Mathematics

12:00 AM

Unfortunately false @Ted

"every undirected graph is connected" @Karlo no, there are disconnected undirected graphs. take for example two triangles with no edges between them

Karlo

@JMoravitz oh yes I forgot that there is edge between 2 verticles.This will mean there are at least 2 edges between 2 verticles which means it is always connected.

@JMoravitz the definition is for every pair ofcourse :)

for connected

JMoravitz

the term is vertices, not verticles,. and the definition of connected is that a graph is connected iff for any two distinct vertices there is a path (not necessarily a path of length 1) between them.

Karlo

@JMoravitz yes typo sorry

JMoravitz

you don't need that every two vertices be adjacent. A dodecagon for example is connected.

Karlo

12:06 AM

@JMoravitz yes if in undirected graph which has 2 edges between every 2 verticies we can go anywhere and go back however for directed this is too strong even for multi directed graph because we might not be able to go back

@JMoravitz that is why in directed we use low connected where we can go fro A to B or from B to A :)

@JMoravitz it is not even needed in undirected to have 2 edges between verticies one will do the work for connected right?

JMoravitz

ignoring the direction on the edges for now, the one on the left is an example of a connected graph

Karlo

but we have path from 5-6 but not from 6-5?

JMoravitz

ignoring the direction on the edges

Karlo

oh yes then the graph

is undirected :)?

JMoravitz

if we were to consider it as a digraph with the directions as pictured, it would not be a strongly connected digraph, but it will be a weakly connected digraph.

Karlo

12:13 AM

yes strong is from A B exists path from B A

and vice versa

and week is when we can go from A to B or from B to A

@JMoravitz so every indirected graph is strong?

@JMoravitz if!!!! there is connection between 2 verticies for sure

@JMoravitz otherwise it will look as the picture on the right

@JMoravitz without the pointing arrows

JMoravitz

every strongly connected digraph is a connected graph when you ignore the directions on the edges., also every weakly connected digraph is a connected graph when ignoring the directions on the edges

Karlo

@JMoravitz exactly ,thanks :)

JMoravitz

Perhaps it is easier to define what it means for the graph to be disconnected. A graph on 2 or more vertices is called disconnected if there exist two vertices with no path between them.

Karlo

@JMoravitz yes

evinda

12:30 AM

Hey @DanielFischer!!! A farmer Α has oranges the 10%of which are sour. A farmer Β has oranges 4% of which are sour. A client chooses per chance ( with propability 1/2) two oranges.
Which is the probabilty, if the first that he chooses is sour, that the second is also sour?

I drawed the following diagram:

Is it right? And could it help?

Julian Rachman

Hello

user159870

youtube.com/watch?v=62XB9IbMnxQ @JulianRachman

Julian Rachman

@user159870 the link is currently not working

nvm. lol. hello

JMoravitz

1:02 AM

@evinda usually for tree diagrams we prefer to have all choices (including which farmer he chooses) as branches. Also, as I am understanding the question, the buyer will choose a single farmer from which to buy, then buy one orange, then buy another orange from the same farmer

evinda

@JMoravitz How could we apply Bayes?

JMoravitz

pardon the bad paintjob, but here is a tree diagram of the situation

evinda

@JMoravitz Could you explain me the diagram?

JMoravitz

$P(\text{second is sour | first is sour}) = \frac{P(\text{both are sour})}{P(\text{first is sour})}$

We begin at start. We then make a choice of either buying from farmer A (denoted by going upright), or buying from farmer B (denoted by going downright)

once there, we buy our first orange. either it is sour or it is not (up right or downright respectively) with probability .1 and .9 respectively if having bought from farmer A, or with probability .04 and .96 respectively if having bought from farmer B

Buying another orange, we either go upright or downright along the diagram if the second orange was also sour or not. (note, the way the problem is worded suggests we buy the orange from the same farmer, so probabilities remain the same)

I've marked which leaves on the diagram correspond to having the first orange be sour

with a red arrow. Considering only those, we look at which have both first and second sour,

By definition of conditional probability, we have what I said above. So, calculate probability both are sour and divide by the probability the first is sour

$\frac{.5\cdot .1\cdot .1 + .5\cdot .04\cdot .04}{.5\cdot .1 + .5\cdot .04}$

bd1251252

2:14 AM

@TedShifrin Haha thanks. By the way -- the formula you mentioned in class. I've taken every course up to differential equations and not once did I see that \begin{align}\sum\limits_{i=1}^{m}n_i=\frac{m\left(m+1\right)}{2}\end{align},\:\‌:\ni\:\:n,m\in\mathbb{Z}\end{align}

Darn...didn't have enough time to edit it

@TedShifrin Only when I bought the book Advanced Calculus by Widder did I study the chapter on Stieltjes Integrals and come across it when simplifying them using the definition (the hard way) s.th. \begin{align}\sum\limits_{n=1}^{\infty}f\left(x_n\right)\left(\alpha\left(x_n\ri‌ght)-\alpha\left(x_{n-1}\right)\right)\end{align}

JMoravitz

2:35 AM

@bd1251252 it appears the mathjax doesn't like it when you do nested \left('s. replacing the "\left("'s inside of the nest fixes your output

$\sum\limits_{n=1}^{\infty}f\left(x_n\right)\left( \alpha (x_n)-\alpha(x_{n-1})\right)$