$2.$Where is the $v_km_k$ from when $k=1,..,K-1$ and $k=K$,because in the formula above,there is only a $-v_1m_1$.there isn't a $v_km_k$

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$mathcal L(s,m,alpha,lambda,v,mu,xi)=\ sum limits_k=1^K[lambda_kR_1,K+(1-lambda_k)R_2,k] \ +muBiggl( 1- sum limits_k=0^Kalpha_kBiggr)+xiBiggl(P-sum limits_k=0^Ks_kBiggr)\ +v_1(eta s_0g_r,1-E^c_1-m_1)\ +sum limits_k=2^Kv_iBiggl(sum limits_k=0^i-1eta s_kg_r,i+sum limits_k=1^i-1eta m_kg_k,i-E^c_i-m_iBiggr)\= $



begincases
-mu_kalpha_k-xi s_k+sum limits_i=1^Keta v_ig_r,is_k , &text k=0 \
lambda_kR_1,k+(1-lambda_ k)R_2,k-mu alpha_k-xi s_k-v_km _k+sum limits_i=k+1^Keta v_i(g_r,is_k+g_k,im_k) &text k=1,...K-1 \lambda_kR_1,k+(1-lambda_k)R_2,k-mu alpha_k-xi s_k- v_km_k &text k=K
endcases



In this complicated calculation,i don't understand two thing,



$1.$Why will the $sum limits_k=2^K$ become $sum limits_k=1^K$ when $k=0 ?$



$2.$Where is the $v_km_k$ from when $k=1,..,K-1$ and $k=K$,because in the formula above,there is only a $-v_1m_1$.there isn't a $v_km_k$




calculus summation




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    $mathcal L(s,m,alpha,lambda,v,mu,xi)=\ sum limits_k=1^K[lambda_kR_1,K+(1-lambda_k)R_2,k] \ +muBiggl( 1- sum limits_k=0^Kalpha_kBiggr)+xiBiggl(P-sum limits_k=0^Ks_kBiggr)\ +v_1(eta s_0g_r,1-E^c_1-m_1)\ +sum limits_k=2^Kv_iBiggl(sum limits_k=0^i-1eta s_kg_r,i+sum limits_k=1^i-1eta m_kg_k,i-E^c_i-m_iBiggr)\= $



    begincases
    -mu_kalpha_k-xi s_k+sum limits_i=1^Keta v_ig_r,is_k , &text k=0 \
    lambda_kR_1,k+(1-lambda_ k)R_2,k-mu alpha_k-xi s_k-v_km _k+sum limits_i=k+1^Keta v_i(g_r,is_k+g_k,im_k) &text k=1,...K-1 \lambda_kR_1,k+(1-lambda_k)R_2,k-mu alpha_k-xi s_k- v_km_k &text k=K
    endcases



    In this complicated calculation,i don't understand two thing,



    $1.$Why will the $sum limits_k=2^K$ become $sum limits_k=1^K$ when $k=0 ?$



    $2.$Where is the $v_km_k$ from when $k=1,..,K-1$ and $k=K$,because in the formula above,there is only a $-v_1m_1$.there isn't a $v_km_k$




    calculus summation




    share|cite|improve this question





















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      down vote

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      up vote
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      down vote

      favorite











      $mathcal L(s,m,alpha,lambda,v,mu,xi)=\ sum limits_k=1^K[lambda_kR_1,K+(1-lambda_k)R_2,k] \ +muBiggl( 1- sum limits_k=0^Kalpha_kBiggr)+xiBiggl(P-sum limits_k=0^Ks_kBiggr)\ +v_1(eta s_0g_r,1-E^c_1-m_1)\ +sum limits_k=2^Kv_iBiggl(sum limits_k=0^i-1eta s_kg_r,i+sum limits_k=1^i-1eta m_kg_k,i-E^c_i-m_iBiggr)\= $



      begincases
      -mu_kalpha_k-xi s_k+sum limits_i=1^Keta v_ig_r,is_k , &text k=0 \
      lambda_kR_1,k+(1-lambda_ k)R_2,k-mu alpha_k-xi s_k-v_km _k+sum limits_i=k+1^Keta v_i(g_r,is_k+g_k,im_k) &text k=1,...K-1 \lambda_kR_1,k+(1-lambda_k)R_2,k-mu alpha_k-xi s_k- v_km_k &text k=K
      endcases



      In this complicated calculation,i don't understand two thing,



      $1.$Why will the $sum limits_k=2^K$ become $sum limits_k=1^K$ when $k=0 ?$



      $2.$Where is the $v_km_k$ from when $k=1,..,K-1$ and $k=K$,because in the formula above,there is only a $-v_1m_1$.there isn't a $v_km_k$




      calculus summation




      share|cite|improve this question











      $mathcal L(s,m,alpha,lambda,v,mu,xi)=\ sum limits_k=1^K[lambda_kR_1,K+(1-lambda_k)R_2,k] \ +muBiggl( 1- sum limits_k=0^Kalpha_kBiggr)+xiBiggl(P-sum limits_k=0^Ks_kBiggr)\ +v_1(eta s_0g_r,1-E^c_1-m_1)\ +sum limits_k=2^Kv_iBiggl(sum limits_k=0^i-1eta s_kg_r,i+sum limits_k=1^i-1eta m_kg_k,i-E^c_i-m_iBiggr)\= $



      begincases
      -mu_kalpha_k-xi s_k+sum limits_i=1^Keta v_ig_r,is_k , &text k=0 \
      lambda_kR_1,k+(1-lambda_ k)R_2,k-mu alpha_k-xi s_k-v_km _k+sum limits_i=k+1^Keta v_i(g_r,is_k+g_k,im_k) &text k=1,...K-1 \lambda_kR_1,k+(1-lambda_k)R_2,k-mu alpha_k-xi s_k- v_km_k &text k=K
      endcases



      In this complicated calculation,i don't understand two thing,



      $1.$Why will the $sum limits_k=2^K$ become $sum limits_k=1^K$ when $k=0 ?$



      $2.$Where is the $v_km_k$ from when $k=1,..,K-1$ and $k=K$,because in the formula above,there is only a $-v_1m_1$.there isn't a $v_km_k$





      calculus summation





      share|cite|improve this question










      share|cite|improve this question




      share|cite|improve this question









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