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Simple
Exercises In The Operational Calculus
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Fibonacci and Lucas
Numbers |
The formulas in the present
paper are all simple exercises in the operational calculus ( once
one knows it ). Formula involving a product of Fibonacci and Lucas
Numbers:
Example
1. L(2n) = L2(n) +
2*(-1)n+1
Solutions:
We use the next
operational equation
(eD+e-D)F(n)=enD-(-1)ne-nD
For
function F(a) we
have:
[F(a+1)+F(a-1)]F(n)=F(a+n)-(-1)nF(a-n)
or
L(a)F(a)=F(a+n)-(-1)nF(a-n)
for
a=2n and
F(3n)=[L2(n)+(-1)n+1]F(n)
L(2n)=L2(n)+2*(-1)n+1
Example
2. F(3a)=F(a)[L2(a)+(-1)a+1]
Solutions:
For
n=2a we have
F(3a)=L(a)F(2a)+F(-a)
as
F(-a)=(-1)a+1F(a) and
F(2a)=L(a)F(a)
F(3a)=F(a)[L2(a)+(-1)a+1]
Example
3. L(n)=2*Phi-sqrt(5)*F(n)
Solutions:
For
function of the Golden Section, Phin, we
have
(Phin+1+Phin-1)F(n)=Phi2n-(-1)n
or
(-1)nphin=Phin-(Phi+phi)F(n)
then
L(n)=2*Phin-(Phi+phi)F(n)
or
L(n)=2*Phi-sqrt(5)F(n)
Example
4. L(n)=F(n+1)+F(n-1)
Solutions:
We
use the next
equation
L(n)=enD+(-1)ne-nD
For
function F(a) we have
L(n)F(a)=F(a+n)+
(-1)nF(a-n)
For
a=1
L(n)=F(n+1)+(-1)nF(1-n)
or
L(n)=F(n+1)+(-1)n(-1)nF(n-1)
So
we have
L(n)=F(n+1)+F(n-1)
Example
5. L(2n)=F2(n+1)+2*F2(n)
+F2(n-1)
Solutions:
We use the next
operational equation for Fibonacci and Lucas
numbers
enD=F(n+1)+F(n)e-nD
For function
L(a) we have
L(a+n)=F(n+1)L(a)+F(n)L(a-1)
for
a=n
L(2n)=F(n+1)L(n)+F(n)L(n-1)
By using basic Fibonacci and
Lucas equations we
get
L(2n)=F2(n+1)+2*F2(n)+F2(n-1)
Example
6. 5*F(n)=L(n+1)+L(n-1)
Solutions:
We
use the next basic
equations:
L(n+1)+L(n-1)=F(n+2)+2*F(n)+F(n-2)=
F(n+1)+4*F(n)-F(n-1)=
=5*F(n)
Example
7. 2*L(2n)=L2(n)+5*F2(n)
Solutions:
We
use the next operational equations for Fibonacci and Lucas
numbers
2*enD=L(n)+F(n)*[eD+e-D]
For
a function L(n) we
have
2*L(2n)=L2(n)+F(n)[L(n+1)+L(n-1)]
or
2*L(2n)=L2(n)+5*F2(n)
Here are formula involving
the Fibonacci numbers and Lucas numbers.May you solve this examples?
Is there a equation you`d like to add? Please, email me your
solutions.( Fibonacci
and Lucas Formula). There is a possibility that I include your
paper in a future www pages.
| F(2n) = F(n) L(n) |
2*L(n+1)=L(n)+5 F(n) |
| 2*F(n+1)=L(n)+F(n) |
L(n + 1)2 + L(n)2 = 5 F(2n + 1) |
| L(n + 1)2 - L(n)2 = 5 F(2n) |
L(n + 1)2 - 5 F(n) = L(2n + 1)2 |
| 3*L(n)=L(n+2)+L(n-2) |
L(n)2 - 2 L(2n) = -5 F(n)2 |
| F(n + 1) L(n) = F(2n + 1) + (-1)n |
F(n) L(m) = F(n + m) + (-1)m F(n - m) |
| L(n + 1) F(n) = F(2n + 1) - (-1)n |
2*F(n+2)=L(n)+3*F(n) |
| F(2n + 1) = F(n + 1) L(n + 1) - F(n) L(n) |
L(n) F(m) = F(n + m) - (-1)m F(n - m) |
| L(2n + 1) = F(n + 1) L(n + 1) + F(n) L(n) |
F(3n)=L(2n)F(n)-F(-n) |
| 5 F(n)2 - L(n)2 = 4 (-1)n +
1 |
5 (F(n)2 + F(n + 1)2) =
L(n)2 + L(n + 1)2 |
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