math - Is there a Solve method in Java for (eq. inside)? -

if want solve variable (p) form following equation, else known :

is there method out there in java let me this?

i can calculator, , sure there out there python - know doable. thank you.

here quick reference graph:

the x-value circle (p) satisfies both sides of equation. calculating values, checking 0 not efficient way of doing this.

you have implement function giving difference of both sides of equation , kind of univariate solver. due convexity of equation on positive axis, classical methods, esp. secant method, should work flawlessly.

class myfunc {     public myfunc(int nn, int cc, double aalpha) {...}      public double eval(double p) {         double sum = 1;         double term = 1;         for(int = 1; i<=c; i++) {              term *= (n*p)/i;              sum += term;          }          return sum*math.exp(-n*p) - alpha;      } }  ..............  public double secant(myfunc f, double a, double b) {     double fa = f.eval(a);     double fb = f.eval(b);     while(math.abs(b-a)>1e-10) {         double c = (a*fb-b*fa)/(fb-fa);         a=b; fa = fb;         b=c; fb = f.eval(b);     }     return b; } 

and call like

p = secant(new myfunc(n,c,alpha), 0, 0.1); 

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it turns out secant method unstable, use modified regula falsi

import java.lang.*;  interface realfunc {     public double eval(double x); }  class myfunc implements realfunc {    int c,n;     double alpha;    public myfunc(int nn, int cc, double aalpha) {     c=cc; n=nn; alpha = aalpha;   }    public double eval(double p) {     double sum = 1;     double term = 1;     for(int = 1; i<=c; i++) {         term *= (n*p)/i;         sum += term;      }      return sum*math.exp(-n*p) - alpha;    } }   public class secantsolverso34980366 {    static public double illinois(realfunc f, double a, double b) {         double fa = f.eval(a);         double fb = f.eval(b);         while(math.abs(b-a) > 1e-10) {         //system.out.printf("a:f(%16.12f) = %16.12f | b: f(%16.12f) = %16.12f \n ",a,fa,b,fb);           double c = (a*fb-b*fa)/(fb-fa);           double fc = f.eval(c);           if( fa*fc < 0 ) {             fa *= 0.5;           } else {             = b; fa = fb;           }           b = c; fb = fc;       }       return b;   }     public static void main(string[] args) {     int n = 1;      for(double alpha = 0.2; alpha <=1.0001; alpha += 0.1) {         system.out.printf("alpha=%4.2f: ", alpha);         for(int c = 2; c < 15; c+=2) {              double p = illinois(new myfunc(n,c,alpha), 0.,(c+13.0)/n);             system.out.printf("(c=%2d):%12.9f  ",c,p);         }         system.out.printf("\n");     }    } }