nyuuzyou/wikis huggingface dataset:
Reciprocal of 89/Mistake

== Source Work ==
:
:The Dictionary
::$89$
== Mistake ==
:$89$ is the $11$th Fibonacci number, and the period of its reciprocal is generated by the Fibonacci sequence: $1 / 89 = 0 \cdotp 11235 \ldots$
== Correction ==
That should read:
:$1 / 89 = 0 \cdotp 01123 \, 5 \ldots$
This mistake is corrected in the of $1997$.
== Sources ==
* : $89$
Category:Mistakes/Curious and Interesting Numbers
Reciprocal of 89 as Sum of Fibonacci Numbers by Negative Powers of 10

== Theorem ==
:$\ds \sum_{k \mathop \ge 0} \dfrac {F_k} {10^{k + 1} } = \dfrac 1 {89}$
where $F_k$ is the $k$th Fibonacci number:
:$F_0 = 0, F_1 = 1, F_k = F_{k - 1} + F_{k - 2}$
That is:
1 / 89 = 0.0
 + 0.01
 + 0.001
 + 0.0002
 + 0.00003
 + 0.000005
 + 0.0000008
 + 0.00000013
 + 0.000000021
 + 0.0000000034
 + 0.00000000055
 + ..............
== Proof ==
First we note that from Reciprocal of $89$:
:$\dfrac 1 {89} = 0 \cdotp \dot 01123 \, 59550 \, 56179 \, 77528 \, 08988 \, 76404 \, 49438 \, 20224 \, 719 \dot 1$
We have that:
:$89 = 10^2 - 10 - 1$
So:
== Sources ==
* : $89$
Category:89
Category:Reciprocals
Category:Fibonacci Numbers
Reciprocal of 909,091

== Theorem ==
The reciprocal of $909 \, 091$ is:
:$\dfrac 1 {909 \, 091} = 0 \cdotp \dot 00000 \, 10999 \, 998 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.00000109999989000001...
 --------------------------------------------
909091)1.00000000000000000000...
 909091
 ------
 9090900
 8181819
 -------
 9090810
 8181819
 -------
 9089910
 8181819
 -------
 9080910
 8181819
 -------
 8990910
 8181819
 -------
 8090910
 7272728
 -------
 8181820
 8181819
 -------
 1000000
 909091
 -------
 ......
Category:909,091
Category:Examples of Reciprocals
Reciprocal of 9091

== Theorem ==
The reciprocal of $9091$ is:
:$\dfrac 1 {9091} = 0 \cdotp \dot 00010 \, 9998 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.00010999890001...
 --------------------------------------------
9091)1.00000000000000...
 9091
 ----
 90900
 81819
 -----
 90810
 81819
 -----
 89910
 81819
 -----
 80910
 72728
 -----
 81820
 81819
 -----
 10000
 9091
 -----
 ...
Category:9091
Category:Examples of Reciprocals
Reciprocal of 97

== Theorem ==
The decimal expansion of the reciprocal of $97$ is as follows:
:$\dfrac 1 {97} = 0 \cdotp \dot 01030 \, 92783 \, 50515 \, 46391 \, 75257 \, 73195 \, 87628 \, 86597 \, 93814 \, 43298 \, 96907 \, 21649 \, 48453 \, 60824 \, 74226 \, 80412 \, 37113 \, 40206 \, 18556 \, \dot 7$
== Proof ==
Performing the calculation using long division:
 0.01030927835051546391752577319587628865979381443298969072164948453608247422680412371134020618556701...
 ------------------------------------------------------------------------------------------------------
97)1.00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
 97 485 873 679 582 679 291 679 776 388 388 291 485
 -- --- --- --- --- --- --- --- --- --- --- --- ---
 300 500 170 310 280 910 290 210 440 720 120 390 650
 291 485 97 291 194 873 194 194 388 679 97 388 582
 --- --- --- --- --- --- --- --- --- --- --- --- ---
 900 150 730 190 860 370 960 160 520 410 230 200 680
 873 97 679 97 776 291 983 97 485 388 194 194 679
 --- --- --- --- --- --- --- --- --- --- --- --- ---
 270 530 510 930 840 790 870 630 350 220 360 600 100
 194 485 485 873 776 776 776 582 291 194 291 582 97
 --- --- --- --- --- --- --- --- --- --- --- --- ---
 760 450 250 570 640 140 940 480 590 260 690 180 ...
 679 388 194 485 582 97 873 388 582 194 679 97
 --- --- --- --- --- --- --- --- --- --- --- ---
 810 620 560 850 580 430 670 920 800 660 110 830
 776 582 485 776 485 388 582 873 776 582 97 776
 --- --- --- --- --- --- --- --- --- --- --- ---
 340 380 750 740 950 420 880 470 240 780 130 540
 291 291 679 679 873 388 873 388 194 776 97 485
 --- --- --- --- --- --- --- --- --- --- --- ---
 490 890 710 610 770 320 700 820 460 400 330 550
 485 873 679 582 679 291 679 776 388 388 291 485
== Historical Note ==
== Sources ==
* : $97$
* : $97$
Category:97
Category:Examples of Reciprocals
Reciprocal of 98

== Theorem ==
The decimal expansion of the reciprocal of $98$ starts with the powers of $2$:
:$\dfrac 1 {98} = 0 \cdotp 0 \dot 1020 \, 40816 \, 32653 \, 06122 \, 44897 \, 95918 \, 36734 \, 69387 \, 75 \dot 5$
== Proof ==
Performing the calculation using long division:
 0.01020408163265306122448979591836734693877551...
 --------------------------------------------------
98)1.00000000000000000000000000000000000000000000000
 98 196 196 490 294 686
 -- --- --- --- --- ---
 200 640 440 900 460 540
 196 588 392 882 392 490
 --- --- --- --- --- ---
 400 520 480 180 680 500
 392 490 392 98 588 490
 --- --- --- --- --- ---
 800 300 880 820 920 100
 784 294 784 784 882 98
 --- --- --- --- --- ---
 160 600 960 360 380 ...
 98 588 882 294 294
 --- --- --- --- ---
 620 120 780 660 860
 588 98 686 588 784
 --- --- --- --- ---
 320 220 940 720 760
 294 196 882 686 686
 --- --- --- --- ---
 260 240 580 340 740
 196 196 490 294 686
We have that:
:$0 \cdotp 01 + 0 \cdotp 0002 + 0 \cdotp 000004 + 0 \cdotp 00000008 + \cdots$
is nothing else but:
:$\ds \dfrac 1 2 \paren {\dfrac 1 {50} + \paren {\dfrac 1 {50} }^2 + \paren {\dfrac 1 {50} }^3 + \paren {\dfrac 1 {50} }^4 + \cdots} = \dfrac 1 2 \sum_{k \mathop \ge 1} \paren {\dfrac 1 {50} }^k$
Hence:
== Sources ==
* : $98$
* : $98$
Category:98
Category:Examples of Reciprocals
Reciprocal of 9801

== Theorem ==
The reciprocal of $9801$ is:
:$\dfrac 1 {9801} = 0 \cdotp \dot 00010 \, 20304 \, 05060 \, 70809 \, 10111 \, 21314 \, ... \, 94959 \, 6979 \dot 9$
== Proof ==
Category:9801
Category:Examples of Reciprocals
Reciprocal of 99

== Theorem ==
The decimal expansion of the reciprocal of $99$ is:
:$\dfrac 1 {99} = 0 \cdotp \dot 0 \dot 1$
== Proof ==
Performing the calculation using long division:
 0.0101...
 ----------
99)1.0000000
 99
 --
 100
 99
 ---
 ...
== Sources ==
* : $99$
* : $99$
Category:99
Category:Examples of Reciprocals
Reciprocal of 99,990,001

== Theorem ==
The reciprocal of $99 \, 990 \, 001$ is:
:$\dfrac 1 {99 \, 990 \, 001} = 0 \cdotp \dot 00000 \, 00100 \, 00999 \, 99998 \, 999 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.00000001000099999998999900000001...
 --------------------------------------------
99990001)1.00000000000000000000000000000000...
 99990001
 --------
 999900000
 899910009
 ---------
 999899910
 899910009
 ---------
 999899010
 899910009
 ---------
 999890010
 899910009
 ---------
 999800010
 899910009
 ---------
 998900010
 899910009
 ---------
 989900010
 899910009
 ---------
 899900010
 799920008
 ---------
 999800020
 899910009
 ---------
 998900110
 899910009
 ---------
 988801010
 899910009
 ---------
 899910010
 899910009
 ---------
 100000000
 99990001
 ---------
 ........
Category:99,990,001
Category:Examples of Reciprocals
Reciprocal of 9901

== Theorem ==
The reciprocal of $9901$ is:
:$\dfrac 1 {9901} = 0 \cdotp \dot 00010 \, 09998 \, 9 \dot 9$
== Proof ==
Performing the calculation using long division:
 0.0001009998990001...
 --------------------------------------------
9901)1.0000000000000000...
 9901
 ----
 99000
 89109
 -----
 98910
 89109
 -----
 98010
 89109
 -----
 89010
 79208
 -----
 98020
 89109
 -----
 89110
 89109
 -----
 10000
 9901
 -----
 ....
Category:9901
Category:Examples of Reciprocals
Reciprocal of Absolutely Convergent Product is Absolutely Convergent

== Theorem ==
Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.
Let $\sequence {1 + a_n}$ be a sequence of nonzero elements of $\mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty \left({1 + a_n}\right)$ converge absolutely to $a \in \mathbb K \setminus \set 0$.
Then $\ds \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$ converges absolutely to $1 / a$.
== Proof ==
By continuity of $x \mapsto 1 / x$:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = 1}^N \frac 1 {1 + a_n} = \frac 1 a$
It remains to prove the absolute convergence.
Because $\ds \prod_{n \mathop = 1}^\infty \paren {1 + a_n}$ converges absolutely, $\ds \sum_{n \mathop = 1}^\infty a_n$ converges absolutely.
By Factors in Absolutely Convergent Product Converge to One, $\norm {a_n} \le \dfrac 1 2 $ for $n$ sufficiently large.
Thus $\norm {\dfrac 1 {a_n + 1} - 1} = \norm {\dfrac {a_n} {a_n + 1} } \le 2 \norm {a_n}$ for $n$ sufficiently large.
By the Comparison Test, $\ds \sum_{n \mathop = 1}^\infty \paren {\frac 1 {a_n + 1} - 1}$ converges absolutely.
Thus $\ds \prod_{n \mathop = 1}^\infty \frac 1 {1 + a_n}$ converges absolutely.
Category:Infinite Products
Reciprocal of Complex Number

== Theorem ==
Let $z = a + i b$ be a complex number.
The reciprocal of $z$ is:
:$\dfrac 1 z = \dfrac {a - i b} {a^2 + b^2}$
== Proof ==
== Also see ==
* Reciprocal of Complex Number in terms of Conjugate and Modulus
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.24$
Category:Complex Numbers
Category:Reciprocals
Reciprocal of Complex Number in terms of Conjugate and Modulus

== Theorem ==
Let $z \in \C$ be a complex number.
The reciprocal of $z$ can be expressed as:
:$\dfrac 1 z = \dfrac {\overline z} {\cmod z^2}$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denotes the modulus of $z$.
== Proof ==
Let $z$ be defined as:
:$z = a = i b$
Then:
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.24$
Category:Complex Modulus
Category:Complex Conjugates
Category:Reciprocals
Reciprocal of Computable Rational Sequence is Computable

== Theorem ==
Let $\sequence {x_n}$ be a computable rational sequence.
Suppose that, for all $n \in \N$:
:$x_n \ne 0$
Then:
:$\sequence {\dfrac 1 {x_n}}$ is a computable rational sequence.
== Proof ==
By definition of computable rational sequence, there exist total recursive $f, g : \N \to \N$ such that:
:$x_n = \dfrac {k_n} {\map g n + 1}$
where $\map f n$ codes the integer $k_n$.
We will define $f', g' : \N \to \N$ as:
:$\map {f'} n = \map {\sgn_\Z} {k_n} \times_\Z \paren {\map g n + 1}_\Z$
:$\map {g'} n = \map {\operatorname{pred}} {\size {k_n}}$
We have that both $f'$ and $g'$ are total recursive by:
* Signum Function on Integers is Primitive Recursive
* Multiplication of Integers is Primitive Recursive
* Code Number for Non-Negative Integer is Primitive Recursive
* Absolute Value of Integer is Primitive Recursive
* Predecessor Function is Primitive Recursive
We now need to show that, for all $n \in \N$:
:$\dfrac 1 {x_n} = \dfrac {k'_n} {\map {g'} n + 1}$
where $\map {f'} n$ codes the integer $k'_n$.
Since $x_n \ne 0$ by assumption, we have:
:$k_n \ne 0$
for all $n \in \N$.
We will now examine the cases of:
:$k_n > 0$
:$k_n < 0$
Suppose $k_n > 0$.
Then:
Suppose $k_n < 0$.
Then:
Thus, by Proof by Cases, the result holds.
Category:Computability Theory
Reciprocal of Computable Real Sequence is Computable

== Theorem ==
Let $\sequence {x_m}$ be a computable real sequence.
Suppose that, for all $m \in \N$:
:$x_m \ne 0$
Then:
:$\sequence {\dfrac 1 {x_m}}$
is a computable real sequence.
== Proof ==
=== Lemma ===
Reciprocal of Computable Real Sequence is Computable/Lemma

== Lemma for Reciprocal of Computable Real Sequence is Computable ==
Let $\sequence {x_m}$ be a computable real sequence.
Suppose that, for all $m \in \N$:
:$x_m \ne 0$
There exists a total recursive function $\psi : \N \to \N$ such that, for all $m, p \in \N$ and $\alpha \in \R$, if:
:$p \ge \map \psi m$
and:
:$\size {x_m - \alpha} \le \dfrac 1 {p + 1}$
it follows that:
:$\size \alpha > \dfrac {\size {x_m}} 2$
== Proof ==
By definition of computable real sequence, there exists a total recursive $f : \N^2 \to \N$ such that, for all $m, n \in \N$:
:$\dfrac {k_{m,n} - 1} {n + 1} < x_m < \dfrac {k_{m,n} + 1} {n + 1}$
where $\map f {m, n}$ codes the integer $k_{m,n}$.
We define $\psi : \N \to \N$ as:
:$\map \psi m = \map {\mu n} {\size {k_{m,n}} > 1}$
which is partial recursive by:
* Absolute Value of Integer is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Constant Function is Primitive Recursive
* Minimization on Relation Equivalent to Minimization on Function
In order to show that $\psi$ is total, we will show that:
:$\size {k_{m,n}} > 1$
is satsified when $\dfrac 1 {n + 1} \le \dfrac {\size {x_m}} 2$.
We have:
Therefore, we have:
:$\size {k_{m,n} } > 1$
By Sequence of Reciprocals is Null Sequence:
:$\dfrac 1 {n + 1} \to 0$ as $n \to \infty$
so there is always $n$ sufficiently large so that:
:$\dfrac 1 {n + 1} \le \dfrac {\size {x_m}} 2$
It follows that $\psi : \N \to \N$ is a total recursive function.
We need to show that the result holds for $\psi$.
Let $m, p \in \N$ and $\alpha \in \R$ satisfy:
:$p \ge \map \psi m$
:$\size {x_m - \alpha} \le \dfrac 1 {p + 1}$
We have that:
:$n = \map \psi m$
satisfies:
:$\size {k_{m,n}} \ge 2$
by the definition of $\psi$.
By assumption:
:$\dfrac {k_{m,n} - 1} {n + 1} < x_m < \dfrac {k_{m,n} + 1} {n + 1}$
Reciprocal of Continuous Complex-Valued Function Vanishing at Infinity does not Vanish at Infinity

== Theorem ==
Let $X$ be a locally compact Hausdorff space. 
Let $f : X \to \C$ be a continuous function that vanishes at infinity such that:
:$\map f x \ne 0$ for each $x \in X$.
Then $\dfrac 1 f : X \to \C$ does not vanish at infinity.
== Proof ==
From Quotient Rule for Continuous Complex Functions:
:$\dfrac 1 f$ is continuous.
 $\dfrac 1 f$ vanishes at infinity.
Then from Continuous Complex-Valued Function Vanishing at Infinity is Bounded and Attains Supremum, $\dfrac 1 f$ is bounded.
Since $f$ vanishes at infinity, for each $n \in \N$ there exists a compact set $K_n \subseteq X$ such that:
:$0 < \cmod {\map f x} < \dfrac 1 n$ for each $x \in X \setminus K_n$ 
That is:
:$\ds \cmod {\frac 1 {\map f x} } > n$ for each $x \in X \setminus K_n$
This contradicts that $\dfrac 1 f$ is bounded.
Hence $\dfrac 1 f$ does not vanish at infinity.
Category:Complex-Valued Functions Vanishing at Infinity
Reciprocal of Difference of Squares as Difference of Reciprocals

== Theorem ==
:$\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 y \paren {x - y} } - \dfrac 1 {2 y \paren {x + y} }$
== Proof ==
== Also see ==
* Reciprocal of Difference of Squares as Sum of Reciprocals
Category:Reciprocal of Difference of Squares
Category:Algebra
Reciprocal of Difference of Squares as Sum of Reciprocals

== Theorem ==
:$\dfrac 1 {x^2 - y^2} = \dfrac 1 {2 x \paren {x + y} } + \dfrac 1 {2 x \paren {x - y} }$
== Proof ==
== Also see ==
* Reciprocal of Difference of Squares as Difference of Reciprocals
Category:Reciprocal of Difference of Squares
Category:Algebra
Reciprocal of Function of Bounded Variation Bounded away from Zero is of Bounded Variation

== Theorem ==
Let $a, b$ be real numbers with $a < b$.
Let $f: \closedint a b \to \R$ be functions of bounded variation.
Let $f$ be bounded away from zero. 
That is, there exists $M \in \R$ such that: 
:$\size {\map f x} \ge M > 0$
for all $x \in \closedint a b$. 
Let $\map {V_f} {\closedint a b}$ be the total variation of $f$ on $\closedint a b$.
Then:
:$\dfrac 1 f$ is of bounded variation
with:
:$\map {V_{1 / f} } {\closedint a b} \le \dfrac {\map {V_f} {\closedint a b} } {M^2}$
where $\map {V_{1 / f} } {\closedint a b}$ denotes the total variation of $\dfrac 1 f$ on $\closedint a b$.
== Proof ==
For each finite subdivision $P$ of $\closedint a b$, write: 
:$P = \set {x_0, x_1, \ldots, x_n }$
with:
:$a = x_0 < x_1 < x_2 < \cdots < x_{n - 1} < x_n = b$
Note that: 
:$\dfrac 1 {\map f x} \le \dfrac 1 M$
for all $x \in \closedint a b$.
We then have: 
Since $f$ is of bounded variation, there exists $K \in \R$ such that: 
:$\map {V_f} {P ; \closedint a b} \le K$
for every finite subdivision $P$. 
Therefore: 
:$\map {V_{1 / f} } {P ; \closedint a b} \le \dfrac K {M^2}$
So $\dfrac 1 f$ is of bounded variation.
We also have: 
== Sources ==
* : $\S 6.4$: Total Variation: Theorem $6.10$
Category:Total Variation of Real Function
Category:Bounded Variation
Reciprocal of Holomorphic Function

== Theorem ==
Let $f: \C \to \C$ be a complex function.
Let $U \subseteq \C$ be an open set such that $f$ has no zeros in $U$.
Suppose further that $f$ is holomorphic in $U$.
Then the complex function
:$\dfrac 1 {f_{\restriction U} } : U \to \C$
is holomorphic.
== Proof ==
Let $g: U \to \C$ be such that $\map g x = 1 / \map f x$.
Since $\map f x$ is nonzero for $x \in U$, $g$ is well-defined.
By Quotient Rule for Continuous Complex Functions, $g$ is continuous.
Let $z_0 \in U$.
As $g$ is continuous:
:$\ds \lim_{h \mathop \to 0} \frac 1 {\map f {z_0 + h} } = \frac 1 {\map f {z_0} }$
As $f$ is holomorphic:
:$\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h = \map {f'} {z_0}$
	
By the Combination Theorem for Limits of Complex Functions:
	
It follows that $g$ is holomorphic.
Category:Holomorphic Functions
Category:Reciprocals
Reciprocal of Logarithm

== Theorem ==
Let $x, y \in \R_{> 0}$ be (strictly) positive real numbers.
Then:
:$\dfrac 1 {\log_x y} = \log_y x$
== Proof ==
== Also presented as ==
This result can also be seen presented as:
:$\paren {\log_x y} \paren {\log_y x} = 1$
== Examples ==
Category:Logarithms
Category:Reciprocals
Category:Reciprocal of Logarithm
Reciprocal of Logarithm/Examples/Logarithm Base 10 of 2

== Example of Reciprocal of Logarithm ==
The reciprocal of $\log_{10} 2$ is $\log_2 10$.
== Proof ==
From Reciprocal of Logarithm:
$\dfrac 1 {\log_x y} = \log_y x$
Let $x = 10, y = 2$:
Then:
$\dfrac 1 {\log_{10} 2} = \log_2 10$
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $20$
Category:Reciprocal of Logarithm
Reciprocal of Null Sequence

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\forall n \in \N: x_n > 0$.
Then:
:$x_n \to 0$ as $n \to \infty$ $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$
=== Corollary ===
== Proof ==
Suppose $x_n \to 0$ as $n \to \infty$.
Let $H > 0$.
So $H^{-1} > 0$.
Since $x_n \to 0$ as $n \to \infty$:
:$\exists N: \forall n > N: \size {x_n} < H^{-1}$
That is:
:$\size {\dfrac 1 {x_n} } > H$.
So:
:$\exists N: \forall n > N: \size {\dfrac 1 {x_n} } > H$
and thus:
:$\sequence {\size {\dfrac 1 {x_n} } }$ diverges to $+\infty$.
Suppose $\size {\dfrac 1 {x_n} } \to \infty$ as $n \to \infty$.
By reversing the argument above, we see that $x_n \to 0$ as $n \to \infty$.
== Also known as ==
Some sources call this the reciprocal rule, but as that name is used throughout mathematical literature for several different concepts, its use is not recommended.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (4)$
Category:Limits of Sequences
Category:Reciprocals
Category:Reciprocal of Null Sequence
Reciprocal of Null Sequence/Corollary

== Theorem ==
Let $\sequence {x_n}$ be a sequence in $\R$.
Let $\forall n \in \N: x_n > 0$.
Then:
:$x_n \to \infty$ as $n \to \infty$ $\size {\dfrac 1 {x_n} } \to 0$ as $n \to \infty$
== Proof ==
Let $\sequence {y_n}$ be the sequence in $\R$ defined as:
:$y_n = \size {\dfrac 1 {x_n} }$
From Reciprocal of Null Sequence:
:$y_n \to 0$ as $n \to \infty$ $\size {\dfrac 1 {y_n} } \to \infty$ as $n \to \infty$
That is:
:$\size {\dfrac 1 {x_n} } \to 0$ as $n \to \infty$ $x_n \to \infty$ as $n \to \infty$
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (4)$
Category:Reciprocal of Null Sequence
Reciprocal of One Minus Cosine

== Theorem ==
:$\dfrac 1 {1 - \cos x} = \dfrac 1 2 \map {\csc^2} {\dfrac x 2}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Reciprocal of One Plus Cosine
* Reciprocal of One Plus Sine
* Reciprocal of One Minus Sine
Category:Reciprocal of One Minus Cosine
Category:Trigonometric Identities
Category:Cosine Function
Reciprocal of One Minus Cosine plus Reciprocal of One Plus Cosine

== Theorem ==
:$\dfrac 1 {1 - \cos x} + \dfrac 1 {1 + \cos x} = 2 \cosec^2 x$
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Exercise $\text {XXXI}$: $1.$
Category:Trigonometric Identities
Category:Cosine Function
Reciprocal of One Minus Sine

== Theorem ==
:$\dfrac 1 {1 - \sin x} = \dfrac 1 2 \map {\sec^2} {\dfrac \pi 4 + \dfrac x 2}$
== Proof ==
== Also see ==
* Reciprocal of One Plus Sine
* Reciprocal of One Plus Cosine
* Reciprocal of One Minus Cosine
Category:Trigonometric Identities
Category:Sine Function
Reciprocal of One Plus Cosine

== Theorem ==
:$\dfrac 1 {1 + \cos x} = \dfrac 1 2 \sec^2 \dfrac x 2$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Reciprocal of One Minus Cosine
* Reciprocal of One Plus Sine
* Reciprocal of One Minus Sine
Category:Reciprocal of One Plus Cosine
Category:Trigonometric Identities
Category:Cosine Function
Reciprocal of One Plus Sine

== Theorem ==
:$\dfrac 1 {1 + \sin x} = \dfrac 1 2 \map {\sec^2} {\dfrac \pi 4 - \dfrac x 2}$
== Proof ==
== Also see ==
* Reciprocal of One Minus Sine
* Reciprocal of One Plus Cosine
* Reciprocal of One Minus Cosine
Category:Trigonometric Identities
Category:Sine Function
Reciprocal of One minus x in terms of Gaussian Hypergeometric Function

== Theorem ==
:$\dfrac 1 {1 - x} = \map F {1, p; p; x}$
where:
:$x$ and $p$ are real numbers with $\size x < 1$
:$\map F {a, b; c; x}$ is the Gaussian hypergeometric function of $x$: $\ds \sum_{k \mathop = 0}^\infty \dfrac { a^{\overline k} b^{\overline k} } { c^{\overline k} } \dfrac {x^k} {k!}$
== Proof ==
== Sources ==
* : $\S 31$: Hypergeometric Functions: Special Cases: $31.10$
Category:Gaussian Hypergeometric Function
Category:Hypergeometric Functions
Reciprocal of Power of Complex Number

== Theorem ==
Let $z \in \C$ be a complex number.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $z^n$ denote $z$ raised to the $n$th power.
The reciprocal of $z^n$ can be expressed as:
:$\dfrac 1 {z^n} = \dfrac {\overline z^n} {\cmod z^{2 n} }$
where:
:$\overline z$ denotes the complex conjugate of $z$
:$\cmod z^2$ denotes the modulus of $z$.
== Proof ==
== Sources ==
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.25$
Category:Reciprocals
Category:Complex Powers
Reciprocal of Quotient of Real Numbers

== Theorem ==
:$\forall x, y \in \R_{\ne 0}: \dfrac 1 {x / y} = \dfrac y x$
== Proof ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(q)}$
Category:Real Division
Category:Reciprocals
Reciprocal of Random Variable with Snedecor's F-Distribution has Snedecor's F-Distribution

== Theorem ==
Let $n, m$ be strictly positive integers. 
Let $X \sim F_{n, m}$ where $F_{n, m}$ is the Snedecor's $F$-distribution with $\tuple {n, m}$ degrees of freedom.
Then: 
:$\dfrac 1 X \sim F_{m, n}$
== Proof == 
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom. 
Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom.
Then, by Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution:
:$X = \dfrac {Y / n} {Z / m} \sim F_{n, m}$
Applying Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has Snedecor's F-Distribution again, we have: 
:$\dfrac 1 X = \dfrac {Z / m} {Y / n} \sim F_{m, n}$
Category:Snedecor's F-Distribution
Reciprocal of Real Number is Non-Zero

== Theorem ==
:$\forall x \in \R: x \ne 0 \implies \dfrac 1 x \ne 0$
== Proof ==
 that:
:$\exists x \in \R_{\ne 0}: \dfrac 1 x = 0$
From Real Zero is Zero Element
:$\dfrac 1 x \times x = 0$
But from :
:$\dfrac 1 x \times x = 1$
The result follows by Proof by Contradiction.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $1 \ \text{(p)}$
Category:Real Numbers
Category:Reciprocals
Reciprocal of Riemann Zeta Function

== Theorem ==
For $\map \Re z > 1$:
:$\ds \frac 1 {\map \zeta z} = \sum_{k \mathop = 1}^\infty \frac{\mu \left({k}\right)} {k^z}$
where:
:$\zeta$ is the Riemann zeta function
:$\mu$ is the Möbius function.
== Proof ==
By Sum of Reciprocals of Powers as Euler Product:
The expansion of this product will be:
:$\ds 1 + \sum_{\text {$n$ prime} } \paren {\frac{-1} {n^z} } + \sum_{n \mathop = p_1 p_2} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} } + \sum_{n \mathop = p_1 p_2 p_3} \paren {\frac {-1} {p_1^z} \frac {-1} {p_2^z} \frac {-1}{p_3^z} } + \cdots$
which is precisely:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map \mu n} {n^z}$
as desired.
Category:Riemann Zeta Function