We have the following function:
$$f(n)=\left(2+\frac{1}{2}\sqrt{2}\right)\left(2+\sqrt{2}\right)^{n}+\left(2-\frac{1}{2}\sqrt{2}\right)\left(2-\sqrt{2}\right)^{n}$$
I'm asked to prove that $f(n)$ is an integer for every integer $n\ge 0$.
I have no idea how to do this, can anyone help?
Asked
Active
Viewed 170 times
0
gymbvghjkgkjkhgfkl
- 8,090
Anabella Woud
- 301
-
1So $f(n)$ is a constant function? – Mar 03 '17 at 09:22
-
1http://math.stackexchange.com/questions/2156316/prove-that-left-frac3-sqrt172-rightn-left-frac3-sqrt172-r – lab bhattacharjee Mar 03 '17 at 09:25
-
Your right hand side expression does not contain n, is that correct? – PM. Mar 03 '17 at 09:26
-
Isn't it clear that however you expand it the odd powers of $\sqrt{2}$ cancel out? And that there are enough $2$'s floating around to kill off the $\frac{1}{2}$? – ancient mathematician Mar 03 '17 at 09:36
-
Try it for $n=1$ then use proof by induction and if that doesnt work try something else :) – Nick Pavini Mar 03 '17 at 09:40
-
The sequence $(f(n))$ solves $$f(0)=4\quad f(1)=10\quad f(n+2)=4f(n+1)-2f(n)$$ – Did Mar 03 '17 at 09:52
-
"I have no idea how to do this" Surely there is some lectures associated to this, what are their main results? – Did Mar 03 '17 at 09:53
-
Glad to see the RHS was edited so it contained n... – PM. Mar 03 '17 at 10:09
2 Answers
2
With a hopefully clear notation,
$$f_n=A+B,$$ and $$f_{n+1}=(2+\sqrt2)A+(2-\sqrt2)B$$ and
$$f_{n+2}=(2+\sqrt2)^2A+(2-\sqrt2)^2B\\ =(6+4\sqrt2)A+(6-4\sqrt2)B\\ =4f_{n+1}-2f_n.$$
As $f_0=4$ and $f_1=10$, all the next $f_n$ are integer.
0
Use the Binomial Theorem and Pascal's Triangle to expand the $n$th powers. Then do the multiplications. ook at the integer terms and the square root terms. What will happen when you add?
victoria
- 1,399
- 6
- 12