the question states
f(x)=(1/2) (a^x+a^-x) if a>0. If f(x+y)+f(x-y)=kf(x)f(y), where k is a constant find that constant k
did you write out the equation to see if you got any ideas? what have you tried?
note that:
$\displaystyle f(x) = \frac 12 \left( a^x + a^{-x} \right)$
$\displaystyle f(y) = \frac 12 \left( a^y + a^{-y} \right)$
$\displaystyle f(x + y) = \frac 12 \left( a^{x + y} + a^{-x - y} \right)$
$\displaystyle f(x - y) = \frac 12 \left( a^{x - y} + a^{y - x} \right)$
now put those together as the equation suggests and see what you get
This is a bit of generalised triganometry
Suppose for some $\displaystyle a$ that:
$\displaystyle f(x)=(1/2) (x^x+a^{-x})$
Then:
$\displaystyle
f(x+y)+f(x-y) = \frac{1}{2}\left[ a^{x+y}+a^{-x-y}\right]+\frac{1}{2}\left[ a^{-x-y}+a^{-x+y}\right]$
.................. $\displaystyle =\frac{1}{2} a^x \left[ a^{y}+a^{-y}\right]+\frac{1}{2} a^{-x} \left[ a^{y}+a^{-y}\right]=2f(x)f(y)$
ZB
How about:
"And perhaps posterity shall thank me for having shown the Ancients did not know everything" (Fermat).
I am not exactly sure what I means, but it sounds like a King James type of quotation which are really eqoulently stated. I wish I could talk like that (in old English) but I am not so good.