Question

A funtion is defined by f(x) = 1/2(10^x + 10^-x) for x in R. Show that:

2f(x)f(y) = f(x+y) + f(x-y)

Answer 1

you have to write it out and you will see that it works

Answer 2

that is, write
\[f(x+y)+f(x-y)=\frac{1}{2}(10^{x+y}+10^{-(x+y)})+\frac{1}{2}(10^{x-y}+10^{-(x-y)})\] and rearrange the terms

Answer 3

i don't really mean "rearrange the terms. when you compute
\[2f(x)f(y)\] you will get the same thing

Answer 4

can you show me it? I'm kind of confused..

Answer 5

let me write it on paper first so i don't make a typo.

Answer 6

ok really there is nothing to do but multiply
\[2f(x)f(y)\] and see that you get the thing i wrote above
here goes

Answer 7

\[2f(x)f(y)=2\times \frac{1}{2}(10^x+10^{-x})\times \frac{1}{2}(10^y+10^{-y})\]

Answer 8

\[=\frac{1}{2}(10^{x+y}+10^{x-y}+10^{-x+y}+10^{-x-y})\]

Answer 9

hope it is clear what i did. just like multiplying
\[(a+b)(c+d)\] you have to do 4 multiplications and since each one has a base of 10 you are adding the exponents

Answer 10

in the top line i wrote the right hand side. it is
\[f(x+y)+f(x-y)=\frac{1}{2}(10^{x+y}+10^{-(x+y)})+\frac{1}{2}(10^{x-y}+10^{-(x-y)})\]

Answer 11

now compare exponents and see that you have the same thing exactly

Answer 12

thank you so much.