Question

Question. (How to proof?)
Advanced math.

Answer 1

\(\color{blue}{ {\rm Prove{\tiny~~~}that{\tiny~~~}for{\tiny~~~}all}{\tiny~~~}n\in\mathbb{N},{\tiny~~~}{\rm Fibonacci{\tiny~~~}numbers}{\tiny~~~}f_n{\tiny~~~}{\rm satisfy}{\tiny~~~} \displaystyle n=\frac{\alpha^n-\beta^n}{\alpha-\beta}\\
\displaystyle {\rm where{\tiny~~~}\alpha{\tiny~~~}and{\tiny~~~}\beta{\tiny~~~}are{\tiny~~~}the{\tiny~~~}two{\tiny~~~}roots{\tiny~~~}of{\tiny~~~}}x^2-x-1.}\)

Answer 2

I was able to find that the roots are:
\(x^2-x-1=0 \quad \Longleftrightarrow \quad x=\displaystyle \frac{1}{2}~\pm~\displaystyle \frac{\sqrt{~5~}}{2}\).

Answer 3

I would try induction
First prove the "base case"

Answer 4

\(f_1=1\)

\(\displaystyle \frac{\alpha^n - \beta^n}{\alpha - \beta}=\frac{(\frac{1}{2}+ \frac{\sqrt{5}}{2})^1 - (\frac{1}{2}- \frac{\sqrt{5}}{2})^1}{(\frac{1}{2}+ \frac{\sqrt{5}}{2})- (\frac{1}{2}- \frac{\sqrt{5}}{2})}=\frac{\frac{\sqrt{5}}{4}}{\frac{\sqrt{5}}{4}}=1 \).

Answer 5

This completes the Basis step, right ?

Answer 6

oh, it is (sqrt 5) on top and bottom ...

Answer 7

not, sqrt(5)/4.

Answer 8

yes.

Now the hard part... so far I've have not gotten it...

Answer 9

\(\displaystyle f_1=1\)

\(\displaystyle \frac{\alpha^n - \beta^n}{\alpha - \beta}=\frac{(\frac{1}{2}+ \frac{\sqrt{5}}{2})^1 - (\frac{1}{2}- \frac{\sqrt{5}}{2})^1}{(\frac{1}{2}+ \frac{\sqrt{5}}{2})- (\frac{1}{2}- \frac{\sqrt{5}}{2})}=\frac{\sqrt{5}}{\sqrt{5}}=1\)

Answer 10

So I will first assume
\(\displaystyle f_k=\frac{\alpha^k - \beta^k}{\alpha - \beta}=\frac{(\frac{1}{2}+ \frac{\sqrt{5}}{2})^k - (\frac{1}{2}- \frac{\sqrt{5}}{2})^k}{\sqrt{~5~}}\),

for all \(k\in\mathbb{N}\).

Answer 11

I think f_k+1 = f_k + f_k-1

Answer 12

\(\displaystyle f_{k+1}= f_k+f_{k-1}=\frac{\alpha^k - \beta^k}{\alpha - \beta}+\frac{\alpha^{k-1} - \beta^{k-1}}{\alpha - \beta}\).

Answer 13

and we have to show the top is equal to a^(k+1) - b^k+1)
(somehow!)

Answer 14

if a= (1- sqr(5))/2
1/a is -b
1/a^2 is b^2 = 1+b

we can use that to show 1/a + 1/a^2 = 1

Answer 15

in other words
a^k + a^(k-1) = a^(k+1) ( 1/a + 1/a^2) = a^(k+1)
similarly for the beta terms

Answer 16

can you finish ?

Answer 17

btw, for the base case, as long as a - b is not zero, you can just do
(a^1 - b^1)/(a-b) = 1
i.e. no need to plug in the actual values

for the 2nd part, we do need to use the actual values of a and b, so that we can show 1/a = -b and 1/a^2 = 1+b

Answer 18

Alright! I think that if I give it some effort I will be able to find the route in my algebra.
Thanks!