YLL討論網

#ed_op#DIV#ed_cl#1.#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#a.x=y   f(2x)=2f(x)+2x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#   →   f(0)=2f(0)+0=0#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#b.x=-y   f(0)=f(x)+f(-x)-2x#ed_op#SUP#ed_cl#2   #ed_op#/SUP#ed_cl#→   f(x)+f(-x)=2x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#c.x=x，y=1   f(x+1)=f(x)+f(1)+2x   →   f(x+1)-f(x)=f(1)+2x#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#由c知f(x)=x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+x[f(1)-1]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#目前只知道f(x)=x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#+x[f(1)-1]，f(0)=0，f(x)+f(-x)=2x#ed_op#SUP#ed_cl#2#ed_op#/SUP#ed_cl#，f(x+1)=f(x)+f(1)+2x#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#就停住了......#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#

設，且滿足條件（１）
（２）當ｘ＝０時，　求ｆ（ｘ）