Derivation needed

$if\quad { x }^{ 2 }=x+1\quad and\quad n\ge 2\quad Then\quad { x }^{ n }={ f }_{ n }x+{ f }_{ n-1 }$

Can anybody provide me a derivation for this Lemma ? I was able to prove this lemma by applying the principle of mathematical induction. However I believe there are variety of proofs and derivations for this lemma. It would be better if all the proofs and derivations are posted in this note. A BIG THANKS to contributors.

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Comments

Well, use Binet's Formula. You will get it. It's easy. Binet's formula is ${f}_{k} = \frac{{\phi}^{k} - {(1-\phi)}^{k}}{\sqrt{5}}$

Kartik Sharma - 6 years, 3 months ago

I presume Binet's formula is derived from this lemma. Today when I asked my cousin about the derivation, he gave a resolved derivation which is understandable for almost everyone. First of all in this lemma GOLDEN RATIO is already given.

Golden ratio: ${ x }^{ 2 }=x+1$

It is also given that $n\ge 2$

Now, $\quad { x }^{ 3 }={ x }^{ 2 }.x\\ \therefore \quad { x }^{ 3 }=(x+1).x\quad ....{ (x }^{ 2 }=x+1)\\ \therefore \quad { x }^{ 3 }={ x }^{ 2 }+x\\ \therefore \quad { x }^{ 3 }=2x+1\quad ....{ (x }^{ 2 }=x+1)$

Similarly it can be proven that,

${ x }^{ 4 }=3x+2\\ { x }^{ 5 }=5x+3\\ { x }^{ 6 }=8x+5$

Thus we can conclude that ${ x }^{ n }={ f }_{ n }x+{ f }_{ n-1 }$

We know the fact that the roots of golden ratio are $\phi$ and $1-\phi$

Now we can deduce two equations:

${ \phi }^{ n }=f_{ n }\phi +{ f }_{ n-1 }\quad \quad \quad \quad \quad \quad \quad \quad ....1)\\ (1-{ \phi ) }^{ n }=f_{ n }(1-\phi )+{ f }_{ n-1 }\quad \quad ....2)$

Subtract equation 2 from 1 to yield Binet's formula

This was taught to me by my cousin. So, from above, I think Binet's formula was derived from this lemma.

Inderjeet Nair - 6 years, 3 months ago

Yeah I thought of that too but I thought Binet's formula would have been better.

Kartik Sharma - 6 years, 3 months ago

@Kartik Sharma – Can you please explain in detail on how we can use this formula to derive the above Lemma?

Inderjeet Nair - 6 years, 3 months ago

Yes you can easily prove using binets formula

To prove the binets formula, i will give a hint

the nth fibonnaci ( $f_n$ ) number is the coefficient of $x^n$ in

$\frac {1}{1-x-x^2}$

now use partial fraction to break it down, you will know your answer as soon as you will do that

btw $\phi \quad$ is one of the roots of the denominator of the infinite polynomial i gave

(note that the denominator is not the same quadratic as the one you gave, but the roots of this one and the one you gave are very common, you will see)

Mvs Saketh - 6 years, 3 months ago

Can you please elaborate.

Inderjeet Nair - 6 years, 3 months ago

can you please tell what $f_n$ stands for?

Mvs Saketh - 6 years, 3 months ago

Fn is the nth fibonacci number

Saigeetha Jayakumar - 6 years, 3 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$