, 2005
6. Nov 10, 2005

### vaishakh

position vector is just a standard way of showing vectors and there is nothing like being attached. just the vector diagram has no physical meaning. the aim of physics is to solve the vector equations and while doing so we sometimes take the help of diagram.

7. Nov 11, 2005

### Ratzinger

thanks for the replies, I found this

An important difference between a position vector R and a general vector such as delta R is that the components of R are x, y and z, whereas for delta R the components are delta x, delta y and delta z. It is important to distinguish between true vectors and position vectors. A true vector does not depend on coordinate system but only on the difference between one end of the vector and the other. A position vector, in contrast, does depend on the coordinate system, because it is used to locate a position relative to a specified reference point.

8. Nov 11, 2005

### pmb_phy

Where did you get this definition from?? A vector is a geometrical object which does not depend on the coordinate system. Any vector may be expressed in terms of other vectors which are related to a particular coordinate system. But that doesn't mean that it is defined by the coordinate system. I thought I made that clear above but I guess not. I did explain that a vector is an arrow. Some vectors are called "Bound vectors" while others are called "free vectors." This is an important distinction that you should learn. The position vector does not depend on any coordinate system whatsoever. It depends on a geometric object, i.e. a "point." This point I speak of is known as the "reference point."
Wrong. You're confusing "reference point" with "coordinate system."
They are very different things.

For details on what I've been talking about see Thorne and Blanchard's online notes at

http://www.pma.caltech.edu/Courses/ph136/yr2004/0401.1.K.pdf

Pete

9. Nov 11, 2005

### Ratzinger

thanks Pete for the great notes you linked

So we have bound and free vectors. That's a nice distinction.

Free vectors can be parallel moved around or can be represented in different coordinate systems, they keep their meaning. Bound vectors do not, because they can clearly not be moved around and keep meaning and when coordinates systems are changed we talk about distance vectors. Bound vectors are bound to a coordinate system. A position vector gives only for his coordinate system information.

10. Nov 11, 2005

### pmb_phy

Bound vectors are not bound to a coordinate system. They're bound to the space that the vector is in and that is independant of the coordinate system. Give me any position vector and I can easily represent it in three different coordinate systems. I clarified this point here as I recall

http://www.geocities.com/physics_world/ma/coord_system.htm

Pete

11. Nov 11, 2005

### Ratzinger

Distance, velocity, force -all coordinate-independent quantities. But to ask what the position is of a point clearly needs a coordinate system. Position is only meaningful with a reference system.

Position vector in position space is a misnomer (in any position coordinates). As much as a momentum vector in momentum space.

12. Nov 11, 2005

### HallsofIvy

Staff Emeritus
Your original question is a very good one. In a Cartesian coordinate system, we can think of the "position vector" as the vector from the origin to the point, so that, at one instant, the tip of that vector is the point and the curve the point moves on is the curve sketched by the tip of the vector.

However, in non-Cartesian coordinates, especially in situations, such as are common in General Relativity, where we have curved surfaces that admit no Cartesian coordinate system, there is no such thing as a "position vector". (I used to worry about what vectors on the surface of a sphere 'looked like"!)
It is much better to think in terms of tangent vectors at every point. I like to think "a vector is a derivative".

13. Nov 12, 2005

### pmb_phy

Given a reference point the position of an arbitrary point requires only a magntitude and a direction - i.e. a vector.

HallsofIvy - I believe that you're confusing coordinate spaces with the space itself. A position vector can be defined in all spaces which have no curvature. Curvature exists independant of a coordinate system. There is no position vector in a curved space.

Pete

14. Nov 12, 2005

### robphy

I think the issue underlying the OP's questions is the distinction between a vector space and an affine space.
http://mathworld.wolfram.com/AffineSpace.html
http://en.wikipedia.org/wiki/Affine_space
A vector space has an "origin", whereas an affine space does not.
Positions are elements of an affine space.
Displacements (the difference of two positions) are elements of a vector space.
(Only after one assigns a norm can one talk of a "magnitude" of a vector.)

15. Nov 12, 2005

### lightgrav

WHY is the distinction between free vectors and bound vectors IMPORTANT?

I can add and subtract position vectors, even tho they're "bound" to origin.
Yet, the Torque due to a Force does NOT treat the Force as a "free" vector.

16. Nov 13, 2005

### pmb_phy

The physical meaning of adding position vectors is to start with one vector whose tail is at the origin. The next position vector added to this one has its tail at the tip of the first one. So this vector is bound as well and is bound at a different place. Tourque is the cross product of a bound vector and a free vector making it a free vector.

Other physical vectors are things like the center of mass vector.

Pete

17. Nov 15, 2005

### pmb_phy

Rob - There is no requirement for a vector space to have an origin. The term "Origin" refers to a particular point that an observer uses as a reference point. The user may also use other points in order to clearly define directions. There is no unique point in any space which demands to be an origin. From your links I don't see what you mean by "affine space." Can you elaborate for me? Thanks.

Pete

18. Nov 15, 2005

### HallsofIvy

Staff Emeritus
Yes, there is a requirement for a vector space to have an "origin". One of the requirements for a vector space is that there be a 0 vector. That is what you are referring to as an "origin". An "affine space" is a set of points such that any line through one of the points is contained in the space. You can think of it as a plane or 3d space or any Rn without a coordinate system. Once you add a coordinate system (so that you have an origin), you can make it a vector space. An affine space is what you seem to be thinking of as a "vector space". You can add vectors, you can't add points in an affine space.

19. Nov 15, 2005

### HallsofIvy

Staff Emeritus
You're right. I referred to "coordinate systems" when I really meant the space itself.

20. Nov 15, 2005

### robphy

Here's some physical motivations on this issue concerning an affine space vs. a vector space.

A space of positions in a plane (or the space of times on a line) is an affine space. With no point being physically distinguished from any other, an affine space is more natural than a vector space. In an affine space, there is no sense of addition of elements of this space. (If you attempt to add two elements, the sum depends on the choice of an "origin" [which does not exist in an affine space]. As was mentioned, one can introduce an origin by introducing a coordinate system. Then, the sum now depends on a choice of coordinate system.) There is, however, a sense of subtraction... the "difference of two positions [in an affine space]" is a vector... the displacement vector. (The difference does not depend on a choice of [or existence of] an origin.)

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# How many roots does x^6 –12x^4 + 32x^2 = 0 have?

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How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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10 Mar 2016, 05:10
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How many roots does x^6 –12x^4 + 32x^2 = 0 have?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

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How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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21 Mar 2017, 19:59
3
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Hi mesutthefail,

The GMAT often tests you on rules/patterns that you know, but sometimes in ways that you're not used to thinking about. This prompt is really just about factoring and Classic Quadratics, but it looks a lot more complicated than it actually is.

A big part of properly dealing with a Quant question on the GMAT is in how you organize the information and 'simplify' what you've been given. This prompt starts us off with....

X^6 –12X^4 + 32X^2 = 0

This certainly looks complex, but if you think about how you can simplify it, then you'll recognize that can 'factor out' X^2 from each term. This gives us...

(X^2)(X^4 - 12X^2 + 32) = 0

Now we have something a bit more manageable. While you're probably used to thinking of Quadratics such as X^2 + 6X + 5 as (X+1)(X+5), that same pattern exists here - it's just the exponents are slightly different (even though the math rules are exactly the SAME). We can further rewrite the above equation as...

(X^2)(X^2 -4)(X^2 - 8) = 0

At this point, you don't really need to calculate much, since each 'piece' of the product should remind you of a pattern that you already know.....

X^2 = 0 --> 1 solution
(X^2 - 4) = 0 --> 2 solutions
(X^2 - 8) = 0 --> 2 solutions

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How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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Updated on: 29 Mar 2017, 08:58
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4
To find the roots we can factor the polynomial down to first order expressions and count how many roots we get.

The first thing to do is to factor out $$x^2$$

Then we have $$x^2(x^4-12x^2+32)=0$$

Now to make things more familiar we can substitute $$y=x^2$$

Now it looks like this
$$y(y^2-12y+32)=0$$

And we can factor it like we are used to

$$y(y-4)(y-8)=0$$

y=0
y=4
y=8

So substituting $$x^2$$ back in for y:

$$x^2=0$$
$$x^2=4$$
$$x^2=8$$

So then the roots are

$$x=0$$
$$x=-2$$
$$x=2$$
$$x=-\sqrt{8}$$
$$x=\sqrt{8}$$

Note, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the $$x^2$$, we have $$x^2$$ and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the $$x^2$$. Total roots = 5.
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Originally posted by davedekoos on 10 Mar 2016, 14:28.
Last edited by davedekoos on 29 Mar 2017, 08:58, edited 1 time in total.
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Joined: 12 Dec 2016
Posts: 10
Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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21 Mar 2017, 02:27
I thought polynoms was not a subject in GMAT ? This questions contains quite the polynomial solutions.
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Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have?  [#permalink]

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21 Mar 2017, 02:31
davedekoos wrote:
To find the roots we can factor the polynomial down to first order expressions and count how many roots we get.

The first thing to do is to factor out $$x^2$$

Then we have $$x^2(x^4-12x^2+32)=0$$

Now to make things more familiar we can substitute $$y=x^2$$

Now it looks like this
$$y(y^2-12y+32)=0$$

And we can factor it like we are used to

$$y(y-4)(y-8)=0$$

y=0
y=4
y=8

So putting it back to x^2

$$x^2=0$$
$$x^2=4$$
$$x^2=8$$

So then the roots are

$$x=0$$
$$x=-2$$
$$x=2$$
$$x=-\sqrt{8}$$
$$x=\sqrt{8}$$

Note, since we are not asked to find the actual roots, we only need to determine the number of roots. After the first step of factoring out the $$x^2$$, we have $$x^2$$ and a 4th order expression. A 4th order expression will have 4 roots (unless it's a perfect square, but our expression is not a perfect square). So the total number of roots will be 4 from the 4th order expression, and one from the $$x^2$$. Total roots = 5.

A question. When you first factored out by x^2, the x expression in 32x^2 disappeared completely, however when you second factored out the "y" expression, 12y^2 became 12y instead of 12. Can you please clarify?
Re: How many roots does x^6 –12x^4 + 32x^2 = 0 have? &nbs [#permalink] 21 Mar 2017, 02:31
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# Hull, Instructional video , ch4 -Duration

#### Branislav

##### Member
Subscriber
Dear David,
Thanks a lot for video lectures they are much inspiring Still I was little bit confused with all these different names duration, modified duration, Macauly duration,.. etc...I will shortly examine mine view of this and kindly ask you to comment ( but without laughing)
According to mine understanding we are methodologically speaking about one risk measure all the time, called duration - how long on average shall i wait as bond holder to receive cash payments, or in terms on formula as you explained ( formula 1):

From this formula we see that that when we say on average, we are referring to time weighted average, so as a result for 3 year maturity bond I will obtain let's say 2.63 - so this is time based measure ( for zero coupon bond it is equal to 3- no cash flows till the very same end, if you can wait that much
If you agree with me on this definition than this is the same thing as Macaulay duration, there is nothing new coming with this new name introduced beside of course great honor and memory on Frederick Macauly who introduced this concept.

So we are still on duration and keep playing further. What if we do the first derivative by yield, just to check what is bond's sensitivity on yield change..

Delta (B)=dB/dY*Delta(y) ( formula 2 ) and dB/dY is "similar" to the right side of the formula 1, just with the minus in front , and we need to "remove" B from the denominator, or put it another way: dB/dY=-B*D and using formula 2 we obtain ( let us call it " yield/price sensy formula"):

Delta (B)=-B*D*Delta(y)

For me this was kind of "magic"..somehow mine year based measure D becomes interest ( yiedl) sensitivity measure!
But basically we are still talking about duration from the beginning of the text, just with this simple "math" transformation we saw that it is also connected to the bonds price sensitivity to yield change
We play further...we assumed above continuous compounding, if we go to the annual compounding, then, bond price is little bit different summation:

( note just the yield y is replaced with i) and duration and its first derivative are little bit different, so their relationship from " yield/price sensy formula"is now transforme to :

Delta (B)=-B*D*Delta(y)/(1+y)
and we introduce new "name" again, "Modified duration" as:
D*=D/(1+y), which transforms previous equation to:
Delta (B)=-B*(D*)*Delta(y)

so again D* is duration from the beginning of the text, just for yearly compounding case, "used" in this formula to express sensitivity of bonds price on yield change.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @Branislav Super glad you found some inspiration in the videos: that's what we are here for In regard to your summary: yes, rIght!! All of that looks solid to me, very well done. I agree with virtually all of your statements and I definitely agree with the substantial point that, to paraphrase, that there is really only one duration. I've posted dozens of times on these concepts but here is how I would summarize, and I believe this summary maps pretty well to your math, and further you can see that I'm agreeing that Mac/mod/Effective duration are three faces of the same single concept. Effective duration simply estimates modified duration, and modified duration is a sort of "adjusted" Macaulay duration which adjusts for the effects of discrete discounting on the price and doesn't matter if discounting is continuous (we tend to refer to Macaulay duration as a maturity and modified duration as a sensitivity, but the mathematical units of both, in fact, are years such that is it a minor mathematical adjustment from one to the other, as you do imply!).

Here is how I would summarize:
• Macaulay duration, I'll denote D_Mac, is the bond's weighted average maturity (as illustrated by your first formula above, where the weights assigned to the maturities are the PV of the cash flows; i.e., the weights, t(i), are weighted by c(i)*exp[-y*t(i)]/B.
• modified duration, I'll denote D_mod, is the linear sensitivity of the bond's price with respect to a small yield change; i.e., if the modified duration is 3.5 years, then we (linearly) approximate a yield change of +Δy will associate with a 3.5*Δy percentage drop in the bond price (knowing that curvature/convexity has been ignored). When the yield is continuously compounded, Mac duration = modified duration. When the yield is discretely rendered, we need to adjust the Mac to retrieve the accurate D_mod = D_mac/(1 + yield/k) where k = number of periods per year; i.e., when discrete, D_mod is always a bit less than D_Mac
• In this way, modified duration is a measure of sensitivity: %ΔP = ΔP/P = -(D_mod)*Δy, solving for D_mod:
• D_mod = -1/P * ΔP/Δy or continuously D_mod = -1/P * ∂P/∂y; i.e., modified duration is the first partial derivative (of bond price) with respect to the yield multiplied by -1/P. If we multiply each side by price, P, then:
• P *D_mod = -ΔP/Δy = "dollar duration;" i.e., dollar duration is the (negative) of the pure first derivative (i.e., the slope of the tangent line, itself negative). Importantly, dollar duration divided by 10,000 is the DV01 because P *D_mod/10,000 = DV01.
• However, if you start with the bond price function (either continuous or discrete) and if you take the first derivative, then you can see that you should end up with (the negative of) the dollar duration: ∂P/∂y = -D_mod*P (this forum has dozens of such actual derivations if you search). Therefore, by definition, if you take the first derivative, you should also (as I think you do imply), equivalently end up with ∂P/∂y = -D_mac*P/(1+y/k). There is an old saying: duration is "infected by price" to acknowledge that 1/P "infects" the pure derivative.
• Effective duration approximates modified duration by shocking the yield and re-pricing in order to retrieve the slope of the nearby tangent. Effective duration is sort of mini-simulation used to estimate the (inherently due to it being Taylor Series) analytical modified duration when it is not analytically available (e.g., MBS with negative convexity throws off the analytics). You will really understand when you can see that the effective duration approximates the modified duration which itself is an exact linear approximation. In this way, effective duration and modified duration, although they differ in approach, are both sensitivities and not conceptually different.

Last edited:

#### Matthew Graves

##### Active Member
Subscriber
I think I would add a few further points with respect to the practical use of Effective Duration and differences between Effective Duration and Modified Duration.

Modified Duration is sensitivity of the price with respect to the Yield to Maturity. These two measures are precisely defined mathematically and not open to interpretation.

Effective Duration, however, is a more practical (and complicated) measure derived through re-valuation of the instrument. It is the price sensitivity of the instrument to a parallel shift in a valuation curve and is therefore model dependent. Depending on the instrument, this can be a deterministic valuation but could equally be based on monte carlo simulation if the instrument has optionality. If the underlying curve is flat at the yield to maturity and the instrument does not have optionality you would expect the Effective Duration to be very close to the Modified Duration. However, in all practical, real-world valuation cases the Effective Duration would not be equal to the Modified Duration due to the shape of the underlying curve and any optionality in the instrument (e.g. callable bonds). Separately (and rather technically), the shift applied for Effective Duration is conventionally applied to the observed market yields comprising the underlying curve before obtaining the zero rate curve. The shift is not applied to the zero curve directly. This has subtle but observable affects on the effective duration also.

#### enjofaes

##### Member
Subscriber
Hi @David Harper CFA FRM . Thanks again for all the material! Was wondering if what you said was correct in the instructional video of duration : around 24'14" Modified duration = dollar duration / 10.000. I thought from the beginning of the video that this is the formula for the DV01.

Kind regards

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @enjofaes If I said that (sorry I'll have to locate the specific video location later), then I misspoke. You are correct: DV01 = (P*D)/10,000 = dollar duration/10,000. I'm actually very happy with my summary note at see https://forum.bionicturtle.com/threads/week-in-financial-education-2021-05-24.23840/post-88846 i.e.,
Note: Durations in CFA and FRM compared

I'd like to clarify duration terminology as it pertains to differences between the CFA and FRM. This forum has hundreds of threads over 12+ years on duration concepts (it's hard to say which links are the best at this point, but I'll maybe come back and curate some best links). Our YouTube channel has a FRM P2.T4 that includes videos on DV01, hedging the DV01, effective duration, modified versus Macaulay duration, and an illustration of all three durations. There are many nuances and further explorations, but here my goal is only to clarify the top-level definitions.

I'll use the simple example of a $100.00 face 20-year zero-coupon bond that currently yields (yield to maturity of) 6.0% per annum. If the yield is 6.0% per annum with continuous compounding, the price is$100.00*exp(-0.060*20) = $30.12. If the yield is 6.0% per annum with annual compounding, the price is$100.00/(1+0.060)^20 = $31.18. Unless otherwise specified, I will assume a continuous compound frequency. Special note: we so often price a bond given the yield (where CPT PV is the final calculator step) that it is easy to forget yield is not actually an input. Yield is the internal rate of return (IRR) assuming the current price. Yield does not determine price; price determines yield. Technical (non-fundamental) factors cause price to fluctuate, therefore yield fluctuates. • ∂P/∂y (or Δp/Δy) is the slope of the tangent line at the selected yield. At 6.0% yield, the slope is -$602.39. How do I know that? Because dollar duration is the negated slope, so in this case dollar duration (DD) = P*D = $30.12 price * 20 years =$602.39. Importantly, the "y" in ∂P/∂y is yield and yield is just one of several interest rate factors.
• Dollar duration (DD; aka, money duration in the CFA) is analytically the product of price and modified duration. Dollar duration (DD) = P*D = $30.12 * 20 =$602.39. Why is it so large? Because it's the (negated) tangent line's slope, so it has the typical first derivative interpretation: DD is the dollar change implied by one unit change in the yield, -∂P/∂y. One unit is 1.0 = 100.0% = 100 * 100 basis points (bps) per 1.0% = 10,000 basis points. So, DD is the dollar change implied by a 100.0% change in yield if we use the straight tangent line which would be a silly thing to do! Recall the constant references to limitations of duration as linear approximation. The linear approximation induces bias at only 5 or 10 or 20 basis points, so 10,000 basis points is literally "off the charts" and not directly meaningful. What is meaningful? The PVBP (aka, DV01) comes to our rescue with a meaningful re-scaling of the DD ...
• Price value of basis point (aka, dollar value of '01, DV01) is the dollar duration ÷ 10,000. It's the tangent line's slope re-scaled from Δy=100.0% to Δy= 0.010% (one basis point). PVBP = P*D/10,000; in this example, PVBP = $30.12 * 20 / 10,000 =$0.06024. It is the dollar change implied by a one basis point decline in the yield. It is still a linear approximation, but much better because we zoomed in to a small change. In this way, the difference between the highly useful PVBP and the dollar duration is merely scale.
• Macaulay duration is the bond's weighted average maturity where the weights are each of the bond's cash flow's present value as a percentage of the bond's price. Macaulay duration is tedious however it is reliable and it is analytical. When we can compute the Macaulay duration, it is accurate; we don't approximate by re-pricing the bond. A zero-coupon bond has a Macaulay duration equal to its maturity because it only has one cash flow (hence the popularity of the zero-coupon bond in exam questions, never mind the zero-coupon bond is a reliable primitive). Our 20-year zero-coupon bond has a Macaulay duration of 20.0 years.
• Modified duration is the measure of interest rate risk. Modified duration is the approximate percentage change in bond price implied by a 1.0% (100 basis point) change in the yield. Just as ∂P/∂y refers to the tangent line's slope which is "infected with price," we divide by price to express the modified duration, D(mod) = -1/P*∂P/∂y. The key relationship between analytical modified and Macaulay duration is the following: modified duration = Macaulay duration / (1 + y/k) where k is the number of compound periods in the year; e.g., k = 1 for annual compounding, k = 2 for semiannual compounding and k = ∞ for continuous compounding. Importantly, if the the compound frequency is continuous then a bond's modified duration equals its Macaulay duration. Notice that T / (1 + y/∞) = T / (1 + 0) = T.
• If the 6.0% yield is annual compounded, our 20-year bond's Macaulay duration is given by 20.0 / (1 + 6.0%) = 18.868 years.
• If the 6.0% yield is continuously compounded, our 20-year bond's modified duration is 20.0 years.
• Effective duration is an approximation of modified duration. Recall the modified duration is a linear approximation, but that's because it is a function of the first derivative; otherwise, modified duration is an exact (analytical or functional) measure of the price sensitivity with respect to the interest rate factor that happens to most often be the yield. We can retrieve it easily whenever we can compute the Macaulay duration, which is the case for any vanilla bond. Otherwise (e.g., bond has an embedded option) we approximate the modified duration by calculating its effective duration. The effective duration approximates the modified duration which itself is a linear approximation. The effective duration is given by [P(-Δy) - P(+Δy)] / (2*Δy) * 1/P. I wrote it this way so you can see that it is essentially similar to ∂P/∂y*1/P where ∂P/∂y ≅ [P(-Δy) - P(+Δy)] / (2*Δy). I've observed that many candidates do not realize that the formula for effective duration is simply slope*1/P. Geometrically, it is the slope of the secant line that is near to the tangent line! Secant's slope approximates the tangent's slope. If you grok the calculus here, I think you'll agree that this is all just one thing! Now we can see how it's not so different. But as you can visualize, there are an almost infinite variety of secants next to the tangent. We arbitrarily choose a nearby secant, but we'd prefer a small delta if the bond is vanilla (i.e., if the bond's cash flows are invariant to rate changes). Although we do not need the effective duration for our example bond, we can compute it:
• If our arbitrary yield shock is10 basis points such that Δy= 0.10%, then P(-Δy)= $100.00*exp(-5.90%*20)=$30.728, and P(+Δy)= $100.00*exp(-6.10%*20)=$29.523. Effective duration= ($30.728 -$29.523)/0.0020 *1/\$30.12= 20.0013 years. Fine approximation!
• On the terminology (CFA versus FRM)
• Interest rate factor: The FRM (informed by Tuckman) starts with a general interest rate factor. This is typically the spot rate, forward rate, par rate, or yield. Importantly, the spot, forward and par rates are term structures, or vectors; the par yield curve is a vector of par rates at various maturities, often at six-month or one-month intervals. Only the yield is a single (aka, scalar) value.
• My above definition of the effective duration is according to the FRM (and to me). The CFA sub-divides this effective duration into either approximate modified duration (if the interest rate factor is the yield) versus effective duration (if the non-vanilla nature of the bond requires a non-yield interest rate factor; i.e., a benchmark yield curve). Personally, I am not keen on this semantic approach because (i) both of these CFA formulas are approximating the modified duration and (ii) I prefer to reserve "effective" for its traditional connotation (e.g., effective convexity is analogous to effective duration), and (iii) we wouldn't anyhow use an inappropriate factor (yield) for certain non-vanilla situations, so we don't really need label-switches to guide us thusly! (the CFA's formula for its approximate modified duration is essentially the same as its effective duration formula). To me, the CFA's approach muddies the terms "approximate" and "effective" where the math gives us natural distinctions. Follow the math, I'd say!


# Definition:Supremum of Set

## Definition

Let $\struct {S, \preceq}$ be an ordered set.

Let $T \subseteq S$.

An element $c \in S$ is the supremum of $T$ in $S$ if and only if:

$(1): \quad c$ is an upper bound of $T$ in $S$
$(2): \quad c \preceq d$ for all upper bounds $d$ of $T$ in $S$.

If there exists a supremum of $T$ (in $S$), we say that:

$T$ admits a supremum (in $S$) or
$T$ has a supremum (in $S$).

### Finite Supremum

If $T$ is finite, $\sup T$ is called a finite supremum.

### Subset of Real Numbers

The concept is usually encountered where $\struct {S, \preceq}$ is the set of real numbers under the usual ordering $\struct {\R, \le}$:

Let $T \subseteq \R$ be a subset of the real numbers.

A real number $c \in \R$ is the supremum of $T$ in $\R$ if and only if:

$(1): \quad c$ is an upper bound of $T$ in $\R$
$(2): \quad c \le d$ for all upper bounds $d$ of $T$ in $\R$.

The supremum of $T$ is denoted $\sup T$ or $\map \sup T$.

## Also known as

Particularly in the field of analysis, the supremum of a set $T$ is often referred to as the least upper bound of $T$ and denoted $\map {\operatorname {lub} } T$ or $\map {\operatorname {l.u.b.} } T$.

Some sources refer to the supremum of a set as the supremum on a set.

## Also defined as

Some sources refer to the supremum as being the upper bound.

Using this convention, any element greater than this is not considered to be an upper bound.

## Also see

• Results about suprema can be found here.

## Linguistic Note

The plural of supremum is suprema, although the (incorrect) form supremums can occasionally be found if you look hard enough.


Largest $n$-vertex polyhedron that fits into a unit sphere

In two dimensions, it is not hard to see that the $n$-vertex polygon of maximum area that fits into a unit circle is the regular $n$-gon whose vertices lie on the circle: For any other vertex configuration, it is always possible to shift a point in a way that increases the area.

In three dimensions, things are much less clear. What is the polyhedron with $n$ vertices of maximum volume that fits into a unit sphere? All vertices of such a polyhedron must lie on the surface of the sphere (if one of them does not, translate it outwards along the vector connecting it to the sphere's midpoint to get a polyhedron of larger volume), but now what? Not even that the polyhedron must be convex for every $n$ is immediately obvious to me.

• If the vertices are on the surface of the sphere the polyhedron will necessarily be convex - it will be the convex hull of the vertices. Because the sphere itself is convex the convex hull will lie entirely within it. Oct 18 '14 at 16:41
• @MarkBennet: Good point, that settles this part at least.
– user139000
Oct 18 '14 at 16:46
• I believe this is an open problem for $n > 8$. Oct 21 '14 at 6:57
• @achille hui: do you know solutions for n = 7, 8? One can check directly that cube is not even a local maximum, having in fact surprisingly poor performance. Oct 23 '14 at 19:50
• A stickler point about your proof for polygons: given that the space of such polygons is compact...
– Max
Oct 23 '14 at 20:42

This is supposed to be a comment but I would like to post a picture.

For any $m \ge 3$, we can put $m+2$ vertices on the unit sphere

$$( 0, 0, \pm 1) \quad\text{ and }\quad \left( \cos\frac{2\pi k}{m}, \sin\frac{2\pi k}{m}, 0 \right) \quad\text{ for }\quad 0 \le k < m$$

Their convex hull will be a $m$-gonal bipyramid which appear below.

Up to my knowledge, the largest $n$-vertex polyhedron inside a sphere is known only up to $n = 8$.

• $n = 4$, a tetrahedron.
• $n = 5$, a triangular bipyramid.
• $n = 6$, a octahedron = a square bipyramid
• $n = 7$, a pentagonal bipyramid.
• $n = 8$, it is neither the cube ( volume: $\frac{8}{3\sqrt{3}} \approx 1.53960$ ) nor the hexagonal bipyramid ( volume: $\sqrt{3} \approx 1.73205$ ). Instead, it has volume $\sqrt{\frac{475+29\sqrt{145}}{250}} \approx 1.815716104224$.
Let $\phi = \cos^{-1}\sqrt{\frac{15+\sqrt{145}}{40}}$, one possible set of vertices are given below: $$( \pm \sin3\phi, 0, +\cos3\phi ),\;\; ( \pm\sin\phi, 0,+\cos\phi ),\\ (0, \pm\sin3\phi, -\cos3\phi),\;\; ( 0, \pm\sin\phi, -\cos\phi).$$ For this set of vertices, the polyhedron is the convex hull of two polylines. One in $xz$-plane and the other in $yz$-plane. Following is a figure of this polyhedron, the red/green/blue arrows are the $x/y/z$-axes respectively.

$\hspace0.75in$

For $n \le 8$, above configurations are known to be optimal. A proof can be found in the paper

Joel D. Berman, Kit Hanes, Volumes of polyhedra inscribed in the unit sphere in $E^3$
Mathematische Annalen 1970, Volume 188, Issue 1, pp 78-84

An online copy of the paper is viewable at here (you need to scroll to image 84/page 78 at first visit).

For $n \le 130$, a good source of close to optimal configurations can be found under N.J.A. Sloane's web page on Maximal Volume Spherical Codes. It contains the best known configuration at least up to year 1994. For example, you can find an alternate set of coordinates for the $n = 8$ case from the maxvol3.8 files under the link to library of 3-d arrangements there.

• That's plenty of information, and I'm willing to give you the bounty since you have answered my question ("Open problem for $n\ge 9$") but I'd like some more information for the $n=8$ case if possible. The cube not being optimal is already surprising since one might expect the 2D argument to somehow transfer to regular polyhedra, but where do the coordinates come from? You say that $n=8$ is known exactly but the coords just look like the result of a numerical optimization run. I'd expect there to be exact polar or cartesian coordinates or at least some formal description of the polyhedron.
– user139000
Oct 25 '14 at 7:44
• That the $n=7$ solution is a pentagonal bipyramid also mildly surprised me. In my mental picture of it, the concentration of vertices in the plane spanned by the pyramids' shared base seems a little too high for the configuration to be optimal. But 3D is hard to imagine accurately, of course...
– user139000
Oct 25 '14 at 7:48
• @pew look at Berman and Hanes paper (linked in updated answer) for a proof. Oct 25 '14 at 9:16


# Find the integral $\int\frac{dx}{\sqrt{x^2-a^2}}$

Evaluate the integral of $\frac{1}{\sqrt{x^2-a^2}}$

Put $x=a\sec\theta\implies dx=a\sec\theta\tan\theta d\theta$ \begin{align} \int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta d\theta}{\sqrt{a^2\sec^2\theta-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\sqrt{\tan^2\theta}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{\tan\theta}}\\&=\int\sec\theta d\theta=\log|\sec\theta+\tan\theta|+C=\log|\frac{x}{a}+\sqrt{\frac{x^2}{a^2}-1}|\\&=\log|\frac{x+\sqrt{x^2-a^2}}{a}|+C=\log|{x+\sqrt{x^2-a^2}}|-\log|a|+C\\&=\log|{x+\sqrt{x^2-a^2}}|+C \end{align} This is how it is solve in my reference. But, $\sqrt{\tan^2\theta}=|\tan\theta|$ right ? Then, does that imply $$\int\frac{dx}{\sqrt{x^2-a^2}}=\int\frac{a\sec\theta\tan\theta d\theta}{a\color{red}{|\tan\theta|}}=\color{red}{\pm}\int\sec\theta d\theta$$ Why am I getting this confusion and is the first solution complete ?

• @TrostAft how can i say $\tan\theta$ is $+$ve from $x^2-a^2>0$ ? – ss1729 May 2 '18 at 7:20
• No first solution is not complete, they silently slipped in the || inside. You know, $\int \sec t = \ln(\sec t + \tan t)$ is valid for $\sec t > 0$ and for $\sec t \lt 0$ the integrand is $\ln(-\sec t - \tan t)$ so the solution $\ln |\sec t + \tan t|$ combines both. – jonsno May 2 '18 at 7:35
• Check by differentiating your solution(s). – Yves Daoust May 2 '18 at 7:35
• Agree with @samjoe. My comment is untrue. – TrostAft May 2 '18 at 7:35
• @samjoe i'm srry dont understand how ur point help me with the doubt in OP ?. Could u pls explain bit more ? – ss1729 May 2 '18 at 7:57

Suppose that $a>0$.

The work is just for the case when $x>a$. The case for $x<-a$ is different, but the finals result is the same.

Let $x=a\sec\theta$, where $\theta\in[0,\frac{\pi}{2})\cup(\frac{\pi}{2},\pi]$. This is the domain of $\textrm{arcsec}$.

For $x< -a$, $\theta\in(\frac{\pi}{2},\pi]$ and so $\tan\theta\le0$.

$$\sqrt{x^2-a^2}=\sqrt{a^2\tan^2\theta}=-a\tan\theta$$

\begin{align*} \int\frac{dx}{\sqrt{x^2-a^2}}&=\int\frac{a\sec\theta\tan\theta}{-a\tan\theta}d \theta\\ &=-\int\sec\theta d\theta\\ &=-\ln|\sec\theta+\tan\theta|+C\\ &=\ln|\sec\theta-\tan\theta|+C\\ &=\ln\left|\frac{x}{a}-\frac{-\sqrt{x^2-a^2}}{a}\right|+C\\ &=\ln\left|x+\sqrt{x^2-a^2}\right|-\ln|a|+C \end{align*}

There are two minus signs and they cancel each other to reach the final result.

• I don't uunderstand how you conclude $\theta \in (\pi/2, \pi]$ because that is the key to the question. – jonsno May 2 '18 at 8:17
• I mean why it can't be $\theta \in (\pi, 3\pi/2)$ where sec is negative but tan is positive – jonsno May 2 '18 at 8:18
• Yes thats what we have to show that tan cannot be positive. – jonsno May 2 '18 at 8:20
• If we take $\theta\in(\pi,3\pi/2)$, then $\tan\theta>0$. The integral is $\int\sec\theta d\theta$. The final answer will be still $\ln|x+\sqrt{x^2-a^2}|-\ln|a|+C$. My point is that we can have $-\int\sec\theta d\theta$ in the work. But then we will have one more minus sign and obtain the same final answer. – CY Aries May 2 '18 at 8:26
• We either take a range so that the work holds for both $x>a$ and $x<-a$, or take a range so that in one case we will have two minus signs to cancel each other. – CY Aries May 2 '18 at 8:29


# Prove that the square root of a positive integer is either an integer or irrational

Is my proof that the square root of a positive integer is either an integer or an irrational number correct?

The proof goes like this:

Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\sqrt{n}$ is rational.

Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.

We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.

Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.

Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n > 1$, then $a>b>1$.

By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes. Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.

However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of $nb^2$ that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.

Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.

Is this sufficient? Or is there any parts I did not explain well?

• I think its correct and very well explained. – Shobhit Jan 30 '17 at 14:25
• It's a bit wordy, but logically you've got a solid proof, the even-ness of the powers of each prime is exactly what you're going for. – Adam Hughes Jan 30 '17 at 14:26
• The conclusion is not precise enough. Instead of "contradicts there is an even number of primes in $a^2$" we want to say that it contradicts the fact that the prime $p$ occurs to odd power in the unique factorization of $nb^2,\,$ but even power in $a^2,\,$ i.e. we are comparing the parity of the count of single prime, not the total of all primes – Bill Dubuque Jan 30 '17 at 14:28
• For example the argument shows that $\,\sqrt{3\cdot 5}\,$ is irrational because $\,3\,$ occurs to odd power in $\,3\cdot 5,\,$ but the total number of primes in $\,3\cdot 5\,$ is even. – Bill Dubuque Jan 30 '17 at 14:37
• On a side note, this result is known as Theaetetus' Theorem, and it's proven in Euclid's Elements here: aleph0.clarku.edu/~djoyce/elements/bookX/propX9.html "[S]quares which do not have to one another the ratio which a square number has to a square number also do not have their sides commensurable in length either." – Keshav Srinivasan Jan 30 '17 at 14:41

Your proof is very good and stated well. I think it can be made shorter and tighter with a little less exposition of the obvious. However, I would prefer students to err on the side of more rather than less so I can't chide you for being thorough. But if you want a critique:

"Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is for some integer a. Therefore, if n is a square number, then $\sqrt{n}$ is rational."

Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.

This can all be said more simply and to argue that if $\sqrt{n}$ is an integer we can conclude $\sqrt{n}$ is rational or that $n$ is therefore a perfect square, is a little heavy handed. Those are definitions and go without saying. However, it shows good insight and understanding to be aware one can assume things and all claims need justification so I can't really call this "wrong".

But it'd be enough to say. "If $n$ is a perfect square then $\sqrt{n}$ is a an integer and therefore rational, so it suffices to prove that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.

We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n}$=ab , where a,b∈Z+,b≠0. We also suppose that a≠0, otherwise ab=0, and n will be a square number, which is rational.

Terminologistically, to say "$n$ is a square number" is to mean $n$ is the square of an integer. If $n = (\frac ab)^2$ we don't usually refer to $n$ as a square (although it is "a square of a rational") We'd never call $13$ a square because $13 = (\sqrt{13})^2$.

Also you don't make the usual specification that $a$ and $b$ have no common factors. As it turns out you didn't need to but it is a standard.

Suppose b=1 . Then $\sqrt{n}$=a , which shows that n is a square number. So b≠1. Since $\sqrt{n}$>1, then a>b>1

This was redundant as $b=1 \implies$ $a/b$ is an integer and we are assuming that $n$ is not a perfect square.

.

By the unique factorization of integers theorem, every positive integer greater than 1 can be expressed as the product of its primes. Therefore, we can write a as a product of primes and for every prime number that exists in a, there will be an even number of primes in a2. Similarly, we can express b as a product of primes and for every prime number that exists in b, there will be an even number of primes in b2

Bill Dubuque in the comments noted what you meant to say was "each prime factor will be raised to any even power".

.

However, we can also express n as a product of primes. Since n is not a square number, then there exist at least one prime number that has an odd number of primes. Therefore, there exists at least one prime in the product of nb2 that has an odd number of primes. Since nb2=a2 , then this contradicts the fact that there is an even number of primes in a2 since a number can neither be even and odd.

Ditto:

Overall I think your proof is very good.

But I should point out there is a simpler one:

Assume $n = \frac {a^2}{b^2}$ where $a,b$ are positive integers with no common factors (other than 1). If $p$ is a prime factor of $b$ and $n$ is an integer, it follows that $p$ is a prime factor of $a^2$ and therefore of $a$. But that contradicts $a$ and $b$ having no common factors. So $b$ can not have any prime factors. But the only positive integer without prime factors is $1$ so $b = 1$ and $n= a^2$ so $\sqrt{n} = a$. So either for any integer either $n$ is a perfect square with an integer square root, or $n$ does not have a rational square root.

And a slight caveat: I'm assuming that your class or text is assuming that all real numbers have square roots (and therefore if there is not rational square root the square root must be irrational). It's worth pointing out, that it is a result of real analysis that speaking of a square root actually makes any sense and that we can claim every positive real number actually does have same square root value. But that's probably beyond the range of this exercise.

But if I want to be completely accurate, you (and I) have actually only proven that positive integer $n$ either has an integer square root or it has no rational square root at all. Which is the same thing as saying if positive integer $n$ has a square root, the root is either integer or irrational. But we have not actually proven that positive integer $n$ actually has any square root at all.

• Thank you for the critique! It definitely helped me a lot! – Icycarus Jan 31 '17 at 11:00
• Can you explain or state a case where we cannot find the square root of a positive integer $n$ (as you state in the last line)? – Hungry Blue Dev Apr 15 '17 at 12:56
• I didn't say there are positive integers without square roots. (There aren't.) I said we haven''t proven that the square root of any integer exists. And we haven't. To prove that we have to prove that $K = \{q \in \mathbb Q| q^2 < n\}$ is not empty, and bounded above and that if $z = \sup K$ then $z^2 - n$. All we have proven so far is that either there is an integer so that $m^2 = n$ or there is no rational $q$ so that $q^2 = n$. We have not proven that there is an irrational $z$ so that $z^2 = n$. – fleablood Apr 15 '17 at 15:54
• If $n$ can not be written as $\frac {a^2}{b^2}$ then, by definition, $n$ does not have a rational square root. If $n$ has a rational square root then that rational square root can be written as $\frac ab$. That is what rational means. And if the square root is $\frac ab$ then $n = (\frac ab)^2 =\frac {a^2}{b^2}$. That is what square root means. So if $n$ can't be written that way then it does not have a rational square root. For example: $3$ can not be written that way and does not have a rational square root. Neither does ANY other non-square. – fleablood Jan 21 at 22:31
• This is a prove that IF a square root is rational then it must be an integer. We are not concerned about irrational square root (not even to the extent of questioning whether they exist). And IF a square root is rational it can be written as $\frac ab$. Which means $n$ can be written as $\frac {a^2}{b^2}$. If it can't be, then it does not have a rational square root. – fleablood Jan 21 at 22:35

We know that $\sqrt{4} = 2$ and $\sqrt{2} = 1.414...$ are rational and irrational respectively, so all we have to do is to show that if $n\in \mathbb{Z+}$ such that $\sqrt{n} = \dfrac{a}{b}$ where $a$ and $b$ are positive integers and the expression $\dfrac{a}{b}$ is in its simplest form then $\sqrt{n}$ is integral. Squaring both sides of the expression we get that $n = \dfrac{a^2}{b^2}$ since $a$ and $b$ have no common factors other than $1$ then $a^2 = n$ and $b^2 = 1$ therefore $b = 1$ hence $\sqrt{n}$ if rational it's an integer.

• How is that a critique of the OP's proof? – fleablood Jan 30 '17 at 18:51
• The op shows (correctly) that the square root of a non square is irrational. There is no need to show that rational roots are integers. Assuming n is not square makes perfect sense and the case for square n's has been covered. Your objections are not valid. Meanwhile you post is just ... weird. What do $\sqrt 4$ and $\sqrt 2$ have to do with anything. And your statement $n = a^2/b^2$ implies $a^2 = n$ and $b^2 =1$ is said without justification when the justification is the entire point. And you are merely pointing out an easier proof. Which is not a valid critique. – fleablood Jan 30 '17 at 22:57


Bertrand's box paradox is a classic paradox of elementary probability theory. It was first posed by Joseph Bertrand in his Calcul des probabilités, published in 1889.

There are three boxes:

1. a box containing two gold coins,
2. a box containing two silver coins,
3. a box containing one gold coin and one silver coin.

After choosing a box at random and withdrawing one coin at random, if that happens to be a gold coin, it may seem that the probability that the remaining coin is gold is 12; in fact, the probability is actually 23. Two problems that are very similar are the Monty Hall problem and the Three Prisoners problem.

These simple but slightly counterintuitive puzzles are used as a standard example in teaching probability theory. Their solution illustrates some basic principles, including the Kolmogorov axioms.

Box version

There are three boxes, each with one drawer on each of two sides. Each drawer contains a coin. One box has a gold coin on each side (GG), one a silver coin on each side (SS), and the other a gold coin on one side and a silver coin on the other (GS). A box is chosen at random, a random drawer is opened, and a gold coin is found inside it. What is the chance of the coin on the other side being gold?

The following reasoning appears to give a probability of 12:

• Originally, all three boxes were equally likely to be chosen.
• The chosen box cannot be box SS.
• So it must be box GG or GS.
• The two remaining possibilities are equally likely. So the probability that the box is GG, and the other coin is also gold, is 12.

The flaw is in the last step. While those two cases were originally equally likely, the fact that you are certain to find a gold coin if you had chosen the GG box, but are only 50% sure of finding a gold coin if you had chosen the GS box, means they are no longer equally likely given that you have found a gold coin. Specifically:

• The probability that GG would produce a gold coin is 1.
• The probability that SS would produce a gold coin is 0.
• The probability that GS would produce a gold coin is 12.

Initially GG, SS and GS are equally likely. Therefore by Bayes rule the conditional probability that the chosen box is GG, given we have observed a gold coin, is:

$\frac { \mathrm{P}(see\ gold \mid GG)} { \mathrm{P}(see\ gold \mid GG)+\mathrm{P}(see\ gold \mid SS)+\mathrm{P}(see\ gold \mid GS) } =\frac{1}{1+0+1/2}= \frac{2}{3}$

The correct answer of 23 can also be obtained as follows:

• Originally, all six coins were equally likely to be chosen.
• The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.
• So it must come from the G drawer of box GS, or either drawer of box GG.
• The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is 23.

Alternatively, one can simply note that the chosen box has two coins of the same type 23 of the time. So, regardless of what kind of coin is in the chosen drawer, the box has two coins of that type 23 of the time. In other words, the problem is equivalent to asking the question "What is the probability that I will pick a box with two coins of the same color?".

Bertrand's point in constructing this example was to show that merely counting cases is not always proper. Instead, one should sum the probabilities that the cases would produce the observed result; and the two methods are equivalent only if this probability is either 1 or 0 in every case. This condition is correctly applied in the second solution method, but not in the first.

The paradox as stated by Bertrand

It can be easier to understand the correct answer if you consider the paradox as Bertrand originally described it. After a box has been chosen, but before a box is opened to let you observe a coin, the probability is 2/3 that the box has two of the same kind of coin. If the probability of "observing a gold coin" in combination with "the box has two of the same kind of coin" is 1/2, then the probability of "observing a silver coin" in combination with "the box has two of the same kind of coin" must also be 1/2. And if the probability that the box has two like coins changes to 1/2 no matter what kind of coin is shown, the probability would have to be 1/2 even if you hadn't observed a coin this way. Since we know his probability is 2/3, not 1/2, we have an apparent paradox. It can be resolved only by recognizing how the combination of "observing a gold coin" with each possible box can only affect the probability that the box was GS or SS, but not GG.

Card version

Suppose there are three cards:

• A black card that is black on both sides,
• A white card that is white on both sides, and
• A mixed card that is black on one side and white on the other.

All the cards are placed into a hat and one is pulled at random and placed on a table. The side facing up is black. What are the odds that the other side is also black?

The answer is that the other side is black with probability 23. However, common intuition suggests a probability of 12 either because there are two cards with black on them that this card could be, or because there are 3 white and 3 black sides and many people forget to eliminate the possibility of the "white card" in this situation (i.e. the card they flipped CANNOT be the "white card" because a black side was turned over).

In a survey of 53 Psychology freshmen taking an introductory probability course, 35 incorrectly responded 12; only 3 students correctly responded 23.[1]

Another presentation of the problem is to say : pick a random card out of the three, what are the odds that it has the same color on the other side? Since only one card is mixed and two have the same color on their sides, it is easier to understand that the probability is 23. Also note that saying that the color is black (or the coin is gold) instead of white doesn't matter since it is symmetric: the answer is the same for white. So is the answer for the generic question 'same color on both sides'.

Preliminaries

To solve the problem, either formally or informally, one must assign probabilities to the events of drawing each of the six faces of the three cards. These probabilities could conceivably be very different; perhaps the white card is larger than the black card, or the black side of the mixed card is heavier than the white side. The statement of the question does not explicitly address these concerns. The only constraints implied by the Kolmogorov axioms are that the probabilities are all non-negative, and they sum to 1.

The custom in problems when one literally pulls objects from a hat is to assume that all the drawing probabilities are equal. This forces the probability of drawing each side to be 16, and so the probability of drawing a given card is 13. In particular, the probability of drawing the double-white card is 13, and the probability of drawing a different card is 23.

In question, however, one has already selected a card from the hat and it shows a black face. At first glance it appears that there is a 50/50 chance (i.e. probability 12) that the other side of the card is black, since there are two cards it might be: the black and the mixed. However, this reasoning fails to exploit all of the information; one knows not only that the card on the table has at least one black face, but also that in the population it was selected from, only 1 of the 3 black faces was on the mixed card.

An easy explanation is that to name the black sides as x, y and z where x and y are on the same card while z is on the mixed card, then the probability is divided on the 3 black sides with 13 each. thus the probability that we chose either x or y is the sum of their probabilities thus 23.

Solutions

Intuition

Intuition tells one that one is choosing a card at random. However, one is actually choosing a face at random. There are 6 faces, of which 3 faces are white and 3 faces are black. Two of the 3 black faces belong to the same card. The chance of choosing one of those 2 faces is 23. Therefore, the chance of flipping the card over and finding another black face is also 23. Another way of thinking about it is that the problem is not about the chance that the other side is black, it's about the chance that you drew the all black card. If you drew a black face, then it's twice as likely that that face belongs to the black card than the mixed card.

Alternately, it can be seen as a bet not on a particular color, but a bet that the sides match. Betting on a particular color regardless of the face shown, will always have a chance of 12. However, betting that the sides match is 23, because 2 cards match and 1 does not.

Labels

One solution method is to label the card faces, for example numbers 1 through 6.[2] Label the faces of the black card 1 and 2; label the faces of the mixed card 3 (black) and 4 (white); and label the faces of the white card 5 and 6. The observed black face could be 1, 2, or 3, all equally likely; if it is 1 or 2, the other side is black, and if it is 3, the other side is white. The probability that the other side is black is 23. This probability can be derived in the following manner: Let random variable B equal the a black face (i.e. the probability of a success since the black face is what we are looking for). Using Kolmogrov's Axiom of all probabilities having to equal 1, we can conclude that the probability of drawing a white face is 1-P(B). Since P(B)=P(1)+P(2) therefore P(B)=13+13=23. Likewise we can do this P(white face)=1-23=13.

Bayes' theorem

Given that the shown face is black, the other face is black if and only if the card is the black card. If the black card is drawn, a black face is shown with probability 1. The total probability of seeing a black face is 12; the total probability of drawing the black card is 13. By Bayes' theorem, the conditional probability of having drawn the black card, given that a black face is showing, is

$\frac{1\cdot1/3}{1/2}=2/3.$

It can be more intuitive to present this argument using Bayes' rule rather than Bayes' theorem[3]. Having seen a black face we can rule out the white card. We are interested in the probability that the card is black given a black face is showing. Initially it is equally likely that the card is black and that it is mixed: the prior odds are 1:1. Given that it is black we are certain to see a black face, but given that it is mixed we are only 50% certain to see a black face. The ratio of these probabilities, called the likelihood ratio or Bayes factor, is 2:1. Bayes' rule says "posterior odds equals prior odds times likelihood ratio". Since the prior odds are 1:1 the posterior odds equals the likelihood ratio, 2:1. It is now twice as likely that the card is black than that it is mixed.

Eliminating the white card

Although the incorrect solution reasons that the white card is eliminated, one can also use that information in a correct solution. Modifying the previous method, given that the white card is not drawn, the probability of seeing a black face is 34, and the probability of drawing the black card is 12. The conditional probability of having drawn the black card, given that a black face is showing, is

$\frac{1/2}{3/4}=2/3.$

Symmetry

The probability (without considering the individual colors) that the hidden color is the same as the displayed color is clearly 23, as this holds if and only if the chosen card is black or white, which chooses 2 of the 3 cards. Symmetry suggests that the probability is independent of the color chosen, so that the information about which color is shown does not affect the odds that both sides have the same color.

This argument is correct and can be formalized as follows. By the law of total probability, the probability that the hidden color is the same as the displayed color equals the weighted average of the probabilities that the hidden color is the same as the displayed color, given that the displayed color is black or white respectively (the weights are the probabilities of seeing black and white respectively). By symmetry, the two conditional probabilities that the colours are the same given we see black and given we see white are the same. Since they moreover average out to 2/3 they must both be equal to 2/3.

Experiment

Using specially constructed cards, the choice can be tested a number of times. Let "B" denote the colour Black. By constructing a fraction with the denominator being the number of times "B" is on top, and the numerator being the number of times both sides are "B", the experimenter will probably find the ratio to be near 23.

Note the logical fact that the B/B card contributes significantly more (in fact twice) to the number of times "B" is on top. With the card B/W there is always a 50% chance W being on top, thus in 50% of the cases card B/W is drawn, the draw affects neither numerator nor denominator and effectively does not count (this is also true for all times W/W is drawn, so that card might as well be removed from the set altogether). Conclusively, the cards B/B and B/W are not of equal chances, because in the 50% of the cases B/W is drawn, this card is simply "disqualified".

Notes and references

1. ^ Bar-Hillel and Falk (page 119)
2. ^ Nickerson (page 158) advocates this solution as "less confusing" than other methods.
3. ^ Bar-Hillel and Falk (page 120) advocate using Bayes' Rule.