this code is for bessel function Y n (x) 2nd kind I'm in the level 4 at computer
engineering my prof. gave me that assignment , I want help to improve it
to get an accurate results thanks!
bessel function 2nd kind Yn(x)
#include<iostream>//bessel function Y n (x) 2nd kind
using namespace std;
#include<cmath>
long double Segma1 (long double,long double);
long double Segmafact (long double);
long double Segma2 (long double,long double);
long double HMMM (long double);
long double factorials (long double,long double);
long double Jn (long double,long double);
int main()
{
long double m=2,c,s,p,b,bessel1,segma1,segma2,Yn;
const long double pi=3.14159265358,EulerConstant=0.577215664901532860606512;
for(double n=0;n<=5;n++)
{
cout<<"n= "<<n<<endl<<endl;
for(double x=0;x<=20;x+=0.5)
{
bessel1= Jn(x,n);
segma1=Segma1(n,x);
segma2=Segma2(n,x);
c=log(x/2)+EulerConstant;
s=m/pi;
p=(pow(x,n))/pi;
b=(pow(x,(-n)))/pi;
Yn=(s*bessel1*c)+(p*segma1)-(b*segma2);
cout<<"Yn("<<x<<") = "<<Yn<<endl;
Yn=0;
}
}
return 0;
}
long double Jn (long double x,long double n)
{
long double k=0,t,z,b,r,m=0,E=0.000001;
for( ; ; ){
t=n+k;
r=n+(2*k);
z=factorials(k,t);
b=z*(pow(-1,k))*(pow((x/2),r));
if(fabs(b)<E)
break;
else
m+=b;
k++;
}
return m;
}
long double Segma1 (long double n,long double x)
{
long double f,g,Hm,Hh,h,Segma1A=0,o=-1,p=2,m=0;
for( ; ;)
{
h=(m+n);
f=Segmafact(h);
g=Segmafact(m);
Hm=HMMM(m);
Hh=HMMM(h);
Segma1A+=(pow(o,(m-1))*(Hm+Hh)*pow(x,(2*m)))/(pow(p,(2*m+n))*g*f);
if(m==99)
break;
m++;
}
return Segma1A;
}
long double Segma2 (long double n,long double x)
{
long double t,f,g ,Segma2B=0;
for(long double m=0;m<=(n-1);m++)
{
t=n-m-1;
f=Segmafact(t);
g=Segmafact(m);
Segma2B+=(f*pow(x,(2*m)))/(pow(2,(2*m-n))*g);
}
return Segma2B;
}
long double factorials (long double k,long double t)
{
long double fact1=1, fact2=1, last;
for(long double j=1;j<=k;++j){
if(k==0)
fact1=1;
else
{fact1*=j;}}
for (long double i=1;i<=t; ++i)
{
fact2*=i;
}
last=1/(fact1*fact2);
return last;
}
long double Segmafact (long double n){
long double fact=1;
if(n!=0)
{
if(n>0){
for(long double j=1; j<=n; j++)
{
fact*=j;
}
}
else
{
for(long double j=n; j<=(fabs(n)); j++)
{
fact*=j;
}
fact*=-1;
}
}
else
fact=1;
return fact;
}
long double HMMM (long double s)
{
long double o=1, l,h=0;
for(long double n=1;n<=s;n++){
l=(o/n);
h+=l;
}
return h;
}
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