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\(2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 \)

Please provide explanation.

 Jul 14, 2018
 #1
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We need to solve \(2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2=0\).

Let \(f(x) = 2(2^x-1)x^2+(2^{x^2}-2)x-2^{x+1}-2\).

If you graph f(x) using any method, you find that it only touches x-axis at one point.

 

Graph link: https://www.desmos.com/calculator/wxie170yvi

 

We will use Newton-Rhapson to find solutions.

\(f'(x) = 4x(2^x-1)+2x^22^x\ln2+2^{x^2}-2+2x^2(2^{x^2}\ln2)-2^{x+1}\ln2\)

As this is too complicated, I solved it with a program.

 

Here is the source code in C++:
#include
using namespace std;
double f(double x){
    return 2*(pow(2,x)-1)*x*x+(pow(2,x*x)-2)*x-pow(2,x+1)-2;
}
double f_prime(double x){
    return 4*x*(pow(2,x)-1)+2*x*x*pow(2,x)*log(2)+pow(2,x*x)-2+2*x*x*(pow(2,x*x))*log(2)-pow(2,x+1)*log(2);
}
int main(){
    double x = 1.1;
    for (int i=0;i<1000;i++){ //iterate 1000 times
        x -= f(x)/f_prime(x);
    }
    printf("Solution: %100g", x);
}

 

Output: Solution: 1.32006

 Jul 15, 2018

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