Marie Neubrander

October 192011

3B

Mrs Johnson

Reverse Fibonacci Segllence

I was prompted to do this project when in math this year we had to write a report about a

mathematician of our choice Some of the other students in my class wrote about Fibonacci I

found the Fibonacci sequence to be interesting and wanted to do something with it First when I

began my research I found that in 1202 Leonardo Pisano known today to the world as

Fibonacci published the first modern algebra book called Liber Abaci For hundreds of years it

was considered the best math textbook that had been written since the end of the ancient world

In it Fibonacci asked the following question How many pairs of rabbits will be produced in a

year beginning with a single pair if in every month each pair bears a new pair which becomes

productive from the second month on and no death occurs The equation for this is an =

an- 1 + an-2 where an denotes the pairs of rabbits in the nth-month and where ao = 0 and a1=

1 Based on this a2 = al + ao = 1 + 0 = 1 a 3 = a2 + al = 1 + 1 = 2 a 4 = a3 + a2 = 2 + 1

=3 and we obtain the original Fibonacci sequence

0 1 123 5 8 1321 34 55 89 144233

where it is easy to compute the first 60 Fibonacci numbers with an Excel spreadsheet (see Table

1 below) The 60th Fibonacci number is a60 = 1548008755920 but shortly after n = 60 Excel

1

runs out of precision and starts rounding the Fibonacci numbers To compute the Fibonacci

numbers an for n ~ 60 one can use the amazing Binet Formula which is

Un = Js (CVSf - C-v5f)middot2

With it one can compute the nth Fibonacci number (starting with 0 and 1) directly without

computing first all the previous ones The number 1+J5 = 16180339 is the golden ratio 2

which can be found in many unexpected places One is in a standard credit card where it

represents the ratio of the lengths of the sides of the card Some other places are your overall

height divided by the height of your navel and the ratio of the length of your elbow to your wrist

to the length of your hand It is found throughout nature in places such as spirals of seed heads

and leaf arrangements

The values of the Fibonacci sequence will change when ao and ai have different starting

values For example if we start the Fibonacci sequence with ao =1 and ai =3 then we obtain the

so-called Lucas numbers

134711 18294776123199

When I started the project I was wondering if there is something like Binets Formula for any

starting values ao and ai The first main result of my project is that the answer is yes and an

explanation of how it works is given below

The second main result of my project concerns the reversal of the Fibonacci sequence

where I start with two consecutive Fibonacci numbers aN-i and aN and then try to find the

beginning numbers ao and ai I was testing the question The values of the Fibonacci sequence

2

deg deg 13 233 26 121393 39 63245986 52 32951280099 65 17167680177565 1 1 14 377 27 196418 40 102334155 53 53316291173 66 27777890035288 2 1 15 610 28 317811 41 165580141 54 86267571272 67 44945570212853 3 2 16 987 29 514229 42 267914296 55 139583862445 middot68 72723460248141 4 3 159717 30 832040 43 433494437 56 225851433717 69 117669030460994 5 5 18 2584 31 1346269 44 701408733 57 365435296162 70 190392490709135 6 8 418119 32 2178309 45 1134903170 58 591286729879 71 308061521170129 7 13 20 6765 33 3524578 46 1836311903 59 956722026041 72 498454011879264 8 21 21 10946 34 5702887 47 2971215073 60 1548008755920 73 806515533049393

9 34 22 17711 35 9227465 48 ~807526976 61 2504730781961 74 1304969544928660 10 55 23 28657 36 14930352 49 7778742049 62 4052739537881 75 2111485077978050

r 11 89 24 46368 37 24157817 1258626902550 63 6557470319842 76 3416454622906710 r 12 144 25 75025 38 39088169 51 20365011074 64 10610209857723 77 5527939700884760

depend on the two initial values chosen Can the Fibonacci sequence always be reversed or will

things such as rounding affect your results From this my hypothesis was that the Fibonacci

sequence cannot always be reversed because after rounding multiple times too much

information is lost about your original numbers I hoped for and finally achieved in finding a

formula that allows the reversal of the Fibonacci sequence for any starting values

In this project the materials needed are simple You will need a pack of paper pencils

and a computer with Excel The paper and pencils win be used to do the mathematics and Excel

is used to compute Fibonacci sequences and test related equations and formulas

STEP 1 Binets Formula for the Fibonacci Sequence Because Exce1 cannot be trusted to

compute Fibonacci numbers an for large n due to a lack of precision (see Table 1 a74 is

incorrect since it is not a73 + a72) the first step in my project is to develop an explicit formula

for the Fibonacci sequence that holds for any starting values (extension of Binets formula)

Table 1

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib

3

Maybe the most important observation when dealing with Fibonacci numbers is that when one

divides two Fibonacci numbers an- i and an (the larger n the better) one obtains the golden

ratio which is ~ i+vs = 16180339 Using Excel spreadsheets I found out that this is an-l 2

true for any Fibonacci sequence independent of the particular choice of when ao and ai

Table 2

n n-th Luc n n-th Luc n n-th Luc n n-th Lucas n n-th Lucas n n-th Lucas

1 13 843 26 439204 39 228826127 52 119218851371 65 62113250390418 deg 1 3 14 1364 27 710647 40 370248451 53 192900153618 66 100501350283429 r 2 4 15 2207 28 1149851 41 599074578 54 312119004989 67 162614600673847 3 7 16 3571 29 1860498 42 969323029 55 505019158607 68 263115950957276 r 4 11 17 5778 30 3010349 43 1568397607 56 817138163596 69 425730551631123 5 18 18 9349 31 4870847 44 2537720636 57 1322157322203 70 688846502588399 6 29 19 15127 32 7881196 45 4106118243 58 2139295485799 71 1114577054219520 7 47 20 24476 33 12752043 46 6643838879 59 3461452808002 72 1803423556807920 8 II

76 21 39603 34 20633239 47 10749957122 60 5600748293801 73 2918000611027440 r 9 123 22 64079 35 33385282 48 17393796X)1 61 9062201101803 74 4721424167835360

r II10 199 23 103682 36 54018521 49 28143753123 62 14662949395604 75 7639424778862810 r11 322 24 167761 37 87403803 50 45537549124 63 23725150497407 76 12360848946698200 l

12 521 25 271443 38 141422324 51 73681302247 64 38388099893011 77 20000273725561000

For example if we look at the Fibonacci sequence that starts with 1 and 3 (the Lucas Numbers

see Table 2) then a45 = 1618034 That is when looking at Fibonacci sequences with different a44

starting values for ao and ai I observed that an always grows exponentially That is it appears

that independent of the starting values ao and ai the quotient ~ is always approximately an-l

equal to the golden ratio i+vs = 16180339 This means that for large numbers like n = 100 2

it appears that

This leads to the idea to look for a number x and a constant c such that

4

Substituting this into the Fibonacci equation an = a n -l + a n -2 gives

nc xn =C x - 1 + c xn-2

After dividing this equation by c and by x n - 2 I obtain the characteristic equation

Now bring x+1 to the other side which gives you the quadratic equation

2x - x -1 = a

1+j5 1-j5whose two solutions are Xl = -- and X2 = -- Now we know that for all numbers Cl C2 the

2 2

sequences an = cl x f and an = C2X~ satisfy the Fibonacci equation an = a n -l + a n -2

Therefore for all Cl C21 the sequence

will also solve the Fibonacci equation an = a n - 1 + a n -2 Next we choose c1 C2 so that

ao = Cl + C2 and al = C1 X l + C2 X2 where ao and al are the given starting values Because

we have two equations for the two unknowns Cl C2 we can solve for Cl and C2 and get

Plug these values into the equation an = Cl xf + C2X~ to get the Generalized Binet Formula

1+ vis d l-vIs wmiddotth thmiddot 1 th Fmiddotb wereh Xl =-- an X2 =-- 1 IS lormu a one can compute e 1 onaccI sequence an 2 2

for any starting values ao and al In order to better understand this formula I will denote from

now on the standard Fibonacci sequence 0 112358 1321345589144233 by

5

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

runs out of precision and starts rounding the Fibonacci numbers To compute the Fibonacci

numbers an for n ~ 60 one can use the amazing Binet Formula which is

Un = Js (CVSf - C-v5f)middot2

With it one can compute the nth Fibonacci number (starting with 0 and 1) directly without

computing first all the previous ones The number 1+J5 = 16180339 is the golden ratio 2

which can be found in many unexpected places One is in a standard credit card where it

represents the ratio of the lengths of the sides of the card Some other places are your overall

height divided by the height of your navel and the ratio of the length of your elbow to your wrist

to the length of your hand It is found throughout nature in places such as spirals of seed heads

and leaf arrangements

The values of the Fibonacci sequence will change when ao and ai have different starting

values For example if we start the Fibonacci sequence with ao =1 and ai =3 then we obtain the

so-called Lucas numbers

134711 18294776123199

When I started the project I was wondering if there is something like Binets Formula for any

starting values ao and ai The first main result of my project is that the answer is yes and an

explanation of how it works is given below

The second main result of my project concerns the reversal of the Fibonacci sequence

where I start with two consecutive Fibonacci numbers aN-i and aN and then try to find the

beginning numbers ao and ai I was testing the question The values of the Fibonacci sequence

2

deg deg 13 233 26 121393 39 63245986 52 32951280099 65 17167680177565 1 1 14 377 27 196418 40 102334155 53 53316291173 66 27777890035288 2 1 15 610 28 317811 41 165580141 54 86267571272 67 44945570212853 3 2 16 987 29 514229 42 267914296 55 139583862445 middot68 72723460248141 4 3 159717 30 832040 43 433494437 56 225851433717 69 117669030460994 5 5 18 2584 31 1346269 44 701408733 57 365435296162 70 190392490709135 6 8 418119 32 2178309 45 1134903170 58 591286729879 71 308061521170129 7 13 20 6765 33 3524578 46 1836311903 59 956722026041 72 498454011879264 8 21 21 10946 34 5702887 47 2971215073 60 1548008755920 73 806515533049393

9 34 22 17711 35 9227465 48 ~807526976 61 2504730781961 74 1304969544928660 10 55 23 28657 36 14930352 49 7778742049 62 4052739537881 75 2111485077978050

r 11 89 24 46368 37 24157817 1258626902550 63 6557470319842 76 3416454622906710 r 12 144 25 75025 38 39088169 51 20365011074 64 10610209857723 77 5527939700884760

depend on the two initial values chosen Can the Fibonacci sequence always be reversed or will

things such as rounding affect your results From this my hypothesis was that the Fibonacci

sequence cannot always be reversed because after rounding multiple times too much

information is lost about your original numbers I hoped for and finally achieved in finding a

formula that allows the reversal of the Fibonacci sequence for any starting values

In this project the materials needed are simple You will need a pack of paper pencils

and a computer with Excel The paper and pencils win be used to do the mathematics and Excel

is used to compute Fibonacci sequences and test related equations and formulas

STEP 1 Binets Formula for the Fibonacci Sequence Because Exce1 cannot be trusted to

compute Fibonacci numbers an for large n due to a lack of precision (see Table 1 a74 is

incorrect since it is not a73 + a72) the first step in my project is to develop an explicit formula

for the Fibonacci sequence that holds for any starting values (extension of Binets formula)

Table 1

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib

3

Maybe the most important observation when dealing with Fibonacci numbers is that when one

divides two Fibonacci numbers an- i and an (the larger n the better) one obtains the golden

ratio which is ~ i+vs = 16180339 Using Excel spreadsheets I found out that this is an-l 2

true for any Fibonacci sequence independent of the particular choice of when ao and ai

Table 2

n n-th Luc n n-th Luc n n-th Luc n n-th Lucas n n-th Lucas n n-th Lucas

1 13 843 26 439204 39 228826127 52 119218851371 65 62113250390418 deg 1 3 14 1364 27 710647 40 370248451 53 192900153618 66 100501350283429 r 2 4 15 2207 28 1149851 41 599074578 54 312119004989 67 162614600673847 3 7 16 3571 29 1860498 42 969323029 55 505019158607 68 263115950957276 r 4 11 17 5778 30 3010349 43 1568397607 56 817138163596 69 425730551631123 5 18 18 9349 31 4870847 44 2537720636 57 1322157322203 70 688846502588399 6 29 19 15127 32 7881196 45 4106118243 58 2139295485799 71 1114577054219520 7 47 20 24476 33 12752043 46 6643838879 59 3461452808002 72 1803423556807920 8 II

76 21 39603 34 20633239 47 10749957122 60 5600748293801 73 2918000611027440 r 9 123 22 64079 35 33385282 48 17393796X)1 61 9062201101803 74 4721424167835360

r II10 199 23 103682 36 54018521 49 28143753123 62 14662949395604 75 7639424778862810 r11 322 24 167761 37 87403803 50 45537549124 63 23725150497407 76 12360848946698200 l

12 521 25 271443 38 141422324 51 73681302247 64 38388099893011 77 20000273725561000

For example if we look at the Fibonacci sequence that starts with 1 and 3 (the Lucas Numbers

see Table 2) then a45 = 1618034 That is when looking at Fibonacci sequences with different a44

starting values for ao and ai I observed that an always grows exponentially That is it appears

that independent of the starting values ao and ai the quotient ~ is always approximately an-l

equal to the golden ratio i+vs = 16180339 This means that for large numbers like n = 100 2

it appears that

This leads to the idea to look for a number x and a constant c such that

4

Substituting this into the Fibonacci equation an = a n -l + a n -2 gives

nc xn =C x - 1 + c xn-2

After dividing this equation by c and by x n - 2 I obtain the characteristic equation

Now bring x+1 to the other side which gives you the quadratic equation

2x - x -1 = a

1+j5 1-j5whose two solutions are Xl = -- and X2 = -- Now we know that for all numbers Cl C2 the

2 2

sequences an = cl x f and an = C2X~ satisfy the Fibonacci equation an = a n -l + a n -2

Therefore for all Cl C21 the sequence

will also solve the Fibonacci equation an = a n - 1 + a n -2 Next we choose c1 C2 so that

ao = Cl + C2 and al = C1 X l + C2 X2 where ao and al are the given starting values Because

we have two equations for the two unknowns Cl C2 we can solve for Cl and C2 and get

Plug these values into the equation an = Cl xf + C2X~ to get the Generalized Binet Formula

1+ vis d l-vIs wmiddotth thmiddot 1 th Fmiddotb wereh Xl =-- an X2 =-- 1 IS lormu a one can compute e 1 onaccI sequence an 2 2

for any starting values ao and al In order to better understand this formula I will denote from

now on the standard Fibonacci sequence 0 112358 1321345589144233 by

5

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

deg deg 13 233 26 121393 39 63245986 52 32951280099 65 17167680177565 1 1 14 377 27 196418 40 102334155 53 53316291173 66 27777890035288 2 1 15 610 28 317811 41 165580141 54 86267571272 67 44945570212853 3 2 16 987 29 514229 42 267914296 55 139583862445 middot68 72723460248141 4 3 159717 30 832040 43 433494437 56 225851433717 69 117669030460994 5 5 18 2584 31 1346269 44 701408733 57 365435296162 70 190392490709135 6 8 418119 32 2178309 45 1134903170 58 591286729879 71 308061521170129 7 13 20 6765 33 3524578 46 1836311903 59 956722026041 72 498454011879264 8 21 21 10946 34 5702887 47 2971215073 60 1548008755920 73 806515533049393

9 34 22 17711 35 9227465 48 ~807526976 61 2504730781961 74 1304969544928660 10 55 23 28657 36 14930352 49 7778742049 62 4052739537881 75 2111485077978050

r 11 89 24 46368 37 24157817 1258626902550 63 6557470319842 76 3416454622906710 r 12 144 25 75025 38 39088169 51 20365011074 64 10610209857723 77 5527939700884760

depend on the two initial values chosen Can the Fibonacci sequence always be reversed or will

things such as rounding affect your results From this my hypothesis was that the Fibonacci

sequence cannot always be reversed because after rounding multiple times too much

information is lost about your original numbers I hoped for and finally achieved in finding a

formula that allows the reversal of the Fibonacci sequence for any starting values

In this project the materials needed are simple You will need a pack of paper pencils

and a computer with Excel The paper and pencils win be used to do the mathematics and Excel

is used to compute Fibonacci sequences and test related equations and formulas

STEP 1 Binets Formula for the Fibonacci Sequence Because Exce1 cannot be trusted to

compute Fibonacci numbers an for large n due to a lack of precision (see Table 1 a74 is

incorrect since it is not a73 + a72) the first step in my project is to develop an explicit formula

for the Fibonacci sequence that holds for any starting values (extension of Binets formula)

Table 1

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib

3

Maybe the most important observation when dealing with Fibonacci numbers is that when one

divides two Fibonacci numbers an- i and an (the larger n the better) one obtains the golden

ratio which is ~ i+vs = 16180339 Using Excel spreadsheets I found out that this is an-l 2

true for any Fibonacci sequence independent of the particular choice of when ao and ai

Table 2

n n-th Luc n n-th Luc n n-th Luc n n-th Lucas n n-th Lucas n n-th Lucas

1 13 843 26 439204 39 228826127 52 119218851371 65 62113250390418 deg 1 3 14 1364 27 710647 40 370248451 53 192900153618 66 100501350283429 r 2 4 15 2207 28 1149851 41 599074578 54 312119004989 67 162614600673847 3 7 16 3571 29 1860498 42 969323029 55 505019158607 68 263115950957276 r 4 11 17 5778 30 3010349 43 1568397607 56 817138163596 69 425730551631123 5 18 18 9349 31 4870847 44 2537720636 57 1322157322203 70 688846502588399 6 29 19 15127 32 7881196 45 4106118243 58 2139295485799 71 1114577054219520 7 47 20 24476 33 12752043 46 6643838879 59 3461452808002 72 1803423556807920 8 II

76 21 39603 34 20633239 47 10749957122 60 5600748293801 73 2918000611027440 r 9 123 22 64079 35 33385282 48 17393796X)1 61 9062201101803 74 4721424167835360

r II10 199 23 103682 36 54018521 49 28143753123 62 14662949395604 75 7639424778862810 r11 322 24 167761 37 87403803 50 45537549124 63 23725150497407 76 12360848946698200 l

12 521 25 271443 38 141422324 51 73681302247 64 38388099893011 77 20000273725561000

For example if we look at the Fibonacci sequence that starts with 1 and 3 (the Lucas Numbers

see Table 2) then a45 = 1618034 That is when looking at Fibonacci sequences with different a44

starting values for ao and ai I observed that an always grows exponentially That is it appears

that independent of the starting values ao and ai the quotient ~ is always approximately an-l

equal to the golden ratio i+vs = 16180339 This means that for large numbers like n = 100 2

it appears that

This leads to the idea to look for a number x and a constant c such that

4

Substituting this into the Fibonacci equation an = a n -l + a n -2 gives

nc xn =C x - 1 + c xn-2

After dividing this equation by c and by x n - 2 I obtain the characteristic equation

Now bring x+1 to the other side which gives you the quadratic equation

2x - x -1 = a

1+j5 1-j5whose two solutions are Xl = -- and X2 = -- Now we know that for all numbers Cl C2 the

2 2

sequences an = cl x f and an = C2X~ satisfy the Fibonacci equation an = a n -l + a n -2

Therefore for all Cl C21 the sequence

will also solve the Fibonacci equation an = a n - 1 + a n -2 Next we choose c1 C2 so that

ao = Cl + C2 and al = C1 X l + C2 X2 where ao and al are the given starting values Because

we have two equations for the two unknowns Cl C2 we can solve for Cl and C2 and get

Plug these values into the equation an = Cl xf + C2X~ to get the Generalized Binet Formula

1+ vis d l-vIs wmiddotth thmiddot 1 th Fmiddotb wereh Xl =-- an X2 =-- 1 IS lormu a one can compute e 1 onaccI sequence an 2 2

for any starting values ao and al In order to better understand this formula I will denote from

now on the standard Fibonacci sequence 0 112358 1321345589144233 by

5

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

Maybe the most important observation when dealing with Fibonacci numbers is that when one

divides two Fibonacci numbers an- i and an (the larger n the better) one obtains the golden

ratio which is ~ i+vs = 16180339 Using Excel spreadsheets I found out that this is an-l 2

true for any Fibonacci sequence independent of the particular choice of when ao and ai

Table 2

n n-th Luc n n-th Luc n n-th Luc n n-th Lucas n n-th Lucas n n-th Lucas

1 13 843 26 439204 39 228826127 52 119218851371 65 62113250390418 deg 1 3 14 1364 27 710647 40 370248451 53 192900153618 66 100501350283429 r 2 4 15 2207 28 1149851 41 599074578 54 312119004989 67 162614600673847 3 7 16 3571 29 1860498 42 969323029 55 505019158607 68 263115950957276 r 4 11 17 5778 30 3010349 43 1568397607 56 817138163596 69 425730551631123 5 18 18 9349 31 4870847 44 2537720636 57 1322157322203 70 688846502588399 6 29 19 15127 32 7881196 45 4106118243 58 2139295485799 71 1114577054219520 7 47 20 24476 33 12752043 46 6643838879 59 3461452808002 72 1803423556807920 8 II

76 21 39603 34 20633239 47 10749957122 60 5600748293801 73 2918000611027440 r 9 123 22 64079 35 33385282 48 17393796X)1 61 9062201101803 74 4721424167835360

r II10 199 23 103682 36 54018521 49 28143753123 62 14662949395604 75 7639424778862810 r11 322 24 167761 37 87403803 50 45537549124 63 23725150497407 76 12360848946698200 l

12 521 25 271443 38 141422324 51 73681302247 64 38388099893011 77 20000273725561000

For example if we look at the Fibonacci sequence that starts with 1 and 3 (the Lucas Numbers

see Table 2) then a45 = 1618034 That is when looking at Fibonacci sequences with different a44

starting values for ao and ai I observed that an always grows exponentially That is it appears

that independent of the starting values ao and ai the quotient ~ is always approximately an-l

equal to the golden ratio i+vs = 16180339 This means that for large numbers like n = 100 2

it appears that

This leads to the idea to look for a number x and a constant c such that

4

Substituting this into the Fibonacci equation an = a n -l + a n -2 gives

nc xn =C x - 1 + c xn-2

After dividing this equation by c and by x n - 2 I obtain the characteristic equation

Now bring x+1 to the other side which gives you the quadratic equation

2x - x -1 = a

1+j5 1-j5whose two solutions are Xl = -- and X2 = -- Now we know that for all numbers Cl C2 the

2 2

sequences an = cl x f and an = C2X~ satisfy the Fibonacci equation an = a n -l + a n -2

Therefore for all Cl C21 the sequence

will also solve the Fibonacci equation an = a n - 1 + a n -2 Next we choose c1 C2 so that

ao = Cl + C2 and al = C1 X l + C2 X2 where ao and al are the given starting values Because

we have two equations for the two unknowns Cl C2 we can solve for Cl and C2 and get

Plug these values into the equation an = Cl xf + C2X~ to get the Generalized Binet Formula

1+ vis d l-vIs wmiddotth thmiddot 1 th Fmiddotb wereh Xl =-- an X2 =-- 1 IS lormu a one can compute e 1 onaccI sequence an 2 2

for any starting values ao and al In order to better understand this formula I will denote from

now on the standard Fibonacci sequence 0 112358 1321345589144233 by

5

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

Substituting this into the Fibonacci equation an = a n -l + a n -2 gives

nc xn =C x - 1 + c xn-2

After dividing this equation by c and by x n - 2 I obtain the characteristic equation

Now bring x+1 to the other side which gives you the quadratic equation

2x - x -1 = a

1+j5 1-j5whose two solutions are Xl = -- and X2 = -- Now we know that for all numbers Cl C2 the

2 2

sequences an = cl x f and an = C2X~ satisfy the Fibonacci equation an = a n -l + a n -2

Therefore for all Cl C21 the sequence

will also solve the Fibonacci equation an = a n - 1 + a n -2 Next we choose c1 C2 so that

ao = Cl + C2 and al = C1 X l + C2 X2 where ao and al are the given starting values Because

we have two equations for the two unknowns Cl C2 we can solve for Cl and C2 and get

Plug these values into the equation an = Cl xf + C2X~ to get the Generalized Binet Formula

1+ vis d l-vIs wmiddotth thmiddot 1 th Fmiddotb wereh Xl =-- an X2 =-- 1 IS lormu a one can compute e 1 onaccI sequence an 2 2

for any starting values ao and al In order to better understand this formula I will denote from

now on the standard Fibonacci sequence 0 112358 1321345589144233 by

5

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

fn instead of anmiddot In the standard Fibonacci sequence one has ao = 0 and al 1 and then the

formula above becomes

fn = -- 1_ [xf -x~] = ~ ((l+VS)ln_(l_VS)n)Xl-X2 2 2 IVS

which is the old formula of Binet that I mentioned already above Going back to the equation

and breaking it into pieces I get

Since Xl X2 = -1 it fonows that

which I can reduce to the following simplified fonn of the Generalized Binet Formula

for the so1ution of the Fibonacci equation an = an-l + an-2 with given starting values ao

and al To see how my fonnula works I need the original Fibonacci numbers fngiven in Table 1

(which are aU correct up to at least n = 60 for a correct list of Fibonacci numbers for n up to 300

see httpwwwmathssurreyacuklhosted-sitesRKnottJFibonaccif) If one wants to find the

Fibonacci sequence that starts with ao =1 and al =3 then according to the fonnula above

Un = 3fn + fn-l

and one can get the sequence an by multiplying one Fibonacci number by 3 and adding to it the

previous one This yields for example that the 60th Lucas number (see Table 2) is three times the

60th plus the 59th Fibonacci number (see Table 1)

6

5600748293801 =31548008755920 + 956722026041

As the next table shows (Table 3) Excel cannot compute a60 if we start with m = 4472 and

[30 = 5477 The result is not a60 = 12757384119926900 as Excel says it is but

a 60=[30 1548008755920 + m 956722026041 = 1275738411992692854265

Table 3

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib

0000 4472 13000 1920181 26000 1000418 844 39000 521220138039 52000 271556692336934 65000 141481557927680000 1000 5477 14000 3106922 27000 1618711693 40000 843351898 967 53000 r 439387958073657 66000 228921969508274000 2000 9949 15000 5027103 28000

r 2619130537 41000 1364572037006 54000 710944650410590 67000 370403527435954000

3000 15427 16000 8134025 29000 4237842230 42000 2207923935973 55000 1150332608484250 68000 599325496944228000 4000 25376 17000 13161127 30000 6856972 767 43000 3572495972979 56000 1861277258894840 69000 969729024380183000

r 5000 r 40803 18000 21295152 31000 11094814998 44000 5780419908952 57000 3011609867379080 70000 1569054521324410000

6000 66178 19000 34456279 3200017951787765 45000 9352915881931 580004872887126273920 71000 2538783545704590000

7000 ~106981 20000 55751431 33000 r 29046602763 46000 15133335790883 59000 7884496993653000 72000 4107838067029010000

8000 173160 2100090207711 34000 46998390528 47000 r 24486251672815 60000 r12757384119926900 73000 6646621612733600000

9000 280141 22000 145959142 35000 76044993291 48000 r 39619587463698 61000 20641881113579900 74000 10754459679762600000

10000 r 453300 23000 236166853 36000 123043383819 49000 64105839136513 62000 33399265233506900 75000 17401081292496200000

11000 733441 24000 382125996 37000 199088377110 SOOOO 103725426600211 63000 54041146347086800 76000 28155540972258800000

12000 1186741 25000 618292849 38000 322131760929 51000 167831265736 723 64000 87440411580593600 77000 45556622264755000000

The result is more striking if we compare Excels a77 = 45556622264755000 with the true

result which is according to my formula a77 = alf77 + aOf76 =[30f77 + m f76 As we will

see Excels number is over 2 trillion larger than the true number Since f77 is known to be

5527939700884757 and f76 =3416454622906707 the true value is

a77 =[3Of77 + mf76 =4343439052998443634

STEP 2 Binets FOTInula for the Reversed Fibonacci Sequence The second step in my

project is to develop an explicit formula for the reversed Fibonacci sequence To do so we must

fIrst have an equation for the reversed Fibonaoci sequence This can be done easily if one looks

at an example If we know that 309 and 500 are two consecutive Fibonacci numbers then we can

go backwards by subtracting 309 from 500 to get 191 Continuing in this manner we get the

7

reverse Fibonacci sequence 500 3091911187345281711651 Now we know that the

Fibonacci numbers 309 and 500 are generated by a Fibonacci sequence with starting values of 1

and 5 and we know that the reverse Fibonacci sequence is given by the equation

with starting values bo = aN and b1 = aN- 1 where aN and aN-1 are two consecutive values of

a Fibonacci sequence This looks simple enough to use an Excel spreadsheet to get the work

done - but what a surprise it is to see how badly Excel messes up IF we look at the Fibonacci

sequence starting with 0 and 02 then we get the numbers in Table 4 which are correct to at least

n= 60

Table 4

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-th Fib

00 00 130 466 260 242786 390 126491972 520 65902560198 650 34335360355130

10 02 140 754 270

392836 400 204668310 530 106632582346 660 55555780070576 20 02 150 1220 280 635622 410

331160282 540 172535142544 670 89891140425706

30 04 160 1974 290 1028458 420 535828592 550 279167724890 680 145446920496282

40 06 170 3194 300 1664080 430 866988874 560 451702867434 690 235338060921988 II rshy50 10 180 5168 310 2692538 440 1402817466 570 730870592324 700 380784981418270

60

16 190 8362 320 4356618 450 2269806340 580 1182573459758 710 616123042340258 70 26 200 13530 330 7049156 460 3672623806 590 1913444052082 720 996908023758528

80 42 210 21892 340 11405n4 470 5942430146 600 3096017511840 730 1613031066098790 90 68 220 35422 350 18454930 480 9615053952 610 5009461563922 740 2609939089857310 100 110 230 57314 360 29860704 490 15557484098 620 8105479075762 750 4222970155956100 110 178 240 92736 370 48315634 500 25172538050 630 13114940639684 760 6832909245813410 120 288 250 150050 380 78176338 510 40730022148 640 21220419715446 no 11055879401769500

As we see a60 = 309601751184 and aS9 = 1913444052082 Subtracting as above we get

aS8 =1182573459758 Handing the next 57 steps over to Excel we see that Excel can

compute aS7 correctly and starts messing up from aS6 on (with the mistakes starting out small

and then getting bigger and bigger)

8

Table 5

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-thFib n n-th Fib

00 3096017511840 130 5942430146 260 11405796 390 10311 520 6033602 650 -3143496116 shy

10 1913444052082 140 3672623806 270 7049120 400 32268 530

-9762478 660 5086283560 20 1182573459758 150 2269806340 280 4356676 410 -21957 540 15796080 670 -8229779676 30 730870592324 160 1402817466 290 2692444 420 54225 550 -25558559 680 13316063236 40 r 451702867434 170866988874 300 1664232 430 -76181 560 41354639 690 -21545842912 50 279167724890 180 535828592 310 1028211 440 130406 570 -66913197 700 34861906148 60 172535142544 190 r 331160281 320 r 636021 450 -206587 580 108267836 710 -56407749060 70 106632582346 200204668311 330 392191 460 336993 590 -175181033 720 91269655207 80 65902560198 210 r 126491970 340 243830 470 -543580 600 283448869 730 -147677404267 90 40730022148 220 78176341 350 148360 480 880572 610 -458629902 740 238947059475 100 25172538050 230 48315629 360 95470 490 -1424152 620 742078771 750 -386624463742

110 15557484098 240 29860712 370 52891 SO 0 2304725 630

-1200708673 760 62557152321 6 120 r 961S053952 250 18454916 380 42579 510 -3728877 640 1942787444 770 -1012195986958

At the end Excel finds that the original Fibonacci sequence started out with ao = 2834488689

and al =-1751810330 This is obviously nonsense since we know that we started out with ao

=0 and al =02

Because Excel spreadsheets obviously do not work at all I explored if there is also a formula

similar to Binets Formula for a reversed Fibonacci sequence bn = -bn - l + bn - 2 As in Step

1 the main idea is to look for a number z and a constant c such that

Substituting this into the reversed Fibonacci equation bn = -bn - l + bn - 2 gives

c zn =-c zn-l + c zn-2

After dividing this equation by c and by zn-2 I obtain the characteristic reverse equation

Now bring -z+1 to the other side which gives you the quadratic equation

9

Table 5

n n-th Fib n n-th Fib n n-th Fib n n-th Fib n n-thFib n n-th Fib

00 3096017511840 130 5942430146 260 11405796 390 10311 520 6033602 650 -3143496116 shy

10 1913444052082 140 3672623806 270 7049120 400 32268 530

-9762478 660 5086283560 20 1182573459758 150 2269806340 280 4356676 410 -21957 540 15796080 670 -8229779676 30 730870592324 160 1402817466 290 2692444 420 54225 550 -25558559 680 13316063236 40 r 451702867434 170866988874 300 1664232 430 -76181 560 41354639 690 -21545842912 50 279167724890 180 535828592 310 1028211 440 130406 570 -66913197 700 34861906148 60 172535142544 190 r 331160281 320 r 636021 450 -206587 580 108267836 710 -56407749060 70 106632582346 200204668311 330 392191 460 336993 590 -175181033 720 91269655207 80 65902560198 210 r 126491970 340 243830 470 -543580 600 283448869 730 -147677404267 90 40730022148 220 78176341 350 148360 480 880572 610 -458629902 740 238947059475 100 25172538050 230 48315629 360 95470 490 -1424152 620 742078771 750 -386624463742

110 15557484098 240 29860712 370 52891 SO 0 2304725 630

-1200708673 760 62557152321 6 120 r 961S053952 250 18454916 380 42579 510 -3728877 640 1942787444 770 -1012195986958

At the end Excel finds that the original Fibonacci sequence started out with ao = 2834488689

and al =-1751810330 This is obviously nonsense since we know that we started out with ao

=0 and al =02

Because Excel spreadsheets obviously do not work at all I explored if there is also a formula

similar to Binets Formula for a reversed Fibonacci sequence bn = -bn - l + bn - 2 As in Step

1 the main idea is to look for a number z and a constant c such that

Substituting this into the reversed Fibonacci equation bn = -bn - l + bn - 2 gives

c zn =-c zn-l + c zn-2

After dividing this equation by c and by zn-2 I obtain the characteristic reverse equation

Now bring -z+1 to the other side which gives you the quadratic equation

9

Z2 + z -1 = 0

-1+vS -1-vSwhose two solutions are Zl =-2- = -X2 and Z2 =-2- =-Xl where Xl2 are as in Step 1

Now we know again that for all constants cl C2 the sequences bn = cl zf and bn = c2zf

satisfy the reversed Fibonacci equation bn = -bn - l + bn - 2 Therefore for all Ch C21 the

sequence

will also solve the reversed Fibonacci equation bn = -bn - l + bn - 2 bull It remains to be shown

that we can choose Cl C2 so that bo = Cl + C2 and bl = C1Zl + C2Z2 where bo =aN and

= aN-l and where aN and aN-l are two consecutive values of a Fibonacci sequence b l

Solving the two equations for the two unknowns Cl C2 one gets

Plug these values into the equation bn = cl zf + c2zf and use again that Xl X2 = -1 to get

the Reversed Binet Formula

bl + boxl -bOX2 - bl = C-l)nC---- xf + xf)

Xl - X2 Xl - X2

10

where Xl = 1+2fS and X2 =1-2fS and where In denotes the standard Fibonacci sequence starting

with 0 and 1 as given in Table 1 To see how my formula works let us go back to the Fibonacci

numbers starting with 0 and 02 as in Table 4 Then

a60 = 309601751184 and a 59 = 1913444052082

159 = 956722026041 and 160 = 1548008755920

Therefore my fOffilula yields that

= -19134440520821548008755920 + 309601751184956722026041= 0

Conclusion

My hypothesis was proved in the sense that rounding did cause Excel to mess up the reversion of

the Fibonacci sequence that is Excel is unable to compute the reverse Fibonacci sequence bn

accurately However I disproved my hypothesis in the sense that the equation

win always work to reverse the Fibonacci sequence if the fonowing three conditions are true

I The numbers aN and aN-l are known without any error

II The original Fibonacci numbers In are known without any error for 0 s n s N

11

III You work with a machine that knows how to mUltiply the large numbers in the formula

bn = (-l)n( -aN-lfn + aNfn-l)

There are many ways to build upon this project The first thing to do is to find out how one

can use Excel to mUltiply large numbers accurately The next step is to investigate the

generalized Fibonacci sequence

where A B are given values and ao and al are given starting values (this equation is also called

a second order linear difference equation)

A practical application of this project is the following banking problem You want to

make sure (for whatever reason) that no one can find out how much money you put in a savings

account where the money gets r interest every month In order to hide from everyone how

much money you put in the bank initially you strike the following deal with the bank

a) The bank never reveals the past of the savings account but only what is in the bank in month N

and month N-l (where N is the number of months the money sits in the savings account)

b) The bank pays interest on what is in the account with a delay of one month That is if an is

the amount of money in the bank at month n then an = an- l + TUn-2

c ) The bank always rounds an to the nearest penny

In this situation because of the rounding of an to the nearest penny it should be impossible to

reverse the sequence an That is given aN and aN-l for sufficiently large N it should be

impossible for anyone to ever find out for sure what ao and al were

12

Works Cited

Burger Edward B and Michael P Starbird Coincidences Chaos and all that Math Jazz

Making Light of Weighty Ideas New York WW Norton 2005

Derbyshire John Unknown Quantity A Real and Imaginary History of Algebra New York

Plume 2007

Fibonacci Leonardo Fibonaccis Liber Abaci A Translation into Modern English of

Leonardo Pisanos Book of Calculation Trans LE Sigler New York Springer 2003

Fibonacci sequence Art of Problem Solving 8 Sept 2011 lt httpwwwartofproblem

solvingcom WikiindexphplFibonacci_sequencegt

Knott Dr Ron Fibonacci Numbers the Golden Section and the Golden String

Mathematics - University of Surrey - Guilford Mathematics Department of the

University of Surrey UK 8 Sept 2011 lt httpwwwmathssurreyacuklhostecishy

sitesRKnottiFibonaccigt

Livio Mario The Golden Ratio The Story of Phi the W orIds Most Astonishing Number

New York Broadway Books 2002

Willers MichaeL A~gebra The x and y of Everyday Math New York Fall River Press 2009

13

where Xl = 1+2fS and X2 =1-2fS and where In denotes the standard Fibonacci sequence starting

with 0 and 1 as given in Table 1 To see how my formula works let us go back to the Fibonacci

numbers starting with 0 and 02 as in Table 4 Then

a60 = 309601751184 and a 59 = 1913444052082

159 = 956722026041 and 160 = 1548008755920

Therefore my fOffilula yields that

= -19134440520821548008755920 + 309601751184956722026041= 0

Conclusion

My hypothesis was proved in the sense that rounding did cause Excel to mess up the reversion of

the Fibonacci sequence that is Excel is unable to compute the reverse Fibonacci sequence bn

accurately However I disproved my hypothesis in the sense that the equation

win always work to reverse the Fibonacci sequence if the fonowing three conditions are true

I The numbers aN and aN-l are known without any error

II The original Fibonacci numbers In are known without any error for 0 s n s N

11

III You work with a machine that knows how to mUltiply the large numbers in the formula

bn = (-l)n( -aN-lfn + aNfn-l)

There are many ways to build upon this project The first thing to do is to find out how one

can use Excel to mUltiply large numbers accurately The next step is to investigate the

generalized Fibonacci sequence

where A B are given values and ao and al are given starting values (this equation is also called

a second order linear difference equation)

A practical application of this project is the following banking problem You want to

make sure (for whatever reason) that no one can find out how much money you put in a savings

account where the money gets r interest every month In order to hide from everyone how

much money you put in the bank initially you strike the following deal with the bank

a) The bank never reveals the past of the savings account but only what is in the bank in month N

and month N-l (where N is the number of months the money sits in the savings account)

b) The bank pays interest on what is in the account with a delay of one month That is if an is

the amount of money in the bank at month n then an = an- l + TUn-2

c ) The bank always rounds an to the nearest penny

In this situation because of the rounding of an to the nearest penny it should be impossible to

reverse the sequence an That is given aN and aN-l for sufficiently large N it should be

impossible for anyone to ever find out for sure what ao and al were

12

Works Cited

Burger Edward B and Michael P Starbird Coincidences Chaos and all that Math Jazz

Making Light of Weighty Ideas New York WW Norton 2005

Derbyshire John Unknown Quantity A Real and Imaginary History of Algebra New York

Plume 2007

Fibonacci Leonardo Fibonaccis Liber Abaci A Translation into Modern English of

Leonardo Pisanos Book of Calculation Trans LE Sigler New York Springer 2003

Fibonacci sequence Art of Problem Solving 8 Sept 2011 lt httpwwwartofproblem

solvingcom WikiindexphplFibonacci_sequencegt

Knott Dr Ron Fibonacci Numbers the Golden Section and the Golden String

Mathematics - University of Surrey - Guilford Mathematics Department of the

University of Surrey UK 8 Sept 2011 lt httpwwwmathssurreyacuklhostecishy

sitesRKnottiFibonaccigt

Livio Mario The Golden Ratio The Story of Phi the W orIds Most Astonishing Number

New York Broadway Books 2002

Willers MichaeL A~gebra The x and y of Everyday Math New York Fall River Press 2009

13

III You work with a machine that knows how to mUltiply the large numbers in the formula

bn = (-l)n( -aN-lfn + aNfn-l)

There are many ways to build upon this project The first thing to do is to find out how one

can use Excel to mUltiply large numbers accurately The next step is to investigate the

generalized Fibonacci sequence

where A B are given values and ao and al are given starting values (this equation is also called

a second order linear difference equation)

A practical application of this project is the following banking problem You want to

make sure (for whatever reason) that no one can find out how much money you put in a savings

account where the money gets r interest every month In order to hide from everyone how

much money you put in the bank initially you strike the following deal with the bank

a) The bank never reveals the past of the savings account but only what is in the bank in month N

and month N-l (where N is the number of months the money sits in the savings account)

b) The bank pays interest on what is in the account with a delay of one month That is if an is

the amount of money in the bank at month n then an = an- l + TUn-2

c ) The bank always rounds an to the nearest penny

In this situation because of the rounding of an to the nearest penny it should be impossible to

reverse the sequence an That is given aN and aN-l for sufficiently large N it should be

impossible for anyone to ever find out for sure what ao and al were

12

Works Cited

Burger Edward B and Michael P Starbird Coincidences Chaos and all that Math Jazz

Making Light of Weighty Ideas New York WW Norton 2005

Derbyshire John Unknown Quantity A Real and Imaginary History of Algebra New York

Plume 2007

Fibonacci Leonardo Fibonaccis Liber Abaci A Translation into Modern English of

Leonardo Pisanos Book of Calculation Trans LE Sigler New York Springer 2003

Fibonacci sequence Art of Problem Solving 8 Sept 2011 lt httpwwwartofproblem

solvingcom WikiindexphplFibonacci_sequencegt

Knott Dr Ron Fibonacci Numbers the Golden Section and the Golden String

Mathematics - University of Surrey - Guilford Mathematics Department of the

University of Surrey UK 8 Sept 2011 lt httpwwwmathssurreyacuklhostecishy

sitesRKnottiFibonaccigt

Livio Mario The Golden Ratio The Story of Phi the W orIds Most Astonishing Number

New York Broadway Books 2002

Willers MichaeL A~gebra The x and y of Everyday Math New York Fall River Press 2009

13

Works Cited

Burger Edward B and Michael P Starbird Coincidences Chaos and all that Math Jazz

Making Light of Weighty Ideas New York WW Norton 2005

Derbyshire John Unknown Quantity A Real and Imaginary History of Algebra New York

Plume 2007

Fibonacci Leonardo Fibonaccis Liber Abaci A Translation into Modern English of

Leonardo Pisanos Book of Calculation Trans LE Sigler New York Springer 2003

Fibonacci sequence Art of Problem Solving 8 Sept 2011 lt httpwwwartofproblem

solvingcom WikiindexphplFibonacci_sequencegt

Knott Dr Ron Fibonacci Numbers the Golden Section and the Golden String

Mathematics - University of Surrey - Guilford Mathematics Department of the

University of Surrey UK 8 Sept 2011 lt httpwwwmathssurreyacuklhostecishy

sitesRKnottiFibonaccigt

Livio Mario The Golden Ratio The Story of Phi the W orIds Most Astonishing Number

New York Broadway Books 2002

Willers MichaeL A~gebra The x and y of Everyday Math New York Fall River Press 2009

13