Fourier's Theorem/Lemma 1

Fourier's Theorem/Lemma 1 0 67846 459093 459092 2020-04-02T21:20:26Z Prime.mover 59 wikitext text/x-wiki == Lemma for Fourier's Theorem == <onlyinclude> Let $\psi$ be a real function defined on a closed interval $\closedint a b$. Let $\psi$ be piecewise continuous with one-sided limits on $\closedint a b$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \map \psi u \sin N u \rd u = 0$ </onlyinclude> == Proof == We are given that $\psi$ is piecewise continuous with one-sided limits on $\closedint a b$. Therefore, there exists a finite subdivision $\set {x_0, x_1, \ldots, x_m}$ of $\closedint a b$, where $x_0 = a$ and $x_m = b$, such that for all $i \in \set {1, 2, \ldots, m}$: :$\psi$ is continuous on $\openint {x_{i - 1} } {x_i}$  :$\displaystyle \lim_{x \mathop \to {x_{i - 1} }^+} \map \psi x$ and $\displaystyle \lim_{x \mathop \to {x_i}^-} \map \psi x$ exist. From the corollary to Sum of Integrals on Adjacent Intervals for Integrable Functions: :$\displaystyle \int_a^b \map \psi u \sin N u \rd u = \sum_{r \mathop = 0}^{m - 1} \int_{x_r}^{x_{r + 1} } \map \psi u \sin N u \rd u$ Then: {{begin-eqn}} {{eqn | n = 1       | l = \int_{x_r}^{x_{r + 1} } \map \psi u \sin N u \rd u       | r = \intlimits {-\map \psi u \frac {\cos N u} N} {x_r} {x_{r + 1} }       | c = Integration by Parts }} {{eqn | o =        | ro= +       | r = \frac 1 N \int_{x_r}^{x_{r + 1} } \map {\psi'} u \cos N u \rd u       | c =  }} {{end-eqn}} The last integral is bounded. Thus $(1)$ is less than $\dfrac {M_r} N$ for $M_r \in \R$. Let $M = \max \set {\size {M_0}, \size {M_1}, \dotsc, \size {M_{m - 1} } }$. Then: :$\displaystyle \size {\int_a^b \map \psi u \sin N u \rd u} < \dfrac {M m} N$ As $M$ and $m$ are finite: :$\displaystyle \lim_{N \mathop \to \infty} \dfrac {M m} N = 0$ Hence the result. {{qed}} == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Definition:Real Left-Hand Derivative|next = Fourier's Theorem/Lemma 1/Mistake 1}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(1)$ Category:Fourier's Theorem l5xmrwqfroh93mu5ef8de049918xstw


Fourier's Theorem/Lemma 2

Fourier's Theorem/Lemma 2 0 67850 459098 348113 2020-04-02T22:11:31Z Prime.mover 59 wikitext text/x-wiki == Lemma for Fourier's Theorem == <onlyinclude> Let $\psi$ be a real function defined on a half-open interval $\hointl 0 a$. Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\hointl 0 a$. Let $\map \psi u$ have a right-hand derivative at $u = 0$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$ where $\map \psi {0^+}$ denotes the limit of $\psi$ at $0$ from the right. </onlyinclude> == Proof == We have: :$\map \psi u = \map \psi {0^+} + \paren {\map \psi u - \map \psi {0^+} }$ from which: :$(1): \quad \displaystyle \int_0^a \map \psi u \frac {\sin N u} u \rd u = \map \psi {0^+} \int_0^a \frac {\sin N u} u \rd u + \int_0^a \map \phi u \sin N u \rd u$ where: :$\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ Let $\xi = N u$. Then: {{begin-eqn}} {{eqn | l = \int_0^a \frac {\sin N u} u \rd u       | r = \int_0^{N a} \frac {\sin \xi} \xi \rd \xi       | c =  }} {{eqn | o = \to       | r = \int_0^\infty \frac {\sin \xi} \xi \rd \xi       | c =  }} {{eqn | r = \frac \pi 2       | c = Integral to Infinity of $\dfrac {\sin \xi} \xi$ }} {{end-eqn}} We have that $\map \psi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$. Hence it follows that $\map \phi u = \dfrac {\map \psi u - \map \psi {0^+} } u$ is also piecewise continuous with one-sided limits on $\hointl 0 a$. We also have that $\map \psi u$ has a right-hand derivative at $u = 0$. It follows that $\map \phi u$ is piecewise continuous with one-sided limits on $\hointl 0 a$. Thus from Lemma 1 for Fourier's Theorem: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \phi u \sin N u \rd u = 0$ and letting $N \to \infty$ in $(1)$ above: :$\displaystyle \lim_{N \mathop \to \infty} \int_0^a \map \psi u \frac {\sin N u} u \rd u = \frac \pi 2 \map \psi {0^+}$ {{qed}} {{Proofread}} == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Fourier's Theorem/Lemma 1/Mistake 2|next = Fourier's Theorem/Lemma 2/Mistake}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(2)$ Category:Fourier's Theorem 5mirn4oim09fohb29j85c0esawwisnn


Fourier's Theorem/Lemma 3

Fourier's Theorem/Lemma 3 0 67853 459101 348117 2020-04-02T22:20:37Z Prime.mover 59 wikitext text/x-wiki == Lemma for Fourier's Theorem == <onlyinclude> Let $\psi$ be a real function defined on an open interval $\openint a b$. Let $\psi$ and its derivative $\psi'$ be piecewise continuous with one-sided limits on $\openint a b$. Let $\map \psi u$ have both right-hand derivative and left-hand derivative at a point $u = x$ where $x \in \openint a b$. Then: :$\displaystyle \lim_{N \mathop \to \infty} \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \frac \pi 2 \paren {\map \psi {x^+} + \map \psi {x^-} }$ where: :$\map \psi {x^+}$ denotes the limit of $\psi$ at $x$ from the right :$\map \psi {x^-}$ denotes the limit of $\psi$ at $x$ from the left. </onlyinclude> == Proof == From Sum of Integrals on Adjacent Intervals for Integrable Functions, we have: :$\displaystyle \int_a^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u + \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u$ Let $u = x - \xi$. Then by Integration by Substitution: :$\displaystyle \int_a^x \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi$ where: :$\map \phi \xi = \map \psi {u - \xi}$ {{explain|It needs to be established how the limits change from $a \to x$ to $0 \to x - a$ as this is not obvious.}} By Fourier's Theorem: Lemma 2: {{begin-eqn}} {{eqn | l = \lim_{N \mathop \to \infty} \int_0^{x - a} \map \phi \xi \frac {\sin N \xi} \xi \rd \xi       | r = \frac \pi 2 \map \phi {0^+}       | c =  }} {{eqn | r = \frac \pi 2 \map \psi {x^-}       | c =  }} {{end-eqn}} Similarly, substituting $u = x + \eta$: :$\displaystyle \int_x^b \map \psi u \frac {\sin N \paren {u - x} } {u - x} \rd u = \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta$ where: :$\map \chi \xi = \map \psi {x + \eta}$ By Fourier's Theorem: Lemma 2: {{begin-eqn}} {{eqn | l = \lim_{N \mathop \to \infty} \int_0^{b - x} \map \chi \eta \frac {\sin N \eta} \eta \rd \eta       | r = \frac \pi 2 \map \phi {0^+}       | c =  }} {{eqn | r = \frac \pi 2 \map \psi {x^+}       | c =  }} {{end-eqn}} The result follows by adding the two limits. {{qed}} {{Proofread}} == Sources == * {{BookReference|Fourier Series|1961|I.N. Sneddon|prev = Fourier's Theorem/Lemma 2/Mistake|next = Fourier's Theorem}}: Chapter Two: $\S 2$. Some Important Limits: Lemma $(3)$ Category:Fourier's Theorem rneh9nk5wbf5h4nyhwnhdivs7nsu33f


Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals/Corollary 0 68118 349357 349294 2018-03-31T20:15:11Z Prime.mover 59 wikitext text/x-wiki == Corollary to Sum of Sequence of Products of Consecutive Odd and Consecutive Even Reciprocals == <onlyinclude> {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \left({j + 2}\right)}       | r = \frac 1 {1 \times 3} + \frac 1 {2 \times 4} + \frac 1 {3 \times 5} + \frac 1 {4 \times 6} + \cdots       | c =  }} {{eqn | r = \frac 3 4       | c =  }} {{end-eqn}} </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \sum_{j \mathop = 1}^\infty \frac 1 {j \left({j + 2}\right)}       | r = \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \left({j + 2}\right)}       | c =  }} {{eqn | r = \lim_{n \mathop \to \infty} \left({\frac 3 4 - \frac {2 n + 3} {2 \left({n + 1}\right) \left({n + 2}\right)} }\right)       | c = Sum of Sequence of Products of Consecutive Odd Reciprocals }} {{eqn | r = \frac 3 4 - \lim_{n \mathop \to \infty} \frac {\frac 2 n + \frac 3 {n^2} } {2 \left({1 + \frac 1 n}\right) \left({1 + \frac 2 n}\right)}       | c = dividing top and bottom by $n^2$ }} {{eqn | r = \frac 3 4       | c = Basic Null Sequences }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Sum of Sequence of Products of Consecutive Odd Reciprocals/Corollary|next = Sum of Sequence of Products of Squares of Consecutive Odd Reciprocals}}: $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.31$ Category:Sums of Sequences Category:Reciprocals o91i9uxaoxvvhgqx6lu7gh97c6hthz3


Binomial Theorem/Examples/4th Power of Sum 0 68172 417824 412152 2019-08-11T12:46:23Z Caliburn 3218 wikitext text/x-wiki == Example of Use of Binomial Theorem == <onlyinclude> :$\paren {x + y}^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$ </onlyinclude> == Proof == Follows directly from the Binomial Theorem: :$\displaystyle \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$ putting $n = 4$. {{qed}} == Sources == * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Difference|next = Fourth Power of Difference}}: $\S 2$: Special Products and Factors: $2.5$ * {{BookReference|Mathematical Handbook of Formulas and Tables|1968|Murray R. Spiegel|prev = Cube of Sum|next = Power Series Expansion of Reciprocal of 1 + x/Proof 2}}: $\S 20$: Binomial Series: $20.7$ * {{BookReference|Elementary Number Theory|1980|David M. Burton|ed = revised|edpage = Revised Printing|prev = Binomial Theorem/Examples/Cube of Sum|next = Definition:Pascal's Triangle/Historical Note}}: Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem Category:Examples of Use of Binomial Theorem Category:Fourth Powers iohl4jm3n8273acnkq9fyb0me4cvzvp


Power Series Expansion of Reciprocal of Square of 1 + x 0 68177 349714 2018-04-03T15:07:34Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $x \in \R$ such that $-1 < x < 1$.  Then:  {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $x \in \R$ such that $-1 < x < 1$. Then: {{begin-eqn}} {{eqn | l = \dfrac 1 {\left({1 + x}\right)^2}       | r = \sum_{k \mathop = 0}^\infty \left({-1}\right)^k \left({k + 1}\right) x^k       | c =  }} {{eqn | r = 1 - 2 x + 3 x^2 - 4 x^3 + 5 x^4 - \cdots       | c =  }} {{end-eqn}} </onlyinclude> == Proof 1 == {{:Power Series Expansion of Reciprocal of Square of 1 + x/Proof 1}} == Proof 2 == {{:Power Series Expansion of Reciprocal of Square of 1 + x/Proof 2}} Category:Examples of Power Series Category:Power Series Expansion of Reciprocal of Square of 1 + x 7robieaxmfj2d4h8hycx1hom9mur1me


Harmonic Numbers/Examples/H1

Harmonic Numbers/Examples/H1 0 69012 355776 355774 2018-05-16T20:19:47Z Prime.mover 59 wikitext text/x-wiki == Example of Harmonic Number == <onlyinclude> :$H_1 = 1$ </onlyinclude> where $H_1$ denotes the first harmonic number. == Proof == {{begin-eqn}} {{eqn | l = H_1       | r = \sum_{k \mathop = 1}^1 \frac 1 k       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \frac 1 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Harmonic Numbers/Examples/H0|next = Harmonic Numbers/Examples/H2}}: $\S 1.2.7$: Harmonic Numbers: Exercise $1$ Category:Examples of Harmonic Numbers atu4v1wo1h1vtcjcwzb01gicuko4wv2


Harmonic Numbers/Examples/H2

Harmonic Numbers/Examples/H2 0 69015 359009 355779 2018-06-15T06:22:29Z Prime.mover 59 wikitext text/x-wiki == Example of Harmonic Number == <onlyinclude> :$H_2 = \dfrac 3 2$ </onlyinclude> where $H_2$ denotes the second harmonic number. == Proof == {{begin-eqn}} {{eqn | l = H_2       | r = \sum_{k \mathop = 1}^2 \frac 1 k       | c = {{Defof|Harmonic Number}} }} {{eqn | r = \frac 1 1 + \frac 1 2       | c =  }} {{eqn | r = \dfrac 3 2       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Harmonic Number H1|next = Upper Bound for Harmonic Number}}: $\S 1.2.7$: Harmonic Numbers: Exercise $1$ Category:Examples of Harmonic Numbers p9ega59o43niojrowf6ddsxvx97aerz


Sum of Arithmetic Sequence/Examples/Sum of j from m to n 0 69146 456256 455622 2020-03-19T10:13:27Z Prime.mover 59 wikitext text/x-wiki == Example of Sum of Arithmetic Sequence == <onlyinclude> {{begin-eqn}} {{eqn | l = \sum_{j \mathop = m}^n j       | r = m \paren {n - m + 1} + \frac 1 2 \paren {n - m} \paren {n - m + 1}       | c =  }} {{eqn | r = \frac {n \paren {n + 1} } 2 - \frac {\paren {m - 1} m} 2       | c =  }} {{end-eqn}} </onlyinclude> == Proof 1 == {{:Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 1}} == Proof 2 == {{:Sum of Arithmetic Sequence/Examples/Sum of j from m to n/Proof 2}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Sum of Geometric Sequence/Examples/One Seventh from 1 to n|next = General Distributivity Theorem/Examples/Sum of j from m to n by Sum of k from r to s}}: $\S 1.2.3$: Sums and Products: Exercise $13$ Category:Examples of Sum of Arithmetic Sequence mn2s6399j6zegqolarfea9gebtt16cc


Factorial/Examples/1

Factorial/Examples/1 0 69333 357503 2018-05-28T07:04:05Z Prime.mover 59 Created page with "== Example of Factorial == <onlyinclude> The factorial of $1$ is $1$: :$1! = 1$ </onlyinclude>  == Proof ==  From the definit..." wikitext text/x-wiki == Example of Factorial == <onlyinclude> The factorial of $1$ is $1$: :$1! = 1$ </onlyinclude> == Proof == From the definition of factorial: :$n! = \displaystyle \prod_{k \mathop = 1}^n k$ where $\prod$ denotes product notation. When $n = 1$ we have: :$1! = \displaystyle \prod_{k \mathop = 1}^1 k$ {{begin-eqn}} {{eqn | l = 1!       | r = \prod_{k \mathop = 1}^1 k       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} Category:Factorials/Examples ozkkkc9bs06qczcuw9c9pcuox40ln6s


Factors of Binomial Coefficient/Corollary 1 0 69484 358308 2018-06-04T21:51:29Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> For all $r \in \R, k \in \Z$: :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ from which: :$\dbinom r k = \dfrac r {r - k} \dbinom {r -..." wikitext text/x-wiki == Theorem == <onlyinclude> For all $r \in \R, k \in \Z$: :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ from which: :$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ (if $r \ne k$) </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = r \binom {r - 1} k       | r = r \frac {\left({r - 1}\right) \left({\left({r - 1}\right) - 1}\right) \cdots \left({\left({r - 1}\right) - k + 2}\right) \left({\left({r - 1}\right) - k + 1}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \frac {r \left({r - 1}\right) \left({r - 2}\right) \cdots \left({r - k + 1}\right) \left({r - k}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \left({r - k}\right) \frac {r \left({r - 1}\right) \left({r - 2}\right) \cdots \left({r - k + 1}\right)} {k \left({k - 1}\right) \left({k - 2}\right) \cdots 1}       | c =  }} {{eqn | r = \left({r - k}\right) \binom r k       | c =  }} {{end-eqn}} {{qed|lemma}} Then: :$\dbinom r k = \dfrac r {r - k} \dbinom {r - 1} k$ follows from the  :$\left ({r - k}\right) \dbinom r k = r \dbinom {r - 1} k$ by dividing both sides by $r - k$. {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Factors of Binomial Coefficient|next = Pascal's Rule/Real Numbers}}: $\S 1.2.6$: Binomial Coefficients: $\text{C}$ Category:Binomial Coefficients pckf7nqveg56hvodmowqbga3in6ilal


Numbers Expressed as Sums of Binomial Coefficients 0 69501 465323 358420 2020-05-02T06:29:16Z RandomUndergrad 3904 wikitext text/x-wiki == Definition == Let $n \in \Z_{> 0}$ be a (strictly) positive integer. Then for all $k \in \Z_{> 0}$, it is possible to express $k$ uniquely in the form: {{begin-eqn}} {{eqn | l = k       | r = \sum_{j \mathop = 1}^n \dbinom {k_j} j       | c =  }} {{eqn | r = \dbinom {k_1} 1 + \dbinom {k_2} 2 + \cdots + \dbinom {k_n} n       | c =  }} {{end-eqn}} such that $0 \le k_1 < k_2 < \cdots < k_n$. == Proof == === Existence of Representation === Proof by induction: For all $k \in \Z_{\ge 0}$, let $\map P k$ be the proposition that it is possible to express $k$ in the form: :$\displaystyle k = \sum_{j \mathop = 1}^n \dbinom {k_j} j$ such that $0 \le k_1 < k_2 < \cdots < k_n$. ==== Basis for the Induction ==== $\map P 0$ is true, as $\displaystyle 0 = \sum_{j \mathop = 1}^n \dbinom {j - 1} j$. This is our basis for the induction. ==== Induction Hypothesis ==== Now we need to show that, if $\map P m$ is true, where $r \ge 2$, then it logically follows that $\map P {m + 1}$ is true. So this is our induction hypothesis: $m$ can be expressed in the form: :$\displaystyle m = \sum_{j \mathop = 1}^n \dbinom {m_j} j$ where $0 \le m_1 < m_2 < \cdots < m_n$. Then we need to show: $m + 1$ can be expressed in the form: :$\displaystyle m = \sum_{j \mathop = 1}^n \dbinom {k_j} j$ where $0 \le k_1 < k_2 < \cdots < k_n$. ==== Induction Step ==== This is our induction step: Suppose the first $c$ $m_j$ are consecutive, that is: :$\forall \, j \in \N: 1 \le j < c: m_{j + 1} - m_j = 1$ and $m_{c + 1} - m_c > 1$ Then: {{begin-eqn}} {{eqn | l = m + 1       | r = \sum_{j \mathop = 1}^n \dbinom {m_j} j + 1       | c = from the induction hypothesis }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {m_j} j + 1       | c =  }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {m_1 + j - 1} j + \dbinom {m_1 + 0 - 1} 0       | c = Binomial Coefficient with Zero }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_1 + c} c       | c = Sum of r+k Choose k up to n }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c - 1} c       | c =  }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c - 1} c + \sum_{j = 1}^{c - 1} \dbinom {j - 1} j       | c =  }} {{end-eqn}} Since $0 \le m_1 < m_2 < \cdots < m_n$, we must have $m_j \ge j - 1$ for each $j \le n$. Hence $0 \le 0 < 1 < \cdots < c - 2 < m_c - 1 < m_{c + 1} < \cdots < m_n$. Thus the expression above satisfy the conditions. So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n \binom {r + k} k = \binom {r + n + 1} n$ {{qed|lemma}} === Uniqueness of Representation === Suppose $k$ can be expressed in the form: :$\displaystyle k = \sum_{j \mathop = 1}^n \dbinom {k_j} j = \sum_{j \mathop = 1}^n \dbinom {m_j} j$ where $0 \le k_1 < k_2 < \cdots < k_n$ and $0 \le m_1 < m_2 < \cdots < m_n$. {{AimForCont}} $k_i$ and $m_i$ are not all equal. Let $c$ be the largest integer such that $k_c \ne m_c$. {{WLOG}} assume $k_c < m_c$. Then: {{begin-eqn}} {{eqn | l = k       | r = \sum_{j \mathop = 1}^n \dbinom {k_j} j }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {k_j} j }} {{eqn | o = \le       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \sum_{j \mathop = 1}^c \dbinom {k_c - c + j} j       | c = $0 \le k_1 < k_2 < \cdots < k_n$ }} {{eqn | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {k_c - c + c + 1} c - \dbinom {k_c - c} 0       | c = Sum of r+k Choose k up to n }} {{eqn | o = \le       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c} c - 1       | c = $k_c + 1 \le m_c$ }} {{eqn | o = <       | r = \sum_{j \mathop = c + 1}^n \dbinom {m_j} j + \dbinom {m_c} c       | c =  }} {{eqn | o = \le       | r = \sum_{j \mathop = 1}^n \dbinom {m_j} j       | c = $\dbinom {m_j} j \ge 0$ }} {{eqn | r = k       | c =  }} {{end-eqn}} which is a contradiction. Thus the representation is unique. {{qed}} == Examples == {{:Numbers Expressed as Sums of Binomial Coefficients/Examples}} == Also see == * Definition:Combinatorial Number System == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Inverse of Stirling's Triangle expressed as Matrix|next = Numbers Expressed as Sums of Binomial Coefficients/Examples/n = 3}}: $\S 1.2.6$: Binomial Coefficients: Exercise $56$ Category:Numbers Expressed as Sums of Binomial Coefficients Category:Binomial Coefficients jgdt82aonbbr8gwtxj44wswha7lvase


Abel's Lemma/Formulation 1

Abel's Lemma/Formulation 1 0 69643 454400 442776 2020-03-14T13:19:32Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$. Then: <onlyinclude> :$\displaystyle \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k} = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}$ </onlyinclude> Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$. === Corollary === {{:Abel's Lemma/Formulation 1/Corollary}} == Proof == {{begin-eqn}} {{eqn | l = \sum_{k \mathop = m}^n a_k \paren {b_{k + 1} - b_k}        | r = \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_k b_k        | c = }} {{eqn | r = \sum_{k \mathop = m}^n a_k b_{k + 1} - \paren {a_m b_m + \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1} - a_{n + 1} b_{n + 1} }        | c = }} {{eqn | r = a_{n + 1} b_{n + 1} - a_m b_m + \sum_{k \mathop = m}^n a_k b_{k + 1} - \sum_{k \mathop = m}^n a_{k + 1} b_{k + 1}        | c = }} {{eqn | r = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_{k + 1}        | c = }} {{end-eqn}} {{qed}} == Also reported as == Some sources give this as: :$\displaystyle \sum_{k \mathop = m}^n \paren {a_{k + 1} - a_k} b_k = a_{n + 1} b_{n + 1} - a_m b_m - \sum_{k \mathop = m}^n a_{k + 1} \paren {b_{k + 1} - b_k}$ which is obtained from the main result by interchanging $a$ and $b$. Others take the upper index to $n - 1$: :$\displaystyle \sum_{k \mathop = m}^{n - 1} \paren {a_{k + 1} - a_k} b_k = a_n b_n - a_m b_m - \sum_{k \mathop = m}^{n - 1} a_{k + 1} \paren {b_{k + 1} - b_k}$ == Also known as == {{:Abel's Lemma/Also known as}} {{Namedfor|Niels Henrik Abel}} Category:Abel's Lemma 9vb9u0cd2cxfgnvmw9v6ewlc4gbrmt3


Abel's Lemma/Formulation 2

Abel's Lemma/Formulation 2 0 69646 454403 413263 2020-03-14T13:21:00Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $\sequence a$ and $\sequence b$ be sequences in an arbitrary ring $R$. <onlyinclude> Let $\displaystyle A_n = \sum_{i \mathop = m}^n {a_i}$ be the partial sum of $\sequence a$ from $m$ to $n$. Then: :$\displaystyle \sum_{k \mathop = m}^n a_k b_k = \sum_{k \mathop = m}^{n - 1} A_k \paren {b_k - b_{k + 1} } + A_n b_n$ </onlyinclude> Note that although proved for the general ring, this result is usually applied to one of the conventional number fields $\Z, \Q, \R$ and $\C$. === Corollary === {{:Abel's Lemma/Formulation 2/Corollary}} == Proof 1 == {{:Abel's Lemma/Formulation 2/Proof 1}} == Proof 2 == {{:Abel's Lemma/Formulation 2/Proof 2}} == Also known as == {{:Abel's Lemma/Also known as}} {{Namedfor|Niels Henrik Abel}} == Sources == * {{BookReference|Dictionary of Mathematics|1989|Ephraim J. Borowski|author2 = Jonathan M. Borwein|prev = Symmetric Group is not Abelian|next = Definition:Abel Summation Method|entry = Abel's partial summation formula}} * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Abel's Limit Theorem|next = Definition:Abel Summation Method|entry = Abel's partial summation formula}} Category:Abel's Lemma rx01s1cgpt2yy19j756dqwulgh7zwhd


Sum of Angles between Straight Lines at Point form Four Right Angles 0 69677 359191 359186 2018-06-18T15:58:12Z Prime.mover 59 wikitext text/x-wiki == Corollary to Two Angles on Straight Line make Two Right Angles == <onlyinclude> If any number of straight lines are drawn from a given point, the sum of the consecutive angles so formed is $4$ right angles. </onlyinclude> == Proof == Let $OA_1, OA_2, \ldots, OA_n$ be straight lines drawn from a point $O$ to points $A_1, A_2, \ldots, A_n$. Let $OA_1$ be produced past $O$ to $B$. Then $OB$ either coincides with $OA_j$ for some $j$ between $1$ and $n$, or $OB$ divides angle $A_j O A_k$ for some $j, k$ between $1$ and $n$. First suppose $OB$ coincides with $OA_j$. {{finish|etc.}} == Sources == * {{BookReference|Problems & Solutions in Euclidean Geometry|1968|M.N. Aref|author2 = William Wernick|prev = Two Angles on Straight Line make Two Right Angles|next = Bisectors of Adjacent Angles between Straight Lines Meeting at Point are Perpendicular}}: Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.1$: Corollary $1$ Category:Angles ekhdayhfzl4qf346826w1frlt9g85hq


Characterization of Strictly Increasing Mapping on Woset 0 69689 360226 360224 2018-07-09T20:07:20Z GFauxPas 522 wikitext text/x-wiki {{tidy}} == Lemma == Let $J$ and $E$ be well-ordered sets. Let $h: J \to E$ be a mapping. Let $S_\alpha$ denote an initial segment determined by $\alpha$. {{TFAE}} :$(1):\quad$ $h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$ :$(2):\quad$ $\forall \alpha \in J: h\left({\alpha}\right) = \min \left({E\setminus h\left[{S_\alpha}\right]}\right)$, and $h[S_\alpha] = S_{h(\alpha)}$ where: :$h\left[{S_\alpha}\right]$ denotes the image of $S_\alpha$ under $h$ :$\min$ denotes the smallest element of the set. == Proof == === $(1)$ implies $(2)$ === Suppose $h$ satisfies: :$h$ is strictly increasing and its image is either all of $E$ or an initial segment of $E$ Then for any $x,y \in J$: {{begin-eqn}} {{eqn | l = x       | o = \prec       | r = y }} {{eqn | ll = \implies       | l = h(x)       | o = \prec       | r = h(y)       | c = {{Defof|Strictly Increasing Mapping|strictly increasing}} }} {{eqn | l = h[S_y]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y) } \right\} }} {{eqn | r = S_{h(y)} }} {{eqn | l = \min\left({E \setminus h\left[{S_y}\right] }\right)       | r = \min\left({E \setminus S_{h(y)} }\right) }} {{eqn | r = h(y)       | c = {{Defof|Smallest Element|smallest}} and of initial segment }} {{end-eqn}} {{qed|lemma}} === $(2)$ implies $(1)$ === Suppose $h$ satisfies: :$h(\alpha)  = \min\left({E \setminus h\left[{S_\alpha}\right] }\right)$ By the Principle of Recursive Definition for Well-Ordered Sets, $h$ is thus uniquely determined. Then: {{begin-eqn}} {{eqn | l = h(y)       | r = \min\left({E \setminus h\left[{S_y}\right] }\right) }} {{eqn | l = h[S_y]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) =              \min\left({E \setminus h\left[{S_y }\right] }\right)} \right\} }} {{eqn | r =  \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(y)} \right\} }} {{eqn | r = S_{h(y)}       | c = {{Defof|Initial Segment|initial segment}} }} {{end-eqn}} Thus for every $x \in S_y$, we have that $h(x) \in S_{h(y)}$. Therefore $h$ is an strictly increasing mapping. Furthermore, the image set of $h$ is the union of initial segments in $E$. By Union of Initial Segments is Initial Segment or All of Woset, $h[J]$ is an initial segment of $E$ or all of $E$. {{qed}} {{proofread}} == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercise $1.2$ Category:Well-Orderings 4nlrguvf1tumug3w9gmy3u63fs0vcml


Strictly Increasing Mapping Between Wosets Implies Order Isomorphism 0 69763 359719 359526 2018-06-28T07:19:25Z Prime.mover 59 wikitext text/x-wiki {{tidy}} == Lemma == Let $J$ and $E$ be well-ordered sets. Let there exist a mapping $k: J \to E$ which is strictly increasing. Then $J$ is order isomorphic to $E$ or an initial segment of $E$. == Proof == If the sets considered are empty or singletons, the theorem holds vacuously or trivially. Suppose $J,E$ both have at least two elements. Let $e_0 = \min E$, the smallest element of $E$. Define the mapping: :$h: J \to E$: :$h(\alpha) =  \begin{cases} \min \left({E \setminus h\left[ {S_\alpha} \right]}\right) & \text{ if } h\left[ {S_\alpha} \right] \ne E \\ e_0 & \text{ if } h\left[ {S_\alpha} \right] = E \end{cases}$ where $S_\alpha$ is the initial segment determined by $\alpha$ and $h\left[ {S_\alpha} \right]$ is the image of $S_\alpha$ under $h$. By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and uniquely determined. Observe that: {{begin-eqn}} {{eqn | l = h[S_\alpha]       | r = \left\{ { h(x) \in E: \exists x \in J: h(x) =              \min\left({E \setminus h\left[{S_\alpha}\right] }\right)} \right\}       | c = {{Defof|Image of Subset under Mapping|image of a subset}} }} {{eqn | r =  \left\{ { h(x) \in E: \exists x \in J: h(x) \prec h(\alpha)} \right\} }} {{eqn | r = S_{h(\alpha)}       | c = {{Defof|Initial Segment|initial segment}} }} {{end-eqn}} This equality also holds if $h(\alpha) = e_0$, by Initial Segment Determined by Smallest Element is Empty. We claim that $h(\alpha) \preceq k(\alpha)$ for all $\alpha \in J$. {{AimForCont}} there is some $a \in J$ such that $h(a) \not \preceq k(a)$. Then $k(a) \prec h(a)$ by the trichotomy law. Because $h(a)$ has an element preceding it, $h(a) \ne e_0$. Thus $k(a) \prec \min \left({E \setminus h\left[ {S_a} \right]}\right)$ by the construction of $k$. Then $k(a) \in h\left[ {S_a} \right]$, because it precedes the smallest element that isn't in $h\left[ {S_a} \right]$. Recall that $h[S_a] = S_{h(a)}$. Then $k(a) \in S_{h(a)}$. This implies that $h(a) \prec k(a)$, contradicting the assumption that $h(a) \not \preceq k(a)$ From this contradiction we can conclude: :$h(\alpha) \preceq k(\alpha)$ for all $\alpha \in J$. {{AimForCont}} there is some $\alpha \in J$ such that $h[S_\alpha] = E$. Recall that $h[S_\alpha] = S_{h(\alpha)}$. Thus, were such an $\alpha$ to exist, then it would succeed all elements in $E$. It particular, it would also succeed $k(\alpha)$. But we showed above that $h(\alpha) \preceq k(\alpha)$. From this contradiction we see that there cannot be any $\alpha \in J$ with $h[S_\alpha] = E$. Thus the definition of $h$ can be simplified: :$h: J \to E$: :$h(\alpha) = \min \left({E \setminus h\left[ {S_\alpha} \right]}\right)$ Then the hypotheses of Characterization of Strictly Increasing Mapping on Woset are satisfied. Thus $h$ is a strictly increasing mapping and its image is $E$ or an initial segment of $E$. From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is also injective. From Injection to Image is Bijection, $h$ is also bijective to its image. We conclude that there is an order isomorphism from $J$ to $E$, or from $J$ to an initial segment of $E$. {{qed}}{{proofread}} == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercises $1.3$ Category:Well-Orderings 26s8gwwcf2jck79dirajwntol68aa7w


Equality to Initial Segment Imposes Well-Ordering 0 69778 483120 359717 2020-08-27T07:42:20Z Prime.mover 59 wikitext text/x-wiki == Lemma == Let $X$ be a set. Let $\AA$ be the set of all ordered pairs $\struct {A, <}$ such that $A$ is a subset of $X$ and $<$ is a strict well-ordering of $A$. Define $\prec$ as: :$\struct {A, <} \prec \struct {A', <'}$ {{iff}}  :$\struct {A, <}$ equals an initial segment of $\struct {A', <'}$. Let $\BB$ be a set of ordered pairs in $\AA$ such that $\BB$ is ordered by $\prec$. Let $B'$ be the union of the sets $B$ for all $\struct {B, <} \in \BB$. Let $<'$ be the union of the relations $<$ for all $\struct {B, <}$. Then $\struct {B', <'}$ is strictly well-ordered set. == Proof == If the set $X$ considered is empty or a singleton, the lemma holds vacuously or trivially. Thus assume $X$ contains at least two elements. We first prove that $\prec$ is a strict partial ordering on $\AA$. From the definition of initial segment, no $\struct {A, <}$ can equal an initial segment of itself. Thus $\prec$ is antireflexive. Suppose $\struct {A, <_A}$ equals an initial segment of $\struct {B, <_B}$ and $\struct {B, <_B}$ equals an initial segment of $\struct {C, <_C}$. Then $\struct {A, <_A}$ equals an initial segment of $\struct {C, <'}$ from Equality is Transitive. Thus $\prec$ is a strict partial ordering on $\AA$. We then prove that any $\struct {B', <'}$ is a strictly well-ordered set. Let $x_1,x_2 \in B'$. That is, let $x_i \in \struct {A_i, <_i}$ for $i = 1, 2$. Suppose $x_2 \prec x_1$. That is, $\struct {A_2, <_2}$ equals an initial segment in $\struct {A_1, <_1}$ By the definition of initial segment, both $x_1$ and $x_2$ are in $\struct {A_1, <_1}$. Thus $<'$ is connected, as all $<_i$ are strict well-orderings by hypothesis. For any $x_i \in \struct {A_i, <_i}$, $x_i \nprec x_i$ as all $<_i$ are strict well-orderings by hypothesis. Thus $<'$ is antireflexive. To show that $<'$ is transitive, consider $x_i \in \struct {B', <'}$ for $i = 1, 2, 3$. Suppose $x_1 <' x_2 <' x_3$. Then $x_1 <_1 x_2$ and $x_2 <_2 x_3$, from the definition of $<'$ as a union of relations $<_i$. Then $\struct {A_j, <_j}$ is an initial segment of $\struct {A_i, <_i}$ for $j  = 1, 2; j < i$ Thus $x_1 <_i x_2 <_i x_3$. Then $x_1 <_i x_3$, as all $<_i$ as all $<_i$ are strict well-orderings by hypothesis. Conclude that $<'$ is itself a strict ordering. It remains to be shown that $<'$ is a well-ordering. Let $A$ be an arbitrary non-empty subset of $B'$. Let $x \in A$ and $x \in \struct {B, <}$ for $\struct {B, <} \in \BB$. Then for all $y \in A$, $y <' x$ {{iff}} $y < x$ and $y \in B$. As $<$ is a well-ordering, $\struct {B \cap A, <}$ has a smallest element $b$. This $b$ is then a smallest element of $<'$ in $A$. Conclude that $<'$ is a strict well-ordering on $B'$. {{qed}} == Sources == * {{BookReference|Topology|2000|James R. Munkres|ed = 2nd|edpage = Second Edition}}: Supplementary Exercises $1.5$ Category:Well-Orderings Category:Set Equality rezg3r5cjiygyv3pkro2xkyvwjxxk0r


Euler-Binet Formula/Corollary 1

Euler-Binet Formula/Corollary 1 0 69795 380982 380978 2018-12-10T07:48:51Z Prime.mover 59 wikitext text/x-wiki == Corollary to Euler-Binet Formula == <onlyinclude> :$F_n = \dfrac {\phi^n} {\sqrt 5}$ rounded to the nearest integer </onlyinclude> where: :$F_n$ denotes the $n$th Fibonacci number :$\phi$ denotes the golden mean. == Proof == By definition of $n$th Fibonacci number, $F_n$ is an integer. From Euler-Binet Formula: :$F_n = \dfrac {\phi^n - \hat \phi^n} {\sqrt 5} = \dfrac {\phi^n } {\sqrt 5} - \dfrac {\hat \phi^n} {\sqrt 5}$ But $\size {\dfrac {\hat \phi^n} {\sqrt 5} } < \dfrac 1 2$ for all $n \ge 0$. Thus $\dfrac {\phi^n } {\sqrt 5}$ differs from $F_n$ by a number less than $\dfrac 1 2$. Thus the nearest integer to $\dfrac {\phi^n } {\sqrt 5}$ is $F_n$. {{qed}} == Sources == * {{BookReference|Curious and Interesting Numbers|1986|David Wells|prev = Euler-Binet Formula/Historical Note|next = Euler-Binet Formula/Corollary 1/Mistake}}: $5$ * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Golden Mean/One Minus Golden Mean/Decimal Expansion|next = Summation over k to n of Product of kth with n-kth Fibonacci Numbers}}: $\S 1.2.8$: Fibonacci Numbers: $(15)$ * {{BookReference|Curious and Interesting Numbers|1997|David Wells|ed = 2nd|edpage = Second Edition|prev = Euler-Binet Formula/Historical Note|next = Cassini's Identity}}: $5$ Category:Euler-Binet Formula qs30a3prs4nh9ygu017vhmfe9ne526h


Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers/Lemma 0 69841 359979 359974 2018-07-03T07:15:12Z Prime.mover 59 wikitext text/x-wiki == Lemma for Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers == <onlyinclude> Let $f \left({n}\right)$ be an arbitrary arithmetic function. Let $\left\langle{a_n}\right\rangle$ be the sequence defined as: :$a_n = \begin{cases} 0 & : n = 0 \\ 1 & : n = 1 \\ a_{n - 1} + a_{n - 2} + f \left({n - 2}\right) & : n > 1 \end{cases}$ Then: :$a_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} f \left({k}\right)$ </onlyinclude> == Proof == Trying out a few values: {{begin-eqn}} {{eqn | l = a_0       | r = 0       | c =  }} {{eqn | r = F_0       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_1       | r = 1       | c =  }} {{eqn | r = F_1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_2       | r = a_1 + a_0 + f \left({0}\right)       | c =  }} {{eqn | r = F_1 + F_0 + f \left({0}\right)       | c =  }} {{eqn | r = F_2 + F_1 f \left({0}\right)       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_3       | r = a_2 + a_1 + f \left({1}\right)       | c =  }} {{eqn | r = F_2 + F_1 f \left({0}\right) + F_1 + f \left({1}\right)       | c =  }} {{eqn | r = F_3 + F_2 f \left({0}\right) + F_1 f \left({1}\right)       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_4       | r = a_3 + a_2 + f \left({2}\right)       | c =  }} {{eqn | r = F_3 + F_2 f \left({0}\right) + F_1 f \left({1}\right) + F_2 + F_1 f \left({0}\right) + f \left({2}\right)       | c =  }} {{eqn | r = F_4 + F_3 f \left({0}\right) + F_2 f \left({1}\right) + F_1 f \left({2}\right)       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = a_5       | r = a_4 + a_3 + f \left({3}\right)       | c =  }} {{eqn | r = F_4 + F_3 f \left({0}\right) + F_2 f \left({1}\right) + F_1 f \left({2}\right) + F_3 + F_2 f \left({0}\right) + F_1 f \left({1}\right) + f \left({3}\right)       | c =  }} {{eqn | r = F_5 + F_4 f \left({0}\right) + F_3 f \left({1}\right) + F_2 f \left({2}\right) + F_1 f \left({3}\right)       | c =  }} {{end-eqn}} The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition: :$a_n = F_n + \displaystyle \sum_{k \mathop = 1}^{n - 1} F_{n - k - 1} f \left({k}\right)$ $P \left({0}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_0 + \sum_{k \mathop = 0}^{0 - 2} F_{0 - k - 1} f \left({k}\right)       | r = F_0       | c = as the summation is vacuous }} {{eqn | r = 0       | c = {{Defof|Fibonacci Number}}: $F_0 = 0$ }} {{eqn | r = a_0       | c = by definition }} {{end-eqn}} Thus $P \left({0}\right)$ is seen to hold. === Basis for the Induction === $P \left({1}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_1 + \sum_{k \mathop = 0}^{1 - 2} F_{1 - k - 1} f \left({k}\right)       | r = F_1       | c = as the summation is vacuous }} {{eqn | r = 1       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{eqn | r = a_1       | c = by definition }} {{end-eqn}} Thus $P \left({1}\right)$ is seen to hold. $P \left({2}\right)$ is the case: {{begin-eqn}} {{eqn | l = F_2 + \sum_{k \mathop = 0}^{2 - 2} F_{2 - k - 1} f \left({k}\right)       | r = F_2 + F_1 f \left({0}\right)       | c =  }} {{eqn | r = F_2 + f \left({0}\right)       | c = {{Defof|Fibonacci Number}}: $F_1 = 1$ }} {{eqn | r = F_1 + F_0 + f \left({0}\right)       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = 1 + 0 + f \left({0}\right)       | c = {{Defof|Fibonacci Number}}: $F_1 = 1, F_0 = 0$ }} {{eqn | r = a_1 + a_0 + f \left({0}\right)       | c = by definition }} {{end-eqn}} Thus $P \left({2}\right)$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 1$, then it logically follows that $P \left({r + 1}\right)$ is true. So this is the induction hypothesis: :$a_r = F_r + \displaystyle \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} f \left({k}\right)$ and: :$a_{r - 1} = F_{r - 1} + \displaystyle \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} f \left({k}\right)$ from which it is to be shown that: :$a_{r + 1} = F_{r + 1} + \displaystyle \sum_{k \mathop = 0}^{r - 1} F_{r - k} f \left({k}\right)$ === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = a_{r + 1}       | r = a_{r - 1} + a_r + f \left({r - 1}\right)       | c =  }} {{eqn | r = \left({F_{r - 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} f \left({k}\right)}\right) + \left({F_r + \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} f \left({k}\right)}\right) + f \left({r - 1}\right)       | c = Induction Hypothesis }} {{eqn | r = F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k - 2} f \left({k}\right) + \sum_{k \mathop = 0}^{r - 2} F_{r - k - 1} f \left({k}\right) + F_1 f \left({r - 1}\right)       | c = {{Defof|Fibonacci Number}} }} {{eqn | r = F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} \left({F_{r - k - 2} + F_{r - k - 1} }\right) f \left({k}\right) + F_{r - \left({r - 2}\right) - 1} f \left({r - 2}\right) + F_1 f \left({r - 1}\right)       | c = factoring out the summation }} {{eqn | r = F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k} f \left({k}\right) + F_1 f \left({r - 2}\right) + F_1 f \left({r - 1}\right)       | c = {{Defof|Fibonacci Number}} and simplifying }} {{eqn | r = F_{r + 1} + \sum_{k \mathop = 0}^{r - 3} F_{r - k} f \left({k}\right) + F_2 f \left({r - 2}\right) + F_1 f \left({r - 1}\right)       | c = {{Defof|Fibonacci Number}}: $F_1 = F_2 = 1$ }} {{eqn | r = F_{r + 1} + \sum_{k \mathop = 0}^{r - 1} F_{r - k} f \left({k}\right)       | c =  }} {{end-eqn}} So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$a_n = F_n + \displaystyle \sum_{k \mathop = 0}^{n - 2} F_{n - k - 1} f \left({k}\right)$ Category:Fibonacci Number plus Arbitrary Function in terms of Fibonacci Numbers q6p3nz26e1plob0y403qqx7k2xtpvru


Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1 0 69897 360163 2018-07-08T11:52:00Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} F_k = F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} {k - 1}_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} = 0$ </onlyinclude> where: :$\dbinom m k_\mathcal F$ denotes a Fibonomial coefficient :$F_{n + k}$ denotes the $n + k$th Fibonacci number :$\left\lceil{\, \cdot \,}\right\rceil$ denotes the ceiling function == Proof == {{ProofWanted}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers|next = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $30$: Solution: $\text{(a)}$ Category:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers dexfwbxjyaq6c8wlvcct67e1mfwm1ik


Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 2 0 69904 360188 2018-07-08T13:24:12Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{..." wikitext text/x-wiki == Theorem == <onlyinclude> :$\displaystyle \sum_{k \mathop \in \Z} \dbinom m k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} \left({-1}\right)^k F_{m - k} = \left({-1}\right)^m F_m \sum_{k \mathop \in \Z} \dbinom {m - 1} k_\mathcal F \left({-1}\right)^{\left\lceil{\left({m - 1 - k}\right) / 2}\right\rceil} {F_{n + k} }^{m - 2} = 0$ </onlyinclude> where: :$\dbinom m k_\mathcal F$ denotes a Fibonomial coefficient :$F_{n + k}$ denotes the $n + k$th Fibonacci number :$\left\lceil{\, \cdot \,}\right\rceil$ denotes the ceiling function == Proof == {{ProofWanted}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Lemma 1|next = Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers/Examples/m = 3}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $30$: Solution: $\text{(b)}$ Category:Recurrence Relation for Sequence of mth Powers of Fibonacci Numbers gaub0lzv715zubeg5xxu0wdrs9ldap5


Addition of 1 in Golden Mean Number System 0 69969 360433 360432 2018-07-13T19:59:10Z Prime.mover 59 wikitext text/x-wiki == Algorithm == Let $x \in \R$ be a real number. The following algorithm performs the operation of addition of $1$ to $x$ in the golden mean number system. Let $S$ be the representation of $x$ in the golden mean number system in its simplest form. :'''Step $1$''': Is the digit immediately to the left of the radix point a zero? :: If '''Yes''', replace that $0$ with $1$. Go to '''Step $4$'''. :: If '''No''', set $m = 2$ and go to '''Step $2$'''. :'''Step $2$''': Does the $m$th place after the radix point contain $0$? :: If '''Yes''', expand the $100$ in the $3$ places ending in the $m$th place with $011$. Subtract $2$ from $m$. Go to '''Step $3$'''. :: If '''No''', add $2$ to $m$. Repeat '''Step $2$'''. :'''Step $3$''': Is $m = 0$? :: If '''Yes''', set the digit immediately to the left of the radix point from $0$ to $1$. Go to '''Step $4$'''. :: If '''No''', go to '''Step $2$'''. :'''Step $4$''': Convert $S$ to its simplest form. '''Stop'''. == Proof == The above constitutes an algorithm, for the following reasons: === Finiteness === The only case in which it is possible for the process not to terminate is if the $m$th place never contains $0$. This can only happen if $S$ ends in an infinite string $01010101 \ldots$ But if this is the case, $S$ is not in its simplest form. === Definiteness === Each step can be seen to be precisely defined. === Inputs === The only input to the algorithm is the representation $S$ of $x$. === Outputs === The only output from the algorithm is the representation $S$ of $x + 1$. All operations that change $S$ are of the following nature: :$(1): \quad$ Simplification of $S$, which does not change $x$, which happens if at all in '''Step $4$. :$(2): \quad$ Expansion of $S$, which does not change $x$, which happens if at all in ''Step $2$. :$(3): \quad$ Setting the digit corresponding to $\phi^0$ to $1$ from $0$, which happens either in '''Step $1$''' or in '''Step $3$'''. :::In either step, it happens only once, after which the algorithm terminates. === Effective === Each step is basic enough to be done exactly and predictably. {{qed}} == Sources == * {{citation|author = George Bergman|title = Number System with an Irrational Base|journal = Mathematics Magazine|abbr = Math. Mag.|volume = 31|issue = 2|date = 1957|startpage = 98|endpage = 110|jstor = 3029218}} Category:Golden Mean Number System 5gqhubgrxtldabakec720486ayfx61p


Fibonacci String/Examples/S3

Fibonacci String/Examples/S3 0 69980 360465 2018-07-14T08:42:36Z Prime.mover 59 Created page with "== Example of Fibonacci Strings == <onlyinclude> The Fibonacci string $S_3$ is $\text{ba}$. </onlyinclude>  ==..." wikitext text/x-wiki == Example of Fibonacci Strings == <onlyinclude> The Fibonacci string $S_3$ is $\text{ba}$. </onlyinclude> == Proof == By definition of Fibonacci string: {{begin-eqn}} {{eqn | l = S_1       | r = \text{a} }} {{eqn | l = S_2       | r = \text{b} }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = S_3       | r = S_2 S_1       | c =  }} {{eqn | r = \text{ba}       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Definition:Fibonacci String|next = Fibonacci String/Examples/S4}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $36$ Category:Fibonacci Strings gj4xfcodh6c7x5k8syqbowp5jacm53y


Fibonacci String/Examples/S4

Fibonacci String/Examples/S4 0 69981 360466 2018-07-14T08:45:06Z Prime.mover 59 Created page with "== Example of Fibonacci Strings == <onlyinclude> The Fibonacci string $S_4$ is $\text{bab}$. </onlyinclude>  ==..." wikitext text/x-wiki == Example of Fibonacci Strings == <onlyinclude> The Fibonacci string $S_4$ is $\text{bab}$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = S_4       | r = S_3 S_2       | c = {{Defof|Fibonacci String}} }} {{eqn | r = \text{ba} \ \text{b}       | c = Definition of Fibonacci String $S_3$ }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|The Art of Computer Programming: Volume 1: Fundamental Algorithms|1997|Donald E. Knuth|ed = 3rd|edpage = Third Edition|prev = Fibonacci String/Examples/S3|next = Fibonacci String/Examples/S5}}: $\S 1.2.8$: Fibonacci Numbers: Exercise $36$ Category:Fibonacci Strings q4xbjhm8j9hrnwcdze0jpptvzajqfnc


Complex Modulus/Examples/i

Complex Modulus/Examples/i 0 70270 362517 361893 2018-08-16T07:11:17Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Modulus == <onlyinclude> :$\cmod i = \cmod {-i} = 1$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \cmod i       | r = \cmod {0 + 1 i}       | c =  }} {{eqn | r = \sqrt {0^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \cmod {-i}       | r = \cmod {0 + \left({-1}\right) i}       | c =  }} {{eqn | r = \sqrt {0^2 + \left({-1}\right)^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 1       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Power of Complex Modulus equals Complex Modulus of Power|next = Complex Modulus/Examples/-5}}: $\S 1.2$. The Algebraic Theory: Examples Category:Examples of Complex Modulus 60nu6xawnfq6iw0ersbqw2abszqxyfw


Complex Modulus/Examples/1+i

Complex Modulus/Examples/1+i 0 70274 496584 361891 2020-10-25T23:13:47Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Modulus == <onlyinclude> :$\cmod {1 + i} = \sqrt 2$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \cmod {1 + i}       | r = \cmod {1 + 1 i}       | c =  }} {{eqn | r = \sqrt {1^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = \sqrt 2       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Modulus/Examples/-5|next = Power of Complex Modulus equals Complex Modulus of Power/Examples/(1+i)^4}}: $\S 1.2$. The Algebraic Theory: Examples Category:Examples of Complex Modulus 2e8ljo6803ducgre7b0p9mds7jhi34i


Polar Form of Complex Number/Examples/i

Polar Form of Complex Number/Examples/i 0 70398 395502 394578 2019-03-13T08:13:05Z Prime.mover 59 wikitext text/x-wiki == Example of Polar Form of Complex Number == <onlyinclude> The imaginary unit $i$ can be expressed in polar form as $\polar {1, \dfrac \pi 2}$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \cmod i       | r = \sqrt {0^2 + 1^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = 1       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \map \cos {\map \arg i}       | r = \dfrac 0 1       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg i       | r = \pm \dfrac \pi 2       | c = Cosine of Half-Integer Multiple of Pi }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \map \sin {\map \arg i}       | r = \dfrac 1 1       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \map \arg i       | r = \dfrac \pi 2       | c = Sine of Half-Integer Multiple of Pi }} {{end-eqn}} Hence: :$\map \arg i = \dfrac \pi 2$ and hence the result. {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Argument of Quotient equals Difference of Arguments|next = Multiplication by Imaginary Unit is Equivalent to Rotation through Right Angle}}: $\S 2$. Geometrical Representations Category:Examples of Polar Form of Complex Number l0ukan4ajbvpnrlz8ckq40uydle6ia3


Euler's Formula/Examples/e^i pi by 2

Euler's Formula/Examples/e^i pi by 2 0 70472 363046 363041 2018-08-20T06:08:50Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler's Formula == <onlyinclude> :$e^{i \pi / 2} = i$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = e^{i \pi / 2}       | r = \cos \frac \pi 2 + i \sin \frac \pi 2       | c = Euler's Formula }} {{eqn | r = 0 + i \times 1       | c = Cosine of $\dfrac \pi 2$, Sine of $\dfrac \pi 2$ }} {{eqn | r = i       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi by 4|next = Euler's Formula/Examples/e^-i pi by 2}}: $\S 2$. Geometrical Representations: $(2.19)$ Category:Examples of Euler's Formula b95q3mt20dmvzmum5n3wqjfcme18b7i


Euler's Formula/Examples/e^-i pi by 2

Euler's Formula/Examples/e^-i pi by 2 0 70473 363045 363043 2018-08-20T06:08:37Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler's Formula == <onlyinclude> :$e^{-i \pi / 2} = -i$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = e^{-i \pi / 2}       | r = \cos \frac {-\pi} 2 + i \sin \frac {-\pi} 2       | c = Euler's Formula }} {{eqn | r = \cos \frac {3 \pi} 2 + i \sin \frac {3 \pi} 2       | c = Cosine of Angle plus Full Angle, Sine of Angle plus Full Angle }} {{eqn | r = 0 + i \times \paren {-1}       | c = Cosine of $\dfrac {3 \pi} 2$, Sine of $\dfrac {3 \pi} 2$ }} {{eqn | r = -i       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi by 2|next = Euler's Formula/Examples/e^i pi}}: $\S 2$. Geometrical Representations: $(2.19)$ Category:Examples of Euler's Formula lha2db5g6v7msclv5qua2zyz4xsk0kh


Euler's Formula/Examples/e^2 i pi

Euler's Formula/Examples/e^2 i pi 0 70477 365064 363049 2018-09-09T00:08:20Z Prime.mover 59 wikitext text/x-wiki == Example of Use of Euler's Formula == <onlyinclude> :$e^{2 i \pi} = 1$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = e^{2 i \pi}       | r = \cos 2 \pi + i \sin 2 \pi       | c = Euler's Formula }} {{eqn | r = 1 + i \times 0       | c = Cosine of $2 \pi$, Sine of $2 \pi$ }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Examples/e^i pi|next = Exponential of Complex Number plus 2 pi i}}: $\S 2$. Geometrical Representations: $(2.19)$ * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Euler's Formula/Proof|next = Period of Complex Exponential Function}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$ Category:Examples of Euler's Formula sy6uwbgoocnlk19yrkru0px3asxjluz


Euler's Formula/Examples/e^2 k i pi

Euler's Formula/Examples/e^2 k i pi 0 70480 363057 2018-08-20T07:07:01Z Prime.mover 59 Created page with "== Example of Use of Euler's Formula == <onlyinclude> :$\forall k \in \Z: e^{2 k i \pi} = 1$ </onlyinclude>  == Proof ==  {{begin-eqn}} {{eqn | l = e^{2 k i \pi}       | r..." wikitext text/x-wiki == Example of Use of Euler's Formula == <onlyinclude> :$\forall k \in \Z: e^{2 k i \pi} = 1$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = e^{2 k i \pi}       | r = \cos 2 k \pi + i \sin 2 k \pi       | c = Euler's Formula }} {{eqn | r = 1 + i \times 0       | c = Cosine of Multiple of $\pi$, Sine of Multiple of $\pi$ }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Exponential of Complex Number plus 2 pi i|next = Period of Complex Exponential Function}}: $\S 2$. Geometrical Representations Category:Examples of Euler's Formula mbx3yw6qpt5at8mltyx1f02mmfwtoca


Euler's Theorem/Corollary 1

Euler's Theorem/Corollary 1 0 70492 374816 363115 2018-11-03T21:59:13Z Prime.mover 59 wikitext text/x-wiki == Corollary to Euler's Theorem == <onlyinclude> Let $p^n$ be a prime power for some prime number $p > 1$. Let $a$ be an integer not divisible by $p: p \nmid a$. Then: : $a^{\paren {p - 1} p^{n - 1} } \equiv 1 \pmod {p^n}$ </onlyinclude> == Proof == We have that Divisor Relation is Transitive. Since $p \divides p^n$, it follows that $p^n \nmid a$. From Euler's Theorem: : $a^{\map \phi {p^n} } \equiv 1 \pmod {p^n}$ From Euler Phi Function of Prime Power: :$\map \phi {p^n} = \paren {p - 1} p^{n - 1}$ Then: : $a^{\paren {p - 1} p^{n - 1} } \equiv 1 \pmod {p^n}$ {{qed}} Category:Prime Numbers Category:Number Theory 0uc9jenglnm484q1xhjndaxfnpeos3l


Argument of Negative Real Number is Pi

Argument of Negative Real Number is Pi 0 70516 363220 2018-08-22T21:43:48Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $x \in \R_{>0}$ be a positive real number.  Then: :$\arg \paren {-x} = \pi$  where $\arg$ denotes the Def..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $x \in \R_{>0}$ be a positive real number. Then: :$\arg \paren {-x} = \pi$ where $\arg$ denotes the argument of a complex number. </onlyinclude> == Proof == We have that: :$-x = -x + 0 i$ and so: {{begin-eqn}} {{eqn | l = \cmod {-x}       | r = \sqrt {\paren {-x}^2 + 0^2}       | c = {{Defof|Complex Modulus}} }} {{eqn | r = x       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \cos \paren {\arg \paren {-x} }       | r = \dfrac {-x} x       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = -1       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {-x}       | r = \pi       | c = Cosine of Multiple of Pi }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \sin \paren {\arg \paren {-x} }       | r = \dfrac 0 x       | c = {{Defof|Argument of Complex Number}} }} {{eqn | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = \arg \paren {-x}       | r = 0 \text { or } \pi       | c = Sine of Multiple of Pi }} {{end-eqn}} Hence: : $\arg \paren {-x} = \pi$ {{qed}} Category:Examples of Arguments of Complex Numbers 10dadpk3wkk9w8kfmpi2lwet2shamug


Equation of Unit Circle in Complex Plane/Corollary 1 0 70543 363328 2018-08-24T22:57:32Z Prime.mover 59 Created page with "== Corollary to Equation of Unit Circle in Complex Plane ==  Consider the unit circle $C$ whose center is at $\l..." wikitext text/x-wiki == Corollary to Equation of Unit Circle in Complex Plane == Consider the unit circle $C$ whose center is at $\left({0, 0}\right)$ on the complex plane. <onlyinclude> The equation of $C$ can be given by: :$z \overline z = 1$ where $\overline z$ denotes the complex conjugate of $z$. </onlyinclude> == Proof == From Equation of Unit Circle in Complex Plane, the equation of $C$ can also be given by: :$\cmod z = 1$ where $\cmod z$ denotes the complex modulus of $z$. Thus: {{begin-eqn}} {{eqn | l = \cmod z       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = \cmod z^2       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = z \overline z       | r = 1       | c = Modulus in Terms of Conjugate }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Equation of Unit Circle|next = Equation of Unit Circle in Complex Plane/Corollary 2}}: $\S 3$. Roots of Unity Category:Equation of Unit Circle in Complex Plane 0khohopk8187y9269w6uf1hjfjdwohv


Complex Roots of Unity/Examples/4th Roots 0 70575 445882 395931 2020-02-03T09:26:52Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Roots of Unity == <onlyinclude> The '''complex $4$th roots of unity''' are the elements of the set: :$U_n = \set {z \in \C: z^4 = 1}$ They are: {{begin-eqn}} {{eqn | l = e^{0 i \pi / 4}       | r = 1 }} {{eqn | l = e^{i \pi / 2}       | r = i }} {{eqn | l = e^{i \pi}       | r = -1 }} {{eqn | l = e^{3 i \pi / 2}       | r = -i }} {{end-eqn}} </onlyinclude> == Proof == By definition, the first complex $4$th root of unity $\alpha$ is given by: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / 4}       | c =  }} {{eqn | r = e^{i \pi / 2}       | c =  }} {{eqn | r = \cos \frac \pi 2 + i \sin \frac \pi 2       | c =  }} {{eqn | r = 0 + i \times 1       | c = Cosine of $\dfrac \pi 2$, Sine of $\dfrac \pi 2$ }} {{eqn | r = i       | c =  }} {{end-eqn}} We have that: :$e^{0 i \pi / 4} = e^0 = 1$ which gives us, as always, the zeroth complex $n$th root of unity for all $n$. The remaining complex $4$th roots of unity can be expressed as $e^{4 i \pi / 4} = e^{i \pi}$ and $e^{6 i \pi / 4} = e^{3 i \pi / 2}$, but it is simpler to calculate them as follows: {{begin-eqn}} {{eqn | l = \alpha^2       | r = i^2       | c =  }} {{eqn | r = -1       | c = {{Defof|Imaginary Unit}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^3       | r = \alpha^2 \times \alpha       | c =  }} {{eqn | r = \paren {-1} \times i       | c =  }} {{eqn | r = -i       | c =  }} {{end-eqn}} :500px {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle|next = Complex 5th Roots of Unity}}: $\S 3$. Roots of Unity: Example $1$. * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Viète's Formulas/Examples/Sum 4, Product 8|next = Complex Roots of Unity/Examples/7th Roots}}: $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $105 \ \text {(a)}$ * {{BookReference|The Concise Oxford Dictionary of Mathematics|2014|Christopher Clapham|author2 = James Nicholson|ed = 5th|edpage = Fifth Edition|prev = Four Squares Theorem|next = Definition:Fractal|entry = fourth root of unity}} Category:Examples of Complex Roots of Unity 626ort6gwxv170mtalcoktj0m3j1ma8


Complex Roots of Unity/Examples/5th Roots 0 70576 437891 395936 2019-12-06T06:38:43Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Roots of Unity == <onlyinclude> The '''complex $5$th roots of unity''' are the elements of the set: :$U_n = \set {z \in \C: z^5 = 1}$ They are: {{begin-eqn}} {{eqn | l = e^{0 \pi / 5}       | r = 1 }} {{eqn | l = e^{2 \pi / 5}       | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4 }} {{eqn | l = e^{4 \pi / 5}       | r = -\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }} {{eqn | l = e^{6 \pi / 5}       | r = -\dfrac {1 + \sqrt 5} 4 - i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }} {{eqn | l = e^{8 \pi / 5}       | r = \dfrac {\sqrt 5 - 1} 4 - i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4 }} {{end-eqn}} </onlyinclude> == Proof == By definition, the first complex $5$th root of unity $\alpha$ is given by: {{begin-eqn}} {{eqn | l = \alpha       | r = e^{2 i \pi / 5}       | c =  }} {{eqn | r = \cos \frac {2 \pi} 5 + i \sin \frac {2 \pi} 5       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = Cosine of $\dfrac {2 \pi} 5$, Sine of $\dfrac {2 \pi} 5$ }} {{end-eqn}} We have that: :$e^{0 i \pi / 5} = e^0 = 1$ which gives us, as always, the zeroth complex $n$th root of unity for all $n$. The remaining complex $5$th roots of unity follow: {{begin-eqn}} {{eqn | l = \alpha^2       | r = e^{4 i \pi / 5}       | c =  }} {{eqn | r = -\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = Cosine of $\dfrac {4 \pi} 5$, Sine of $\dfrac {4 \pi} 5$ }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^3       | r = \overline {\alpha^{5 - 3} }       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = \overline {\alpha^2}       | c =  }} {{eqn | r = \overline {-\dfrac {1 + \sqrt 5} 4 + i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8} }       | c =  }} {{eqn | r = -\dfrac {1 + \sqrt 5} 4 - i \sqrt {\dfrac 5 8 - \dfrac {\sqrt 5} 8}       | c = {{Defof|Complex Conjugate}} }} {{end-eqn}} {{begin-eqn}} {{eqn | l = \alpha^4       | r = \overline {\alpha^{4 - 1} }       | c = Complex Roots of Unity occur in Conjugate Pairs }} {{eqn | r = \overline \alpha       | c =  }} {{eqn | r = \overline {\dfrac {\sqrt 5 - 1} 4 + i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4}       | c =  }} {{eqn | r = \dfrac {\sqrt 5 - 1} 4 - i \dfrac {\sqrt {10 + 2 \sqrt 5} } 4       | c = {{Defof|Complex Conjugate}} }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex 4th Roots of Unity|next = Complex Roots of Unity/Examples/6th Roots/Illustration}}: $\S 3$. Roots of Unity: Example $2$. * {{BookReference|Theory and Problems of Complex Variables|1981|Murray R. Spiegel|ed = SI|edpage = SI (Metric) Edition|prev = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 2|next = Sum of Cosines of Fractions of Pi}}: $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $37$ Category:Examples of Complex Roots of Unity Category:Complex 5th Roots of Unity s7scye8xf3d942lgzdqkq63ufkrm3ja


Polynomial Factor Theorem/Corollary/Complex Numbers 0 70604 399146 363764 2019-04-05T22:07:27Z Prime.mover 59 wikitext text/x-wiki == Corollary to Polynomial Factor Theorem == <onlyinclude> Let $\map P z$ be a polynomial in $z$ over the complex numbers $\C$ of degree $n$. Suppose there exists $\zeta \in \C: \map P \xi = 0$. Then: :$\map P z = \paren {x - \zeta} \map Q z$ where $\map Q z$ is a polynomial of degree $n - 1$. Hence, if $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \dotsb = \map P {\zeta_n} = 0$, then: :$\displaystyle \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$ where $k \in \C$. </onlyinclude> == Proof == Recall that Complex Numbers form Field. The result then follows from the Polynomial Factor Theorem. {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Corollary/Examples/Cube Roots of 8i|next = Power of Complex Number minus 1}}: $\S 3$. Roots of Unity Category:Polynomial Theory 5a1t2sdcgh059t3b0jzevzvi5rwhxlk


Complex Algebra/Examples/z^6 + z^3 + 1

Complex Algebra/Examples/z^6 + z^3 + 1 0 70620 456565 363910 2020-03-19T23:14:08Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Algebra == <onlyinclude> :$z^6 + z^3 + 1 = \paren {z^2 - 2 z \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 z \cos \dfrac {8 \pi} 9 + 1}$ </onlyinclude> == Proof == From Sum of Geometric Sequence or Difference of Two Cubes: :$z^6 + z^3 + 1 = \dfrac {z^9 - 1} {z^3 - 1}$ Then from Factorisation of $x^{2 n + 1} - 1$ in Real Domain: :$z^9 - 1 = \paren {z - 1} \displaystyle \prod_{k \mathop = 1}^4 \paren {z^2 - 2 \cos \dfrac {2 \pi k} 9 + 1}$ and: {{begin-eqn}} {{eqn | l = z^3 - 1       | r = \paren {z - 1} \prod_{k \mathop = 1}^1 \paren {z^2 - 2 \cos \dfrac {2 \pi k} 3 + 1}       | c =  }} {{eqn | r = \paren {z - 1} \paren {z^2 - 2 \cos \dfrac {2 \pi} 3 + 1}       | c =  }} {{eqn | r = \paren {z - 1} \paren {z^2 - 2 \cos \dfrac {6 \pi} 9 + 1}       | c =  }} {{end-eqn}} Thus: :$\paren {z^3 - 1} \paren {z^6 + z^3 + 1} = \paren {\paren {z - 1} \paren {z^2 - 2 \cos \dfrac {6 \pi} 9 + 1} } \paren {\paren {z^2 - 2 \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {8 \pi} 9 + 1} }$ from which: :$\paren {z^6 + z^3 + 1} = \paren {z^2 - 2 \cos \dfrac {2 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {4 \pi} 9 + 1} \paren {z^2 - 2 \cos \dfrac {8 \pi} 9 + 1}$ {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Factorisation of z^n-a|next = Triple Angle Formulas/Cosine/2 cos 3 theta + 1}}: $\S 3$. Roots of Unity: Example $6$: $(3.14)$ Category:Complex Roots Category:Examples of Complex Algebra 7rexsnnxj4bf04rohyd7kfxfpkltee3


Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration 0 70649 364035 2018-09-02T06:13:38Z Prime.mover 59 Created page with "== Illustration of Roots of $z^8 + 1 = 0$ == <onlyinclude> The roots of the Definition:Pol..." wikitext text/x-wiki == Illustration of Roots of $z^8 + 1 = 0$ == <onlyinclude> The roots of the polynomial: :$z^8 + 1 = 0$ are illustrated below: :620px </onlyinclude> == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Examples/z^8 + 1 = 0|next = Complex Algebra/Examples/z^8 + 1}}: $\S 3$. Roots of Unity: Exercise $8$ Category:Examples of Complex Roots 576jivb1jsfmev9aqjwjg3o0oeu5huo


Complex Algebra/Examples/z^8 + 1

Complex Algebra/Examples/z^8 + 1 0 70650 364074 364042 2018-09-02T10:06:57Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Algebra == <onlyinclude> :$z^8 + 1 = \paren {z^2 - 2 z \cos \dfrac \pi 8 + 1} \paren {z^2 - 2 z \cos \dfrac {3 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {5 \pi} 8 + 1} \paren {z^2 - 2 z \cos \dfrac {7 \pi} 8 + 1}$ </onlyinclude> == Proof == From Roots of $z^8 + 1 = 0$ and the corollary to the Polynomial Factor Theorem: :$z^8 + 1 = \displaystyle \prod_{k \mathop = 0}^7 \paren {z - \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8} }$ Hence: {{begin-eqn}} {{eqn | l = z^8 + 1       | r = \prod_{k \mathop = 0}^7 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z - \exp \dfrac {\paren {2 k + 1} i \pi} 8} \paren {z - \exp \dfrac {-\paren {2 k + 1} i \pi} 8}       | c = Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + \exp \dfrac {\paren {2 k + 1} i \pi} 8 \exp \dfrac {-\paren {2 k + 1} i \pi} 8}       | c = multiplying out }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\exp \dfrac {\paren {2 k + 1} i \pi} 8 + \exp \dfrac {-\paren {2 k + 1} i \pi} 8} + 1}       | c = simplifying }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - z \paren {\cos \dfrac {\paren {2 k + 1} \pi} 8 + i \sin \dfrac {\paren {2 k + 1} \pi} 8 + \cos \dfrac {\paren {2 k + 1} \pi} 8 - i \sin \dfrac {\paren {2 k + 1} \pi} 8} + 1}       | c = {{Defof|Exponential Form of Complex Number}} }} {{eqn | r = \prod_{k \mathop = 0}^3 \paren {z^2 - 2 z \cos \dfrac {\paren {2 k + 1} \pi} 8 + 1}       | c = simplifying }} {{end-eqn}} Hence the result. {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Roots of Complex Number/Examples/z^8 + 1 = 0/Illustration|next = Quadruple Angle Formulas/Cosine/Factor Form}}: $\S 3$. Roots of Unity: Exercise $8$ Category:Complex Roots Category:Examples of Complex Algebra tcb2reog2caqly9vaw987knentssqgp


Euler's Formula/Corollary

Euler's Formula/Corollary 0 70856 365723 365141 2018-09-14T22:06:16Z Prime.mover 59 wikitext text/x-wiki == Corollary to Euler's Formula == Let $z \in \C$ be a complex number. Then: <onlyinclude> :$e^{-i z} = \cos z - i \sin z$ </onlyinclude> where: : $e^{-i z}$ denotes the complex exponential function : $\cos z$ denotes the complex cosine function : $\sin z$ denotes complex sine function : $i$ denotes the imaginary unit. === Corollary === This result is often presented and proved separately for arguments in the real domain: {{:Euler's Formula/Real Domain/Corollary}} == Proof == {{begin-eqn}} {{eqn | l = e^{-i z}       | r = \cos \paren {-z} + i \sin \paren {-z}       | c = Euler's Formula }} {{eqn | r = \cos z + i \sin \paren {-z}       | c = Cosine Function is Even }} {{eqn | r = \cos z - i \sin z       | c = Sine Function is Odd }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Sine Function is Odd|next = Cosine Exponential Formulation/Proof 3}}: $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$: $\text{(ii)}$: $(4.16)$ Category:Euler's Formula od8drfqw7i7eigin9lgee53gjmm4c2s


Complex Natural Logarithm/Examples/i

Complex Natural Logarithm/Examples/i 0 70907 365496 365494 2018-09-11T22:18:14Z Prime.mover 59 wikitext text/x-wiki == Example of Complex Natural Logarithm == <onlyinclude> :$\ln \paren i = \paren {4 k + 1} \dfrac {\pi i} 2$ for all $k \in \Z$. </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = i       | r = \exp \paren {\dfrac {i \pi} 2}       | c = Polar Form of  $i$ }} {{eqn | ll= \leadsto       | l = \ln \paren i       | r = \ln \paren {\exp \paren {\dfrac {i \pi} 2 + 2 k \pi i} }       | c =  }} {{eqn | r = \dfrac {i \pi + 4 k \pi i} 2       | c = {{Defof|Complex Natural Logarithm}} }} {{eqn | r = \paren {4 k + 1} \dfrac {\pi i} 2       | c =  }} {{end-eqn}} {{qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Natural Logarithm/Examples/-2|next = Complex Cosine Function/Examples/4 cos z = 3+i}}: $\S 4.6$. The Logarithm: Examples: $\text {(ii)}$ Category:Examples of Complex Natural Logarithms 4vjtql85ggcsbb4s8ujl8vs7hcm55md


Square Root of Complex Number in Cartesian Form/Examples/-8+6i 0 70916 365526 365505 2018-09-12T06:24:03Z Prime.mover 59 wikitext text/x-wiki == Example of Square Root of Complex Number in Cartesian Form == <onlyinclude> :$\sqrt {-8 + 6 i} = \pm \paren {1 + 3 i}$ </onlyinclude> == Proof == {{begin-eqn}} {{eqn | l = \paren {x + i y}^2       | r = -8 + 6 i       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | r = \dfrac {-8 + \sqrt {\paren {-8}^2 + 6^2} } 2       | c = Square Root of Complex Number in Cartesian Form }} {{eqn | r = \dfrac {-8 + \sqrt {100} } 2       | c =  }} {{eqn | r = \dfrac {-8 + 10} 2       | c =  }} {{eqn | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = \pm 1       | c =  }} {{eqn | ll= \leadsto       | l = y       | r = \pm \dfrac 6 {2 \times 1}       | c =  }} {{eqn | r = \pm 3       | c =  }} {{end-eqn}} As $2 x y = 6$ it follows that the two solutions are: :$1 + 3 i$ :$-1 - 3 i$ {{Qed}} == Sources == * {{BookReference|Complex Numbers|1960|Walter Ledermann|prev = Complex Cosine Function/Examples/4 cos z = 3+i|next = Inverse Tangent of i}}: $\S 4.6$. The Logarithm: Examples: $\text {(iii)}$ Category:Examples of Square Roots lgeddlb38hbut62yl9p26gre49nmkar


Proof by Contraposition

Proof by Contraposition 0 70942 370643 370486 2018-10-12T07:57:47Z KarlFrei 3474 fixed this as well wikitext text/x-wiki == Proof Technique == '''Proof by contraposition''' is a rule of inference used in proofs.   This rule infers a conditional statement from its contrapositive. It is based on the Rule of Transposition, which says that a conditional statement and its contrapositive have the same truth value: :$p \implies q \dashv \vdash \neg q \implies \neg p$ In other words, the conclusion "if A, then B" is drawn from the single premise "if not B, then not A."  == Explanation == '''Proof by Contraposition''' can be expressed in natural language as follows: If we know that by making an assumption  :$\neg q$  we can deduce  :$\neg p$ then it must be the case that  :$p \implies q$. Thus it provides a means of proving a logical implication. This proof is often confused with Reductio ad Absurdum, which also starts with an assumption $\neg q$. Reductio ad Absurdum itself is often confused with Proof by Contradiction. Unlike Reductio ad Absurdum however, Proof by Contraposition ''can'' be a valid proof in intuitionistic logic, just like Proof by Contradiction. Specifically, suppose :$p \implies q$ is true.  Suppose furthermore that we have a proof of :$\neg q$. Then if we had a proof of $p$, it could be turned into a proof of $q$. This would imply :$q\land \neg q$ which is impossible. Therefore :$\neg p$. However, now suppose :$\neg q \implies \neg p$ is true. Suppose furthermore that we have a proof of :$p$. Then if we had a proof of $\neg q$, it could be turned into a proof of $\neg p$. This would imply :$p\land\neg p$ which is impossible. Thus it is not possible to prove $\neg q$. That means in this case we only know :$\neg \neg q$. Category:Proof Techniques dfilcz1dhbzddxx88jkwh9kofb4jzyy


Hyperbolic Sine of Complex Number

Hyperbolic Sine of Complex Number 0 70971 365755 2018-09-15T09:06:14Z Prime.mover 59 Created page with "== Theorem == <onlyinclude> Let $a$ and $b$ be real numbers.  Let $i$ be the imaginary unit.  Then: :$\sinh \paren {a..." wikitext text/x-wiki == Theorem == <onlyinclude> Let $a$ and $b$ be real numbers. Let $i$ be the imaginary unit. Then: :$\sinh \paren {a + b i} = \sinh a \cos b + i \cosh a \sin b$ where: :$\sin$ denotes the real sine function :$\cos$ denotes the real cosine function :$\sinh$ denotes the hyperbolic sine function :$\cosh$ denotes the hyperbolic cosine function. </onlyinclude> == Proof 1 == {{:Hyperbolic Sine of Complex Number/Proof 1}} == Proof 2 == {{:Hyperbolic Sine of Complex Number/Proof 2}} == Also see == * Hyperbolic Cosine of Complex Number * Hyperbolic Tangent of Complex Number * Hyperbolic Cosecant of Complex Number * Hyperbolic Secant of Complex Number * Hyperbolic Cotangent of Complex Number Category:Hyperbolic Sine Function Category:Complex Numbers Category:Hyperbolic Sine of Complex Number a6nti2dzyb53lz3n90ykjnnj4vsqqjc


Hyperbolic Cotangent of Complex Number/Formulation 1 0 70986 365788 2018-09-15T10:49:12Z Prime.mover 59 Created page with "== Theorem ==  Let $a$ and $b$ be real numbers.  Let $i$ be the imaginary unit.   Then: <onlyinclude> :$\coth \paren {..." wikitext text/x-wiki == Theorem == Let $a$ and $b$ be real numbers. Let $i$ be the imaginary unit. Then: <onlyinclude> :$\coth \paren {a + b i} = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}$ </onlyinclude> where: :$\coth$ denotes the hyperbolic cotangent function :$\sin$ denotes the real sine function :$\cos$ denotes the real cosine function :$\sinh$ denotes the hyperbolic sine function :$\cosh$ denotes the hyperbolic cosine function. == Proof == {{begin-eqn}} {{eqn | l = \coth \paren {a + b i}       | r = \frac {\cosh \paren {a + b i} } {\sinh \paren {a + b i} }       | c = {{Defof|Hyperbolic Cotangent}} }} {{eqn | r = \dfrac {\cosh a \cos b + i \sinh a \sin b} {\sinh a \cos b + i \cosh a \sin b}       | c = Hyperbolic Sine of Complex Number and Hyperbolic Cosine of Complex Number }} {{end-eqn}} {{qed}} == Also see == * Hyperbolic Sine of Complex Number * Hyperbolic Cosine of Complex Number * Hyperbolic Tangent of Complex Number * Hyperbolic Cosecant of Complex Number * Hyperbolic Secant of Complex Number Category:Hyperbolic Cotangent of Complex Number dpmhmf06af3ewir8mn3uwbn7sinxmtn


Cardinality/Examples/0 less than x less than 6 0 71072 366321 366318 2018-09-18T07:13:08Z Prime.mover 59 wikitext text/x-wiki == Example of Cardinality == Let $S_2 = \set {x \in \Z: 0 < x < 6}$. <onlyinclude> The cardinality of $S_2$ is given by: :$\card {S_2} = 5$ </onlyinclude> == Proof == The elements of $S_2$, by definition, are those integers greater than $0$ and less than $6$. That is: :$S_2 = \set {1, 2, 3, 4, 5}$ Thus $S_2$ has $5$ elements: $1, 2, 3, 4, 5$. Hence the result by definition of cardinality. {{qed}} == Sources == * {{BookReference|An Introduction to Abstract Algebra|1978|Thomas A. Whitelaw|prev = Cardinality/Examples/-1,0,1|next = Cardinality/Examples/x^2-x}}: Chapter $1$: Sets and Logic: Exercise $4$ Category:Examples of Cardinality 2pb843cw778el57i367ci63qz0be32w


Fermat's Little Theorem/Corollary 3

Fermat's Little Theorem/Corollary 3 0 71091 392010 366591 2019-02-10T22:10:48Z Prime.mover 59 wikitext text/x-wiki {{Proofread}} == Corollary to Fermat's Little Theorem == <onlyinclude> Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$. Then: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$ </onlyinclude> == Proof == The proof proceeds by induction. For all $k \in \Z_{\ge 1}$, let $P \paren {k}$ be the proposition: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$ === Basis for the Induction === $P \paren {1}$ is the case: :$\forall n \in \Z_{\gt 0}: n^p \equiv n \pmod p$ which follows from the corollary 1 to Fermat's Little Theorem. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that, if $P \paren{k-1}$ is true, where $k \ge 2$, then it logically follows that $P \paren {k}$ is true. So this is the induction hypothesis: :$\forall n \in \Z_{\gt 0}: n^{p^{k - 1} } \equiv n \pmod p$ from which it is to be shown that: :$\forall n \in \Z_{\gt 0}: n^{p^k} \equiv n \pmod p$ === Induction Step === This is the induction step: For any $n \in \Z_{\gt 0}$ then: {{begin-eqn}} {{eqn | l = n^{p^k}       | r = \paren {n^{p^{k - 1} } }^p }} {{eqn | r = n^{p^{k - 1} } \pmod p       | o = \equiv       | c = Corollary 1 to Fermat's Little Theorem }} {{eqn | r = n \pmod p       | o = \equiv       | c = Induction Hypothesis }} {{end-eqn}} {{qed}} == Sources == Category:Fermat's Little Theorem 84bvsuw33h9sxbgpby1ep4ig4816wh1


Fermat's Little Theorem/Corollary 4

Fermat's Little Theorem/Corollary 4 0 71092 392011 366414 2019-02-10T22:11:15Z Prime.mover 59 wikitext text/x-wiki {{Proofread}} == Corollary to Fermat's Little Theorem == <onlyinclude> Let $p^k$ be a prime power for some prime number $p$ and $k \in \Z_{\gt 0}$. Let $n \in \Z_{\gt 0}$ with $p \nmid n$. Then: :$n^{p^k - 1} \equiv 1 \pmod p$ </onlyinclude> == Proof == By corollary 3 of Fermat's Little Theorem: :$n^{p^k} \equiv n \pmod p$ That is: :$p \divides \paren {n^{p^k} - n} = n \paren {n^{p^k - 1} - 1}$ Since $p \nmid n$, by Corollary to Divisors of Product of Coprime Integers: :$p \divides \paren {n^{p^k - 1} - 1}$ That is: :$n^{p^k-1} \equiv 1 \pmod p$ {{qed}} == Sources == Category:Fermat's Little Theorem q0leb0mj7psa4cwwhoy40z4g2khedcq


Total Number of Set Partitions/Examples

Total Number of Set Partitions/Examples 0 71112 366790 366778 2018-09-22T08:10:30Z Prime.mover 59 wikitext text/x-wiki == Examples of Total Number of Set Partitions == <onlyinclude> === Example: $\card S = 2$ === {{:Total Number of Set Partitions/Examples/2}} === Example: $\card S = 3$ === {{:Total Number of Set Partitions/Examples/3}} === Example: $\card S = 4$ === {{:Total Number of Set Partitions/Examples/4}}</onlyinclude> Category:Examples of Set Partitions to3rp6r8nn76o1kly0i4xiycq66lq29