Show that $begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix} =...

1

$begingroup$

The Fibonacci numbers $F_n$ are recursively defined by

$F_0 = 0, F_1 = 1$

$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$

i) Show that $begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix} = begin{bmatrix}1&1\1&0end{bmatrix}^n$ for all $n ∈ N$.

ii) Show that $F_0^2 + F_1^2 + ...+F_n^2 = F_n F_{n+1}$ for all $n ∈ N$.

iii) Show that $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for all $n ∈ N$.

discrete-mathematics fibonacci-numbers

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edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title and question i) are not the same. Also what have you tried?
  $endgroup$
  – mathreadler
  Jan 30 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Converting recursive equations into matrices
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, only a duplicate for (i).
  $endgroup$
  – lhf
  Jan 30 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
  $endgroup$
  – Diger
  Jan 30 at 15:02
  

  
  
  
  

add a comment | 

1

$begingroup$

The Fibonacci numbers $F_n$ are recursively defined by

$F_0 = 0, F_1 = 1$

$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$

i) Show that $begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix} = begin{bmatrix}1&1\1&0end{bmatrix}^n$ for all $n ∈ N$.

ii) Show that $F_0^2 + F_1^2 + ...+F_n^2 = F_n F_{n+1}$ for all $n ∈ N$.

iii) Show that $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for all $n ∈ N$.

discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title and question i) are not the same. Also what have you tried?
  $endgroup$
  – mathreadler
  Jan 30 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Converting recursive equations into matrices
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, only a duplicate for (i).
  $endgroup$
  – lhf
  Jan 30 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
  $endgroup$
  – Diger
  Jan 30 at 15:02
  

  
  
  
  

add a comment | 

1

1

1

$begingroup$

The Fibonacci numbers $F_n$ are recursively defined by

$F_0 = 0, F_1 = 1$

$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$

i) Show that $begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix} = begin{bmatrix}1&1\1&0end{bmatrix}^n$ for all $n ∈ N$.

ii) Show that $F_0^2 + F_1^2 + ...+F_n^2 = F_n F_{n+1}$ for all $n ∈ N$.

iii) Show that $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for all $n ∈ N$.

discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

$endgroup$

The Fibonacci numbers $F_n$ are recursively defined by

$F_0 = 0, F_1 = 1$

$F_{n+2} = F_{n+1} + F_n, n = 0,1,...$

i) Show that $begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix} = begin{bmatrix}1&1\1&0end{bmatrix}^n$ for all $n ∈ N$.

ii) Show that $F_0^2 + F_1^2 + ...+F_n^2 = F_n F_{n+1}$ for all $n ∈ N$.

iii) Show that $F_{n-1} F_{n+1} - F_n^2 = (-1)^n$ for all $n ∈ N$.

discrete-mathematics fibonacci-numbers

discrete-mathematics fibonacci-numbers

share|cite|improve this question

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

share|cite|improve this question

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

share|cite|improve this question

share|cite|improve this question

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

edited Jan 30 at 12:56

Jyrki Lahtonen

110k13172389

110k13172389

asked Jan 30 at 12:38

SomethingSomething

817

asked Jan 30 at 12:38

SomethingSomething

817

asked Jan 30 at 12:38

SomethingSomething

817

817

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title and question i) are not the same. Also what have you tried?
  $endgroup$
  – mathreadler
  Jan 30 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Converting recursive equations into matrices
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, only a duplicate for (i).
  $endgroup$
  – lhf
  Jan 30 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
  $endgroup$
  – Diger
  Jan 30 at 15:02
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  The title and question i) are not the same. Also what have you tried?
  $endgroup$
  – mathreadler
  Jan 30 at 12:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Converting recursive equations into matrices
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:00
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @JyrkiLahtonen, only a duplicate for (i).
  $endgroup$
  – lhf
  Jan 30 at 13:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
  $endgroup$
  – Jyrki Lahtonen
  Jan 30 at 13:04
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
  $endgroup$
  – Diger
  Jan 30 at 15:02
  

  
  
  
  

1

1

$begingroup$
The title and question i) are not the same. Also what have you tried?
$endgroup$
– mathreadler
Jan 30 at 12:39

$begingroup$
The title and question i) are not the same. Also what have you tried?
$endgroup$
– mathreadler
Jan 30 at 12:39

$begingroup$
Possible duplicate of Converting recursive equations into matrices
$endgroup$
– Jyrki Lahtonen
Jan 30 at 13:00

$begingroup$
Possible duplicate of Converting recursive equations into matrices
$endgroup$
– Jyrki Lahtonen
Jan 30 at 13:00

$begingroup$
@JyrkiLahtonen, only a duplicate for (i).
$endgroup$
– lhf
Jan 30 at 13:02

$begingroup$
@JyrkiLahtonen, only a duplicate for (i).
$endgroup$
– lhf
Jan 30 at 13:02

$begingroup$
Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 13:04

$begingroup$
Well, A) it answers part (iii) as well, B) there should be only one question.per post. It is highly likely that the other parts have been covered earlier as well.
$endgroup$
– Jyrki Lahtonen
Jan 30 at 13:04

$begingroup$
I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
$endgroup$
– Diger
Jan 30 at 15:02

$begingroup$
I guess the easiest way it to multiply the equation by $$begin{bmatrix}1&1\1&0end{bmatrix}$$ and then the RHS is $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix}$$ by assumption and you just need to evaluate $$begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}begin{bmatrix}1&1\1&0end{bmatrix}=begin{bmatrix}F_{n+1}+F_n&F_{n+1}\F_n+F_{n-1}&F_{n}end{bmatrix}$$ which by the Fibonacci recursion however matches $$begin{bmatrix}F_{n+2}&F_{n+1}\F_{n+1}&F_{n}end{bmatrix} , .$$ For iii) just take the determinant.
$endgroup$
– Diger
Jan 30 at 15:02

add a comment | 

3 Answers
3

active

oldest

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0

$begingroup$

Just try to prove those by induction on $ninmathbb{N}$.

For example, the "induction step" for the second exercise is:

Suppose that $F_0^2+cdots+F_n^2=F_nF_{n+1}$, then $$F_0^2+cdots+F_n^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$$

Try the same approach in the other exercises. Don't forget the base cases.

share|cite|improve this answer

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems that (ii) and (iii) are meant to be solved using (i).
  $endgroup$
  – lhf
  Jan 30 at 12:51
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Hint:

(i) Easy induction.

(ii)
Note that
$$
begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}^2 =
begin{bmatrix}F_{n+1}^2+F_n^2&*\*&*end{bmatrix}
$$

(iii) Take determinants.

share|cite|improve this answer

answered Jan 30 at 12:59

lhflhf

167k11172404

$endgroup$

add a comment | 

0

$begingroup$

i) Diagonalize the RHS matrix:
$$begin{pmatrix}1&1\ 1&0end{pmatrix}^n=
begin{pmatrix}psi&phi\ 1&1end{pmatrix}
begin{pmatrix}psi^n &0\ 0&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}psi^{n+1}&phi^{n+1}\ psi^n&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}frac{phi^{n+1}-psi^{n+1}}{sqrt{5}}&frac{phi^n-psi^n}{sqrt{5}}\ frac{phi^n-psi^n}{sqrt{5}}&frac{phi^{n-1}-psi^{n-1}}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}F_{n+1}&F_n\ F_n&F_{n-1}end{pmatrix}.$$
Note: $phicdot psi=-1$.

share|cite|improve this answer

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

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3 Answers
3

active

oldest

votes

3 Answers
3

active

oldest

votes

active

oldest

votes

active

oldest

votes

0

$begingroup$

Just try to prove those by induction on $ninmathbb{N}$.

For example, the "induction step" for the second exercise is:

Suppose that $F_0^2+cdots+F_n^2=F_nF_{n+1}$, then $$F_0^2+cdots+F_n^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$$

Try the same approach in the other exercises. Don't forget the base cases.

share|cite|improve this answer

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems that (ii) and (iii) are meant to be solved using (i).
  $endgroup$
  – lhf
  Jan 30 at 12:51
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Just try to prove those by induction on $ninmathbb{N}$.

For example, the "induction step" for the second exercise is:

Suppose that $F_0^2+cdots+F_n^2=F_nF_{n+1}$, then $$F_0^2+cdots+F_n^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$$

Try the same approach in the other exercises. Don't forget the base cases.

share|cite|improve this answer

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems that (ii) and (iii) are meant to be solved using (i).
  $endgroup$
  – lhf
  Jan 30 at 12:51
  
  
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

Just try to prove those by induction on $ninmathbb{N}$.

For example, the "induction step" for the second exercise is:

Suppose that $F_0^2+cdots+F_n^2=F_nF_{n+1}$, then $$F_0^2+cdots+F_n^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$$

Try the same approach in the other exercises. Don't forget the base cases.

share|cite|improve this answer

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

$endgroup$

Just try to prove those by induction on $ninmathbb{N}$.

For example, the "induction step" for the second exercise is:

Suppose that $F_0^2+cdots+F_n^2=F_nF_{n+1}$, then $$F_0^2+cdots+F_n^2+F_{n+1}^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}$$

Try the same approach in the other exercises. Don't forget the base cases.

share|cite|improve this answer

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

share|cite|improve this answer

share|cite|improve this answer

edited Jan 30 at 12:50

edited Jan 30 at 12:50

edited Jan 30 at 12:50

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

answered Jan 30 at 12:44

Don FanucciDon Fanucci

1,320421

1,320421

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems that (ii) and (iii) are meant to be solved using (i).
  $endgroup$
  – lhf
  Jan 30 at 12:51
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It seems that (ii) and (iii) are meant to be solved using (i).
  $endgroup$
  – lhf
  Jan 30 at 12:51
  
  
  

  
  
  
  

$begingroup$
It seems that (ii) and (iii) are meant to be solved using (i).
$endgroup$
– lhf
Jan 30 at 12:51

$begingroup$
It seems that (ii) and (iii) are meant to be solved using (i).
$endgroup$
– lhf
Jan 30 at 12:51

add a comment | 

0

$begingroup$

Hint:

(i) Easy induction.

(ii)
Note that
$$
begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}^2 =
begin{bmatrix}F_{n+1}^2+F_n^2&*\*&*end{bmatrix}
$$

(iii) Take determinants.

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answered Jan 30 at 12:59

lhflhf

167k11172404

$endgroup$

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0

$begingroup$

Hint:

(i) Easy induction.

(ii)
Note that
$$
begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}^2 =
begin{bmatrix}F_{n+1}^2+F_n^2&*\*&*end{bmatrix}
$$

(iii) Take determinants.

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answered Jan 30 at 12:59

lhflhf

167k11172404

$endgroup$

add a comment | 

0

0

0

$begingroup$

Hint:

(i) Easy induction.

(ii)
Note that
$$
begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}^2 =
begin{bmatrix}F_{n+1}^2+F_n^2&*\*&*end{bmatrix}
$$

(iii) Take determinants.

share|cite|improve this answer

answered Jan 30 at 12:59

lhflhf

167k11172404

$endgroup$

Hint:

(i) Easy induction.

(ii)
Note that
$$
begin{bmatrix}F_{n+1}&F_n\F_n&F_{n-1}end{bmatrix}^2 =
begin{bmatrix}F_{n+1}^2+F_n^2&*\*&*end{bmatrix}
$$

(iii) Take determinants.

share|cite|improve this answer

answered Jan 30 at 12:59

lhflhf

167k11172404

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 12:59

lhflhf

167k11172404

answered Jan 30 at 12:59

lhflhf

167k11172404

answered Jan 30 at 12:59

lhflhf

167k11172404

167k11172404

add a comment | 

add a comment | 

0

$begingroup$

i) Diagonalize the RHS matrix:
$$begin{pmatrix}1&1\ 1&0end{pmatrix}^n=
begin{pmatrix}psi&phi\ 1&1end{pmatrix}
begin{pmatrix}psi^n &0\ 0&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}psi^{n+1}&phi^{n+1}\ psi^n&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}frac{phi^{n+1}-psi^{n+1}}{sqrt{5}}&frac{phi^n-psi^n}{sqrt{5}}\ frac{phi^n-psi^n}{sqrt{5}}&frac{phi^{n-1}-psi^{n-1}}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}F_{n+1}&F_n\ F_n&F_{n-1}end{pmatrix}.$$
Note: $phicdot psi=-1$.

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answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

0

$begingroup$

i) Diagonalize the RHS matrix:
$$begin{pmatrix}1&1\ 1&0end{pmatrix}^n=
begin{pmatrix}psi&phi\ 1&1end{pmatrix}
begin{pmatrix}psi^n &0\ 0&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}psi^{n+1}&phi^{n+1}\ psi^n&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}frac{phi^{n+1}-psi^{n+1}}{sqrt{5}}&frac{phi^n-psi^n}{sqrt{5}}\ frac{phi^n-psi^n}{sqrt{5}}&frac{phi^{n-1}-psi^{n-1}}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}F_{n+1}&F_n\ F_n&F_{n-1}end{pmatrix}.$$
Note: $phicdot psi=-1$.

share|cite|improve this answer

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

$endgroup$

add a comment | 

0

0

0

$begingroup$

i) Diagonalize the RHS matrix:
$$begin{pmatrix}1&1\ 1&0end{pmatrix}^n=
begin{pmatrix}psi&phi\ 1&1end{pmatrix}
begin{pmatrix}psi^n &0\ 0&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}psi^{n+1}&phi^{n+1}\ psi^n&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}frac{phi^{n+1}-psi^{n+1}}{sqrt{5}}&frac{phi^n-psi^n}{sqrt{5}}\ frac{phi^n-psi^n}{sqrt{5}}&frac{phi^{n-1}-psi^{n-1}}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}F_{n+1}&F_n\ F_n&F_{n-1}end{pmatrix}.$$
Note: $phicdot psi=-1$.

share|cite|improve this answer

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

$endgroup$

i) Diagonalize the RHS matrix:
$$begin{pmatrix}1&1\ 1&0end{pmatrix}^n=
begin{pmatrix}psi&phi\ 1&1end{pmatrix}
begin{pmatrix}psi^n &0\ 0&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}psi^{n+1}&phi^{n+1}\ psi^n&phi^nend{pmatrix}
begin{pmatrix}-frac{1}{sqrt{5}}&frac{phi}{sqrt{5}}\ frac1{sqrt{5}}&-frac{psi}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}frac{phi^{n+1}-psi^{n+1}}{sqrt{5}}&frac{phi^n-psi^n}{sqrt{5}}\ frac{phi^n-psi^n}{sqrt{5}}&frac{phi^{n-1}-psi^{n-1}}{sqrt{5}}end{pmatrix}=\
begin{pmatrix}F_{n+1}&F_n\ F_n&F_{n-1}end{pmatrix}.$$
Note: $phicdot psi=-1$.

share|cite|improve this answer

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

share|cite|improve this answer

share|cite|improve this answer

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

answered Jan 30 at 13:51

farruhotafarruhota

21.8k2842

21.8k2842

add a comment | 

add a comment | 

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