Question

证明：若f(x)=2分之1（a的x次幂+a的-x次幂）,其中a>0,则 f(x+y)+f(x-y)=2f(x)f(y).

Answer 1

f(x)=[a^x+a^(-x)]/2,

f(y)=[a^y+a^(-y)]/2,

f(x+y)=[a^(x+y)+a^(-x-y)]/2,

f(x-y)=[a^(x-y)+a^(-x+y)]/2

f(x+y)+f(x-y)=[a^(x+y)+a^(-x-y)]/2+[a^(x-y)+a^(-x+y)]/2

=[a^x*a^y+a^(-x)*a^(-y)+a^x*a^(-y)+a^(-x)*a^y]/2

={a^y[a^x+a^(-x)]+a^(-y)[a^x+a^(-x)]}/2

=[a^x+a^(-x)]*[a^y+aY(-y)]/2

=2f(x)*2f(y)/2

=2f(x)f(y).

证毕.