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I think infinitesimal can be expressed in some of this way:
"(0"
(which means close to null).
Here's some definition I figured out:
Let have a sequence
I forgot... Hi.
Some people think it is a waste of time because they are so smart in other areas.
Or not...
HA-HA-HA!!!
x ≈ 2.5070186440929762986607999237156780290259
764201303696751265821783529769648210199715760
034086194090715665720271011885426265690772588
1312692288635583724
(150d)
And now, the exact result:
Another related question:
Let
100!=93326215443944152681699
2388562667004907159682643816
2146859296389521759999322991
5608941463976156518286253697
920827223758251185210916864 0000000000 0000000000 0000
Maths rocks, isn't it?
Here's a more general fomula:
If p is prime the the gratest power of p which divides n! is given by the formula:
For the limes thingy:
If
and...again...NO SOLUTION!!!
(If there was some limmy solution, then all math theory would have been contradictory-and this would have been TERRIBLE!!)
By what do you mean it solved 87^87? Surely you aren't doing it in C++. Are you using mathimatica?
No,no- with the c-program:
87^87=5472364007515806092890840962213361933646
5578673599554575543693463433762205742631692905
6636192499927745119880215695036404581245556681
7070274944448633167362192918054601383.
Entering this in my program gives 87.
Try it out!
Another copy of the command prompt:
Input y (y>1)(0 for quit):112349123746128387461279838741628934876129341237463924
The solution of x^x=1.12349e+053 is:34.4979.
Input y (y>1)(0 for quit):23452432324957346598724365234987623945623475238476239847623984762394862
348962349
The solution of x^x=2.34524e+079 is:47.3708.
Input y (y>1)(0 for quit):1102374123084712304821930487213498071234098732094712340123478901732081
The solution of x^x=1.10237e+069 is:42.4204.
Input y (y>1)(0 for quit):12039487123804981392847382905651208936410293874123098471238490382714809
23874019238741809238472189568293047213094
The solution of x^x=1.20395e+111 is:61.9786.
Input y (y>1)(0 for quit):12349812634906785610238470238109471324098465182394712380947238409734863
2840923174819028347120398416581372410938471234896358903471823409183409178236581293047128390472138
94
The solution of x^x=1.23498e+169 is:87.1489.
Input y (y>1)(0 for quit):19837469812374598123712894716328941723649127834537891263419283746192378
4126937845128374619234691287345617858912364198287346981273649182374612389476374573846132947123946
1273461923784519326493812874162398476397851374612384912347612398745713984612938764238914761937846
193246918
The solution of x^x=1.98375e+273 is:129.405.
Input y (y>1)(0 for quit):91236549123461289374612398451723469123874612987346123984762357894321034
7507236401823748941287346172389475197813264123784587346871346129847364982734512379847612348912734
6178235412783461298351364781923746129837643294872367461392841763489726349183247612839746328941723
64918237461298374612938476123795793649182387416923746127893476127983461932
The solution of x^x=1.#INF is:-1.#IND.
Input y (y>1)(0 for quit):
It works for all doubles.
No, that's not the limit. It solved 87^87!!!
The limit is: 47^47=3877924263464448622666648186154330754898344901344205917642325627886496385062863
What would you say?
It crashed for y=50^50. I'll see what's the maximum.
Here's a copy of the command prompt:
Input y (y>1)(0 for quit):12089258196146291747061760000000000000000000000000000000000000000
The solution of x^x=1.20893e+064 is:40.
Input y (y>1)(0 for quit):24806364445134114549464918239541268974453058149265416432172060012817382
8125
The solution of x^x=2.48064e+074 is:45.
Input y (y>1)(0 for quit):24889437491384718234
The solution of x^x=2.48894e+019 is:16.0794.
Input y (y>1)(0 for quit):123408471204832479128412342406123046213749812462891384761234
The solution of x^x=1.23408e+059 is:37.5324.
Input y (y>1)(0 for quit):13240127341289348712634981271478293647123058734122348329472198034861239
4712349
The solution of x^x=1.32401e+077 is:46.3026.
Input y (y>1)(0 for quit):
Here's my program-based on requrent computation of the power log omega.
#include<iostream>
#include<math.h>
using namespace std;
int const pr=10;
double const e=2.7182818284590452354;
double const l2=0.69314718055994530942;
double omega(int n,double x);
int main(){
st: cout<<"Input y (y>1)(0 for quit):";
double y;
cin>>y;
if(y==0.) goto en;
cout<<"The solution of x^x="<<y<<" is:"<<pow(e,omega(pr,log(y)))<<".\n";
goto st;
en: cout<<"Press any key to continue..."<<endl;
char c;
cin>>c;
return 0;
}
double omega(int n,double x){
if(n==0){
return 0;
}
else{
double om=omega(n-1,x);
double omp=pow(e,om);
double result=om-(om*omp-x)/(omp*(om+1)-((om+2)*(om*omp-x))/(2*om+2));
return result;
}
}
I'm making tests now. The presiesility may be changed by changing pr.
What is y? And can you write out W?
Sorry. I assumed that:
The answer is the function:
edit: sorry but the first function was incorrect.
Only a note- the minimum of the function is at
Can't we got some series for f^-1(x)?
Using series:
A plot of x^x:
Interesting theme.
I'll post something after a while...
For example if you want to produce
He wants to say:
[math]