id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-14600 | Vajda's Identity/Formulation 2 | :$F_{n + k} F_{m - k} - F_n F_m = \paren {-1}^n F_{m - n - k} F_k$ | We have:
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = \paren {x^{n + k} - y^{n + k} } \paren {x^{m - k} - y^{m - k} } - \paren {x^n - y^n} \paren {x^m - y^m}
| c =
}}
{{eqn | r = x^{n + m} + y^{n + m} - x^{m - k} y^{n + k} - x^{n + k} y^{m - k} - x^{n + m} - y^{n + m} + x^n y^m + x^m y^n
| c =
}}
... | :$F_{n + k} F_{m - k} - F_n F_m = \paren {-1}^n F_{m - n - k} F_k$ | We have:
{{begin-eqn}}
{{eqn | n = 1
| o =
| r = \paren {x^{n + k} - y^{n + k} } \paren {x^{m - k} - y^{m - k} } - \paren {x^n - y^n} \paren {x^m - y^m}
| c =
}}
{{eqn | r = x^{n + m} + y^{n + m} - x^{m - k} y^{n + k} - x^{n + k} y^{m - k} - x^{n + m} - y^{n + m} + x^n y^m + x^m y^n
| c =
}}... | Vajda's Identity/Formulation 2 | https://proofwiki.org/wiki/Vajda's_Identity/Formulation_2 | https://proofwiki.org/wiki/Vajda's_Identity/Formulation_2 | [
"Vajda's Identity"
] | [] | [
"Definition:Golden Mean",
"Reciprocal Form of One Minus Golden Mean",
"Euler-Binet Formula",
"Euler-Binet Formula"
] |
proofwiki-14601 | Sum of Squares of Consecutive Fibonacci Numbers | :${F_n}^2 + {F_{n + 1} }^2 = F_{2 n + 1}$
where $F_n$ denotes the $n$th Fibonacci number. | {{begin-eqn}}
{{eqn | l = {F_n}^2 + {F_{n + 1} }^2
| r = F_{n + 1} F_{n - 1} - \paren {-1}^n + F_{n + 2} F_n - \paren {-1}^{n + 1}
| c = Cassini's Identity
}}
{{eqn | r = F_n F_{n + 2} + F_{n - 1} F_{n + 1}
| c =
}}
{{eqn | r = F_{n + n + 1}
| c = Honsberger's Identity
}}
{{eqn | r = F_{2 n + 1... | :${F_n}^2 + {F_{n + 1} }^2 = F_{2 n + 1}$
where $F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | {{begin-eqn}}
{{eqn | l = {F_n}^2 + {F_{n + 1} }^2
| r = F_{n + 1} F_{n - 1} - \paren {-1}^n + F_{n + 2} F_n - \paren {-1}^{n + 1}
| c = [[Cassini's Identity]]
}}
{{eqn | r = F_n F_{n + 2} + F_{n - 1} F_{n + 1}
| c =
}}
{{eqn | r = F_{n + n + 1}
| c = [[Honsberger's Identity]]
}}
{{eqn | r = F_... | Sum of Squares of Consecutive Fibonacci Numbers | https://proofwiki.org/wiki/Sum_of_Squares_of_Consecutive_Fibonacci_Numbers | https://proofwiki.org/wiki/Sum_of_Squares_of_Consecutive_Fibonacci_Numbers | [
"Fibonacci Numbers"
] | [
"Definition:Fibonacci Number"
] | [
"Cassini's Identity",
"Honsberger's Identity"
] |
proofwiki-14602 | Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem | Let Hausdorff's Maximal Principle hold.
Then Zermelo's Well-Ordering Theorem holds. | {{tidy|Work in progress}}
Let $X$ be a non-empty set.
Let $X$ contain at least two elements; otherwise, $X$ can be trivially well-ordered.
We use Hausdorff's Maximal Principle to construct a well-ordering on $X$.
We note that there exists at least the following strict well-orderings on subsets of $X$:
For any singleto... | Let [[Hausdorff's Maximal Principle]] hold.
Then [[Zermelo's Well-Ordering Theorem]] holds. | {{tidy|Work in progress}}
Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $X$ contain at least two [[Definition:Element|elements]]; otherwise, $X$ can be trivially [[Definition:Well-Ordering|well-ordered]].
We use [[Hausdorff's Maximal Principle]] to construct a well-ordering on $X$.
We note that there ... | Hausdorff's Maximal Principle implies Zermelo's Well-Ordering Theorem | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_implies_Zermelo's_Well-Ordering_Theorem | https://proofwiki.org/wiki/Hausdorff's_Maximal_Principle_implies_Zermelo's_Well-Ordering_Theorem | [
"Zermelo's Well-Ordering Theorem",
"Hausdorff's Maximal Principle"
] | [
"Hausdorff's Maximal Principle",
"Zermelo's Well-Ordering Theorem"
] | [
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:Well-Ordering",
"Hausdorff's Maximal Principle",
"Definition:Strict Well-Ordering",
"Definition:Subset",
"Definition:Singleton",
"Definition:Vacuous Truth",
"Definition:Doubleton",
"Finite Totally Ordered Set is Well-Ordered",
"Defini... |
proofwiki-14603 | Cosine of 36 Degrees | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the golden mean. | Let $u = \cos 72 \degrees, v = \cos 36 \degrees$.
Recall from Cosine of Complement equals Sine that $\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \cos {90 \degrees - 18 \degrees}
| r = \map \sin {18 \degrees}
| c =
}}
{{eqn | ll= \leadsto
| l = \map \cos ... | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | Let $u = \cos 72 \degrees, v = \cos 36 \degrees$.
Recall from [[Cosine of Complement equals Sine]] that $\map \cos {\dfrac \pi 2 - \theta} = \sin \theta$
Therefore:
{{begin-eqn}}
{{eqn | l = \map \cos {90 \degrees - 18 \degrees}
| r = \map \sin {18 \degrees}
| c =
}}
{{eqn | ll= \leadsto
| l = \map... | Cosine of 36 Degrees/Proof 1 | https://proofwiki.org/wiki/Cosine_of_36_Degrees | https://proofwiki.org/wiki/Cosine_of_36_Degrees/Proof_1 | [
"Cosine of 36 Degrees",
"Cosine Function",
"Golden Mean"
] | [
"Definition:Golden Mean"
] | [
"Cosine of Complement equals Sine",
"Square of Golden Mean equals One plus Golden Mean"
] |
proofwiki-14604 | Cosine of 36 Degrees | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the golden mean. | From Complex Algebra: $z^4 - 3z^2 + 1 = 0$, the roots of $z^4 - 3z^2 + 1 = 0$ are:
:$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$
Then:
{{begin-eqn}}
{{eqn | l = z^4 - 3z^2 + 1
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = z^2
| r = \dfrac {3 \pm \sqrt {\paren... | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | From [[Complex Algebra/Examples/z^4 - 3z^2 + 1 = 0|Complex Algebra: $z^4 - 3z^2 + 1 = 0$]], the roots of $z^4 - 3z^2 + 1 = 0$ are:
:$2 \cos 36 \degrees, 2 \cos 72 \degrees, 2 \cos 216 \degrees, 2 \cos 252 \degrees$
Then:
{{begin-eqn}}
{{eqn | l = z^4 - 3z^2 + 1
| r = 0
| c =
}}
{{eqn | ll= \leadsto
... | Cosine of 36 Degrees/Proof 2 | https://proofwiki.org/wiki/Cosine_of_36_Degrees | https://proofwiki.org/wiki/Cosine_of_36_Degrees/Proof_2 | [
"Cosine of 36 Degrees",
"Cosine Function",
"Golden Mean"
] | [
"Definition:Golden Mean"
] | [
"Complex Algebra/Examples/z^4 - 3z^2 + 1 = 0",
"Solution to Quadratic Equation",
"Shape of Cosine Function",
"Square of Sum",
"Definition:Square Number"
] |
proofwiki-14605 | Cosine of 36 Degrees | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the golden mean. | {{begin-eqn}}
{{eqn | l = \sin 108 \degrees
| r = 3 \sin 36 \degrees - 4 \sin^3 36 \degrees
| c = Triple Angle Formula for Sine
}}
{{eqn | l = \sin 72 \degrees
| r = 3 \sin 36 \degrees - 4 \sin^3 36 \degrees
| c = Sine of Supplementary Angle
}}
{{eqn | l = 2 \sin 36 \degrees \cos 36 \degrees
... | :$\cos 36 \degrees = \cos \dfrac \pi 5 = \dfrac \phi 2 = 2^{-1} \phi = \dfrac {1 + \sqrt 5} 4$
where $\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | {{begin-eqn}}
{{eqn | l = \sin 108 \degrees
| r = 3 \sin 36 \degrees - 4 \sin^3 36 \degrees
| c = [[Triple Angle Formula for Sine]]
}}
{{eqn | l = \sin 72 \degrees
| r = 3 \sin 36 \degrees - 4 \sin^3 36 \degrees
| c = [[Sine of Supplementary Angle]]
}}
{{eqn | l = 2 \sin 36 \degrees \cos 36 \deg... | Cosine of 36 Degrees/Proof 3 | https://proofwiki.org/wiki/Cosine_of_36_Degrees | https://proofwiki.org/wiki/Cosine_of_36_Degrees/Proof_3 | [
"Cosine of 36 Degrees",
"Cosine Function",
"Golden Mean"
] | [
"Definition:Golden Mean"
] | [
"Triple Angle Formulas/Sine",
"Sine of Supplementary Angle",
"Double Angle Formulas/Sine",
"Sum of Squares of Sine and Cosine",
"Solution to Quadratic Equation",
"Definition:Square Root/Negative"
] |
proofwiki-14606 | Sine of 36 Degrees | :$\sin 36 \degrees = \sin \dfrac \pi 5 = \dfrac {\sqrt {\sqrt 5 / \phi} } 2 = 2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2}= \dfrac {\sqrt {3 - \phi} } 2 = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}$ | {{begin-eqn}}
{{eqn | l = \sin 36 \degrees
| r = \sqrt {1 - \cos^2 36 \degrees}
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | r = \sqrt {1 - \frac {\phi^2} 4}
| c = {{cos|36}}
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \phi^2}
| c =
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 2 + \frac {\... | :$\sin 36 \degrees = \sin \dfrac \pi 5 = \dfrac {\sqrt {\sqrt 5 / \phi} } 2 = 2^{-1} 5^{\frac 1 4} \phi^{-\frac 1 2}= \dfrac {\sqrt {3 - \phi} } 2 = \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}$ | {{begin-eqn}}
{{eqn | l = \sin 36 \degrees
| r = \sqrt {1 - \cos^2 36 \degrees}
| c = [[Sum of Squares of Sine and Cosine]]
}}
{{eqn | r = \sqrt {1 - \frac {\phi^2} 4}
| c = {{cos|36}}
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \phi^2}
| c =
}}
{{eqn | r = \frac 1 2 \sqrt {4 - \paren {\frac 1 2 + \fra... | Sine of 36 Degrees | https://proofwiki.org/wiki/Sine_of_36_Degrees | https://proofwiki.org/wiki/Sine_of_36_Degrees | [
"Sine Function",
"Golden Mean"
] | [] | [
"Sum of Squares of Sine and Cosine",
"Square of Sum"
] |
proofwiki-14607 | Partial Sums of Power Series with Fibonacci Coefficients | :<nowiki>$\ds \sum_{k \mathop = 0}^n F_k x^k = \begin {cases}
\dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\
\dfrac {\paren {n + 1} x^n F_{n + 1} + \paren {n + 2} x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end {cases}$</nowiki>
where $F_n$ denotes the $n$th Fibonacci numb... | Multiplying the summation by $x^2 + x - 1$:
{{begin-eqn}}
{{eqn | r = \sum_{k \mathop = 0}^n F_k x^k \paren {x^2 + x - 1}
| o =
| c =
}}
{{eqn | r = F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots + F_{n - 2} x^n + F_{n - 1} x^{n + 1} + F_n x^{n + 2}
| c =
}}
{{eqn | o =
| ro= +
| r = F_0 x + F_1... | :<nowiki>$\ds \sum_{k \mathop = 0}^n F_k x^k = \begin {cases}
\dfrac {x^{n + 1} F_{n + 1} + x^{n + 2} F_n - x} {x^2 + x - 1} & : x^2 + x - 1 \ne 0 \\
\dfrac {\paren {n + 1} x^n F_{n + 1} + \paren {n + 2} x^{n + 1} F_n - 1} {2 x + 1} & : x^2 + x - 1 = 0 \end {cases}$</nowiki>
where $F_n$ denotes the $n$th [[Definition:... | Multiplying the [[Definition:Summation|summation]] by $x^2 + x - 1$:
{{begin-eqn}}
{{eqn | r = \sum_{k \mathop = 0}^n F_k x^k \paren {x^2 + x - 1}
| o =
| c =
}}
{{eqn | r = F_0 x^2 + F_1 x^3 + F_2 x^4 + \cdots + F_{n - 2} x^n + F_{n - 1} x^{n + 1} + F_n x^{n + 2}
| c =
}}
{{eqn | o =
| ro=... | Partial Sums of Power Series with Fibonacci Coefficients | https://proofwiki.org/wiki/Partial_Sums_of_Power_Series_with_Fibonacci_Coefficients | https://proofwiki.org/wiki/Partial_Sums_of_Power_Series_with_Fibonacci_Coefficients | [
"Fibonacci Numbers",
"Real Power Series"
] | [
"Definition:Fibonacci Number"
] | [
"Definition:Summation",
"Definition:Fraction/Denominator",
"Definition:Fraction/Numerator",
"Definition:Differentiation",
"Definition:Fraction/Numerator",
"Definition:Fraction/Denominator",
"L'Hôpital's Rule",
"Power Rule for Derivatives"
] |
proofwiki-14608 | Sum over k of n Choose k by Fibonacci Number with index m+k | :$\ds \sum_{k \mathop \ge 0} \binom n k F_{m + k} = F_{m + 2 n}$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$F_n$ denotes the $n$th Fibonacci number. | From Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k:
:$(1): \quad \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
Letting $t = 2$ in $(1)$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k {F_2}^k {F_{2 - 1} }^{n - k} ... | :$\ds \sum_{k \mathop \ge 0} \binom n k F_{m + k} = F_{m + 2 n}$
where:
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | From [[Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k]]:
:$(1): \quad \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
Letting $t = 2$ in $(1)$:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop \ge 0} \binom n k {F_2}^k {F_{2 - 1} }^{n... | Sum over k of n Choose k by Fibonacci Number with index m+k | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_Fibonacci_Number_with_index_m+k | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_Fibonacci_Number_with_index_m+k | [
"Fibonacci Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Fibonacci Number"
] | [
"Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k"
] |
proofwiki-14609 | Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k | :$\ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
where:
:$\dbinom n k$ denotes a binomial coefficient
:$F_n$ denotes the $n$th Fibonacci number. | The proof proceeds by induction on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\forall m, t \in \N: \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
$\map P 0$ is the case:
:$\ds \binom 0 0 {F_t}^0 {F_{t - 1} }^0 F_{m + 0} = F_m$
Thus $\map P 0$ is se... | :$\ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
where:
:$\dbinom n k$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]]
:$F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | The proof proceeds by [[Principle of Mathematical Induction|induction]] on $n$.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\forall m, t \in \N: \ds \sum_{k \mathop \ge 0} \binom n k {F_t}^k {F_{t - 1} }^{n - k} F_{m + k} = F_{m + t n}$
$\map P 0$ is the case:
:$\ds \... | Sum over k of n Choose k by Fibonacci t to the k by Fibonacci t-1 to the n-k by Fibonacci m+k | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_Fibonacci_t_to_the_k_by_Fibonacci_t-1_to_the_n-k_by_Fibonacci_m+k | https://proofwiki.org/wiki/Sum_over_k_of_n_Choose_k_by_Fibonacci_t_to_the_k_by_Fibonacci_t-1_to_the_n-k_by_Fibonacci_m+k | [
"Fibonacci Numbers",
"Binomial Coefficients"
] | [
"Definition:Binomial Coefficient",
"Definition:Fibonacci Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14610 | Matrix whose Determinant is Fibonacci Number | The $n \times n$ determinant:
:<nowiki>$D_n = \begin{vmatrix}
1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 1 & 1 & -1 \\
0 & 0 & 0 & 0 & \cdots & 0 & ... | The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$D_n = F_{n + 1}$ | The $n \times n$ [[Definition:Determinant of Matrix|determinant]]:
:<nowiki>$D_n = \begin{vmatrix}
1 & -1 & 0 & 0 & \cdots & 0 & 0 & 0 \\
1 & 1 & -1 & 0 & \cdots & 0 & 0 & 0 \\
0 & 1 & 1 & -1 & \cdots & 0 & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 1 & 1... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$D_n = F_{n + 1}$ | Matrix whose Determinant is Fibonacci Number | https://proofwiki.org/wiki/Matrix_whose_Determinant_is_Fibonacci_Number | https://proofwiki.org/wiki/Matrix_whose_Determinant_is_Fibonacci_Number | [
"Fibonacci Numbers",
"Determinants"
] | [
"Definition:Determinant/Matrix"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14611 | Determinant with Unit Element in Otherwise Zero Column | Let $D$ be the determinant:
:<nowiki>$D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$</nowiki>
Then:
:<nowiki>$D = \begin{vmatrix}
b_{22} & \cdots & b_{2n} \\
\vdots & \ddots & \vdots \\
b_{n2}... | We note that:
:<nowiki>$D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$</nowiki>
is the transpose of:
:<nowiki>$D^\intercal = \begin{vmatrix}
1 & 0 & \cdots & 0 \\
b_{12} & b_{22} & \cdots & b_... | Let $D$ be the [[Definition:Determinant of Matrix|determinant]]:
:<nowiki>$D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$</nowiki>
Then:
:<nowiki>$D = \begin{vmatrix}
b_{22} & \cdots & b_{2n}... | We note that:
:<nowiki>$D = \begin{vmatrix}
1 & b_{12} & \cdots & b_{1n} \\
0 & b_{22} & \cdots & b_{2n} \\
\vdots & \vdots & \ddots & \vdots \\
0 & b_{n2} & \cdots & b_{nn}
\end{vmatrix}$</nowiki>
is the [[Definition:Transpose of Matrix|transpose]] of:
:<nowiki>$D^\intercal = \begin{vmatrix}
1 & 0 & \cdots ... | Determinant with Unit Element in Otherwise Zero Column | https://proofwiki.org/wiki/Determinant_with_Unit_Element_in_Otherwise_Zero_Column | https://proofwiki.org/wiki/Determinant_with_Unit_Element_in_Otherwise_Zero_Column | [
"Determinants"
] | [
"Definition:Determinant/Matrix"
] | [
"Definition:Transpose of Matrix",
"Determinant with Unit Element in Otherwise Zero Row",
"Determinant of Transpose"
] |
proofwiki-14612 | Zermelo's Well-Ordering Theorem implies Hausdorff's Maximal Principle | Let Zermelo's Well-Ordering Theorem hold.
Then Hausdorff's Maximal Principle holds. | Let $X$ be a non-empty set.
Let $X$ contain at least two elements.
Otherwise, any non-empty ordering on $X$ is trivially a maximal chain.
By Zermelo's Well-Ordering Theorem, $X$ can be well-ordered set.
Fix such a well-ordering.
Let $\le$ be any ordering on $X$.
Let $\map P {a, Y}$ be the predicate:
:$a$ is $\le$-compa... | Let [[Zermelo's Well-Ordering Theorem]] hold.
Then [[Hausdorff's Maximal Principle]] holds. | Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $X$ contain at least two [[Definition:Element|elements]].
Otherwise, any [[Definition:Non-Empty Set|non-empty]] [[Definition:Ordering|ordering]] on $X$ is trivially a [[Definition:Maximal Chain|maximal chain]].
By [[Zermelo's Well-Ordering Theorem]], $X$ c... | Zermelo's Well-Ordering Theorem implies Hausdorff's Maximal Principle | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem_implies_Hausdorff's_Maximal_Principle | https://proofwiki.org/wiki/Zermelo's_Well-Ordering_Theorem_implies_Hausdorff's_Maximal_Principle | [
"Hausdorff's Maximal Principle",
"Zermelo's Well-Ordering Theorem"
] | [
"Zermelo's Well-Ordering Theorem",
"Hausdorff's Maximal Principle"
] | [
"Definition:Non-Empty Set",
"Definition:Element",
"Definition:Non-Empty Set",
"Definition:Ordering",
"Definition:Maximal Chain",
"Zermelo's Well-Ordering Theorem",
"Definition:Well-Ordered Set",
"Definition:Well-Ordering",
"Definition:Ordering",
"Definition:Predicate",
"Definition:Comparable Ele... |
proofwiki-14613 | Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p | Let $p = 2 n + 1$ be an odd prime.
Then:
:$F_p \equiv 5^n \pmod p$ | From Fibonacci Number by Power of 2:
:$2^p F_p = 2 \ds \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k$
From Binomial Coefficient of Prime:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
and from Fermat's Little Theorem: Corollary 1:
:$2^p \equiv 2 \pmod p$
Hence:
{{begin-eqn}}
{{eqn | l = 2^p F_p
|... | Let $p = 2 n + 1$ be an [[Definition:Odd Prime|odd prime]].
Then:
:$F_p \equiv 5^n \pmod p$ | From [[Fibonacci Number by Power of 2]]:
:$2^p F_p = 2 \ds \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k$
From [[Binomial Coefficient of Prime]]:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
and from [[Fermat's Little Theorem/Corollary 1|Fermat's Little Theorem: Corollary 1]]:
:$2^p \equiv 2 \pmod p... | Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p | https://proofwiki.org/wiki/Fibonacci_Number_with_Prime_Index_2n+1_is_Congruent_to_5^n_Modulo_p | https://proofwiki.org/wiki/Fibonacci_Number_with_Prime_Index_2n+1_is_Congruent_to_5^n_Modulo_p | [
"Fibonacci Numbers",
"Binomial Coefficients"
] | [
"Definition:Odd Prime"
] | [
"Fibonacci Number by Power of 2",
"Binomial Coefficient of Prime",
"Fermat's Little Theorem/Corollary 1"
] |
proofwiki-14614 | Fibonacci Number whose Index is Plus or Minus Prime p is Multiple of p | Let $p$ be a prime number distinct from $5$.
Let $F_n$ denote the $n$th Fibonacci number.
Then either $F_{p - 1}$ or $F_{p + 1}$ (but not both) is a multiple of $p$. | First we consider the edge cases:
For $p = 2$ we have that:
:$F_{2 + 1} = F_3 = 2$ which is a multiple of $2$.
:$F_{2 - 1} = F_1 = 1$ which is not a multiple of $2$.
For $p = 3$ we have that:
:$F_{3 + 1} = F_4 = 3$ which is a multiple of $3$.
:$F_{3 - 1} = F_2 = 1$ which is not a multiple of $3$.
Let $p$ be an odd prim... | Let $p$ be a [[Definition:Prime Number|prime number]] [[Definition:Distinct|distinct]] from $5$.
Let $F_n$ denote the $n$th [[Definition:Fibonacci Number|Fibonacci number]].
Then either $F_{p - 1}$ or $F_{p + 1}$ (but not both) is a [[Definition:Multiple of Integer|multiple]] of $p$. | First we consider the edge cases:
For $p = 2$ we have that:
:$F_{2 + 1} = F_3 = 2$ which is a [[Definition:Multiple of Integer|multiple]] of $2$.
:$F_{2 - 1} = F_1 = 1$ which is not a [[Definition:Multiple of Integer|multiple]] of $2$.
For $p = 3$ we have that:
:$F_{3 + 1} = F_4 = 3$ which is a [[Definition:Multiple ... | Fibonacci Number whose Index is Plus or Minus Prime p is Multiple of p | https://proofwiki.org/wiki/Fibonacci_Number_whose_Index_is_Plus_or_Minus_Prime_p_is_Multiple_of_p | https://proofwiki.org/wiki/Fibonacci_Number_whose_Index_is_Plus_or_Minus_Prime_p_is_Multiple_of_p | [
"Fibonacci Numbers"
] | [
"Definition:Prime Number",
"Definition:Distinct",
"Definition:Fibonacci Number",
"Definition:Multiple/Integer"
] | [
"Definition:Multiple/Integer",
"Definition:Multiple/Integer",
"Definition:Multiple/Integer",
"Definition:Multiple/Integer",
"Definition:Odd Prime",
"Cassini's Identity",
"Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p",
"Fermat's Little Theorem",
"Definition:Multiple/Integer",
... |
proofwiki-14615 | Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n | :$F_{n + 1} - \phi F_n = \hat \phi^n$
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\phi$ denotes the golden mean. | {{begin-eqn}}
{{eqn | l = F_{n + 1} - \phi F_n
| r = \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} } - \dfrac \phi {\sqrt 5} \paren {\phi^n - \hat \phi^n}
| c = Euler-Binet Formula
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} - \phi^{n + 1} + \phi \hat \phi^n}
... | :$F_{n + 1} - \phi F_n = \hat \phi^n$
where:
:$F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]. | {{begin-eqn}}
{{eqn | l = F_{n + 1} - \phi F_n
| r = \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} } - \dfrac \phi {\sqrt 5} \paren {\phi^n - \hat \phi^n}
| c = [[Euler-Binet Formula]]
}}
{{eqn | r = \dfrac 1 {\sqrt 5} \paren {\phi^{n + 1} - \hat \phi^{n + 1} - \phi^{n + 1} + \phi \hat \phi^n}... | Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n/Proof 1 | https://proofwiki.org/wiki/Fibonacci_Number_n+1_Minus_Golden_Mean_by_Fibonacci_Number_n | https://proofwiki.org/wiki/Fibonacci_Number_n+1_Minus_Golden_Mean_by_Fibonacci_Number_n/Proof_1 | [
"Fibonacci Numbers",
"Golden Mean",
"Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n"
] | [
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Euler-Binet Formula",
"Euler-Binet Formula"
] |
proofwiki-14616 | Recurrence Relation where n+1th Term is A by nth term + B to the n | Let $\sequence {a_n}$ be the sequence defined by the recurrence relation:
:$a_n = \begin {cases} 0 & : n = 0 \\ A a_{n - 1} + B^{n - 1} & : n > 0 \end {cases}$
for numbers $A$ and $B$.
Then the closed form for $\sequence {a_n}$ is given by:
:$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & ... | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end {cases}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \dfrac {A^0 - B^0} {A - B}
| r = \dfrac {1 - 1} {A - B}
| c =... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined by the [[Definition:Recurrence Relation|recurrence relation]]:
:$a_n = \begin {cases} 0 & : n = 0 \\ A a_{n - 1} + B^{n - 1} & : n > 0 \end {cases}$
for [[Definition:Number|numbers]] $A$ and $B$.
Then the [[Definition:Closed-Form Expression|close... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$a_n = \begin {cases} \dfrac {A^n - B^n} {A - B} & : A \ne B \\ n A^{n - 1} & : A = B \end {cases}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = \d... | Recurrence Relation where n+1th Term is A by nth term + B to the n | https://proofwiki.org/wiki/Recurrence_Relation_where_n+1th_Term_is_A_by_nth_term_+_B_to_the_n | https://proofwiki.org/wiki/Recurrence_Relation_where_n+1th_Term_is_A_by_nth_term_+_B_to_the_n | [
"Closed Forms"
] | [
"Definition:Sequence",
"Definition:Recursive Sequence/Recurrence Relation",
"Definition:Number",
"Definition:Closed Form Expression"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14617 | Fibonomial Coefficient is Integer | Let $\dbinom n k_\FF$ be a Fibonomial coefficient.
Then $\dbinom n k_\FF$ is an integer. | Recurrence Relation for Fibonomial Coefficients gives:
:$\dbinom n k_\FF = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF$
where $F_{k - 1}$ etc. denote Fibonacci numbers.
It follows that each Fibonomial coefficient is the sum of integers, and so an integer.
{{qed}} | Let $\dbinom n k_\FF$ be a [[Definition:Fibonomial Coefficient|Fibonomial coefficient]].
Then $\dbinom n k_\FF$ is an [[Definition:Integer|integer]]. | [[Recurrence Relation for Fibonomial Coefficients]] gives:
:$\dbinom n k_\FF = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF$
where $F_{k - 1}$ etc. denote [[Definition:Fibonacci Number|Fibonacci numbers]].
It follows that each [[Definition:Fibonomial Coefficient|Fibonomial coefficient... | Fibonomial Coefficient is Integer | https://proofwiki.org/wiki/Fibonomial_Coefficient_is_Integer | https://proofwiki.org/wiki/Fibonomial_Coefficient_is_Integer | [
"Fibonomial Coefficients"
] | [
"Definition:Fibonomial Coefficient",
"Definition:Integer"
] | [
"Recurrence Relation for Fibonomial Coefficients",
"Definition:Fibonacci Number",
"Definition:Fibonomial Coefficient",
"Definition:Addition/Integers",
"Definition:Integer",
"Definition:Integer"
] |
proofwiki-14618 | Recurrence Relation for Fibonomial Coefficients | :$\dbinom n k_\FF = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF$
where:
:$\dbinom n k_\FF$ denotes a Fibonomial coefficient
:$F_{k - 1}$ etc. denote Fibonacci numbers. | {{begin-eqn}}
{{eqn | r = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF
| o =
| c =
}}
{{eqn | r = F_{k - 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} F_{n - k} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1} + F_{n - k + 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k... | :$\dbinom n k_\FF = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF$
where:
:$\dbinom n k_\FF$ denotes a [[Definition:Fibonomial Coefficient|Fibonomial coefficient]]
:$F_{k - 1}$ etc. denote [[Definition:Fibonacci Number|Fibonacci numbers]]. | {{begin-eqn}}
{{eqn | r = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF
| o =
| c =
}}
{{eqn | r = F_{k - 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k + 1} F_{n - k} } {F_k F_{k - 1} F_{k - 2} F_{k - 3} \cdots F_1} + F_{n - k + 1} \dfrac {F_{n - 1} F_{n - 2} \cdots F_{n - k... | Recurrence Relation for Fibonomial Coefficients | https://proofwiki.org/wiki/Recurrence_Relation_for_Fibonomial_Coefficients | https://proofwiki.org/wiki/Recurrence_Relation_for_Fibonomial_Coefficients | [
"Fibonomial Coefficients"
] | [
"Definition:Fibonomial Coefficient",
"Definition:Fibonacci Number"
] | [
"Honsberger's Identity"
] |
proofwiki-14619 | Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers | Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom m k_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 1} = 0$
where:
:$\dbinom m k_\FF$ denotes a Fibonomial coefficient
:$F_{n + k}$ denotes the $n + k$th Fibonacci number
:$\ceiling {\, \cdot \,}$ denote... | === Lemma 1 ===
{{:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1}} | Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Then:
:$\ds \sum_{k \mathop \in \Z} \dbinom m k_\FF \paren {-1}^{\ceiling {\paren {m - k} / 2} } {F_{n + k} }^{m - 1} = 0$
where:
:$\dbinom m k_\FF$ denotes a [[Definition:Fibonomial Coefficient|Fibonomial coefficient]]
:$F... | === [[Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1|Lemma 1]] ===
{{:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1}} | Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers | https://proofwiki.org/wiki/Recurrence_Relation_for_Sequence_of_mth_Powers_of_Fibonacci_Numbers | https://proofwiki.org/wiki/Recurrence_Relation_for_Sequence_of_mth_Powers_of_Fibonacci_Numbers | [
"Fibonomial Coefficients",
"Fibonacci Numbers",
"Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Fibonomial Coefficient",
"Definition:Fibonacci Number",
"Definition:Ceiling Function"
] | [
"Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1"
] |
proofwiki-14620 | Golden Mean by One Minus Golden Mean equals Minus 1 | :$\phi \hat \phi = -1$
where:
:$\phi$ denotes the golden mean
:$\hat \phi := 1 - \phi$ | {{begin-eqn}}
{{eqn | l = \phi
| r = \frac 1 {\phi - 1}
| c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | ll= \leadsto
| l = \phi \paren {\phi - 1}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \phi \paren {1 - \phi}
| r = -1
| c =
}}
{{eqn | ll= \leadsto
| l = \phi... | :$\phi \hat \phi = -1$
where:
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]
:$\hat \phi := 1 - \phi$ | {{begin-eqn}}
{{eqn | l = \phi
| r = \frac 1 {\phi - 1}
| c = {{Defof|Golden Mean|index = 3}}
}}
{{eqn | ll= \leadsto
| l = \phi \paren {\phi - 1}
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \phi \paren {1 - \phi}
| r = -1
| c =
}}
{{eqn | ll= \leadsto
| l = \phi... | Golden Mean by One Minus Golden Mean equals Minus 1 | https://proofwiki.org/wiki/Golden_Mean_by_One_Minus_Golden_Mean_equals_Minus_1 | https://proofwiki.org/wiki/Golden_Mean_by_One_Minus_Golden_Mean_equals_Minus_1 | [
"Golden Mean"
] | [
"Definition:Golden Mean"
] | [
"Category:Golden Mean"
] |
proofwiki-14621 | Fibonacci Number of Even Index by Golden Mean Modulo 1 | Let $n \in \Z$ be an integer.
Then:
:$F_{2 n} \phi \bmod 1 = 1 - \phi^{-2 n}$
:$F_n$ denotes the $n$th Fibonacci number
:$\phi$ is the golden mean: $\phi = \dfrac {1 + \sqrt 5} 2$ | From definition of$\bmod 1$, the statement above is equivalent to the statement:
:$F_{2 n} \phi - 1 + \phi^{-2 n}$ is an integer
We have:
{{begin-eqn}}
{{eqn | l = F_{2 n} \phi - 1 + \phi^{-2 n}
| r = -1 + F_{2 n + 1} - \hat \phi^{2 n} + \phi^{-2 n}
| c = Fibonacci Number n+1 Minus Golden Mean by Fibonacci ... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Then:
:$F_{2 n} \phi \bmod 1 = 1 - \phi^{-2 n}$
:$F_n$ denotes the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]]
:$\phi$ is the [[Definition:Golden Mean|golden mean]]: $\phi = \dfrac {1 + \sqrt 5} 2$ | From definition of[[Definition:Modulo Operation/Modulo One|$\bmod 1$]], the statement above is equivalent to the statement:
:$F_{2 n} \phi - 1 + \phi^{-2 n}$ is an [[Definition:Integer|integer]]
We have:
{{begin-eqn}}
{{eqn | l = F_{2 n} \phi - 1 + \phi^{-2 n}
| r = -1 + F_{2 n + 1} - \hat \phi^{2 n} + \phi^{-2 ... | Fibonacci Number of Even Index by Golden Mean Modulo 1 | https://proofwiki.org/wiki/Fibonacci_Number_of_Even_Index_by_Golden_Mean_Modulo_1 | https://proofwiki.org/wiki/Fibonacci_Number_of_Even_Index_by_Golden_Mean_Modulo_1 | [
"Fibonacci Numbers",
"Golden Mean"
] | [
"Definition:Integer",
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Definition:Modulo Operation/Modulo One",
"Definition:Integer",
"Fibonacci Number n+1 Minus Golden Mean by Fibonacci Number n",
"Golden Mean by One Minus Golden Mean equals Minus 1",
"Definition:Integer"
] |
proofwiki-14622 | Fibonacci Number of Odd Index by Golden Mean Modulo 1 | Let $n \in \Z$ be an integer.
Then:
:$F_{2 n + 1} \phi \bmod 1 = \phi^{-2 n - 1}$
where:
:$F_n$ denotes the $n$th Fibonacci number
:$\phi$ is the golden mean: $\phi = \dfrac {1 + \sqrt 5} 2$ | From definition of$\bmod 1$, the statement above is equivalent to the statement:
:$F_{2 n + 1} \phi - \phi^{-2 n - 1}$ is an integer
We have:
{{begin-eqn}}
{{eqn | l = \phi^2 - \phi \sqrt 5
| r = \paren {\frac {1 + \sqrt 5} 2}^2 - \paren {\frac {1 + \sqrt 5} 2} \sqrt 5
}}
{{eqn | r = \frac {6 + 2 \sqrt 5} 4 - \fr... | Let $n \in \Z$ be an [[Definition:Integer|integer]].
Then:
:$F_{2 n + 1} \phi \bmod 1 = \phi^{-2 n - 1}$
where:
:$F_n$ denotes the $n$th [[Definition:Fibonacci Numbers|Fibonacci number]]
:$\phi$ is the [[Definition:Golden Mean|golden mean]]: $\phi = \dfrac {1 + \sqrt 5} 2$ | From definition of[[Definition:Modulo Operation/Modulo One|$\bmod 1$]], the statement above is equivalent to the statement:
:$F_{2 n + 1} \phi - \phi^{-2 n - 1}$ is an [[Definition:Integer|integer]]
We have:
{{begin-eqn}}
{{eqn | l = \phi^2 - \phi \sqrt 5
| r = \paren {\frac {1 + \sqrt 5} 2}^2 - \paren {\frac {1... | Fibonacci Number of Odd Index by Golden Mean Modulo 1 | https://proofwiki.org/wiki/Fibonacci_Number_of_Odd_Index_by_Golden_Mean_Modulo_1 | https://proofwiki.org/wiki/Fibonacci_Number_of_Odd_Index_by_Golden_Mean_Modulo_1 | [
"Fibonacci Numbers",
"Golden Mean"
] | [
"Definition:Integer",
"Definition:Fibonacci Number",
"Definition:Golden Mean"
] | [
"Definition:Modulo Operation/Modulo One",
"Definition:Integer",
"Euler-Binet Formula",
"Euler-Binet Formula",
"Definition:Integer"
] |
proofwiki-14623 | Remainder of Fibonacci Number Divided by Fibonacci Number is Plus or Minus Fibonacci Number | Let $F_n$ and $F_m$ be Fibonacci numbers.
By the Division Theorem, let:
:$F_n = q F_m + r$
where:
:$q \in \Z$
:$r \in \Z: 0 \le r < \size {F_m}$
Then either $r$ or $\size {F_m} - r$, or both, is a Fibonacci number. | Follows directly from Residue of Fibonacci Number Modulo Fibonacci Number.
{{qed}} | Let $F_n$ and $F_m$ be [[Definition:Fibonacci Number|Fibonacci numbers]].
By the [[Division Theorem]], let:
:$F_n = q F_m + r$
where:
:$q \in \Z$
:$r \in \Z: 0 \le r < \size {F_m}$
Then either $r$ or $\size {F_m} - r$, or both, is a [[Definition:Fibonacci Number|Fibonacci number]]. | Follows directly from [[Residue of Fibonacci Number Modulo Fibonacci Number]].
{{qed}} | Remainder of Fibonacci Number Divided by Fibonacci Number is Plus or Minus Fibonacci Number | https://proofwiki.org/wiki/Remainder_of_Fibonacci_Number_Divided_by_Fibonacci_Number_is_Plus_or_Minus_Fibonacci_Number | https://proofwiki.org/wiki/Remainder_of_Fibonacci_Number_Divided_by_Fibonacci_Number_is_Plus_or_Minus_Fibonacci_Number | [
"Fibonacci Numbers"
] | [
"Definition:Fibonacci Number",
"Division Theorem",
"Definition:Fibonacci Number"
] | [
"Residue of Fibonacci Number Modulo Fibonacci Number"
] |
proofwiki-14624 | Residue of Fibonacci Number Modulo Fibonacci Number | Let $F_n$ denote the $n$th Fibonacci number.
Let $m, r$ be non-negative integers.
Then:
:<nowiki>$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\
\paren {-1}^n F_r & : m \bmod 4 = 2 \\
\paren {-1}^{r + 1 + n} F_{n - r} & : m \bmod 4 = 3 \end{cases} } ... | === Lemma ===
{{:Residue of Fibonacci Number Modulo Fibonacci Number/Lemma}}{{qed|lemma}}
We prove the result for all $r$ by induction on $r$:
For all $r \in \N$, let $\map P r$ be the proposition:
:<nowiki>$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = ... | Let $F_n$ denote the $n$th [[Definition:Fibonacci Number|Fibonacci number]].
Let $m, r$ be [[Definition:Non-Negative Integer|non-negative integers]].
Then:
:<nowiki>$F_{m n + r} \equiv \paren {\begin{cases} F_r & : m \bmod 4 = 0 \\
\paren {-1}^{r + 1} F_{n - r} & : m \bmod 4 = 1 \\
\paren {-1}^n F_r & : m \bmod 4 = ... | === [[Residue of Fibonacci Number Modulo Fibonacci Number/Lemma|Lemma]] ===
{{:Residue of Fibonacci Number Modulo Fibonacci Number/Lemma}}{{qed|lemma}}
We prove the result for all $r$ by [[Principle of Mathematical Induction|induction]] on $r$:
For all $r \in \N$, let $\map P r$ be the [[Definition:Proposition|propo... | Residue of Fibonacci Number Modulo Fibonacci Number | https://proofwiki.org/wiki/Residue_of_Fibonacci_Number_Modulo_Fibonacci_Number | https://proofwiki.org/wiki/Residue_of_Fibonacci_Number_Modulo_Fibonacci_Number | [
"Fibonacci Numbers",
"Proofs by Induction",
"Residue of Fibonacci Number Modulo Fibonacci Number"
] | [
"Definition:Fibonacci Number",
"Definition:Positive/Integer"
] | [
"Residue of Fibonacci Number Modulo Fibonacci Number/Lemma",
"Principle of Mathematical Induction",
"Definition:Proposition",
"Residue of Fibonacci Number Modulo Fibonacci Number/Lemma",
"Residue of Fibonacci Number Modulo Fibonacci Number",
"Principle of Mathematical Induction"
] |
proofwiki-14625 | Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean | Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the golden mean
:$F_n$ denotes the $n$th Fibonacci number. | {{begin-eqn}}
{{eqn | l = \sin n z
| r = \map \sin {\dfrac {n \pi} 2 + i n \ln \phi}
| c =
}}
{{eqn | r = \frac {e^{i \paren {\paren {n \pi / 2} + i n \ln \phi} } - e^{-i \paren {\paren {n \pi / 2} + i n \ln \phi} } } {2 i}
| c = Euler's Sine Identity
}}
{{eqn | r = \frac {e^{i n \pi / 2} e^{-n \ln \... | Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]
:$F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | {{begin-eqn}}
{{eqn | l = \sin n z
| r = \map \sin {\dfrac {n \pi} 2 + i n \ln \phi}
| c =
}}
{{eqn | r = \frac {e^{i \paren {\paren {n \pi / 2} + i n \ln \phi} } - e^{-i \paren {\paren {n \pi / 2} + i n \ln \phi} } } {2 i}
| c = [[Euler's Sine Identity]]
}}
{{eqn | r = \frac {e^{i n \pi / 2} e^{-n \... | Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean/Proof 1 | https://proofwiki.org/wiki/Sine_of_Multiple_of_Pi_by_2_plus_i_by_Natural_Logarithm_of_Golden_Mean | https://proofwiki.org/wiki/Sine_of_Multiple_of_Pi_by_2_plus_i_by_Natural_Logarithm_of_Golden_Mean/Proof_1 | [
"Fibonacci Numbers",
"Golden Mean",
"Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean"
] | [
"Definition:Golden Mean",
"Definition:Fibonacci Number"
] | [
"Euler's Sine Identity",
"Euler's Formula",
"Euler's Formula/Corollary",
"Cosine of Half-Integer Multiple of Pi",
"Sine of Half-Integer Multiple of Pi",
"Exponential of Natural Logarithm",
"Reciprocal Form of One Minus Golden Mean",
"Euler-Binet Formula"
] |
proofwiki-14626 | Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean | Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the golden mean
:$F_n$ denotes the $n$th Fibonacci number. | {{begin-eqn}}
{{eqn | l = \cos z
| r = \map \cos {\dfrac \pi 2 + i \ln \phi}
| c =
}}
{{eqn | r = \frac {e^{i \paren {\paren {\pi / 2} + i \ln \phi} } + e^{-i \paren {\paren {\pi / 2} + i \ln \phi} } } 2
| c = Euler's Cosine Identity
}}
{{eqn | r = \frac {e^{i \pi / 2} e^{-\ln \phi} + e^{-i \pi / 2} ... | Let $z = \dfrac \pi 2 + i \ln \phi$.
Then:
:$\dfrac {\sin n z} {\sin z} = i^{1 - n} F_n$
where:
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]]
:$F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | {{begin-eqn}}
{{eqn | l = \cos z
| r = \map \cos {\dfrac \pi 2 + i \ln \phi}
| c =
}}
{{eqn | r = \frac {e^{i \paren {\paren {\pi / 2} + i \ln \phi} } + e^{-i \paren {\paren {\pi / 2} + i \ln \phi} } } 2
| c = [[Euler's Cosine Identity]]
}}
{{eqn | r = \frac {e^{i \pi / 2} e^{-\ln \phi} + e^{-i \pi /... | Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean/Proof 2 | https://proofwiki.org/wiki/Sine_of_Multiple_of_Pi_by_2_plus_i_by_Natural_Logarithm_of_Golden_Mean | https://proofwiki.org/wiki/Sine_of_Multiple_of_Pi_by_2_plus_i_by_Natural_Logarithm_of_Golden_Mean/Proof_2 | [
"Fibonacci Numbers",
"Golden Mean",
"Sine of Multiple of Pi by 2 plus i by Natural Logarithm of Golden Mean"
] | [
"Definition:Golden Mean",
"Definition:Fibonacci Number"
] | [
"Euler's Cosine Identity",
"Euler's Formula",
"Cosine of Half-Integer Multiple of Pi",
"Sine of Half-Integer Multiple of Pi",
"Exponential of Natural Logarithm",
"Square of Golden Mean equals One plus Golden Mean",
"Werner Formulas/Sine by Cosine"
] |
proofwiki-14627 | Equality of Towers in Set | Let $X$ be a non-empty set.
Let $\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ be towers in $X$.
Then either:
:$\struct {T_1, \preccurlyeq_1} = \struct {T_2, \preccurlyeq_2}$
or:
:$\struct {T_1, \preccurlyeq_1}$ is an initial segment of $\struct {T_2, \preccurlyeq_2}$
or:
:$\struct {T_2, \preccurl... | By the definition of tower:
:$\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ are well-ordered sets.
By Wosets are Isomorphic to Each Other or Initial Segments, either:
:$(1): \quad$ the towers are order isomorphic to each other
or:
:$(2): \quad$ one is order isomorphic to an initial segment in the o... | Let $X$ be a [[Definition:Non-Empty Set|non-empty set]].
Let $\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ be [[Definition:Tower in Set|towers in $X$]].
Then either:
:$\struct {T_1, \preccurlyeq_1} = \struct {T_2, \preccurlyeq_2}$
or:
:$\struct {T_1, \preccurlyeq_1}$ is an [[Definition:Init... | By the definition of [[Definition:Tower in Set|tower]]:
:$\struct {T_1, \preccurlyeq_1}$ and $\struct {T_2, \preccurlyeq_2}$ are [[Definition:Well-Ordered Set|well-ordered sets]].
By [[Wosets are Isomorphic to Each Other or Initial Segments]], either:
:$(1): \quad$ the [[Definition:Tower in Set|towers]] are [[Definiti... | Equality of Towers in Set | https://proofwiki.org/wiki/Equality_of_Towers_in_Set | https://proofwiki.org/wiki/Equality_of_Towers_in_Set | [
"Well-Orderings"
] | [
"Definition:Non-Empty Set",
"Definition:Tower in Set",
"Definition:Initial Segment",
"Definition:Initial Segment"
] | [
"Definition:Tower in Set",
"Definition:Well-Ordered Set",
"Wosets are Isomorphic to Each Other or Initial Segments",
"Definition:Tower in Set",
"Definition:Order Isomorphism",
"Definition:Order Isomorphism",
"Definition:Initial Segment",
"Definition:Order Isomorphism",
"Definition:Initial Segment",
... |
proofwiki-14628 | Representations for 1 in Golden Mean Number System | Then there are infinitely many ways to express the number $1$ in the golden mean number system. | We have that:
:$\phi^0 = 1$
and so $1$ has the representation $S_1$ as:
:$S_1 := \left[{1 \cdotp 00}\right]_\phi$
By inspection it is seen that $S_1$ is the simplest form of $1$.
Expanding $S_1$:
:$S_2 := \left[{0 \cdotp 11}\right]_\phi$
which can then be expressed as:
:$S_2 := \left[{0 \cdotp 1100}\right]_\phi$
Expand... | Then there are [[Definition:Infinite Set|infinitely many]] ways to express the number $1$ in the [[Definition:Golden Mean Number System|golden mean number system]]. | We have that:
:$\phi^0 = 1$
and so $1$ has the representation $S_1$ as:
:$S_1 := \left[{1 \cdotp 00}\right]_\phi$
By inspection it is seen that $S_1$ is the [[Definition:Simplest Form of Number in Golden Mean Number System|simplest form]] of $1$.
[[Definition:Expansion of Number in Golden Mean Number System|Expandin... | Representations for 1 in Golden Mean Number System | https://proofwiki.org/wiki/Representations_for_1_in_Golden_Mean_Number_System | https://proofwiki.org/wiki/Representations_for_1_in_Golden_Mean_Number_System | [
"Golden Mean Number System",
"Representations for 1 in Golden Mean Number System"
] | [
"Definition:Infinite Set",
"Definition:Golden Mean Number System"
] | [
"Definition:Golden Mean Number System/Simplest Form",
"Definition:Golden Mean Number System/Expansion",
"Definition:Golden Mean Number System/Expansion",
"Definition:Sequence/Infinite Sequence"
] |
proofwiki-14629 | Golden Mean as Root of Quadratic | The golden mean $\phi$ is one of the roots of the quadratic equation:
:$x^2 - x - 1 = 0$
The other root is $\hat \phi = 1 - \phi$. | By Solution to Quadratic Equation:
{{begin-eqn}}
{{eqn | l = x
| r = \frac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times \paren {-1} } } {2 \times 1}
| c = Solution to Quadratic Equation
}}
{{eqn | r = \frac {1 \pm \sqrt 5} 2
| c =
}}
{{end-eqn}}
Thus
:$x = \dfrac {1 + \sqrt 5} 2$
and:
:... | The [[Definition:Golden Mean|golden mean]] $\phi$ is one of the [[Definition:Root of Polynomial|roots]] of the [[Definition:Quadratic Equation|quadratic equation]]:
:$x^2 - x - 1 = 0$
The other root is $\hat \phi = 1 - \phi$. | By [[Solution to Quadratic Equation]]:
{{begin-eqn}}
{{eqn | l = x
| r = \frac {-\paren {-1} \pm \sqrt {\paren {-1}^2 - 4 \times 1 \times \paren {-1} } } {2 \times 1}
| c = [[Solution to Quadratic Equation]]
}}
{{eqn | r = \frac {1 \pm \sqrt 5} 2
| c =
}}
{{end-eqn}}
Thus
:$x = \dfrac {1 + \sqrt 5... | Golden Mean as Root of Quadratic | https://proofwiki.org/wiki/Golden_Mean_as_Root_of_Quadratic | https://proofwiki.org/wiki/Golden_Mean_as_Root_of_Quadratic | [
"Golden Mean"
] | [
"Definition:Golden Mean",
"Definition:Root of Polynomial",
"Definition:Quadratic Equation"
] | [
"Solution to Quadratic Equation",
"Solution to Quadratic Equation",
"Definition:Golden Mean",
"Closed Form of One Minus Golden Mean",
"Category:Golden Mean"
] |
proofwiki-14630 | Proper Subtower is Initial Segment | Let $\struct {T_1, \preccurlyeq}$ be a proper subtower of $\struct {T_2, \preccurlyeq}$.
Then $\struct {T_1, \preccurlyeq}$ is an initial segment of $\struct {T_2, \preccurlyeq}$.
{{explain|What sort of object is $\struct {T_2, \preccurlyeq}$ -- an ordered set, a totally ordered set, a well-ordered set, an ordinal, wha... | Define the set:
:$Y = \set {y \in T_1: S_y \text { is an initial segment of } \struct {T_2, \preccurlyeq} }$.
Then:
{{begin-eqn}}
{{eqn | l = \map {S_x} {T_1}
| r = \set {b \in T_1, x \in T_1: b \prec x}
| c = {{Defof|Initial Segment}}
}}
{{eqn | r = \set {b \in T_2, x \in T_2: b \prec x}
| c = {{Defo... | Let $\struct {T_1, \preccurlyeq}$ be a [[Definition:Proper Subtower in Set|proper subtower]] of $\struct {T_2, \preccurlyeq}$.
Then $\struct {T_1, \preccurlyeq}$ is an [[Definition:Initial Segment|initial segment]] of $\struct {T_2, \preccurlyeq}$.
{{explain|What sort of object is $\struct {T_2, \preccurlyeq}$ -- an ... | Define the set:
:$Y = \set {y \in T_1: S_y \text { is an initial segment of } \struct {T_2, \preccurlyeq} }$.
Then:
{{begin-eqn}}
{{eqn | l = \map {S_x} {T_1}
| r = \set {b \in T_1, x \in T_1: b \prec x}
| c = {{Defof|Initial Segment}}
}}
{{eqn | r = \set {b \in T_2, x \in T_2: b \prec x}
| c = {{D... | Proper Subtower is Initial Segment | https://proofwiki.org/wiki/Proper_Subtower_is_Initial_Segment | https://proofwiki.org/wiki/Proper_Subtower_is_Initial_Segment | [
"Well-Orderings"
] | [
"Definition:Proper Subtower in Set",
"Definition:Initial Segment"
] | [
"Principle of Mathematical Induction/Well-Ordered Set",
"Definition:Initial Segment"
] |
proofwiki-14631 | Power of Golden Mean as Sum of Smaller Powers | Let $\phi$ denote the golden mean.
Then:
:$\forall z \in \C: \phi^z = \phi^{z - 1} + \phi^{z - 2}$ | Let $z \in \C$.
Let $w \in \C$ such that $w + 2 = z$.
Then:
{{begin-eqn}}
{{eqn | l = \phi^z
| r = \phi^{w + 2}
| c =
}}
{{eqn | r = \phi^w \phi^2
| c =
}}
{{eqn | r = \phi^w \left({\phi + 1}\right)
| c = Square of Golden Mean equals One plus Golden Mean
}}
{{eqn | r = \phi^{w + 1} + \phi^w
... | Let $\phi$ denote the [[Definition:Golden Mean|golden mean]].
Then:
:$\forall z \in \C: \phi^z = \phi^{z - 1} + \phi^{z - 2}$ | Let $z \in \C$.
Let $w \in \C$ such that $w + 2 = z$.
Then:
{{begin-eqn}}
{{eqn | l = \phi^z
| r = \phi^{w + 2}
| c =
}}
{{eqn | r = \phi^w \phi^2
| c =
}}
{{eqn | r = \phi^w \left({\phi + 1}\right)
| c = [[Square of Golden Mean equals One plus Golden Mean]]
}}
{{eqn | r = \phi^{w + 1} + \p... | Power of Golden Mean as Sum of Smaller Powers | https://proofwiki.org/wiki/Power_of_Golden_Mean_as_Sum_of_Smaller_Powers | https://proofwiki.org/wiki/Power_of_Golden_Mean_as_Sum_of_Smaller_Powers | [
"Golden Mean"
] | [
"Definition:Golden Mean"
] | [
"Square of Golden Mean equals One plus Golden Mean"
] |
proofwiki-14632 | 100 in Golden Mean Number System is Equivalent to 011 | Consider the golden mean number system.
Let $p$ and $q$ be arbitrary strings in $\set {0, 1}$.
Let $x \in \R_{\ge 0}$ have a representation which includes the string $100$, say:
:$x = p100q$
Then $x \in \R_{\ge 0}$ also has the representation:
:$x = p011q$
Similarly, let $x \in \R_{\ge 0}$ have a representation which i... | Let $100$ appear anywhere within $x$.
Then:
:$x = \phi^r + \ds \sum_{c \mathop \in C} \phi^c$
where:
:$C \subset \Z$
:$r \in \Z$
:$r \notin C, r - 1 \notin C, r - 2 \notin C$
That is, the instance of $100$ corresponds to the indices $r, r - 1, r - 2$.
From Power of Golden Mean as Sum of Smaller Powers:
:$\phi^r = \phi^... | Consider the [[Definition:Golden Mean Number System|golden mean number system]].
Let $p$ and $q$ be arbitrary [[Definition:String|strings]] in $\set {0, 1}$.
Let $x \in \R_{\ge 0}$ have a representation which includes the [[Definition:String|string]] $100$, say:
:$x = p100q$
Then $x \in \R_{\ge 0}$ also has the re... | Let $100$ appear anywhere within $x$.
Then:
:$x = \phi^r + \ds \sum_{c \mathop \in C} \phi^c$
where:
:$C \subset \Z$
:$r \in \Z$
:$r \notin C, r - 1 \notin C, r - 2 \notin C$
That is, the instance of $100$ corresponds to the indices $r, r - 1, r - 2$.
From [[Power of Golden Mean as Sum of Smaller Powers]]:
:$\phi^r ... | 100 in Golden Mean Number System is Equivalent to 011 | https://proofwiki.org/wiki/100_in_Golden_Mean_Number_System_is_Equivalent_to_011 | https://proofwiki.org/wiki/100_in_Golden_Mean_Number_System_is_Equivalent_to_011 | [
"Golden Mean Number System"
] | [
"Definition:Golden Mean Number System",
"Definition:String",
"Definition:String",
"Definition:String",
"Definition:Positive/Real Number",
"Definition:Golden Mean Number System/Equivalent Representations",
"Definition:Number Base/Radix Point",
"Definition:Golden Mean Number System/Equivalent Representa... | [
"Power of Golden Mean as Sum of Smaller Powers",
"Definition:String"
] |
proofwiki-14633 | Conversion of Number in Golden Mean Number System to Simplest Form | Let $x \in \R_{\ge 0}$ have a representation $S$ in the golden mean number system.
Then $S$ can be converted to its simplest form as follows:
:$(1): \quad$ Replace any infinite string on the right hand end of $S$ of the form $010101 \ldots$ with $100$
:$(2): \quad$ Repeatedly replace the leftmost instance of $011$ with... | Note that step $(2)$ is an instance of a simplification of $S$.
From 100 in Golden Mean Number System is Equivalent to 011, it has been established that $011$ is equivalent to $100$.
The following need to be established:
:$010101 \ldots$ is equivalent to $100$
:Replacing the leftmost $011$ with $100$ reduces the overal... | Let $x \in \R_{\ge 0}$ have a representation $S$ in the [[Definition:Golden Mean Number System|golden mean number system]].
Then $S$ can be converted to its [[Definition:Simplest Form of Number in Golden Mean Number System|simplest form]] as follows:
:$(1): \quad$ Replace any [[Definition:Infinite String|infinite stri... | Note that step $(2)$ is an instance of a [[Definition:Simplification of Number in Golden Mean Number System|simplification]] of $S$.
From [[100 in Golden Mean Number System is Equivalent to 011]], it has been established that $011$ is [[Definition:Equivalent Representations in Golden Mean Number System|equivalent]] to... | Conversion of Number in Golden Mean Number System to Simplest Form | https://proofwiki.org/wiki/Conversion_of_Number_in_Golden_Mean_Number_System_to_Simplest_Form | https://proofwiki.org/wiki/Conversion_of_Number_in_Golden_Mean_Number_System_to_Simplest_Form | [
"Golden Mean Number System"
] | [
"Definition:Golden Mean Number System",
"Definition:Golden Mean Number System/Simplest Form",
"Definition:String/Infinite"
] | [
"Definition:Golden Mean Number System/Simplification",
"100 in Golden Mean Number System is Equivalent to 011",
"Definition:Golden Mean Number System/Equivalent Representations",
"Definition:Golden Mean Number System/Equivalent Representations",
"Definition:Real Number",
"Sum of Infinite Geometric Sequenc... |
proofwiki-14634 | Representation of Integers in Golden Mean Number System | The positive integers $n$ are represented in the golden mean number system in their simplest form $S_n$ as follows:
:{| border="1"
|-
! align="right" style = "padding: 2px 10px" | $n$
! align="left" style = "padding: 2px 10px" | $S_n$
|-
| align="right" style = "padding: 2px 10px" | $1$
| align="left" style = "paddin... | :$1$ is represented by $\left[{1}\right]_\phi = \phi^0$.
From there, the algorithm for Addition of 1 in Golden Mean Number System is run.
{{begin-eqn}}
{{eqn | n = 2
| l = 1 + \left[{1}\right]_\phi
| r = 1 + \left[{0 \cdotp 11}\right]_\phi
| c = Expansion
}}
{{eqn | r = \left[{1 \cdotp 11}\right]_\phi... | The [[Definition:Positive Integer|positive integers]] $n$ are represented in the [[Definition:Golden Mean Number System|golden mean number system]] in their [[Definition:Simplest Form of Number in Golden Mean Number System|simplest form]] $S_n$ as follows:
:{| border="1"
|-
! align="right" style = "padding: 2px 10px" ... | :$1$ is represented by $\left[{1}\right]_\phi = \phi^0$.
From there, the [[Definition:Algorithm|algorithm]] for [[Addition of 1 in Golden Mean Number System]] is run.
{{begin-eqn}}
{{eqn | n = 2
| l = 1 + \left[{1}\right]_\phi
| r = 1 + \left[{0 \cdotp 11}\right]_\phi
| c = [[Definition:Expansion of... | Representation of Integers in Golden Mean Number System | https://proofwiki.org/wiki/Representation_of_Integers_in_Golden_Mean_Number_System | https://proofwiki.org/wiki/Representation_of_Integers_in_Golden_Mean_Number_System | [
"Golden Mean Number System"
] | [
"Definition:Positive/Integer",
"Definition:Golden Mean Number System",
"Definition:Golden Mean Number System/Simplest Form"
] | [
"Definition:Algorithm",
"Addition of 1 in Golden Mean Number System",
"Definition:Golden Mean Number System/Expansion",
"Definition:Golden Mean Number System/Simplification",
"Definition:Golden Mean Number System/Simplification",
"Definition:Golden Mean Number System/Expansion",
"Definition:Golden Mean ... |
proofwiki-14635 | Length of Fibonacci String is Fibonacci Number | Let $S_n$ denote the $n$th Fibonacci string.
Let $\map \len {S_n}$ denote the length of $S_n$.
Then:
:$\map \len {S_n} = F_n$
where $F_n$ denotes the $n$th Fibonacci number. | The proof proceeds by strong induction.
For all $n \in \Z_{>0}$, let $\map P n$ be the proposition:
:$\map \len {S_n} = F_n$ | Let $S_n$ denote the $n$th [[Definition:Fibonacci String|Fibonacci string]].
Let $\map \len {S_n}$ denote the [[Definition:Length of String|length]] of $S_n$.
Then:
:$\map \len {S_n} = F_n$
where $F_n$ denotes the $n$th [[Definition:Fibonacci Number|Fibonacci number]]. | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map \len {S_n} = F_n$ | Length of Fibonacci String is Fibonacci Number | https://proofwiki.org/wiki/Length_of_Fibonacci_String_is_Fibonacci_Number | https://proofwiki.org/wiki/Length_of_Fibonacci_String_is_Fibonacci_Number | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String",
"Definition:Length of String",
"Definition:Fibonacci Number"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Second Principle of Mathematical Induction"
] |
proofwiki-14636 | Incidence of Double Letters in Fibonacci String | Let $S_n$ denote the $n$th Fibonacci string.
Then:
:$(1):\quad$ There are no instances of $2$ $\text a$'s together
:$(2):\quad$ There are no instances of $3$ $\text b$'s together
in $S_n$. | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$.
$\map P 1$ is the case:
:$S_1 = \text a$ | Let $S_n$ denote the $n$th [[Definition:Fibonacci String|Fibonacci string]].
Then:
:$(1):\quad$ There are no instances of $2$ $\text a$'s together
:$(2):\quad$ There are no instances of $3$ $\text b$'s together
in $S_n$. | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:There are no instances of $2$ $\text a$'s or $3$ $\text b$'s together in $S_n$.
$\map P 1$ is the case:
:$S_1 = \text a$ | Incidence of Double Letters in Fibonacci String | https://proofwiki.org/wiki/Incidence_of_Double_Letters_in_Fibonacci_String | https://proofwiki.org/wiki/Incidence_of_Double_Letters_in_Fibonacci_String | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Second Principle of Mathematical Induction"
] |
proofwiki-14637 | Fibonacci String Begins with ba | Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ begins with $\text {ba}$. | The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$S_n$ begins with $\text {ba}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$. | Let $S_n$ be a [[Definition:Fibonacci String|Fibonacci string]] of [[Definition:Length of String|length]] $n$.
Then for $n \ge 3$, $S_n$ begins with $\text {ba}$. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$S_n$ begins with $\text {ba}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$. | Fibonacci String Begins with ba | https://proofwiki.org/wiki/Fibonacci_String_Begins_with_ba | https://proofwiki.org/wiki/Fibonacci_String_Begins_with_ba | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String",
"Definition:Length of String"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-14638 | Fibonacci String Ends with ab or ba | Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ ends either with $\text{ba}$ or with $\text{ab}$. | The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
:$S_n$ ends either with $\text{ba}$ or with $\text{ab}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these ends either with $\text{ba}$ or with $\text{ab}$. | Let $S_n$ be a [[Definition:Fibonacci String|Fibonacci string]] of [[Definition:Length of String|length]] $n$.
Then for $n \ge 3$, $S_n$ ends either with $\text{ba}$ or with $\text{ab}$. | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:$S_n$ ends either with $\text{ba}$ or with $\text{ab}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these ends ei... | Fibonacci String Ends with ab or ba | https://proofwiki.org/wiki/Fibonacci_String_Ends_with_ab_or_ba | https://proofwiki.org/wiki/Fibonacci_String_Ends_with_ab_or_ba | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String",
"Definition:Length of String"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14639 | Count of a's and b's in Fibonacci String | Let $S_n$ denote the $n$th Fibonacci string.
Then for $n \ge 3$, $S_n$ has:
:$F_{n - 2}$ instances of $\text a$
:$F_{n - 1}$ instances of $\text b$. | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$. | Let $S_n$ denote the $n$th [[Definition:Fibonacci String|Fibonacci string]].
Then for $n \ge 3$, $S_n$ has:
:$F_{n - 2}$ instances of $\text a$
:$F_{n - 1}$ instances of $\text b$. | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$S_n$ has $F_{n - 2}$ instances of $\text a$ and $F_{n - 1}$ instances of $\text b$. | Count of a's and b's in Fibonacci String | https://proofwiki.org/wiki/Count_of_a's_and_b's_in_Fibonacci_String | https://proofwiki.org/wiki/Count_of_a's_and_b's_in_Fibonacci_String | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Second Principle of Mathematical Induction"
] |
proofwiki-14640 | Prefix of Fibonacci String | Let $n \in \Z_{>1}$.
Let $S_n$ denote the $n$th Fibonacci string.
Let $m \in \Z$ such that $1 < m \le n$.
Let $F_m$ denote the $m$th Fibonacci number.
The prefix of $S_n$ of length $F_m$ is the Fibonacci string $S_m$. | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:for all $m \in \Z$ such that $1 < m \le n$, the prefix of $S_n$ of length $F_m$ is $S_m$. | Let $n \in \Z_{>1}$.
Let $S_n$ denote the $n$th [[Definition:Fibonacci String|Fibonacci string]].
Let $m \in \Z$ such that $1 < m \le n$.
Let $F_m$ denote the $m$th [[Definition:Fibonacci Number|Fibonacci number]].
The [[Definition:Prefix|prefix]] of $S_n$ of [[Definition:Length of String|length]] $F_m$ is the [[D... | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:for all $m \in \Z$ such that $1 < m \le n$, the [[Definition:Prefix|prefix]] of $S_n$ of [[Definition:Length of String|length]] $F_m$ is $S_m... | Prefix of Fibonacci String | https://proofwiki.org/wiki/Prefix_of_Fibonacci_String | https://proofwiki.org/wiki/Prefix_of_Fibonacci_String | [
"Fibonacci Strings"
] | [
"Definition:Fibonacci String",
"Definition:Fibonacci Number",
"Definition:Prefix",
"Definition:Length of String",
"Definition:Fibonacci String"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Definition:Prefix",
"Definition:Length of String",
"Definition:Prefix",
"Definition:Length of String",
"Definition:Prefix",
"Definition:Length of String",
"Definition:Prefix",
"Definition:Length of String",
"Definition:Pref... |
proofwiki-14641 | Optimal Strategy for Fibonacci Nim | Consider a game of Fibonacci nim with $n$ counters.
Let it be the turn of player $\text A$.
Let the maximum number of counters that can be taken by $\text A$ be $q$.
Let $n$ be expressed in Zeckendorf representation as:
:$n = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$
Then $\text A$ can force a win {{iff}}:
:$F_{k_r} \le q$... | {{ProofWanted|working on it, it's fiddly}} | Consider a game of [[Definition:Fibonacci Nim|Fibonacci nim]] with $n$ counters.
Let it be the turn of [[Definition:Player|player]] $\text A$.
Let the maximum number of counters that can be taken by $\text A$ be $q$.
Let $n$ be expressed in [[Definition:Zeckendorf Representation|Zeckendorf representation]] as:
:$n =... | {{ProofWanted|working on it, it's fiddly}} | Optimal Strategy for Fibonacci Nim | https://proofwiki.org/wiki/Optimal_Strategy_for_Fibonacci_Nim | https://proofwiki.org/wiki/Optimal_Strategy_for_Fibonacci_Nim | [
"Fibonacci Nim"
] | [
"Definition:Fibonacci Nim",
"Definition:Game/Player",
"Definition:Zeckendorf Representation"
] | [] |
proofwiki-14642 | Recursively Defined Sequence/Examples/Term is Term of Index less 1 plus 6 times Term of Index less 2 | Consider the integer sequence $\left\langle{a_n}\right\rangle$ defined recursively as:
:$a_n = \begin{cases} 0 & : n = 0 \\
1 & : n = 1 \\
a_{n - 1} + 6 a_{n - 2} & : \text{otherwise} \end{cases}$
$a_n$ has a closed-form expression:
:$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$ | The proof proceeds by strong induction.
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
:$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$ | Consider the [[Definition:Integer Sequence|integer sequence]] $\left\langle{a_n}\right\rangle$ defined recursively as:
:$a_n = \begin{cases} 0 & : n = 0 \\
1 & : n = 1 \\
a_{n - 1} + 6 a_{n - 2} & : \text{otherwise} \end{cases}$
$a_n$ has a [[Definition:Closed-Form Expression|closed-form expression]]:
:$a_n = \dfrac... | The proof proceeds by [[Second Principle of Mathematical Induction|strong induction]].
For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the [[Definition:Proposition|proposition]]:
:$a_n = \dfrac {3^n - \left({-2}\right)^n} 5$ | Recursively Defined Sequence/Examples/Term is Term of Index less 1 plus 6 times Term of Index less 2 | https://proofwiki.org/wiki/Recursively_Defined_Sequence/Examples/Term_is_Term_of_Index_less_1_plus_6_times_Term_of_Index_less_2 | https://proofwiki.org/wiki/Recursively_Defined_Sequence/Examples/Term_is_Term_of_Index_less_1_plus_6_times_Term_of_Index_less_2 | [
"Integer Sequences"
] | [
"Definition:Integer Sequence",
"Definition:Closed Form Expression"
] | [
"Second Principle of Mathematical Induction",
"Definition:Proposition",
"Second Principle of Mathematical Induction",
"Second Principle of Mathematical Induction",
"Second Principle of Mathematical Induction",
"Second Principle of Mathematical Induction"
] |
proofwiki-14643 | Recursively Defined Sequence/Examples/Minimum over k of Maximum of 1 plus Function of k and 2 plus Function of n-k | Consider the integer sequence $\sequence {\map f n}$ defined recusrively as:
:<nowiki>$\map f n = \begin{cases} 0 & : n = 1 \\
\ds \min_{0 \mathop < k \mathop < n} \map \max {1 + \map f k, 2 + \map f {n - k} } & : n > 1 \end{cases}$</nowiki>
$\map f n$ has a closed-form expression:
:$\map f n = m$ for $F_m < n \le F_{m... | The proof proceeds by induction.
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
:$\map f n = m$ for $F_m < n \le F_{m + 1}$ | Consider the [[Definition:Integer Sequence|integer sequence]] $\sequence {\map f n}$ defined recusrively as:
:<nowiki>$\map f n = \begin{cases} 0 & : n = 1 \\
\ds \min_{0 \mathop < k \mathop < n} \map \max {1 + \map f k, 2 + \map f {n - k} } & : n > 1 \end{cases}$</nowiki>
$\map f n$ has a [[Definition:Closed-Form Ex... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 1}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map f n = m$ for $F_m < n \le F_{m + 1}$ | Recursively Defined Sequence/Examples/Minimum over k of Maximum of 1 plus Function of k and 2 plus Function of n-k | https://proofwiki.org/wiki/Recursively_Defined_Sequence/Examples/Minimum_over_k_of_Maximum_of_1_plus_Function_of_k_and_2_plus_Function_of_n-k | https://proofwiki.org/wiki/Recursively_Defined_Sequence/Examples/Minimum_over_k_of_Maximum_of_1_plus_Function_of_k_and_2_plus_Function_of_n-k | [
"Integer Sequences",
"Fibonacci Numbers"
] | [
"Definition:Integer Sequence",
"Definition:Closed Form Expression",
"Definition:Fibonacci Number"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-14644 | Zeckendorf Representation of Integer shifted Left | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$
where:
:$\floor {\, \cdot \,}$ denotes the floor function
:$\phi$ denotes the golden mean.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $n$ be expressed in Zeckendorf representation:
:$n = F_{k_1} + F_{k_2} ... | We have:
{{begin-eqn}}
{{eqn | l = F_{k_j + 1} \hat \phi + F_{k_j}
| r = \hat \phi^{k_j + 1}
| c = Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less
}}
{{eqn | ll= \leadsto
| l = F_{k_j + 1}
| r = \hat \phi^{k_j} - \frac {F_{k_j} } {\hat \phi}
| c =
}}
{{eq... | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$
where:
:$\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]].
Let $n \in \Z_{\ge 0}$ be a [[Def... | We have:
{{begin-eqn}}
{{eqn | l = F_{k_j + 1} \hat \phi + F_{k_j}
| r = \hat \phi^{k_j + 1}
| c = [[Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less]]
}}
{{eqn | ll= \leadsto
| l = F_{k_j + 1}
| r = \hat \phi^{k_j} - \frac {F_{k_j} } {\hat \phi}
| c =
}}... | Zeckendorf Representation of Integer shifted Left | https://proofwiki.org/wiki/Zeckendorf_Representation_of_Integer_shifted_Left | https://proofwiki.org/wiki/Zeckendorf_Representation_of_Integer_shifted_Left | [
"Zeckendorf Representation"
] | [
"Definition:Real Function",
"Definition:Floor Function",
"Definition:Golden Mean",
"Definition:Positive/Integer",
"Definition:Zeckendorf Representation"
] | [
"Fibonacci Number by One Minus Golden Mean plus Fibonacci Number of Index One Less",
"Reciprocal Form of One Minus Golden Mean",
"Sum of Infinite Geometric Sequence",
"Golden Mean by One Minus Golden Mean equals Minus 1",
"Square of Golden Mean equals One plus Golden Mean",
"Reciprocal Form of One Minus G... |
proofwiki-14645 | Zeckendorf Representation of Integer shifted Right | Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$
where:
:$\floor {\, \cdot \,}$ denotes the floor function
:$\phi$ denotes the golden mean.
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $n$ be expressed in Zeckendorf representation:
:$n = F_{k_1} + F_{k_2} ... | Follows directly from Zeckendorf Representation of Integer shifted Left, substituting $F_{k_j - 1}$ for $F_{k_j}$ throughout.
{{qed}} | Let $f: \R \to \R$ be the [[Definition:Real Function|real function]] defined as:
:$\forall x \in \R: \map f x = \floor {x + \phi^{-1} }$
where:
:$\floor {\, \cdot \,}$ denotes the [[Definition:Floor Function|floor function]]
:$\phi$ denotes the [[Definition:Golden Mean|golden mean]].
Let $n \in \Z_{\ge 0}$ be a [[Def... | Follows directly from [[Zeckendorf Representation of Integer shifted Left]], substituting $F_{k_j - 1}$ for $F_{k_j}$ throughout.
{{qed}} | Zeckendorf Representation of Integer shifted Right | https://proofwiki.org/wiki/Zeckendorf_Representation_of_Integer_shifted_Right | https://proofwiki.org/wiki/Zeckendorf_Representation_of_Integer_shifted_Right | [
"Zeckendorf Representation"
] | [
"Definition:Real Function",
"Definition:Floor Function",
"Definition:Golden Mean",
"Definition:Positive/Integer",
"Definition:Zeckendorf Representation"
] | [
"Zeckendorf Representation of Integer shifted Left"
] |
proofwiki-14646 | Two Non-Negative Integers have Zeckendorf Representations of which one is Shifted Representation of the Other | Let $m, n \in \Z_{\ge 0}$ be non-negative integers.
Then there exists a unique set of integers:
:$\left\{ {k_1, k_2, \ldots, k_r}\right\}$
where:
: $k_1 \gg k_2 \gg \cdots \gg k_r$
where $a \gg b$ denotes that $a - b > 1$
such that:
:$m = F_{k_1} + F_{k_2} + \cdots + F_{k_r}$
and:
:$n = F_{k_1 + 1} + F_{k_2 + 1} + \cdo... | {{ProofWanted|When I have the concentration necessary. It's too hot to think at the moment.}} | Let $m, n \in \Z_{\ge 0}$ be [[Definition:Non-Negative Integer|non-negative integers]].
Then there exists a [[Definition:Unique|unique]] set of [[Definition:Integer|integers]]:
:$\left\{ {k_1, k_2, \ldots, k_r}\right\}$
where:
: $k_1 \gg k_2 \gg \cdots \gg k_r$
where $a \gg b$ denotes that $a - b > 1$
such that:
:$m... | {{ProofWanted|When I have the concentration necessary. It's too hot to think at the moment.}} | Two Non-Negative Integers have Zeckendorf Representations of which one is Shifted Representation of the Other | https://proofwiki.org/wiki/Two_Non-Negative_Integers_have_Zeckendorf_Representations_of_which_one_is_Shifted_Representation_of_the_Other | https://proofwiki.org/wiki/Two_Non-Negative_Integers_have_Zeckendorf_Representations_of_which_one_is_Shifted_Representation_of_the_Other | [
"Zeckendorf Representation"
] | [
"Definition:Positive/Integer",
"Definition:Unique",
"Definition:Integer",
"Definition:Negative/Integer"
] | [] |
proofwiki-14647 | Existence of Interval of Convergence of Power Series/Corollary 2 | Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n x^n$ be a power series.
Let $\map S x$ be convergent at $x = x_0$.
Then $\map S x$ is convergent for all $x$ such that $\sequence x < \sequence {x_0}$. | Let $\map S x$ converge when $x = x_0$.
Then by Existence of Interval of Convergence of Power Series, $x_0$ is in the interval of convergence of $\map S x$ whose midpoint is $0$.
The result follows by definition of interval of convergence.
{{qed}} | Let $\ds \map S x = \sum_{n \mathop = 0}^\infty a_n x^n$ be a [[Definition:Power Series|power series]].
Let $\map S x$ be [[Definition:Convergent Series|convergent]] at $x = x_0$.
Then $\map S x$ is [[Definition:Convergent Series|convergent]] for all $x$ such that $\sequence x < \sequence {x_0}$. | Let $\map S x$ [[Definition:Convergent Series|converge]] when $x = x_0$.
Then by [[Existence of Interval of Convergence of Power Series]], $x_0$ is in the [[Definition:Interval of Convergence|interval of convergence]] of $\map S x$ whose [[Definition:Midpoint of Interval|midpoint]] is $0$.
The result follows by defin... | Existence of Interval of Convergence of Power Series/Corollary 2 | https://proofwiki.org/wiki/Existence_of_Interval_of_Convergence_of_Power_Series/Corollary_2 | https://proofwiki.org/wiki/Existence_of_Interval_of_Convergence_of_Power_Series/Corollary_2 | [
"Existence of Interval of Convergence of Power Series"
] | [
"Definition:Power Series",
"Definition:Convergent Series",
"Definition:Convergent Series"
] | [
"Definition:Convergent Series",
"Existence of Interval of Convergence of Power Series",
"Definition:Interval of Convergence",
"Definition:Real Interval/Midpoint",
"Definition:Interval of Convergence"
] |
proofwiki-14648 | Linear Combination of Generating Functions | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
and $\map H z$ be the generating function for the sequence $\sequence {b_n}$.
Then $\alpha \map G z + \beta \map H z$ is the generating function for the sequence $\sequence {\alpha a_n + \beta b_n}$. | By definition:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Let $\map G z$ and $\map H z$ converge to $x$ and $y$ respectively for some $z_0 \in \R_{>0}$.
Then from Linear Combination of Convergent Series:
:$\ds \sum_{n \mathop \ge 0} \paren {\alpha a_n + \beta b_n} ... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
and $\map H z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {b_n}$.
Then $\alpha \map G z + \beta \map H z$ is ... | By definition:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Let $\map G z$ and $\map H z$ [[Definition:Convergent Series|converge]] to $x$ and $y$ respectively for some $z_0 \in \R_{>0}$.
Then from [[Linear Combination of Convergent Series]]:
:$\ds \sum_{n \mathop ... | Linear Combination of Generating Functions | https://proofwiki.org/wiki/Linear_Combination_of_Generating_Functions | https://proofwiki.org/wiki/Linear_Combination_of_Generating_Functions | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Convergent Series",
"Linear Combination of Convergent Series"
] |
proofwiki-14649 | Generating Function by Power of Parameter | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a non-negative integer.
Then $z^m \map G z$ is the generating function for the sequence $\sequence {a_{n - m} }$. | {{begin-eqn}}
{{eqn | l = z^m \map G z
| r = z^m \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n + m}
| c =
}}
{{eqn | r = \sum_{n + m \mathop \ge 0} a_{n - m} z^n
| c = Translation of Index Variable of Summation
}}
{{eqn | r = \... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then $z^m \map G z$ is the [[Definition:Generating Function|generating function]] for the [[De... | {{begin-eqn}}
{{eqn | l = z^m \map G z
| r = z^m \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n + m}
| c =
}}
{{eqn | r = \sum_{n + m \mathop \ge 0} a_{n - m} z^n
| c = [[Translation of Index Variable of Summation]]
}}
{{eqn | r... | Generating Function by Power of Parameter | https://proofwiki.org/wiki/Generating_Function_by_Power_of_Parameter | https://proofwiki.org/wiki/Generating_Function_by_Power_of_Parameter | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Positive/Integer",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Translation of Index Variable of Summation"
] |
proofwiki-14650 | Generating Function Divided by Power of Parameter | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a non-negative integer.
Then $\dfrac 1 {z^m} \paren {\map G z - \ds \sum_{k \mathop = 0}^{m - 1} a_k z^k}$ is the generating function for the sequence $\sequence {a_{n + m} }$. | {{begin-eqn}}
{{eqn | l = z^{-m} \map G z
| r = z^{-m} \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n - m}
| c =
}}
{{eqn | r = \sum_{n + m \mathop \ge 0} a_{n + m} z^n
| c = Translation of Index Variable of Summation
}}
{{eqn |... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Let $m \in \Z_{\ge 0}$ be a [[Definition:Non-Negative Integer|non-negative integer]].
Then $\dfrac 1 {z^m} \paren {\map G z - \ds \sum_{k \mathop = 0}^{m - 1} a_k z^k}$ is the [[De... | {{begin-eqn}}
{{eqn | l = z^{-m} \map G z
| r = z^{-m} \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} a_n z^{n - m}
| c =
}}
{{eqn | r = \sum_{n + m \mathop \ge 0} a_{n + m} z^n
| c = [[Translation of Index Variable of Summation]]
}}
{{e... | Generating Function Divided by Power of Parameter | https://proofwiki.org/wiki/Generating_Function_Divided_by_Power_of_Parameter | https://proofwiki.org/wiki/Generating_Function_Divided_by_Power_of_Parameter | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Positive/Integer",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Translation of Index Variable of Summation",
"Translation of Index Variable of Summation"
] |
proofwiki-14651 | Generating Function for Linearly Recurrent Sequence | Let $\sequence {a_n}$ be a linearly recurrent sequence defined as:
:$a_n = \begin{cases} b_n & : 1 \le n \le m \\ c_1 a_{n - 1} + c_2 a_{n - 2} + \cdots + c_m a_{n - m} & : n > m \end{cases}$
where:
:$m \in \Z_{>0}$ is a (strictly) positive integer
:$b_1, \ldots, b_m$ are constants.
Then the generating function for $\s... | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 1} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop > m} a_n z^n + \sum_{n \mathop = 1}^m a_n z^n
| c =
}}
{{eqn | r = \sum_{n \mathop > m} \paren {\sum_{n \mathop = 1}^m c_k a_{n-k} } z^n + \sum_{n \mathop = 1}^m ... | Let $\sequence {a_n}$ be a [[Definition:Linearly Recurrent Sequence|linearly recurrent sequence]] defined as:
:$a_n = \begin{cases} b_n & : 1 \le n \le m \\ c_1 a_{n - 1} + c_2 a_{n - 2} + \cdots + c_m a_{n - m} & : n > m \end{cases}$
where:
:$m \in \Z_{>0}$ is a [[Definition:Strictly Positive Integer|(strictly) posit... | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 1} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop > m} a_n z^n + \sum_{n \mathop = 1}^m a_n z^n
| c =
}}
{{eqn | r = \sum_{n \mathop > m} \paren {\sum_{n \mathop = 1}^m c_k a_{n-k} } z^n + \sum_{n \mathop = 1}^m ... | Generating Function for Linearly Recurrent Sequence | https://proofwiki.org/wiki/Generating_Function_for_Linearly_Recurrent_Sequence | https://proofwiki.org/wiki/Generating_Function_for_Linearly_Recurrent_Sequence | [
"Generating Functions"
] | [
"Definition:Linearly Recurrent Sequence",
"Definition:Strictly Positive/Integer",
"Definition:Constant",
"Definition:Generating Function",
"Definition:Polynomial"
] | [
"Exchange of Order of Summation/Finite and Infinite Series",
"Generating Function by Power of Parameter"
] |
proofwiki-14652 | Product of Generating Functions | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $\map H z$ be the generating function for the sequence $\sequence {b_n}$.
Then $\map G z \map H z$ is the generating function for the sequence $\sequence {c_n}$, where:
:$\forall n \in \Z_{\ge 0}: c_n = \ds \sum_{k \mathop = 0}^n a_k b_{n... | By definition of generating function:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Then:
$\quad \begin {array}{c|ccccc}
\times & b_0 & b_1 z & b_2 z^2 & b_3 z^3 & \cdots \\
\hline
a_0 & a_0 b_0 & a_0 b_1 z & a_0 b_2 z^2 & a_0 b_3 z^3 & \cdots \\
a_1 z & a_1 b_0 z & a... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Let $\map H z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {b_n}$.
Then $\map G z \map H z$ is the [[Definitio... | By definition of [[Definition:Generating Function|generating function]]:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
:$\map H z = \ds \sum_{n \mathop \ge 0} b_n z^n$
Then:
$\quad \begin {array}{c|ccccc}
\times & b_0 & b_1 z & b_2 z^2 & b_3 z^3 & \cdots \\
\hline
a_0 & a_0 b_0 & a_0 b_1 z & a_0 b_2 z^2 & a_0 b_... | Product of Generating Functions | https://proofwiki.org/wiki/Product_of_Generating_Functions | https://proofwiki.org/wiki/Product_of_Generating_Functions | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Generating Function",
"Definition:Generating Function"
] |
proofwiki-14653 | Generating Function for Sequence of Partial Sums of Series | Let $s$ be the the series:
:$\ds s = \sum_{n \mathop = 1}^\infty a_n = a_0 + a_1 + a_2 + a_3 + \cdots$
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let $\sequence {c_n}$ denote the sequence of partial sums of $s$.
Then the generating function for $\sequence {c_n}$ is given by:
:$\ds \df... | By definition of sequence of partial sums of $s$:
{{begin-eqn}}
{{eqn | l = \sequence {c_n}
| r = a_0 + \paren {a_0 + a_1} + \paren {a_0 + a_1 + a_2} + \cdots
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n a_k}
| c =
}}
{{end-eqn}}
Consider the sequence $\sequence {b_n}$ ... | Let $s$ be the the [[Definition:Series|series]]:
:$\ds s = \sum_{n \mathop = 1}^\infty a_n = a_0 + a_1 + a_2 + a_3 + \cdots$
Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Let $\sequence {c_n}$ denote the [[Definition:Sequence o... | By definition of [[Definition:Sequence of Partial Sums|sequence of partial sums]] of $s$:
{{begin-eqn}}
{{eqn | l = \sequence {c_n}
| r = a_0 + \paren {a_0 + a_1} + \paren {a_0 + a_1 + a_2} + \cdots
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {\sum_{k \mathop = 0}^n a_k}
| c =
}}
{{end-eqn}}... | Generating Function for Sequence of Partial Sums of Series | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Partial_Sums_of_Series | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Partial_Sums_of_Series | [
"Generating Functions",
"Series"
] | [
"Definition:Series",
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Series/Sequence of Partial Sums",
"Definition:Generating Function"
] | [
"Definition:Series/Sequence of Partial Sums",
"Definition:Sequence",
"Definition:Generating Function",
"Generating Function for Constant Sequence",
"Product of Generating Functions",
"Definition:Generating Function"
] |
proofwiki-14654 | Product of Generating Functions/General Rule | Let $\map {G_0} z, \map {G_1} z, \map {G_2} z, \ldots$ be any number of generating functions (up to countably infinite) for the sequences $\sequence {a_0 n}, \sequence {a_1 n}, \sequence {a_2 n}, \ldots$
Then:
{{begin-eqn}}
{{eqn | l = \prod_{j \mathop \ge 0} \map {G_j} z
| r = \prod_{j \mathop \ge 0} \sum_{k \ma... | {{ProofWanted|There may be some general convolution result that can be used. Good luck Jim.}} | Let $\map {G_0} z, \map {G_1} z, \map {G_2} z, \ldots$ be any number of [[Definition:Generating Function|generating functions]] (up to [[Definition:Countably Infinite Set|countably infinite]]) for the [[Definition:Sequence|sequences]] $\sequence {a_0 n}, \sequence {a_1 n}, \sequence {a_2 n}, \ldots$
Then:
{{begin-eqn... | {{ProofWanted|There may be some general convolution result that can be used. Good luck Jim.}} | Product of Generating Functions/General Rule | https://proofwiki.org/wiki/Product_of_Generating_Functions/General_Rule | https://proofwiki.org/wiki/Product_of_Generating_Functions/General_Rule | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Countably Infinite/Set",
"Definition:Sequence"
] | [] |
proofwiki-14655 | Product of Exponential Generating Functions | Let $\map G z$ be the exponential generating function for the sequence $\sequence {\dfrac {a_n} {n!} }$.
Let $\map H z$ be the exponential generating function for the sequence $\sequence {\dfrac {b_n} {n!} }$.
Then $\map G z \map H z$ is the generating function for the sequence $\sequence {\dfrac {c_n} {n!} }$, where:
... | Let $\map G z \map H z$ be the generating function for the sequence $\sequence {c'_n}$.
By definition of generating function:
{{begin-eqn}}
{{eqn | l = \map G z \map H z
| r = \sum_{k \mathop \ge 0} \dfrac {a_k} {k!} z^k \sum_{k \mathop \ge 0} \dfrac {b_k} {k!} z^k
| c =
}}
{{eqn | r = \paren {\dfrac {a_0}... | Let $\map G z$ be the [[Definition:Exponential Generating Function|exponential generating function]] for the [[Definition:Sequence|sequence]] $\sequence {\dfrac {a_n} {n!} }$.
Let $\map H z$ be the [[Definition:Exponential Generating Function|exponential generating function]] for the [[Definition:Sequence|sequence]] $... | Let $\map G z \map H z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {c'_n}$.
By definition of [[Definition:Generating Function|generating function]]:
{{begin-eqn}}
{{eqn | l = \map G z \map H z
| r = \sum_{k \mathop \ge 0} \dfrac {a_k} {k!} ... | Product of Exponential Generating Functions/Proof | https://proofwiki.org/wiki/Product_of_Exponential_Generating_Functions | https://proofwiki.org/wiki/Product_of_Exponential_Generating_Functions/Proof | [
"Generating Functions",
"Product of Exponential Generating Functions"
] | [
"Definition:Exponential Generating Function",
"Definition:Sequence",
"Definition:Exponential Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Product of Generating Functions"
] |
proofwiki-14656 | Generating Function of Multiple of Parameter | Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$.
Let $c$ be a constant.
Then $G \left({c z}\right)$ be the generating function for the sequence $\left\langle{b_n}\right\rangle$ where:
:$\forall n \in \Z_{\ge 0}: b_n = c^n a_n$ | {{begin-eqn}}
{{eqn | l = G \left({c z}\right)
| r = \sum_{n \mathop \ge 0} a_n \left({c z}\right)^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} \left({a_n c^n}\right) z^n
| c =
}}
{{end-eqn}}
Hence the result by definition of generating function.
{{qed}} | Let $G \left({z}\right)$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\left\langle{a_n}\right\rangle$.
Let $c$ be a [[Definition:Constant|constant]].
Then $G \left({c z}\right)$ be the [[Definition:Generating Function|generating function]] for the [[Definiti... | {{begin-eqn}}
{{eqn | l = G \left({c z}\right)
| r = \sum_{n \mathop \ge 0} a_n \left({c z}\right)^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{n \mathop \ge 0} \left({a_n c^n}\right) z^n
| c =
}}
{{end-eqn}}
Hence the result by definition of [[Definition:Generating Function|generating... | Generating Function of Multiple of Parameter | https://proofwiki.org/wiki/Generating_Function_of_Multiple_of_Parameter | https://proofwiki.org/wiki/Generating_Function_of_Multiple_of_Parameter | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Constant",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Generating Function"
] |
proofwiki-14657 | Generating Function for Sequence of Powers of Constant | Let $c \in \R$ be a constant.
Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \Z_{\ge 0}: a_n = c^n$
That is:
:$\sequence {a_n} = 1, c, c^2, c^3, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - c z}$ | Consider the sequence $\sequence {b_n}$ defined as:
: $\forall n \in \Z_{\ge 0}: b_n = 1$
Let $\map H z$ be the generating function for $\sequence {b_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \map H z
| r = \dfrac 1 {1 - z}
| c = Generating Function for Constant Sequence
}}
{{eqn | ll= \leadsto
| l =\map H... | Let $c \in \R$ be a [[Definition:Constant|constant]].
Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \Z_{\ge 0}: a_n = c^n$
That is:
:$\sequence {a_n} = 1, c, c^2, c^3, \ldots$
Then the [[Definition:Generating Function|generating function]] for $\sequence {a_n}$ is given as... | Consider the [[Definition:Sequence|sequence]] $\sequence {b_n}$ defined as:
: $\forall n \in \Z_{\ge 0}: b_n = 1$
Let $\map H z$ be the [[Definition:Generating Function|generating function]] for $\sequence {b_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \map H z
| r = \dfrac 1 {1 - z}
| c = [[Generating Functio... | Generating Function for Sequence of Powers of Constant | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Powers_of_Constant | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Powers_of_Constant | [
"Generating Function for Sequence of Powers of Constant",
"Examples of Generating Functions"
] | [
"Definition:Constant",
"Definition:Sequence",
"Definition:Generating Function"
] | [
"Definition:Sequence",
"Definition:Generating Function",
"Generating Function for Constant Sequence",
"Generating Function of Multiple of Parameter",
"Definition:Generating Function"
] |
proofwiki-14658 | Generating Function for Even Terms of Sequence | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Consider the subsequence $\sequence {b_n} := \tuple {a_0, a_2, a_4, \ldots}$
Then the generating function for $\sequence {b_n}$ is:
:$\dfrac 1 2 \paren {\map G z + \map G {-z} }$ | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}
| c = separating out odd and even integers
}}
{{eqn | l = \map G {-z}
| r = \sum_{n \... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Consider the [[Definition:Subsequence|subsequence]] $\sequence {b_n} := \tuple {a_0, a_2, a_4, \ldots}$
Then the [[Definition:Generating Function|generating function]] for $\sequen... | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}
| c = separating out [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]] in... | Generating Function for Even Terms of Sequence | https://proofwiki.org/wiki/Generating_Function_for_Even_Terms_of_Sequence | https://proofwiki.org/wiki/Generating_Function_for_Even_Terms_of_Sequence | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Subsequence",
"Definition:Generating Function"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer"
] |
proofwiki-14659 | Generating Function for Odd Terms of Sequence | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Consider the subsequence $\sequence {b_n} := \tuple {a_1, a_3, a_5, \ldots}$
Then the generating function for $\sequence {b_n}$ is:
:$\dfrac 1 2 \paren {\map G z - \map G {-z} }$ | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}
| c = separating out odd and even integers
}}
{{eqn | l = \map G {-z}
| r = \sum_{n \... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Consider the [[Definition:Subsequence|subsequence]] $\sequence {b_n} := \tuple {a_1, a_3, a_5, \ldots}$
Then the [[Definition:Generating Function|generating function]] for $\sequen... | {{begin-eqn}}
{{eqn | l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{r \mathop \ge 0} a_{2 r} z^{2 r} + \sum_{r \mathop \ge 0} a_{2 r + 1} z^{2 r + 1}
| c = separating out [[Definition:Odd Integer|odd]] and [[Definition:Even Integer|even]] in... | Generating Function for Odd Terms of Sequence | https://proofwiki.org/wiki/Generating_Function_for_Odd_Terms_of_Sequence | https://proofwiki.org/wiki/Generating_Function_for_Odd_Terms_of_Sequence | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Subsequence",
"Definition:Generating Function"
] | [
"Definition:Odd Integer",
"Definition:Even Integer",
"Definition:Odd Integer",
"Definition:Even Integer"
] |
proofwiki-14660 | Generating Function for mth Terms of Sequence | Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\omega = e^{2 i \pi / m} = \cos \dfrac {2 \pi} m + i \sin \dfrac {2 \pi} m$.
Then for $r \in \Z$ such that $0 \le r < m$:
:$\displaystyle \sum_{n \bmod m \mat... | {{begin-eqn}}
{{eqn | l = \omega^{-k r} \map G {\omega^k z}
| r = \sum_{j \mathop \ge 0} a_j \omega^{k \paren {j - r} } z^j
| c = {{Defof|Generating Function}}
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 m \sum_{0 \mathop \le k \mathop < m} \omega^{-k r} \map G {\omega^k z}
| r = \dfrac 1 m \sum_{0 \... | Let $G \left({z}\right)$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\left\langle{a_n}\right\rangle$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let $\omega = e^{2 i \pi / m} = \cos \dfrac {2 \pi} m + i \si... | {{begin-eqn}}
{{eqn | l = \omega^{-k r} \map G {\omega^k z}
| r = \sum_{j \mathop \ge 0} a_j \omega^{k \paren {j - r} } z^j
| c = {{Defof|Generating Function}}
}}
{{eqn | ll= \leadsto
| l = \dfrac 1 m \sum_{0 \mathop \le k \mathop < m} \omega^{-k r} \map G {\omega^k z}
| r = \dfrac 1 m \sum_{0 \... | Generating Function for mth Terms of Sequence/Proof | https://proofwiki.org/wiki/Generating_Function_for_mth_Terms_of_Sequence | https://proofwiki.org/wiki/Generating_Function_for_mth_Terms_of_Sequence/Proof | [
"Generating Functions",
"Generating Function for mth Terms of Sequence"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Strictly Positive/Integer"
] | [
"Sum of Geometric Sequence",
"Definition:Iverson's Convention"
] |
proofwiki-14661 | Derivative of Generating Function | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map G z
| r = \sum_{k \mathop \ge 0} \left({k + 1}\right) a_{k + 1} z^k
| c =
}}
{{eqn | r = a_1 + 2 a_2 z + 3 a_3 z^3 + \cdots
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map G z
| r = \frac \d {\d z} \paren {\sum_{k \mathop \ge 0} a_k z^k}
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\frac \d {\d z} a_k z^k}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} k a_k z^{k - 1}
| c = Power Rul... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map G z
| r = \sum_{k \mathop \ge 0} \left({k + 1}\right) a_{k + 1} z^k
| c =
}}
{{eqn | r = a_1 + 2 a_2 z + 3 a_3 z^3 ... | {{begin-eqn}}
{{eqn | l = \frac \d {\d z} \map G z
| r = \frac \d {\d z} \paren {\sum_{k \mathop \ge 0} a_k z^k}
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\frac \d {\d z} a_k z^k}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} k a_k z^{k - 1}
| c = [[Power R... | Derivative of Generating Function | https://proofwiki.org/wiki/Derivative_of_Generating_Function | https://proofwiki.org/wiki/Derivative_of_Generating_Function | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Power Rule for Derivatives",
"Definition:Zeroth",
"Translation of Index Variable of Summation"
] |
proofwiki-14662 | Generating Function of Sequence by Index | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Then:
:$z \map {G'} z$ is the generating function for the sequence $\sequence {n a_n}$
where $\map {G'} z$ is the derivative of $\map G z$ {{WRT|Differentiation}} $z$. | {{begin-eqn}}
{{eqn | l = \map {G'} z
| r = \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^k
| c = Derivative of Generating Function
}}
{{eqn | ll= \leadsto
| l = z \map {G'} z
| r = \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^{k + 1}
| c =
}}
{{eqn | r = \sum_{k + 1 \mathop \ge ... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Then:
:$z \map {G'} z$ is the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {n a_n}$
where $\map {G'} z$ is the [[Defini... | {{begin-eqn}}
{{eqn | l = \map {G'} z
| r = \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^k
| c = [[Derivative of Generating Function]]
}}
{{eqn | ll= \leadsto
| l = z \map {G'} z
| r = \sum_{k \mathop \ge 0} \paren {k + 1} a_{k + 1} z^{k + 1}
| c =
}}
{{eqn | r = \sum_{k + 1 \mathop ... | Generating Function of Sequence by Index | https://proofwiki.org/wiki/Generating_Function_of_Sequence_by_Index | https://proofwiki.org/wiki/Generating_Function_of_Sequence_by_Index | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Derivative"
] | [
"Derivative of Generating Function",
"Translation of Index Variable of Summation",
"Definition:Generating Function"
] |
proofwiki-14663 | Integral of Generating Function | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^z \map G t \rd t
| r = \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k
| c =
}}
{{eqn | r = a_0 z + \dfrac {a_1 z^2} 2 + \dfrac {a_2 z^3} 3 + \dfrac {a_3 z^4} 4 + \cdots
| c =
}}
{{en... | {{begin-eqn}}
{{eqn | l = \int_0^z \map G t \rd t
| r = \int_0^z \paren {\sum_{k \mathop \ge 0} a_k t^k} \rd t
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\int_0^z a_k t^k \rd t}
| c = Tonelli's Theorem
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\intlimits {a_k... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Then:
{{begin-eqn}}
{{eqn | l = \int_0^z \map G t \rd t
| r = \sum_{k \mathop \ge 1} \dfrac {a_{k - 1} z^k} k
| c =
}}
{{eqn | r = a_0 z + \dfrac {a_1 z^2} 2 + \dfrac {... | {{begin-eqn}}
{{eqn | l = \int_0^z \map G t \rd t
| r = \int_0^z \paren {\sum_{k \mathop \ge 0} a_k t^k} \rd t
| c = {{Defof|Generating Function}}
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\int_0^z a_k t^k \rd t}
| c = [[Tonelli's Theorem]]
}}
{{eqn | r = \sum_{k \mathop \ge 0} \paren {\intlimits ... | Integral of Generating Function | https://proofwiki.org/wiki/Integral_of_Generating_Function | https://proofwiki.org/wiki/Integral_of_Generating_Function | [
"Generating Functions"
] | [
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Tonelli's Theorem",
"Primitive of Power",
"Translation of Index Variable of Summation"
] |
proofwiki-14664 | Generating Function for Sequence of Reciprocals of Natural Numbers | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = \dfrac 1 n$
That is:
:$\sequence {a_n} = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \map \ln {\dfrac 1 {1 - z} }$ | Take the sequence:
:$S_n = 1, 1, 1, \ldots$
From Generating Function for Constant Sequence, this has the generating function:
:$\ds \map G z = \sum_{n \mathop = 0}^\infty z^n = \frac 1 {1 - z}$
By Integral of Generating Function:
{{begin-eqn}}
{{eqn | l = \int_0^z \map G t \rd t
| r = \sum_{k \mathop \ge 1} \dfra... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N_{> 0}: a_n = \dfrac 1 n$
That is:
:$\sequence {a_n} = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
Then the [[Definition:Generating Function|generating function]] for $\sequence {a_n}$ is given as:
:$\map G z = \map \ln {\d... | Take the [[Definition:Sequence|sequence]]:
:$S_n = 1, 1, 1, \ldots$
From [[Generating Function for Constant Sequence]], this has the [[Definition:Generating Function|generating function]]:
:$\ds \map G z = \sum_{n \mathop = 0}^\infty z^n = \frac 1 {1 - z}$
By [[Integral of Generating Function]]:
{{begin-eqn}}
{{eq... | Generating Function for Sequence of Reciprocals of Natural Numbers | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Reciprocals_of_Natural_Numbers | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Reciprocals_of_Natural_Numbers | [
"Reciprocals",
"Examples of Generating Functions"
] | [
"Definition:Sequence",
"Definition:Generating Function"
] | [
"Definition:Sequence",
"Generating Function for Constant Sequence",
"Definition:Generating Function",
"Integral of Generating Function",
"Definition:Power Series",
"Definition:Coefficient",
"Primitive of Reciprocal",
"Integration by Substitution",
"Definition:Generating Function"
] |
proofwiki-14665 | Generating Function for Sequence of Harmonic Numbers | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th harmonic number.
That is:
:$\sequence {a_n} = 1, 1 + \dfrac 1 2, 1 + \dfrac 1 2 + \dfrac 1 3, \ldots$
Then the generating function for $\sequence {a_n}$ is given as:
:$\map G z = \dfrac 1 {1 - z} \map \l... | Take the sequence:
:$S_n = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
From Generating Function for Sequence of Reciprocals of Natural Numbers, this has the generating function:
:$\map H z = \map \ln {\dfrac 1 {1 - z} }$
By definition, $\sequence {a_n}$ is the sequence of partial sums of $\sequence {a_n}$.
The resul... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]].
That is:
:$\sequence {a_n} = 1, 1 + \dfrac 1 2, 1 + \dfrac 1 2 + \dfrac 1 3, \ldots$
Then the [[Definition:Generating Function|g... | Take the [[Definition:Sequence|sequence]]:
:$S_n = 1, \dfrac 1 2, \dfrac 1 3, \dfrac 1 4, \ldots$
From [[Generating Function for Sequence of Reciprocals of Natural Numbers]], this has the [[Definition:Generating Function|generating function]]:
:$\map H z = \map \ln {\dfrac 1 {1 - z} }$
By definition, $\sequence {a_... | Generating Function for Sequence of Harmonic Numbers | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Harmonic_Numbers | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Harmonic_Numbers | [
"Harmonic Numbers",
"Examples of Generating Functions"
] | [
"Definition:Sequence",
"Definition:Harmonic Numbers",
"Definition:Generating Function"
] | [
"Definition:Sequence",
"Generating Function for Sequence of Reciprocals of Natural Numbers",
"Definition:Generating Function",
"Definition:Series/Sequence of Partial Sums",
"Generating Function for Sequence of Partial Sums of Series"
] |
proofwiki-14666 | Binomial Theorem for Negative Index and Negative Parameter | Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $z \in \R$ be a real number such that $\size z < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 - z}^{n + 1} }
| r = \sum_{k \mathop \ge 0} \binom {-n - 1} k \paren {-z}^k
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {n + k} n z^k
| c =... | {{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 - z}^{n + 1} }
| r = \paren {1 + \paren {-z} }^{- n - 1}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {- n - 1} k \paren {-z}^k
| c = General Binomial Theorem
}}
{{eqn | r = \sum_{k \mathop \ge 0} \dbinom {n + 1 + k - 1} k \paren {-1}^k \paren {-z}^... | Let $n \in \Z_{\ge 0}$ be a [[Definition:Positive Integer|positive integer]].
Let $z \in \R$ be a [[Definition:Real Number|real number]] such that $\size z < 1$.
Then:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 - z}^{n + 1} }
| r = \sum_{k \mathop \ge 0} \binom {-n - 1} k \paren {-z}^k
| c =
}}
{{eq... | {{begin-eqn}}
{{eqn | l = \dfrac 1 {\paren {1 - z}^{n + 1} }
| r = \paren {1 + \paren {-z} }^{- n - 1}
| c =
}}
{{eqn | r = \sum_{k \mathop \ge 0} \binom {- n - 1} k \paren {-z}^k
| c = [[General Binomial Theorem]]
}}
{{eqn | r = \sum_{k \mathop \ge 0} \dbinom {n + 1 + k - 1} k \paren {-1}^k \paren {... | Binomial Theorem for Negative Index and Negative Parameter | https://proofwiki.org/wiki/Binomial_Theorem_for_Negative_Index_and_Negative_Parameter | https://proofwiki.org/wiki/Binomial_Theorem_for_Negative_Index_and_Negative_Parameter | [
"Binomial Theorem"
] | [
"Definition:Positive/Integer",
"Definition:Real Number",
"Definition:Binomial Coefficient/Real Numbers"
] | [
"Binomial Theorem/General Binomial Theorem",
"Negated Upper Index of Binomial Coefficient/Corollary 1"
] |
proofwiki-14667 | Power Series Expansion for Integer Power of Exponential Function minus 1 | Let $e^z$ denote the exponential function.
Then:
{{begin-eqn}}
{{eqn | l = \paren {e^z - 1}^n
| r = z^n + \dfrac 1 {n + 1} {n + 1 \brace n} z^{n + 1} + \cdots
| c =
}}
{{eqn | r = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}
| c =
}}
{{end-eqn}}
where $\ds {k \brace n}$ denotes a Stirlin... | The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\ds \paren {e^z - 1}^n = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = 0! \sum_{k \mathop \in \Z} {k \brace 0} \frac {z^k} {k!}
| r = \sum_{k \mathop \in ... | Let $e^z$ denote the [[Definition:Real Exponential Function|exponential function]].
Then:
{{begin-eqn}}
{{eqn | l = \paren {e^z - 1}^n
| r = z^n + \dfrac 1 {n + 1} {n + 1 \brace n} z^{n + 1} + \cdots
| c =
}}
{{eqn | r = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}
| c =
}}
{{end-eqn}}... | The proof proceeds by [[Principle of Mathematical Induction|induction]].
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\ds \paren {e^z - 1}^n = n! \sum_{k \mathop \in \Z} {k \brace n} \frac {z^k} {k!}$
$\map P 0$ is the case:
{{begin-eqn}}
{{eqn | l = 0! \sum_{k \mathop... | Power Series Expansion for Integer Power of Exponential Function minus 1/Proof | https://proofwiki.org/wiki/Power_Series_Expansion_for_Integer_Power_of_Exponential_Function_minus_1 | https://proofwiki.org/wiki/Power_Series_Expansion_for_Integer_Power_of_Exponential_Function_minus_1/Proof | [
"Exponential Function",
"Examples of Power Series",
"Power Series Expansion for Integer Power of Exponential Function minus 1"
] | [
"Definition:Exponential Function/Real",
"Definition:Stirling Numbers of the Second Kind"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Stirling Number of the Second Kind of n+1 with 1",
"Power Series Expansion for Exponential Function",
"Definition:Basis for the Induction",
"Definition:Induction Hypothesis",
"Definition:Induction Step",
"Power Series Expansion for Inte... |
proofwiki-14668 | Ultraproduct is Well-Defined | Ultraproduct is well-defined. | Specifically, following the definitions on ultraproduct, it is to be proved that:
:$(1) \quad f^\MM$ is well-defined
:$(2) \quad R^\MM$ is well-defined
First of all, we need to prove: | [[Definition:Ultraproduct|Ultraproduct]] is well-defined. | Specifically, following the definitions on [[Definition:Ultraproduct|ultraproduct]], it is to be proved that:
:$(1) \quad f^\MM$ is well-defined
:$(2) \quad R^\MM$ is well-defined
First of all, we need to prove: | Ultraproduct is Well-Defined | https://proofwiki.org/wiki/Ultraproduct_is_Well-Defined | https://proofwiki.org/wiki/Ultraproduct_is_Well-Defined | [
"Model Theory for Predicate Logic"
] | [
"Definition:Ultraproduct"
] | [
"Definition:Ultraproduct",
"Definition:Ultraproduct"
] |
proofwiki-14669 | Cauchy's Integral Formula/General Result/Corollary | Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Let the coefficient of $z^n$ extracted from $\map G z$ be denoted:
:$\sqbrk {z^n} \map G z := a_n$
Let $\map G z$ be convergent for $z = z_0$ and $0 < r < \cmod {z_0}$.
Then:
:$\sqbrk {z^n} \map G z = \ds \frac 1 {2 \pi i} \oint_{\cmod z \mat... | {{begin-eqn}}
{{eqn | l = \sqbrk {z^n} \map G z
| r = \frac 1 {n!} \map {G^\paren n} 0
| c = {{Corollary|Derivative of Generating Function/General Result|disp = Derivative of Generating Function: General Result}}
}}
{{eqn | r = \frac 1 {n!} \paren {\frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map G z} {z... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {a_n}$.
Let the [[Definition:Generating Function/Extraction of Coefficient|coefficient of $z^n$ extracted from $\map G z$]] be denoted:
:$\sqbrk {z^n} \map G z := a_n$
Let $\map G z$ be [[D... | {{begin-eqn}}
{{eqn | l = \sqbrk {z^n} \map G z
| r = \frac 1 {n!} \map {G^\paren n} 0
| c = {{Corollary|Derivative of Generating Function/General Result|disp = Derivative of Generating Function: General Result}}
}}
{{eqn | r = \frac 1 {n!} \paren {\frac {n!} {2 \pi i} \oint_{\partial D} \frac {\map G z} {z... | Cauchy's Integral Formula/General Result/Corollary | https://proofwiki.org/wiki/Cauchy's_Integral_Formula/General_Result/Corollary | https://proofwiki.org/wiki/Cauchy's_Integral_Formula/General_Result/Corollary | [
"Generating Functions",
"Cauchy's Integral Formula"
] | [
"Definition:Generating Function",
"Definition:Sequence",
"Definition:Generating Function/Extraction of Coefficient",
"Definition:Convergent Mapping/Complex Function"
] | [
"Cauchy's Integral Formula/General Result"
] |
proofwiki-14670 | Summation of Products of n Numbers taken m at a time with Repetitions | Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \par... | === Lemma 1 ===
{{:Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1}} | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $b \ge a$.
Let $U$ be a [[Definition:Set|set]] of $n = b - a + 1$ [[Definition:Number|numbers]] $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let:
{{begin-eqn}}
{{eq... | === [[Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1|Lemma 1]] ===
{{:Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1}} | Summation of Products of n Numbers taken m at a time with Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | [
"Summations",
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Integer",
"Definition:Set",
"Definition:Number",
"Definition:Strictly Positive/Integer",
"Definition:Multiplication",
"Definition:Ordered Tuple",
"Definition:Element"
] | [
"Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1"
] |
proofwiki-14671 | Summation of Products of n Numbers taken m at a time with Repetitions | Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \par... | Let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j
| c =
}}
{{eqn | r = \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j
| c = Summation of i from 1 to n of Summation of j from 1 to i
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \mathop... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $b \ge a$.
Let $U$ be a [[Definition:Set|set]] of $n = b - a + 1$ [[Definition:Number|numbers]] $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let:
{{begin-eqn}}
{{eq... | Let:
{{begin-eqn}}
{{eqn | l = S_1
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j
| c =
}}
{{eqn | r = \sum_{j \mathop = a}^b \sum_{i \mathop = j}^b x_i x_j
| c = [[Summation of i from 1 to n of Summation of j from 1 to i]]
}}
{{eqn | r = \sum_{i \mathop = a}^b \sum_{j \ma... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2/Proof_1 | [
"Summations",
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Integer",
"Definition:Set",
"Definition:Number",
"Definition:Strictly Positive/Integer",
"Definition:Multiplication",
"Definition:Ordered Tuple",
"Definition:Element"
] | [
"Summation of i from 1 to n of Summation of j from 1 to i",
"Change of Index Variable of Summation",
"Sum of Summations over Overlapping Domains/Example",
"Sum of Summations equals Summation of Sum",
"General Distributivity Theorem",
"Change of Index Variable of Summation"
] |
proofwiki-14672 | Summation of Products of n Numbers taken m at a time with Repetitions | Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \par... | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $b \ge a$.
Let $U$ be a [[Definition:Set|set]] of $n = b - a + 1$ [[Definition:Number|numbers]] $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let:
{{begin-eqn}}
{{eq... | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i x_i x_j = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le b} x_{j_1} x_{j_2}$
From [[Summation of Products of n Numbers taken m at a time with Repetitions]]:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} ... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 2/Proof 2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_2/Proof_2 | [
"Summations",
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Integer",
"Definition:Set",
"Definition:Number",
"Definition:Strictly Positive/Integer",
"Definition:Multiplication",
"Definition:Ordered Tuple",
"Definition:Element"
] | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] |
proofwiki-14673 | Summation of Products of n Numbers taken m at a time with Repetitions | Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \par... | Let:
{{begin-eqn}}
{{eqn | n = a
| l = A
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k
| c =
}}
{{eqn | r = \sum_{a \mathop \le i \mathop \le j \mathop \le k \mathop \le b} x_i x_j x_k
| c =
}}
{{eqn | n = b
| r = \sum_{i \mathop = ... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $b \ge a$.
Let $U$ be a [[Definition:Set|set]] of $n = b - a + 1$ [[Definition:Number|numbers]] $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let:
{{begin-eqn}}
{{eq... | Let:
{{begin-eqn}}
{{eqn | n = a
| l = A
| o = :=
| r = \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k
| c =
}}
{{eqn | r = \sum_{a \mathop \le i \mathop \le j \mathop \le k \mathop \le b} x_i x_j x_k
| c =
}}
{{eqn | n = b
| r = \sum_{i \mathop =... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3/Proof 1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3/Proof_1 | [
"Summations",
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Integer",
"Definition:Set",
"Definition:Number",
"Definition:Strictly Positive/Integer",
"Definition:Multiplication",
"Definition:Ordered Tuple",
"Definition:Element"
] | [] |
proofwiki-14674 | Summation of Products of n Numbers taken m at a time with Repetitions | Let $a, b \in \Z$ be integers such that $b \ge a$.
Let $U$ be a set of $n = b - a + 1$ numbers $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let:
{{begin-eqn}}
{{eqn | l = h_m
| r = \sum_{a \mathop \le j_1 \mathop \le \mathop \cdots \mathop \le j_m \mathop \le b} \par... | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$
From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cd... | Let $a, b \in \Z$ be [[Definition:Integer|integers]] such that $b \ge a$.
Let $U$ be a [[Definition:Set|set]] of $n = b - a + 1$ [[Definition:Number|numbers]] $\set {x_a, x_{a + 1}, \ldots, x_b}$.
Let $m \in \Z_{>0}$ be a [[Definition:Strictly Positive Integer|(strictly) positive integer]].
Let:
{{begin-eqn}}
{{eq... | We have that:
:$\ds \sum_{i \mathop = a}^b \sum_{j \mathop = a}^i \sum_{k \mathop = a}^j x_i x_j x_k = \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le b} x_{j_1} x_{j_2} x_{j_3}$
From [[Summation of Products of n Numbers taken m at a time with Repetitions]]:
:$\ds \sum_{a \mathop \le j_1 \mathop... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 3/Proof 2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_3/Proof_2 | [
"Summations",
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Integer",
"Definition:Set",
"Definition:Number",
"Definition:Strictly Positive/Integer",
"Definition:Multiplication",
"Definition:Ordered Tuple",
"Definition:Element"
] | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] |
proofwiki-14675 | Derivative of Generating Function for Sequence of Harmonic Numbers | Let $\sequence {a_n}$ be the sequence defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th harmonic number.
Let $\map G z$ be the generating function for $\sequence {a_n}$:
:$\map G z = \dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} }$
from Generating Function for Sequence of Harmonic Numbers.
Th... | {{begin-eqn}}
{{eqn | l = \map {G'} z
| r = \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} } }
| c =
}}
{{eqn | r = \dfrac 1 {1 - z} \map {\dfrac \d {\d z} } {\map \ln {\dfrac 1 {1 - z} } } + \map \ln {\dfrac 1 {1 - z} } \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} }
| c = Produc... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] defined as:
:$\forall n \in \N_{> 0}: a_n = H_n$
where $H_n$ denotes the $n$th [[Definition:Harmonic Number|harmonic number]].
Let $\map G z$ be the [[Definition:Generating Function|generating function]] for $\sequence {a_n}$:
:$\map G z = \dfrac 1 {1 - z} ... | {{begin-eqn}}
{{eqn | l = \map {G'} z
| r = \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} \map \ln {\dfrac 1 {1 - z} } }
| c =
}}
{{eqn | r = \dfrac 1 {1 - z} \map {\dfrac \d {\d z} } {\map \ln {\dfrac 1 {1 - z} } } + \map \ln {\dfrac 1 {1 - z} } \map {\dfrac \d {\d z} } {\dfrac 1 {1 - z} }
| c = [[Prod... | Derivative of Generating Function for Sequence of Harmonic Numbers | https://proofwiki.org/wiki/Derivative_of_Generating_Function_for_Sequence_of_Harmonic_Numbers | https://proofwiki.org/wiki/Derivative_of_Generating_Function_for_Sequence_of_Harmonic_Numbers | [
"Generating Functions",
"Harmonic Numbers"
] | [
"Definition:Sequence",
"Definition:Harmonic Numbers",
"Definition:Generating Function",
"Generating Function for Sequence of Harmonic Numbers",
"Definition:Derivative"
] | [
"Product Rule for Derivatives",
"Derivative of Logarithm Function",
"Derivative of Composite Function",
"Power Rule for Derivatives"
] |
proofwiki-14676 | Generating Function for Sequence of Sum over k to n of Reciprocal of k by n-k | Let $\sequence {a_n}$ be the sequence whose terms are defined as:
:$\forall n \in \Z_{\ge 0}: a_n = \ds \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k \paren {n - k} }$
Then $\sequence {a_n}$ has the generating function $\map G z$ such that:
:$\map G z = \paren {\ln \dfrac 1 {1 - z} }^2$
and whose terms are:
:$a_n = \dfrac {... | From Product of Generating Functions:
:$\map G z = \paren {\map {G_1} z}^2$
where $\map {G_1} z$ is the generating function for $\ds \sum_{k \mathop \ge 1} \dfrac 1 k$.
From Generating Function for Sequence of Reciprocals of Natural Numbers:
:$\map {G_1} z = \map \ln {\dfrac 1 {1 - z} }$
Hence:
:$\map G z = \paren {\ln... | Let $\sequence {a_n}$ be the [[Definition:Sequence|sequence]] whose [[Definition:Term of Sequence|terms]] are defined as:
:$\forall n \in \Z_{\ge 0}: a_n = \ds \sum_{k \mathop = 1}^{n - 1} \dfrac 1 {k \paren {n - k} }$
Then $\sequence {a_n}$ has the [[Definition:Generating Function|generating function]] $\map G z$ su... | From [[Product of Generating Functions]]:
:$\map G z = \paren {\map {G_1} z}^2$
where $\map {G_1} z$ is the [[Definition:Generating Function|generating function]] for $\ds \sum_{k \mathop \ge 1} \dfrac 1 k$.
From [[Generating Function for Sequence of Reciprocals of Natural Numbers]]:
:$\map {G_1} z = \map \ln {\dfra... | Generating Function for Sequence of Sum over k to n of Reciprocal of k by n-k | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Sum_over_k_to_n_of_Reciprocal_of_k_by_n-k | https://proofwiki.org/wiki/Generating_Function_for_Sequence_of_Sum_over_k_to_n_of_Reciprocal_of_k_by_n-k | [
"Examples of Generating Functions"
] | [
"Definition:Sequence",
"Definition:Term of Sequence",
"Definition:Generating Function",
"Definition:Term of Sequence"
] | [
"Product of Generating Functions",
"Definition:Generating Function",
"Generating Function for Sequence of Reciprocals of Natural Numbers",
"Definition:Differentiation",
"Derivative of Composite Function",
"Power Rule for Derivatives",
"Derivative of Composite Function",
"Derivative of Logarithm Functi... |
proofwiki-14677 | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 4 | :$\ds \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} + \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 + \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$. | From Summation of Products of n Numbers taken m at a time with Repetitions:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^... | :$\ds \sum_{a \mathop \le j_1 \mathop \le j_2 \mathop \le j_3 \mathop \le j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} + \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 + \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$. | From [[Summation of Products of n Numbers taken m at a time with Repetitions]]:
:$\ds \sum_{a \mathop \le j_1 \mathop \le \cdots \mathop \le j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \cdots \mathop + m k_m \mathop = m} } \dfrac { {... | Summation of Products of n Numbers taken m at a time with Repetitions/Examples/Order 4 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_4 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Examples/Order_4 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] |
proofwiki-14678 | Summation of Products of n Numbers taken m at a time with Repetitions/Recurrence Formula | A recurrence relation for $h_n$ can be given as:
{{begin-eqn}}
{{eqn | l = h_n
| r = \sum_{k \mathop = 1}^n \dfrac {S_k h_{n - k} } n
| c =
}}
{{eqn | r = \dfrac 1 n \paren {S_1 h_{n - 1} + S_2 h_{n - 2} + \cdots S_n h_0}
| c =
}}
{{end-eqn}}
for $n \ge 1$. | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k
| c = Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2
}}
{{eqn | ll= \leadsto
| l = \map {\dfrac \d {\d z} } {\map \ln {\map G z} }
| r = \map {\dfrac \d {\d z} } {\sum_{k... | A [[Definition:Recurrence Relation|recurrence relation]] for $h_n$ can be given as:
{{begin-eqn}}
{{eqn | l = h_n
| r = \sum_{k \mathop = 1}^n \dfrac {S_k h_{n - k} } n
| c =
}}
{{eqn | r = \dfrac 1 n \paren {S_1 h_{n - 1} + S_2 h_{n - 2} + \cdots S_n h_0}
| c =
}}
{{end-eqn}}
for $n \ge 1$. | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k
| c = [[Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2|Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2]]
}}
{{eqn | ll= \leadsto
| l = \map {\dfra... | Summation of Products of n Numbers taken m at a time with Repetitions/Recurrence Formula | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Recurrence_Formula | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Recurrence_Formula | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Recursive Sequence/Recurrence Relation"
] | [
"Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2",
"Derivative of Logarithm Function",
"Derivative of Composite Function",
"Derivative of Generating Function",
"Power Rule for Derivatives",
"Product of Generating Functions"
] |
proofwiki-14679 | Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1 | Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}
| c =
}}
{{eqn | r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \cdots \paren {1 - x_b z} }
| c =
}}
{{end-eqn}} | For each $k \in \set {a, a + 1, \ldots, b}$, the product of $x_k$ taken $m$ at a time is simply ${x_k}^m$.
Thus for $n = 1$ we have:
:$h_m = {x_k}^m$
Let the generating function for such a $\sequence {h_m}$ be $\map {G_k} z$.
From Generating Function for Sequence of Powers of Constant:
:$\map {G_k} z = \dfrac 1 {1 - x_... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {h_m}$.
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \prod_{k \mathop = a}^b \dfrac 1 {1 - x_k z}
| c =
}}
{{eqn | r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \c... | For each $k \in \set {a, a + 1, \ldots, b}$, the [[Definition:Product (Algebra)|product]] of $x_k$ taken $m$ at a time is simply ${x_k}^m$.
Thus for $n = 1$ we have:
:$h_m = {x_k}^m$
Let the [[Definition:Generating Function|generating function]] for such a $\sequence {h_m}$ be $\map {G_k} z$.
From [[Generating Funct... | Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Lemma_1 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Lemma_1 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Multiplication/Product",
"Definition:Generating Function",
"Generating Function for Sequence of Powers of Constant",
"Product of Summations",
"Product of Generating Functions/General Rule"
] |
proofwiki-14680 | Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2 | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \map G z
| r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \cdots \paren {1 - x_b z} }
| c = Lemma 1
}}
{{eqn | ll= \leadsto
| l = \map \ln {\map G z}
| r = \ln \dfrac 1 {1 - x_a z} + \ln \dfrac 1 {1 - x_{a + 1} z} + \cdots + \ln \dfrac 1 {1 - x_b z}
| c... | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \dfrac {S_k z^k} k
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \map G z
| r = \dfrac 1 {\paren {1 - x_a z} \paren {1 - x_{a + 1} z} \cdots \paren {1 - x_b z} }
| c = [[Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1|Lemma 1]]
}}
{{eqn | ll= \leadsto
| l = \map \ln {\map G z}
| r = \ln \dfrac 1 {1 - x_a... | Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Lemma_2 | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Lemma_2 | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 1",
"Sum of Logarithms",
"Generating Function for Sequence of Reciprocals of Natural Numbers"
] |
proofwiki-14681 | Generating Function for Elementary Symmetric Function | Let $U$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the elementary symmetric function of degree $m$ on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} \paren {\prod_{i \mathop = 1}^m x_{... | The summation for $\map G z$ is a finite sum $m = 0, 1, \ldots, n$, which settles convergence issues.
Begin with Viète's Formulas:
:$\ds \prod_{k \mathop = a}^b \paren {x - x_k} = x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, x^m$
Change variables $x = -1 / z$:
{{begin-eqn}}
{{eqn | l = ... | Let $U$ be a [[Definition:Set|set]] of $n$ [[Definition:Number|numbers]] $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the [[Definition:Elementary Symmetric Function|elementary symmetric function of degree $m$]] on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \... | The summation for $\map G z$ is a finite sum $m = 0, 1, \ldots, n$, which settles convergence issues.
Begin with [[Viète's Formulas]]:
:$\ds \prod_{k \mathop = a}^b \paren {x - x_k} = x^n + \sum_{m \mathop = 0}^{n - 1} \paren {-1}^{n - m} \map {e_{n - m} } U \, x^m$
Change variables $x = -1 / z$:
{{begin-eqn}}
{{eq... | Generating Function for Elementary Symmetric Function/Proof 1 | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function/Proof_1 | [
"Generating Function for Elementary Symmetric Function",
"Elementary Symmetric Functions",
"Examples of Generating Functions",
"Summations"
] | [
"Definition:Set",
"Definition:Number",
"Definition:Symmetric Function/Elementary",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Viète's Formulas"
] |
proofwiki-14682 | Generating Function for Elementary Symmetric Function | Let $U$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the elementary symmetric function of degree $m$ on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} \paren {\prod_{i \mathop = 1}^m x_{... | Apply mathematical induction on $n$.
Let $\map P n$ be the statement:
{{begin-eqn}}
{{eqn | l = \map G z
| o = \equiv
| r = \sum_{m \mathop = 0}^{n + 1} \map {e_m} {\set {x_1, \ldots, x_n} } z^m
}}
{{eqn | r = \prod_{k \mathop = 1}^n \paren {1 + x_k z}
}}
{{end-eqn}}
Basis for the induction:
Set $U = \set {... | Let $U$ be a [[Definition:Set|set]] of $n$ [[Definition:Number|numbers]] $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the [[Definition:Elementary Symmetric Function|elementary symmetric function of degree $m$]] on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \... | Apply [[Definition:Principle of Mathematical Induction|mathematical induction]] on $n$.
Let $\map P n$ be the statement:
{{begin-eqn}}
{{eqn | l = \map G z
| o = \equiv
| r = \sum_{m \mathop = 0}^{n + 1} \map {e_m} {\set {x_1, \ldots, x_n} } z^m
}}
{{eqn | r = \prod_{k \mathop = 1}^n \paren {1 + x_k z}
}}... | Generating Function for Elementary Symmetric Function/Proof 2 | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function/Proof_2 | [
"Generating Function for Elementary Symmetric Function",
"Elementary Symmetric Functions",
"Examples of Generating Functions",
"Summations"
] | [
"Definition:Set",
"Definition:Number",
"Definition:Symmetric Function/Elementary",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Principle of Mathematical Induction",
"Definition:Basis for the Induction",
"Definition:Induction Step",
"Recursion Property of Elementary Symmetric Function"
] |
proofwiki-14683 | Generating Function for Elementary Symmetric Function | Let $U$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the elementary symmetric function of degree $m$ on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \mathop \cdots \mathop < j_m \mathop \le n} \paren {\prod_{i \mathop = 1}^m x_{... | We have by definition of generating function that:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
We have that:
:$a_0 = 1$
{{explain|We need $a_0 {{=}} 1$ but I can't get my head around as to why at the moment}}
Suppose $n = 1$.
Let $\map G z$ be the generating function for $\sequence {a_m}$ under this condition.
The... | Let $U$ be a [[Definition:Set|set]] of $n$ [[Definition:Number|numbers]] $\set {x_1, x_2, \ldots, x_n}$.
Let $\map {e_m} U$ be the [[Definition:Elementary Symmetric Function|elementary symmetric function of degree $m$]] on $U$:
{{begin-eqn}}
{{eqn | l = \map {e_m} U
| r = \sum_{1 \mathop \le j_1 \mathop < j_2 \... | We have by definition of [[Definition:Generating Function|generating function]] that:
:$\map G z = \ds \sum_{n \mathop \ge 0} a_n z^n$
We have that:
:$a_0 = 1$
{{explain|We need $a_0 {{=}} 1$ but I can't get my head around as to why at the moment}}
Suppose $n = 1$.
Let $\map G z$ be the [[Definition:Generating Func... | Generating Function for Elementary Symmetric Function/Proof 3 | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function | https://proofwiki.org/wiki/Generating_Function_for_Elementary_Symmetric_Function/Proof_3 | [
"Generating Function for Elementary Symmetric Function",
"Elementary Symmetric Functions",
"Examples of Generating Functions",
"Summations"
] | [
"Definition:Set",
"Definition:Number",
"Definition:Symmetric Function/Elementary",
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Definition:Generating Function",
"Definition:Generating Function",
"Definition:Kronecker Delta",
"Product of Generating Functions"
] |
proofwiki-14684 | Newton-Girard Identities/Lemma 1 | Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \prod_{k \mathop = a}^b \paren {1 + x_k z}
| c =
}}
{{eqn | r = \paren {1 + x_a z} \paren {1 + x_{a + 1} z} \cdots \paren {1 + x_b z}
| c =
}}
{{end-eqn}} | This is an instance of Generating Function for Elementary Symmetric Function.
{{qed}}
Category:Newton-Girard Identities
gtks9s10xucbklmenzhlazwqcxzg73t | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {h_m}$.
Then:
{{begin-eqn}}
{{eqn | l = \map G z
| r = \prod_{k \mathop = a}^b \paren {1 + x_k z}
| c =
}}
{{eqn | r = \paren {1 + x_a z} \paren {1 + x_{a + 1} z} \cdots \paren ... | This is an instance of [[Generating Function for Elementary Symmetric Function]].
{{qed}}
[[Category:Newton-Girard Identities]]
gtks9s10xucbklmenzhlazwqcxzg73t | Newton-Girard Identities/Lemma 1 | https://proofwiki.org/wiki/Newton-Girard_Identities/Lemma_1 | https://proofwiki.org/wiki/Newton-Girard_Identities/Lemma_1 | [
"Newton-Girard Identities"
] | [
"Definition:Generating Function",
"Definition:Sequence"
] | [
"Generating Function for Elementary Symmetric Function",
"Category:Newton-Girard Identities"
] |
proofwiki-14685 | Newton-Girard Identities/Lemma 2 | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \paren {-1}^{k + 1} \dfrac {S_k z^k} k
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \map G z
| r = \paren {1 + x_a z} \paren {1 + x_{a + 1} z} \cdots \paren {1 + x_b z}
| c = Lemma 1
}}
{{eqn | ll= \leadsto
| l = \map \ln {\map G z}
| r = \map \ln {1 + x_a z} + \map \ln {1 + x_{a + 1} z} + \cdots + \map \ln {1 + x_b z}
| c = Sum of Logarithms
}}
... | {{begin-eqn}}
{{eqn | l = \map \ln {\map G z}
| r = \sum_{k \mathop \ge 1} \paren {-1}^{k + 1} \dfrac {S_k z^k} k
| c =
}}
{{end-eqn}} | {{begin-eqn}}
{{eqn | l = \map G z
| r = \paren {1 + x_a z} \paren {1 + x_{a + 1} z} \cdots \paren {1 + x_b z}
| c = [[Newton-Girard Formulas/Lemma 1|Lemma 1]]
}}
{{eqn | ll= \leadsto
| l = \map \ln {\map G z}
| r = \map \ln {1 + x_a z} + \map \ln {1 + x_{a + 1} z} + \cdots + \map \ln {1 + x_b z... | Newton-Girard Identities/Lemma 2 | https://proofwiki.org/wiki/Newton-Girard_Identities/Lemma_2 | https://proofwiki.org/wiki/Newton-Girard_Identities/Lemma_2 | [
"Newton-Girard Identities"
] | [] | [
"Newton-Girard Identities/Lemma 1",
"Sum of Logarithms",
"Power Series Expansion for Logarithm of 1 + x",
"Category:Newton-Girard Identities"
] |
proofwiki-14686 | Newton-Girard Identities/Examples/Order 2 | :$\ds \sum_{a \mathop \le i \mathop < j \mathop \le b} x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 - \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-... | :$\ds \sum_{a \mathop \le i \mathop < j \mathop \le b} x_i x_j = \dfrac 1 2 \paren {\paren {\sum_{i \mathop = a}^b x_i}^2 - \paren {\sum_{i \mathop = a}^b {x_i}^2} }$ | From [[Newton-Girard Identities]]:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\par... | Newton-Girard Identities/Examples/Order 2 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_2 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_2 | [
"Examples of Newton-Girard Identities"
] | [] | [
"Newton-Girard Identities"
] |
proofwiki-14687 | Newton-Girard Identities/Examples/Order 3 | :$\ds \sum_{a \mathop \le i \mathop < j \mathop < k \mathop \le b} x_i x_j x_k = \dfrac { {S_1}^3} 6 - \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-... | :$\ds \sum_{a \mathop \le i \mathop < j \mathop < k \mathop \le b} x_i x_j x_k = \dfrac { {S_1}^3} 6 - \dfrac {S_1 S_2} 2 + \dfrac {S_3} 3$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | From [[Newton-Girard Identities]]:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\par... | Newton-Girard Identities/Examples/Order 3 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_3 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_3 | [
"Examples of Newton-Girard Identities"
] | [] | [
"Newton-Girard Identities"
] |
proofwiki-14688 | Newton-Girard Identities/Examples/Order 4 | :$\ds \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < j_3 \mathop < j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} - \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 - \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$. | From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\... | :$\ds \sum_{a \mathop \le j_1 \mathop < j_2 \mathop < j_3 \mathop < j_4 \mathop \le b} x_{j_1} x_{j_2} x_{j_3} x_{j_4} = \dfrac { {S_1}^4} {24} - \dfrac { {S_1}^2 S_2} 4 + \dfrac { {S_2}^2} 8 + \dfrac {S_1 S_3} 3 - \dfrac {S_4} 4$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$. | From [[Newton-Girard Identities]]:
:$\ds \sum_{a \mathop \le j_1 \mathop < \mathop \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfr... | Newton-Girard Identities/Examples/Order 4 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_4 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_4 | [
"Examples of Newton-Girard Identities"
] | [] | [
"Newton-Girard Identities",
"Definition:Ordered Tuple as Ordered Set/Ordered Quadruple"
] |
proofwiki-14689 | Newton-Girard Identities/Examples/Order 1 | :$\ds \sum_{a \mathop \le i \mathop \le b} x_i = S_1$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | From Newton-Girard Identities:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\paren {-... | :$\ds \sum_{a \mathop \le i \mathop \le b} x_i = S_1$
where:
:$\ds S_r := \sum_{k \mathop = a}^b {x_k}^r$ | From [[Newton-Girard Identities]]:
:$\ds \sum_{a \mathop \le j_1 \mathop < \cdots \mathop < j_m \mathop \le b} x_{j_1} \cdots x_{j_m} = \sum_{\substack {k_1, k_2, \ldots, k_m \mathop \ge 0 \\ k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} } \dfrac { {S_1}^{k_1} } {1^{k_1} k_1 !} \dfrac {\par... | Newton-Girard Identities/Examples/Order 1 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_1 | https://proofwiki.org/wiki/Newton-Girard_Identities/Examples/Order_1 | [
"Examples of Newton-Girard Identities"
] | [] | [
"Newton-Girard Identities"
] |
proofwiki-14690 | Newton-Girard Identities | Let $X$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Define:
{{begin-eqn}}
{{eqn | l = \mathbf S_m
| r = \set {\paren {j_1, \ldots, j_m} : 1 \le j_1 < \cdots < j_m \le n}
| c = $1 \le m \le n$
}}
{{eqn | l = \map {e_m} X
| r = \begin {cases} 1 & : m = 0 \\ \ds \sum_{\mathbf S_m} x_{j_1} \c... | === Outline ===
The proof is divided into three cases: $k < n$, $k = n$ and $k > n$.
The tools are Viète's Formulas, Recursion Property of Elementary Symmetric Function, telescoping sums and homogeneous functions of degree $k$. | Let $X$ be a [[Definition:Set|set]] of $n$ [[Definition:Number|numbers]] $\set {x_1, x_2, \ldots, x_n}$.
Define:
{{begin-eqn}}
{{eqn | l = \mathbf S_m
| r = \set {\paren {j_1, \ldots, j_m} : 1 \le j_1 < \cdots < j_m \le n}
| c = $1 \le m \le n$
}}
{{eqn | l = \map {e_m} X
| r = \begin {cases} 1 & :... | === Outline ===
The proof is divided into three cases: $k < n$, $k = n$ and $k > n$.
The tools are [[Viète's Formulas]], [[Recursion Property of Elementary Symmetric Function]], [[Definition:Telescoping Series|telescoping sums]] and [[Definition:Homogeneous Function|homogeneous functions]] of degree $k$. | Newton-Girard Identities/Proof 1 | https://proofwiki.org/wiki/Newton-Girard_Identities | https://proofwiki.org/wiki/Newton-Girard_Identities/Proof_1 | [
"Newton-Girard Identities",
"Elementary Symmetric Functions"
] | [
"Definition:Set",
"Definition:Number",
"Newton-Girard Identities"
] | [
"Viète's Formulas",
"Recursion Property of Elementary Symmetric Function",
"Definition:Telescoping Series",
"Definition:Homogeneous Function"
] |
proofwiki-14691 | Newton-Girard Identities | Let $X$ be a set of $n$ numbers $\set {x_1, x_2, \ldots, x_n}$.
Define:
{{begin-eqn}}
{{eqn | l = \mathbf S_m
| r = \set {\paren {j_1, \ldots, j_m} : 1 \le j_1 < \cdots < j_m \le n}
| c = $1 \le m \le n$
}}
{{eqn | l = \map {e_m} X
| r = \begin {cases} 1 & : m = 0 \\ \ds \sum_{\mathbf S_m} x_{j_1} \c... | === Outline ===
Calculus is used to prove identities (1) and (2) in a single effort.
The tools are Viète's Formulas, the calculus derivative of powers $x^n$ and logarithm $\ln \size x$, Maclaurin series expansion coefficients, mathematical induction, and Leibniz's Rule in One Variable.
=== Lemma 1 ===
{{begin-eqn}}
{{... | Let $X$ be a [[Definition:Set|set]] of $n$ [[Definition:Number|numbers]] $\set {x_1, x_2, \ldots, x_n}$.
Define:
{{begin-eqn}}
{{eqn | l = \mathbf S_m
| r = \set {\paren {j_1, \ldots, j_m} : 1 \le j_1 < \cdots < j_m \le n}
| c = $1 \le m \le n$
}}
{{eqn | l = \map {e_m} X
| r = \begin {cases} 1 & :... | === Outline ===
Calculus is used to prove identities (1) and (2) in a single effort.
The tools are [[Viète's Formulas]], the [[Definition:Derivative/Notation|calculus derivative]] of powers $x^n$ and logarithm $\ln \size x$, [[Definition:Maclaurin Series|Maclaurin series expansion]] coefficients, [[Definition:Mathem... | Newton-Girard Identities/Proof 2 | https://proofwiki.org/wiki/Newton-Girard_Identities | https://proofwiki.org/wiki/Newton-Girard_Identities/Proof_2 | [
"Newton-Girard Identities",
"Elementary Symmetric Functions"
] | [
"Definition:Set",
"Definition:Number",
"Newton-Girard Identities"
] | [
"Viète's Formulas",
"Definition:Derivative/Notation",
"Definition:Maclaurin Series",
"Definition:Mathematical Induction",
"Leibniz's Rule/One Variable",
"Viète's Formulas",
"Generating Function for Elementary Symmetric Function",
"Definition:Derivative/Higher Derivatives/Higher Order",
"Derivative o... |
proofwiki-14692 | Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula | Let $S_m$ be expressed in the form:
:$S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} A_m {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$
for $k_1, k_2, \ldots, k_m \ge 0$.
Then :
:$A_m = \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! } ... | Let $\map G z$ be the generating function for the sequence $\sequence {h_m}$.
{{begin-eqn}}
{{eqn | l = \sum_{m \mathop \ge 1} \dfrac {S_m z^m} m
| r = \map \ln {\map G z}
| c = Summation of Products of n Numbers taken m at a time with Repetitions: Lemma 2
}}
{{eqn | r = \map \ln {1 + h_1 z + h_2 z^2 + \cdo... | Let $S_m$ be expressed in the form:
:$S_m = \ds \sum_{k_1 \mathop + 2 k_2 \mathop + \mathop \cdots \mathop + m k_m \mathop = m} A_m {h_1}^{k_1} {h_2}^{k_2} \cdots {h_m}^{k_m}$
for $k_1, k_2, \ldots, k_m \ge 0$.
Then :
:$A_m = \paren {-1}^{k_1 + k_2 + \cdots + k_m - 1} \dfrac {m \paren {k_1 + k_2 + \cdots + k_m - 1}! ... | Let $\map G z$ be the [[Definition:Generating Function|generating function]] for the [[Definition:Sequence|sequence]] $\sequence {h_m}$.
{{begin-eqn}}
{{eqn | l = \sum_{m \mathop \ge 1} \dfrac {S_m z^m} m
| r = \map \ln {\map G z}
| c = [[Summation of Products of n Numbers taken m at a time with Repetition... | Summation of Products of n Numbers taken m at a time with Repetitions/Inverse Formula | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Inverse_Formula | https://proofwiki.org/wiki/Summation_of_Products_of_n_Numbers_taken_m_at_a_time_with_Repetitions/Inverse_Formula | [
"Summation of Products of n Numbers taken m at a time with Repetitions"
] | [] | [
"Definition:Generating Function",
"Definition:Sequence",
"Summation of Products of n Numbers taken m at a time with Repetitions/Lemma 2",
"Power Series Expansion for Logarithm of 1 + x"
] |
proofwiki-14693 | Doubly Sequenced Generating Function for Binomial Coefficients | Let $\sequence {a_{m n} }$ be the doubly subscripted sequence defined as:
:$\forall m, n \in \N_{\ge 0}: a_{m n} = \dbinom n m$
where $\dbinom n m$ denotes a binomial coefficient.
Then the generating function for $\sequence {a_{m n} }$ is given as:
:$\map G {w, z} = \dfrac 1 {1 - z - w z}$ | {{begin-eqn}}
{{eqn | l = \map G {w, z}
| r = \sum_{m, \, n \mathop \ge 0} a_{m n} w^m z^n
| c = {{Defof|Generating Function for Doubly Subscripted Sequence}}
}}
{{eqn | r = \sum_{m, \, n \mathop \ge 0} \dbinom n m w^m z^n
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {1 + w}^n z^n
| c = B... | Let $\sequence {a_{m n} }$ be the [[Definition:Doubly Subscripted Sequence|doubly subscripted sequence]] defined as:
:$\forall m, n \in \N_{\ge 0}: a_{m n} = \dbinom n m$
where $\dbinom n m$ denotes a [[Definition:Binomial Coefficient|binomial coefficient]].
Then the [[Definition:Generating Function|generating funct... | {{begin-eqn}}
{{eqn | l = \map G {w, z}
| r = \sum_{m, \, n \mathop \ge 0} a_{m n} w^m z^n
| c = {{Defof|Generating Function for Doubly Subscripted Sequence}}
}}
{{eqn | r = \sum_{m, \, n \mathop \ge 0} \dbinom n m w^m z^n
| c =
}}
{{eqn | r = \sum_{n \mathop \ge 0} \paren {1 + w}^n z^n
| c = [... | Doubly Sequenced Generating Function for Binomial Coefficients | https://proofwiki.org/wiki/Doubly_Sequenced_Generating_Function_for_Binomial_Coefficients | https://proofwiki.org/wiki/Doubly_Sequenced_Generating_Function_for_Binomial_Coefficients | [
"Generating Functions",
"Binomial Coefficients"
] | [
"Definition:Sequence/Doubly Subscripted",
"Definition:Binomial Coefficient",
"Definition:Generating Function"
] | [
"Binomial Theorem",
"Sum of Infinite Geometric Sequence"
] |
proofwiki-14694 | Laplace Transform of Generating Function of Sequence | Let $\sequence{a_n}$ be a sequence which has a generating function which is convergent.
Let $\map G z$ be the generating function for $\sequence{a_n}$.
Let $\map f x$ be the step function:
:$\map f x = \ds \sum_{k \mathop \in \Z} a_k \sqbrk{0 \le k \le x}$
where $\sqbrk{0 \le k \le x}$ is Iverson's convention.
Then the... | We note that:
{{begin-eqn}}
{{eqn | n = 1
| l = \laptrans {\map f s}
| r = \int_0^\infty e^{-s t} \map f t \rd t
| c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | n = 2
| l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}}
}... | Let $\sequence{a_n}$ be a [[Definition:Sequence|sequence]] which has a [[Definition:Generating Function|generating function]] which is [[Definition:Convergent Real Function|convergent]].
Let $\map G z$ be the [[Definition:Generating Function|generating function]] for $\sequence{a_n}$.
Let $\map f x$ be the [[Definit... | We note that:
{{begin-eqn}}
{{eqn | n = 1
| l = \laptrans {\map f s}
| r = \int_0^\infty e^{-s t} \map f t \rd t
| c = {{Defof|Laplace Transform}}
}}
{{end-eqn}}
and:
{{begin-eqn}}
{{eqn | n = 2
| l = \map G z
| r = \sum_{n \mathop \ge 0} a_n z^n
| c = {{Defof|Generating Function}... | Laplace Transform of Generating Function of Sequence | https://proofwiki.org/wiki/Laplace_Transform_of_Generating_Function_of_Sequence | https://proofwiki.org/wiki/Laplace_Transform_of_Generating_Function_of_Sequence | [
"Generating Functions",
"Laplace Transforms"
] | [
"Definition:Sequence",
"Definition:Generating Function",
"Definition:Convergent Mapping/Real Function",
"Definition:Generating Function",
"Definition:Step Function",
"Definition:Iverson's Convention",
"Definition:Laplace Transform"
] | [
"Primitive of Exponential Function"
] |
proofwiki-14695 | Limit of Rational Sequence is not necessarily Rational | Let $S = \sequence {a_n}$ be a rational sequence.
Let $S$ be convergent to a limit $L$.
Then it is not necessarily the case that $L$ is itself a rational number. | Proof by Counterexample:
By definition, Euler's number $e$ can be defined as:
:$e = \ds \sum_{n \mathop = 0}^\infty \frac 1 {n!}$
Each of the terms of the sequence of partial sums is rational.
However, from Euler's Number is Irrational, $e$ itself is not.
{{qed}} | Let $S = \sequence {a_n}$ be a [[Definition:Rational Sequence|rational sequence]].
Let $S$ be [[Definition:Convergent Rational Sequence|convergent]] to a [[Definition:Limit of Rational Sequence|limit]] $L$.
Then it is not necessarily the case that $L$ is itself a [[Definition:Rational Number|rational number]]. | [[Proof by Counterexample]]:
By definition, [[Definition:Euler's Number|Euler's number]] $e$ can be defined as:
:$e = \ds \sum_{n \mathop = 0}^\infty \frac 1 {n!}$
Each of the [[Definition:Term of Sequence|terms]] of the [[Definition:Sequence of Partial Sums|sequence of partial sums]] is [[Definition:Rational Number... | Limit of Rational Sequence is not necessarily Rational | https://proofwiki.org/wiki/Limit_of_Rational_Sequence_is_not_necessarily_Rational | https://proofwiki.org/wiki/Limit_of_Rational_Sequence_is_not_necessarily_Rational | [
"Rational Sequences"
] | [
"Definition:Rational Sequence",
"Definition:Convergent Sequence/Rational Numbers",
"Definition:Limit of Sequence/Rational Numbers",
"Definition:Rational Number"
] | [
"Proof by Counterexample",
"Definition:Euler's Number",
"Definition:Term of Sequence",
"Definition:Series/Sequence of Partial Sums",
"Definition:Rational Number",
"Euler's Number is Irrational"
] |
proofwiki-14696 | Product of Negative Real Numbers is Positive | Let $a, b \in \R_{\le 0}$ be negative real numbers.
Then:
:$a \times b \in \R_{\ge 0}$
That is, their product $a \times b$ is a positive real number. | From Real Numbers form Ring, the set $\R$ of real numbers forms a ring.
The result then follows from Product of Ring Negatives.
{{qed}} | Let $a, b \in \R_{\le 0}$ be [[Definition:Negative Real Number|negative real numbers]].
Then:
:$a \times b \in \R_{\ge 0}$
That is, their [[Definition:Real Multiplication|product]] $a \times b$ is a [[Definition:Positive Real Number|positive real number]]. | From [[Real Numbers form Ring]], the [[Definition:Set|set]] $\R$ of [[Definition:Real Number|real numbers]] forms a [[Definition:Ring (Abstract Algebra)|ring]].
The result then follows from [[Product of Ring Negatives]].
{{qed}} | Product of Negative Real Numbers is Positive | https://proofwiki.org/wiki/Product_of_Negative_Real_Numbers_is_Positive | https://proofwiki.org/wiki/Product_of_Negative_Real_Numbers_is_Positive | [
"Real Multiplication"
] | [
"Definition:Negative/Real Number",
"Definition:Multiplication/Real Numbers",
"Definition:Positive/Real Number"
] | [
"Real Numbers form Ring",
"Definition:Set",
"Definition:Real Number",
"Definition:Ring (Abstract Algebra)",
"Product of Ring Negatives"
] |
proofwiki-14697 | Powers of Imaginary Unit | The (integer) powers of the imaginary unit $i$ are:
{{begin-eqn}}
{{eqn | l = i^0
| r = 1
| c =
}}
{{eqn | l = i^1
| r = i
| c =
}}
{{eqn | l = i^2
| r = -1
| c =
}}
{{eqn | l = i^3
| r = -i
| c =
}}
{{eqn | l = i^4
| r = 1
| c =
}}
{{end-eqn}}
{{qed}} | By definition:
:$i^2 = -1$
Then we have:
{{begin-eqn}}
{{eqn | l = i^0
| r = e^{0 \ln i}
| c = {{Defof|Power to Complex Number}}
}}
{{eqn | r = e^0
| c =
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = i^1
| r = e^{1 \ln i}
| c = {{Defof|Power to Complex Numbe... | The [[Definition:Integer Power|(integer) powers]] of the [[Definition:Imaginary Unit|imaginary unit]] $i$ are:
{{begin-eqn}}
{{eqn | l = i^0
| r = 1
| c =
}}
{{eqn | l = i^1
| r = i
| c =
}}
{{eqn | l = i^2
| r = -1
| c =
}}
{{eqn | l = i^3
| r = -i
| c =
}}
{{eqn | ... | By definition:
:$i^2 = -1$
Then we have:
{{begin-eqn}}
{{eqn | l = i^0
| r = e^{0 \ln i}
| c = {{Defof|Power to Complex Number}}
}}
{{eqn | r = e^0
| c =
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = i^1
| r = e^{1 \ln i}
| c = {{Defof|Power to Complex... | Powers of Imaginary Unit | https://proofwiki.org/wiki/Powers_of_Imaginary_Unit | https://proofwiki.org/wiki/Powers_of_Imaginary_Unit | [
"Imaginary Unit"
] | [
"Definition:Power (Algebra)/Integer",
"Definition:Complex Number/Imaginary Unit"
] | [] |
proofwiki-14698 | Real Number Multiplied by Complex Number | Let $a \in \R$ be a real number.
Let $c + d i \in \C$ be a complex number.
Then:
:$a \times \paren {c + d i} = \paren {c + d i} \times a = a c + i a d$ | $a$ can be expressed as a wholly real complex number $a + 0 i$.
Then we have:
{{begin-eqn}}
{{eqn | l = a \times \paren {c + d i}
| r = \paren {a + 0 i} \times \paren {c + d i}
| c = {{Defof|Wholly Real}}
}}
{{eqn | r = \paren {a c - 0 d} + \paren {a d + 0 c} i
| c = {{Defof|Complex Multiplication}}
}... | Let $a \in \R$ be a [[Definition:Real Number|real number]].
Let $c + d i \in \C$ be a [[Definition:Complex Number|complex number]].
Then:
:$a \times \paren {c + d i} = \paren {c + d i} \times a = a c + i a d$ | $a$ can be expressed as a [[Definition:Wholly Real|wholly real]] [[Definition:Complex Number|complex number]] $a + 0 i$.
Then we have:
{{begin-eqn}}
{{eqn | l = a \times \paren {c + d i}
| r = \paren {a + 0 i} \times \paren {c + d i}
| c = {{Defof|Wholly Real}}
}}
{{eqn | r = \paren {a c - 0 d} + \paren {... | Real Number Multiplied by Complex Number | https://proofwiki.org/wiki/Real_Number_Multiplied_by_Complex_Number | https://proofwiki.org/wiki/Real_Number_Multiplied_by_Complex_Number | [
"Complex Multiplication"
] | [
"Definition:Real Number",
"Definition:Complex Number"
] | [
"Definition:Complex Number/Wholly Real",
"Definition:Complex Number",
"Complex Multiplication is Commutative"
] |
proofwiki-14699 | Complex Number equals Negative of Conjugate iff Wholly Imaginary | Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Then $\overline z = -z$ {{iff}} $z$ is wholly imaginary. | Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \overline z
| r = -z
| c =
}}
{{eqn | ll= \leadsto
| l = x - i y
| r = -\left({x + i y}\right)
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | ll= \leadsto
| l = +x
| r = -x
| c =
}}
{{eqn | ll= \leadsto
| l = x
... | Let $z \in \C$ be a [[Definition:Complex Number|complex number]].
Let $\overline z$ be the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
Then $\overline z = -z$ {{iff}} $z$ is [[Definition:Wholly Imaginary|wholly imaginary]]. | Let $z = x + i y$.
Then:
{{begin-eqn}}
{{eqn | l = \overline z
| r = -z
| c =
}}
{{eqn | ll= \leadsto
| l = x - i y
| r = -\left({x + i y}\right)
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | ll= \leadsto
| l = +x
| r = -x
| c =
}}
{{eqn | ll= \leadsto
| l = x
... | Complex Number equals Negative of Conjugate iff Wholly Imaginary | https://proofwiki.org/wiki/Complex_Number_equals_Negative_of_Conjugate_iff_Wholly_Imaginary | https://proofwiki.org/wiki/Complex_Number_equals_Negative_of_Conjugate_iff_Wholly_Imaginary | [
"Complex Conjugates"
] | [
"Definition:Complex Number",
"Definition:Complex Conjugate",
"Definition:Complex Number/Wholly Imaginary"
] | [
"Definition:Complex Number/Wholly Imaginary",
"Definition:Complex Number/Wholly Imaginary"
] |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.