/sci/ - Science & Math

[Anonymous No.5640305]

Let's do this /sci/

[Anonymous No.5640320]

How about I operate my foot up your ass?

>redforman.jpg

[Anonymous No.5640324]

>>5640320

10/10.

[Anonymous No.5640325]

Try this one out:

<span class="math">a * b = a^{\ln(b)}[/spoiler]

Can you see if it has any special properties?

[Anonymous No.5640326]

/g/ here

++

it adds 1 to the number

[Anonymous No.5640328]

>>5640326
face-palm

CS
>nor science
>nor computer

[Anonymous No.5640339]

Given a group (G, *), we define an operation ° : G x G -> G, with °(a, b) = a*(b/b), or a ° b.

Examples: On (Z, +), 5 ° 8 = 5 + (8 - 8) = 5.
Where da Fields Medal at?

[Anonymous No.5640343]

>>5640328
tell that to my computer that is executing sciency code

[Anonymous No.5640348]

>>5640339
I don't see how this is useful, at all

[Anonymous No.5640353]

a + b = a plus b

I wouldn't be surprised if this catches on.

[Anonymous No.5640357]

0.9999999999... is simply written as 1

[Anonymous No.5640358]

>>5640348
Yep

[Anonymous No.5640360]

>>5640339
>2013
>not using the universal property of the projection

[Anonymous No.5640361]

>>5640339

I deem this potent

[Anonymous No.5640365]

>>5640357
No, fuck you. 0.999999999999999999999999999999999999999999999999999999999.. does not equal 1

[Anonymous No.5640371]

>>5640365
it has the same value though so it doesn't matter

this has been discussed countless times and unless you have OCD or something you wouldn't care about that 1 at the end

[Anonymous No.5640370]

a| =a + arbitrary constant

[Anonymous No.5640376]

>>5640360
0/10 no even trold, fugging bleb :D

[Anonymous No.5640383]

(-a)-(-b)=-1(a+b)

[Anonymous No.5640384]

>>5640357
This.

[Anonymous No.5640386]

An operation, to do comparison on fractals. And in general operations, that deal with fractals. Fusing them together, subtracting them from each other, recognizing sub fractals... that be great.

[Anonymous No.5640390]

<div class="math">a \smile b \Leftrightarrow \frac{a}{b + \frac{a}{b}}</div>

<div class="math">a \frown b \Leftrightarrow \frac{b}{a + \frac{b}{a}}</div>

[Anonymous No.5640394]

>>5640305
>Rep
An operation, to do comparison on fractals. And in general, operations, that deal with fractals. Fusing them together, subtracting them from each other, recognizing sub fractals... that be great.

[Anonymous No.5640404]

superduper exponents, represented by ~ or whatever you want
x~y
x^x^x.... y times

[Anonymous No.5640412]

A/2+7=X
Where A is your age and X is the acceptable minimum dating age.
>brocode?

[Anonymous No.5640411]

>>5640383
LOL yes!

[Anonymous No.5640418]

8○1=2
It calculates how mani curcles the left number hsd vs the irgt nunmver

[Anonymous No.5640419]

>>5640390
I like this, I like it a lot

7.5/10

[Anonymous No.5640420]

>>5640404
This may yet have some use.

[Anonymous No.5640425]

here's a brain tickler: invent a new operator that cannot be expressed in terms of any other known operator. For example, multiplication is just a combination of the addition operator and a counter operator(the counter being resonsible for counting the number of times the addition is done) Addition itself though cannot be expressed into anything more axiomatic than itself (as far as im aware). So far, the most axiomatic operators i can think of are addition, counter, and limits(since subtraction is just the addition of negative numbers. the set of numbers in question exist INDIPENDANTLY of the operator)

[Anonymous No.5640427]

>>5640390
Also, b != 0 for first one and a != 0 for second one.

[Anonymous No.5640428]

>>5640371
applied math cares,
quit being a retarded uneducated faggot

[Anonymous No.5640430]

>>5640418
To gibe more exösmles
10○1=1
0○0=0
667○6=1
000007º4=5
Maybr not useful bit preyt cool xd

[Anonymous No.5640431]

>>5640418
love it

[Anonymous No.5640432]

>>5640430
you sound drunk off your tits

[Anonymous No.5640433]

>>5640425

P(n) = nth prime

[Anonymous No.5640437]

>>5640404
>>5640419
what is uparrow-notation

[Anonymous No.5640439]

>>5640425
Ada Lovelace apparently proved you can't do this. Fucked if I know how though

[Anonymous No.5640444]

x#y
x cannot be expressed in terms of y

[Anonymous No.5640450]

>>5640439
Cant find anything about this while reaserching. Are u sure this is legit? sauce pls

[Anonymous No.5640452]

>>5640444
Example?

[Anonymous No.5640454]

testing tex e^{x}

[Anonymous No.5640459]

how the fuck do you use tex on here

[Anonymous No.5640460]

>>5640425
P(n) = the probability that an n-bit program will halt

[Anonymous No.5640472]

>>5640425
A + B = take the successor of the A term B times

[Anonymous No.5640474]

pic related equals
<span class="math">e^{\int{P(x) dx}} [/spoiler]

[Anonymous No.5640478]

>>5640459
<div class="math"></div>

[Anonymous No.5640479]

>>5640452
for two functions
F(x)=sin(x)
G(x)=Ln(x)
F(x)#G(x)

[Anonymous No.5640485]

>>5640459

[Anonymous No.5640489]

fixp(f(x)) returns a fix-point of f

[Anonymous No.5640491]

>>5640479

sin(x)=sin(x)ln(x)/ln(x)

[Anonymous No.5640493]

fuckit(x) = replace x with something more convenient

[Anonymous No.5640501]

<span class="math">
a \wr b = \frac{1}{\frac{1}{a} + \frac{1}{b} }
[/spoiler]
please tell me something like this already exists

[metric !1tensorX8Q No.5640503]

I've used this in the past as a shorthand for exponential operators. Given an operator <span class="math">\hat{A}[/spoiler] and an object <span class="math">f[/spoiler] (it could be a function, or a vector, etc.), then:

<div class="math">\hat{A}\circledast f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f</div>
For example, for the differential operator on a function f(x):

<div class="math">\frac{\partial }{\partial x}\circledast f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)</div>
Writing <span class="math">e^\hat{A}[/spoiler] or <span class="math">exp(\hat{A})[/spoiler] can get sloppy-looking for more complicated operators.

[Anonymous No.5640510]

>>5640491
damn. I don't think I've thought this through.
Been looking at transcendental numbers you see.

[Anonymous No.5640511]

>>5640501
stupid font size
i mean pic related

[metric !1tensorX8Q No.5640516]

**Last one didn't work. Let's try this:

I've used this in the past as a shorthand for exponential operators. Given an operator <span class="math">\hat{A}[/spoiler] and an object <span class="math">f[/spoiler] (it could be a function, or a vector, etc.), then:

<div class="math">\hat{A}\blacklozenge f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f</div>
For example, for the differential operator on a function f(x):

<div class="math">\frac{\partial }{\partial x}\blacklozenge f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)</div>
Writing <span class="math">e^\hat{A}[/spoiler] or <span class="math">exp(\hat{A})[/spoiler] can get sloppy-looking for more complicated operators.

[Anonymous No.5640518]

>>5640516
noope

[metric !1tensorX8Q No.5640521]

**Last one didn't work. Let's try this:

I've used this in the past as a shorthand for exponential operators. Given an operator <span class="math">\hat{A}[/spoiler] and an object <span class="math">f[/spoiler] (it could be a function, or a vector, etc.), then:

<div class="math">\hat{A}\lozenge f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f</div>
For example, for the differential operator on a function f(x):

<div class="math">\frac{\partial }{\partial x}\lozenge f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)</div>
Writing <span class="math">e^\hat{A}[/spoiler] or <span class="math">exp(\hat{A})[/spoiler] can get sloppy-looking for more complicated operators.

[metric !1tensorX8Q No.5640523]

**Last one didn't work. Let's try this:

I've used this in the past as a shorthand for exponential operators. Given an operator <span class="math">\hat{A}[/spoiler] and an object <span class="math">f[/spoiler] (it could be a function, or a vector, etc.), then:

<div class="math">\hat{A}\star f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f</div>
For example, for the differential operator on a function f(x):

<div class="math">\frac{\partial }{\partial x}\star f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)</div>
Writing <span class="math">e^\hat{A}[/spoiler] or <span class="math">exp(\hat{A})[/spoiler] can get sloppy-looking for more complicated operators.

[Anonymous No.5640526]

>>5640523
>>5640521
>>5640516
>>5640503

hahahaha wat

[Anonymous No.5640528]

>>5640357
0.999999999?=?1
It means it definitely is equal to it, but people will argue about it anyway.
see also ?≠?
as in 1/0?≠?∞

[metric !1tensorX8Q No.5640530]

**Goddammit

I've used this in the past as a shorthand for exponential operators. Given an operator <span class="math">\hat{A}[/spoiler] and an object <span class="math">f[/spoiler] (it could be a function, or a vector, etc.), then:

<div class="math">\hat{A}\star f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f</div>
For example, for the differential operator on a function f(x):

<div class="math">\frac{\partial }{\partial x}\star f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)</div>
Writing <span class="math">e^{\hat{A}}[/spoiler] or <span class="math">exp(\hat{A})[/spoiler] can get sloppy-looking for more complicated operators.

[metric !1tensorX8Q No.5640535]

>>5640530
should be:

<div class="math">\hat{A} \star f=e^{\hat{A}} f= \left [1+\hat{A}+\frac{1}{2} \hat{A}^2+\frac{1}{6} \hat{A}^3+... \right ]f</div>

[Anonymous No.5640537]

>>5640530
<span class="math">\hat{A}\star f=e^{\hat{A}}f=\left [ 1+\hat{A}+\frac{1}{2}\hat{A}^2+\frac{1}{6}\hat{A}^3+... \right ]f

For example, for the differential operator on a function f(x):

\frac{\partial }{\partial x}\star f(x)=e^{\partial /\partial x}f(x)=\left [ 1+\frac{\partial }{\partial x}+\frac{1}{2}\frac{\partial^2 }{\partial x^2}+\frac{1}{6}\frac{\partial^3 }{\partial x^3}+... \right ]f(x)

Writing e^{\hat{A}} or exp(\hat{A}) can get sloppy-looking for more complicated operators.[/spoiler]

[Anonymous No.5640538]

>>5640305
Q(x)=x^x^x^x^x^x

[Anonymous No.5640534]

>>5640530
duuuude you got 'fra c' instead of 'frac'

[Anonymous No.5640540]

>>5640493
I've been doing this on tests for years.
>mfw instructors think I just misread the problem

[metric !1tensorX8Q No.5640542]

>>5640534
I didn't actually. The shitty tex on this site just decided to interpret it that way.

[Anonymous No.5640545]

>>5640365
0.99999.....= 1
If you haven't seen the proof for that by know you have no business on sci

[Anonymous No.5640549]

The stochastic operator <span class="math">\heartsuit[/spoiler]

<span class="math">a \heartsuit b[/spoiler] means <span class="math">a = b[/spoiler] with probability 33.34%

Example: <span class="math">0.999\ldots \heartsuit 1[/spoiler]

[Anonymous No.5640550]

>>5640545
>know

[Anonymous No.5640556]

>>5640550
>Snow
I'm sleep deprived, and this isn't /lit, you don't get points for spelling

[Anonymous No.5640658]

>>5640305
Let <span class="math">||:\mathbb{C}^+\times \mathbb{C}^+\to \mathbb{C}[/spoiler], where <span class="math">{\mathbb{C}^+} [/spoiler] is the set of complex numbers with positive real part, be an operator defined as follows, where <span class="math">Z_1,Z_2\in \mathbb{C}^+:[/spoiler]
If <span class="math">Z_1\neq 0[/spoiler] or <span class="math"> Z_2\neq 0[/spoiler], then <div class="math">
Z_1||Z_2 = \frac{Z_1 Z_2}{Z_1+Z_2}.
</div>
If <span class="math">Z_1 = 0[/spoiler] and <span class="math"> Z_2 = 0[/spoiler], then <div class="math">
Z_1 || Z_2 = 0.
</div>
I used this operator all the time when I took electronics.

[Anonymous No.5640685]

>>5640325
associative
commutative
the neutral element is e
the inverse element to an element a is e^(1/ln(a))
if we define "." as standard multiplication, we have distributive property:
a*(b.c)=(a*b).(a*c)
and
(a.b)*c=(a*c).(a*b)

my summary: good operation/10 would use again

[I should be sleeping by now. No.5640690]

Y1=x^2-1
Y2=x

§Y means find the area between the curves.

[Anonymous No.5640698]

%: a%b=ab/100

[Anonymous No.5640745]

I once invented an "imaginary" number <span class="math">\tau[/spoiler], it had the property that <span class="math">\tau^x = x[/spoiler].

Apparently it broke pretty much every common axiom so my teacher didn't deem it useful.

[Anonymous No.5640749]

>>5640404

[Anonymous No.5640753]

OP &&& =

OP &&& = FAG

[Anonymous No.5640757]

>>5640404
>tetration
Nothing new, sorry

[Anonymous No.5640763]

>>5640745
so you invented the x-th root?

[Anonymous No.5640767]

>>5640305

The number line is extended to a maximum value of 2pi thus everything can be expressed within the confines of a circle.

[Anonymous No.5640781]

>>5640412

>minimum acceptable dating age is one ninth your age

[Anonymous No.5640805]

>>5640528
why does this not have more notes this operation is perfect for /sci/

[Anonymous No.5640829]

The trivial operation: a *0* b

5 *0* 3 = 0
190 *0* -32.8 = 0
g *0* h = 0 (identity), g,h in group G
Z2 *0* V4 = 0 (trivial group)
(Structure 1) *0* (Structure 2) = 0 (trivial structure)

what do i win

[Anonymous No.5640844]

>>5640781
>one ninth your age
Lrn2PEMDAS, dumbass.

[Anonymous No.5640851]

>>5640428
>applied math
>mfw

[Anonymous No.5640856]

>>5640472
This, succession, counter, limit are the only legit ones

[Anonymous No.5640860]

>>5640510
No fucking dip it simplifies

[Anonymous No.5640863]

>>5640844
>expressing an expression linearly without the liberal use of brackets for the sake of ambiguity

[Anonymous No.5640864]

>>5640390
It looks happy and then it looks sad

[Anonymous No.5640873]

>>5640501
>>5640511
... equals ((ab)/(a+b)), why do we need an operator for this

[Anonymous No.5640883]

>>5640863
>((((expressing (an expression)) linearly) (without ((the (liberal use)) (of brackets)))) (for ((the sake) (of ambiguity))))

[Anonymous No.5640886]

>>5640390
First one equals ((ab)/((b^2)+a))
Second one equals ((ab)/((a^2)+b))

[Anonymous No.5640902]

>>5640501

this is just half of the harmonic mean of a and b

[Anonymous No.5640906]

I'm on my phone so I can't post a thread and this is the first math thread I saw, so what's the fastest way to find out 2 numbers that multiply to give you another number? For example if you had to find what #/fraction multiplied by 5 gives you 8, what's the quickest method?

[sage No.5640909]

>>5640906

5x = 8

[Anonymous No.5640916]

>>5640906

this is actually an alternative statement to one of the millennium prize problems

[Anonymous No.5640920]

>>5640916
Which

[Anonymous No.5640928]

>>5640383
B-but that's wrong...

[Anonymous No.5640934]

>>5640493
wait what? what do you mean

[Anonymous No.5640943]

>>5640909
sage goes in the email field newfag

[Anonymous No.5640948]

I propose an "imaginary" number i with the property i^2 = -1.
I know it sounds crazy because actually we cannot square root negative numbers.

[Anonymous No.5640997]

>>5640948

Yes we can, all we have to do it take the square root of a real number then add a squiggle to the indice e.g. x^~1 === sqrt(-x)

I think it's more pertinent that someone develops a way for us mathematicians to express our ideas in a more natural way, almost handwritten, on a bulletin board system much like this.

[Anonymous No.5641014]

>>5640501
I usually express this like || when I have to calculate the total resistance of a paralel circuit

R1||R2

[Anonymous No.5641016]

>>5641014
>>5640658

fist bro

[Anonymous No.5641019]

>>5640658
How much college do I have to do before I learn this stuff?

[Anonymous No.5641021]

x ◘ y

"Force base" give value of x if it were interpreted as base y.

12 ◘ 4 = 6
820 ◘ 16 = 2080
etc..

[Anonymous No.5641027]

>>5641019
like a week? there really arent any complicated concepts in there.

[Anonymous No.5641028]

if x!=x*(x-1)*(x-2)*...*1
then I'll create
x?=x+(x-1)+(x-2)+...+1+0

[Anonymous No.5641030]

>>5641027
you're right that it's not that complicated, but they pussyfoot around the issue for a couple of years before giving it to you.

You can learn it on your own with little difficulty if you have the right books, but they won't teach it in the first week of college unless you skip out of the first few years of math.

[Anonymous No.5641033]

A ⁂ B = A + 1/A + 1/B + 1

[Anonymous No.5641038]

>>5641030
what? the only thing that is not completely obvious to pretty much everyone with highschool education is maybe the definition of complex numbers and the real part and the notation at the start, which you learn in the first week of any math class ever. The whole post is readable like a regular english sentence.
bro.

[Anonymous No.5641041]

>>5641038
You give the US high school education system too much credit.

[Anonymous No.5641044]

>>5641038
>math class
university math class

[Anonymous No.5641048]

>>5640658

Maximum power transfer?

[Anonymous No.5641049]

>>5641044
>implying babby's first sets should be learned in university

[Anonymous No.5641053]

>>5641049
we're not denying that's how it should be, we're simply saying that it's not how it is.

[metric !1tensorX8Q No.5641058]

>>5641028
By your notation x?=x(x+1)/2. So that's not a very useful operation, i.e. there's no need for a shorthand notation. Sigma summation works just fine.

[Anonymous No.5641061]

>>5641049
nope, but any uni math class will cover the absolute basics in the first week to make sure everybody is on the same base level
Thats why you need at most a week of college to know what the operation is about, dude

[Anonymous No.5641369]

>>5641019

The concepts are really simple, it's just the notation which is used for convenience and clarity.


Basically you're defining a function, saying the domain and codomain (range in highschool) giving some definitions and then saying how the function works.

[Anonymous No.5641387]

>>5640305
derivatives are ghey f'''''(x) now equals f(5')x

[Anonymous No.5641394]

I thought about something along the lines of the limit of a Riemann product. We know (using sloppy notation and provided the function we're talking about behaves properly, choosing the correct range for <span class="math">i[/spoiler] and so on and so on):
<div class="math">
\lim_{\Delta x \to 0} \sum_{i} f\left( x_i \right) \Delta x = \int_{a}^{b} \! f\left( x\right) \,\mathrm{d}x
</div>
We might as well have:
<div class="math">
\lim_{\Delta x \to 0} \prod_{i} f\left( x_i \right) \Delta x = \left[ \text{herp} \right]_{a}^{b} \! f\left( x\right) \,\mathrm{d}x
</div>

[Anonymous No.5641395]

>>5641394
Ugh, stupid LATEX plugin...
<div class="math">
\lim_{\Delta x \to 0} \prod_{i} f\left( x_i \right) \Delta x = \left[ \mathrm{herp} \right]_{a}^{b} \! f\left( x\right) \,\mathrm{d}x
</div>

[Anonymous No.5641398]

>>5641387
Most people just put 5 in brackets as an exponent for the fifth derivative.

[Anonymous No.5641399]

>>5641395
That would just approach zero as delta x does.

[Anonymous No.5641404]

>>5640501
In electric engineering typical notation for impedance in parallel is what you have except with || (ie double bars)

A in parallel with B can be written shorthand with A||B

[metric !1tensorX8Q No.5641405]

>>5641394
http://en.wikipedia.org/wiki/Product_integral

Also somewhat related is a path integral:
http://en.wikipedia.org/wiki/Functional_integration

[Anonymous No.5641407]

>>5641399
>derp

[Anonymous No.5641409]

>>5641405
Ghe, cool. Thanks!

[Anonymous No.5641415]

>>5640357
1.0000000000000.. Is simply written as 1.1

[Anonymous No.5641420]

>>5640427
also
a != -b^2 for the first one
b != -a^2 for the second one

[Anonymous No.5641421]

>>5640658
>>5641019
Paralel resitors

[Anonymous No.5641434]

>>5640460
I think this is impossible, unless you specify the context more

[Anonymous No.5641436]

>>5641021
nice

[Anonymous No.5641448]

>>5641019
None. It's middle school math.

[Anonymous No.5641468]

>>5641021
nice

[Anonymous No.5641555]

>>5640459
look beneath the post form

[Anonymous No.5641556]

Recursive fractional division:
classically:
100/10=10; 10/10=1; 1/10=.1; etc
but what if:
100#&={10,1,0.1...}

[Anonymous No.5641558]

>>5641556
100#&10={10,1,0.1...}
If I could not forget parts of it...

[Anonymous No.5641601]

>>5640863
>order of operations
>ambiguous
Sure is middle school in here

[Anonymous No.5641605]

>>5641601

<div class="math">\frac{A}{7+2} = Sexytime</div>

[Anonymous No.5641650]

>>5641605
Then he would have used brackets, dumbass.

What is order of operations?

[Nona No.5641660]

ff(x)=f(f(x))

f(x)=x^2
ff(x)=(x^2)^2

[Anonymous No.5641662]

>>5641660
Pretty sure this already exists

[Nona No.5641667]

>>5641662

really? example?

[Anonymous No.5641674]

>>5641667
http://en.wikipedia.org/wiki/Function_composition

(f o f) (x) = f(f(x) = f^2(x)

[Anonymous No.5641675]

>>5641674
f(f(x)) *

[Nona No.5641687]

>>5641674
okay, thanks.

damn.

[Anonymous No.5641690]

~

x~ means x*x

example
2~ = 2*2 = 4

[Anonymous No.5641718]

>>5640430
complete fail. your operator isn't even well-defined.

7=07
7º3=0
07º3=1

>hurr

[Anonymous No.5641739]

x ° y

y{int}
x{int}

1234567897 ° 3 = 4567897123
1234567897 ° -2 = 9712345678

[Anonymous No.5641743]

>>5641739
I would say this is pretty useless, unless you came up with a way of actually applying it to something worthwhile.

[Anonymous No.5641751]

>>5641690
Congratulations on making a new symbol for taking something to the power of 2.

[Anonymous No.5641752]

>>5641751
>x^x = x^2
/sci/ i am dissapoint

[Anonymous No.5641754]

>>5641752

[Anonymous No.5641756]

>>5641743

combine it with conditional batman arithmetic:
((((((x-0.5)/sqrt((x-0.5)*(x-0.5)))/2)+0.5)*A)+(((((x-0.5)/sqrt((x-0.5)*(x-0.5)))/-2)+0.5)*B))

selector arithmetic
((((((x-(offset))/sqrt((x-(offset))*(x-(offset))))/2)+0.5)*f(x))+(((((x-(offset))/sqrt((x-(offset))*(x-(offset))))/-2)+0.5)*g(x)))

and regular logic operators (+, * [NOT would be If 1 then 0 if 0 then 1])

and you can have an algebraic touring machine.

obviously it'd be massively inefficient for doing actual computer arithmetic, but you can use the principles of algebra to reduce and optimize the function.

if you model something such as the SHA-256/512/1024 algorithm you could relatively easily swap the input and the output, concentrate on an arbitrary zero and generate fake hashes.

in the scope of galactic evolution it's pretty irrelevant.

[Anonymous No.5641757]

>>5641752
He said x*x, not x^x.
Multiplying a number by itself is squaring.

[Anonymous No.5641759]

>>5641756
>batman arithmetic

[Anonymous No.5641760]

>>5641757
Ah. My bad. Point taken.

[Anonymous No.5641763]

>>5641760
You had me doubting my knowledge of squares for a second, there.

[Anonymous No.5641764]

>>5641757

>implying x^x =/= x^2 if x=2

sort your ruselage, my friend.

[Anonymous No.5641765]

For CS
inb4 not science
x* = <span class="math">2^{x}[/spoiler]
16* = 65536

also
<span class="math">\hat{x}[/spoiler]* = <span class="math">2^{x}-1[/spoiler]

[Anonymous No.5641767]

>>5641764
If x=2, then x^x = x*x, obviously.
But the rule he gave was x~ = x*x. Where is everyone getting this x^x nonsense from?

[Anonymous No.5641771]

>>5641767
We're all autistic.

[Anonymous No.5641773]

>>5641765

that operator already exists, it's three letters long.

2bit
8bit
16bit

[Anonymous No.5641777]

>>5641765
Are bit shifts too complicated for you?
1 << x

[Anonymous No.5641802]

I came up with this lately (it's not really an operator but I'll post anyway). When you have a (bijective) relation, for example:
<span class="math">y=x^2[/spoiler]
for positive x, y, and you want the function that gives y from a given x, you can denote it with:
<span class="math">y_x[/spoiler]
(read: the function that gives y from x)
So that:
<span class="math">y_x(x) = x^2 = y[/spoiler]
<span class="math">x_y(y) = sqrt(y) = sqrt(x^2) = x[/spoiler]
and of course
<span class="math">x_x(z)/math] and <span class="math">y_y[/spoiler] are the identity functions: f(x) = x for every x.
(it seemed to be less vonfusing to me, and useful to explain the concept of taking the derivative of another variable.
for example, when you have
<span class="math">y = x^2[/spoiler]
<span class="math">f(x) = x^4[/spoiler]
then you can define the derivative with respect to a variable y (which is related to some variable x, so you can see it as a function from x to y as well, denoted with) and a function f (which has some relation to y, so also with x) in the point z as
the limit of <span class="math">\frac{f_x(z+h)-f_x(z)}{y_x(z+h)-y_x(z)}[/spoiler] as h goes to 0
with seemed more intuitive, and seemed more appropriate because you can read it really as the limit of a ratio between two functions. It makes the relation between variables and functions clearer, IMO.

This is a shitty explanation. I hope you get the picture.[/spoiler]

[Anonymous No.5641811]

>>5641802
Fixed version

I came up with this lately (it's not really an operator but I'll post anyway). When you have a (bijective) relation, for example:
<span class="math">y=x^2[/spoiler]
for positive x, y, and you want the function that gives y from a given x, you can denote it with:
<span class="math">y_x[/spoiler]
(read: the function that gives y from x)
So that:
<span class="math">y_x(x) = x^2 = y[/spoiler]
<span class="math">x_y(y) = sqrt(y) = sqrt(x^2) = x[/spoiler]
and of course
<span class="math">x_x(z)[/spoiler] and <span class="math">y_y[/spoiler] are the identity functions: f(x) = x for every x.
(it seemed to be less vonfusing to me, and useful to explain the concept of taking the derivative of another variable.
for example, when you have
<span class="math">y = x^2[/spoiler]
<span class="math">f(x) = x^4[/spoiler]
then you can define the derivative with respect to a variable y (which is related to some variable x, so you can see it as a function from x to y as well, denoted with) and a function f (which has some relation to y, so also with x) in the point z as
the limit of <span class="math">\frac{f_x(z+h)-f_x(z)}{y_x(z+h)-y_x(z)}[/spoiler] as h goes to 0
with seemed more intuitive, and seemed more appropriate because you can read it really as the limit of a ratio between two functions. It makes the relation between variables and functions clearer, IMO.

This is a shitty explanation. I hope you get the picture.

[Anonymous No.5641818]

I also had something to work out the distributions of stochasts when you add them, but this is probably something that already exists (it is similiar to convolution, but then with + instead of *).
Something similiar could be done for division and subtraction.

[Anonymous No.5641828]

If you have a length of wood, or metal or something that makes a tone when struck. Cutting it to a / b its length with result in a tone change of b / a higher.

For example, cutting it in half will double its frequency (1 /2 to 2 / 1)

[Anonymous No.5641851]

Fourier transform.
Pic related

[Anonymous No.5641856]

xy = |(x-y)/x * (y-x)/y|
reads "x attracts y". the result is the attraction between x and y. whatever that means.

38 = |(3-8)/3 * (8-3)/8|
= 1.0416

64 = |(6-4)/6 * (4-6)/4|
= 0.16666...

1514 = |(15-14)/15 * (14-15)/14|
= 0.00476

(32.7)2 = |(3-2.7)/3 * (2.7-3)/2.7|
= 0.012
= 198.005

[Anonymous No.5641874]

>>5640745
While I don't think that there can be one number with that property, that reminds me of the roots of unity; which when taken to the nth power, become one. Just multiply them by x^(1/n) and they should act similarly to your number.

[Anonymous No.5641928]

>>5640745
Yeah that kinda gives you an inconsistent system.
tau ^ 2 = 2
So tau = sqrt(2)
but tau ^ 3 = 3
So tau is also 3^(1/3)

Kinda like tau + x = 2x
tau + 10 = 20 so tau = 10
tau + 1 = 2 so tau = 1

[Anonymous No.5641932]

x >=> y = "I'm not exactly sure how but I'm pretty that x equals y."

[Anonymous No.5641935]

>>5641932
pretty sure*

[Anonymous No.5641988]

>>5641874
>While I don't think that there can be one number with that property
Yeah it's basically always 1 since <span class="math">\tau = \tau^1 = 1[/spoiler], so it wouldn't work. It's actually just a silly operator or something to that effect.

>>5641928
>tau + x = 2x
I realize that the whole tau thing is bullshit but why would tau + x become 2x?

[Anonymous No.5641989]

>>5640305
The Busch transform

1) take any natural number
2) create a 1*n array or 1s
3) rearrange the 1s into a bush

the advantages of this transform is that you get a picture of a bush at the end

[Anonymous No.5641990]

In <div class="math">R[x]/(x^2),</div>

x^2 = 0, but x =/= 0. Therefore we can interpret it as 0.00...1, so 1-x = 0.99999...

[Anonymous No.5642004]

does this count?

[Anonymous No.5642039]

>>5641989
I lol'd.

>>5642004
What is the use of this?

[Anonymous No.5642043]

>>5642039
it kept me busy when i was at work. with these rules you can take any function of sigmas and turn it into a linear combination of sigmas.

[Anonymous No.5642046]

>>5642043
Interedasting. I still don't get it though.

[Anonymous No.5642051]

>>5640328
Motherfucker, every CS Student is more scientist than you!

[Anonymous No.5642057]

>>5642043
Also, does this operation have a name? I mean did you name it?

[Anonymous No.5642070]

>>5641990
I'd rather retain the fact that R is a field than be able to calculate with infinitesimals. 5/18

[Anonymous No.5642073]

>>5642043
also, i played around with it , putting it into that conjecture that if you devide by 2 and multiply by 3 then +1 you eventually get 1. so that a use i guess.

>>5642057
i called it sigma p of x.

[Anonymous No.5642079]

>>5640425
If you have + and *, and work in base 2, then you can do any computable operation since you can make a nand gate from 1+b*a, nand gates are universal

[Anonymous No.5642092]

>>5641028
triangular numbers. already been done.

[Anonymous No.5642094]

>>5640370

simple and appliable

9/10

[Anonymous No.5642104]

O + P Massive douche = OP IS a MASSIVE douche solve for OP

[Anonymous No.5642127]

>>5642094
Not as good as f(x)+C.

Remember that you're only interested in f(x)+C because you care about things like (f(x2)+C)-(f(x1)+C)=f(x2)-f(x1).

It's important to label your arbitrary constant because in some operations the same arbitrary constant cancels itself out. It's not generally true that any two arbitrary constants are equal.

[Anonymous No.5642163]

>>5642127
>It's not generally true that any two arbitrary constants are equal
Explain

[Anonymous No.5642166]

>>5641988
No, it wouldn't: I just give a similiar example that doesn't work, though in the example I give it's more intuitive to see why it doesn't work (that is, it gives you an inconsistent system).

[Anonymous No.5642174]

>>5642163

Arbitrary means "could be anything." If two numbers each could be anything, then they're probably not the same.

[Anonymous No.5642203]

>>5642174
But..but... basic group theory? Every element maps to some other element, there is no single element that maps to two elements.
>mfw I realize you're probably trolling

[Anonymous No.5642210]

>>5642073
>i called it sigma p of x.

Hehe.

[Anonymous No.5642214]

1&1= 11
1&2 = 12
3&4 = 43
8&3&9 = 839

etc...

[Anonymous No.5642213]

>>5641420
Hmm ur smart

[Anonymous No.5642217]

>>5641033
=1+ab

[Anonymous No.5642222]

d(a, b) = sqrt(a^2 + b^2)
a||b = ab/(a+b)

are ones i use all the time

[Anonymous No.5642225]

>>5642214

[Anonymous No.5642237]

people, can you actually come up with something original, and not just shit that's a combination of classical operators?

the tau^x=x was pretty cool, the rest was meh-to-lame.

[Anonymous No.5642242]

xay= |x-y|*|x+y|

4a7=33
2a5=21
11a10=21
254a254=0

meh.

[Anonymous No.5642510]

>>5642237
⊔(f(x)) ~ infinitely small values of f(x) on which you can do operations

⊔(f(x)) - 1 = ⊔(f(x))
⊔(f(x) - 1) = indeterminate

⊔(f(x)) + ⊔(g(x)) = lim (x->0) (f(x) * g(x))
⊔(f(x)) * ⊔(g(x)) = lim (x->0) (f(x) / g(x))

[Anonymous No.5642526]

(->

When applied to any problem it always outputs a solution or set of solutions.

Noble Prize please.

[Anonymous No.5642556]

x|a <-- x to the power of itself a times.

[Anonymous No.5642562]

>>5642556
not a new function

the up arrow symbol defines that, it's called tetration

[Anonymous No.5642570]

>>5642556
Already been invented

http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

[Anonymous No.5642648]

>>5640343
>tell that to my computer that is executing sciency code
Which you have EEs and CompEs to thank for.

[Anonymous No.5642665]

>>5641021
hey guys, we like this right? This makes sense?

[Anonymous No.5642670]

>>5642214
>3&4 = 43
>>5642225
>Brilliant

[Anonymous No.5642696]

>>5641061
Heh, my college classes have barely even mentioned the basics of set theory, while still using it. I'm usually the only one who understands what's going on.

[Anonymous No.5642730]

>>5642051
0/10.

Go kill yourself, pleb.

[Anonymous No.5642777]

x // y takes away all the occurrences of the number y in x.

Example :
1234 // 4 = 123
524136426 // 413 = 526426
34 // 235436 = 34

[Anonymous No.5642789]

>>5642777
34343 // 343 = ?

[Anonymous No.5642793]

>>5642665
it is not always defined

612 ◘ 4 = ?

[Anonymous No.5642804]

a%b for a and b integers, divides a by the highest posible power of b while still giving an integer result

12%2 = 3
252%6 = 7

[Anonymous No.5642812]

>>5642648

implying computational engineering is not a submodule of computer science.

[Anonymous No.5642814]

>>5642793
612 written in base 4 is 21210, I don't see the problem

[Anonymous No.5642821]

>>5642804

that's interesting. could you come up with an algebraic formula that does that?

[Anonymous No.5642824]

>>5642821

[Anonymous No.5642828]

>>5642789
Nice one. Well I'd say 34.
If we define the operator recursively starting from the right, we get 34.

>>5642814
That's not what the operator does, though.
x ◘ y converts x in base 10 as if it was in base y.
12 ◘ 4 = 1*4 + 2*1 = 6

[Anonymous No.5642844]

>>5642821
>>5642804
it makes some interesting plots

y = x%2

[Anonymous No.5642855]

>>5642844
that graph is wrong

the first few values should be 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15

[Anonymous No.5642853]

>>5642844

huh reminded me of one of my plots, but it doesn't look so similar now.

actually, your shit is pretty epic man. if you.. ..hm. can I write a paper about this? lol.

[Anonymous No.5642856]

>>5640404
Soopersupedruper exponentes:
x?z=((x~y)~y)~y....+i
>z times + i

[Anonymous No.5642858]

>>5642855
oh fuck you're right, I just plotted the largest power

here you go

[Anonymous No.5642870]

>>5642858
What's this software?

[Anonymous No.5642876]

let f(x1,x2,x3...xn) and g(x1,x2,x3....xn) be two functions over a set of variables. Then
f✈g defines f(g(x1,0,0,0,0....), g(0,x2,0,0,0,...)....)

Ta-dah

[Anonymous No.5642878]

>>5642858
this kind of function, and more complex ones but similar, is all over number theory

it's why i love number theory

[Anonymous No.5642881]

>>5642858

OOOOOooooooooooooooh

That looks cool.

[Anonymous No.5642890]

4-i^2=5

[Anonymous No.5642892]

>>5642870
Mathematica

looks like <span class="math">y = x\%n[/spoiler] will have "lines" of the form <span class="math">y = \frac{1}{n^k}x[/spoiler] where k is a positive integer

[Anonymous No.5642894]

lines of the form <div class="math">y = \frac{1}{n^k}x</div>

[Anonymous No.5642898]

>>5642004
That's pretty slick man.

[Anonymous No.5642920]

+^
its like addition except for negative numbers
2 +^ -2 = 4

A +^ B = abs(A) + abs(B)

[Anonymous No.5642931]

>>5642920
So instead of writing |A+B|, you write A+^B, saving a total of zero lines.

[Anonymous No.5642932]

>>5642821
It looks like some kind of integer logarithm.

<span class="math"> 12 = 3 \times 2^2 [/spoiler]
<span class="math"> \log_{2}(12) = \log_{2}(3) + 2 \log_{2}(2) [/spoiler]
And we're trying to extract the 3
If we divide by <span class="math"> \log_2 (2) [/spoiler] take the fractional part and put everything back to the power of 2, it works, but not in general.

So it would be,
<div class="math"> k\%n = n^{\mathrm{frac} \left(\frac{\log_n (k)}{\log_n (n)} \right) } </div> but doesn't work

[Anonymous No.5642941]

stern-brocot(d,p)

a function which creates an image of the stern brocot tree starting from a path p to depth d

[Anonymous No.5642942]

>>5642931
No, instead of writing |A| + |B| you write A+^B.

[Anonymous No.5642947]

goddamnit, this is not about programmed functions. I'm a computer scientist myself but CS =/= math (and in my opinion, programming itself isn't even science), some computer scientists will never understand this

[Anonymous No.5642956]

>>5642932
the problem is you're also taking the fractional part of <span class="math">\log_{2}(3)[/spoiler] when you actually just want to leave it alone

[Anonymous No.5642959]

>>5642947
The theory of computer science is basically applied pure math. If the 'phrase applied pure math' makes any sense.

[Anonymous No.5642961]

>>5642947
>what is lambda calculus

[Anonymous No.5642962]

>>5642947

why tho

pointer arithmetic is math, boolean algebra is math, all other programming is just an abstraction thereof. not math? basically saying like algebra or calculus isn't math.

[Anonymous No.5642963]

>>5642961

lambda calculus is to computer science as string theory is to physics.

[Anonymous No.5642979]

>>5642947
> implying

[Anonymous No.5642988]

>>5642959
Haha, that doesn't really make sense, but I know what you mean.
Anyway, I agree with what you say, it just doesn't affect my point. Mathematical functions are just a whole other thing then programmed functions.

[Anonymous No.5642990]

>>5642947
please give an example of a programming language function which is not a math function

[Anonymous No.5642998]

>>5642990
im on your side
but how about: sleep(1000)

[/r/SRS No.5643009]

>>5642998
<div class="math">\sum_{n=0}^{C*1000}0</div>
c= clock cycles per second

[Anonymous No.5643020]

>>5643009
fair nuf

[Anonymous No.5643041]

>>5643009

...isn't that always =0?

[Anonymous No.5643045]

x¡
Inverse of x!.
(x¡ * x!) / x = 1

[Anonymous No.5643048]

>>5643047
meant for
>>5643045

[Anonymous No.5643047]

if (x¡ * x!) / x = 1 doesn't that mean x¡ = 1/(x-1)!

[Anonymous No.5643050]

>>5643047
No clue. I just made something up. As in:
5¡ = 5/(5!)

[Anonymous No.5643051]

>>5643047
well spotted

[Anonymous No.5643053]

>>5640450
Not 100% sure myself, will do a bit of research.

[Anonymous No.5643055]

>>5640805
thank you buddy

[Anonymous No.5643056]

>>5643045
that's not an inverse

you mean x!¡ = x

[Anonymous No.5643062]

>>5643056
Yea, I meant:
x¡ = x/(x!)

[Anonymous No.5643066]

>>5643062
Its derivative:
<span class="math">
d/dx(1/(x!)) = -(Gamma(x+1) polygamma(0, x+1))/(x!)^2
[\math][/spoiler]

[Anonymous No.5643070]

>>5643062
so x¡ = 1/(x-1)!

[Anonymous No.5643071]

>>5643062
Its derivative:

<span class="math">
d/dx(1/(x!)) = -(Gamma(x+1) polygamma(0, x+1))/(x!)^2
\<span class="math">[/spoiler][/spoiler]

[Anonymous No.5643078]

>>5643062
that's still wrong

[Anonymous No.5643079]

>>5643078
No, because that's my definition. Now go away.

[Anonymous No.5643081]

To the hyperpower of:

10>^4 = 100000000 (10*10 -> 100*100 -> 10000*100000 = 100000000)

[Anonymous No.5643083]

>>5642941
what fucks with me is that since an image is just a binary number this is an actual function

[Anonymous No.5643085]

>>5643079
sure, but it isn't an inverse function, as you claimed

[Anonymous No.5643087]

>>5643085
Nah mang, it's an identity function now.

[Anonymous No.5643092]

>>5643070
Well no. Because I can do 1¡, equaling 1/1! = 1
You'll get something undefined. If we limit x > 1 then it works.

[Anonymous No.5643091]

>>5643087
there is only one identity function f(x) = x

[Anonymous No.5643094]

>>5643092
no, i get 1/0! = 1

they are the same

[Anonymous No.5643098]

>>5643094
Yea, sorry. Here, have the sum:
http://www.wolframalpha.com/input/?i=sum+of+x%2Fx%21

[Anonymous No.5643119]

>>5642844
pascal triangle?

[Anonymous No.5643162]

>>5640305
<span class="math"> a xD b = a e^{-\omega b \tau}
So...?[/spoiler]

[Anonymous No.5643208]

>>5640425
Multiplication isn't just repeated addition+counting (easy example, 1.1 * 1.1). Addition is reducible to a successor operation, I'm not sure if you'd consider them different ops though.

[Anonymous No.5643211]

>>5640540
I usually do a line like "For convenience, let t = x (fuck t)." Then at the end if there's still xs in the answer I say "returning to our original notation..." and express my answer in t.

Other things that can fuck off:
x if cross products are involved
P or rho if both appear
Capital and lower case phi
s

[Anonymous No.5643212]

>>5643208
explain how you define rational numbers, including multiplication. Then I'll believe your example.

[Anonymous No.5643230]

>>5643208
1.1+1.1[sup]-1

[Anonymous No.5643309]

>>5641928
f(x) g(x) = 2h(x)
Averages two functions.

✿F✿ = G where F X G resembles a flower.

=î0_~ĵ=
It's the cat curve on the xy-plane. It isn't so much an operator as it is a special parameterization of a curve that looks like a cat.

[Anonymous No.5643315]

>>5643309
first one should be f(x) (libra sign) g(x) = 2h(x)

for some reason that symbol isn't showing up.

[Anonymous No.5643325]

a 8==> b

it's a boolean operator that returns true if 8a >= b and false if 8a < b

[Anonymous No.5643360]

>>5642203
Maybe it's clearer if I say it this way: two different constants should not be mistaken for two references to the same constant.

If you have one expression f(x)+C where C is an arbitary constant, and another expression g(x)+C where C is an arbitrary constant, you should assume that these are not the same C and may be unequal unless you have a definite reason for them to be equal.

So (f(x2)+C) - (f(x1)+C) = f(x2)-f(x1), but (f(x)+C) - (g(x)+C) = f(x) - g(x) + C, if the f(x)+C and g(x)+C expressions are unrelated and each is supposed to contain an "arbitrary constant". When you perform an operation which adds an arbitrary constant, it's a good idea to give it a unique label in your work. It's a lot clearer if you start with f(x)+C1 and g(x)+C2.

[Anonymous No.5643377]

>>5640348
>the joke
>your head
One of these has at a passed at a higher altitude than the other

[Anonymous No.5643383]

>>5640404
Tetration. We already have notation for that.

[Anonymous No.5643380]

>>5640370
Actual good one

[Anonymous No.5643388]

>>5640501
I like this one.

[Anonymous No.5643391]

>>5640549
Fuck you.

[Anonymous No.5643398]

>>5641021
Cool.

[Anonymous No.5643413]

>>5640493

Is this new physics notation?

[Anonymous No.5643429]

The prime factorial function, <span class="math">x!_p[/spoiler]
Like the regular function, except only multiply by prime numbers below x. For example, <span class="math">20!_p = 20 * 19 * 17 * 13 * 11 * 7 * 5 * 3 * 2[/spoiler]

[Anonymous No.5643432]

>>5641856
I can't imagine more confusing notation.

[Anonymous No.5643438]

>>5641932
>>5641935

Might be the most useful thing in here.

Not sure if it's an operation though.

[Anonymous No.5643440]

>>5643429
Actually useful.
>FACT

[Anonymous No.5643442]

>>5640370
Hate the notation, but love the idea.

[Anonymous No.5643444]

>>5643429
Yet you still multiplied by 20 anyway

[Anonymous No.5643448]

>My favourites so far:
>>5641021
This is cool, and probably useful.

>>5643429
That's cool I like that, anything involving prime relations is interesting because it's not something you can formalise in any equation.

[Anonymous No.5643446]

x =~= y

It means x = y because mathematica said so.

[Anonymous No.5643452]

>>5642793
If you assume the number is just a polynomial then:

>6*4^2 + 1*4^1 + 2*4^0 = 102

[Anonymous No.5643453]

>>5643444
Yes, that's what a factorial function does. I guess if you wanted to you could define a prime gamma function <span class="math">\Gamma_p(x)[/spoiler] for the positive integers where <span class="math">\Gamma_p(20) = 19 * 17 * 13 * 11 * 7 * 5 * 3 * 2[/spoiler]

[Anonymous No.5643460]

>>5641021

would people adopt this if it was written in a paper (and obviously explained)? isn't that just what every notation was spurred by?

[Anonymous No.5643469]

>>5642237
I trolled through my AA final by using a V4 ring that had a second operation which mapped any element to the 0 element. Gets confusing when you try to talk about 1s and 0s.

[Anonymous No.5643473]

>>5642665
seems an awful like like a fourier transform

[Anonymous No.5643475]

>>5640305

x°y = ((x-y)*(xy))^(x+y)

[Anonymous No.5643482]

>>5643453
That's not what you said though:
>except only multiply by prime numbers below x
You multiplied by a composite number initially, you actually meant the factorial that you just wrote.

[Anonymous No.5643491]

>>5643050
Why not just make it a straight up inverse?

Right now you have 5¡ = 1/4!, it seems simpler to just make it so n¡=(n!)^-1

[Anonymous No.5643504]

>>5643460
If it was being used in analysis in a helpful way I assume people would adopt it, that's the criteria for any annotation.

[Anonymous No.5643538]

>>5641021

I don't get it.

[Anonymous No.5643624]

>>5640412
I think you meant to say (A/2)+7

[Anonymous No.5643639]

>>5640472
wouldn't that just be (a+x)*b where x > a?

[Anonymous No.5643643]

>>5640474
e to the integral of the probability of x with respect to x?

what's it supposed to do, turn a statistical distribution curve into a log chart?

[Anonymous No.5643685]

>>5642990
Functions like date() and any kind random(). Given the same arguments a mathematical function must always return the same result.

[Anonymous No.5643712]

>>5640745
define a sequence
tau = { n^(1/n) ; n in reals }

[Anonymous No.5643922]

>>5643429
http://en.wikipedia.org/wiki/Primorial

[Anonymous No.5643968]

>>5643685
You're a computer scientist, you should know about monads.
date : State Date ⊤ -> Date = Date -> Date
random : State Seed ⊤ -> Nat = Seed -> Nat

[Anonymous No.5643989]

>>5642990
void swap(int *a, int *b)
{ int *temp = a;
a = b;
b = temp;
}

Also, all functions that change state and/or print anything (you consider this as a change of state, though).

[Anonymous No.5643999]

>>5643989
You consider it as a change of state. I consider it a function that accepts a world as an argument and returns a new world with a change.

[Anonymous No.5644030]

>>5643989
Function accepting a list of integers representing the memory, the pointers are indices of that list, dereference is indexing the list, assignment replaces the old value with the new value and returns the new list.
Yep, still a State.

[Anonymous No.5644057]

>>5640370

Hmm not bad

[Anonymous No.5644065]

>>5640404

Nice try, but it already exists as: x↑y

A useful concept which can be expanded to x↑↑y, same with x↑...↑y and this is how Graham's number is defined (largest useful number)

[Anonymous No.5644079]

>>5641021

Fuuuuark, strong operation. I can see how this could be useful.

So 31◘8=25◘12

(taken from popular maths joke)

[Anonymous No.5644101]

>>5641021
what is 8 ◘ 2?

[GOMADWarrior No.5644115]

exp(x)
gives the result of the infinite sum x^k/k!
ln(x)
gives the inverse of exp(x), i.e. ln is such that ln(exp(x)) = x

[Anonymous No.5645109]

>>5640530
yeah but the way the \f comprises itself means that it's not actually hindered by the introduction of the operation, and at, which it is right off the bat, I mean, x cant equal 1 here otherwise it's 1.0.1.0.1... everytime an execution of addition of the equation, which is indefinitely undefined, which goes as far to say that r is r in every equation, which its not

tl;dr ( ) \f (ex)+(x)

>>5640883
fckn lold
Let linear be y
Express be r
Use be x
Reason be x^2
And ambiguity be \f ((x)z)

Care to simplify anon?

------

*(*)*
Or * \f *

i call it "Dividing"

Alternatively /(/)/ is something I like to call "Multiplying"

shd I lev PO Box 4 Nobel in email?

[Anonymous No.5645126]

>>5645109
fyi everybody
Sage, even its part of a word, if in the email field, does not bump a thread

there was a point in time where /sci/ coders had way to much time

[Anonymous No.5645174]

In group theory, given A and B are groups, instead of having AxB=(a,b) where a is any element of A and b is any element of B be the way you do direct product, I define a different function so that A#B=ab where a is any element of A and b is any element of B.

It works out the same, but looks cleaner. For example, if you have two C2 groups (one generated with a, and one with b) and a C4 group generated with c, then
C2#C2#C4 = {e,a,b,ab,c,ca,cb,cab,c^2,c^2a,c^2b,c^2ab,c^3,c^3a,c^3b,c^3ab,}