由 danny 於 星期六 七月 14, 2007 1:39 am
#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#