將\(L(y_i,f(x_i))\)在\(f(x_i)=f_{m-1}(x_i)\)處泰勒展開到一階（捨去餘項，故為近似）

\[L(y_i,f(x_i))\approx L(y_i,f_{m-1}(x_i))+\left. \frac{\partial L(y_i,f(x_i))}{\partial f(x_i)} \right|_{f(x_i)=f_{m-1}(x_i)}\cdot (f(x_i)-f_{m-1}(x_i)) \]

令\(f(x_i) = f_{m-1}(x_i)\)且\(f_m(x_i) = f_{m-1}(x_i)+T_m(x_i;\theta _m)\)帶入上式並移項

\[L(y_i,f_m(x_i))-L(y_i,f_{m-1}(x_i))\approx \left. \frac{\partial L(y_i,f(x_i))}{\partial f(x_i)} \right|_{f(x_i)=f_{m-1}(x_i)}\cdot T_m(x_i;\theta _m) \]

左式需小於0（每輪得到的強學習器需要比上一輪強學習器在損失函式更小，不然優化無意義），故令\(T_m(x_i;\theta _m)\)去擬合\(-\left. \frac{\partial L(y_i,f(x_i))}{\partial f(x_i)} \right|_{f(x_i)=f_{m-1}(x_i)}\)

使得右式小於0。
混淆點：\(f(x_i)\)是一個變數，代表最終求得的強學習器在第\(i\)個樣本\(x_i\)上的預測，\(f_{m-1}(x_i)\)和\(f_m(x_i)\)是常量，即\((m-1)\)輪和\(m\)輪得到的強學習器在樣本\(x_i\)上的預測