Nauseated

This equation relates the luminosity of a star to various vectors and physical Moduli.

It’s concepts are taught in basic astrophysics classes –usually in grade 6 or 7, depending on which school you attend.

This site calculator will solve symbolic math. You can use it to test your solutions.

This returns the solution for the question above.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, c3}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{6}}}}\\
{\mathtt{c2}}={\frac{{\mathtt{\pi}}{\mathtt{\,\times\,}}{{\mathtt{d}}}^{{\mathtt{3}}}}{{\mathtt{4}}}}\\
{\mathtt{c3}}=\left({\mathtt{1}}{\mathtt{\,-\,}}\left({\frac{{\mathtt{c1}}}{{\mathtt{c2}}}}\right)\right)\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{0}}\\
{\mathtt{d}} = {\mathtt{0}}\\
{\mathtt{c2}} = {\mathtt{0}}\\
{\mathtt{c3}} = {\mathtt{r12}}\\
{\mathtt{c1}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{6}}}}\\
{\mathtt{d}} = {\mathtt{r13}}\\
{\mathtt{c2}} = {\frac{{{\mathtt{r13}}}^{{\mathtt{3}}}{\mathtt{\,\times\,}}{\mathtt{\pi}}}{{\mathtt{4}}}}\\
{\mathtt{c3}} = {\frac{{\mathtt{1}}}{{\mathtt{3}}}}\\
\end{array} \right\}$$

The result is the same as CPhill’s solution.

Note that all the Cs & Ds are not affecting the result. This calculator has a high resistance to CDD

Here’s another one

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c1, d, c2, x}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{c1}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}{\mathtt{0.5}}{d}\\
{\mathtt{c2}}={\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\times\,}}\left({\mathtt{0.5}}{d}{\mathtt{\,\small\textbf+\,}}{\mathtt{x}}\right)\\
{\mathtt{c2}}={\mathtt{c1}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{c1}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\\
{\mathtt{d}} = {\mathtt{r9}}\\
{\mathtt{c2}} = {\mathtt{r9}}{\mathtt{\,\times\,}}{\mathtt{\pi}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\\
{\mathtt{x}} = {\frac{{\mathtt{1}}}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{\mathtt{\pi}}\right)}}\\
\end{array} \right\}$$ 

^ - - - Does anyone know what it is for? It’s easy.

This one is totally symbolic.

$$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{c2}}}}}{{solve}}{\left(\begin{array}{l}{\frac{{\mathtt{c2}}{\mathtt{\,\times\,}}{\mathtt{d2}}}{{\mathtt{t2}}}}={\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}}{{\mathtt{t1}}}}\end{array}\right)} \Rightarrow {\mathtt{c2}} = {\frac{{\mathtt{c1}}{\mathtt{\,\times\,}}{\mathtt{d1}}{\mathtt{\,\times\,}}{\mathtt{t2}}}{\left({\mathtt{d2}}{\mathtt{\,\times\,}}{\mathtt{t1}}\right)}}$$

If the boxes were distinguishable then your solution would be correct. In effect, this is distributing indistinguishable b***s (k) into distinguishable boxes (N), forming a combination of size k, taken from a set of size n.

 $$\displaystyle \left( {\begin{array}{*{20}c} 4 \\3 \\ \end{array}} \right)\; = \; \dfrac{(4)!}{3!(4-3)!} \; = \;\hspace{5pt} \Text {4 \ ways} \\\$$

When both b***s and boxes are indistinguishable, empty boxes are forbidden, if they were allowed then they are distinguishable, and the above formula would apply.

The reason for this is the mathsgod said so. Mathsgod, please come and verify the veracity of this statement.

These types of questions are not intuitive. Consider your statement: “All the b***s and all the boxes are identical . . .” Though it would seem so, this is not the same as indistinguishable in a mathematical sense.

If identical boxes are present at the same time then they are identifiable via their positions in space. Like many hypothetical math questions, it becomes difficult -- sometimes impossible-- to create a real situation where these concepts can be examined. Many math realities exist only in the mathematics and cannot exist in our normal space-time. Torricelli's trumpet is a classic example of this.

This question can exist in normal space but probably not with boxes. An approximation might be to place the boxes behind a frosted glass that distorts their size, shape and position in space. This approximation still falls short. To be true to the question, the boxes cannot exist until the b***s are placed in them. This is why it is a partition question. There are no empty partitions because the b***s themselves make the partitions.

As complex as this may seem, it becomes more so when the b***s are distinguishable. These are royal ball-busters. Sterling numbers of the second kind are used for these partitions. 

When the both the boxes and the b***s are indistinguishable then it is a partition question.

The partition function is defined as the number of k-element partitions of N.

In this case, the b***s are the k-element and the boxes are N.

Partitions of 4 constraint of 3. Sometimes denoted as Part(4,3)

$$\displaystyle \text {List: Part(4,k)}

4=4

2+2=4

2+1+1 = 4 \leftarrow \text {This is the only solution}

1+3=4$$

There is only one (1) way to Partition 4 b***s into 3 boxes

Wolframalpha scripted link :

What is it with all the box and ball questions? Have the new text books arrived or are we are all in a billiards tournament getting our b***s racked?

We’ll find out after the break . . .

$$\displaystyle \left( {\begin{array}{*{20}c} 11 \\ 8 \\ \end{array}} \right)\; = \; \dfrac{11!}{8!(11-8)!} \; = \;\hspace{5pt} \Text {165 \ ways} \\\$$

This formula is rare. I have yet to find it after searching several statistical and combinatorics sites. I am sure it is on the web somewhere. How do you search for a formula that does not have a name?

This formula was given to me years ago by a classmate. I am unsure if he found it in a publication or independently discovered it –he definitely had the math skills to do that. I wrote a Fortran program to test it. All the tests demonstrated (not proved) it distributes the counts such that N contains at least one (or more) k elements.

I’ll continue to look for the theory behind this formula, either on the web or by dissecting it. It looks very similar to the Binomial CDF that should give a starting point.

How do you know which number is N and which one is k. Should N be the smallest one, it was for your other example.

Yes, the (k) has to be equal to or greater than the (N). The (i) functions as the k in the binomial.

Has anyone noticed odd changes -- like space-time distortions where yesterday is tomorrow, and the universe seems a little more salty?

Maybe it’s just me :)

Geno’s solution applies if the setup instructions restrict a computer to one graphic device at a time and allows a computer to be disconnected from all graphic devices.

The set up in the question requires a computer connection to at least one peripheral graphics device and allows for more than one connection. The peripheral graphics device require at least one connection to a computer and allows for more than one connection. Because of the bidirectional nature of this communication, if a computer is connected to a peripheral, the peripheral is also connected to the computer. This may seem obvious but the connection requirements would not work if this were not so.

 This kind of software is usually installed in a print server. The computers poll the peripherals to find the status and the peripherals broadcast their status to the computers. Similar software controls network connections. The connection protocols are relatively trivial, while the algorithms that optimize the routing hops through nodes are complex and intricate. The solution to this is found using this formula. N = number of computers and k = number of peripheral graphics devices.

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \hspace{15pt}| \hspace{15pt} \Text {N=19 \; k=23} \\\$$

Wolfram Alpha: scripted link

Technically, the equal sign made it true, but it’s confusing:  Remove errant Binomial from my mind and this post.

A little CDD here. The formula is correct: (4+3-1)!/3! = (6!/3!); the binomial is in error.

Now corrected and noted.