ROOTS OF FIBONACCI POLYNOMIALS
V. E.HOGGATT,JR„
San Jose State University, San Jose, California
and
MARJORIE BICKNELL
A. C. Wilcox High School, Santa Clara, California
Usually the roots of polynomial equations of degree
n become m o r e difficult to find
exactly as n i n c r e a s e s , and for n ^ 5, no general formula can be applied.
But, for c e r -
tain c l a s s e s of polynomials, the roots can be derived by using hyperbolic trigonometric functions.
Here, we solve for the roots of Fibonacci and Lucas polynomials of degree n.
The Fibonacci polynomials F (x), defined by
F t (x) = 1,
F2(x) = x,
F n + 1 (x) = xF n (x) + F n _ 1 (x) ,
and the Lucas polynomials L (x),
L2(x) = x 2 + 2,
Li(x) = x,
L n + 1 (x) = xL n (x) + L ^ f c s ) ,
have the auxiliary equation
Y2 = xY + 1
which a r i s e s from the recurrence relation, and which has roots
/nX
(1)
a
=
x + 's/ x 2 + 4
s
,
x Q
P
= ~
N/ x
9
2
+ 4
It can be shown by mathematical induction that
n
(2)
_n
L
x
n(
> = «n + ^
n
F"n 1(x)
= 2-^4•
"'
a - B '
— v
The first few Fibonacci and Lucas polynomials are given in Table 1.
x = 1,
F (x) = F
and L (x) = L , the n
Observe that, when
Fibonacci and Lucas numbers, respectively,
See [1] for an introductory article on Fibonacci polynomials.
Now, we develop formulae for finding the roots of any Fibonacci or Lucas polynomial
equation using hyperbolic functions defined by
sinh z = (e
- e" ) / 2 ,
cosh z = (e
271
+ e" )/2
272
ROOTS OF FIBONACCI POLYNOMIALS
[Oct.
Table 1
Fibonacci and Lucas Polynomials
F
n
1
2
L (x)
n
n
1
x* + 2
X
X3
+ 1
+ 2x
x 3 + 3x
x 4 + 4x 2 + 2
5
X4
+ 3x2 + 1
x 5 + 5x3 + 5x
6
X5
+ 4x 3 + 3x
3
X
4
2
7
X6
8
X7
9
4
x 6 + 6x4 + 9x2 + 2
2
x 7 + 7x5 + 14x3 + 7x
+ 5x + 6x + 1
+ 6x5 + 10x3 + 4x
6
4
x 8 + 8x6 + 20x4 + 16x2 + 2
2
x 9 + 9x7 + 25x 5 + 30x3 + 9x
+ 7x + 15x + 10x + 1
X8
which satisfy, among many other identities,
cosh 2 z - sinh 2 z = 1
cosh iy = cos y,
If we let x = 2 sinh z,
e
sinh iy = i sin y
then v x 2 + 4 = 2 cosh z, and from (1), a = cosh z + sinh z =
while jS = sinh z - cosh z = - e " .
F (x) =
a
n
a
Then,
-g
fln
- P
=
e
zn
/
e
/ v
n , 0n
L (x) = a + j8
T
= e
lX n
-z <-»
e
-nz
+ e"z
nz , / lV n -nz
+ (-1) e
Thus
p
J
/v\
2nw
=
sinh 2nz
coshz '
J
, ,
2n+lw
=
cosh (2n + l ) z
cosh z
(3)
L 2 / v = 2 cosh 2nz,
L
(x) = 2 sinh (2n + l)z .
Now, clearly the polynomial equation equals zero when the corresponding hyperbolic
function vanishes. For z = x + iy (see [ 2 ] , p. 55)
I sinh z | 2 = sinh 2 x + sin 2 y
|cosh z | 2 = sinh 2 x + cos 2 y
Thus, since for real x,
sinh x = 0 if and only if x = 0,
.
this implies that the zeroes of
sinh z a r e those of sinh iy = i sin y, and the zeroes of cosh z a r e the zeroes of cosh iy =
cosy.
Thus, we can easily find the z T s n e c e s s a r y and sufficient for
be zero.
F (x) and
L (x)
to
1973]
ROOTS OF FIBONACCI POLYNOMIALS
Example.
cos y i
273
F 2 n (x) = 0 implies that sinh 2nz = 0, cosh z ^ 0,
0, so 2ny = k77 and z = iy.
Thus,
x = + 2i sin k77/2n,
so that
sin 2ny = 0,
k = 0, 1, 2, • •• , n - 1.
Specifically, the zeroes of F6(x) are given by x = +2i sin k77/6, k = 0, 1, 2, so that x =
F6(x) = x(x2 + i)( x 2 + 3),
0, ±i, ±iN/3. As a check, since
we can see that the formula is
working.
F2
1 (x)
= 0
only if
cosh (2n + l)z = 0,
cos (2n + l)y = 0, cos y f 0.
Then,
cosh z f 0,
or when cosh (2n + l)iy =
(2n + l)y = (2k + l)7r/2, SO that
.
Z
i(2k + 1)77
(2n + 1)2 '
ly
so that
,0.
.
/2k
x = +2i sm I
+ l\
,
,
77
J • rr ,
A
-,
T
k = 0, 1, • •• , n - 1 .
To summarize, taking x = 2 sinh z leads to the following solutions:
F 2 n (x) = 0 :
x = ± 2i sin —• ,
F2n+1(x) = 0 :
x = ± 2i sin f | ^ 4
L 2 n (x) = 0 :
x = ± 2i sin \I — 52n
7 — j) '• -2 :,
L
2n+l(x)
=
0 :
k = 0, 1, • • • , n - 1
/2k + l \
X
=
± 2 i s i n
77
J • ^ :,
"^2n + i y
2
/2k + l \
£
k = 0, 1, • • • , n - 1
k = 0, 1, • • • , n - 1
k7T
2^TT'
k = 0, l , . . - . n - l
.
Compare with Webb and P a r b e r r y [3].
Suppose that, on the other hand, we s t a r t over again with x = 2i cosh z
2
N/ x + 4 = 2i sinh z,
z
and a = i e , /3 = i e " .
^ . v
(4)
so that
Then, by (2),
- / zn
-zn\
/
1) sinh nz
.n-11 e
- e
\ _ .(nsinh z
\ e - e
/
L (x) = enz
n
Now this looks better.
Z
+ e"nz = 2-incoshnz
.
F o r the Fibonacci polynomials, F (x) = 0 when sinh nz = 0, sinh z
^ 0. Since sinh nz = 0 if and only if sin ny = 0 or when z = iy, we must have ny = ±k77
so that z = ±ik77/n. Since i cosh iy = i cos y, x = 2i cosh z = 2i cos k77/n, k = 1, 2,
•••,
n-1.
Now, for the Lucas polynomials,
when ny is an odd multiple of 77/2,
L (x) = cosh nz = 0 if and only if
and again
z = iy,
cos ny = 0,
so that x = 2i cosh z
x = 2i cos (2k + l)77/2n, k = 0, 1, • • • , n - 1.
To summarize, taking x = 2 cosh z leads to the following solutions:
or
becomes
274
ROOTS OF FIBONACCI POLYNOMIALS
F (x) = 0 :
n
x = 2i cos —
n
L (x) = 0 :
n
x = 2i cos
Oct. 1973
k = 1, 2, • * * , n - 1
(2k
0
+
1)7T
2n
Actually, there is another way, using
,
k = 0, 1, • • • , n - 1
F 0 (x) = F (x)L (x).
n
aXi
Now, if we can solve
n
F (x) = 0, then the roots of L (x) a r e those roots of F 9 (x) which a r e not roots of F (x)
m
uW
n
n
Please note how this agrees with our r e s u l t s :
F 2 n (x) = 0
x = 2i cos ^
,
k = 1, 2, ' ' • , 2n - 1
F n (x) = 0
x = 2i cos %& ,
j = 1, 2, • • • , n - 1
L n (x) = 0
x = 2i cos
(2j + 1)7r
2 n
,
j = 0, 1, • • • , n - 1 .
Thus the roots separate each other.
REFERENCES
1. Marjorie Bicknell, TA P r i m e r for the Fibonacci Numbers: P a r t VII — An Introduction to
Fibonacci Polynomials and Their Divisibility P r o p e r t i e s , " Fibonacci Quarterly, Vol. 8,
No. 4, October, 1970, pp. 407-420. Also available as p a r t of A P r i m e r for the Fibonacci
Numbers, The Fibonacci Association, Dec. , 1972.
2.
Ruel V. Churchill, Complex Variables and Applications, McGraw-Hill, New York, 1960.
3.
W. A. Webb and E. A. P a r b e r r y , "Divisibility P r o p e r t i e s of Fibonacci Polynomials, M
Fibonacci Quarterly, Vol. 7, No. 5, Dec. , 1969, pp. 457-463.
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