Proving a statement involving the rational root theorem The rational root theorem states that, given $P \in \mathbb{Z}[x], P = \displaystyle\sum_{j=0}^n a_j x^j$ with $a_0 \neq 0$ and $a_n \neq 0$, if $P(p/q)=0, p\perp q$ then $p|a_0$ and $q|a_n$.

As a follow-up to this question, I'd like to know how can I prove the following:

$$a_0 = a_n \displaystyle\prod_{j=1}^n (-r_j)$$, with $r_j$ being roots of P(x)

This is the same as saying:

$$a_0 = a_n \frac{-p}{q} \displaystyle\prod_{j=2}^n (-r_j)$$

Rearranging this, we get:

$$a_n = q\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)}$$

I want to prove that $$\frac{a_0}{-p \displaystyle\prod_{j=2}^n (-r_j)} \in \mathbb{Z}$$ without using the rational root theorem (actually, I want to prove the rational root theorem using this identity).

So far I've tried to prove it by contradiction, assuming $a_0 = k(-p)\displaystyle\prod_{j=2}^n (-r_j) + \epsilon$ with $\epsilon \in (0, (-p)\displaystyle\prod_{j=2}^n (-r_j))$ but I don't seem to find anything weird rearranging that equation.

If you manage to find a proof please just post a small hint. Thanks in advance.

Those are just Vieta's formulas for the coefficients in terms of the roots.


How is it called a minimal morphism of category Rel? How is it called a Rel-morphism $(f;A;B)$ such that: a. $f=\varnothing$; b. $f\ne\varnothing$?

Is there any special term for this?

First things first: $\textbf{Rel}$ is an enriched category, and in more than one way.

$\textbf{Rel}$ is a pointed category: that means it has a zero object and is enriched over the category of pointed sets. The empty relation $\emptyset : A \to B$ can be characterised as the unique relation $A \to B$ that factors through the zero object $\emptyset$, so we call it a zero morphism. Obviously, this property is preserved by composition on the left and on the right.
$\textbf{Rel}$ is a locally posetal category: that means each hom-set is partially ordered, and composition is monotonic. The empty relation is clearly the bottom element of each hom-poset.
Extending the above in a trivial way, $\textbf{Rel}$ is a 2-category: that means each hom-set is a category, and composition is functorial. The empty relation is clearly the initial object of each hom-category.
$\textbf{Rel}$ is in fact enriched over itself, as discussed here, but to make this concrete, we should say that $\textbf{Rel}$ is enriched over the category of cocomplete join semilattices. (I say this instead of boolean algebra because composition only preserves joins.) As in (2), the empty relation is distinguished here as the bottom element of each hom-semilattice.


Find the limit points of these sequences $1,2,3,1\cdots$ and $(-1)^n$ I have two sequences
$$1,2,3,1,2,3,1,2,3,1,2,3,\cdots$$
and the second sequence is
$(-1)^n$.

I am saying that both the sequence have no limit point but my friend is saying both the sequence have limit point

Which is correct, please help...

The definition is: A point $l$ is said to be a limit point of a sequence $\{a_n\}$ if every neighborhood of $l$ contains infinite terms of the sequence. The limit points of first sequence are $1,2,3$ and for the 2nd sequence the limit points are $-1$ and $1$. As every neighborhood of $1$ contains infinite terms: namely $a_{3k+1},~ k=0,1,2...$ so $1$ is a limit point of the first sequence. Similarly you can do for other points

Neither sequence has a limit because for both sequences, the claim

$$\exists L: \forall \epsilon > 0 \exists N\in\mathbb N\forall n\in\mathbb N: n>N\implies |L-a_n|<\epsilon$$

or, in writing:

  There exists an $L$ such that, for all $\epsilon>0$, there exists some $N$ such that $|L-a_n|<\epsilon$ is true for all $n>N$

is false.

You can show the claim is false by showing that its negation,

$$\forall L\exists \epsilon > 0\forall N\in \mathbb N\exists n>N: |L-a_n|\geq\epsilon$$

is true.

An easier way to disprove the claim, of course, is to show that both sequence have two subsequences with different limits, and that is only possible if the sequence is not convergent.


Find an efficient algorithm on shortest path $l$ while at most $C$ cost a graph $G$, each edge $e$ has two parameters: l(e) is the length and c(e) is the cost; both are nonnegative. Given a starting vertex $s$, find the shortest total length path to a destination vertex $t$ such that the total cost along this path is at most C.

My attempt:
Use Bellman-Ford algorithm to, iterate $V$ times to get the short path from $s$ to $t$. Meanwhile store every path that costs less than $C$ when updating $t$. After iteration, I compare those paths to get the shortest paths with at most $C$ cost.

This runs $O(V \times E)$. I feel it's not elegant because of its complexity.

Is there a more efficient way or new algorithm to do this?

I know in case the cost is enumerable (e.g. $C$ and the cost on every edge are integers), one can do it by splitting every point into $C$ points and then applying Dijkstra's algorithm. But in case the cost is not enumerable, or if it's computationally expensive to quantize the cost and split the vertices, we would have to try something else.

I tried two heuristics and failed. Number one was to minimize $l+\lambda C$ using Dijkstra and then find the smallest $\lambda>0$ that produces a path that is within budget. The problem with this algorithm is that sometimes the possibility frontier of $(l,C)$ is not concave. Therefore, as one tunes the value of $\lambda$, it's possible to jump from one extremely long path with low cost to another extremely costly path with short distance without going through the middle grounds.

Then I tried the network flow idea. Let the in-flow be $1$ unit into the starting point and the out-flow be $1$ unit out of the destination. The classical shortest path problem would automatically give a flow that only follows one path - the shortest path. In case there are multiple shortest paths, only those flows that follow one path would be on a vertex of the feasible region. So any linear programming (LP) solver that only lands on vertices would give a flow that follows only one path, which is at the same time shortest. But once there's the cost budget, although the LP solver still finds a solution, it need not follow only one path any more.

So I'm not sure if a polynomial-time algorithm exists for this shortest path problem with an upper bound for the cost budget.

This problem is NP-hard as one may reduce the Partition Problem to it.

For a set $S=\{a_1, \dots, a_n\}$ and a map function $s:S\rightarrow N^+$ and $A=\sum_{a\in S} s(a)$, the partition problem asks whether whether there is $S' \subseteq S $ s.t. $\sum_{a\in S'} s(a) = A/2$.

Create n+1 node, $B_1, \dots, B_{n+1}$, add two edges from $B_i$ to $B_{i+1}$, one edge costs $s(a_i) + 1$ with distance $1$, the other costs $1$ with distance $s(a_i) + 1$ (of course one may replace the parallel edges by adding an intemediate node).  Let $C$ be $A/2 + n$, and if the minimum distance from $B_1$ to $B_n$ can reach $A/2 + n$ within cost of $C$, there will be a valid partition.


Finding the speed of a person After hours of struggling I'm posting it here. I hope someone will solve it.

  David and Jess start at the same time from points $A$ and $B$ and travel towards $B$ and $A$ respectively. After they meet, David takes $40$ minutes to reach $B$ and Jess takes $1.5$ hours to reach $A$. If David's speed is $36$ km/h, what is Jess's speed?

My Work:
if two people P and Q start at the same same time from A and B, after crossing each other they take X and Y seconds to reach B and A then P's speed : Q's speed = squareroot(y) : squareroot(x). i.e squareroot(1.5*60):squareroot(40) which is 3:2. so Jess's speed 36*2/3= 24Kmph. But i dont understand how P's speed : Q's speed = squareroot(y) : squareroot(x).

Let $S_D$ be Dave's speed, and $S_J$ be Jess's. Say they meet at time $t$ at distance $d_1$ from $A$ and $d_2$ from $B$. We can write $t=d_1/S_D$  and $t = d_2/S_J$ and combine them into
$$\frac{d_1}{d_2} = \frac{S_D}{S_J}.$$

On the latter part of the journey,
we can write similar equations $40=d_2/S_D$ and $90=d_1/S_J$ and combine these into

$$\frac{d_1}{d_2} = \frac{90S_J}{40S_D}.$$

Combining this with the equation above we get

$$90S_J^2 = 40S_D^2,$$

or
$$S_J = \frac23 S_D.$$

$S_D = 36/60$ km/minute so $S_J=24/60$ km/minute, or $24$ kph.

David needs x+2/3 hours at 36 km/h, so the full distance is 36*(x+2/3) km, right?
Jess needs x+3/2 hours for the 36*(x+2/3) km, so he travels 36*(x+2/3)/(x+3/2).

David will need x hours for 36x km
Jess will need x hours for 36*(x+2/3)/(x+3/2) km.

So the full distance is 36x+36*(x+2/3)/(x+3/2).

36*(x+2/3) = 36x+36*(x+2/3)/(x+3/2)

or shorter

(x+2/3) = x+(x+2/3)/(x+3/2)

which then is

2/3 = (x+2/3)/(x+3/2)
(x+3/2)*2/3 = (x+2/3)
2x/3+1 = x+2/3
2x/3-x = 2/3-1
x = 1

Quod errat demonstrator.


Show that two $R$-algebras are not isomorphic. Consider the $\mathbb{R}$-algebras $A := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{a^2+b^2+c^2+d^2}\right]$ and $B := \mathbb{R}\left[a,b,c,d\right]\left[\frac{1}{ad-bc}\right]$ (so we are adjoining the inverses of $a^2 + b^2 + c^2 + d^2$ and $ad-bc$, respectively).
I claim that these are not isomorphic as $\mathbb{R}$-algebras, but I'm not sure how to formalize this.

My intuition says that this is because $$a^2 + b^2 + c^2 + d^2 = 0 \iff a=b=c=d=0,$$ while $ad-bc=0$ has nonzero solutions. I tried to formalize this using evaluation homomorphisms, but to no avail. I also tried assuming they were isomorphic and obtaining a contradiction, but haven't been able to.
Does anyone have any pointers on how to approach this?

I don't know a purely algebraic proof. The following proof involves a little algebraic geometry and algebraic topology, so it may not be the intended answer to your question. In that case, please specify your context (i.e. where did you get the problem from).
Suppose $A$ and $B$ are isomorphic as $\Bbb R$-algebras. Then their corresponding $\Bbb R$-points, viewed as real manifolds, are isomorphic, as polynomial maps are continuous (and even $C^\infty$).
However, these two manifolds are: $A(\Bbb R) = \Bbb R^4 \backslash \{0\}$ and $B(\Bbb R) = \operatorname{GL}(2, \Bbb R)$. It is obvious that they are not isomorphic as topological spaces, by e.g. comparing their simplicial (co)homology groups.

Question: "Does anyone have any pointers on how to approach this?"
Answer: Your argument is correct: Define
$$A := \mathbb{R}[a,b,c,d]/(a^2+b^2+c^2+d^2), B := \mathbb{R}[a,b,c,d]/(ad-bc).$$
Formally, if there is an isomorphism of affine schemes (over $k$ - the real numbers) $\phi: T:=Spec(A) \cong S:=Spec(B)$ it would follow that $T$ and $S$ have the same $k$-rational points: The map $\phi$ would induce a 1-1 correspondence
$$\phi(k): T(k) \cong S(k)$$
and an equality of sets
$$T(k) \cong Hom_{k-alg}(A,k) \cong Hom_{k-alg}(B,k)\cong S(k),$$
but $T$ has only one $k$-rational point: $x:=(0,0,0,0)$. $S$ has a rational point $x:=(a,b,c,d)$ for any $x$ with $det(x)=0$.
Formally: If $T\cong S$ are isomorphic schemes (over $k$) it follows $h_T\cong h_S$
are isomorphic as functors (over $k$). Hence an isomorphism $\phi$ of affine schemes over $k$ does not exist.


What is wrong with this fake proof $e^i = 1$? $$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$

Obviously, one of my algebraic manipulations is not valid.

As said in the comments, the expressions $(a^b)^c$ and $a^{bc}$ are in fact multivalued functions; they are not a uniquely determined complex number. A classic example of a multivalued function is the complex logarithm denoted by $\log(z)$, $z\in\mathbb{C}$. The complex logarithm $\log(z)$ is any complex number $w$ satisfying $e^w=z$ (which has several solutions, see e.g. this), and hence $\log(z)$ gives rise to a whole set of complex numbers instead of just a single complex number.

Complex exponentiation such as $z^w$ for $z,w\in\mathbb{C}$ is usually defined as
$$
z^w=\exp(w\log(z)),
$$
where $\log(z)$ is the complex logarithm, and hence this is also a multivalued function. I hope this sheds some light on the problems with doing manipulations on complex numbers as if they were real numbers. See also this for other examples of identities which fail when using complex numbers as they were real numbers.

$$e^{i} = e^{i + i2\pi k} = e^{(i2\pi + i(2\pi)^2k)\frac{1}{2\pi}}=\bigg(e^{(i2\pi + i(2\pi)^2k)}\bigg)^{\frac{1}{2 \pi}} = \bigg(e^{i2\pi} e^{i(2\pi)^2k}\bigg)^{\frac{1}{2 \pi}} = \bigg( e^{i(2\pi)^2k}\bigg)^{\frac{1}{2 \pi}} = e^{i2\pi k}$$

$$e^{i} \ne e^{i2\pi k}$$

maybe you are right, its still a paradox.

Could it be that the algebraic property:

$$a^{pq} = (a^p)^q$$

is not a valid manipulation for complex exponentials?

$$e^{\mathbf{i}pq} \ne (e^{\mathbf{i}p})^q$$

pulling a power out of the complex exponential has an effect of changing the frequency of oscillation of the complex exponential...

I would change your complex algebra rule book to state that pulling a power out of a complex exponential can lead to inconsistent equations and should be avoided.

complex exponential anti-rule #1:  Algebraically pulling a power out of the complex exponential leads to inconsistent equations.  While $a^{pq} = (a^p)^q$  is true for normal algebra, for complex exponentials:  $e^{\mathbf{i}pq} \ne (e^{\mathbf{i}p})^q$.  For example: $e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$, but $e^{i} \ne 1$.

But,  you can find it with Euler's formula:

$$e^x = cos x + i sin x$$

$$e^i = e^{1i}$$

$$e^i = \cos 1 + i \sin 1$$


Given point A(-4,2,3) and B(4,0,1) what conditions is the line: [x,y,z] = [4,0,1] + t[m,n,1] perpendicular to AB? Then determine a vector equation either in terms of m or n, of the line that satisfies the condition.

Attempt:

AB = [8,-2,-2]

Therefore, the dot product of [8,-2,-2] and [m,n,1] must be zero.

8m-2n -2 = 0

To find the direction vector of the new vector equation they are asking for, should i take cross product of AB and [m,n,1]

and then do it all in terms of n or m, using 8m-2n-2 = 0?

Thank you!


How to calculate: (Number of sets in power set size n) - (Number of sets in power set size n that have less than or equal to m items) I know that power set size is 2^n, but how may one find the size 2^n - f(m), where f(m) is the size of all sets less than or equal to m (in powerset 2^n).

Example:

when n = 4, m = 2; size = 5

{a,b,c},{a,b,d},{a,c,d},{b,c,d},{a,b,c,d}

* Deleted: {},{a},{b},{c},{d},{a,b},{a,c},{a,d},{b,c},{b,d},{c,d}*

Any suggestions on how to do this? I am aware than as n gets larger, m will become less and less significant compared to the size of the powerset.

Hint:  how many subsets are there of size $0$? Of size $1$? Of size $k$?  Now the number removed is the sum from $0$ to $m$ of the number of subsets.

There is an explicit formula: the number of subsets of $\{1,\ldots,n\}$ of size more than $m$ is
$$
\sum_{k=m+1}^n \binom{n}{k}.
$$


probability of drawing a 5 card flush given n cards Given $n$ random cards from a standard $52$ card deck, what is the probability of getting at least a 5 card flush within those $n$ cards? $n$ would be 5 <= $n$ < 17. The probability would get closer and closer to 1 as $n$ approaches 17. Basically, I need an equation to compute the increasing probability of getting at least a 5 card flush as you draw more and more cards from a 52 card deck.

Is it simply $$\frac {(^4C_1*   ^{13}C_5)}{^{52}C_n}$$

I've got kind of a dumb answer. I just wrote a program that counts the number $f$ of non-flush hands for a hand of $n$ cards and the total number of possible hands $t$ to arrive at the probability $P=1-f/t$.

! flush.f90 -- Calculates probability of getting a flush
module FlushFun
   use ISO_FORTRAN_ENV
   implicit none
   integer choose(0:13)
   contains
      function nfree(n)
         integer n
         integer(INT64) nfree
         integer n1, n2, n3
         integer(INT64) L1, L2, L3
         nfree = 0
         do n1 = max(0,n-12), min(4,n)
            L1 = choose(n1)
            do n2 = max(0,n-n1-8), min(4,n-n1)
               L2 = L1*choose(n2)
               do n3 = max(0,n-n1-n2-4), min(4,n-n1-n2)
                  L3 = L2*choose(n3)
                  nfree = nfree+L3*choose(n-n1-n2-n3)
               end do
            end do
         end do
      end function nfree
      function ntotal(n)
         integer n
         integer(INT64) ntotal
         integer i
         ntotal = 1
         do i = 1, N
            ntotal = ntotal*(52-i+1)/i
         end do
      end function ntotal
      subroutine init_params
         integer i
         integer(INT64) j
         integer(INT64) fact(0:13)
         fact = [(product([(j,j=1,i)]),i=0,13)]
         choose = [(fact(13)/fact(i)/fact(13-i),i=0,13)]
      end subroutine init_params
end module FlushFun

program flush
   use FlushFun
   implicit none
   integer(INT64) free, total
   real(REAL64) probability
   integer n
   call init_params
   write(*,'(a)') '\begin{array}{rrrl}'
   write(*,'(a)') '\text{Cards} & \text{Non-Flush} & \text{Total} & \text{Probability}\\'
   write(*,'(a)') '\hline'
   do n = 4,17
      free = nfree(n)
      total = ntotal(n)
      probability = 1-real(free,REAL64)/total
      write(*,'(*(g0))') n, ' & ', free,' & ',total,' & ',probability,' \\'
   end do
   write(*,'(a)') '\end{array}'
end program flush

$$\begin{array}{rrrl}
\text{Cards} & \text{Non-Flush} & \text{Total} & \text{Probability}\\
\hline
4 & 270725 & 270725 & 0.0000000000000000 \\
5 & 2593812 & 2598960 & 0.19807923169267161E-002 \\
6 & 20150884 & 20358520 & 0.10198973206303807E-001 \\
7 & 129695332 & 133784560 & 0.30565769323455561E-001 \\
8 & 700131510 & 752538150 & 0.69639844837102283E-001 \\
9 & 3187627300 & 3679075400 & 0.13357924113216058 \\
10 & 12234737086 & 15820024220 & 0.22662968679070705 \\
11 & 39326862432 & 60403728840 & 0.34893320019744667 \\
12 & 104364416156 & 206379406870 & 0.49430799449026441 \\
13 & 222766089260 & 635013559600 & 0.64919475199817445 \\
14 & 364941033600 & 1768966344600 & 0.79369814767023128 \\
15 & 418161601000 & 4481381406320 & 0.90668912929163414 \\
16 & 261351000625 & 10363194502115 & 0.97478084575449575 \\
17 & 0 & 21945588357420 & 1.0000000000000000 \\
\end{array}$$
Not carefully checked, but at least it gives the right answers for $n\in\{4,5,17\}$.

EDIT: To show how you could solve this problem by hand I wrote a program that really does find all the partitions of $n$ into $4$ integers in $[0,4]$. The most partitions you get is $8$ for $n=8$. Consider the partition $8=4+2+2+0$. In the table below this is represented as $4$ clubs, $2$ diamonds, $2$ hearts and no spades, but there are actually
$$\frac{4!}{1!2!1!}=12$$
Ways to choose the suits. For the given choice of suits, there are $\binom{13}{4}=715$ ways to select $4$ clubs, $\binom{13}{2}=78$ ways to select $2$ diamonds,  $\binom{13}{2}=78$ ways to select $2$ hearts, and $\binom{13}{0}=1$ way to select $0$ spades, so there are $12\times715\times78\times78\times1=52200720$ possible non-flush hands with the $4-2-2-0$ distribution. Here is the program that shows these calculations:

! flush2.f90 -- Calculates probability of getting a flush
module FlushFun
   use ISO_FORTRAN_ENV
   implicit none
   integer choose(0:13)
   contains
      function nfree(n)
         integer n
         integer(INT64) nfree
         integer n1, n2, n3, n4
         integer(INT64) L1, L2, L3, L4
         integer(INT64) term
         integer current, w1, w2, w3, w4
         nfree = 0
         write(*,'(a)') '$$\begin{array}{rrrr|r|rrrr|r}'
         write(*,'(a)') '\rlap{\text{Number}}&&&&'// &
            '&\rlap{\text{Hands in Suit}}&&&&\\'
         write(*,'(a)') '\clubsuit&\diamondsuit&\heartsuit&'// &
            '\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&'// &
            '\heartsuit&\spadesuit&\text{Total}\\'
         write(*,'(a)') '\hline'
         do n1 = min(4,n), (n+3)/4, -1
            L1 = choose(n1)
            w1 = 1
            current = 1
            do n2 = min(n1,n-n1), (n-n1+2)/3, -1
               L2 = choose(n2)
               current = merge(current+1,1,n2 == n1)
               w2 = w1*current
               do n3 = min(n2,n-n1-n2), (n-n1-n2+1)/2, -1
                  L3 = choose(n3)
                  current = merge(current+1,1,n3 == n2)
                  w3 = w2*current
                  n4 = n-n1-n2-n3
                  L4 = choose(n4)
                  current = merge(current+1,1,n4 == n3)
                  w4 = w3*current
                  term = 24/w4*L1*L2*L3*L4
                  write(*,'(*(g0))') n1,'&',n2,'&',n3,'&', &
                     n4,'&',24/w4,'&',L1,'&',L2,'&',L3,'&', &
                     L4,'&',term,'\\'
                  nfree = nfree+term
               end do
            end do
         end do
         write(*,'(*(g0))') '\hline&&&&&&&&\llap{\text{Hands for ',n,' cards:}}&',nfree
         write(*,'(a)') '\end{array}$$'
      end function nfree
      subroutine init_params
         integer i
         integer(INT64) j
         integer(INT64) fact(0:13)
         fact = [(product([(j,j=1,i)]),i=0,13)]
         choose = [(fact(13)/fact(i)/fact(13-i),i=0,13)]
      end subroutine init_params
end module FlushFun

program flush2
   use FlushFun
   implicit none
   integer(INT64) free
   integer n
   call init_params
   do n = 4,17
      free = nfree(n)
   end do
end program flush2

And here are the tables in prints out:
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&0&0&0&4&715&1&1&1&2860\\
3&1&0&0&12&286&13&1&1&44616\\
2&2&0&0&6&78&78&1&1&36504\\
2&1&1&0&12&78&13&13&1&158184\\
1&1&1&1&1&13&13&13&13&28561\\
\hline&&&&&&&&\llap{\text{Hands for 4 cards:}}&270725
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&1&0&0&12&715&13&1&1&111540\\
3&2&0&0&12&286&78&1&1&267696\\
3&1&1&0&12&286&13&13&1&580008\\
2&2&1&0&12&78&78&13&1&949104\\
2&1&1&1&4&78&13&13&13&685464\\
\hline&&&&&&&&\llap{\text{Hands for 5 cards:}}&2593812
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&2&0&0&12&715&78&1&1&669240\\
4&1&1&0&12&715&13&13&1&1450020\\
3&3&0&0&6&286&286&1&1&490776\\
3&2&1&0&24&286&78&13&1&6960096\\
3&1&1&1&4&286&13&13&13&2513368\\
2&2&2&0&4&78&78&78&1&1898208\\
2&2&1&1&6&78&78&13&13&6169176\\
\hline&&&&&&&&\llap{\text{Hands for 6 cards:}}&20150884
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&3&0&0&12&715&286&1&1&2453880\\
4&2&1&0&24&715&78&13&1&17400240\\
4&1&1&1&4&715&13&13&13&6283420\\
3&3&1&0&12&286&286&13&1&12760176\\
3&2&2&0&12&286&78&78&1&20880288\\
3&2&1&1&12&286&78&13&13&45240624\\
2&2&2&1&4&78&78&78&13&24676704\\
\hline&&&&&&&&\llap{\text{Hands for 7 cards:}}&129695332
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&0&0&6&715&715&1&1&3067350\\
4&3&1&0&24&715&286&13&1&63800880\\
4&2&2&0&12&715&78&78&1&52200720\\
4&2&1&1&12&715&78&13&13&113101560\\
3&3&2&0&12&286&286&78&1&76561056\\
3&3&1&1&6&286&286&13&13&82941144\\
3&2&2&1&12&286&78&78&13&271443744\\
2&2&2&2&1&78&78&78&78&37015056\\
\hline&&&&&&&&\llap{\text{Hands for 8 cards:}}&700131510
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&1&0&12&715&715&13&1&79751100\\
4&3&2&0&24&715&286&78&1&382805280\\
4&3&1&1&12&715&286&13&13&414705720\\
4&2&2&1&12&715&78&78&13&678609360\\
3&3&3&0&4&286&286&286&1&93574624\\
3&3&2&1&12&286&286&78&13&995293728\\
3&2&2&2&4&286&78&78&78&542887488\\
\hline&&&&&&&&\llap{\text{Hands for 9 cards:}}&3187627300
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&2&0&12&715&715&78&1&478506600\\
4&4&1&1&6&715&715&13&13&518382150\\
4&3&3&0&12&715&286&286&1&701809680\\
4&3&2&1&24&715&286&78&13&4976468640\\
4&2&2&2&4&715&78&78&78&1357218720\\
3&3&3&1&4&286&286&286&13&1216470112\\
3&3&2&2&6&286&286&78&78&2985881184\\
\hline&&&&&&&&\llap{\text{Hands for 10 cards:}}&12234737086
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&3&0&12&715&715&286&1&1754524200\\
4&4&2&1&12&715&715&78&13&6220585800\\
4&3&3&1&12&715&286&286&13&9123525840\\
4&3&2&2&12&715&286&78&78&14929405920\\
3&3&3&2&4&286&286&286&78&7298820672\\
\hline&&&&&&&&\llap{\text{Hands for 11 cards:}}&39326862432
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&4&0&4&715&715&715&1&1462103500\\
4&4&3&1&12&715&715&286&13&22808814600\\
4&4&2&2&6&715&715&78&78&18661757400\\
4&3&3&2&12&715&286&286&78&54741155040\\
3&3&3&3&1&286&286&286&286&6690585616\\
\hline&&&&&&&&\llap{\text{Hands for 12 cards:}}&104364416156
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&4&1&4&715&715&715&13&19007345500\\
4&4&3&2&12&715&715&286&78&136852887600\\
4&3&3&3&4&715&286&286&286&66905856160\\
\hline&&&&&&&&\llap{\text{Hands for 13 cards:}}&222766089260
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&4&2&4&715&715&715&78&114044073000\\
4&4&3&3&6&715&715&286&286&250896960600\\
\hline&&&&&&&&\llap{\text{Hands for 14 cards:}}&364941033600
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&4&3&4&715&715&715&286&418161601000\\
\hline&&&&&&&&\llap{\text{Hands for 15 cards:}}&418161601000
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
4&4&4&4&1&715&715&715&715&261351000625\\
\hline&&&&&&&&\llap{\text{Hands for 16 cards:}}&261351000625
\end{array}$$
$$\begin{array}{rrrr|r|rrrr|r}
\rlap{\text{Number}}&&&&&\rlap{\text{Hands in Suit}}&&&&\\
\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Ways}&\clubsuit&\diamondsuit&\heartsuit&\spadesuit&\text{Total}\\
\hline
\hline&&&&&&&&\llap{\text{Hands for 17 cards:}}&0
\end{array}$$

A simple approach!
You draw say 10 cards.
The probability that any of these cards is a particular suite is 1/4. Conversely the prob that any card in not say a diamond is 3/4.
Then what is probability that 5 cards are the same suite is inverse that any 5 cards are Not the same suite. This is simply 3/4 ^ 5 = 23.7%. So prob of a simple 5 card flush from 10 cards is  76.3% which to a card player  feels about right!
Cheers


Non-linear systems help! I have a non-linear system of equations, $$\left\{ \begin{array}{rcl} x^2 - xy + 8 = 0 \\ x^2 - 8x + y = 0 \\ \end{array} \right.$$
I have tried equating the expressions (because both equal 0), which tells me: $$x^2 - xy + 8 = x^2 - 8x + y$$
Moving all expressions to the right yields: $$0 = xy - 8x + y - 8$$
Factoring the equation: $$0 = x(y-8) + 1(y-8)$$
$$0 = (x+1)(y-8)$$
$$x=-1$$
$$y=8$$
Problem solved, right? No. When you plug in the values into the equations above, you get a false statement. Allow me to demonstrate:
$$x^2 - 8x + y = 0$$
$$(-1)^2 - 8(-1) + (8) = 0$$
$$1 - (-8) + 8 = 0$$
$$1 + 8 + 8 = 0$$
$$17 = 0$$
Can someone please help me solve this system of nonlinear equations? I am stuck.

Hint: You got $x=-1$  OR $y=8$. BTW, cool idea to equate the expressions.

Выражаешь из второго $y$. Потом подставляешь в первое вместо $y$, далее решаешь квадратное уравнение относительно $x$. Находишь два корня. Потом для каждого из корней находишь $y$. В итоге получится два корня $x$ и два корня $y$. Всё банально просто.

Quoth google translate:

  Expresses the second $y$. Then you substitute the first instead of $y$, continue to solve quadratic equation with respect to $x$. There are two root. Then for each of the roots are $y$. That will have two roots and two root $x$ $y$. All corny easy.


number of ways to select a committee of 3 students with at least 1 girl Suppose a class comprises eight girls and six boys. How many ways can a committee of three students of among 8 girls and 6 boys be formed if the committee must include at least one girl?

My solution:

Ways to form a committee with exactly $g$ girls:

$g = 1$: $\displaystyle\binom 8 1 \binom 6 2 = 120$

$g = 2$: $\displaystyle\binom 8 2 \binom 6 1 = 168$

$g = 3$: $\displaystyle\binom 8 3 = 56$

I have no idea what to do next.

I will assume from your work that there are $\bf 8$ girls AND $\bf 6$ boys (I will take your "9" in the second computation to be a typo), and you need to determine how many committees can be formed from those $14$ students that have at least one girl on the committee of three.

The "simpler approach" to this would be to take the total number of possible ways of selecting any committee of 3 students from among $ 8 + 6 = 14$ students, and then subtract from that total the number of committees/combinations that can be formed without girls. The result will give you the number of committees which then contain at least one girl.

$$\binom{14}{3} - \binom 63 = \text{total number of committees of 3 students with $\bf at\;least\; one\; girl$.}$$

Alternatively, you can add the results of your posted calculations, and arrive at the same number of valid combinations as you would using this approach. Both approaches give you a total of $344$ possible valid combinations.


For which natural numbers $n$ is $\sqrt n$ irrational? How would you prove your answer? Can someone help me out here? I'm trying to prove this the same way as whether if $\sqrt 2$ is irrational, but i'm not sure what am i doing.
Can someone show me how to prove this?

In fact, every integer which is not a square has an irrational square root. Can you see how to generalize the result for $\sqrt{2}$ to prove this?

Take $\sqrt a=\frac{p}{q}$ with gcd(p,q)=1 and a is prime square both sides...
$$
a=\frac{p^2}{q^2}
$$
then
$$
2q^2=p^2
$$
then
$$
a|p^2
$$
a contradiction

then think about what would happen if a was not prime


How to explain your area of study to non-math people I'm sorry in advance if here is not the suitable place to ask this question, and people can feel free to vote to close this if that's the case. However, since I'm not sure about this issue, I'll ask it in any way:

I was reading Hungerford's Algebra, and my friend working on food science saw the title "Fundamental Theorem of Galois Theory", and asked me what it's about. (Seeing the word "fundamental", he thought it should be a very basic, simple concept.) Then, I said something like "there are structures called groups, fields, and they have some special subsets called subgroup, subfields. In order to understand subfields, mathematicans makes a correspondence between subgroups of some associated group, which is easier to work with and better understood, compared to fields."

He said it made some sense, but I'm sure it was a terribly bad explanation. My question is how do people generally deal with this issue? Is this only my problem (which really might be the case), or is this a general problem with pure math people?

This issue really annoys me in the sense that, for example, when you consider some person working on food science, psychology, chemistry etc., they're generally able to tell roughly what they're trying to do. But, I'm always having very hard time when I'm asked the same question. I'm a PhD student now, and I plan on spending whole my life with mathematics. But, I'm afraid that, say in my 50 (if I live that much), if someone asks me what I did in my entire 50 years, and I fail to answer, that would be a shameful situation for me, I believe.

I tried my best to express my question in a good way, but still I'm sorry if it's not good enough, and don't hesitate to let me know if I should write some parts more clearly/explicitly.

The difficult task of explaining what you do or study as a mathematician in layman terms is a common experience.

I think that the main mistake is to try to be precise nonetheless. In your case of Galois theory, terms as "fields" or "groups" are useless and may even lead to more confusion (I remember once a physicist getting really confused at a talk because he had never been exposed to the algebraic meaning of the term "field"). I would have said something along the following lines:

"There is a large class of numbers called algebraic numbers which need to be considered when solving simple equations in just one unknown. They include most numbers people are familiar with, like usual fractions or their roots, but don't include others, like $\pi$. The set of algebraic numbers has symmetries, which are in some sense like spatial symmetries, but less easy to visualize. Galois theory is the theory of these symmetries."

From here one could move on (for instance telling how the $\pm$ sign in the formula for the quadratic equations--something that everybody has seen in his/her school years--is a manifestation of these symmetries) according to the interest shown by the person we are talking to.

My six pence:

Tell them that you are studying how the four elementary operations ($+,-,\times,\div$) generalize to other things than ordinary numbers. For instance, when you are limited to a range like $[0,255]$. It turns out that this theory is extremely rich and powerful, and we are still far from understanding all of it. I has numerous real-world applications like cryptology (avoiding that your data be stolen on the cloud) or error correction (ensuring that your scratched DVD is still readable).


Find numbers of the form $\frac{n(n+1)}{2}$ such that the digits of the number are all same Find all $a \in \{1, 2, \ldots, 8, 9\}$ such that $\exists n \in \mathbb{N}$ and the digits of $\dfrac{n(n+1)}{2}$ are all $a$.

In other words, we have to find the values of $a$ such that $\dfrac{n(n+1)}{2} = a \cdot (1111\ldots1111)$ where $a$ is a Numerical Digit of Base $10$, i.e. $a \in \{1, 2, \ldots, 8, 9\}$.

I made the following progress:

Let's take $(1111\ldots1111) = R$ for convenience.
So we get the following equation: $n^2 + n - 2aR = 0$.
Taking the roots of $n$ by using the Quadratic Formula, we get $n = \dfrac{-1\pm\sqrt{1+8aR}}{2}$ .

As $n \in \mathbb{N}$, $-1\pm\sqrt{1+8aR}$ must be even, so $\sqrt{1+8aR}$ must be odd and hence $1+8aR$ must also be odd.
We also know that $1+8aR$ must be a perfect square.

So, our problem reduces down to $1+8aR = f^2$ where $f \in \mathbb{N}$.

I couldn't proceed ahead from here.

Series of natural numbers which has all same digits also talks about pretty much the same problem. However, no conclusive solution was reached in that thread. I want to do the above question with an analytical approach. Hence, I beleive that my question is not a duplicate.

[Source: As far as I know, this problem is from IMOTC India]

Let $T_n$ be the $n^{th}$ triangular number.

Modulo $10$, $T_n$ is $0,1,3,5,6$ or $8$.

Clearly $8$ is not a triangle number.

Modulo $100$, $T_n$ is never $88$.

Therefore $8$ does not work, and the only solutions are
$1,3,5,6$ (as shown in the comments)


How to prove the inequality $ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$ How to prove the inequality $$ \frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2}$$
for $a,b,c>0$ and $abc=1$?

I have tried prove $\frac{a}{\sqrt{1+a}}\ge \frac{3a+1}{4\sqrt{2}}$

Indeed,$\frac{{{a}^{2}}}{1+a}\ge \frac{9{{a}^{2}}+6a+1}{32}$

$\Leftrightarrow 32{{a}^{2}}\ge 9{{a}^{2}}+6a+1+9{{a}^{3}}+6{{a}^{2}}+a$
$\Leftrightarrow 9{{a}^{3}}-17{{a}^{2}}+7a+1\le 0$
$\Leftrightarrow 9{{\left( a-1 \right)}^{2}}\left( a+\frac{1}{9} \right)\le 0$ (!)
It is wrong. Advice on solving this problem.

Let $a=\frac{x}{y}$, $b=\frac{y}{z}$ and $c=\frac{z}{x}$, where $x$, $y$ and $z$ are positives.

Hence, we need to prove that
$$\sum_{cyc}\frac{\frac{x}{y}}{\sqrt{1+\frac{x}{y}}}\geq\frac{3}{\sqrt2}$$ or
$$\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\geq\frac{3}{\sqrt2}.$$
Now, by Holder
$$\left(\sum_{cyc}\frac{x}{\sqrt{y(x+y)}}\right)^2\sum_{cyc}xy(x+y)(2z+x+y)^3\geq\left(\sum_{cyc}x(2z+x+y)\right)^3.$$
Thus, it remains to prove that
$$2\left(\sum_{cyc}(x^2+3xy)\right)^3\geq9\sum_{cyc}xy(x+y)(2z+x+y)^3$$ or
$$\sum_{sym}(x^6+9x^5y+24x^4y^2+18x^3y^3+9x^4yz-36x^3y^2z-25x^2y^2z^2)\geq0,$$
which is obviously true.

Done!

Some observations:

$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \tag{1}$$

$$A=\sqrt{1+a}+\sqrt{1+b}+\sqrt{1+c} +\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} -\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)$$
$$A\ge(2+2+2)-\left(\frac{2}{\sqrt{1+a}}+\frac{2}{\sqrt{1+b}}+\frac{2}{\sqrt{1+c}} \right)\tag{2}$$

Comparing with the original definition:
$$A=\frac{a}{\sqrt{1+a}}+\frac{b}{\sqrt{1+b}}+\frac{c}{\sqrt{1+c}} \ge \frac{3\sqrt{2}}{2},\tag{3}$$

We conclude that we need to prove:

$$\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}} \le 3-\frac{3\sqrt{2}}{4}\tag{4}$$


Geometrical meaning of the Column Space Suppose I have $2$ planes in $R^3$ and they form a system $Ax=b$.

I know the NullSpace of $A$ represents geometrically the vectors that form the intersection between the 2 planes shifted to the origin.

I also know that the Row Space of A represents the span of the normal vectors to the 2 planes.

But i was looking for some geometrical meaning for the Column Space of $A$.
They contain all the vectors $b$ that make $Ax=b$ have at least one solution, that i know.

What I'm trying to see is what does the vectors from the Column Space mean in relating to the $2$ planes in $R^3$ or relating to the possible shiftings the system of planes could suffer to still yield a solution.

Well, they live in a separate, 'transpose' world. Now, as you have $2$ equations (of planes), $b$ and the columns will be elements of $\Bbb R^2$ and not $\Bbb R^3$.

I doubt they would represent anything special, related to the $2$ planes in $\Bbb R^3$, a strong reason is that if we multiply the equation of a plane (a row) by a scalar, we get the same plane, but the column space suffers a radical change..


Solving a system of 3 variables How to solve or what is the algorithm to solve a system of equations like this:
$$\eqalign{
(x +\phantom{3} z)^2 + (y +\phantom{3} w)^2 &= 52\cr
(x + 3z)^2 + (y + 3w)^2 &= 296\cr
(x -\phantom{3} z)^2 + (y -\phantom{3} w)^2 &= 8\cr
(x -2z)^2 + (y - 2w)^2 &= 61?
}$$

By trying to solve this equation (adding and subtracting first and third equations, putting results to fourth equation) I eventually got these:

$$\begin{cases}z^2+w^2=25\\
  x^2+y^2=5\\
 zx+wy=11\end{cases}$$

All of the given equations are the results of the equations I have provided. (You can check it).
And I found that the only integers which can solve the equation are:

$$x=2, y=1, z=4, w=3\quad\text{or}\quad
x=-2, y=-1, z=-4, w=-3$$


Euler formula, polyhedron with a hole I'd like to ask help for this question. I know that I need to have a little more concept on what kind of polyhedra that Euler's formula is valid.
Choose the false alternative:
a) F = 12
b) V = 20
c) E = 30
d) Euler's formula F + V - E = 2 applies
e) All above are true
Actually, by looking at this picture we can conclude that E is correct, since all of them seem to be true. However, shouldn't this formula be F + V - E = -2, since the genus is 2 ?

The number of faces is listed as 12.  There are ten vertical faces and two "horizontal" faces (namely, top level and bottom level of the polyhedron).  However, the latter pair are not admissible as faces because a face has to be a convex polygon for the formula to apply.

See: https://en.wikipedia.org/wiki/Polygon_with_holes#In_polyhedra
$\chi =V-E+F-\text{holes}=2(1-g)$
$V=20, E=30, F=12, H=4$ (2 faces each with 2 holes)
$\chi=20-30+12-4=-2=2(1-g)
\implies g=2$ so it's $2$-holed polyhedron.


Understanding implications I have a set of questions where to task is to add the correct implication arrow.
I was able to make sense of most of them, but for two problems, I can't figure out why I was wrong.
Here is what I put:

Statement $S1$
Statement $S2$
I answered
The answer in the book is

$\frac{x}{x+1} = 0$
$x=0$
$S1\iff S2$
$S1 \Rightarrow S2$

$(a,b)$ is a point on the line $y=2x-1$
$b=2a-1$
$S1 \iff S2$
$S1 \Rightarrow S2$

For the first one, I can't see how when $x=0$, $\frac{x}{x+1}$ can equal anything other than $0$, so surely the implication works both ways? I presume I'm missing something obvious.
For the second, the point $(a, b)$ is on line $y = 2x -1$, so $b$ must equal $2a - 1$. Again I presume I've missing something here.
Really would appreciate any help here, I've run out of ideas.

First one:
Sufficiency ($\Rightarrow)$: Suppose $x/(x+1) = 0$. This expression already implies $x\neq -1$. So we can multiply it with $x+1$ to get $x = 0$. (True)
Necessity ($\Leftarrow)$: Suppose $x = 0$. It is a trivial result that $x/(x+1) = 0$. (True)
Therefore, the first one is an N-S condition.
Second one:
In set-builder notation, 'the point is on the line' if and only if
$$
(a,b) \in \{(x,y) \in \mathbb{R}^2~|~y = 2x -1\}
$$
and it is equivalent to say that $b = 2a - 1$ given that $(a,b) \in \mathbb{R}^2$ due to Axiom schema of specification.
That is, the second one is also an N-S condition.

The symbols represent the direction of the implication. It means you can read back and/or forwards substituting by "then". The one with two arrows is "similar" to the equivalence symbol.
In the first, x = 0 does not imply only that x/(x+1)=0, but much other expressions. The same for the line.
Strictly speaking, the statement has a direction which must be rigidly read.


Kolmogorov's Truncation Lemma (ii) Probability with Martingales:

How exactly do we have the part in the $\color{red}{\text{red}}$ box?

What I tried:

$$E\left[ \sum_{n=1}^{\infty} 1_{|X| > n} \right]$$

$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{|X| > n} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\lfloor|X|\rfloor > n} \right]$$

$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1_{\Omega} + \sum_{n=\lfloor|X|\rfloor}^{\infty} 1_{\emptyset} \right]$$

$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 + \sum_{n=\lfloor|X|\rfloor}^{\infty} 0 \right]$$

$$ = E\left[ \sum_{n=1}^{\lfloor|X|\rfloor - 1} 1 \right]$$

$$ = E\left[ \sum_{n=\color{blue}{1}}^{\color{red}{\lfloor|X|\rfloor - 1}} \color{green}{1} \right]$$

$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1} - \color{blue}{1} + 1) (\color{green}{1}) \right]$$

$$ = E\left[ (\color{red}{\lfloor|X|\rfloor - 1}) (\color{green}{1}) \right]$$

$$ = E\left[ \color{red}{\lfloor|X|\rfloor - 1} \right]$$

$$ = E\left[ \lfloor|X|\rfloor - 1 \right]$$

$$ \le E\left[ |X|  \right]$$

Is that right?


How does the notation of PGF read in plain English? How does the above notation of PGF read in plain English?

Is it $\alpha$ to the power $X$? What does it even mean?

Start here: https://en.wikipedia.org/wiki/Probability-generating_function

In general, be careful with generating functions. They're very useful for doing theoretic work, but you can't treat them like a normal finite polynomial. They're an infinite series with useful properties, but the "infinite" part of infinite series can often lead people astray who aren't very practiced with them.

To translate, as you asked, into plain English, it says that the Probability Generating function of the random variable $X$ is written as $g_X(\alpha)$ and is equal to the Expected Value of $\alpha^X$, defined for $\alpha$ between 0 and 1.

$\alpha$ is a dummy variable. What is missing is the summation form, given at the link above. The last part of your picture is the beginning of a property or alternate definition for pgfs that only works if $X$ is integer-valued, but isn't part of the first two lines.

As for what the formula means, since $X$ is a random variable, any formula with $X$ in it can have an expected value calculated by the usual formula of summing over every possible result of $X$, multiplying the formula by the probability of that result occurring.

IE, $$g_X(\alpha)=E(\alpha^X) = \sum_{X} p(X)\alpha^X$$ If $X$ only takes on integer values, this becomes a standard power series.

Yes, that is a number, E, times the number, $\alpha$, to the X power.  We can write $\alpha^x= e^{ln(\alpha^x}= e^{xln(\alpha)}= \left(e^{ln(\alpha)}\right)^x$


Tough inverse Laplace transform I know what the solution is to this inverse Laplace transform, I just have NO idea how to get there.

$$\mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right)$$

Basically, my question is what modification do I have to do to the equation above?

Using a table, note the form:

$$f(t) = t\sin(at)$$

$$F(s) = \frac{2as}{(s^2+a^2)^2}$$

Using this:
$$f(t) = \mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right) = \mathcal{L}^{-1}\left(\frac{2*2*s}{\left(s^2+2^2\right)^2}*4\right) =4t\sin(2t)$$

Omitting $s$ rule

Omit $s$
Then we get $\frac d{dT}[L $ inverse of $\frac 16{(s^2+4)^2}]$
Take $16$(const out)
Take $l$ inverse and differentiate with respect to $t$


What does the "closed over"/"closed under" terminology mean exactly and where did it come from? I've been trying to teach my partner some set theory, and I got thrown for a loop while trying to give her a precise definition of some basic terminology.
So we've heard of a set being described as "closed under" an operation, as well as an operation being "closed over" a set.

First, correct me if I'm wrong, but my impression is that these things mean the same thing.

Second, does this terminology have a precise definition?
For instance we'd normally say that set $A$ is closed under $f:X\rightarrow Y$ if $A\subseteq X$ and for any $x\in A$, $f(x)\in A$.  But it's also common to extend the definition such that $X$ isn't a superset of $A$ but of $A^n$ for some $n$.  For instance, we'd say $A$ is closed under addition if the addition operator maps elements of $A^2$ to $A$.  Does this extend to anything bigger than $n$-tuples?  Infinite sequences, for instance?

Third, what's the motivation for this terminology and is it in any way related to the normal topological definition of closedness?  There's a way that closed sets are related to sets that are closed under a particular operation in certain topological spaces (if you allow the usage that includes infinite sequences as above), but I haven't come up with a general relation between the two concepts.  What's the motivation for saying an operation is closed over a set though, or is that just a corruption of the former?  Does anyone know the historical reasoning behind any of this terminology?

In an ordinary mathematical context I would never say that an operation $\mathcal O$ is closed over a set $A$; I consider that an error for ‘$A$ is closed under $\mathcal O$’. The latter terminology can be properly applied very generally. For starters, if $A \subseteq S$, $n \in \omega$, and $f:S^n \to S$, ‘$A$ is closed under $f$’ means precisely that for every $\langle a_1,\dots,a_n \rangle \in A^n$, $f(a_1,\dots,a_n) \in A$.

This is the most common algebraic usage, I think, but it’s just the beginning. For instance, the non-negative integers $n$ can be replaced by any ordinal $\alpha$, and the operation need not be defined on all members of $S^\alpha$. Suppose that $X$ is a topological space and $A \subseteq X$. ‘$A$ is closed under (the operation of taking) limits of convergent sequences’ then means that if $\langle a_n:n \in \omega \rangle$ is a sequence of points in $A$ that converges as a sequence in the space $X$, its limit point is actually in $A$. Here $\alpha = \omega$, and the operation is defined only for those elements of $X^\omega$ that converge in $X$. This example at least starts to show the relationship between the general notion and that of topological closedness.

Moreover, the terminology is still used when the input to the operation isn’t ordered: it’s perfectly correct to say that a family $\mathcal A$ of sets is closed under (taking) finite intersections, for instance, meaning that if $\mathcal F$ is any finite subfamily of $\mathcal A$, $\bigcap \mathcal F \in \mathcal A$. If $S$ is the underlying set, the operation is $\mathcal O:[(\mathcal P(S)]^{< \omega} \to \mathcal P(S):\mathcal F \mapsto \bigcap \mathcal F$, and $\mathcal A \subseteq \mathcal P(S)$ is closed under it because $\mathcal O$ maps $[\mathcal A]^{< \omega}$ into $\mathcal A$. (Here $[X]^{< \omega}$ denotes the set of finite subsets of $X$.)

It would be a bit difficult to formulate an exhaustive formal definition of the usage, and I’m not at all sure that it would be particularly helpful; it seems more useful to present a variety of examples showing the flexibility of the usage.

First, closed under and closed over mean the same thing.

Second, your definition is correct except its usually done with an operation.  An operation $*$ is a function from $G\times G \to H.$ An operation is closed over a set $G$ if if an only if, for every $x,y\in G$ $f(x,y)=x*y\in G.$


Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$ $$a_1 =  2$$
$$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$

I tried to use induction but it looks as if the sequence isn't monotonic.

If you JUST want to find the limit the easiest thing to do is solve
$$L=\frac{1}{2}\left(L+\frac{1}{L}\right)$$ you get $L=\pm 1$, and since $a_n$ is positive, $L=1$.

If you want to show the limit exists, it is decreasing.
$$a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)\leq a_n$$
is equivalent to $1\leq a_n^2$ and this is true because $$1\leq \frac{1}{2}\left(x+\frac{1}{x}\right)$$ for all $x$.


Is (A ∨ B) → (A ∧ B) possible? Hi still fairly new to propositional logic, and I'm attempting to prove
(A ∨ B) → (A ∧ B)

I have tried via De Morgan Laws but to no success.

If not, would this work?
A → (A ∧ B)

Any help will be appreciated.

You can't prove it because it isn't true.  If $A$ is true and $B$ is false, $A\vee B$ is true and $A \wedge B$ is false, so the whole statement is false.  Just read it in English as (A or B) implies (A and B) and it doesn't make sense that it would be true.

That proposition that you are trying to prove is NOT true: $$(A\lor B) \to (A\land B)\;\;\color{red}{\Huge \checkmark}$$

When we have the antecedent $A\lor B$, it is true when $A$ is true, but not $B$, or when $B$  is true, but not $A$.  However, in either of those cases, $A\land B$ is not true, because $A \land B$ is true only when both A and B are true.  For example, If I say "I'll have pie, or cake for desert", I will not be lying if I have either pie or cake, but not both.  However, if that's my promise, no one can accuse me of claiming therefore "I'll have pie AND cake for desert."

Whenever the antecedent can be true, but the consequent false, the implication is fails.

For your second question, it is not true that $A\to (A\land B).$  If A is true, but B is false, then the antecedent is true, but the consequent is false, because (true $\land$ false$ evaluates to false.

What is true is that $$[(A\land B) \to A] \land [(A\land B) \to B]$$

What is true is that $(A\land B) \to (A\lor B)$.


How does Cramer's rule work? I know Cramer's rule works for 3 linear equations. I know all steps to get solutions. But I don't know why (how) Cramer's rule gives us solutions?

Why do we get $x=\frac{\Delta_1}\Delta$ and $y$ and $z$ in the same way?

I want to know how these steps give us solutions?

It's actually simple; I explain it here in two variables, but the principle is the same.

Say you have an equation
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}p\\q \end{pmatrix}$$

Now you can see that the following holds

$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}x&0\\y&1\end{pmatrix}=\begin{pmatrix}p&b\\q &d\end{pmatrix}$$

Finally just take the determinant of this last equation; $\det$ is multiplicative so you get $$\Delta x=\Delta_1$$

Given a linear system $\rm A x = b$, where $\mathrm A \in \mathbb R^{n \times n}$ is invertible and $\mathrm b \in \mathbb R^n$, let $\rm \bar{x} := A^{-1} b$ denote the (unique) solution. Replacing the $k$-th column of matrix $\rm A$ (denoted by $\mathrm a_k$) with $\rm b$ and computing the determinant of this new matrix using the matrix determinant lemma,

$$\begin{aligned} \det \left( \mathrm A + (\mathrm b - \mathrm a_k) \mathrm e_k^\top \right) &= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm A^{-1} (\mathrm b - \mathrm a_k) \right)\\ &= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm A^{-1} \mathrm b - \mathrm e_k^\top \mathrm A^{-1} \mathrm a_k \right)\\ &= \det (\mathrm A) \cdot \left( 1 + \mathrm e_k^\top \mathrm{\bar{x}} - \mathrm e_k^\top \mathrm e_k \right)\\ &= \det (\mathrm A) \cdot \bar{x}_k\end{aligned}$$

and, thus, we obtain Cramer's rule

$$\bar{x}_k = \frac{\det \left( \mathrm A + (\mathrm b - \mathrm a_k) \mathrm e_k^\top \right)}{\det (\mathrm A)}$$


Double differentiation of characteristic function of Normal random variable Knowing that a one-dimensional random variable $\Gamma$ is Gaussian if it has the characteristic function
$$\mathbb{E}\hspace{0.15cm}e^{i\xi\Gamma}=e^{im\xi-\frac{1}{2}\sigma^2\xi^2}\tag{1}$$
for some real numbers $m\in\mathbb{R}$ and $\sigma\geq0$. If we differentiate $(1)$ two times with respect to $\xi$ and set $\xi=0$, we see that
$$m=\mathbb{E}\hspace{0.15cm}\Gamma\hspace{0.3cm}\sigma^2=\mathbb{V}\hspace{0.15cm}\Gamma\tag{2}$$

I cannot understand how to get to $(2)$ by double differentiating $(1)$ with respect to $\xi$, setting $\xi=0$. If I differentiate r.h.s. of $(1)$ two times with respect to $\xi$ and I set $\xi=0$, I get $-m^2$.

This question, the other answer, and the comment discussion are muddled.
For a random variable $X$ with finite second moment, with characteristic function  $\varphi_X(t)=E[e^{itX}]$, the following are true:
$$  \varphi_X(t)=1+ iE[X]t -\frac{E[X^2]}2 t^2+o(t^2)\\E[X] =\mu = -i\varphi_X'(0)\\E[X^2] = \mu^2+\sigma^2= -\varphi_X''0).$$
If you define $\psi=\log\varphi_X$, then
$$\psi(t)=\mu it-\sigma^2\frac{t^2}2+o(t^2)\\
\mu = -i \psi'(0)\\
\sigma^2 = -\psi''(0),$$
which can be connected with the previous equations by application of the chain rule and product rule:
$$\varphi_X(t)=e^{\psi(t)}\\
\varphi_X'(t) = e^{\psi(t)} \psi'(t)\\
\varphi_X''(t) = e^{\psi(t)}( \psi'(t))^2 + e^{\psi(t)}\psi''(t)$$$$
\varphi_X'(0) = e^{\psi(0)} \psi'(0)=\psi'(0)=-i\mu\\
\varphi_X''(0) = e^{\psi(0)}( \psi'(0))^2 + e^{\psi(0)}\psi''(0)=-\mu^2-\sigma^2,$$
and so on.

I will treat $\mathbb{E}$ as a constant.
$$\mathbb{E} \cdot e^{iξ \Gamma}=e^{imξ−\frac{1}{2}σ^2ξ^2}$$
So then
$$\mathbb{E} \cdot e^{iξ \Gamma}=\frac{e^{imξ}}{e^{\frac{1}{2}σ^2ξ^2}}$$
Differentiating once w.r.t $ξ$,
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot  e^{imξ} - e^{imξ} \cdot 2 \cdot\frac{1}{2}\sigma^2 \cdot ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{\bigg( e^{\frac{1}{2}σ^2ξ^2} \bigg)^2}$$
Simplifying, we obtain:
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \cdot im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \cdot e^{\frac{1}{2}σ^2ξ^2}}{e^{σ^2ξ^2}}$$
Furthermore,
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{e^{\frac{1}{2}σ^2ξ^2} \bigg( im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ \bigg) }{\bigg (e^{\frac{1}{2} \cdot σ^2ξ^2} \bigg)^2}$$
So
$$\mathbb{E} \cdot i\Gamma \cdot e^{iξ \Gamma}=\frac{im \cdot  e^{imξ} - e^{imξ} \cdot \sigma^2 ξ}{e^{\frac{1}{2} \cdot σ^2ξ^2}}$$
Which means that
$$\mathbb{E} i\Gamma \cdot e^{iξ \Gamma} \cdot e^{\frac{1}{2} \cdot σ^2ξ^2} =e^{imξ} (im - \sigma^2 ξ)$$
Multiplying both sides by $e^{-im ξ}$,
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} =im - \sigma^2 ξ$$
Differentiating again w.r.t  $ξ$ we obtain:
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} \cdot \frac{d}{dξ}\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)= - \sigma^2$$
Thus,
$$\mathbb{E} i\Gamma \cdot e^{\bigg( iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ \bigg)} \cdot\bigg( i \Gamma + 2\frac{1}{2} \cdot σ^2ξ - im\bigg)= - \sigma^2$$
Finally, we obtain that if we differentiate the first equation twice by $ξ$ we obtain that:
$$\mathbb{E} i\Gamma \cdot e^{ iξ \Gamma + \frac{1}{2} \cdot σ^2ξ^2 - im  ξ} \cdot\bigg( i \Gamma + σ^2ξ - im\bigg)= - \sigma^2$$
Now setting $$ξ=0,$$
We obtain that:
$$\mathbb{E} i\Gamma \cdot e^{ i(0) \Gamma + \frac{1}{2} \cdot σ^2(0)^2 - im  (0)} \cdot\bigg( i \Gamma + σ^2(0) - im\bigg)= - \sigma^2$$
So then
$$\mathbb{E} i\Gamma \cdot e^{0} \cdot\bigg( i \Gamma- im\bigg)= - \sigma^2$$
Thus,
$$\mathbb{E} i\Gamma \cdot\bigg( i \Gamma - im\bigg)= - \sigma^2$$
Simplifying our expression,
$$\mathbb{E} i^2\Gamma^2 - \mathbb{E}i^2 \Gamma m= - \sigma^2$$
$$-\mathbb{E}\Gamma^2 + \mathbb{E}\Gamma m= - \sigma^2$$
Then
$$\mathbb{E}\Gamma m= \mathbb{E}\Gamma^2- \sigma^2$$
$$ m= \frac{\mathbb{E}\Gamma^2- \sigma^2}{\mathbb{E}\Gamma}$$
So
$$ m= \Gamma - \frac{\sigma^2}{\mathbb{E}\Gamma}$$
You should note that I treat $\mathbb{E}$ as a constant. If it is a function of $ξ$, I will edit my answer.
But since RHS isn't dependent on $\mathbb{E}$, I can safely say that the second derivative of the RHS is
\begin{align*}
\frac{d^2}{dξ^2} \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) &=
\frac{d}{dξ}\bigg[ \bigg( e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \cdot \bigg( im - σ^2ξ  \bigg) \bigg] \\
&= \frac{d}{dξ}\bigg(  im \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - σ^2ξ \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg) \\
&= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \bigg[ \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} + \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \bigg] \\
&= im \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2ξ \big( im - \sigma^2 ξ \big) \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} - \sigma^2 \cdot e^{imξ−\frac{1}{2}σ^2ξ^2} \\
&= e^{imξ−\frac{1}{2}σ^2ξ^2} \cdot \bigg[ im \big( im - \sigma^2 ξ \big) -  \sigma^2ξ \big( im - \sigma^2 ξ \big) - \sigma^2 \bigg]
\end{align*}
Now setting $ ξ=0$ we obtain that:
$$\frac{d^2}{dξ^2} \big( e^{imξ−\frac{1}{2}σ^2ξ^2} \big) \bigg|_{ξ=0} =
i^2m^2 - \sigma^2 = -m^2 - \sigma^2$$


In the ring $\mathbb{Z}_m$ the ideal $(n) \cong n\mathbb{Z}/m\mathbb{Z}$? How does one show in the ring $\mathbb{Z}_m$ that the ideal $(n) \cong n\mathbb{Z}/m\mathbb{Z}$?

I only know this for $n=1$ but it’s meant to be true for all $n \in \mathbb{Z}_m$

Actually this is true only if $n\mid m$. Indeed
$$n\cdot\mathbf Z/m\mathbf Z=(n\mathbf Z+m\mathbf Z)/m\mathbf Z=\gcd(m,n)\mathbf Z/m\mathbf Z.$$

$(n)/(m)$ doesn't even make sense unless $(m) \subseteq (n)$, which is equivalent to $n | m$, so I'll assume this. But your statement is false unless $n = 0$ or $m = 0$. If neither is 0, then $(n)/(m) \cong \mathbb Z/(m/n)\mathbb Z$. This is a finite group, as $m/n \neq 0$. However, $(n)$ is infinite as $n \neq 0$ so these are not isomorphic.


Usage of LU decomposition over UL decomposition What are some advantages of using LU decomposition over UL decomposition? From what I see, they can both be used equally well to solve $MX=V$ through forward and backward substitution, and since the determinant is commutative for triangular matricies, they can both be used. Some googling returns nothing on the UL decomposition, leading me to believe that LU has some advantages over UL. So why is LU decomposition preferred?

Source: Linear Algebra by Cherney, Denton, Waldron

In an otherwise banded matrix, sparseness is preserved by:
(1) LU decomposition when there are full rows at the bottom and/or
columns on the right;
(2) UL decomposition when there are full rows at the top and/or columns on
the left.


How can I find the sum of $\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$ How can I find the sum of the series:
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$$
I did this:
$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$
But I am not sure how to proceed.


How to show the following for Planar Graphs-Proof Verification Please tell me where I am wrong:
Consider this graph $G$:

Show that this graph is not planar.

My answer:
Let us consider the vertices $\{0,1,2,3,4\}$.
Then the sub-graph induced by these vertices is a complete graph on $5$ vertices and hence is isomorphic to $K_5$.
We know that by Kuratowski's Theorem ,

A graph $G$ is planar if and only $G$ contains no subgraph homeomorphic to $K_5$ or $K_{3,3}$.

Since $G$ contains $K_5$ as a sub-graph , so by Kuratowski's Theorem $G$ is not planar.
NOTE:I wrote this answer in my exam and got $0$.

My professor wrote in comments "Kuratowski Theorem not applied properly , no use of homeomorphic property"

Can someone kindly say where did I go wrong in my answer? I cant figure out it myself.
If someone could kindly explain what I did wrong in my problem in the exam, I will be extremely grateful to him/her

Your professors comment seems to imply that using Kuratowski's theorem is valid. You wrote that $G$ contains a subgraph that is isomorphic to $K_5$. Kuratowski's theorem says homeomorphic. Now if two graphs are isomorphic they are automatically homeomorphic. I would have given you full credit for your solution but maybe your professor wanted you to explicitly write down the last sentence showing that you knew the meaning of the different concepts.

Perhaps the professor wanted you to explicitly show an isomorphism between $K_5$ and the induced subgraph you picked out in $K_6$?

Here's a way to prove it without using the theorem.

A planar graph with $n$ vertices can have at most $3n - 6$ edges. Since $K_6$ has 6 vertices it would have at most 12 edges if it were planar. However, $K_6$ has 15 edges so it cannot be planar.


A question on conditional expectation A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain
  a 6 and $Y$ the number of rolls needed to obtain a 5.
  I need help in computing $E(X |Y = 1)$ and $E(X|Y=5)$.

I know that $X$ is a geometric random variable, so $E(X)=6.$ My guess for the first one is 7, but I'm not sure. Given $Y=1$, it means that the first roll can't be a 6, so it will take at least 2 rolls to obtain a 6. Is my reasoning right?

EDIT : Thanks to Andre, he gave me a slap in the face right when I needed it, i.e. before my applied analysis exam tomorrow. XD I'll make this answer right before I bring anymore shame on me.

My ex-approach for the case $Y=5$ is indeed making everything more complicated. The more easy and non-complicated way to do this is just use the definition of the expectation :
$$
\mathbb E(X \, | \, Y = 5) = \sum_{n=1}^{\infty} \, n \, \mathbb P(X = n \, | \, Y = 5).
$$
Now we just need to compute those probabilities. We know that $P(X = 5 \, | \, Y = 5) = 0$, so that removes this term. For the first four terms, note that
$$
\mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^{n-1} \left( \frac 15 \right) = \frac{4^{n-1}}{5^n}, n=1, 2, 3, 4.
$$
because the condition $Y=5$ only means for $X$ that the first four rolls cannot take the value $5$, hence the value of those rolls become independent and uniformly distributed over $\{1,2,3,4,6\}$.

For the next rolls, $Y=5$ gives information on the first five rolls but no information on the ones after. Thus
$$
\mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^4 \left( \frac 56 \right)^{n-6} \left( \frac 16 \right).
$$
The $n-6$ stands for the number of rolls after the first five rolls which are not a 6 before you actually get your first $6$ (which gives the $1/6$ term). Therefore, the series we look at in the expectation can be computed, after the first 5 terms, as the derivative of a geometric series in $\left( \frac 56 \right)$. I don't wanna compute it right now because I am going to mess it up, I am definitively too tired for this.


Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:

$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$

I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other?

Another way (by Euler, I think), from the geometric sum:

$$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1}-1}{x-1} \tag{1}$$

Differentiate both sides and multiply by $x$:

$$x + 2 x^2 + 3 x^3 + \cdots + n x^{n} = \frac{n x^{n+2}-(n+1) x^{n+1} +x}{(x-1)^2} \tag{2}$$

Differentiating once more, we get on the LHS

$$1 + 2^2 x + 3^2 x^2 + \cdots + n^2 x^{n-1} \tag{3}$$

which, evaluated at $x=1$ gives our desired sum $\sum_{k=1}^n k^2$. Hence, we just need to compute the derivative on the RHS in $(2)$ (eg, with L'Hopital rule) and evaluate it at $x \to 1$.

It should be evident that this procedure also can be applied (though it also turns more cumbersome) for sums of higher powers.

(Update) here's a concrete computation, applying the binomial theorem on the RHS of $(1)$ and doing a series expansion around $x=1$. Let

$$\begin{align}
g(x)&=\frac{x^{n+1}-1}{x-1}\\
&=\frac{\left(1+(x-1)\right)^{n+1}-1}{x-1}\\
&={n+1 \choose 1}+{n+1 \choose 2}(x-1)+{n+1 \choose 3}(x-1)^2+O\left((x-1)^3\right) \tag{4}
\end{align}$$

Deriving, multiplying by $x$ and deriving again: $$(g'(x) \, x)'={n+1 \choose 2}+{n+1 \choose 3}2 \, x + O(x-1) \tag{5}$$
which evaluated at $x=1$ gives the desired answer:
$${n+1 \choose 2}+2{n+1 \choose 3} =\frac{n(n+1)(2n+1)}{6}  $$

We can apply the same procedure for higher powers. For example:
$$ \sum_{k=1}^n k^3={n+1 \choose 2}+{n+1 \choose 3}6+{n+1 \choose 4}6$$

Another method by using telescoping sum :-
 We know $(a+b)^3-a^3-b^3=3ab(a+b)$ , take

$a=k-1 , b=2$ , then $a+b=k+1$ and $(k+1)^3-(k-1)^3-2^3=6(k-1)(k+1)=6k^2-6$ ,

hence $(k+1)^3-(k-1)^3-8+6=(k+1)^3-k^3+k^3-(k-1)^3-2=6k^2$ , taking sum over $k$

from $1$ to $n$ we get , $\sum_{k=1}^n [(k+1)^3-k^3] + \sum_{k=1}^n [k^3-(k-1)^3] -\sum_{k=1}^n 2 = 6 \sum_{k=1}^nk^2$ , the first

sum on the left hand side is telescoping resulting in $(n+1)^3-1$ , the second sum is also telescoping resulting in $n^3-(1-1)^3=n^3$ , and the third sum is simply $2n$ , hence

$6 \sum_{k=1}^nk^2=(n+1)^3-1+n^3-2n=2n^3+3n^2+n=n(n+1)(2n+1)$


Derivation of Variance of Discrete Uniform Distribution over custom interval I'm trying to prove that the variance of a discrete uniform distribution is equal to $\cfrac{(b-a+1)^2-1}{12}$. I've looked at other proofs, and it makes sense to me that in the case where the distribution starts at 1 and goes to n, the variance is equal to $\cfrac{(n)^2-1}{12}$. I want to find the variance of $unif(a,b)$, that is a uniform distribution that goes from $a$ to $b$, where $a<b$ and $a$ does not necessarily equal 1.

I also realize that you can add / subtract to the distribution, and the variance will not change; hence, you can simply plug in the value $n=b-a+1$. However, I was wondering if there was any way to directly derive the first form of the variance using the property that $V(X) = E(X^2)-\mu^2$. This is the work I've done thus far, but I'm at an impasse. Additionally, is there a easier way to extend the result for the variance of $unif(1,n)$ to $unif(a,b)$? I suppose an easier way would be to show that the $\mathbb{V}(X) = \mathbb{V}(X+a)$, where $a$ is a constant, and then plug in the new range $(b-a+1)$

$
    \begin{split}
      \mathbb{V}(X) &= \mathbb{E}(X-\mu)^2
      \\
      &= \mathbb{E}(X^2) - \mu^2
      \\
      &= \frac{1}{b-a+1}\sum_{x=a}^{b}x^2 - \frac{(a+b)^2}{4}
      \\
      &= \frac{1}{b-a+1} \left(\frac{(b)(b+1)(2b+1) - (a-1)((a-1)+1)(2(a-1)+1)}{6}\right) -
      \frac{(a+b)^2}{4}
      \\
      &= \frac{1}{b-a+1} \left(\frac{(b)(b+1)(2b+1)-(a-1)(a)(2a-1)}{6}\right)-
      \frac{(a+b)^2}{4}
      \\
      &= \frac{1}{b-a+1} \left(\frac{(2b^3+3b^2+b)-(2a^3-3a^2+a)}{6}\right) -
      \frac{(a+b)^2}{4}
      \\
      &= \frac{1}{b-a+1} \frac{(4b^3+6b^2+2b)-(4a^3-6a^2+2a)-(3a^2+6ab+3b^2)}{12}
      \\
      &= \frac{1}{b-a+1} \frac{4b^3+3b^2+2b-4a^3+3a^2-2a-6ab}{12}
      \\
      &= \dots
      \\
      &= \frac{(b-a+1)^2-1}{12}
    \end{split}
$

So much time has passed, but at least now we know it can be done: https://en.wikibooks.org/wiki/Statistics/Distributions/Discrete_Uniform


g.c.d.{$m^p-m:m=2$ to $n$} $= ?$ Let $p$ be an odd prime and $n>2$ is an integer , then what is the $g.c.d.$ of the numbers
{$m^p-m:m=2$ to $n$} ? (by Fermat's little theorem it is easy to see , $2p$ divides the g.c.d. , but I can not proceed further)

Let me expand on benh's answer, which somewhat answers the question for sufficiently large $m$. In particular, we have an exact answer when we set $m\geq \lceil \sqrt p\rceil$. I have yet to consider the case when $n>\lceil \sqrt p\rceil$, although it should be similar.

Let $D$ be the GCD that we seek.

Our first goal is reduce the types of $m$ we need to test.

1. It suffices to consider $m^p-m$ for prime $m$

Suppose we have $q$ any prime and $n$ any integer such that
$$D|q^p-q$$
$$D|n^p-n$$
Then
\begin{align*}
(qn)^p-(qn) &= q^p(n^p)-q^p(n)+q^p(n)-(qn)\\
&=q^p(n^p-n)+n(q^p-q)
\end{align*}
Since $D$ divides $n^p-n$ and $q^p-q$, $D$ divides $q^p(n^p-n)+n(q^p-q)=(qn)^p-(qn)$.

This shows that it suffices to compute $D$ from all $m$ that are prime. Let us denote
$$S_B=\{p\leq B\;|\;p\text{ is prime}\}$$

2. $D$ is squarefree if $B=\lceil\sqrt{p}\rceil$

Let $D$ be obtained from GCDs of $m^p-m$ with $m\in S_B$. Suppose we have some $q|D$. We are interested in whether $q^2|D$. Suppose it does, then we know that
$$q^2|D, D|m^p-m$$
therefore
$$q^2|m^p-m\Longleftrightarrow m^p-m\equiv 0\pmod{q^2}$$
However, suppose we have $q\in S_B$. Then when we test GCD with $m=q$, we have
$$m^p-m=q^p-q=q^2q^{p-2}-q\equiv 0-q\equiv q^2-q\pmod{q^2}$$
which is not $0$. We restate this as follows:

  $\textbf{Proposition.}$ Let $q\in S_B$ and let $D=\gcd(m^p-m\;|\;m\in S_B)$. Then $q^2\nmid D$.

From benh's answer, we know that for any $q|D$, we must have
$$(q-1)|(p-1)$$
from which we can deduce
$$q-1\leq \sqrt {p-1}$$
$$q\leq \sqrt {p-1}+1<\sqrt p + 1$$
$$q\leq \lceil \sqrt p\rceil$$

Therefore if we set $B=\lceil \sqrt p\rceil$, then any prime $q|D$ must lie in $S_B$. Now by our proposition above we must have $q^2\nmid D$. This shows that $D$ is squarefree.

3. Primes $l>p$ does not reduce $D$

Now we consider what happens when we take gcd of $D$ with some prime $l>\lceil \sqrt p\rceil$. Recall that we are not concerned with composite $>p$, since they are always divisible by $D$ by section 1. Suppose we have some integer $q$ such that $q|D$. Then since $D$ is squarefree, $q$ is in fact a prime.

We observe that
\begin{align*}
l^p-l &\equiv (l-q)^p-(l-q)\pmod {q}\\
&\equiv (l-2q)^p-(l-2q)\pmod {q}\\
&\dots\\
&\equiv (l-kq)^p-(l-kq)\pmod {q}
\end{align*}
So that $q|l^p-l$ if and only if $q|(l-kq)^p-(l-kq)$, i.e. when $m=l-kq$, for any integer $k$. But this means we can scale $m=l-kq$ such that $m\leq B$, which we then know that $q|m^p-m$.

Therefore in fact $q|l^p-l$ for any $l>\lceil \sqrt p\rceil$. By consider prime by prime, this shows that in fact $D|l^p-l$. Therefore $D$ does not change when taking GCD with $l>\lceil \sqrt p\rceil$.

4. Exact answer when $B=\lceil \sqrt p\rceil$

We summarize the results so far:
(1) We can find the minimum GCD $D$ by taking $B=\lceil \sqrt p\rceil$ and setting
$$D=\gcd(m^p-m\;|\;m\in S_B)$$
taking GCD with larger $m$ does not change the result.

(2) Every prime factor $q$ of $D$ satisfies
$$(q-1)|(p-1)$$
$$q^2\nmid D$$

Therefore, to find $D$, it suffices to consider all divisors of $p-1$:
$$T=\{x\in\Bbb N\;|\; x|p-1\}$$
If $e\in T$ and $e+1$ is prime, then $e|D$. We can check this 1 by 1, through the list $T$.
(This is the same function as in benh's.)

5. Example

Let $p=7919$, the $1000$-th prime. We have the factorization
$$p-1=2 \cdot 37 \cdot 107$$
so that our set $T$, integers that divide $p-1$, is
$$T=\{1, 2, 37, 74, 107, 214, 3959, 7918\}$$
Note that $q-1\in T$, so we are interested in primes of $T+1$:
$$T+1=\{2, 3, 38, 75, 108, 215, 3960, 7919\}$$
The primes are
$$\{2,3,7191\}$$
Which tells us that
$$D=2\cdot 3\cdot 7191=47514$$

A computer check tells us that this is correct, along with all other primes $\leq 7191$.

There is only one case to consider i.e $n<2p$, otherwise,Fermat's theorem breaks down.
Consider the first two consecutive terms of the set;
from Fermat's theorem,
$m^p-m=kp$ and $(m+1)^p-(m+1)=k'p$, divide the two equations to get;
$(m^p-m)k'={(m+1)^p-(m+1)}k$, now either $(m^p-m)|{(m+1)^p-(m+1)} or $(m^p-m)|k$.
Case 1: $(m^p-m)|{(m+1)^p-(m+1)},
it follows that
$m(m^(p-1)-1)|(m+1){(m+1)^(p-1)-1}, now m does not divide m+1, else m=1, a contradiction, further, if ${m^(p-1)-1}|{(m+1)^(p-1)-1}, then $m|(m+1)$, again a contradiction showing that case 1 does not hold afterall
case 2:  $(m^p-m)|k$
but since  $(m^p-m)=kp$, it follows that either $k|(m^p-m)$ in which case k=1, or k is a multiple of (m^p-m) showing that $p=1$ a contradiction.
the same argument follows for other consecutive values of the given set. Hence, the required $gcd=p$


$\lim\limits_{x\to0}\frac{e^{2x}-1-2x}{x^2}$ without L'Hôpital Can someone help me to find the following limit without using L'Hôpital ?
$$\lim_{x\to0}\left(\frac{e^{2x}-1-2x}{x^2}\right)$$

this is my try for help you to find the following limit without using L'Hôpital


Problem with concepts of circular permutation. I am having problem in understanding this concept:

  Circular permutation : The definition in my book goes like that ' Arrangements of things in a circle or a ring are called circular permutations. The fundamental difference between linear and that of circular permutation is that in the former, there are always two separate ends but in circular permutations we cannot distinguish the two ends. For this,in linear permutations ,arrangements depend on the absolute position while in the case of circular permutations,we shall be concerned with relative positions of the things. Thus, no. of circular permutations of $n$ different things taken all at a time is $$(n-1)!$$ ways taking one of the $n$ things fixed.

This is what is written in the book. Now, it gave two questions ; 1. In how many ways can 6 boys form a ring? . The answer was,according to the formula, was $5!$ ways. 2. In how many ways can 6 men be seated at a round table? The answer, I thought,would be again $5!$ . But book gave the answer $6!$ giving reason that since $6$ men were to be arranged with respect to the table and not with themselves,hence the problem is equivalent to linear permutation.

But I didn't understand their reasoning; 6 men were forming a ring and sitting around a round table,what is the difference? And what the book wanted to tell by saying relative position ?

Another problem is the necklace problem: In how many ways can 6 beads of different colours be arranged to form necklace? The answer,again to me, was $5!$ . But the correct answer ,the book said,was $\frac{5!}{2}$ reasoning that clockwise & anti-clockwise arrangement can't be distinguished. But I couldn't understand this brief reasoning.

Plz help explaining these problems.

The "ring" question asks "how many ways are there to put 6 people in a ring", where two "ways" are identical if each individual has the same left and right neighbours.

The "table" question asks "how many ways are there to put 6 people at a table", where two "ways" are identical if each individual has the same left and right neighbours and the same place at the table.

It is easy to see that since the possible rotations are 6, the answer to the first question will be 6 times the answer to the second question, and indeed 6! = 6 x 5!.

Concerning the "necklace" problem, the difference from the "ring" problem is that you cannot put people upside down without noticing - while you can to so with a necklace, obtaining a pattern which is symmetric. So the number of combination is halved, because two symmetric combinations are considered as one.

Circular permutation of n distinct object actually means shifting for n objects. Their sequence/order will always remain same.

E.g. Consider the following round table.
A is seated besides B and D.
A, in circular permutation, will never be seated besides C.

Here, A can be shifted 3 times in clockwise directions and 3 times
in anti-clockwise direction.
Total=6


Find all entire functions that satisfy following inequality Find all entire functions that satisfy following inequality:

$$ |f(z)| \leq |z| e^{\Re(z)} $$ for all $ z \in \Bbb C $

With

$z = x + iy, \tag 1$

since $\vert e^{iy} \vert = 1$ we have

$e^{\Re(z)} = e^x = \vert e^x \vert = \vert e^x \vert \vert e^{iy} \vert = \vert e^{x + iy} \vert = \vert e^z \vert; \tag 2$

thus

$\vert f(z) \vert \le \vert z \vert e^{\Re(z)} = \vert z \vert \vert e^z \vert = \vert z e^z \vert, \tag 3$

which implies

$f(0) = 0; \tag 4$

it follows then that both $z^{-1}f(z)$ and $e^{-z}z^{-1}f(z)$  are also entire and that (3) yields

$\vert z^{-1} e^{-z} f(z) \vert \le 1, \forall z \in \Bbb C; \tag 5$

therefore, via Liouville's theorem (a bounded entire function is constant),

$z^{-1} e^{-z} f(z) = c \in \Bbb C, \tag 6$

where in fact

$\vert c \vert \le 1, \tag 7$

whence

$f(z) = cze^z, c \in \Bbb C, \; \vert c \vert \le 1. \tag 8$

.

$|\frac{f(z)}{e^z}|\leq|z|\implies f(z)=e^z(az+b)$(by Liouville theorem) $\implies f(z)=e^z(az)$ as $f(0)=0$.


Detail about the definition of orientability I am a bit struggling with one of the many definitions of orientability. In what follows $M$ will always denote a smooth, connected manifold, $T_{m}M$ will be the tangent space at $m$, $\mathcal{B}_{m}$ will be the set of ordered bases of $T_{m}M$, i.e. $$\mathcal{B}_{m}:=\{(v_{1},\ldots,v_{n})\in (T_{m}M)^{n};\operatorname{span}_{\mathbb{R}}\{v_{1},\ldots,v_{n}\}=T_{m}M\}$$$\sim_{m}$ will denote the equivalence relation on $\mathcal{B}_{m}$ obtained by $v\sim_{m}w:\Leftrightarrow \det(A(v,w))>0$, where $v=(v_{1},\ldots,v_{n})$ and $w=(w_{1},\ldots,w_{n})$ are elements in $\mathcal{B}_{m}$ and $A(v,w)$ is the unique element in $\operatorname{GL}(T_{m}M)$ satisfying $A(v,w)v_{i}=w_{i}$ for all $1\leq i\leq n$. Finally $\Lambda_{n}(T_{m}M)$ denotes the space of alternating $n$-forms on $T_{m}M$ and $\Lambda_{n}^{\ast}(M):=\bigcup_{m\in M}\{m\}\times\Lambda_{n}(T_{m}M)$. I have topologized $\Lambda_{n}^{\ast}(M)$ by requiring that if $(U,\rho)$ is a chart in the maximal atlas on $M$, then the map $\tilde{\rho}:\tilde{U}\to\mathbb{R}^{n+1}$ is smooth, where:
$$\tilde{U}:=\bigcup_{m\in U}\Lambda_{n}(T_{m}M) \text{ and }\tilde{\rho}(m,\omega):=\left(\rho(m),\omega\left(\frac{\partial}{\partial x_{1}}\bigg|_{m},\ldots,\frac{\partial}{\partial x_{n}}\bigg|_{m}\right)\right)$$ Indeed the image of $\tilde{\rho}$ is open in $\mathbb{R}^{n+1}$ and so on. The main goal is to put more sense into the following statement:

  Definition

  M is orientable iff $\Lambda_{n}^{\ast}(M)\setminus N$ has two connected components, where $N$ is the range of the map $\omega:M\to\Lambda_{n}^{*}(M): \omega(m):=(m,0)$.

I am struggling with the following statement:

  Problem

  $\Lambda_{n}^{\ast}(M)\setminus N$ has at most two components.

That is it has either one or two. I approached it by comparing different notions of orientability. So far I have managed to prove the following statement:

Theorem

Let $M$ be a smooth manifold and $\dim M=n$. Then the following are equivalent:

There exists a smooth, no-where vanishing differential $n$-form on $M$
There exists an atlas $\mathcal{A}$ on $M$ such that for all $(U,\rho),(V,\psi)\in\mathcal{A}$ holds $\det(\operatorname{d}_{m}(\rho\circ\psi^{-1}))>0$ for all $m\in U\cap V$.
There exists a map $\epsilon:M\to\bigcup_{m\in M}\mathcal{B}_{m}/\sim_{m}$ and for all $m\in M$ there exists $U\subseteq M$ an open neighbourhood of $m$ and smooth vector fields $X_{1},\ldots,X_{n}$ on $U$ such that for all $p\in U$ holds $(X_{1}(p),\ldots,X_{n}(p))\in\epsilon(p)$.
$\Lambda_{n}^{\ast}(M)\setminus N$ has two components.

This does not tell me why we have either one or two components (and definitely not more). I have no idea what to do (I even tried a direct proof).

Edit: It seems to me to be most promising to prove that non-3 implies that $\Lambda_{n}^{\ast}(M)$ has only one component, i.e. I would try to show that if for every $\epsilon:M\to\bigcup_{m\in M}\mathcal{B}_{m}/\sim_{m}$ there exists $m\in M$ such that for every open neighbourhood $m\in U\subseteq M$ and all smooth vector fields $X_{1},\ldots,X_{n}$ on $U$ satisfying $(X_{1}(p),\ldots,X_{n}(p))\in\mathcal{B}_{p}$ for all $p\in U$ we can find $p,q\in U$ such that $(X_{1}(p),\ldots,X_{n}(p))\in\epsilon(p)$ and $(X_{1}(q),\ldots,X_{n}(q))\not\in\epsilon(q)$, then $\Lambda_{n}^{\ast}(M)\setminus N$ has one component. What do you think of this? Is this feasible?

Thank you so much

Manuel

A sketch of a possible proof:

let $\pi: \Lambda_n^*(M) \to M$ be the natural projection map.

Let $C$ be a connected component of $\Lambda_n^*(M)$. Show that $\pi(C) = M$.
Let $U$ be a nonempty connected trivializing open subset of $M$, then $\pi^{-1}(U) - U = U \times \mathbb{R}_{> 0} \cup U \times \mathbb{R}_{< 0}$. (1) shows that $C$ contains at least one of these two components.

a. If for some $U$, $C$ contains both components, prove that $C = \Lambda_n^*(M)$.

b. If for all $U$, $C$ contains only one component, let $D$ be the union of all the other components. Show that $D$ is connected and $\Lambda_n^*(M) = C \cup D$.

You can easily show that $\dim\Lambda_n^*(V) = 1$ (as a vector bundle), so you only need to show, that for every section $\omega\neq 0$ and $\lambda > 0$, $\lambda\omega\in\Lambda_n^*(V)\setminus\{0\}$. That way, for all $\omega\neq 0$ you'd get path-connectedness to all other $\omega'$ on the same “side” of $0$ (always remembering that you are working on the real axis here), so you have at most 2 connected components.


How can I prove the sum of squares and the sum of cubes with binomial coefficients? My main problem is starting. I can't "see" anything that might give me an idea to find a relationship between these two things

Thank you :)

If I may hazard a guess, perhaps you are expected to use the identities
$$\sum_{k=0}^n\binom k2=\binom{n+1}3$$
and
$$\sum_{k=0}^n\binom k3=\binom{n+1}4$$
to derive formulas for
$$\sum_{k=0}^nk^2$$
and
$$\sum_{k=0}^nk^3.$$
For instance,
$$\binom{n+1}3=\sum_{k=0}^n\binom k2=\sum_{k=0}^n\frac{k^2-k}2=\frac12\sum_{k=0}^nk^2-\frac12\sum_{k=0}^nk$$
so
$$\sum_{k=0}^nk^2=2\binom{n+1}3+\sum_{k=0}^nk=2\binom{n+1}3+\binom{n+1}2.$$

To prove $\sum_{k=1}^n k^3 = \binom{n+1}{2}^2$, use Pascal's law to find

$$\binom{n+1}{2}^2 = \left( \binom{n}{2} + \binom{n}{1} \right)^2 = \binom{n}{2}^2 + n^3  $$

and use induction. You're likely to find the formula for sums of squares in a similar fashion.


If a collection on $\Omega$ is closed under finite intersections, then can we conclude that it contains $\Omega$? Here an example that illustrates my question.

A $\pi$-system $\mathcal P$ on set $\Omega$ is defined here to be a collection that is not empty and satisfies the rule $A,B\in\mathcal P\implies A\cap B\in\mathcal P$.

Then they go on by saying that the collection is closed under finite intersections.

But wait a minute... Isn't $\varnothing$ a finite set?

Interpreting the empty intersection as $\Omega$, this can make people think that $\Omega\in\mathcal P$, which is not a consequence of the definition.

($\varnothing\subseteq\mathcal P$ and it is vacuously true that $\omega\in P$ for every $\omega\in\Omega$ and every $P\in\varnothing$)

Personally I would rather go for "closed under binary intersection" keeping in mind of course that this implies "closed under non-empty finite intersections".

Thanks in advance for sharing your thoughts on this.

I agree that for maximal naturality and elegance we should demand that $\Omega$ is in every $\pi$-system.

However, the notion of a $\pi$-system is already not a particularly natural one and is only really used in the $\pi-\lambda$ theorem that says that any $\lambda$-system contains a $\pi$-system then it also contains the $\sigma$-algebra generated by that $\pi$-system. Since $\sigma$-algebras and $\lambda$-systems already have to contain $\Omega$ this theorem would only be made weaker by demanding that $\Omega$ be in the $\pi$-system.

As it is we have made the $\pi-\lambda$ theorem slightly stronger, and at not much expense of elegance, since the notion of a $\pi$-system would is pretty ugly anyway.

They don't mean intersections of finite sets, they mean intersections of finite numbers of sets. For instance:
if A,B, and C are sets in P, then $A\cap B\cap C\in P$
In general, this is true because, the base case follows from the definition of P and if $\cap$ {$A_1,A_2,....A_k$} $\in P$, for $k\in \mathbb N$, and $A_{k+1}$ is any set, ($\cap$ {$A_1,A_2,....A_k$})$\cap A_{k+1}$ is just an intersection of 2 sets, which we know is in P by the definition of P. Thus, by induction on the number of sets in an intersection P is closed under the intersection of any finite number n of sets: i.e. closed under finite intersection.

Also, from the information you provided, it doesn't even follow that the empty set has to belong to a given pi system of a set omega, although there are surely some pi systems that do; for instance, the pi system {O}. Clearly this is a pi system of any set omega since it's a non-empty collection of subsets of omega that is closed under intersections of any finite number of sets within it.


2nd order differential equation with variable coefficients Can someone give advice on how i can solve the following,

$$
\xi\Phi''(\xi ) + ( 1+2\xi^2)\Phi'(\xi)+4\xi\Phi(\xi) = 0
$$

Thanks!

Hint:

$\xi\Phi''(\xi)+(1+2\xi^2)\Phi'(\xi)+4\xi\Phi(\xi)=0$

$\xi^2\dfrac{d^2\Phi}{d\xi^2}+\xi(2\xi^2+1)\dfrac{d\Phi}{d\xi}+4\xi^2\Phi=0$

Let $x=\xi^2$ ,

Then $\dfrac{d\Phi}{d\xi}=\dfrac{d\Phi}{dx}\dfrac{dx}{d\xi}=2\xi\dfrac{d\Phi}{dx}$

$\dfrac{d^2\Phi}{d\xi^2}=\dfrac{d}{d\xi}\left(2\xi\dfrac{d\Phi}{dx}\right)=2\xi\dfrac{d}{d\xi}\left(\dfrac{d\Phi}{dx}\right)+2\dfrac{d\Phi}{dx}=2\xi\dfrac{d}{dx}\left(\dfrac{d\Phi}{dx}\right)\dfrac{dx}{d\xi}+2\dfrac{d\Phi}{dx}=2\xi\dfrac{d^2\Phi}{dx^2}2\xi+2\dfrac{d\Phi}{dx}=4\xi^2\dfrac{d^2\Phi}{dx^2}+2\dfrac{d\Phi}{dx}$

$\therefore\xi^2\left(4\xi^2\dfrac{d^2\Phi}{dx^2}+2\dfrac{d\Phi}{dx}\right)+2\xi^2(2\xi^2+1)\dfrac{d\Phi}{dx}+4\xi^2\Phi=0$

$4\xi^4\dfrac{d^2\Phi}{dx^2}+4\xi^2(\xi^2+1)\dfrac{d\Phi}{dx}+4\xi^2\Phi=0$

$x\dfrac{d^2\Phi}{dx^2}+(x+1)\dfrac{d\Phi}{dx}+\Phi=0$

It has a particular solution $\Phi=e^{-x}$ .

Three hints:

Note that: $4\xi=\dfrac{d}{d\xi}\left[1+2\xi^2\right]$
You could add (and substract) $\dfrac{d\Phi\left(\xi\right)}{d\xi}\dfrac{d\xi}{d\xi}$
Apply the (reverse) product rule: $u\dfrac{dv}{d\xi}+\dfrac{du}{d\xi}v=\dfrac{d(uv)}{d\xi}$


How to find the number of four letter words that can be formed with letter a,b,c such that all three letters occur How to find the number of four letter words that can be formed with letter a,b,c such that all three letters occur

Do we need to make cases :

Case I : a,b,c,a

Case II : a,b,c,a

Case III : a,b,c,c

Please guide what is the correct approach to solve this.

They way I'd go about this is as follows.
First, we choose one of the three letters to be the letter that is repeated.
That's three possibilities.

Next, we find all distinct permutations of our four letters.
There is a letter that appears twice, and two other letters that each appear once.
To count these permutations, we use the multinomial coefficient $\binom{4}{2,1,1} = \frac{4!}{2!1!1!} = 12$.

That would mean the total is $3\times 12 = 36$.

(https://i.stack.imgur.com/JUhMI.jpg)
Copy the link and open the image you will get the solution.


Testing if a function has an inverse. I was just wondering how you apply the rule:
$$f(x_1) = f(x_2) => x_1 = x_2 $$
on the function:
$$f(x) = x^3 - 9x^2 +33x +45$$

Any suggestions on how to proceed would be appreciated.

EDIT: Yes there should have been a $$9x^2$$ in there.

The derivative of $f$ is
$$f'(x)=3x^2-18x+33=3(x^2-6x+11)=3[(x-3)^2+2]$$
which is positive. Therefore, $f$ is strictly increasing and hence injective. Nevertheless, an algebraic expression for the inverse it is complicated, because it involves solving of a cubic equation.

Very simple assume that $f(x_1)=f(x_2) \implies x_1^3+24x_1+45 = x_2^3+24x_2+45$ then we can simplyfy this by saying that $x_1^3-x_2^3 +24(x_1-x_2) =0$

If $x_1 \neq x_2$ then can not be true,thus they must be equal,thus function is injective thus has inverse


Prove that if the $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$ I need to prove that:

If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1.$

So far I used what's given so I have:

$ax+by=1$ (I wrote the gcd of a and b as a linear combination)

and

$(a+b)u+ab(v)=1$ (I also wrote this as a linear combination)

where do I go from here?

Below is a proof that is messier but more primitive than Yiorgos' proof. By Bezout's Identity, we have that there exists $x,y \in \mathbb{Z}$ such that

$$ax + by = 1.$$

Squaring both sides, we see that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = 1.$$

But notice that

$$a^2 x ^ 2 + 2abxy + b^2 y^2 = ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2).$$

And hence, those same $x,y$ as above give

$$ ab(2xy-x^2-y^2) + (a+b)(ax^2+by^2) = 1.$$

So it must be that $\gcd{(a+b,ab)} = 1$.

Proof broken into lemmas:

Lemma 1:  If $n$ is rel. prime to $a$ and $n$ is rel. prime to $b$, then $n$ is rel. prime to $ab$.

Proof:  multiply together the Bezout identities for the two hypotheses.

Lemma 2:  If $a$ and $b$ are rel. prime, then both are rel. prime to $a+b$.

Proof:  Its enough to prove if  $a$ and $b$ are rel. prime then $b$ is rel. prime to $a+b$.
We have the Bezout identity:  $a x + by = 1$.  Now add and subtract $bx$ on the left.

Conclusion:  We are given $a$ and $b$ rel. prime.  By the 2nd lemma, $a+b$ is rel. prime to both $a$ and $b$.  By the first lemma, $a+b$ is rel. prime to $ab$.


trouble in evaluating a double integral using polar coordinates. Here I have:
$$\int _0^a\int _{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}\sqrt{a^2-\left(x^2+y^2\right)}\:\:dydx$$
which can also be expressed as:
$$2\int \:\int \:r\sqrt{a^2-r^2}drdθ$$
over a region which should be bounded by:
$$θ=0\:,\:θ=\frac{\pi }{2}\:,\:r=2\:,\:r=asecθ$$
I tried sketching the region but couldn't still find the intervals of $dr,dθ$

It seems to me that this region should be split to two parts:

$|\theta| \leq \cos^{-1}(\frac {2}{a})$, in this part r goes from 0 to $\frac{a}{\cos \theta}$

The second region(s) are for $\frac{\pi}{2} \geq \theta \geq \cos^{-1}(\frac {2}{a})$
and $-\frac{\pi}{2} \leq \theta \leq -\cos^{-1}(\frac {2}{a})$, for which r goes from 0 to 2.


Factors of a number? I'm studying for the SAT on collegeboard.org, and I came across the problem: What is the least positive integer that has the same number of positive factors as 175?

The answer was 12, explained as:

The number 175 is equal to 5 x 7 x 5. It has a total of 6 positive factors: 1, 5, 7, 25, 35, and 175. The numbers with 6 positive factors are of the form$p_1^2*p_2$or$p^5$. The least number of the first form is 12. The least number of the second form is 32. Thus, the answer is 12, which has the  positive factors 1, 2, 3, 4, 6, and 12.

I've never learned anything about the number of factors for a number, so this line: The numbers with 6 positive factors are of the form$p_1^2*p_2$or$p^5$ doesn't make sense to me. Out of context, I understand that $p_1$ is 2, but I don't understand where those two formulas came from or what p is supposed to be the set of.

$p_1$, $p_2$, and $p$ in your quote stand for primes - arbitrary primes. That is, if $p$ is any prime whatsoever, then $p^5$ has exactly 6 factors; and if $n$ has exactly 6 factors, then either $n=p^5$ for some prime $p$, or $n=p_1^2p_2$ for some primes $p_1$ and $p_2$. Everything rests on a formula for the number of factors: if $$n=p_1^{a_1}p_2^{a_2}\cdots p_r^{a_r}$$ where the $p_i$ are distinct primes and the $a_1$ are positive integers, then the number of factors of $n$ is $$(a_1+1)(a_2+1)\times\cdots\times(a_r+1)$$

You can find this explained in any good introductory number theory textbook.

This seems a little hard for the SATs.  I assume there were multiple choices?  This might have been one case where you might have tried working backwards, counting the factors of each answer starting with the least.


Why do we use decimal system? Why do we use decimal system?

Some people give the answer that we have 10 fingers and it is therefore convenient for us to use decimal system, but I am not satisfy with this as I think this is too rough. Are there more reasonable or scientific explanation about why do we use decimal system or what are the advantages of using decimal system? Thank you.

There is no definite reason as to why the system chosen was the decimal system. A long time ago, every civilization had their own numeral system. The Egyptians used the decimal system, the Babylonians used a sexagesimal (base-60) system, some used a base-20 system (fingers and toes), and others counted 1, 2, 3, and anything 4 or greater was just "many". The consensus upon the decimal system was probably due to either the ease of use or that it was the first numeral system to come up with easily used symbols for arithmetic.

interesting question ! :)

I THINK THIS A POSITIONAL NUMBER SYSTEM .and that's of great advantage ..simple shifting the position of decimal . this is the most conventional answer ..! but somehow it seems , not that strong . we could have tried to construct a fractional system in other number systems as well !

but there are certain points i have made out :

if the base is more than 10 , say 20 . then  if you want to use our 10 symbols (i.e. 0, 1..,9) it not possible for us to write numbers more than 9 and less than 20 or numbers more than 189 and less than 200 and numbers of similar  form ) , so in that case for each unit either you will have to allot two places for digits (which will make it difficult to read a number) or you will have introduce more 10 symbols , which will make it more complex and we wont get any advantage. cause 20 is just a a multiple of 10 by 2 ..and 2 is a factor of 10 ( if the base would be 30 then , we could take it as an advantage that the problem in fixing the decimal we face while handling division by 3 could be simplified . but in that case just imagine how many symbols you will have to introduce !!! the base could be 12 . in that case dividing by three would not face any approximation problem..or if the base would be 14 then we would easily be able to divide by 7 ..but in both 12 and 14 base the problem in dividing by 5 arises. and in ancient days as our hand and toe consists of 5 fingers , i think divisibility ease for 5 was given preference ) so among systems with 10 or more than ten , the base 10 system , i.e. decimal system was given preference in different civilization , as a human mind always and everywhere think in the same direction and following our ancestors we set this as our ruling international number system mostly used !

now , for bases less than 10 : there is no number which has three prime factors ! among numbers 10 or less than 10 , 10 has the highest number of one digit-ed prime factors ! and for that reason in decimal system , we can properly divide any number by most number of numbers compared to all the other systems with base less than 10 .!(except 6 . which was again given less priority because 10 is a multiple of 5 )

NOTE : some say , "you can order the numbers with specific space between in decimal system .. If we can't order the number with equal interval among them , then it wont be possible to get any number line,and whole number theory will possibly be collapse then . the way co-ordinate geometry ,..vector algebra work , where ordering is very important will collapse ..! e.g in octal system 63 =77 where 64 = 100 and 65 = 101 ..so , then what's the measure of unit countable increase ? .." but , that's not a proper argument in my opinion . there  we will have to consider 77+1 = 100 , 7+1=10 , 777+1=1000...and so on ,..