Why do commuting operators imply that A=A(a) will commute with b?

[Nikitin]

Hi. Say a, A(a) and b are well behaving functions. Then say [a,b] = 0, i.e. a and b commute.

Why will this automatically mean that A=A(a) will commute with b? Can somebody give me an intuitive explanation, or link me to some proof?

 

[ShayanJ]

I think one way is taylor expanding A(a). Then $[A(a),b]=c_0 [1,b]+c_1 [a,b]+c_2[a^2,b]+c_3[a^3,b]+...=0$.