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klikaa

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Lepton

Lepton (1/13)

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  1. thanks for you help... i figured it out.. now its all working
  2. and how can i change that? that i doesn't give nan ?
  3. The part for(int i=1; i<n; i++) { x=sqrt(y*i); cout<<"x"<<i<<": "<<x<<endl; } isn't working wll, for some elements the result is non, i thought that i because the numbers are complex, but not with include complex i get the same resut.. can anyone help me #include<iostream> #include<math.h> #include <complex> using namespace std; int main() { int n,elem[20]; float y[20],A[20]; complex<float> x[20]; cout<<"Number of elements?"<<endl; cin>>n; cout<<"Insert elements:"<<endl; for(int i=0; i<n; i++) { cin>>elem[i]; } A[0]=pow(elem[0],2); for(int j=1; j<n; j++) { A[j]=pow(elem[j],2); int i=1; while((i+j)<=n-1&&(i<=j)) { A[j]=A[j]+pow((-1),i)*2*(elem[j-i]*elem[j+i]); i=i+1; } cout<<"A"<<j<<":"<<A[j]<<endl; } for(int i=1; i<n; i++) { y[i]=A[i]/-A[i-1]; cout<<"y"<<i<<": "<<y[i]<<endl; } for(int i=1; i<n; i++) { x[i]=sqrt(y[i]*i); cout<<"x"<<i<<": "<<x[i]<<endl; } }
  4. So i have to write a c++ program for the Graeffe's square root method I have am stuck here when i have this formula transform into c++ code, the formula is on the link The code works particulary, the (elem[j-1]*elem[j+i]) doesn't work, it's beeing ignored and i don't know why... can any one help me? http://latex.codecogs.com/gif.latex?A_{k}=a_{k}^{2}+2\sum_{s=1}^{k}(-1)^{s}a_{k-s}*a_{k+s},&space;k=0,....n cout<<"How many elements?"<<endl; cin>>n; cout<<"Insert x:"<<endl; for(int i=0; i<n; i++) { cin>>elem; } C[0]=pow(elem[0],2); for(int j=1; j<n; j++) { C[j]=pow(elem[j],2); int i=1; while((i+j)<=n-1&&(i<=j)) { C[j]=C[j]+(-1)^i*2*(elem[j-1]*elem[j+i]); i=i+1; } cout<<"C"<<j<<":"<<C[j]<<endl; }
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