C evaluating a recursive sequence - expected expression error -

i'm trying code recursive sequence values n = 60. know x0 = 1.

xn+1 = 2^(n+1) * (sqrt( 1 + 2^-n*xn) - 1)

so far have

int main () {     // first term x0 = 1     double xn = 1;     double x;     double squareroot = 1 + (pow(2, -x) * xn);      (x = 1; x <= 60; x++)     {         double xn = pow(2, x) * double sqrt(squareroot) - 1);         printf("%f\n", xn);     }  } 

but expected expression error on line there double sqrt.

as mentioned in other answers, code had syntax errors in line:

double xn = pow(2, x) * double sqrt(squareroot) - 1); 

here parenthesis not balanced have 1 more ) (. in addition cannot place double did (it not needed sqrt returns double)

below placed code comments of how interpreted equation along notes on found ambiguous how written:

#include <math.h> #include <stdio.h>  int main() {     int n;      // mentioned x0 1.0 , variable updated initialized 1.0     double x = 1.0;      // goal: xn+1 = 2^(n+1) * (sqrt( 1 + 2^-n*xn) - 1)     // notes:     // - named xn x separate updated variable iteration variable     // - interpreted "2^-n*xn" "pow(2.0, -n) * x" (rather "pow(2.0, -n * x)")     (n = 1; n < 60; n++) {         // term a: 2^(n+1)         double terma = pow(2.0, n + 1.0);          // term b: (sqrt( 1 + 2^-n*xn) - 1)         // note involves updated variable, must in loop         double termb = sqrt(1.0 + pow(2.0, -n) * x) - 1.0;          // update x: terma * termb         x = terma * termb;         printf("%f\n", x);     }      return 0; } 

added clarity:

int main() {     int n;      double x = 1.0;     (n = 1; n < 60; n++) {         // these temporary values assigned every iteration         // note unlike code "pow(2.0, -n) * x" evaluated         // every iteration , updates reflect desired equation         // in code used "squareroot" did not update every iteration         double terma = pow(2.0, n + 1.0);         double termb = sqrt(1.0 + pow(2.0, -n) * x) - 1.0;          // updates value of "x" every iteration         // "x" corresponds called "xn" in equation         x = terma * termb;         printf("%f\n", x);     }      return 0; }