avewright/proofwiki-math huggingface dataset:
Theorem: Scalar Multiple of Function of Exponential Order

Let $f: \R \to \F$ be a function, where $\F \in \set {\R, \C}$.
Let $\lambda$ be a complex constant.
Suppose $f$ is of exponential order $a$.
Then $\lambda f$ is also of exponential order $a$.

Proof:
If $\lambda = 0$, the theorem holds trivially.
Let $\lambda \ne 0$.
{{begin-eqn}}
{{eqn | l = \size {\map f t}
      | o = <
      | r = K e^{a t}
      | c = {{Defof|Exponential Order to Real Index}}
}}
{{eqn | ll= \leadsto
      | l = \size \lambda \size {\map f t}
      | o = <
      | r = \size \lambda K e^{a t}
}}
{{eqn | ll= \leadsto
      | l = \size {\lambda \, \map f t}
      | o = <
      | r = K' e^{a t}
      | c = Modulus of Product, $K' = \size \lambda K$
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
mp7qgw2si88xgwoe25abb571f11yrtc
Theorem: Function of Exponential Order of Scalar Multiple

Let $f: \R \to \F$ be a function, where $\F \in \set {\R, \C}$.
Let $\lambda$ be a real constant.
Let $\map f t$ be of exponential order $a$.
Then the function defined by $t \mapsto \map f {\lambda t}$ is of exponential order $a\lambda$.

Proof:
{{begin-eqn}}
{{eqn | l = \size {\map f t}
      | o = <
      | r = K e^{a t}
      | c = {{Defof|Exponential Order to Real Index}}
}}
{{eqn | ll= \leadsto
      | l = \size {\map f {\lambda t} }
      | o = <
      | r = K e^{a \lambda t}
      | c = replacing $t$ with $\lambda t$
}}
{{end-eqn}}
The result follows by the definition of exponential order.
{{qed}}
Category:Exponential Order
ddyn9d7knsd6ra2jdgv0fe7a5nmip37
Theorem: Identity is of Exponential Order Epsilon

Let $I_\R: t \mapsto t$ be the identity mapping on $\R_{\ge 0}$.
Then $I_\R$ is of exponential order $\epsilon$ for any  $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
{{begin-eqn}}
{{eqn | l = e^{\epsilon t}
      | o = \ge
      | r = 1 + \epsilon t
      | c = Exponential of $t$ not less than $1 + t$
}}
{{eqn | o = >
      | r = \epsilon t
}}
{{eqn | ll= \leadsto
      | l = K e^{\epsilon t}
      | o = >
      | r = t
      | c = $K = \dfrac 1 \epsilon$
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
id26q54j2usaswssa7j8ydooc0fbkys
Theorem: Product of Functions of Exponential Order

Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Let $f$ be of exponential order $a$ and $g$ be of exponential order $b$.
Then $f g: t \mapsto \map f t \map g t$ is of exponential order $a+b$.

Proof:
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
{{begin-eqn}}
{{eqn | l = \size {\map f t}
      | o = <
      | r = K_1 e^{a t}
}}
{{eqn | l = \size {\map g t}
      | o = <
      | r = K_2 e^{b t}
}}
{{eqn | ll= \leadsto
      | l = \size {\map f t} \size {\map g t}
      | o = <
      | r = K_1 K_2 e^{a t} e^{b t}
      | c = Positive Real Number Inequalities can be Multiplied
}}
{{eqn | ll= \leadsto
      | l = \size {\map f t \map g t} 
      | o = <
      | r = K' e^{\paren {a + b} t}
      | c = Modulus of Product, Exponential of Sum of Real Numbers, $K' = K_1 K_2$
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
hldgyyoafpyrbrfj8s9wk7dg5sp6lg8
Theorem: Sum of Functions of Exponential Order

Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $f + g: t \mapsto \map f t + \map g t$ is of exponential order $\max \set {a, b}$.

Proof:
Let $t$ be sufficiently large so that both $f$ and $g$ are of exponential order on some shared unbounded closed interval.
By the definition of exponential order:
{{begin-eqn}}
{{eqn | l = \size {\map f t}
      | o = <
      | r = K_1 e^{a t}
}}
{{eqn | l = \size {\map g t}
      | o = <
      | r = K_2 e^{b t}
}}
{{eqn | l = \size {\map f t} + \size {\map g t}
      | o = <
      | r = K_1 e^{a t} + K_2 e^{b t}
      | c = Real Number Inequalities can be Added
}}
{{eqn | ll= \leadsto
      | l = \size {\map f t + \map g t}
      | o = < 
      | r = K_1 e^{\max \size {a, b} t} + K_2 e^{\max \size {a, b} t}
      | c = Triangle Inequality for Real Numbers, Exponential is Strictly Increasing
}}
{{eqn | r = K' e^{\max \size {a, b} t}
      | c = $K' = K_1 + K_2$
}}
{{end-eqn}}
{{qed}}
Category:Exponential Order
ev9cglofxub0yhotwdqt0tv8ec14ek9
Theorem: Linear Combination of Functions of Exponential Order

Let $f, g: \R \to \F$ be functions, where $\F \in \set {\R, \C}$.
Let $\lambda, \mu$ be complex numbers.
Suppose $f$ is of exponential order $a$ and $g$ is of exponential order $b$.
Then $\map {\paren {\lambda f + \mu g} } t = \lambda \, \map f t + \mu \, \map g t$ is of exponential order $\max \set {a, b}$.

Proof:
Follows from:
* Scalar Multiple of Function of Exponential Order
* Sum of Functions of Exponential Order
{{qed}}
Category:Exponential Order
l0rjg4v33skvpaty328lncl3s7n5zut
Theorem: Constant Function is of Exponential Order Zero

Let $f_C: \R \to \GF: t \mapsto C$ be a constant function, where $\GF \in \set {\R, \C}$.
Then $f_C$ is of exponential order $0$.

Proof:
{{begin-eqn}}
{{eqn | q = \forall t \ge 1
      | l = \size C
      | o = <
      | r = \size C + 1
}}
{{eqn | r = \paren {\size C + 1} e^{0 t}
      | c = Exponential of Zero
}}
{{end-eqn}}
The result follows from the definition of exponential order, with $M = 1$, $K = \size C + 1$, and $a = 0$.
{{qed}}
Category:Exponential Order
aziej8xmlc47bhcq0tlf69wt947phhb
Theorem: Polynomial is of Exponential Order Epsilon

Let $P: \R \to \mathbb F$ be a polynomial, where $\mathbb F \in \set {\R, \C}$.
Then $P$ is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
If $P = 0$, the theorem holds trivially. 
Let $P_n$ be a polynomial of degree $n$, where $n \ge 0$.
The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial.
Theorem: Fibonacci Number by Power of 2

{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
      | l = 2^{n - 1} F_n
      | r = \sum_k 5^k \dbinom n {2 k + 1}
      | c = 
}}
{{eqn | r = \dbinom n 1 + 5 \dbinom n 3 + 5^2 \dbinom n 5 + \cdots
      | c = 
}}
{{end-eqn}}
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\dbinom n {2 k + 1} \ $ denotes a binomial coefficient.

Proof:
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds 2^{n - 1} F_n = \sum_k 5^k \dbinom n {2 k + 1}$
First note the bounds of the summation.
By definition, $\dbinom n k = 0$ where $k < 0$ or $k > n$.
Thus in all cases in the following, terms outside the range $0 \le k \le n$ can be discarded.
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = 2^{-1} F_0
      | r = 0
      | c = {{Defof|Fibonacci Numbers}}: $F_0 = 0$
}}
{{eqn | q = \forall k \in \Z
      | r = 5^k \dbinom 0 {2 k + 1}
      | c = Zero Choose n
}}
{{end-eqn}}
Thus $\map P 0$ is seen to hold.
=== Basis for the Induction ===
$\map P 1$ is the case:
{{begin-eqn}}
{{eqn | l = 2^0 F_1
      | r = 1
      | c = {{Defof|Fibonacci Numbers}}: $F_1 = 1$
}}
{{eqn | r = 5^0 \dbinom 1 {2 \times 0 + 1}
      | c = One Choose n
}}
{{end-eqn}}
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:$\ds 2^{r - 1} F_r = \sum_k 5^k \dbinom r {2 k + 1}$
from which it is to be shown that:
:$\ds 2^r F_{r + 1} = \sum_k 5^k \dbinom {r + 1} {2 k + 1}$
=== Induction Step ===
This is the induction step:
{{begin-eqn}}
{{eqn | l = 2^r F_{r + 1}
      | r = 2^r \paren {F_{r - 1} + F_r}
      | c = {{Defof|Fibonacci Numbers}}
}}
{{eqn | r = 4 \times 2^{r - 2} F_{r - 1} + 2 \times 2^{r - 1} F_r
      | c = 
}}
{{eqn | r = 4 \times \sum_k 5^k \dbinom {r - 1} {2 k + 1} + 2 \sum_k 5^k \dbinom r {2 k + 1}
      | c = Induction Hypothesis
}}
{{eqn | r = 2 \times \sum_k 5^k \dbinom {r - 1} {2 k + 1} + 2 \sum_k 5^k \paren {\dbinom {r - 1} {2 k + 1} + \dbinom r {2 k + 1} }
      | c = 
}}
{{end-eqn}}
{{finish|To handle the awkward boundary cases, consider splitting the cases up into odd and even $r$}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \ds 2^{n - 1} F_n = \sum_k 5^k \dbinom n {2 k + 1}$
{{qed}}
Theorem: Fibonacci Number by Power of 2

{{begin-eqn}}
{{eqn | q = \forall n \in \Z_{\ge 0}
      | l = 2^{n - 1} F_n
      | r = \sum_k 5^k \dbinom n {2 k + 1}
      | c = 
}}
{{eqn | r = \dbinom n 1 + 5 \dbinom n 3 + 5^2 \dbinom n 5 + \cdots
      | c = 
}}
{{end-eqn}}
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\dbinom n {2 k + 1} \ $ denotes a binomial coefficient.

Proof:
{{begin-eqn}}
{{eqn | l = 2^{n - 1} F_n
      | r = \dfrac {2^n} {2 \sqrt 5} \paren {\phi^n - \hat \phi^n}
      | c = Euler-Binet Formula
}}
{{eqn | r = \dfrac {\paren {1 + \sqrt 5}^n - \paren {1 - \sqrt 5}^n} {2 \sqrt 5}
      | c = {{Defof|Golden Mean|index = 2}}
}}
{{eqn | r = \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \binom n j \sqrt 5^j - \dfrac 1 {2 \sqrt 5} \sum_{j \mathop = 0}^n \paren {-1}^j \binom n j \sqrt 5^j
      | c = Binomial Theorem
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \sum_{\substack {0 \mathop \le j \mathop \le n \\ j \text { odd} } } \binom n j \sqrt 5^j
      | c = even terms vanish, odd terms double up
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \sum_{j \text { odd} } \binom n j 5^{j / 2}
      | c = {{Defof|Binomial Coefficient}}: $\dbinom n j = 0$ for $j < 0$ and $j > n$
}}
{{eqn | r = \sum_{j \text { odd} } \binom n j 5^{\paren {j - 1} / 2}
      | c = gathering the spare $\sqrt 5$ into the index
}}
{{end-eqn}}
Setting $j = 2 k + 1$ for $0 \le k \le \paren {j - 1} / 2$ gives:
:$\ds \sum_{k \mathop \ge 0} \binom n {2 k + 1} 5^k$
and the limits of the index of the summation are irrelevant, as $\dbinom n {2 k + 1} = 0$ for $j < 0$ and $j > n$.
Hence the result.
Theorem: Ratio of 2016 to Aliquot Sum

$2016$ has the property that its ratio to its aliquot sum is $4 : 9$.

Proof:
The aliquot sum of an integer $n$ is the integer sum  of the aliquot parts of $n$.
That is, the aliquot sum of $2016$ is the divisor sum of $2016$ minus $2016$.
Thus:
{{begin-eqn}}
{{eqn | l = \map {\sigma_1} {2016} - 2016
      | r = 6552 - 2016
      | c = {{DSFLink|2016}}
}}
{{eqn | r = 4536
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 4 \times \paren {\map {\sigma_1} {2016} - 2016}
      | r = 18 \, 144
      | c = 
}}
{{eqn | r = 9 \times 2016
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Approximation to Golden Rectangle using Fibonacci Squares

An approximation to a golden rectangle can be obtained by placing adjacent to one another squares with side lengths corresponding to consecutive Fibonacci numbers in the following manner:
:800px
It can also be noted, as from Sequence of Golden Rectangles, that an equiangular spiral can be approximated by constructing quarter circles as indicated.

Proof:
Let the last two squares to be added have side lengths of $F_{n - 1}$ and $F_n$.
Then from the method of construction, the sides of the rectangle generated will be $F_n$ and $F_{n + 1}$.
From Continued Fraction Expansion of Golden Mean it follows that the limit of the ratio of the side lengths of the rectangle, as $n$ tends to infinity, is the golden section $\phi$.
Hence the result.
{{qed}}
Theorem: Approximation to Golden Rectangle using Fibonacci Squares

An approximation to a golden rectangle can be obtained by placing adjacent to one another squares with side lengths corresponding to consecutive Fibonacci numbers in the following manner:
:800px
It can also be noted, as from Sequence of Golden Rectangles, that an equiangular spiral can be approximated by constructing quarter circles as indicated.

Proof:
From Sum of Sequence of Squares of Fibonacci Numbers:
:$\forall n \ge 1: \ds \sum_{j \mathop = 1}^n {F_j}^2 = F_n F_{n + 1}$
Hence the result.
{{qed}}
Theorem: Fibonacci Number of Index 2n as Sum of Squares of Fibonacci Numbers

Let $F_n$ denote the $n$th Fibonacci number.
Then:
:$F_{2 n} = {F_{n + 1} }^2 - {F_{n - 1} }^2$

Proof:
From Honsberger's Identity:
:$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Setting $m = n$:
{{begin-eqn}}
{{eqn | l = F_{2 n}
      | r = F_{n - 1} F_n + F_n F_{n + 1}
      | c = 
}}
{{eqn | r = F_n \paren {F_{n + 1} + F_{n - 1} }
      | c = 
}}
{{eqn | r = \paren {F_{n + 1} - F_{n - 1} } \paren {F_{n + 1} + F_{n - 1} }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | r = {F_{n + 1} }^2 - {F_{n - 1} }^2
      | c = Difference of Two Squares
}}
{{end-eqn}}
{{qed}}
Theorem: Fibonacci Number of Index 3n as Sum of Cubes of Fibonacci Numbers

Let $F_n$ denote the $n$th Fibonacci number.
Then:
:$F_{3 n} = {F_{n + 1} }^3 + {F_n}^3 - {F_{n - 1} }^3$

Proof:
From Honsberger's Identity:
:$\forall m, n \in \Z_{>0}: F_{m + n} = F_{m - 1} F_n + F_m F_{n + 1}$
Setting $m = 2 n$:
{{begin-eqn}}
{{eqn | l = F_{3 n}
      | r = F_{2 n + n}
      | c = 
}}
{{eqn | r = F_{2 n - 1} F_n + F_{2 n} F_{n + 1}
      | c = Honsberger's Identity
}}
{{eqn | r = F_{n + \paren{n - 1} } F_n + F_{2 n} F_{n + 1}
      | c = 
}}
{{eqn | r = \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + F_{2 n} F_{n + 1}
      | c = Honsberger's Identity
}}
{{eqn | r = \paren { {F_{n - 1} }^2 + {F_n}^2} F_n + \paren { {F_{n + 1} }^2 - {F_{n - 1} }^2} F_{n + 1}
      | c = Fibonacci Number of Index 2n as Sum of Squares of Fibonacci Numbers
}}
{{eqn | r = {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 F_{n + 1}
      | c = rearranging
}}
{{eqn | r = {F_{n - 1} }^2 F_n + {F_n}^3 + {F_{n + 1} }^3 - {F_{n - 1} }^2 \paren {F_n + F_{n - 1} }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | r = {F_{n + 1} }^3 + {F_n}^3 - {F_{n - 1} }^3
      | c = simplifying
}}
{{end-eqn}}
Theorem: Pythagorean Triangle from Fibonacci Numbers

Take $4$ consecutive Fibonacci numbers:
:$F_n, F_{n + 1}, F_{n + 2}, F_{n + 3}$
Let:
:$a := F_n F_{n + 3}$
:$b := 2 F_{n + 1} F_{n + 2}$
:$c := F_{2 n + 3}$
Then:
:$a^2 + b^2 = c^2$
and:
:$\dfrac {a b} 2 = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}$
That is, if the legs of a right triangle are the product of the outer terms and twice the inner terms, then:
:the hypotenuse is the Fibonacci number whose index is half the sum of the indices of the four given Fibonacci numbers.
:the area is the product of the four given Fibonacci numbers.

Proof:
By definition of Fibonacci numbers:
:$F_n = F_{n + 2} - F_{n + 1}$
and:
:$F_{n + 3} = F_{n + 2} + F_{n + 1}$
Then $a$ can be expressed as:
{{begin-eqn}}
{{eqn | l = a
      | r = \paren {F_{n + 2} - F_{n + 1} } \paren {F_{n + 2} + F_{n + 1} }
      | c = 
}}
{{eqn | r = {F_{n + 2} }^2 - {F_{n + 1} }^2
      | c = 
}}
{{end-eqn}}
Now consider:
{{begin-eqn}}
{{eqn | r = F_{n + 2} F_{n + 3} - F_n F_{n + 1}
      | o = 
      | c = 
}}
{{eqn | r = F_{n + 2} \paren {F_{n + 2} + F_{n + 1} } - \paren {F_{n + 2} - F_{n + 1} } F_{n + 1}
      | c = 
}}
{{eqn | r = {F_{n + 2} }^2 + {F_{n + 1} }^2
      | c = multiplying out
}}
{{end-eqn}}
and it can be seen that the ordered triple:
:$\paren {{F_{n + 2} }^2 - {F_{n + 1} }^2, 2 F_{n + 1} F_{n + 2}, {F_{n + 2} }^2 + {F_{n + 1} }^2}$
is in the form of Solutions of Pythagorean Equation.
It remains to be shown that:
:${F_{n + 2} }^2 + {F_{n + 1} }^2 = F_{2 n + 3}$
We have:
{{begin-eqn}}
{{eqn | l = F_{2 n + 3}
      | r = F_{\paren {n + 1} + \paren {n + 2} }
      | c = 
}}
{{eqn | r = F_{\paren {n + 2} } F_{\paren {n + 1} + 1} + F_{\paren {n + 2} - 1} F_{\paren {n + 1} }
      | c = Honsberger's Identity
}}
{{eqn | r = {F_{n + 2} }^2 + {F_{n + 1} }^2
      | c = 
}}
{{end-eqn}}
and the proof is complete.
{{qed|lemma}}
The area is trivial:
{{begin-eqn}}
{{eqn | l = a
      | r = F_n F_{n + 3}
      | c = 
}}
{{eqn | l = b
      | r = 2 F_{n + 1} F_{n + 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {a b} 2
      | o = =
      | r = \frac {F_n F_{n + 3} \times 2 F_{n + 1} F_{n + 2} } 2
      | c = 
}}
{{eqn | r = F_n \times F_{n + 1} \times F_{n + 2} \times F_{n + 3}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: No 4 Fibonacci Numbers can be in Arithmetic Sequence

Let $a, b, c, d$ be distinct Fibonacci numbers.
Then, except for the trivial case:
:$a = 0, b = 1, c = 2, d = 3$
it is not possible that $a, b, c, d$ are in arithmetic sequence.

Proof:
Let:
:$a = F_i, b = F_j, c = F_k, d = F_l$
where $F_n$ denotes the $n$th Fibonacci number.
{{WLOG}}, further suppose that;
:$a < b < c < d$
or equivalently:
:$i < j < k < l$
Since $i, j, k, l$ are integers, the inequality could be written as:
:$i \le j - 1 \le k - 2 \le l - 3$
Now consider:
{{begin-eqn}}
{{eqn | l = d - c
      | r = F_l - F_k
      | c = 
}}
{{eqn | o = \ge
      | r = F_l - F_{l - 1}
      | c = By assumption, $k - 2 \le l - 3$
}}
{{eqn | r = F_{l - 2}
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | o = \ge
      | r = F_j
      | c = By assumption, $j - 1 \le l - 3$
}}
{{eqn | o = \ge
      | r = F_j - F_i
      | c = 
}}
{{eqn | r = b - a
      | c = 
}}
{{end-eqn}}
For $a, b, c, d$ be in arithmetic sequence:
:$d - c = b - a$
This means that the all the inequalities above must be equalities:
:$F_l - F_k = F_l - F_{l - 1}$
:$F_{l - 2} = F_j$
:$F_j = F_j - F_i$
So:
:$F_i = 0$
and:
:$F_k = F_{l - 1}$
:$F_j = F_{l - 2}$ 
The only Fibonacci numbers having different index but have the same value is $F_1 = F_2 = 1$.
So one of the following is true:
:$F_k = F_{l - 1} = 1$
:$F_j = F_{l - 2} = 1$ 
:$j - 1 = k - 2 = l - 3$
Suppose the third statement is true.
Write $k = j + 1$, $l = j + 2$.
Then:
{{begin-eqn}}
{{eqn | l = F_{j + 2} - F_{j + 1}
      | r = F_{j + 1} - F_j
      | c = $F_j, F_{j + 1}, F_{j + 2}$ form an arithmetic sequence
}}
{{eqn | ll = \leadsto
      | l = F_j
      | r = F_{j - 1}
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | ll = \leadsto
      | l = F_j - F_{j - 1}
      | r = 0
      | c = 
}}
{{eqn | ll = \leadsto
      | l = F_{j - 2}
      | r = 0
      | c = {{Defof|Fibonacci Number}}
}}
{{end-eqn}}
The only zero term of the Fibonacci numbers is $F_0$.
This gives $j = 2$.
Therefore the only arithmetic sequence among Fibonacci numbers satisfying the condition above is:
:$F_0, F_2, F_3, F_4$
which corresponds to:
:$0, 1, 2, 3$
Now suppose $F_j = 1$.
Since $F_i, F_j, F_k, F_l$ form an arithmetic sequence:
:$F_k = F_j + \paren {F_j - F_i} = 2$ 
:$F_l = F_k + \paren {F_j - F_i} = 3$ 
Which again gives the arithmetic sequence $0, 1, 2, 3$.
Finally suppose $F_k = 1$. 
Since $F_i, F_j, F_k$ form an arithmetic sequence:
:$F_j = \dfrac 1 2 \paren {F_i + F_k} = \dfrac 1 2$
which is not an integer.
So $F_k \ne 1$.
All cases have been accounted for, and the only arithmetic sequence that can be formed is $0, 1, 2, 3$.
{{qed}}
Theorem: Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 2} \left({-1}\right)^k \dfrac 1 {F_k F_{k + 1} }
      | r = \dfrac 1 {1 \times 2} - \dfrac 1 {2 \times 3} + \dfrac 1 {3 \times 5} - \dfrac 1 {5 \times 8} + \cdots
      | c = 
}}
{{eqn | r = \phi^{-2}
      | c = 
}}
{{end-eqn}}
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 2} \paren {-1}^k \dfrac 1 {F_k F_{k + 1} }
      | r = \sum_{k \mathop \ge 2} \paren {F_{k + 1} F_{k - 1} - F_k^2} \dfrac 1 {F_k F_{k + 1} }
      | c = Cassini's Identity
}}
{{eqn | r = \sum_{k \mathop \ge 2} \paren {\dfrac {F_{k - 1} } {F_k} - \dfrac {F_k} {F_{k + 1} } }
      | c = 
}}
{{eqn | r = \dfrac {F_1} {F_2} - \lim_{k \mathop \to \infty} \dfrac {F_k} {F_{k + 1} }
      | c = Telescoping Series
}}
{{eqn | r = 1 - \dfrac 1 \phi
      | c = Ratio of Consecutive Fibonacci Numbers
}}
{{eqn | r = \phi^{-1} \paren {\phi - 1}
      | c = 
}}
{{eqn | r = \phi^{-2}
      | c = {{Defof|Golden Mean}}
}}
{{end-eqn}}
{{qed}}
Theorem: Sum of Reciprocals of Sequence of Pairs of Even Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared

{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k} F_{2 k + 2} }
      | r = \dfrac 1 {1 \times 3} + \dfrac 1 {3 \times 8} + \dfrac 1 {8 \times 21} + \dfrac 1 {21 \times 55} + \cdots
      | c = 
}}
{{eqn | r = \phi^{-2}
      | c = 
}}
{{end-eqn}}
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.

Proof:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k} F_{2 k + 2} }
      | r = \sum_{k \mathop \ge 1} \dfrac 1 {F_{2 k} F_{2 k + 2} } \paren {\dfrac {F_{2 k + 2} - F_{2 k} } {F_{2 k + 2} - F_{2 k} } }
      | c = multiplying by $1$
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} } - \dfrac 1 {F_{2 k + 2} } } \paren {\dfrac 1 {F_{2 k + 2} - F_{2 k} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} } - \dfrac 1 {F_{2 k + 2} } } \paren {\dfrac 1 {\paren {F_{2 k + 1} + F_{2 k} } - F_{2 k} } }
      | c = {{Defof|Fibonacci Number}}
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} } - \dfrac 1 {F_{2 k + 2} } } \paren {\dfrac 1 {F_{2 k + 1} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 1} \paren {\dfrac 1 {F_{2 k} F_{2 k + 1} } - \dfrac 1 {F_{2 k + 1} F_{2 k + 2} } }
      | c = 
}}
{{eqn | r = \sum_{k \mathop \ge 2} \paren {-1}^k \dfrac 1 {F_k F_{k + 1} }
      | c = 
}}
{{eqn | r = \phi^{-2}
      | c = Sum of Alternating Sign Reciprocals of Sequence of Pairs of Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared
}}
{{end-eqn}}
{{qed}}
Theorem: Number of Fibonacci Numbers between n and 2n

Let $n \in \Z_{> 0}$ be a (strictly) positive integer.
Then there exists either one or two Fibonacci numbers between $n$ and $2 n$ inclusive.

Proof:
First existence is demonstrated.
Let $F_m \ge n$ such that $F_{m - 1} < n$.
{{begin-eqn}}
{{eqn | l = F_m
      | r = F_{m - 1} + F_{m - 2}
      | c = Definition of Fibonacci Numbers
}}
{{eqn | o = <
      | r = 2 F_{m - 1}
      | c = as $F_{m - 2} < F_{m - 1}$
}}
{{eqn | o = <
      | r = 2 n
      | c = as $F_{m - 1} < n$
}}
{{end-eqn}}
This shows that the smallest Fibonacci number greater than $n$ is less than $2 n$.
Thus there exists at least one Fibonacci number between $n$ and $2 n$.
{{AimForCont}} there exist $3$ Fibonacci numbers between $n$ and $2 n$.
Let $F_m \ge n$ be the smallest of those Fibonacci numbers.
Then:
{{begin-eqn}}
{{eqn | l = F_{m + 2}
      | r = F_m + F_{m + 1}
      | c = Definition of Fibonacci Numbers
}}
{{eqn | o = >
      | r = 2 F_m
      | c = as $F_m < F_{m + 1}$
}}
{{eqn | o = >
      | r = 2 n
      | c = as $F_m > n$
}}
{{end-eqn}}
But $F_{m + 2} < 2 n$ by hypothesis.
Hence by Proof by Contradiction there can be no more than $2$ Fibonacci numbers between $n$ and $2 n$.
Let $n = 2$.
Then between $2$ and $4$ there exist $F_3 = 2$ and $F_4 = 3$.
Let $n = 10$.
Then between $10$ and $20$ there exists $F_7 = 13$ and no other Fibonacci numbers.
Thus it has been demonstrated:
:There always exists at least one Fibonacci number between $n$ and $2 n$
:There never exist more than $2$ Fibonacci number between $n$ and $2 n$
:There exist $n$ such that there exists exactly one Fibonacci number between $n$ and $2 n$
:There exist $n$ such that there exist exactly $2$ Fibonacci numbers between $n$ and $2 n$.
The result is complete.
{{qed}}
Theorem: Square Root is of Exponential Order Epsilon

The positive square root function:
:$t \mapsto \sqrt t$
is of exponential order $\epsilon$ for any  $\epsilon > 0$ arbitrarily small in magnitude.

Proof:
{{begin-eqn}}
{{eqn | l = \sqrt t
      | o = <
      | r = K e^{a t}
      | c = an Ansatz
}}
{{eqn | ll = \impliedby
      | l = t
      | o = <
      | r = \paren {K e^{a t} }^2
      | c = Square Root is Strictly Increasing
}}
{{eqn | r = K^2 e^{2 a t}
      | c = Exponential of Product
}}
{{eqn | r =  K' e^{a' t}
      | c = $K^2 = K', 2 a = a'$
}}
{{end-eqn}}
Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a' t}$ for ''any'' $a' > 0$, arbitrarily small in magnitude.
Therefore the inequality $\sqrt t < K e^{a t}$ has solutions of the same nature.
{{qed}}
Category:Exponential Order
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