# Discrete Question

• Apr 18th 2010, 08:48 PM
lozts
Discrete Question
f(0) ≠ 0
f(1) = 3
f(x) * f(y) = f(x+y) + f(x-y)

1. Determine f(7).

Thanks for the help!
• Apr 18th 2010, 10:15 PM
emakarov
The same problem was discussed here.
• Apr 19th 2010, 05:28 AM
lozts
Didn't quite understand the responses
Yes, I saw this post and tried to do it out myself out of pure curiosity but had problems trying to do it, which is why I am asking. I am curious how it is done. Could you please elaborate on your answer?
• Apr 19th 2010, 06:23 AM
Plato
Quote:

Originally Posted by lozts
I looked over the responses on that post already but still couldn't really figure out the answer to either question. Could you please elaborate?

What about the other discussion did you not follow?
I consider those two replies a complete solution.
I wonder what more short of a complete ready to turn is reply do you need.
• Apr 19th 2010, 07:33 AM
lozts
I don't quite get how to determine f(7). I don't know if I am just over looking something or what. I tried to find f(7) going off of tonio's post, but didn't have much luck.
• Apr 19th 2010, 08:02 AM
Plato
It is easy to see that:
\$\displaystyle f(n+1)=3f(n)-f(n-1)\$
\$\displaystyle f(2)=3f(1)-f(0)=9\$
\$\displaystyle f(3)=3f(2)-f(1)=?\$
\$\displaystyle {\color{white}.}~~~~~~~ \vdots \$
\$\displaystyle f(7)=3f(6)-f(5)\$.
• Apr 19th 2010, 08:11 AM
lozts
Thanks a lot, I completely overlooked that but wouldn't
\$\displaystyle
f(2) = 3f(1) - f(0) = 9 - f(0) ??
\$

since f(0) isn't equal to 0.
• Apr 19th 2010, 08:19 AM
Plato
Quote:

Originally Posted by lozts
\$\displaystyle
f(2) = 3f(1) - f(0) = 9 - f(0) ??
\$

Yes that is correct?
\$\displaystyle f(1)f(0)=f(1+0)+f(1-0)=6\$ so \$\displaystyle f(0)=?\$