# Thread: The centroid of a triangle with coordinates

1. ## The centroid of a triangle with coordinates

Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.

I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.

2. Originally Posted by darksoulzero
Triangle DEF has vertices D(1,3) and E(6,1), and centroid at C(3,4). Determine the coordinates of point F.

I know that the medians in a triangle intersect at the centroid and that the centroid divides each median in a ration of 2:1. Also, I know that each median intersects a side at its midpoint with the shorter part of the median. What I did was find the midpoint between point E and F by taking half of the line D to C and adding onto C to give point G. The coordinates of point G was found to be (4, 4.5). Since I knew that the change in the coordinates between points E and G, (-2, 3.5), was only half of the line EF I doubled it and added it point E to give the answer (2, 8). I wanted to know if there was an easier way to do this.
Read Centroid - AoPSWiki

3. ## centroid of triangle

Hi darksoulzero,

Your solution is confusing.Here is a suggested method.

connect M midpoint of DE and C (centroid) with extended lenght.F lies on this line. FC = 2CM. Slope diagram of C and M = 2/1/2.Slope diagram of FC is twice that or 4/1. Working from point C one point left and 4 points up gives F (2,8)

bjh

4. If ABC is a triangle with $A\left( {{x_1},{y_1}} \right),B\left( {{x_2},{y_2}} \right),C\left( {{x_3},{y_3}} \right)$ then its centroid is
$G\left( {\tfrac{{{x_1} + {x_2} + {x_3}}}{3},\tfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)$

5. Hello, darksoulzero!

$\text{Triangle }D{E}F\text{ has vertices: }D(1,3)\text{ and }E(6,1)\text{, and centroid at }C(3,4).$
$\text{Determine the coordinates of point }F.$
Code:

1
F o-----+
\    :
\   :4
\  :
\ :
\:0.5
(3,4)o---+
D       C\  :
o         \ :2
(1,3)   *    \:
M o
(3.5,2)   *     E
o
(6,1)

We have vertices $D(1,3)$ and $E(6,1)$, and centroid $C(3,4).$

The midpoint of $DE$ is: $M(3\tfrac{1}{2},\,2).$

The median to side $DE$ starts at $\,M$, passes through $\,C,$
. . and extends to $\,F$, where: . $FC \,=\,2\!\cdot\!CM.$

Going from $\,M$ to $\,C$, we move up 2 and left $\frac{1}{2}$
Hence, going from $\,C$ to $\,F$, we move up 4 and left 1.

Therefore, we have: . $F(2,8).$


# F(x) symmetric about the line x=2

## Main Question or Discussion Point

Why is the function: y = f(x) = a(x-1)(x-2)(x-3) symmetrical about the line x = 2? I mean how can we be sure that it is? Is there any method to check it?

## Answers and Replies

ShayanJ
Gold Member
Change the variable to y=x-2 and see whether the resulting function is even or odd or none!

You mean x = x-2 right? It doesn't stay the same on doing that. So it should not be symmetrical about x = 2, but it is. And I am not able to see how that happens.

ShayanJ
Gold Member
$f(x)=a(x-1)(x-2)(x-3) \rightarrow f(y)=a(y+1)y(y-1)$
$f(-y)=a(-y+1)(-y)(-y-1)=-a(y-1)y(y+1)=-f(y)$
So f(y) is odd which means f(x) is not symmetric around x=2 but is not that much asymmetric because we have f(2-x)=-f(x-2).

Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1

And this function is said to be symmetrical about the line x = 2. But I am unable to see how?

ShayanJ
Gold Member
Its the same trick. Just do the tranformation $x\rightarrow x+2$ and check whether the resulting function is even. If its even, then the original function is symmetric around x=2.

They are not the same, i.e. after changing x to x+2 in f(x), f(x) ≠ f(-x). But what I am reading, it says that f(x) is symmetrical about x=2 and I am still wondering that how would I go about proving it?

ShayanJ
Gold Member
You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.

PeroK
Homework Helper
Gold Member
Oh no. I posted the wrong function. The function is: y = f(x) = a(x4/4 -2x3 + 11x2/2 - 6x) + 1

And this function is said to be symmetrical about the line x = 2. But I am unable to see how?
Another way to look at it is as follows:

Imagine starting at x = 2 and moving the same distance, d, to the right and left (d > 0). So, to the right we have 2 + d and to the left we have 2 - d.

Now, if f is symmetrical about x = 2, then f(2-d) = f(2+d) for all d. You could try that approach.

You're doing something wrong. You should be able to reduce f(x+2) to $\frac 1 2 a[x^2(\frac 1 2 x^2-1)-4]+1$ which is even.
How did you reduce it down to that? Can you show me? I am unable to get to that point.

ShayanJ
Gold Member
$f(x+2)=a[ \frac 1 4 (x+2)^4-2(x+2)^3+\frac{11}{2} (x+2)^2-6(x+2)]+1=\\ a(x+2)[ \frac 1 4 (x+2)^3-2(x+2)^2+\frac{11}{2} (x+2)-6]+1=\\ a(x+2)(\frac 1 4 x^3+\frac 3 2 x^2+3x+2-2x^2-8x-8+\frac{11}{2}x+5)+1=\\ \frac 1 2 a(x+2)(\frac 1 2 x^3-x^2+x-2)+1=\frac 1 2 a (\frac 1 2 x^4+x^3-x^3-2x^2+x^2+2x-2x-4)+1=\\ \frac 1 2 a (\frac 1 2 x^4-x^2-4)+1$

And what is the reason that we transformed x to x+2?

ShayanJ
Gold Member
Simple. When we say a function is even, we mean its symmetric around x=0. So if a function is symmetric around $x=a \neq 0$, it means if we move the origin to x=a, the resulting function would be even.

Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?

ShayanJ
Gold Member
Check here!
But we aren't shifting the curve as in your link but instead we are shifting the coordinate axes. The curve stays where it was. The axes are what shift.

PeroK
Homework Helper
Gold Member
Right. So if we move the coordinates of origin from (0,0) to (2,0) shouldn't the abscissa of a point in the new axes, change from x to x-2 and not x+2?
This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.

If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.

But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.

In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.

This is one reason why it's much better to change the variable name - until, like Shyan, you've mastered this. Let's use z.

If we have z = x + 2, then x = 2 maps to z = 4, which is not what we want.

But, if we have z = x - 2, then x = 2 maps to z = 0, which is what we want. The origin of the z-variable is at x = 2.

In fact, if I'm honest, I always prefer to change the name of the variable to avoid the mistake you just made.
Wait. So you are saying that f(x) should change to a new variable so that it becomes f(z), right? But that would again mean f(x-2) and not f(x+2).. got me confused

ShayanJ
Gold Member
$z=x-2 \Rightarrow x=z+2$ so $f(x)=f(z+2)$!(Forget the confusing $x\rightarrow x+2$!)

It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?

For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.

ShayanJ
Gold Member
It just simply fails to make sense to me because of the simple logic that in older coordinate plane, suppose x (any number in domain of f(x)) mapped to f(x) = y (say). Now, in the new coordinate system, f(xold) = f(xnew) provided that xnew = xold - 2. So shouldn't we evaluate f(x-2) instead of f(x+2)?

For example, earlier xold = 2 gave f(2) = c. Now xnew = 0 (i.e. xold-2, where xold = 2 ) would give the result c.
Let's start from the beginning.
At first we have a coordinate system which we call xy. Now I define a function y=f(x). Then I move the origin to x=a and name the new coordinate system zy. But this doesn't change the function, only the coordinate system has moved. But if I insist that the function f has the same form in terms of both x and z, then this means that the function has changed which isn't right.(Imagine y=x^2. z=x-a so x=z+a. But if I say that y=z^2, this function would have its minimum at z=0 so x=a which means the function has changed!) So I should have y=g(z). But g should be related to f somehow that we actually get the same function. So let's see what's the relationship. At x=0, f gives f(0), so at z=-a, g should give f(0). Then at x=a, f gives f(a), so at z=0, g should give f(a). Now we have two relationships g(-a)=f(0) and g(0)=f(a) and so we can deduce that g(z)=f(z+a)=f(x).


# Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$

We all know that $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^s}$ converges for $s>1$ and diverges for $s \leq 1$ (Assume $s \in \mathbb{R}$).

I was curious to see till what extent I can push the denominator so that it will still diverges.

So I took $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n\log n}$ and found that it still diverges. (This can be checked by using the well known test that if we have a monotone decreasing sequence, then $\displaystyle \sum_{n=2}^{\infty} a_n$ converges iff $\displaystyle \sum_{n=2}^{\infty} 2^na_{2^n}$ converges).

No surprises here. I expected it to diverge since $\log n$ grows slowly than any power of $n$.

However, when I take $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n(\log n)^s}$, I find that it converges $\forall s>1$.

(By the same argument as previous).

This doesn't make sense to me though.

If this were to converge, then I should be able to find a $s_1 > 1$ such that

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

Doesn't this mean that in some sense $\log n$ grows faster than a power of $n$?

(or)

How should I make sense of (or) interpret this result?

(I am assuming that my convergence and divergence conclusions are right).

• They are right. You don't seem to have completed one of your sentences: "if this were to converge, then I should be able to find a s_1 > 1 such that"...? – Qiaochu Yuan Nov 11 '10 at 21:24
• There seems to be some bug with this. It doesn't display the entire thing. – user17762 Nov 11 '10 at 21:25
• @Sivaram: Yes even i tried editing it, and it still has not worked – anonymous Nov 11 '10 at 21:27
• There seems to be a bug/feature! when dealing with <. Probably confusion with html tags. Sivaram, please use \log n instead of just logn. – Aryabhata Nov 11 '10 at 21:32
• @ Chandru and Moron: Thanks for editing it. Now it looks fine. – user17762 Nov 11 '10 at 21:33

## 6 Answers

Yes $\displaystyle \sum_{n=2}^{\infty} \dfrac{1}{n (\log n)^s}$ is convergent if $\displaystyle s > 1$, we can see that by comparing with the corresponding integral.

As to your other question, if

$\displaystyle \sum_{n=2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{n=2}^{\infty} \frac{1}{n (\log n)^s}$

does not imply $\log n$ grows faster than a power of $\displaystyle n$. You cannot compare them term by term.

What happens is that the first "few" terms of the series dominate (the remainder goes to 0). For a small enough $\displaystyle \epsilon$, we have that $\log n > n^{\epsilon}$ for a sufficient number of (initial) terms, enough for the series without $\log n$ to dominate the other.

• Sorry. I meant $\exists s_1 > 1$ such that $\displaystyle \sum_{2}^{\infty} \frac{1}{n^{s_1}}$ is greater than $\displaystyle \sum_{2}^{\infty} \frac{1}{n(\log n)^{s}}$ – user17762 Nov 11 '10 at 21:56

This is a special case of classic general logarithmic convergence tests. Some early work in asymptotics was motivated by attempts to determine the "boundary of convergence" in terms of various functions, e.g. log-exp functions. Below is a a very interesting excerpt from Hardy's classic "A course in pure mathematics". For further work see his "Orders of infinity" and see especially this very interesting paper by G. Fisher on Paul du Bois-Reymond's work on the boundary between convergence and divergence (where he discovered diagonalization before Cantor).

Note in particular the following very general result that is presented in the excerpt: the following series and integral are convergent if $\rm\ s > 1\$ and divergent if $\rm\: s\le 1\$ where $\rm\: n_0,\ a\$ are any numbers large enough to ensure positivity of $\rm\ log_k x := \log \log \cdots \log x\:,\$ iterated $\rm\:k\:$ times.

$$\rm \sum_{n_0}^\infty \frac{1}{n \log n\ \log_2 n\ \cdots\ \log_{k-1} n\ (\log_k n)^s }$$

$$\rm \int_{a}^\infty \frac{dx}{x \log x\ \log_2 x\ \cdots\ \log_{k-1} x\ (\log_k x)^s }$$

• Off-topic question: What text is the excerpt from? Thank you! – Ayesha Jan 29 '14 at 1:04
• @Ayesha I say that in the text above it: Hardy's .... – Bill Dubuque Jan 29 '14 at 1:09
• Sorry, sorry, I somehow glanced over that. Thank you again for responding to my rather asinine question. – Ayesha Jan 29 '14 at 1:43

Du Bois-Reymond was interested in this question, positing some sort of series "between" all convergent series and all divergent series. Hausdorff didn't like that idea, and showed that given any countable sequences of convergence series converging "ever more slowly" and divergent series diverging "ever more slowly", you can find a convergent and a divergent series "in between". Here the order relation is the one given by the (conclusive) ratio test.

Bill's remark is related to the recursive Elias codes. The first few codes correspond to the convergent series

$\displaystyle \frac{1}{n(\log n)^2}, \frac{1}{n\log n (\log \log n)^2}, \ldots$

He then diagonalizes to obtain his $\omega$ code (Elias omega code, if you want to look it up), which corresponds to

$\displaystyle \frac{1}{n\log n\log \log n \cdots (\log^* n)^2},$

where $\log^* n$ is the number of times you need to apply $\log$ until you get below some constant (the other logs continue up to this constant); confusingly, coding theorists use $\log^* n$ to mean the $\log n \log\log n \cdots$ part. We can continue the iteration like this:

$\displaystyle \frac{1}{n\log n\log\log n \cdots \log^*n (\log \log^* n)^2}, \frac{1}{n\log n\log\log n \cdots \log^*n \log \log^* n (\log\log \log^* n)^2}, \cdots$

and so on. We can diagonalize again to obtain an $\omega + \omega$ code, and so on at least for $n \omega + m$. We an probably continue even to $\omega^2$ and beyond.

The corresponding divergent series are $\displaystyle \frac{1}{n\log n}, \frac{1}{n\log n\log\log n}, \ldots, \frac{1}{n\log n\log\log n \cdots \log^* n}, \frac{1}{n\log n\log\log n \cdots \log^* n \log\log^* n}, \ldots$

One can ask whether there is a scale for convergent series, that is a chain of (mutually comparable) series converging more and more slowly, such that for each convergent series there's one in the chain converging even more slowly; you can ask the same question about divergent series. Surprisingly, the existence of these is independent of ZFC (they exist given CH, and models in which they don't exist can be constructed using forcing).

• Indeed, it deserves to be better known that Du Bois Reymond - not Cantor - is the true discoverer of diagonalization - in this context, i.e. asymptotic growth rates of functions. – Bill Dubuque Nov 11 '10 at 22:53
• Thanks Bill and Yuval. I was looking for series which diverges slower and slower and your answer helps me on that note. – user17762 Nov 11 '10 at 23:09

Another test that applies to series of positive decreasing terms (and in this particular one in a rather elegant fashion) is the following: $$\sum_{n=1}^\infty a_n<\infty \quad\Longleftrightarrow\quad\sum_{k=1}^\infty 2^ka_{2^k}<\infty.$$ In our case $$\sum_{k=1}^\infty 2^ka_{2^k}=\sum_{k=1}^\infty 2^k\frac{1}{2^k(\log 2^k)^s}= \frac{1}{(\log 2)^s}\sum_{k=1}^\infty \frac{1}{k^s},$$ and thus $$\sum_{n=2}^\infty \frac{1}{n(\log n)^s}\quad\Longleftrightarrow\quad \sum_{k=1}^\infty \frac{1}{k^s}\quad\Longleftrightarrow\quad s>1.$$

• Monotonicity (as is in this case) is required? – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:40
• @Martín-BlasPérezPinilla: I do not know the most general version of this criterion, but the version I know requires $a_n$ to be non-negative and decreasing. – Yiorgos S. Smyrlis Jan 31 '14 at 19:46
• Yes, now I remember I've seen in Baby Rudin. – Martín-Blas Pérez Pinilla Jan 31 '14 at 19:56

By the integral comparison test the given series is convergent if and only if the integral $$\int_2^\infty\frac{dx}{x\log^s(x)}$$ is convergent and notice that $\log'(x)=\frac1x$ so

• if $s>1$ $$\int_2^\infty\frac{dx}{x\log^s(x)}=\frac1{1-s}\log^{1-s}(x)\Big|_2^\infty<+\infty$$ so the series is convergent
• if $s=1$ $$\int_2^\infty\frac{dx}{x\log(x)}=\log(\log(x))\Big|_2^\infty=+\infty$$ so the series is divergent.
• The exponent $s$ is just on the log term, not the entire term – user2566092 Jan 31 '14 at 19:38
• Ah I'll edit my answer! – user63181 Jan 31 '14 at 19:41
• @user2566092 I edited my answer. – user63181 Jan 31 '14 at 19:50
• What about $s < 1$? – mavavilj Sep 19 '15 at 17:25

Convergence does not change if you switch to base-2 logs. There is a theorem that says that if $a_n$ is a decreasing positive sequence, then $\sum_n a_n$ converges if and only if $\sum_k 2^k a_{2^k}$ converges. If you apply this theorem's transformation to your sequence, you get $\sum_k 1/k^s$, which obviously converges if $s > 1$ and diverges if $s \leq 1$. If you'd like a proof of that theorem (called the Cauchy condensation test), I can provide it.


What's new

# Difference between probability density function and inverse cumulative distribution function?

#### bpdulog

##### Active Member
What is the difference between probability density function and inverse cumulative distribution function, if any? They both look the same

#### ShaktiRathore

##### Well-Known Member
Subscriber
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks

#### brian.field

##### Well-Known Member
Subscriber
David also covers this extensively in his Study Notes for Miller. Have you read them?

#### bpdulog

##### Active Member
Hi,
No they are not the same.
The inverse cumulative distribution function is the quantile function it gives the value of the quantile(z) at which the probability of the random variable is <=the given probability value or the cumulative probability of random variable is = the given probability value.For e.g.at 95% cumulative probability the value of quantile is z=1.645,at 99% cumulative probability z=2.33 and so on so that the plot of the values of z Vs the cumulative probability values gives the inverse cumulative distribution function ICDF. ICDF(95%)=1.645, ICDF(99%)=2.33.
Whereas probability density function P(z) gives the value of probability at a given quantile,so that when you integrate the function over a quantile range shall give the value of the cumulative distribution function,integration of P(z) over -inf to inf is=1,integration of P(z) over -inf to 1.645 is=95%,at any point z on the P(z), probability(z)=0.
thanks
If I understand what are you saying, in the inverse distribution the 1.645 represents 95% and less. And in the probability density function the 1.645 represents 95% only?

#### ShaktiRathore

##### Well-Known Member
Subscriber
Hi
in the inverse distribution the 1.645(quantile) represents 95%(cumulative probability). And in the probability density function P(x) ,when x=1.645 ,P(x) shall give the probability density at x,and we integrate the P(x) from -inf to x=1.645 to get the cumulative probability of 95%.
thanks

#### bpdulog

##### Active Member

Ah ok! I think I get it now. It just 2 different ways to display a distribution of data. The top is the PDF and the PPF is basically the CDF with the axes transposed.

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Yuu definitely want to understand @ShaktiRathore 's above relationship. Miller's Chapter 2 does a decent job. We don't use the continuous pdf too much. The pdf integrates to the CDF, and we're arguably more interested in the relationships around the CDF, as Shakti illustrates. For example, if probability (p) = 5.0%, then:
• p = 5% = F(q), where F(.) is the cumulative distribution function, so if we are given the probability of 5% because we want a 95% confident normal VaR, then we use the inverse CDF to retrieve the quantile:
• q = F^(-1)(p) = F^(-1)(5%) = -1.645, where F^(-1) is the inverse CDF. "Inverse" implies that F[F^(-1)(p)] = p; e.g., F[F^(-1)(5%)] = F[-1.645] = 5%, just like also F^(-1)[F(q)] = q. This is why Dowd says "VaR is just a quantile;" i.e., because -1.645 is the quantile (q) retrieved by using the inverse CDF given a probability (p) of 5%. This is an essential FRM building block. I hope that adds something.

#### theapplecrispguy

##### New Member
In Question 300.2, the final price of the bond (i.e. the value of "x") is calculated using the inverse CDF such that 5% of the distribution is less than or equal to "x". This price is calculated to be $1.842 using p=5%. My interpretation of this is that 5% of the time the bond will be priced below$1.842 and the other 95% of the time the bond will be priced above $1.842. Is this correct? The question then states that the 95% VAR is given by$5.00 (the face value of the bond) minus $1.842 (q(.05)) equals$3.158. I don't understand conceptually how the formula for 95% VAR was determined to be $5.00 - q(.05). Can you please explain what the 95% VAR ($3.158) represents in the context of the area under the cumulative distribution graph?

#### David Harper CFA FRM

##### David Harper CFA FRM
Staff member
Subscriber
Hi @theapplecrispguy FYI, the source Q&A (with follow on discussion) is here at https://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/ but briefly:
• Re: "My interpretation of this is that 5% of the time the bond will be priced below $1.842 and the other 95% of the time the bond will be priced above$1.842. Is this correct?" Yes, exactly correct. The integral (anti-derivative) of the given f(x) is the CDF which is F(x) = x^3/125. When we have the CDF, it means that cumulative p = x^3/125; i.e., p is the probability that the random variable will be less than or equal to x. Because 1.842^3/125 = 5.0%, under this distribution, there is a 5.0% probability that X will be less than $1.842; and, as you say, 95.0% of the time X will be greater than$1.842 (but less than $5.00, as the function domain is$0.00 < X < 5.00). As the question is testing the associated Miller, this is really the essential point of the question ....
• Because the VaR part here is really simplistic. VaR is a loss relative to something. This question gives the assumption that the expected future value is $5.00 so that the VaR can be computed as a simple "worst expected loss" of$5.00 - $1.842 =$3.158; ie, because there is a 5.0% probability of the final price falling below $1.842, there is a 5.0% probability of a loss worse than$3.158. If we were not given this assumption, the proper approaches are classically either:
• Compute the loss relative to the expected future value as the proper mean of the future distribution. In this case, as stated in the question, this relative VaR (it is called because it is relative to the expected future value) would be $3.75 -$1.842 = $1.908. I just didn't want to make the question too difficult; or, • Compute the loss relative to the current value, which is properly termed the absolute VaR (relative to the initial value). I don't think we have information to do that, but it would be given by discounted present value [$5.00 par] - $1.842 = absolute VaR. This shows why a good question needs to be specific about the VaR it looks for, VaR is a worst expected loss over a horizon (choice #1) with a confidence level (choice #2) relative to something (definitional #3). I hope that helps! #### theapplecrispguy ##### New Member Thank you David for the explanation which is very helpful. If the question had asked for the 5% VAR would the answer have been the same? #### David Harper CFA FRM ##### David Harper CFA FRM Staff member Subscriber Hi @theapplecrispguy Yes, but only because occasionally (or rarely) the 95.0% VaR will be represented by an author as the 0.050 VaR, treating them as the same idea. In any realistic use case, we are referring to the losses that happen in the worst 5.0% of times, the losing tail. It would not be realistic/sensible to speak of the losses that occur in the worst 95.0% of times (although mathematically it works of course). But, again, the idea is that we have a cumulative distribution function here, given by F(x) = p = x^3/125. Solving for x, we have x = (p*125)^(1/3). So for example, • at p = 5%, x = (5%*125)^(1/3) =$1.84
• at p = 10%, x = (10%*125)^(1/3) = $2.32 • at p = 20%, x = (20%*125)^(1/3) =$2.92
• at p = 50%, x = (50%*125)^(1/3) = $3.97; i.e., the median • at p = 95%, x = (95%*125)^(1/3) =$4.92
It's not a distribution of losses, it's a distribution of future values. We always want to be mindful of the difference, sometimes the distribution is losses. As this is future values, the risk tail is "down" at p = 5%, it is not "up" at p = 95%, so that either a 95% or 0.05 VaR would be realistically referring to the p = 5%. At the same time, there are valid usages of VaR for lower confidence levels, so we could have a 90% VaR (given by the loss of $5.00 minus$2.32) or a even an 80% VaR (given by the loss of $5.00 minus$2.92). I hope that helps!

Last edited:

#### theapplecrispguy

##### New Member
Thank you. Clear and much appreciated!


# what happens to i^(square root of 2)?

I always thought what be in store for us if take irrational number to power of a complex number? we already have natural number, integers, rational numbers, real numbers and complex numbers. Will it create a class of its own?

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9 months, 2 weeks ago

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yes sir but when the value of k is increased in polar form of (i)^(square root of 2), we do not get the solutions that superimpose the previous solutions. so we instead get infinite no of solutions. it form a circle with magnitude equal to one and center as origin in Argand Plane after k becomes infinite. so does that mean (i)^(any irrational number) forms a circle with magnitude of value one and center at origin.and get same set of solutions. does this mean (i)^any irrational number has same set of values.

- 9 months, 2 weeks ago

The basic way to define powers of complex numbers is via the formula $z^w \; = \; e^{w\log z}$ and so the whole business revolves around the definition of the logarithm of complex numbers. We need $\log z \; = \; \ln |z| + i\mathrm{Arg}\,z$ and here is the real problem. There is no way of defining the argument function continuously (let along differentiably) on the whole complex plane.

Since we want the argument (and hence the logarithm) to be differentiable, it has to be defined on an open set (so that we can consider its derivative at all points), and complex functions are considered on open connected domains. The standard way of doing this for the argument is to cut the plane. This means considering the domain formed by the complex numbers with a straight line out from $$0$$ to infinity removed. The argument can be defined uniquely (and differentiably) on any such domain, and so can the natural logarithm of the modulus (since we have removed $$0$$).

For example, we could consider the cut plane $$\mathbb{C} \backslash (-\infty,0]$$ consisting of the complex numbers except for the nonpositive reals. This means that $$|z| > 0$$ and $$-\pi < \mathrm{Arg}\,z < \pi$$ for all $$z$$ in the cut plane, and we can define $$\log z$$ uniquely (and differentiably) now. This choice is called using the principal branch of the argument. In this case we would have $$\ln i = \tfrac12\pi i$$, and so $$i^\sqrt{2} = e^{\frac{\pi i}{\sqrt{2}}}$$.

A different cut would be to exclude the nonnegative reals $$[0,\infty)$$, and decide that $$2\pi < \mathrm{Arg}\,z < 4\pi$$ for all $$z$$ in this cut plane. This would make $$\ln i = \tfrac52\pi$$, giving a totally different value for $$i^{\sqrt{2}}$$.

This is what @Pi Han Goh means about this function being multivalued. There is no way to have a single function that works everywhere.

The other option is to allow all these values to work at the same time! This means we have to leave the ordinary complex plane, and work with a Riemann surface. Imagine an infinite number of complex planes stacked one on top of another. Cut them all along the positive real axis, and stick the "first quadrant edge" of each sheet to the "fourth quadrant edge" of the sheet below. The resulting helicoidal shape allows you to consider complex numbers with positive moduli and all possible arguments. The number $$i$$ would lie in one sheet, the number $$ie^{2\pi i}$$ would lie in the sheet above, the number $$ie^{-6\pi i}$$ would lie in three sheets below, and so one. You could then define a logarithm on the whole Riemann surface at one go...

- 9 months, 1 week ago

Ah crap. I can't answer this question either. My first though is Equidistribution theorem, but I'm not entirely confident.

Summoning the great @Mark Hennings again!

- 9 months, 1 week ago

No, $$i^{\sqrt 2}$$ is just a multivalued complex numbers.

Let $$x = i^{\sqrt 2}$$, then $$\ln x = \sqrt 2 \cdot i = \sqrt 2 ( 0 + 1i )$$.

With $$0 + 1i = \cos \left (\frac\pi 2 + 2\pi k \right) + i \sin \left ( \frac\pi 2 + 2\pi k \right)$$, where $$k$$ is any integer.

Can you finish it off from here?

- 9 months, 2 weeks ago


# $\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$ converges?

Determine whether this series converges or diverges: $$\sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$

Thought about using the limit theorem or by comparison but am so stuck. any pointers would be appreciated guys

• What tests do you know for convergence? – Qiaochu Yuan Oct 20 '10 at 21:32
• Look at the absolute value of the ratio of a term in the series to the subsequent term. You should already know what the possible results imply... – Brandon Carter Oct 20 '10 at 23:35

Another way is

If $\displaystyle S_n = 1 + \frac{1}{3!} + \dots + \frac{1}{(2n+1)!}$

We have that

$\displaystyle S_n \le 1 + \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n+1)}$

$\displaystyle = 1 + (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n+1}) = 2 - \frac{1}{n+1} < 2$

Thus $S_n < 2$

thus we have the $\displaystyle S_n$ is monotonically increasing and bounded above and so is convergent.

I think svenkatr's response is correct. He is using the comparison test, in particular, comparing with the exponential function for $x=1$, that is obviously a number, so he doesn't have to prove that the series for e converges.

Maybe you can prove the same by using the ratio test $\lim_{n \rightarrow \infty} \displaystyle |\frac{a_{n+1}}{a_{n}}|$. For example, you have $a_{n}=\displaystyle \frac{1}{(2n+1)!}$ and $a_{n+1}=\displaystyle \frac{(2n+1)!}{(2n+3)!}$, then using the definition for the factorial you have $\lim_{n \rightarrow \infty} \displaystyle \frac{1}{(2n+3)(2n+2)}$ which is 0. According to the ratio test:

If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.

Therefore, the series converges.

• Just out of the curiosity, I plugged the series into Wolfram|Alpha. It gives the very nice result of sinh(1). If anyone knows how to prove that, it would be wonderful to know it. By the way, how can I add an hyperlink in the comment section? – Robert Smith Oct 21 '10 at 0:18
• $\sinh(x)=\frac{\exp(x)-\exp(-x)}{2}$. Also, [this](http://functions.wolfram.com/ElementaryFunctions/Sinh/) gives this. – J. M. is a poor mathematician Oct 21 '10 at 1:21
• @Robert: It is the Taylor series for sinh evaluated at 1. – GEdgar Jun 16 '12 at 16:28
• Thanks. I didn't know that. – Robert Smith Jun 16 '12 at 18:06

The series you have is

$1 + \frac{1}{3!} + \frac{1}{5!} \ldots$

If you add the even factorial terms, you get an upper bound i.e.,

$1 + \frac{1}{3!} + \frac{1}{5!} \ldots < \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!}+ \frac{1}{5!} \ldots$

This can be written more compactly as

$\sum_{n=0}^\infty \frac{1}{(2n+1)!} < \sum_{n=0}^\infty \frac{1}{n!} = e^1$

Therefore the series converges.

• I don't see the point of doing this when proving that the series for e converges is exactly as hard. – Qiaochu Yuan Oct 20 '10 at 22:23
• In short, $e=\cosh\;1+\;sinh\;1$. :) – J. M. is a poor mathematician Oct 21 '10 at 0:19
• @ Qiaochu Yuan. You make a valid point. I guess the Ratio test(which Robert Smith has mentioned) is the rigorous answer :). – svenkatr Oct 21 '10 at 3:53

We have $$e^{1} = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots$$ and $$e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + \cdots$$

Subtracting these two we get $$e - e^{-1} = 2 \cdot \Bigl( 1 + \frac{1}{3!} + \frac{1}{5!} + \cdots \Bigr)$$ Therefore the series converges to $$\frac{e-e^{-1}}{2} = \sum\limits_{n=0}^{\infty} \frac{1}{(2n+1)!}$$

• The question was «does the series converge?» Your answer begins by asserting convergence of a series of equivalent difficulty :) – Mariano Suárez-Álvarez Oct 21 '10 at 14:12

Can you bound the series from above by one that you know converges? The factorials grow very fast, so you should be able to.

To elaborate on the first answer given to this question by Ross Millikan.

$$\sum_{n=0}^\infty \frac{1}{(2n+1)!} = 1 + \sum_{n=1}^\infty \frac{1}{(2n+1)!}$$

$$< 1 + \sum_{n=1}^\infty \frac{1}{4^n} = \frac{4}{3}, \quad \textrm{ as } \frac{1}{(2n+1)!} < \frac{1}{4^n} \textrm{ for } n \ge1.$$

Hence by the comparison test the series converges.

Comparing with another more manageable series could be useful in this case for possible follow-on questions as, with this approach, it's not much extra work to prove that it converges to an irrational number. Such a proof might include: Let $S$ be the series and $S_N$ the $N$th partial sum and $R_N$ the remainder then $S=S_N + R_N,$ where we note that

$$R_N < \frac{1}{(2n+3)!} \left( 1 + \frac{1}{(2n+3)^2} + \frac{1}{(2n+3)^4} + \cdots \right).$$

METHOD I

We may simply resort to the Basel problem and get the inequality: $$0<\sum_{k=0}^{\infty}\frac{1}{(1+2k)!}\leq\sum_{k=0}^{\infty}\frac{1}{(1+k)^2}=\frac{\pi^2}{6}$$

METHOD II

According to Taylor's expansion we have that:

$$\sinh(x) = \sum_{k=0}^{\infty}\frac{x^{1+2k}}{(1+2k)!}$$

For $x=1$ we get that the value of the series is $\sinh(1)$. The series converges.

Q.E.D.

This is how sometimes we can extract the exact closed form of some series :

$f(x) = \sum_{n=0}^{\infty} \frac{x^{2n+1}}{(2n+1)!}$

The radius of convergence is $\infty$

$f \in C^{\infty}(\mathbb{R})$

$f''(x) = f(x)$

solving our equation, we get $f(x) = ae^{\lambda_1x}+be^{\lambda_2x}$

$\lambda^2-1=0$

$\lambda \in \{-1,1\}$

$f(x) = ae^{-x}+be^{x}$

$f(0)= 0 \implies a+b=0 \implies a=-b \implies f(x) = a(e^{x}- e^{-x})$

$f'(0) = 1 \implies a=\frac{1}{2}$

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

so our series $\sum_{n=0}^{\infty} \frac{1}{(2n+1)!}$ does converge to $\sinh(1)$

• This only proes that if the series converges, then its sum is $\sinh(1)$. – José Carlos Santos Jul 26 '18 at 20:24
• Yes that is true, but I wanted to say, regarding to other answers, that I would find the limit. So I didn't repeat any other done process here, that is namely true. Anyway that is the same, effectively the sum itself implies the convergence of that series to $\sinh(x)$. So you had not always to prove the convergence of series, since it does have a closed form. the closed form itself implies the convergence, that is so understandable. –  Ахмед Jul 26 '18 at 20:26
• your implication is not true for closed form and finite values. that does mean, the sum does converge for all $x\in \mathbb{R}$ to $\sinh(x)$ which is finite. –  Ахмед Jul 26 '18 at 20:33


# Statement Proof.

1. Jul 11, 2007

### linuxux

My question is this:

Theorem 1.4.13 part (ii) says: If $$A_n$$ is a countable set for each $$n \in \mathbf{N}$$, then $$\cup^{\infty}_{n=1} A_n$$ is countable.

I can't use induction to prove the validity of the theorem, but the question does say how does arranging $$\mathbf{N}$$ into a 2-d array:

1 3 6 10 15 ...
2 5 9 14 ...
4 8 13 ...
7 12 ...
11 ...

lead to the proof of part (ii) of Theorem 1.4.13?

so obviously it has something to do with the (x,y) co-ordinate system of the array, but I nt sure how it leads to the proof.

Last edited: Jul 11, 2007
2. Jul 11, 2007

### CRGreathouse

You can just show a function $f: A\to\mathbb{N}$.

$$f(0)=A_1(0)$$
$$f(1)=A_1(1)$$
$$f(2)=A_2(0)$$
$$f(3)=A_1(2)$$
$$f(4)=A_2(1)$$
$$f(5)=A_3(0)$$
$$f(6)=A_1(3)$$
. . .

3. Jul 11, 2007

### linuxux

but what is the significance of putting it into a 2-d array? i could just as easily say: 1, 2, 3, 4, ... and call it a 1-d array.

4. Jul 11, 2007

### CRGreathouse

$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...
... ... ... ...

5. Jul 11, 2007

### mathwonk

this proof is due to cantor, 100 years ago. just start couting from the upper left corner, counting by sw/ne diagonals.

6. Jul 12, 2007

### matt grime

To demonstrate a bijection between N (the indices you're inserting) and the countable union of N (the locations in the doubly infinite array). I.E the whole point of the question.

7. Jul 14, 2007

### linuxux

but how is this a 1-1 and onto function? how is it that $$A_n$$ is a function itself? the cantor theorem is the next section in my text so i can't apply that theory now. i understand that f(n) is a function, but why also $$A_n$$?

Last edited: Jul 14, 2007
8. Jul 14, 2007

### CompuChip

Here's a different explanation (sometimes I see that the understanding of the principle depends on the way it is explained, so maybe this will help you - maybe it will make it less clear to you, then please ignore this )
Suppose we put two copies of the natural numbers next to each other. We want to show that, doing this, the total number we get is still countable (that there are "as many" elements in this string of numbers as there are natural numbers, by assigning a natural number to each of the elements).
Clearly, just counting them in a line won't work. Then we assign each natural number to itself in the first copy, but when we're through with that one, we don't have any natural numbers left for the second copy. So what we do is: assign 1 to 1 from the first copy, 2 to 1 from the second copy, 3 to 2 from the first copy, 4 to 2 from the second copy, etc. - so
$$2n - 1 \mapsto n \text{ from the first copy of } \textbf{N}, 2n \mapsto n \text{ from the second copy of } \textbf{N}.$$.

Now consider the Cartesian product $$\textbf N \times \textbf N$$, consisting of all pairs of natural numbers (a, b). How do you prove they are countable. One way would be to start by assigning n to (n, 1), but when we're done, we won't have any natural numbers left to assign to (n, 2) and (n, 3), etc. So here's the trick: we organize the pairs like
(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)
....
and now we can map them like
1 2 6 7 ...
3 5 8 ...
4 9 10
without running out of numbers too soon.

In the same way, you can prove that $\textbf{N}^k$ is countable for any natural number k, as well as $\frac{1}{2} \textbf{N}$ (all halfintegers 0, 1/2, 1, 3/2, ...) and the integers $\textbf{Z}$ (it's basically two copies of the naturals, like in my first example, with a zero added) and even that the rationals $\textbf{Q}$ are countable. And, also the question from your initial post can be solved like this.

9. Jul 14, 2007

### linuxux

thanks, i understand the example you showed. and to think, this is just an introductory text and i'm self-studying!...

Last edited: Jul 14, 2007
10. Jul 14, 2007

### linuxux

(1,1) (1,2) (1,3) ...
(2,1) (2,2) (2,3)
(3,1) (2,3) (3,3)

this part i understand. each (a,b) has a particular n mapped to it, thus (a,b)~N. so i got that.

but, looking at this:
$A_1(0)$ $A_1(1)$ $A_1(2)$ $A_1(3)$&...
$A_2(0)$ $A_2(1)$ $A_2(2)$ $A_2(3)$&...
$A_3(0)$ $A_3(1)$ $A_3(2)$ $A_3(3)$&...
$A_4(0)$ $A_4(1)$ $A_4(2)$ $A_4(3)$&...
$A_5(0)$ $A_5(1)$ $A_5(2)$ $A_5(3)$&...

i understand this to mean that the number of sets of $$A_n$$ is countable, not that once the union is performed on the sets the elements of the union will be countable. I had a previous problem where i had to show that the union of $$A_1$$ and $$A_2$$ was countable, i did this by defining a function that chose the minimum element first in $$A_1$$ and then in $$A_2$$, then choosing the next smallest element in both, and so on, thus i was able to show $$A_1\cup{}A_2$$~$$\mathbb{N}$$. So in that exercise i addressed the elements in the sets, not the sets themselves. how does the section in red address the elements in the sets? is that even necessary?

**********

wait a minute: i know each $$A_n$$ is countable, so each $$A_n(a\in\mathbb{N})$$ addresses the elements in $$A_n$$. now i see.

Last edited: Jul 14, 2007
11. Jul 15, 2007

### matt grime

By construction: do you write down the same number twice or fail to fill in a box?

It isn't a function.

12. Jul 15, 2007

### linuxux

wouldn't it qualify as a function since $$A_n$$ is countable thus $$A_n \mapsto \mathbb{N}$$?
i thought $$\mapsto$$ represents some type of function. does it not?

i also have another question, what happens if a particular $$A_n$$ is finite?

Last edited: Jul 15, 2007
13. Jul 15, 2007

### CompuChip

Yeah, so for each n we can assign a natural number to each element from $$A_n$$, and in the end, have neither elements of $$A_n$$ nor natural numbers "left" (that is, there is a bijection). The claim is, that the union is countable, so there is a bijection from the naturals to the union. What the 2-D array tries to visualize, is the following: since each $$A_n$$ is countable, we can denote the elements by $$A_n(k)$$ (with k = 1, 2, 3, ...). Now you can order the elements in the union like this:
$$A_1(1) \, A_2(1) \, A_3(1) \, \cdots$$
$$A_1(2) \, A_2(2) \, A_3(2) \, \cdots$$
$$A_1(3) \, A_2(3) \, A_3(3) \, \cdots$$
and use a diagonal counting argument, like before.

Another way is this: each element from the union is indexed by two numbers (n, k) (assuming the union is disjoint, as we implicitly did before). So it's really trivial to write down a bijection from this union to $$\textbf{N} \times \textbf{N}$$. As shown in my earlier post, the latter is countable (there is a bijection to $$\textbf{N}$$, so by composing the bijections you get one from the union to $$\textbf{N}$$. Actually, it also uses the diagonal counting argument (to show that $$\textbf{N} \times \textbf{N}$$ is countable) but perhaps this approach is more intuitive (and once you've proven this result about $$\textbf{N} \times \textbf{N} \times \cdots \times \textbf{N}$$ you can use it and prove countability of other things, like this union of $$A_n$$, without explicitly invoking a diagonal counting argument).

Last edited: Jul 15, 2007
14. Jul 15, 2007

### CompuChip

Does it matter? If you have a set in the union, that is not countably infinite, but finite, will the union become bigger (so big, it will become uncountable?)

If you want to do it officially, consider this: we said a set is countable if we can assign the numbers 1, 2, 3, ... to it. But of course, we can use another offset, and assign 2, 3, 4, ... to it. Or 154842, 154843, 154844, ...
So you could of course first count off the finite sets in the union (in order, no need for diagonals here), and then the infinite ones with the diagonal argument, but using a certain offset.

15. Jul 15, 2007

### linuxux

thanks a million man!

16. Jul 15, 2007

### CompuChip

Most welcome, this stuff is fun and moreover, it's important basics.

Later you will encounter more diagonal arguments, for example: if you want to prove that the real numbers are uncountable, or if you get to work with converging sequences.

17. Jul 16, 2007

### matt grime

You said A_n was a function, not that |-> was a function. Don't be sloppy.


# The converse of “nilpotent elements are zero-divisors”

For commutative rings $$A$$ with identity $$1\ne0$$, nilpotent elements are zero-divisors. The converse is false, i.e. there is a commutative ring $$A$$ with identity $$1\ne0$$ and a zero-divisor $$x$$ in $$A$$ which is not nilpotent. Where is such an example?

Is it meaningful to ask the "percentage" of commutative rings $$A$$ with identity $$1\ne0$$ for which the converse holds, i.e. zero-divisors are nilpotent?

• Have you thought about the integers modulo $n$? – Lord Shark the Unknown Dec 22 '18 at 12:45
• You can also search yourself at this site, e.g. here, for useful links. – Dietrich Burde Dec 22 '18 at 12:51
• @Lord Shark the Unknown: Thank you. Right, in $\mathbb{Z}/10\mathbb{Z}$, $2$ is a zero-divisor but not nilpotent. How did you came up with this example? How about the next question? – user584333 Dec 22 '18 at 12:52
• Sai, have a look at this post, which gives an answer which rings have this property. – Dietrich Burde Dec 22 '18 at 12:56
• @Dietrich Burde: Thank you for the reference. – user584333 Dec 22 '18 at 13:01

An example is in the integers mod $$6$$, where $$2\cdot 3=0$$, but no power of either of these individually is zero.

If we had some sort of measure on a set of rings (which could not possibly be all rings, because there are too many) with finite total measure, we could ask about proportion. I know of no such commonly used measure, but it's possible such a thing has been considered.

In my intuition, the proportion would be small, but I can't think of a quick reason to see why, other than that being nilpotent seems like a special property while being a zero divisor seems very common. For example, the only time this is true for rings of integers modulo $$n$$ is when $$n$$ is a prime power, and there are vanishingly few of those compared to all integers.

If you look at the answer by rschwieb in the question Under what conditions does a ring R have the property that every zero divisor is a nilpotent element? linked above in a comment by Dietrich Burde, which is not the accepted answer, it characterizes these rings. A ring has this property if and only if it is the quotient of an arbitrary commutative ring by a primary ideal. An ideal $$I$$ is said to be primary if whenever we have that $$a,b\in R$$ satisfy $$ab\in I$$, then we have either $$a\in I$$ or $$b^n\in I$$ for some $$n>0$$. This answers part of your question. It is nicely illustrated in the $$\Bbb Z_n$$ case: an ideal $$n\Bbb Z$$ of $$\Bbb Z$$ is primary if and only if $$n=p^k$$ for some prime $$p$$ and integer $$k>0$$.

So rings like these are actually "close" to integral domains, which are quotients of arbitrary commutative rings by prime ideals. Again, no quantitative argument for why they should be rare, only an intuitive one.

• Couldn't one equip a ring $R$ with the Zariski topology, then produce a Borel sigma algebra, and equip it with a pre-measure, which can then be extended to a measure by Caratheodory's extension theorem, to at least begin discussing "percentages" when the ring itself is of finite measure? – Chickenmancer Dec 22 '18 at 21:44
• @Chickenmancer that wouldn't give a percentage of rings, I suspect you mean to give a percentage of zero divisors that are nilpotent? – jgon Dec 22 '18 at 21:50
• Ah, I misinterpreted the question as "what percentage of a ring $A$." Thanks for clarifying, @jgon. – Chickenmancer Dec 22 '18 at 22:15
• Curious about the downvote. An explanation would be helpful. – Matt Samuel Dec 23 '18 at 17:43

One perspective on what proportion of rings have no nonnilpotent zero divisors is the following. Note that this is really rough first thought.

First let's be careful though, rings with no zero-divisors trivially have this property, so I'll try to consider what rings with zero-divisors have the property that all zero-divisors are nilpotent.

Step 1: Set up a space to parametrize a nice class of rings

Let $$k$$ be an algebraically closed field. Let $$n,d_1,d_2 > 0$$. Let $$x=(x_1,x_2,\ldots,x_n)$$ be coordinates on $$\AA^n_k$$. Let $$r_1=\binom{n+d_1}{d_2}$$, which is the number of monomials in $$k[x_1,\ldots,x_n]$$ of degree at most $$d_1$$. Thus there are $$r_1$$ multiindices $$I$$ of degree at most $$d_1$$. Let $$r_2=\binom{n+d_2}{d_2}$$ as well. Let $$c=(c_I)$$ be coordinates on $$\PP^{r_1-1}_k$$, and let $$d=(d_J)$$ be coordinates on $$\PP^{r_2-1}_k$$ as $$I$$ ranges over multiindices of degree at most $$d_1$$ and $$J$$ ranges over multiindices of degree at most $$d_2$$.

Then consider the subvariety, $$V$$, of $$\AA^n_k\times_k\PP^{r_1-1}_k\times_k \PP^{r_2-1}_k$$ cut out by the polynomial $$f_{c,d}(x)=g_c(x)g_d(x):=\left(\sum_I c_Ix^I\right)\left(\sum_J d_Jx^J\right).$$

$$V$$ is equipped with a natural map $$V\to \PP^{r_1-1}_k\times_k\PP^{r_2-1}$$, and the fiber over a point $$(a,b)$$ in $$\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$ is the subvariety of $$\AA^n_k$$ cut out by the degree at most $$d_1+d_2$$ reducible polynomial $$f_{a,b}(x)=g_a(x)g_b(x).$$

This subvariety of $$\AA^n_k$$ can be thought of as corresponding to the ring $$k[x_1,\ldots,x_n]/(g_ag_b)$$, so we can think of our variety $$V$$ as parametrizing a certain nice class of rings all (except for a Zariski closed subset where $$g_a=0$$ or $$g_b=0$$) of which are guaranteed to have zero-divisors, since $$g_a$$ and $$g_b$$ are zero divisors.

Step 2: Investigate the points corresponding to rings with the desired property and related properties

We can then ask if we can characterize the subset of $$V$$ corresponding to rings in which all zero-divisors are nilpotent.

Well, what are the zero-divisors in $$k[x_1,\ldots,x_n]/(f_{a,b})$$? Since $$k[x_1,\ldots,x_n]$$ is a UFD, handily enough, the zero-divisors are precisely the factors of $$f_{a,b}$$. Moreover $$k[x_1,\ldots,x_n]/(f_{a,b})$$ has nilpotents if and only if $$f_{a,b}$$ is not square free (take the radical of $$f_{a,b}$$, it will be nilpotent if and only if it is nonzero, if and only if $$f_{a,b}$$ is not square free). However the ring $$k[x_1,\ldots,x_n]/(f_{a,b})$$ satisfies our property that all zero-divisors are nilpotent if and only if $$f_{a,b}$$ is a power of an irreducible polynomial.

Step 3: Conclude

Note that if $$f_{a,b}$$ is square-free if and only if it is relatively prime to $$f_{a,b}'$$, which is true if and only if $$\Disc(f_{a,b})\ne 0$$. Thus every point $$(a,b)\in\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$ corresponding to a ring with any nilpotents at all, let alone one in which every zero-divisor is nilpotent, satisfies a polynomial equation, $$\Disc(f_{a,b})=0$$. Thus rings in our parametrized class with any nilpotents at all correspond to a (proper) Zariski closed subset of $$\PP^{r_1-1}_k\times_k\PP^{r_2-1}_k$$, which is irreducible (see here).

Thus rings in our parametrized class satisfying your property (all zero-divisors are nilpotent) are contained in a codimension 1 class of rings (any nilpotents). Hence almost all rings in our parametrized class have zero-divisors that are not nilpotent.

Notes

This is really rough. The parametrized class of rings is a very specific, very nice subset of $$k$$-algebras. Nonetheless, hopefully it will give you (or other readers) intuition on why very few rings should have the property you want (as long as they have some zero divisors of course).

I needed to eliminate integral domains because I chose to parametrize hypersurfaces, and most hypersurfaces are irreducible. (Check out the link, Qiaochu Yuan gives a really nice quick proof of this fact). There's a reasonable chance that I wouldn't have needed to eliminate integral domains if I'd chosen e.g. codimension 2 subvarieties, but those are much harder to characterize.

• I was thinking of excluding integral domains as well, but notice for integers that isn't necessary. A ring of integers modulo $n$ is an integral domain only if $n$ is prime. – Matt Samuel Dec 23 '18 at 0:41
• @MattSamuel, I originally didn't think to do so, but unfortunately, the particular class of rings I chose to parametrize requires me to do so, since most hypersurfaces in dimensions $\ge 2$ are irreducible. I suspect I wouldn't have to eliminate integral domains if I'd chosen e.g. codimension 2 subspaces, but those are harder to work with. – jgon Dec 23 '18 at 1:07
• I'll add a link to a great answer by Qiaochu Yuan justifying the claim that most hypersurfaces are irreducible when I eventually get back on a computer and I remember. – jgon Dec 23 '18 at 1:08

Another class of examples are matrix rings over a field. A $$n\times n$$ matrix $$A$$ is a zero-divisor if and only if $$A$$ is not invertible.

Certainly if $$A$$ is a zero divisor then $$A$$ cannot be invertible. If $$A$$ is not invertible then $$\ker A \ne 0$$ and taking any non-zero matrix $$B$$ with $$\operatorname{col}B \subseteq \ker A$$ you have $$AB = 0$$. For example, let $$v$$ be a non-zero vector with $$Av = 0$$ and let $$B$$ be the matrix whose columns are all $$v$$.

The nilpotent matrices are rare among zero-divisors. Specifically, nilpotent matrices are solutions to the equation $$A^n = 0$$. As a rule, there are always more non-solutions to a polynomial equation than there are solutions.

In precise terms, we look at the zero set $$\{A : \det A = 0\}$$ (the set of zero-divisors). This zero-set is a manifold if the underlying field is $$\mathbf{C}$$ and is a variety in the general case. The non-nilpotents are dense in the Zariski topology looking over a general ring, and dense in the classical topology looking over $$\mathbf C$$.


# Positive or negative remainder

Is 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?

fresh_42
Mentor
Is 23 = 5(-4)-3 gives a remainder -3 when divided by 5 ? is this statement true ? some of my colleagues said that remainder cannot be negative numbers as definition but I am doubt that can -3 be a remainder too?
Usually we consider entire equivalence classes in such cases: Every single element of ##\{\ldots -13, -8, -3, 2, 7 , 12, \ldots\}## belongs to the same remainder of a division by ##5##. We then define all five possible classes
##\{\ldots -15, -10, -5, 0, 5, 10, \ldots\}##
##\{\ldots -14, -9, -4, 1, 6, 11, \ldots\}##
##\{\ldots -13, -8, -3, 2, 7, 12, \ldots\}##
##\{\ldots -12, -7, -2, 3, 8, 13, \ldots\}##
##\{\ldots -11, -6, -1, 4, 9, 14, \ldots\}##
as elements of a new set with five elements ##\{ \; \{\ldots -15, -10, -5, 0, 5, 10, \ldots\}\, , \, \{\ldots -14, -9, -4, 1, 6, 11, \ldots\}\, , \, \ldots \}##.

This notation is a bit nasty to handle, so we choose one representative out of every set. E.g. ##\{[-15],[-9],[12],[3],[-1]\}## could be chosen, but this is still a bit messy to do calculations with. So the most convenient representation is ##\{[0],[1],[2],[3],[4]\}## with the non-negative remainders smaller than ##5##. However, this is only a convention. ##-3## is a remainder, too, belonging to the class ##[2]##. So the answer to your questions is: The statement is true, as all integers are remainders.

mfb
Mentor
The remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.

The remainder is usually required to be between 0 and N-1 inclusive. 23 and -2 (not -3) are in the same equivalence class. This can also be written as 23 = -2 mod 5.
Sorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?

jbriggs444
Homework Helper
Sorry it should be -23 = 5(-4) - 3 , so in conclusion is, this statement true ?
"-23 divided by 5 is -4 with a remainder of -3". I would consider that statement true.
"-23 divided by 5 is -5 with a remainder of 2". I would also consider that statement to be true.

The convention you use for integer division will determine which of those statements is conventional and which is unconventional.

In many programming languages, integer division follows a "truncate toward zero" convention. For instance, in Ada, -23/5 = -4. The "rem" operator then gives the remainder. So -23 rem 5 = -3.

If one adopts a convention that integer division (by a positive number) truncates toward negative infinity then one would get a different conventional remainder. -23/5 would be -5 and -23 mod 5 would be +2. The Ada "mod" operator uses this convention.

In mathematics, one typically adopts the line of reasoning given by @fresh_42 in post#2 above. The canonical exemplar in the equivalence class of possible remainders is normally the one in the range from 0 to divisor - 1.


Homework Help: Help with circular motion

1. Dec 15, 2004

newcool

Hi, I recieved this problem today. If you put a small marble of mass m on top of a large round object that has a radius r, and then release the marble, when will it come off of the side so that the large round object will not be in contact with the marble?

The answer is arccosine of 2/3. I have tried making free body diagrams of the object when the normal force is 0 and got this equation: a * cosin(theta) = g.

I am stuck as to how to get the right answer.

Thanks for any help

2. Dec 15, 2004

Staff: Mentor

As the marble rolls down the sphere, it will maintain contact as long as there is sufficient force to produce the required centripetal acceleration. Ask yourself: What provides the centripetal force? How does the required centripetal acceleration depend on the the angle that the marble makes with the vertical?

3. Dec 20, 2004

newcool

I made a free body diagram of when the sphere has a normal force of 0. I got one force,F_c pointing in towards the center and another force mg pointing down. Taking the y value of F_c, i got that:
$$F_c*cos(\theta) + mg = ma$$
$$mv^2/r * cos(\theta) + mg = mv^2/r$$
$$v^2/r * cos(\theta) + g = v^2/r$$

I tried a different approach using the fact that the sum of all energies is equal to 0.
$$0 = \Delta K + \Delta U_g$$
$$0 = 1/2 *mv^2 + -mgh$$
$$mgh = 1/2 *mv^2$$
$$gh = 1/2 *v^2$$
$$h = r - r*cos(\theta)$$

so $$gr(1 - cos(\theta)) = 1/2 * v^2$$

However, I have gotten nowhere with these equations. Any input on what I did wrong would be appreciated.

Thanks

4. Dec 20, 2004

Pyrrhus

Ok the answer is acrosinus of 2/3 = 48.2 degrees.

Force analysis

$$n - mg \cos \theta = -m \frac{v^2}{R}$$

The object will fall when n = 0 so

$$v^2 = Rg \cos \theta$$

when it loses contact with the surface

Assuming isolated system

I can use Conservation of mechanical energy

$$K + \Omega = K_{0} + \Omega_{0}$$

$$\frac{1}{2}mv^2 + mgR \cos \theta = 0 + mgR$$

Plugging our speed when it loses contact

$$\frac{1}{2}mRg \cos \theta + mgR \cos \theta = mgR$$

which gives:

$$\cos \theta = \frac{2}{3}$$

$$\theta = 48.2^{o}$$

which is the angle when it will lose contact with the surface.

That's of course assuming an isolated system.

5. Dec 20, 2004

Staff: Mentor

One big problem here: You seem to be treating the "centripetal force" as though it were a separate force (like friction or weight). Not so! "Centripetal" just means "towards the center": The centripetal force is just those real forces that act towards the center. The only forces acting on the mass m are: (1) mg, acting down, and (2) N, the normal force, acting normal to the surface.

So what's the centripetal force? Just the components of those forces acting towards the center of the circular motion, thus producing the centripetal acceleration:
$mg cos\theta - N = F_c = mv^2/r$
Of course, you'll set the normal force to zero, so:
$mg cos\theta = mv^2/r$

Exactly correct. But that's not a "different approach"--it's a necessary part of solving this problem!

Now just combine the two equations and solve for $\theta$.

6. Dec 20, 2004

Pyrrhus

Too bad, i just did the work

Well, also read Doc Al's explanation, it's quite good

7. Dec 20, 2004

newcool

Thanks for the help everyone. Just one question, Doc, how did you get that
$$mg cos\theta - N = F_c = mv^2/r$$

Isn't mg pointing straight down so the component that points towards the center is
$$mg/cos\theta$$

8. Dec 20, 2004

Pyrrhus

We got a vector

$$\vec{R}$$

with y component

$$R \cos \theta$$

and x component

$$R \sin \theta$$

where do you get $\frac{mg}{\cos \theta}$ ???

9. Dec 20, 2004

Pyrrhus

I think you got a misconception with centripetal force, Centripetal force is a role asigned to forces, because they act towards the center. The forces acting on the body are normal and the weight, and the components acting towards the center are equal to $m \frac{v^2}{r}$.

10. Dec 21, 2004

Staff: Mentor

Yes, mg points straight down. Therefore its component towards the center will be $mg cos\theta$, not $mg/cos\theta$. (Draw yourself a picture.)

I believe you are thinking like this: That mg is the vertical component of the "centripetal force" ($F_c$), so $F_c cos\theta = mg$ ==> $F_c = mg/cos\theta$. This is incorrect thinking.

One thing you must realize is that "centripetal" is just a decription of the direction that a force has: it just means "towards the center". (Another term used is "radial".) It is not a kind of force. It's just like describing a force as a horizontal force.

Think like this: The mass m must have an acceleration towards the center since it moves in a circle. So, let's apply Newton's 2nd Law in that radial (or centripetal) direction. As Cyclovenom and I have explained, the only forces acting on the mass are gravity and the normal force. We know that N points away from the center. What's the component of gravity (mg) towards the center? $mg cos\theta$. So:
$$F_{towards-center} = ma_{towards-center}$$
$$mg cos\theta - N = mv^2/r$$

Make sense?

11. Dec 21, 2004

newcool

Thanks for all the help, Doc and Cyclove, I understand it now


# Combination notation vs. Binomial Coefficient Formula

I'm studying probability and statistics and had a question regarding notation.

I noticed that combinations and the binomial coefficient are essentially the same thing, that is:

$$\binom{n}{k}\ =\ _nC_k\ =\ \frac{n!}{(n-k)!k!}$$

But I was wondering, is there a particular difference between the two that people should be aware of? For example, are there certain use cases where one is preferred over the other?

Thank you.

• They are exactly the same thing, just different notation. I think the most popular (at least from what I see) is the $\binom{n}{k}$ notation, however I have seen $C^n_r$ used in probability courses. – Dave Jun 29 '18 at 2:48
• Also ${^n\mathrm C_r}$ – Graham Kemp Jun 29 '18 at 3:31
• Also $_r\text{C}^n$ – Dzoooks Jun 29 '18 at 4:04

All three expressions mean when dealing with combinations the same. But there are some aspects which should be considered.

We often find the binomial coefficients $\binom{n}{k}$ resp. $_nC_k$ defined by factorials. \begin{align*} \binom{n}{k}:=\frac{n!}{k!(n-k)!}\qquad\qquad\text{resp.}\qquad\qquad _nC_k:=\frac{n!}{k!(n-k)!} \end{align*} From this point of view the factorials $n!$ can be seen as basic building blocks for the shorthand notations $\binom{n}{k}$ and $_nC_k$. Since using factorials is more fundamental than the other two representations I will consider in the following only $\binom{n}{k}$ and $_nC_k$.

Historical aspects:

• C. Jordan writes in his classic Calculus of Finite Differences (1939)

• ($\mathrm{\S}$ 22): Since the binomial coefficient is without doubt the most important function of the Calculus of Finite Differences it was necessary to adopt some brief notation for this function. We accepted above the notation of J. L. Raabe [Journal für reine und angewandte Mathematik 1851, Vol. 42, p. 350] which is most in use now, putting

\begin{align*} \binom{x}{n}=\frac{x(x-1)(x-2)\cdots(x-n+1)}{1\cdot2\cdot3\cdots n} \end{align*}

• ($\mathrm{\S}$ 22, footnote 18): Euler first used the notation $\left[\frac{x}{n}\right]$ in Acta Acad. Petrop.V, 1781 and then $\left(\frac{x}{n}\right)$ in Nova Acta Acad. Petrop. XV. 1799-1802. Raabe's notation $\binom{x}{n}$ is a slight modification of the second. It is used for instance in:
• Bierens de Haan, Tables d'Int$\mathrm{\acute{e}}$grales d$\mathrm{\acute{e}}$finies, Leide, 1867
• Hagen, Synopsis Vol. I. p. 57, Leipzig, 1891,
• Pascal, Repertorium Vol. I, p. 47, Leipzig, 1910,
• Encyclopädie der Math. Wissenschaften, 1898-1930,
• L. M. Milne Thomson, Calculus of Finite Differences, 1933,
• G. H. Hardy, Course of Pure Mathematics, p. 256, 1908

Euler's notation is also cited in

• A History of Mathematical Notations by F. Cajori, which also provides some information about the use of $_nC_k$-like notations in $\mathrm{\S}$ 451:

• George Peacock (Treatise of Algebra, 1830) introduces $C_r$ for the combinations of $n$ things taken $r$ at a time.
• Robert Potts (Elementary Algebra, 1880) begins his treatment by letting the number of combinations of $n$ different things taken $r$ at a time be denoted by $^nC_r$
• W.A. Whitworth uses $C_r^n$ in Choice and Change (1886).
• G. Chrystal writes $_nC_r$ in Algebra, Part II (1899).

Note the different variations $C_r, {^{n}C}_r, C^n_r$ of $_nC_r$-like notations.

Here is a selection of some classics from the 20th century.

Some classics:

• An Introduction to Combinatorial Analysis (1958) by J. Riordan

• (3.1): ... and \begin{align*} C(n,r)=\frac{n(n-1)\cdots(n-r+1)}{r!}=\frac{n!}{r!(n-r)!}=\binom{n}{r} \end{align*} where the last symbol is that usual for binomial coefficients, that is, the coefficients in the expansion of $(a+b)^n$. ($C_r^n, C_n^r, _nC_r$ and $(n,r)$ are alternative notations; ...
• Combinatorial Identities (1968) by J. Riordan

• (1.1): Perhaps the simplest combinatorial entities are the binomial coefficients, that is, the combinations, for example of $n$ things, $k$ at a time. They take their name from the generating function for combinations, which is a power of a binomial, namely \begin{align*} (1+x)^n=\sum_{k=0}^n\binom{n}{k}x^k \end{align*} where, of course, $\binom{n}{K}=C(n,k)=\frac{n!}{k!(n-k)!}$ is the usual notation for a binomial coefficient.
• (II.4): ... the number of subpopulations of size $r$ is therefore given by $(n)_r/r!$. Expressions of this kind are known as binomial coefficients and the standard notation for them is \begin{align*} \binom{n}{r}=\frac{(n)_r}{r!}=\frac{n(n-1)\cdots(n-r+1)}{1\cdot 2\cdots (r-1)\cdot r} \end{align*}
• Advanced Combinatorics (1974) by L. Comtet

• (1.4): ... We will adopt the notation $\binom{n}{k}$, used almost in this form by Euler and fixed by Raabe, with the exclusion of all other notations, as this notation is used in the great majority of the present literature, and its use is even so still increasing. This symbol has all the qualities of a good notation: economical (no new letters introduced), expressive (it is very close in appearence to the explicit value $\frac{(n)_k}{k!}$, typical (no risk of being confused with others), and beautiful.
• Enumerative Combinatorics, Volume I (1986) by R. P. Stanley

• (I.1.4): ... where $d_n=\sum_{i=0}^n\binom{n}{i}a_ib_{n-i}$, with $\binom{n}{i}=n!/i!(n-i)!$.

and the author continues to use the notation $\binom{n}{k}$ without any more citations, indicating this notation being commonly used.

• D. E. Knuth who gave us $\TeX$ is besides being a great mathematician an extraordinary expert in typography and mathematical writing. His Mathematical Typography (1979) gives us a glimpse of his deep thoughts about these issues. Another one being the report Mathematical Writing (1987) written together with T. Larrabee and P. M. Roberts.

In $\TeX$ we use the command "$\text{n \choose k}$" giving us $\binom{n}{k}$.

The enhanced readability of the notation $\binom{n}{k}$ becomes rather obvious in more complex expressions. Compare for instance formula (5.32) in Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik which is stated for integers $l,m,n$; $n\geq 0$ as \begin{align*} \sum_{j,k}(-1)^{j+k}\binom{j+k}{k+l}\binom{r}{j}\binom{n}{k}\binom{s+n-j-k}{m-j}=(-1)^l\binom{n+r}{n+l}\binom{s-r}{m-n-l} \end{align*} with the representation \begin{align*} \sum_{j,k}(-1)^{j+k}\,_{j+k}C_{k+l}\,_rC_j\,_nC_k\,_{s+n-j-k}C_{m-j}=(-1)^l\,_{n+r}C_{n+l}\,_{s-r}C_{m-n-l} \end{align*}

• @BruceET: You're welcome. – Markus Scheuer Jun 29 '18 at 8:51
• Thank you for taking the time and effort to write such a nice answer! :) – Seankala Jun 29 '18 at 15:42
• @Sean: You're welcome. Thanks for your nice comment. :-) – Markus Scheuer Jun 29 '18 at 15:45

In books printed before $\LaTeX$ came to be widely used, setting ${n \choose r}$ into type was difficult (and expensive if used throughout a book). I 'know' this because of conversations with editors of math books over the years. By contrast, typesetting is not difficult for the variants of $C_r^n,$ including $C(n, r)$ not yet mentioned in this discussion.

Nowadays, I think there is a trend to use ${n \choose r},$ except for authors who have a strong preference for C-notations and those writing in Microsoft Word and trying to avoid using its 'equation editor'.

Curiously, a generally-accepted convention for permutations $P_r^n = r!{n \choose r}$ does not seem to have emerged. Feller used $(n)_r$ in his famous probability book, which one would have thought might have set a trend 60 years ago, but apparently not. His book may have set a record for the density per page of big parentheses for various uses of ${n \choose r}$-notation--even before $\LaTeX.$

There is no difference at all. Personally I prefer using the shortest notation aCb for obvious reasons.

• Are the "a" and "b" significant, there, or is it more the ${}_{\cdot}C_{\cdot}$ notation that you prefer? – Cameron Buie Jun 29 '18 at 2:56
• a and b are just numbers . Nothing more than that . Yes I prefer that notation because it's shorter and I see it very often – Kwnstantinos Nikoloutsos Jun 29 '18 at 2:59


# Unify the sampling of NIntegrate[ {f, g, h} w ]

I'm trying to numerically integrate a function which has a vector-valued slow part and a much faster component which is shared by all the components, i.e. an integral of the form $$\int_a^b\begin{pmatrix}f(x)\\ g(x) \\ h(x)\end{pmatrix}w(x)\,\text dx.$$ Because NIntegrate is nicely Listable on its first argument, I can feed it a list-valued argument without a problem. However, it appears to be doing each component as a completely separate integral, which results in a lot of work being re-calculated.

For example, the (simplified) example integral

samplePointsList = Reap[
NIntegrate[
{x, x^2, x^3} Cos[10 x + Cos[x]]
, {x, 0, 5}
, EvaluationMonitor :> Sow[x]
]
][[2, 1]];

gives the sample point diagram (through ListPlot[Transpose[{samplePointsList, Range[Length[samplePointsList]]}]])

It is clear that the kernel is doing the initial sampling, and then the further refinements, separately for each component. While the required sample points are not identical, there is a lot of shared work and I feel there is a fair bit of room for optimization there.

I am aware that, since the sampling requirements of each component are slightly different, binding them completely will require more evaluations in some components than would be necessary. (For instance, in the example above, the extra detail required by the first component around $2\leq x\leq 3$ would be slowed down slightly if it was required to also calculate the second and third components there, though they do not require it.) However, the components in my case are similar enough that I do not think this would be an issue.

Is there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?

One idea is to integrate once to get the sample points then compute the remaining integrals as sums:

 sample = Transpose@
SortBy[First@Last@Reap[
NIntegrate[x (c = Cos[10 x + Cos[x]]), {x, 0, 5},
EvaluationMonitor :> Sow[{x, c}]]], #[[1]] &];
wt = ((#[[3]] - #[[1]])/2) & /@ Partition[Join[{0}, sample[[1]], {5}], 3, 1];
wt.(sample[[2]] #) & /@ {sample[[1]], sample[[1]]^2, sample[[1]]^3}

{0.0133333, 0.133275, 0.861541}

Note the accuracy is not terribly good, NIntegrate gives:

{0.0125266, 0.131514, 0.855716}

Somethings a bit off in my quick&dirty trapezoid integration but i think this can be made to work.

For this example there really is little benefit to NIntegrate's adaptive sampling so we might as well just use a uniform sampling:

 np = 651;(*assumed odd for simpsons rule*)
a=5
b=0
wt = ( (a-b)/(np - 1) )/3 Join[{1}, Flatten@ConstantArray[{4, 2}, (np - 1)/2 - 1], {4, 1}] // N;
x = b + (Range[0, np - 1] (a-b)/(np - 1)) // N;
fast = Cos[10 # + Cos[#]] & /@ x // N;
(# fast).wt & /@ {x, x^2, x^3}

{0.0125266, 0.131514, 0.855716}

Is there a way to force Mathematica into this sort of separation of the integrand and unification of the sampling? If not, is there a way to make it aware of the previously calculated values and sampling points which will speed up the process?

are "yes" and "yes" as demonstrated and discussed in this post to "How to calculate the numerical integral more efficiently?".

More concretely:

Import["https://raw.githubusercontent.com/antononcube/\
MathematicaForPrediction/master/Misc/ArrayOfFunctionsRule.m"]

funcExpr = {x, x^2, x^3} Cos[10 x + Cos[x]]

(* {x Cos[10 x + Cos[x]], x^2 Cos[10 x + Cos[x]],
x^3 Cos[10 x + Cos[x]]} *)

funcMat = Table[i*funcExpr, {i, 100}];

AbsoluteTiming[
res0 = NIntegrate[funcMat, {x, 0, 5}];
]
res0[[12]]

(* {1.711, Null} *)
(* {0.150319, 1.57817, 10.2686} *)

AbsoluteTiming[
res1 = NIntegrate[1, {x, 0, 5},
Method -> {"GlobalAdaptive", "SingularityHandler" -> None,
Method -> {ArrayOfFunctionsRule, "Functions" -> funcMat}}];
]
res1[[12]]

(* {0.029007, Null} *)
(* {0.150319, 1.57817, 10.2686} *)

Norm[res0 - res1, 2]

(* 2.03398*10^-12 *)

• That's pretty interesting; I hope I can find time to test it out soon. In the meantime, can you comment on the relative gains as a function of the size of the matrix? I'm interested in $n\leq3$ but the other answer looks like it targets bigger arrays. – Emilio Pisanty Jan 9 '17 at 19:22
• @EmilioPisanty I am not sure what you mean -- performance of the ArrayOfFunctionsRule with respect to the size of the integrand? I did a similar comparison during the experiments for the linked MSE answer... – Anton Antonov Jan 10 '17 at 1:30


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# A coin is tossed 7 times. Find the probability of getting

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A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Jan 2012, 13:47
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A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH

Then why in the below problem - we use combinations and not permutations ?
I am so confused.

A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

B. 63/128
C. 4/7
D. 61/256

Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)
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Posts: 47112

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12 Jan 2012, 05:34
7
6
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Assuming the coin is fair - P(H)=P(T)=1/2

We can do as proposed by the explanation in your initial post:

Total outcomes: 2^7

Favorable outcomes:
4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);
5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;
6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;
7 heads --> combination of HHHHHHH --> 1;

P(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.

BUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))

If it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?

Now, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\frac{1-P(H=T)}{2}=\frac{1-\frac{70}{2^8}}{2}=\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.

Hope it's clear.

Similar questions for practice:
probability-question-100222.html?hilit=coin%20tossed#p772756
hard-probability-99478.html?hilit=coin%20tossed
some-ps-questions-need-explanation-99282.html?hilit=coin%20tossed
probability-question-gmatprep-85802.html?hilit=coin%20tossed
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Manager
Joined: 25 Dec 2011
Posts: 54
GMAT Date: 05-31-2012

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07 Jan 2012, 13:49
1
Hi Sorry ...1 head out of 7 can be arranged in following ways :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHTT
TTTTTHT
TTTTTTH
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Jun 2013, 07:03
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Jun 2013, 08:48
2
Probability of getting Head, P(H) = 1/2

For No. of heads > no. of Tails, there can be 4,5,6 or 7 Heads.

As per GMAT Club math book, probability of occurrence of an event k times in a n time sequence, for independent and mutually exclusive events:

$$P=C(n,k)*p^k*(1-p)^n^-^k$$
$$P(H=4) = C(7,4)*(1/2)^4*(1/2)^3 = C(7,4) * (1/2)^7$$
$$P(H=5) = C(7,5) * (1/2)^7$$
$$P(H=6) = C(7,6) * (1/2)^7$$
$$P(H=7) = C(7,7) * (1/2)^7$$

Total $$P(H > 3) = P(H=4) + P(H=5) + P(H=6) + P(H=7)$$
$$= [ C(7,4) + C(7,5) + C(7,6) + C(7,7)] * (1/2)^7$$
$$= (35 + 21 + 7 + 1) / 128$$
$$= 64/128$$
$$= 1/2$$
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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05 May 2014, 05:47
My approach

7 toss
coin has 2 out comes H or T

now basically if i see this a game where there are 2 teams one selects Heads and other team selects Tails. If in 7 toss whichever comes maximum (heads or Tails) the corresponding team wins..

Clearly the probability is 50% for both the cases max Heads or max Tails............
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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01 May 2015, 02:54
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.

I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.
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A coin is tossed 7 times. Find the probability of getting  [#permalink]

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01 May 2015, 05:11
2
DipikaP wrote:
For those who have trouble grasping the concept behind this - check out Khan Academy lessons on probability and combinatorics. I'm not allowed to post urls as a newbie but a simple Google search will throw up the link.

I couldn't make head or tail of these questions before seeing them, tried memorizing the formulae and always messed up. I invested 3 hours in going through those videos and can now solve these questions without knowing any formulae - it's all conceptual. Sal's great with breaking down concepts to simple, relatable stuff.

Hi Dipika,

You are totally right when you say "it is all conceptual". Trying to memorise formulae is not the right approach.

For example in this question:
First we should think what are the possible outcomes when we toss a coin: head or tail (2 outcomes)
Now as the coin is fair, the probability that we will get a head or a tail is 1/2

To illustrate, let's take a smaller version of the above question:

What is the probability of getting more heads in 3 tosses?

1st case: We can get 3 heads: HHH
Probability of HHH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 3 heads = 1/8

2nd case: We can get two heads: HHT, HTH, THH
Probability of HHT = 1/2 * 1/2 * 1/2 = 1/8
Probability of HTH = 1/2 * 1/2 * 1/2 = 1/8
Probability of THH = 1/2 * 1/2 * 1/2 = 1/8
Probability of getting 2 heads = 3* 1/8

As you can see HHT, HTH and THH are different arrangements of HHT
Probability of getting 2 heads = (No. of arrangements of HHT)* (Probability of getting HHT) = 3!/2! * (1/2 * 1/2 * 1/2) = 3/8

Total probability of getting more heads in 3 tosses = 1/8 + 3/8 = 4/8 = 1/2

Thinking on these lines you can solve the above question easily.

But if you think a little further, we are talking about odd number of tosses (3 and 7). So, either there will be more heads or more tails. It is not possible to get equal number of heads and tails. Hence, in half of the outcomes we will get more heads than tails and in the the other half we will have more tails than heads. Thus, the probability of getting more heads = probability of getting more tails = 1/2.
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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04 Aug 2015, 12:50
morya003 wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH

Then why in the below problem - we use combinations and not permutations ?
I am so confused.

A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

B. 63/128
C. 4/7
D. 61/256

Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)

Since number of times coin has been tossed is 7, either number of heads will be more than tails or vice versa. There is no way number of heads become equal to number of tails. Since head and tails are equally favourable outcome of a coin, possibility of getting more heads = possiblity of getting more tails = 1/2. In other words there are only 2 equally likely events (event 1: heads more than tails, event2: tails more than heads) constituting all the outcomes.

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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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07 Dec 2016, 20:46
Total outcome = 2^7 = 128

HHHHHHH = 7!/7! = 1
HHHHHHT = 7!/1!6! = 7
HHHHHTT = 7!/2!5! = 21
HHHHTTT = 7!/3!4! = 35

1+7+21+35 = 36
36/128 = 1/2

A.
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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02 Feb 2017, 04:42
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain

thank u..
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Posts: 47112
Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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02 Feb 2017, 06:12
manojhanagandi wrote:
hi bunnel...
As stated in the question" Find the probability of getting more heads than tails in all 7 tosses?"
tails in all the tosses means no heads ie 0 heads and more than that means at least one heads.
so we have to find the the solution for atleast one heads in 7 tosses...
can we restate the question as above?
pls explain

thank u..

If you read the solutions above you'll see that this is not correct. More heads than tails in all 7 tosses means at least 4 heads (so more than half must be heads):

HHHHTTT
HHHHHTT
HHHHHHT
HHHHHHH
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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03 Feb 2017, 04:22
Thank You bunuel...
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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06 Sep 2017, 00:48
Ans is A

HHHHTTT= $$(0.5)^4$$ x $$(0.5)^3$$ {7C4}
similarly for 5 H , 6H and 7 H we will calculate only 7C5 , 7C6 and 7C7 will change other part (0.5)^7 remains same which is multiplied, so add all 4 cases of 4H 5H 6H 7H
$$(0.5)^7$$ x {7C4+7C5+7C6 +7C7}
$$\frac{64}{128}$$= $$\frac{1}{2}$$
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Re: A coin is tossed 7 times. Find the probability of getting  [#permalink]

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06 Sep 2017, 03:39
morya003 wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Hi
I want to understand why combination has been used in the below problem.
I thought permutation is used for order and - for probability one needs to find the number of outcomes as well as order of the outcomes.
For example if the question was - probability of getting 1 head then we would have put it as follows :-
HTTTTTT
THTTTTT
TTHTTTT
TTTHTTT
TTTTHHH
TTTTTHH
TTTTTTH

Then why in the below problem - we use combinations and not permutations ?
I am so confused.

A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?

B. 63/128
C. 4/7
D. 61/256

Explanation
ANS. (a) ( Total outcomes= 2^7 = 128, Number outcomes for which heads are more than tails = 7 combination 4 (Heads=4 & Tails=3) + 7 combination 5 + 7 combination 6 + 7 combination 7) = 35+21+7+1= 64, so probability of getting more heads = 64/128 = ½)

Now this is a very simple question and can be solved without doing all the long calculations.
Here important thing to notice is P(H) = P(T) = 1/2 in a single toss.

Now 7 Toss can have following possible ways
0H 7T say x way
1H 6T say y way
2H 5T say z way
3H 4T say w way
--------------------------
4H 3T w way
5H 2T z way
6H 1T x way
7H 0T x way

So 4 out of 8 ways will have heads greater than Tail. Also P(H) = P(T) = 1/2 = i.e. equal in a single toss. So, we don't need to calculate the number of ways for each case.

Require probability = (w+z+x+y)[/(x+y+z+w) +(w+z+x+y)] = 1/2

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A coin is tossed 7 times. Find the probability of getting  [#permalink]

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20 Dec 2017, 09:22
My approach: either Number of heads can be more or less than tails. They can't be equal as no of toss is 7 which is odd.

Hence probability is (no of cases when no of head is more) /((no of cases when no of head is more)+(no of cases when no of head is more))= 1/2

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Bunuel wrote:
A coin is tossed 7 times. Find the probability of getting more heads than tails in all 7 tosses?
A. 1/2
B. 63/128
C. 4/7
D. 61/256
E. 63/64

Assuming the coin is fair - P(H)=P(T)=1/2

We can do as proposed by the explanation in your initial post:

Total outcomes: 2^7

Favorable outcomes:
4 heads --> combination of HHHHTTT --> 7!/(4!*3!)=35 (# of permutation of 7 letters out of which 4 H's and 3 T's are identical);
5 heads --> combination of HHHHHTT --> 7!/(5!*2!)=21;
6 heads --> combination of HHHHHHT --> 7!/(6!*1!)=7;
7 heads --> combination of HHHHHHH --> 1;

P(H>T)=Favorable outcomes/Total outcomes=(35+21+7+1)/2^7=1/2.

BUT: there is MUCH simpler and elegant way to solve this question. Since the probability of getting either heads or tails is equal (1/2) and a tie in 7 (odd) tosses is not possible then the probability of getting more heads than tails = to the probability of getting more tails than heads = 1/2. How else? Does the probability favor any of tails or heads? (The distribution of the probabilities is symmetrical: P(H=7)=P(T=7), P(H=5)=P(T=5), ... also P(H>4)=P(T>4))

If it were: A fair coin is tossed 8 times. Find the probability of getting more heads than tails in all 8 tosses?

Now, almost the same here: as 8 is even then a tie is possible but again as distribution is symmetrical then $$P(H>T)=\frac{1-P(H=T)}{2}=P(T>H)$$ (so we just subtract the probability of a tie and then divide the given value by 2 as P(H>T)=P(H<T)). As $$P(H=T)=\frac{8!}{4!*4!}=70$$ (# of permutation of 8 letters HHHHTTTT, out of which 4 H's and H T's are identical) then $$P(H>T)=\frac{1-P(H=T)}{2}=\frac{1-\frac{70}{2^8}}{2}=\frac{93}{256}$$. You can check this in following way: total # of outcomes = 2^8=256, out of which in 70 cases there will be a tie, in 93 cases H>T and also in 93 cases T>H --> 70+93+93=256.

Hope it's clear.

Similar questions for practice:
http://gmatclub.com/forum/probability-q ... ed#p772756
http://gmatclub.com/forum/hard-probabil ... n%20tossed
http://gmatclub.com/forum/some-ps-quest ... n%20tossed
http://gmatclub.com/forum/probability-q ... n%20tossed

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