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# In converting Continuous rate to Discrete, does it matter how often interest is paid?

#### Steve Jobs

##### Active Member
Assume the following:

An asset is quoted at 12% annually with continuous rate.
Interest is paid quarterly.

Is this correct for equivalent rate with monthly compounding?
r = 12 * [ e^(.12/12)) - 1] = 12.06%

Does it matter whether interest is paid quarterly, monthly or annually? What about doing the reverse convert from continuous to discrete?

#### ShaktiRathore

##### Well-Known Member
Subscriber
hi,
let i deposit 1 unit today,
then from monthly compounding my deposit value after 1 yr is if r is the annual rate,
(1+r/12)^12
also from continuous compounding assuming continous rate rc is in annual terms after 1 yr is,
e^rc
pays from both should be equal,
(1+r/12)^12=1.e^rc
(1+r/12)=e^(rc/12)
(r/12)=[e^(rc/12)-1]
r=12*[e^(rc/12)-1]
for n periods compounding,
r=n*[e^(rc/n)-1]
for e.g. n=4(12/3) quarterly, r= 4* [ e^(.12/4)) - 1]
hence it matters whether interest is paid quarterly, monthly or annually
for reverse convert from continuous to discrete, following the same logic as above
(1+r/n)^n=e^rc
ln((1+r/n)^n)=rc
rc=n ln(1+r/n)

thanks

#### Steve Jobs

##### Active Member
Thanks Shakti,

When we say compounded monthly, are we assuming that the interest income is withdrawn each month too?

Otherwise there will be two periods, one period specifying how often the rate is compounded and the second one for how often the interest is actually withdrawn and paid. My confusion is regarding these two periods and whether I should be worry about this in the exam?

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
That's interesting, Steve (to me, because I write so many questions).

With regard to "An asset is quoted at 12% annually with continuous rate. Interest is paid quarterly." Note three timeframes are invoked:
1. Interest paid quarterly (4 per year)
2. The rate curve used to compound or discount (FV or PV more likely) should always be expressed "per annum" which is independent of compound frequency; i.e., even if the "annually" were omitted, we would assume the 12.0% is per annum
3. Compounding frequency is continuous
A modern version of the question is more likely (imo) to rephrase, in a manner typical of Hull, as follows (eg):
"An asset pays interest quarterly and the [spot | zero | discount | swap rate curve] is flat at 12.0% per annum with continuous compounding"
... Note in a carefully phrased question, how we can easily see that purpose of the 12% is to discount to price (or compound forward to an expected future price)

So, as phrased above, the continuous would not be converted;
For example, assume a bond with a 12% quarterly coupon when the spot rate curve is flat at 12.0% per annum with continuous compounding.
It's price is slightly less than par:
($100*12%/4)*exp(-0.25*12%); first quarterly coupon discounted continuously + ($100*12%/4)*exp(-0.50*12%); i.e. coupon paid quarterly but discounted continuously
....
+ $100+($100*12%/4)*exp(-n*12%)
Deposit no. 2: The bank quotes the interest rate on a $1,000 as 9.0% per annum with continuous compounding but the interest is paid monthly. For each deposit, calculate the equivalent discrete rate per annum with monthly compounding. I think in the above cases: a. the equivalent rates will be the same for both deposits b. but there could be differences in the amount of interest received by the customer for one year, dependent on whether the the interest that is paid monthly is being re-invested again or withdrawn by the customer. So the difference is not in the equivalent rates but in deposit terms/conditions and customer decision. #### Steve Jobs ##### Active Member The question was asking for the amount of interest, I changed it to equivalent rate according to the point wanted to raise. Actually, the only similarity now is the interest rate. #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber I can see that I did write a question just like Hull's BT 157.3 Suppose a lender quotes the interest rate on a$1,000 loan as 9.0% per annum with continuous compounding but the interest is actually paid monthly. What are the monthly interest payments?

Hull 4.10 A deposit account pays 12% per annum with continuous compounding, but interest is actually paid quarterly. How much interest will be paid each quarter on a $10,000 deposit? If the question seems awkward (i.e., we could skip the continuous rate and quote the corresponding monthly or annual rate), that's only because the whole point of the question is to test the equivalence-translation from one rate to another. So, the meaning of 9% per annum with continuous compounding is: regardless of how frequently the interest is paid, the terminal value at the end of one year will be$1,000*exp(9%) = $1,094.17; it is also same as saying, regardless of the interest payout frequency, the effective annual return = exp(9%)-1 = 9.417%. Any conversion-to-discrete will assume reinvestment (we generally do assume reinvestment). As below, i input your questions, the first scenario is quarterly interest (first two columns), which implies$22.76 interest paid per quarter , but those interest payments are reinvested, so they accumulate to the same $94.17 at the end of the year. If the interest pays annual (k =1), then the$94.17 is just paid once, at the end of the year. I didn't detail the monthly, but each $7.5282 will get reinvested and at the end of the year, this will also equal$94.17. So the profile of cash flow varies, but the very design of the question means to treat 9.0% continuous as the rate earned regardless of interest payout frequency. Regardless of (k), the terminal value of \$1,094.17 is unchanged, is how i look at this. thanks,

#### Steve Jobs

##### Active Member
Thanks David,

So in the exam, for the equivalent rate questions, I should ignore any details provided regarding the frequency of interest payout because it's assumed that the interest is reinvested. In reality I guess such details should be documented.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
It's safe to assume reinvestment (which is different than assuming we know the reinvestment rate: yield assumes it knows. Reinvestment risk refers to the reality we don't know. But the existence of some re-investment assumption is tantamount, I think, to assuming time value of money, and we always assume TVM).

Yes, details will be documented. If there is a #1 topic upon which GARP has received feedback with respect to specificity, i think it probably is careful specification of compound frequency assumptions. It's one of the more seasoned elements such that they know to give details, thanks,


# Help with vector notation

## Homework Statement

Hi. Its not really a problem but just a general question. When doing graph translations, such as move the parabola x units right or y units up etc, Is it okay to use vector format. So instead of saying move this equation 4 units left, could you write it like this -> <-4,0>

## Homework Equations

I know this is used for translations of shapes but my main question is can it also be used for graphs

## The Attempt at a Solution

I asked my teacher but he just shouted at me for asking him concepts he didnt talk about.

Related Precalculus Mathematics Homework Help News on Phys.org
fresh_42
Mentor

## Homework Statement

Hi. Its not really a problem but just a general question. When doing graph translations, such as move the parabola x units right or y units up etc, Is it okay to use vector format. So instead of saying move this equation 4 units left, could you write it like this -> <-4,0>

## Homework Equations

I know this is used for translations of shapes but my main question is can it also be used for graphs

## The Attempt at a Solution

I asked my teacher but he just shouted at me for asking him concepts he didnt talk about.
A graph $\mathcal{G}$ of usually a function $x \mapsto f(x)$ is a set of points: $\mathcal{G}=\{\,(x,f(x)\,|\,x \in \operatorname{domain}(f)\,\}$. This is a subset of $\mathbb{R}^2$ and to move it means to move it within this plane $\mathbb{R}^2$. Thus you need a direction in which it is moved and a distance by which it is moved. But this exactly defines a vector $v=(v_x,v_y)$. So the moved object will be $v+\mathcal{G}=\{\,(x+v_x,f(x)+v_y)\,|\,x \in \operatorname{domain}(f)\,\}$. So in a sense this is even better than just to say move by $"v_0"$ in direction $"x"$ or $"y"$ since you allow more than two directions.

However, whether you should argue with your teacher is a complete different question. It's not always important to be right. Sometimes it's smarter not to be.

Last edited:
A graph $\mathcal{G}$ of usually a function $x \mapsto f(x)$ is a set of points: $\mathcal{G}=\{\,(x,f(x)\,|\,x \in \operatorname{domain}(f)\,\}$. This is a subset of $\mathbb{R}^2$ and to move it means to move it within this plane $\mathbb{R}^2$. Thus you need a direction in which it is moved and a distance by which it is moved. But this exactly defines a vector $v=(v_x,v_y)$. So the moved object will be $v+\mathcal{G}=\{\,(x+v_x,f(x)+v_y\,|\,x \in \operatorname{domain}(f)\,\}$. So in a sense this is even better than just to say move by $"v_0"$ in direction $"x"$ or $"y"$ since you allow more than two directions.

However, whether you should argue with your teacher is a complete different question. It's not always important to be right. Sometimes it's smarter not to be.

So is it okay to write like this ? If i wrote like this In a paper would it be correct ?

fresh_42
Mentor
So is it okay to write like this ? If i wrote like this In a paper would it be correct ?
I'm not quite sure what you mean by this. That's why I wrote the formulas, which are correct. 4 to the left would be $v_x=-4$ and $v_y=0$. It would be correct, if it would be o.k. depends on a lot of human factors. I wouldn't argue just to be right. 4 to the left is usually as good.

Ray Vickson
If I were your teacher, what I would prefer to see are either of the following (assuming you are graphing a parabola of the form $y = a + b x^2$). If you want to move the whole graph $A$ units to the right (to the left if $A < 0$) and $B$ units up (down if $B < 0$), you could either say that (i) the vertex moves from $(a,0)$ to $(a+A, 0+B)$ but the shape remains unchanged; or (ii) explain that the new graph has equation $y-B = a + b(x-A)^2 \Rightarrow y = a+B+b(x-A)^2.$ Since you are plotting graphs in two-dimensional cartesian coordinate systems, I don't think you could be yelled at for doing it using method (i). However, my personal preference would be that you show your understanding by using method (ii).


# Toronto Math Forum

## APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Yifei Hu on February 21, 2022, 11:29:41 AM

Title: Textbook Chapter 2.5 Example 2
Post by: Yifei Hu on February 21, 2022, 11:29:41 AM
Hi Professor,

I think one coefficient is wrong in the solution.

In this problem the solution to the inhomogeneous equation with homogeneous initial condition should start with coefficient $\frac{1}{2c} = \frac{1}{6}$. After cancelling the 36 in the $f(x,t)$, we should arrive at $6\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$ instead of $3\int_0^t \int_{x-3(t-t')}^{x+3(t-t')}\frac{1}{t^2+1}dx'dt'$.

Can you help me confirm if I missed something or the textbook has a typo?


# Function, f, with domain (-infinity, + infinity)

1. Jan 30, 2010

### ajassat

I was working on the following problem from a textbook. The textbook has no answer. I have included my solution - I am not sure whether it is correct Any ideas and or solutions? (guidance)

Question:
Suppose that f is any function with domain (-infinity, +infinity)

a) Does the function g defined by g(x) = f(x) + f(-x) have any special symmetry?

My solution

A function f which satisfies:
f(-x) = -f(x) is symmetrical about the horizontal axis. This is also:
f(x) + f(-x) = 0
or we could substitute values of x giving non zero solutions for:
f(x) + f(-x) = g(x) (some other function)

Therefore g(x) = f(x) + f(-x) has symmetry about horizontal axis?

Any guidance appreciated.

Regards,
Adam

2. Jan 30, 2010

### JSuarez

No. f(x)+f(-x) is an even function: it's symmetric about the vertical axis.

3. Jan 30, 2010

### some_dude

Try and work backwards: take an arbitrary function $$g$$ defined on $$\mathbb{R}$$ and try to create a function $$f$$ s.t. $$f(x) + f(-x) = g(x)$$ for all x. What assumptions must be made about $$g$$ in order to do this?

4. Jan 30, 2010

### uart

Just use the definition of odd and even functions for this type of problem. Without making any assumptions about f(x) (apart from the given domain) simply write g(-x) and compare it with g(x). That is,

g(x) = f(x) + f(-x) : given

So g(-x) = f(-x) + f(-(-x))
= f(-x) + f(x)
= g(x).

And g(-x) = g(x) is the definition of an even function.

Last edited: Jan 30, 2010
5. Jan 31, 2010

### ajassat

I figured it was an even function after I had posted this.

Thanks to both some_dude and uart for posting two different approaches to the problem. I understand this problem a lot better and have solved some similar now.

Thanks again,
Adam

6. Feb 1, 2010

### Landau

To test yourself, a new question: do the same for g(x)=f(x)-f(-x).

7. Feb 2, 2010

### HallsofIvy

Staff Emeritus
By the way, f(x)+ f(-x) and f(x)- f(-x) are almost the "even and odd parts" of the function f(x). e(x)= (f(x)+ f(-x))/2 and o(x)= (f(x)- f(-x))/2 are even and odd functions that add to give f(x). Of course, if f(x) is an even function to begin with, o(x) will be 0 and e(x)= f(x). If f(x) is an odd function, e(x)= 0 and o(x)= f(x).

The exponential function, ex, is neither even nor odd. Its even and odd parts are $\frac{e^x+ e^{-x}}{2}= cosh(x)$ and $\frac{e^x- e^{-x}}{2}= sinh(x)$.

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# Homework Help: Linear Algebra Money Question

1. Sep 14, 2012

### bleedblue1234

1. The problem statement, all variables and given/known data

I have 32 bills in my wallet in the denominations $1,$5, and $10, worth$100 in total. How many of each denomination do I have?

2. Relevant equations

A= # $1 bills B= #$5 bills
C= # $10 bills A+B+C = 32 1A+5B+10C = 100 3. The attempt at a solution So I attempted to solve for C in terms of A and B in terms of A but I'm getting nowhere. 2. Sep 14, 2012 ### E_M_C Hi bleedblue1234, You can only solve for n variables when you have n linearly independent equations. In this case, you have 3 variables and 2 linearly independent equations, so you're one equation short. But if you choose a value of zero for A, B or C then you reduce the problem to 2 variables and 2 linearly independent equations. What do you get when you try out the different combinations? Be careful: There is more than one solution. 3. Sep 14, 2012 ### azizlwl You can narrow the selection.$1 can only be in a group of 5.

4. Sep 14, 2012

### HallsofIvy

This not, strictly speaking, a "linear algebra" problem, but a "Diophantine equation" because the "number of bills" of each denomination must be integer. Letting "O", "F", and "T" be, respectively, the number of "ones", "fives" and "tens", we must have O+ F+ T= 32 and O+ 5F+ 10T= 100. Subtracting the first equation from the second, 4F+ 9T= 68.
Now you can use the standard "Eucidean algorithm" to find all possible integer values for F and T and then find O.

Last edited by a moderator: Sep 14, 2012
5. Sep 14, 2012

### Ray Vickson

You can solve for A and B in terms of C, just by solving the two simple equations
A + B = 32 - C
A + 5C = 100 - 10C.

Now you can plug in C = 0, 1, 2, ... and see which values (if any) give you non-negative integer values of A and B.

RGV

6. Sep 14, 2012

### HallsofIvy

Oh, well- if you want to do it the easy way!

7. Sep 14, 2012

### bleedblue1234

Ya I just set the equations equal and restricted B and C to be natural numbers and just checked which B would give me the correct C, which in tern gave me the correct A. Thank you.

8. Sep 14, 2012

### E_M_C

Be careful with that. Most people (in my experience) define the natural numbers as N = {1, 2, 3, ...} which doesn't include zero. So if you're restricting B and C to natural numbers, as defined above, you may be cheating yourself out of a solution. As I mentioned earlier, there is more than one solution.


# Finding & Plotting equation of hyperbola given foci, and difference in distances between them.

I have to plot the hyperbola (3 of them actually) in MATLAB, and so it'd be good if I could find some sort of general formula.

The foci do not necessarily have to be on the axes (e.g. $(5,3)$ and $(4,8)$ can also be the foci). I have the difference in distances between them at a point which lies on the hyperbola. Will there be a general equation describing this hyperbola? If yes, what will it be? (e.g. say the points are $(x_1,y_1)$ & $(x_2,y_2)$ and difference in distance is $d$).

I know that I can shift and rotate the axes so that I get what I want, but I wanted to know if there was a better method, since while plotting i'll have to reconvert the values into the normal axes.

I'm going to try trilateration, which is why i'm asking for the equation.

-
The answers to this question math.stackexchange.com/questions/31756/… provide what you are looking for, I think. – Ellie Kesselman Jan 22 '12 at 15:16
Actually, one of the answers specifically addresses the case where the foci are not on the axes, and just about everything else possible for a hyperbola. It is a remarkably fine answer math.stackexchange.com/a/31761/9260 – Ellie Kesselman Jan 22 '12 at 15:21
@FeralOink - Thanks! :) – Khushman Patel Jan 22 '12 at 17:16
The equation of the hyperbola is an implicit one. Can't you plot implicit equations? – Américo Tavares Jan 22 '12 at 22:54
Simply put, if you need to use parametric equations as opposed to an implicit Cartesian equation, rotate/shift is really the only way to go. Otherwise, if your computing environment supports implicit plotting, then even the obvious approach of using the distance formula works nicely. – J. M. Jul 24 '12 at 21:49

say the points are $(x_1,y_1)$ & $(x_2,y_2)$ and difference in distance is $d$

Could you use the literal translation of this statement into algebra? $$\left|\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_2)^2+(y-y_2)^2}\right|=d$$ or $$\left(\sqrt{(x-x_1)^2+(y-y_1)^2}-\sqrt{(x-x_2)^2+(y-y_2)^2}\right)^2=d^2$$ or $$(x-x_1)^2+(y-y_1)^2+2\sqrt{(x-x_1)^2+(y-y_1)^2}\sqrt{(x-x_2)^2+(y-y_2)^2}+(x-x_2)^2+(y-y_2)^2=d^2$$

-
Typically one does $\sqrt{(x-x_1)^2+(y-y_1)^2} = \sqrt{(x-x_2)^2+(y-y_2)^2} \pm d$, so that (after squaring both sides, simplifying, and squaring again) one gets a quadratic equation, not a quartic one. – Rahul Feb 20 '12 at 8:42
@ℝⁿ. - Taking the +d case, squaring both sides gives $$(x-x_1)^2 + (y-y_1)^2 = (x-x_2)^2 + (y-y_2)^2 + 2d\sqrt((x-x_2)^2 + (y-y_2)^2) + d^2$$. Only a little bit of simplification seems possible: $$-2xx_1 + x_1^2 -2yy_1 + y_1^2 = -2xx_2 + x_2^2 -2yy_2 + y_2^2 + 2d\sqrt((x-x_2)^2 + (y-y_2)^2) + d^2$$. Now from here if you move everything but the $$2d\sqrt((x-x_2)^2 + (y-y_2)^2)$$ to the left side so you can square both sides again to get rid of the sqrt, I count 9 additive terms on the left that all have to be multiplied by each other... is that really what needs to be done? – David Doria Feb 20 '13 at 18:36
@David: Well, you ought to group them into the form $ax+by+c$ before squaring. – Rahul Feb 20 '13 at 18:42
@ℝⁿ. See my answer below - any more suggestions for improvement :)? – David Doria Feb 20 '13 at 20:57

Perhaps someone can verify the result below, or suggest cleaner ways to expand/write this?

Starting from the definition of a hyperbola as the locus of points X = (x,y) that have a constant distance difference to two points (the foci, A = (a_x, a_y) and B = (b_x, b_y)): $$|A-X| - |B-X| = d$$

We can write this (at least half of the solution) as:

$$\sqrt{(a_x-x)^2 + (a_y-y)^2} - \sqrt{(b_x-x)^2 + (b_y-y)^2} = d$$

There is tremendous manipulation necessary to write this equation in general quadratic form ($Ax^2 + By^2 + Cxy + Dx + Ey + F = 0$):

Move one of the square roots to the other side of the equation so that when we square both sides we will not eventually introduce quartic terms: $$\sqrt{(a_x-x)^2 + (a_y-y)^2} = \sqrt{(b_x-x)^2 + (b_y-y)^2} + d$$

Square both sides: $$(a_x-x)^2 + (a_y-y)^2 = (b_x-x)^2 + (b_y-y)^2 + 2d \sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$

Expand: $$a_x^2 -2xa_x + x^2 + a_y^2 - 2a_y y + y^2 = b_x^2 - 2xb_x + x^2 + b_y^2 -2b_y y + y^2 + 2d \sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$

Cancel $x$ and $y$ quadratic terms (that occur on both sides of the equation): $$a_x^2 -2xa_x + a_y^2 - 2a_y y = b_x^2 - 2xb_x + b_y^2 -2b_y y + 2d \sqrt{(b_x-x)^2 + (b_y-y)^2} + d^2$$

Move everything not connect to the square root to the left side so we can eventually square both sides again to get rid of the square root: $$a_x^2 - 2xa_x + a_y^2 - 2a_y y - b_x^2 = 2xb_x - b_y^2 + 2b_y y - d^2 = 2d \sqrt{(b_x-x)^2 + (b_y-y)^2}$$

Group terms on the left side into the form Jx + Ky + L to make squaring easier: $$\left((-2a_x + 2b_x)x + (-2a_y + 2b_y)y + (a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)\right) = 2d \sqrt{(b_x-x)^2 + (b_y-y)^2}$$

Assign $J = -2a_x + 2b_x$, $k = -2a_y + 2b_y$, $L = a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2$ so this can be written as

$$Jx + Ky + L = 2d \sqrt{(b_x-x)^2 + (b_y-y)^2}$$

Square both sides:

$$(Jx + Ky + L)^2 = 4d^2 \left((b_x-x)^2 + (b_y-y)^2\right)$$

$$J^2x^2 + JxKy + JxL + KyJx + K^2y^2 + KyL + LJx + LKy + L^2 = 4d^2 \left(b_x^2 - 2xb_x +x^2 + b_y^2 - 2yb_y + y^2\right)$$

Combine terms on the left, and multiply through on the right:

$$J^2x^2 + K^2y^2 + 2JxKy + 2JxL + 2KyL + L^2 = 4d^2 b_x^2 - 8d^2xb_x + 4d^2x^2 + 4d^2b_y^2 - 8d^2yb_y + 4d^2y^2$$

Combine all terms ($x^2, y^2, xy, x, y$):

$$(J^2-4d^2)x^2 + (K^2-4d^2)y^2 + (2JK)xy + (2JL+8d^2b_x)x + (2KL + 8d^2b_y)y + (L^2 - 4d^2 b_x^2 - 4d^2 b_y^2) = 0$$

Substituting back in the values of J, K, and L:

$$((-2a_x + 2b_x)^2-4d^2)x^2 + ((-2a_y + 2b_y)^2-4d^2)y^2 + (2(-2a_x + 2b_x)(-2a_y + 2b_y))xy \\+ (2(-2a_x + 2b_x)(a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)+8d^2b_x)x + (2(-2a_y + 2b_y)(a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2) + 8d^2b_y)y \\+ ((a_x^2 + a_y^2 - b_x^2 - b_y^2 - d^2)^2 - 4d^2 b_x^2 - 4d^2 b_y^2) = 0$$

Expand: $$(4a_x^2 - 8a_x b_x + 4b_x^2)x^2 + \\ (4a_y^2 - 8a_y b_y + 4b_y^2)y^2 + \\ (8a_xa_y - 8a_xb_y - 8a_yb_x + 8b_xb_y)xy +\\ (-4a_x^3 -4a_xa_y^2 + 4 a_x b_x^2 + 4 a_x b_y^2 + 4 a_x d^2 + 4b_xa_x^2 + 4b_xa_y^2 + 4b_x^3 - 4b_xb_y^2 + 4b_xd^2)x +\\ (-4a_ya_x^2 - 4a_y^3 + 4 a_y b_x^2 + 4 a_y b_y^2 + 4 a_y d^2 + 4b_ya_x^2 + 4b_ya_y^2 - 4b_yb_x^2 - 4b_y^3 + 4b_yd^2)y +\\ a_x^4 + a_x^2 a_y^2 - a_x^2b_x^2 - a_x^2b_y^2 - a_x^2d^2\\ + a_y^2 a_x^2 + a_y^4 - a_y^2b_x^2 - a_y^2b_y^2 - a_y^2 d^2\\ - b_x^2 a_x^2 - b_x^2 a_y^2 + b_x^4 + b_x^2 b_y^2 -3 b_x^2 d_2\\ - b_y^2 a_x^2 - b_y^2a_y^2 + b_y^2b_x^2 + b_y^4 + b_y^2 d^2\\ - d^2 a_x^2 - d^2 a_y^2 + d^2 b_x^2 -3 d^2 b_y^2 + d^4\\ = 0$$

-
A slightly cleaner way would be to keep everything in vector notation. $\sqrt{\lVert\mathbf a-\mathbf x\rVert^2} = \sqrt{\lVert\mathbf b-\mathbf x\rVert^2} + d$, $\lVert\mathbf a\rVert^2-2\mathbf a\cdot\mathbf x+\lVert\mathbf x\rVert^2=\lVert\mathbf b\rVert^2-2\mathbf b\cdot\mathbf x+\lVert\mathbf x\rVert^2+2d\sqrt{\lVert\mathbf b-\mathbf x\rVert^2}+d^2$, and so on. This has the additional advantage that the derivation doesn't change at all in higher dimensions. – Rahul Mar 6 '13 at 11:33
In this derivation we assumed |A−X|>|B−X|. Even though we only solved "half" of the absolute value equation given by the definition of the hyperbola, we still seem to obtain the entire hyperbola. What is the relation of the object obtained if |B−X|>|A−X| to the one we derived above? – David Doria Mar 18 '13 at 17:04


# Probability distribution of number of times required to roll all of a certain set of k numbers?

Consider a 12-sided fair die. What is the distribution of the number T of rolls required to roll a 1, a 2, a 3, and a 4?

Taking inspiration from the Coupon Collector's Problem, I believe that the expected number of rolls to achieve the goal would be

$$E[T] = \sum\limits_{i=0}^3 \frac{12}{4-i} = 25$$

Similarly, the variance would be

$$Var[T] = \sum\limits_{i=0}^3 \frac{1-\frac{4-i}{12}}{(\frac{4-i}{12})^2} = 180$$

But applying Chebyshev here does not yield very useful bounds. My question is therefore, how would you compute, for example, $P(T=16)$ or $P(T<30)$?

Ideally this could be generalized to a set of k required numbers, not just 4 as in the example.

-

Personally I would call $P(n,b,k,j)$ the probability that after rolling $n$ dice with $b$ sides, $j$ of the target $k$ sides had been found, with the formula

$$P(n+1,b,k,j)= \frac{b-k+j}{b} P(n,b,k,j) + \frac{k+1-j}{b} P(n,b,k,j-1)$$

starting from $P(0,b,k,j)=0$ for $j \not = 0$ and $P(0,b,k,0)=1$.

Then $$\Pr(T \le t) = P(t,b,k,k)$$ and $$\Pr(T = t) = P(t,b,k,k)-P(t-1,b,k,k).$$

So in your example it is not difficult to calculate $\Pr(T = 16) \approx 0.0380722$ and $\Pr(T \le 30) \approx 0.7305294$. Your expected value of $T$ and variance appear to be correct.

-
Thanks for the response. A recursive definition is a good idea. I feel though that in your response, in the recurrence relation, one of the P's needs to have a j-1. – sffc Dec 27 '12 at 13:31
@vote539: your final point is correct and I will edit – Henry Dec 27 '12 at 13:48
Thanks for the clarification! The results yielded by your solution match those yielded by a simulation of the dice rolls. A note to anyone who wants to implement this: the recursion stack grows very quickly, so it is necessary to use dynamic programming; that is, cache the results of the P function and only perform the computation when there is no cached value. – sffc Dec 27 '12 at 19:46
Or you could use a spreadsheet – Henry Dec 27 '12 at 23:26
Our answers agree on $\textsf{Pr}(T=16)$ and $\textsf{Pr}(T\le30)$. (The example in the question was $\textsf{Pr}(T\lt30)$.) – joriki May 18 at 19:15

You can get this distribution by conditioning on the number $N$ of rolls up to $4$ required:

\begin{align} \textsf{Pr}(T\le t) &=\left(\frac23\right)^t\sum_{n=0}^t\binom tn2^{-n}\textsf{Pr}(N\le n) \\ &=\left(\frac23\right)^t\sum_{n=0}^t\binom tn2^{-n}\frac{4!}{4^n}\left\{n\atop4\right\} \\ &=\left(\frac23\right)^t\sum_{n=0}^t\binom tn8^{-n}\sum_{j=0}^4(-1)^j\binom4jj^n \\ &=\left(\frac23\right)^t\sum_{j=0}^4(-1)^j\binom4j\sum_{n=0}^t\binom tn\left(\frac j8\right)^n \\ &=\left(\frac23\right)^t\sum_{j=0}^4(-1)^j\binom4j\left(1+\frac j8\right)^t \\ &=\sum_{j=0}^4(-1)^j\binom4j\left(\frac23+\frac j{12}\right)^t \\ &=1-4\left(\frac{11}{12}\right)^t+6\left(\frac56\right)^t-4\left(\frac34\right)^t+\left(\frac23\right)^t\;, \end{align}

where $\left\{n\atop4\right\}$ is a Stirling number of the second kind and the distribution $P(N\le n)$ is given at Probability distribution in the coupon collector's problem.

The expectation comes out right as

\begin{align} E[T]&=\sum_{t=0}^\infty\textsf{Pr}(T\gt t)\\ &=\sum_{t=0}^\infty\left(4\left(\frac{11}{12}\right)^t-6\left(\frac56\right)^t+4\left(\frac34\right)^t-\left(\frac23\right)^t\right)\\ &=12\left(4\cdot\frac11-6\cdot\frac12+4\cdot\frac13-1\cdot\frac14\right)\\ &=25\;. \end{align}

Here are plots of the cumulative distribution function and the probability mass function.

$$\textsf{Pr}(T=16)=\textsf{Pr}(T\gt15)-\textsf{Pr}(T\gt16)=\frac{293289532268461}{7703510787293184}\approx3.8\%$$

and

$$\textsf{Pr}(T\lt30)=\textsf{Pr}(T\le29)=\frac{292029927548835623394780280045}{412111655902378135987760922624}\approx70.9\%\;.$$

For general $k$ instead of $4$ numbers required and general $m$ instead of $12$ numbers on the die, the result is

$$\textsf{Pr}(T\le t)=\sum_{j=0}^k(-1)^j\binom kj\left(1-\frac{k-j}m\right)^t\;.$$

-

I think it's the sum of four different geometric distributions.

For example, you start off and you haven't rolled any of the numbers. The probability of rolling one of them is 4/12 so you start a geometric distribution with that parameter. Then you suddenly roll one of them. Now you need to roll one of the other three, each roll has a probability of 3/12 of getting one of them, so it's another geometric distribution.

By this logic I think T ~ Geo(4/12) + Geo(3/12) + Geo(2/12) + Geo(1/12) which of course is probably some hideous ditribution define in terms of convolutions (or you know, could miraculously be nice and of closed-form, I don't know what.)

-
This is sort-of what I was thinking, but what are the next steps to get to the point when you can actually compute the probability densities? – sffc Dec 27 '12 at 10:01


# Math Help - Meeting point between two vehicles

1. ## Meeting point between two vehicles

Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.
Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.
Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

a) How long does it take driver B to catch up with driver A?
b) How long have they driven when they meet?
c) How fast do each driver drive when they meet?

I have tried the four formulas:

v = at+v0

s = 1/2 (v+v0) * t

s = 1/2 a * t^2 + v0 * t

2 * a * s = v^2 - v0^2

With no luck at all. I am confident that some calculations might be wrong.

All help is appreciated, hopefully a quick response.

2. ## Re: Meeting point between two vehicles

Originally Posted by roahau
Driver A starts his car and accelerates with a constant accelerasion of 0.8 m/s^2.
Driver B starts and follows on a motorcycle from the same spot, 6,0 sec later.
Driver B has got a constant acceleration of 1.8 m/s^2. (Neither of them accelerate higher than 80 km/h)

a) How long does it take driver B to catch up with driver A?
b) How long have they driven when they meet?
c) How fast do each driver drive when they meet?
for driver A ...

$v_A = 0.8t$

$s_A = 0.4t^2$

for driver B ...

$v_B = 1.8(t-6)$

$s_B = 0.9(t-6)^2$

for (a) and (b) , set $s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $(t-6)$. once you get the time, find the value of $s_A$ and $s_B$ (which should be the same, right?)

for (c) , determine the values of $v_A$ and $v_B$ at the time found above.

3. ## Re: Meeting point between two vehicles

Originally Posted by skeeter
for driver A ...

$v_A = 0.8t$

$s_A = 0.4t^2$

for driver B ...

$v_B = 1.8(t-6)$

$s_B = 0.9(t-6)^2$

for (a) and (b) , set $s_A = s_B$ , solve for t to find the time driver A's run ... don't forget that driver B's run is $(t-6)$. once you get the time, find the value of $s_A$ and $s_B$ (which should be the same, right?)

for (c) , determine the values of $v_A$ and $v_B$ at the time found above.
How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity

4. ## Re: Meeting point between two vehicles

Originally Posted by roahau
How am I supposed to determine t without the values of vA and vB ? I can't use the first formula without time, and I can't use the time forumula without the final velocity
when the two vehicles meet, they are at the same position.

$s_A = s_B$

$0.4t^2 = 0.9(t-6)^2$

solve for t , then proceed as I stated in my previous post.


# $S_n$ is a group

This is an extremely simple fact, but I want to make sure that the assertions I am making are valid. In particular, I assert several times that a function is a permutation if and only if it is a bijection from a finite set to itself. If this is fine, I believe my arguments below are likely fine. I cannot think of a good counterexample to this.

There are four requirements we need to verify: closed under product operations, associative, has an identity, and closed under inverses.

Closed under product operation: An element of $$S_n$$ is a permutation of the elements $$1, 2, \ldots, n$$. This is a bijection $$\alpha: \{1, 2, \ldots, n\} \to \{1, 2, \ldots, n\}$$. A composition of these bijections, $$\alpha$$ and $$\beta$$, is another bijection from $$\{1, 2, \ldots, n\}$$ to $$\{1, 2, \ldots, n\}$$, and hnce $$\alpha \beta$$ is another permutation of the elements $$\{1, 2, \ldots, n\}$$.

Associativity. An element of $$S_n$$ is a permutation, i.e., a function from a set to itself, and function composition can easily be shown to be associative. For simplicity, take $$f, g, h$$ to be well-defined functions that map $$A$$ to $$A$$. For any $$a \in A$$, we have: $$((f \circ g) \circ h)(a) = (f \circ g)(h(a)) = f(g(h(a))$$ $$(f \circ (g \circ h))(a) = f((g \circ h)(a)) = f(g(h(a)).$$ Hence, composing permutations is associative because permutations are just functions.

Identity. The identity permutation, $$\text{id}$$, sends every element to itself. For any $$\alpha \in S_n$$ and any $$x \in \{1, 2, 3, \ldots,n\}$$, we have: \begin{align*} (\alpha \circ \text{id})(x) = \alpha(\text{id}(x)) = \alpha(x) \\ (\text{id} \circ \alpha)(x) = \text{id}(\alpha(x)) = \alpha(x). \end{align*} Hence, the identity permutation is the identity element in the group.

Inverses. Let $$\sigma \in S_n$$. Then, $$\sigma$$ is a bijection from $$\{1, 2, \ldots, n\}$$ to $$\{1, 2, \ldots, n\}$$. A function is bijective if and only if it is invertible. Hence, another bijection, $$\sigma^{-1}$$, also exists. Since this is a bijection from a set of finite elements to itself, it is also a permutation of $$\{1, 2, \ldots, n\}$$, and hence lives in $$S_n$$.

• Yes, it is fine. Mar 9 '20 at 23:33
• "There are four requirements we need to verify". I believe the things went historically the other way around: after having realized that the set of bijections on a set does fulfill the four properties, where "Inverses" is actually the qualifying one (e.g. associativity holds for any map on a set), someone thought to define "group" an algebric structure patterned upon the "4 properties" of the bijections.
– user750041
Mar 10 '20 at 11:24

With regards to this problem, yes, it is true that a function is a permutation if and only if it is a bijection from a finite set to itself. This is true essentially by definition of the word "permutation". In other contexts, the definition of permutation might be extended to allow bijections from any arbitrary (finite or infinite) set to itself (e.g. $$x\mapsto 1/x$$ is a bijection from $$(0,\infty)$$ to itself, and so is a permutation of $$(0,\infty)$$), but of course this is not relevant here since we are only looking at finite sets $$\{1,2,\dots,n\}$$.


1. ## Chinese Remainder Theorem

For the question,

Find the smallest x which solves the system of congruence equations:

x =+ 3 mod 15
x =+ 6 mod 224, where =+ stands for congruence modulo operation.

I gave an argument oriented solution as follows.

Any multiple of 224 is even.
So, 6+ any multiple of 224 is even.
=> x is even.

Also, any multiple of 15 ends with 5 or 0.
3+any multiple of 15 ends with 8 or 3.

Since X is even, the number should end with 8.
Since X = A number ending with 8,
Also, X = A multiple of 224 + 6
This means the multiple of 224 ends with 2.

Checking the multiples of 224, we get 224, 448, 672.

The smallest one is 672 ==> X is 678.

This solves both the equations as 678 = 224*3 +6
= 45*15 +3

My friend argued about a general solution for the same problem using Chinese Remainder Theorem. I browsed results for the same in various sites.
I find only a general solution found in most cases.

Can anyone describe how to get the particular solution for this problem through Chinese Remainder Theorem?

Also, please mention if there are any mistakes in my argument-oriented solution.

2. yes you can....

so we have

3 mod 15
6 mod 224

gcd(224,15)=1 so we use chinese thm.

first thing we need is integers a and b s.t...

a15+b224=1

and after we find a and b using euclid's algorithm we compute

6a15+3b224 (mod 15.224=3360)

euclid's algorithm gives a=15 and b=-1

so 6(15)15+3(-1)224=678 mod 3360

and adding and subtracting 3360 would give you many solutions.

you can also use this method to solve many problems which your method would be increasingly difficult to use for example....

Find a number which is 3 mod 7, 2 mod 5 and 1 mod 2.

7,5 and 2 are coprime so use chinese thm...

first concentrate on 3 mod 7 and 2 mod 5

find a,b so that a7+b5=1

a=-2 and b=3 using euclid

compute 2a7+3b5 (mod 7.5=35)=2(-2)7+3(3)5=17 mod 35

now do the same with 17 mod 35 and 1 mod 2

i need to compute 1a35+17b2 where a35+b2=1

a=1 and b=-17

so 1a35+17b2=17 mod 70=87mod70

Note: the reason why we need the gcd to be 1 is because the chinese remainder theorem is based on this.

but you have a nice little argument there, nice one.

3. Here's a similar argument:

x= 6 mod 224 so x= 224k+ 6 for some integer k. But 224= 14 mod 15 so that is the same as x= 14k+ 6 mod 15. We have x= 14k+ 6= 3 mod 15 or 14k= -3 mod 15 so 14k= 15j- 3 for some integer j. That is, 15j- 14k= 3. It is obvious that 15- 14= 1 (shortcutting the Euclidean division algorithm!) so 15(3)- 14(3)= 3. k= j= 3 is a solution. Taking k= 3, x= 224(3)+ 6= 678.

But if you can come up with a valid argument of your own, that's always best.

4. @Krahl
Thanks for your descriptive reply for the question; I get the solution part right till the general solution; You had mentioned that adding and subtracting 3360 would yield other solutions; It is this part where I do mistakes some times.

Can you please throw some more light on one or two of the particular solutions?

MAX,

5. @HallsofIvy

That's another interesting aliter for the same problem; Good one

6. ## CRT

I was just thinking it might actually help you to see the actual proof of the theorem so you can see why these solution methods actually work. I am not sure if you are familiar with rings and ideals and stuff, so I will leave it in terms of the integers, but if you are interested, the same proof idea works in general.

You assume that (m,n)=1 (this is enough to show in a ring (m)+(n)=R).
Then you have the following isomorphism:
$\frac{\mathbb{Z}}{mn\mathbb{Z}}\cong \frac{\mathbb{Z}}{m\mathbb{Z}} \times \frac{\mathbb{Z}}{n\mathbb{Z}}$

Here is the isomorphism:
$\phi(x)=(x$ (mod m), $x$ (mod n)).
It is a homomorphism:
$\phi(x+y)=(x+y$ (mod m), $x+y$ (mod n)) $=(x$ (mod m), $x$ (mod n)) $+(y$ (mod m), $y$ (mod n))= $\phi(x)+\phi(y)$.

It is pretty clear that $ker(\phi)=mn\mathbb{Z}$

So it remains to show that it is surjective, and this is the key part of the argument for what you are looking for.

Let $(a$ (mod m), $b$ (mod n)) be an arbitrary element of $\frac{\mathbb{Z}}{m\mathbb{Z}} \times \frac{\mathbb{Z}}{n\mathbb{Z}}$

This is basically saying we need some integer that will satisfy
$x\equiv a$ (mod m) and
$x\equiv b$ (mod n).

So this is the trick. Because we know (m, n) by the euclidean algorithm, there exists $s,t \in \mathbb{Z}$ such that $sm+tn=1$.

Just a side note: In the more general setting you say that the ring contains a unit and since we have (m)+(n)=R, there exists $i\in (m)$ and $j\in (n)$ such that $i +j =1$. Actually the ring need not even be a PID, it is true for any ideals that are comaximal in a commutative ring.

So back to the surjectivity. We want to sort of cross multiply here and take the value $bsm+atn$ and see that it maps where we want it.

$\phi(bsm+atn)=(bsm+atn$ (mod m), $bsm+atn$ (mod n)) $=(atn$ (mod m), $bsm$ (mod n))= $(a(1-sm)$ (mod m), $b(1-tn)$ (mod n))= $(a-asm)$ (mod m), $b-btn$ (mod n))= $(a)$ (mod m), $b$ (mod n)).

So this proves the isomorphism. This shows you $bsm+atn$ (mod mn) is the number you want to pick to solve the system of equations above.

The fact that the kernel is $mn\mathbb{Z}$ just tells you that you can actually just add kmn for any integer k to your solution and you will get all the solutions to this system of equations.

7. Originally Posted by MAX09
@Krahl
Thanks for your descriptive reply for the question; I get the solution part right till the general solution; You had mentioned that adding and subtracting 3360 would yield other solutions; It is this part where I do mistakes some times.

Can you please throw some more light on one or two of the particular solutions?

MAX,
Hi max

not sure what part of the solution you're asking for but i'll try writing a few things on the solutions.

well you agree that 678 is a solution right?

Now the useful thing about using the chinese rem thm is that you end up not just with the answer 678 but also mod 3360.

This mean that any number which is modulo 3360 would satisfy your equations,

for example if we add and subtract 3360 we get the following alternative solutions;

........, -2682,678,4038,7398.......,

Now this means that all these numbers are 678 modulo 3360
i.e. have remainder 678 when divided by 3360.

so 4038=3360+678, 7398=3360*2+678 etc...

And the theorem claims that all these solutions also satisfy, so let's check them;

7398=15*493+3 so it is 3 mod 15

7398=224*33+6 so it is 6 mod 224

That is why the question asks for the smallest positive solution.

I did the same with the other example and got the solutions 17 mod 70 and 87 mod 70.

Also notice this step i did;

6a15+3b224

its 6 mod 224 that is why 6 is next to 15. since 224 = 0 mod 224 so that you are left with 6a15 + 0 mod224

and so a*15 must give me 1 mod 224 so that
6a15+0 mod 224=6 mod 224

and same with mod 15.

any problems dont hesitate to post

8. @Gamma

That wasn't exactly the procedure I was looking for. Anyways, thanks for the effort.. It helped understanding the relation the CRT had with isomorphisms. Thanks a bunch !!!

9. ## @krahl

Yes, your posted help me understand the procedure completely I had to get my basics on the Extended Euclidean Algorithm right. The following url helped me.

http://marauder.millersville.edu/~bikenaga/absalg/euc/euclidex.html

Thanks again, Krahl!!

Max


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This thinking process can be very helpful to improve our we are given conditional probabilities in a chain format. %PDF-1.6 %���� possibilities, $GG=(\textrm{girl, girl}), GB, BG, BB$, and $P(GG)=P(GB)=P(BG)=P(BB)=\frac{1}{4}$. I win if probability of $GG$ is higher. Figure 1.27 shows a tree diagram for this problem. Here you can assume that if a child is a girl, her name will be Lilia with probability $\alpha \ll 1$ win. We already know that As it is seen from the problem statement, Second, after obtaining counterintuitive results, you are encouraged to think deeply Let $C_1, C_2,\cdots,C_M$ be a partition of the sample space $S$, and $A$ and $B$ be two events. we know that the family has at least one girl. endstream endobj 1171 0 obj <>/Metadata 74 0 R/Names 1199 0 R/Outlines 196 0 R/PageMode/UseNone/Pages 1166 0 R/StructTreeRoot 233 0 R/Type/Catalog>> endobj 1172 0 obj <>/ProcSet[/PDF/Text]>>/Rotate 0/StructParents 0/Tabs/S/Type/Page>> endobj 1173 0 obj <>stream In Then we have total probability for $A \cap B$: In my town, it's rainy one third of the days. equally likely outcomes: $GG, GB, BG$, thus the conditional probability of $GG$ is I toss a coin repeatedly. We can use the Venn diagram in Figure 1.26 to better On the other hand, if the outcome is $THTHT\underline{HH}$, I win.

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# Can squares of side 1/2, 1/3, 1/4, … be packed into three quarters of a unit square?

My question is prompted by this illustration from Eugenia Cheng’s book Beyond Infinity, where it appears in reference to the Basel problem.

Is it known whether the infinite set of squares of side $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},...$ can be packed into three quarters of a unit square?

(It doesn't seem obvious that the illustrated packing could be continued to infinity.)

• Answers regarding this similar problem might suggest some possible approaches: mathoverflow.net/questions/34145/… – Yoav Kallus Aug 8 '17 at 14:41
• @YoavKallus Thanks! The comment by Timothy Chow referencing Marc Paulus's paper An algorithm for packing squares seems particularly relevant. sciencedirect.com/science/article/pii/S0097316597928363 – Robin Houston Aug 8 '17 at 14:53
• I think this actually should be much easier than the other problem, just because $\frac{\pi^2}{6}-1<\frac{3}{4},$ which gives you a chance of reducing this to a finite computation. In particular if there is an $n$ so that you can pack squares of side lengths $1/2\cdots, 1/(n-1)$ and $n$ right triangles of side length $2/n$, then you can also pack the squares in the question. This is because you can pack squares of side length $1,1/2,\cdots$ in a right triangle of side length $2$. – dhy Aug 8 '17 at 15:04

## 3 Answers

The standard simple proof that $\sum_{n=1}^\infty \frac1{n^2}$ converges is to round each $n$ down to the nearest $2^k$; this rounds each $\frac1{n^2}$ up to the nearest $\frac1{2^{2k}}$. In fact, one gets $2^k$ copies of $\frac1{2^{2k}}$ for each $k$; hence $$\sum_{n=1}^\infty n^{-2} < \sum_{k=0}^\infty 2^k\frac1{2^{2k}} = \sum_{k=0}^\infty \frac1{2^{k}} = 2.$$

This is a sloppy estimate, and one way of improving it is to round each $n$ down to the nearest $2^k$ OR $3\cdot2^{k}$. The first cluster of $\frac1{2^{2k}}$s, comprising a single $\frac1{2^0}$, is unaffected; all subsequent clusters cleave into $2^{k-1}$ copies of $\frac1{2^{2k}}$ and $2^{k-1}$ copies of $\frac1{(3\cdot2^{k-1})^2}$. In this way, we get $$\sum_{n=1}^\infty n^{-2} < 1+\sum_{k=1}^\infty 2^{k-1}\frac1{2^{2k}} + \sum_{k=1}^\infty 2^{k-1}\frac1{(3\cdot2^{k-1})^2} \\ = 1+\sum_{k=1}^\infty \frac1{2^{k+1}} + \frac19\sum_{k=1}^\infty \frac1{2^{k-1}} = 1 + \frac12+\frac29.$$

This sum (minus the irrelevant first term) can be represented geometrically as in the following image, which is reasonably similar to the one you posted. (But the shaded region has area $\frac16+\frac19=\frac5{18}>\frac14$.)

• That's a nice solution. I might add that the rows that start with 1/12, 1/24, 1/48, and so on can be rotated to columns and moved into the shaded region between 1/2 and 1/3 to obtain a solution to the problem in the original post. – Yoav Kallus Aug 8 '17 at 19:55
• You might observe that the "packing area" in your image can be dissected and rearranged (without cutting any interior squares) into (a continuation of) the image in the question. Gerhard "This Makes 'Yes' Appropriately Explicit" Paseman, 2017.08.08. – Gerhard Paseman Aug 8 '17 at 19:58
• Thank you! You are, of course, both correct---my main excuse is that I drew what was easiest for me to draw. (Note how the paper has been strategically folded.) I also wanted to emphasise the two geometric series (one up the left-hand side, the other up the right-hand side). If I weren't a dunce at computer graphics, I'd make an animation to show how the various pieces can be rearranged. – Jeff Egger Aug 8 '17 at 20:53
• Instead of an animation, cut your square in half and rotate the right half. Then take the part that "sticks out" of the unit square and duplicate it to the appropriate place in the unit square, and draw an arrow showing the correspondence. Gerhard "We Can Imagine The Motion" Paseman, 2017.08.08. – Gerhard Paseman Aug 8 '17 at 21:51
• If I didn't overlook anything, this does prove that total area is less than 3/4, but how does this prove that it is actually possible to put the squares? Perhaps there might be some "leftover space between squares" that can't be dealt with... Although my intuition says it is indeed possible, I think this part should also be explicitly addressed in an answer to this question. For example, I doubt this could be done with circles of areas $\frac{1}{n^2}$. – Pedro A Aug 9 '17 at 10:36

Note: This answer is wrong. There are two problems:

1. The claim of Lemma 1 should presumably be read in the context of the global assumption in Paulhus's paper that $w\leq l$. This assumption invalidates the instance I wanted to use (i.e. $l=\frac12$, $w=1$).

2. The proof of the lemma seems flawed, as mentioned below and succinctly summarised in Yoav Kallus’s comment. Another example where the packing constructed for the proof does not fit is $l = \frac{9}{5}$, $w = \frac{7}{9}$, $n=2$, which I mention only since (unlike the example below) it has $w<l$.

The answer as written is next, followed by the note of doubt I added after trying to understand the proof.

The answer is yes.

Thanks to the comment by Yoav Kallus above, and a comment by Timothy Chow on the question he linked to, I found the paper An algorithm for packing squares by Marc M. Paulhus. (Journal of Combinatorial Theory, Series A, Volume 82, Issue 2, 1998, Pages 147-157)

Lemma 1 of that paper shows: All the reciprocal squares from n to infinity can be packed in a rectangle of length $l$ and width $w$ provided $$n\geq \frac{1+l}{wl}$$

With $l=1/2$ and $w=1$, this shows that we could place the square of side 1/2 in such a way that the remaining space is a rectangle of dimensions 1 x 1/2, and then pack the remaining squares into that rectangle using Paulhus's procedure.

It doesn't obviously answer the question of whether the packing could be completed if started as shown. In particular, can the squares of side 1/3, 1/4, etc. be packed into two unconnected squares of side 1/2?

Added previously: I am not convinced that the proof of Lemma 1 used above is correct. As I understand it (specialising to the case $l=\frac 12$, $w=1$ for concreteness) the idea is to pack the squares in rows like this:

If the first square in row $i$ has side $1/{n_i}$, then $n_0=3$ and $n_{i+1}=\lfloor \frac{3}{2}n_i\rfloor$.

But then the total height of the first seven rows is already $$\frac 13 + \frac 14 + \frac 16 + \frac 19 + \frac 1{13} + \frac 1{19} + \frac 1 {28} > 1.$$

On the other hand this arrangement is not optimal, since the sixth row has squares $\frac 1{19}, \dots, \frac 1{27}$ but we could fit up to $\frac 1{30}$. So perhaps the claim is still true, notwithstanding the unconvincing proof?

• The top-left square in the illustration in the question includes a free rectangle of dimensions $1/4\times3/8$. By Lemma 1, this should be enough to hold reciprocal squares for $n\ge14$. I believe 12 and 13 comfortably fit in the free space in the bottom-right square, next to the 11. – Emil Jeřábek supports Monica Aug 8 '17 at 16:26
• You seem to be correct that the proof is in error. Particularly, I see a false deduction in the last set of inequalities: the proof claims that if $w\ge(1+l)/(n_0 l)$ ($=(1/n_0)\sum_{i=0}^\infty (1/(1+l))^i$), then $w\ge \sum_{i=0}^{\infty} 1/n_i$ follows. That would make sense if we had $n_i\ge n_0 (1+l)^i$, but the inequality shown earlier is actually reversed. – Yoav Kallus Aug 8 '17 at 19:48
• Perhaps this could also be due to an assumption in the paper earlier that width is considered smaller (i.e., $l\geq w$)? – LegionMammal978 Aug 8 '17 at 21:12
• I think your argument is somewhat useful, though. – Takahiro Waki Aug 8 '17 at 23:42
• @RobinHouston Even if you put as many squares as possible into each row, the height of the first seven rows will be $\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{9}+\frac{1}{14}+\frac{1}{22}+\frac{1}{35}>1$. – Peter Mueller Aug 9 '17 at 7:29

Using Clive Tooth's heuristic for this similar (but more difficult) problem https://math.stackexchange.com/questions/2645195/can-the-squares-with-side-1-n-be-packed-into-a-1-times-zeta2-rectangle

I was able to pack the first 10^9 squares into the 3/4 unit square. Not a proof, of course, but evidence, at least, that no "bottleneck" occurs when placing the smallest sized squares.


# What is the VC dimension of the hypothesis class $H=\left\{f_{\theta_{1}, \theta_{2}}: R^{2} \rightarrow\{0,1\} \mid 0<\theta_{1}<\theta_{2}\right\}$?

I would like to know what is the VC dimension of the following hypothesis class.

$$H=\left\{f_{\theta_{1}, \theta_{2}}: R^{2} \rightarrow\{0,1\} \mid 0<\theta_{1}<\theta_{2}\right\}$$

where $$f_{\theta_{1}, \theta_{2}}(x, y)=1$$ if $$\theta_{1} x \leqslant y \leqslant \theta_{2} x,$$ else $$f_{\theta_{1}, \theta_{2}}(x, y)=0$$.

I am not really sure how to prove it. What do you think?

The VC-dimension of your hypothesis class $$\mathcal H$$ is 2. To see this, we begin by showing that $$\mathcal H$$ shatters any 2-element set $$\{(a_1 a_2), (b_1, b_2)\}$$ of real numbers where all components of the pairs are positive:

1. $$\emptyset$$ is accounted for by $$f_{c, c + \varepsilon}$$ for any real $$c$$ such that $$ca_1 \neq a_a$$ and $$cb_1 \neq b_2$$ and some sufficiently small $$\varepsilon > 0$$.
2. $$\{a\}$$ (and similarly $$\{b\}$$) is accounted for by $$f_{c, c + \varepsilon}$$ where $$c = a_2/a_1$$ and $$\varepsilon > 0$$ is sufficiently small.
3. $$\{a, b\}$$ is accounted for by $$f_{\varepsilon, c}$$ for some sufficiently small $$\varepsilon > 0$$ and some sufficiently large $$c$$ (specifically, one can set $$\varepsilon = \min \{a_2 / a_1, b_2 / b_1\} / 2$$ and $$c = 1 + \max \{a_2 / a_1, b_2 / b_1\}$$)

This yields $$\operatorname{VC}(\mathcal H) \geq 2$$.

Now consider some arbitrary set $$X = \{(a_1, a_2), (b_1, b_2), (c_1, c_2)\}$$ of pairwise distinct points in $$\mathbb R^2$$. If the points in $$X \cup \{(0, 0)\}$$ are not in general position then $$\mathcal H$$ cannot shatter $$X$$ as it means that there are at least two points $$s, t \in X$$ which fall onto a line with the origin and as every classifier in $$\mathcal H$$ has linear decision boundaries it must always label $$s$$ and $$t$$ the same way, preventing it from shattering any set containing these points.

So let us assume that $$X \cup \{(0, 0)\}$$ is a set of points in general position and shattered by $$\mathcal H$$. As all functions $$f_{\theta, \varphi} \in \mathcal H$$ represent areas between two lines with positive slopes (since we require $$0 < \theta < \varphi$$) we can infer that all the lines connecting points in $$X$$ with the origin also must have positive slopes (note that $$X$$ is not in general position when the origin is an element of $$X$$). Hence we can order the points in $$X$$ ascendingly by these slopes, i.e. we may write $$a_2 / a_1 < b_2/ b_1 < c_2 / c_1$$ and since this requires all $$x$$-coordinates of the points in $$X$$ to be nonzero, we can simplify the expression for the preimage of $$1$$ of any $$f_{\theta, \varphi} \in \mathcal H$$ restricted to $$X$$ to $$f_{\theta, \varphi}|_X(x, y)^{-1} = \{(x, y) \in X \mid \theta \leq y / x \leq \varphi\}$$ by dividing by $$x$$. Now consider the subset $$X' = \{a, c\}$$ of $$X$$ and suppose $$f_{\theta, \varphi} \in \mathcal H$$ accounts for $$X'$$, i.e. $$f_{\theta, \varphi}|_X^{-1}(1) = X'$$. But then we must have $$\theta \leq a_2 / a_1 < b_2 / b_ 1 < c_2 / c_1 \leq \varphi$$ which yields $$f_{\theta, \varphi}|_X(b) = 1$$ and thus $$f_{\theta, \varphi}$$ does not account for $$X'$$ which is a contradiction. Therefore $$\mathcal H$$ does not shatter $$X$$ and and thus no set of size $$\geq 3$$.

It follows that $$2 \leq \operatorname{VC}(\mathcal H) < 3$$ and thereby $$\operatorname{VC}(\mathcal H) = 2.$$

• Thanks for the amazing explanation. Is it easy to make an example with 3 samples that give an intuition of how you proved that it the VC(H) cannot be more than 3? – hinduml Jun 15 '20 at 10:39
• For $\operatorname{VC}(\mathcal H) < 3$ we need to show that no such sample exists, so considering examples is not enough. However, the idea here is a geometric one: Note that the set where some $f \in \mathcal H$ is 1 is the area between two lines with positive slopes, so if $f$ is 1 on the two "outer" samples then it must also be 1 on the "middle" sample so it cannot shatter any such set. I wrote a formal proof for this (which is a bit more involved and technical) but the idea is rather simple if you draw up a picture :) – Watercrystal Jun 15 '20 at 11:20
• Thanks a lot for the explanation, its really helpful. – hinduml Jun 15 '20 at 11:57
• Glad I could help out :) – Watercrystal Jun 15 '20 at 12:02
• The two linear functions that bound the area are $\theta x$ and $\varphi x$ with $0 < \theta < \varphi$ required by your definition. If the slope is $> 0$, then it must be positive. The value of $x$ does not play a role for that. However, a linear function $ax$ with a positive slope (i.e. $a > 0$) attains negative values when $x$ is negative. – Watercrystal Jun 15 '20 at 15:30


# Solving an Inequality

1. Aug 26, 2009

### S_David

Hello,

Where is the mistake in the following solution of the inequality:

\begin{align*} &\frac{2x-5}{x-2}<1\\ &2x-5<x-2\\ &x<3, x\ne 2 \end{align}

2. Aug 26, 2009

### Hurkyl

Staff Emeritus
The following is not true:
If a<b, then ac<bc

In fact you have three separate cases, depending on c -- do you know what they are?

3. Aug 26, 2009

### HallsofIvy

Staff Emeritus
Hurkyl's point is that you multiplied both sides of the inequality by x- 2 and, since you don't know what x is, you don't know if x- 2 is positive or negative. (Oops, I just gave you two of the three cases Hurkyl asked about!)

My preferred method for solving anything more than linear inequalities is to first solve the corresponding equation. Here, solve $$\frac{2x-5}{x-2}= 1[/itex]. The "x" that satisfies that and the "x" that makes the denominator 0 are the only places where the inequality can "change". They divide the real line into three intervals- check one point in each interval to see which give ">". 4. Aug 26, 2009 ### Дьявол And why don't you try: [tex]\frac{2x-5}{x-2} - 1 < 0$$

?

Then consider these cases

$$a/b <0$$

should "a" and /or "b" be positive or negative so that the the fraction a/b would be negative?

Regards.

5. Aug 26, 2009

### arildno

You may split your analysis into two cases:

1) x-2>0,
and

2) x-2<0

You will find that for 2), there are NO solutions, meaning that x must be greater than 2.

6. Aug 26, 2009

### uart

Another excelent method in this case is to multiply both sides by (x-2)^2, with the squared assuring that we haven't multiplied by a negative number, and then just re-arrange it to give a quadratic inequality : x^2 - 5x + 6 < 0

7. Aug 26, 2009

### S_David

Ok, I can see where I did mistake the inequality. So, we have two cases:

\begin{align} x-2&>0\\ x-2&<0 \end{align}

In the first case the multiplication does not change the direction of the inequality, so:

\begin{align*} 2x-5&<x-2\\ x&<3 \end{align}

Then the open interval $$(2,3)$$ is the solution set. In the second case, the direction of the inequality changed, so:

\begin{align*} 2x-5&>x-2\\ x&>3 \end{align}

But $$x<2$$, then there is no solution.

The solution provided by uart is a good one, too. Where he eliminated the problem of negative sign possibility in the unknown $$x$$.

Actually, I am reviewing the precalculus and calculus books, and such things I forgot because I don't practice it continousely.

Anyway, thank you all guys.

Best regards

8. Aug 26, 2009

### Дьявол

You could solve it this way:
$$\frac{2x-5}{x-2} - 1 < 0$$

$$\frac{x-3}{x-2}<0$$

$$\begin{bmatrix} \left\{\begin{matrix} x-3<0\\ x-2>0 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x-3>0\\ x-2<0 \end{matrix}\right. \end{matrix}$$

$$\begin{bmatrix} \left\{\begin{matrix} x<3\\ x>2 \end{matrix}\right. \\ \\ \left\{\begin{matrix} x>3\\ x<2 \end{matrix}\right. \end{matrix}$$

Because 3<x<2 is not valid,

the only solution is 2<x<3 or (2,3).

Regards.

9. Aug 27, 2009

### S_David

Dear Дьявол,

This method is as described in my calculus book, and yes it is easier than the one I explained earlier. But I wanted to know the different methods to solve the inequality.

Thanks