nyuuzyou/wikis huggingface dataset:
Sum of Reciprocals of Sequence of Pairs of Even Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared

== Theorem ==
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.
== Proof ==
== Historical Note ==
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Golden Mean
Sum of Reciprocals of Sequence of Pairs of Even Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared/Historical Note

== Historical Note on Sum of Reciprocals of Sequence of Pairs of Even Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared ==
This result is attributed to .
== Sources ==
* : $5$
:::(although erroneously given as Schue)
* : $5$
Category:Historical Notes
Sum of Reciprocals of Sequence of Pairs of Odd Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean

== Theorem ==
where:
:$F_k$ denotes the $k$th Fibonacci number
:$\phi$ denotes the golden mean.
== Proof ==
== Also see ==
*Sum of Reciprocals of Sequence of Pairs of Even Index Consecutive Fibonacci Numbers is Reciprocal of Golden Mean Squared
Category:Fibonacci Numbers
Category:Golden Mean
Sum of Reciprocals of Sixth Powers Alternating in Sign

== Theorem == 
== Proof == 
== Also see ==
* Sum of Reciprocals of Even Powers of Integers Alternating in Sign
* Sum of Reciprocals of Squares Alternating in Sign
* Sum of Reciprocals of Fourth Powers Alternating in Sign
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.24$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.24.$
Category:Formulas for Pi
Category:Sums of Sequences
Category:Sixth Powers
Sum of Reciprocals of Sixth Powers of Odd Integers

== Theorem ==
== Proof ==
 
== Also presented as ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.27$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.27.$
Category:Sum of Reciprocals of Sixth Powers of Odd Integers
Category:Formulas for Pi
Category:Sixth Powers
Sum of Reciprocals of Sixth Powers of Odd Integers/Also presented as

== Sum of Reciprocals of Sixth Powers of Odd Integers: Also presented as == 
This result can also be presented as:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^6} = \dfrac {\pi^6} {960}$
Category:Sum of Reciprocals of Sixth Powers of Odd Integers
Sum of Reciprocals of Squares Alternating in Sign

== Theorem == 
== Proof 1 == 
 
== Proof 2 ==
 
== Proof 3 ==
 
== Proof 4 ==
 
== Also see ==
* Sum of Reciprocals of Even Powers of Integers Alternating in Sign
* Sum of Reciprocals of Fourth Powers Alternating in Sign
* Sum of Reciprocals of Sixth Powers Alternating in Sign
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.22$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.22.$
Category:Sum of Reciprocals of Squares Alternating in Sign
Category:Formulas for Pi
Category:Sums of Sequences
Sum of Reciprocals of Squares Alternating in Sign/Proof 1

== Theorem == 
== Proof ==
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:$\map f x = \begin {cases} \paren {x - \pi}^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end {cases}$
From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:
:$(1): \quad \map f x \sim \ds \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \paren {\frac {2 \cos n x} {n^2} + \paren {\frac {\paren {-1}^n \pi} n + \frac {2 \paren {\paren {-1}^n - 1} } {\pi n^3} } \sin n x}$
We have that:
where $\map f {\pi - 0}$ and $\map f {\pi + 0}$ denote the limit from the left and limit from the right respectively of $\map f \pi$.
It is apparent that $\map f x$ satisfies the Dirichlet conditions:
:$(\text D 1): \quad f$ is bounded on $\openint 0 {2 \pi}$
:$(\text D 2): \quad f$ has a finite number of local maxima and local minima.
:$(\text D 3): \quad f$ has $1$ discontinuity, which is finite.
Hence from Fourier's Theorem:
Thus setting $x = \pi$ in $(1)$:
== Sources ==
* : Chapter One: $\S 2$. Fourier Series: Example $1$
Category:Sum of Reciprocals of Squares Alternating in Sign
Sum of Reciprocals of Squares Alternating in Sign/Proof 3

== Theorem == 
== Proof ==
Let $\map f x$ be the real function defined on $\openint {-\pi} \pi$ as:
:$\map f x = \pi^2 - x^2$
By Fourier Series: $\pi^2 - x^2$ over $\openint {-\pi} \pi$:
:$\ds \pi^2 - x^2 \sim \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \paren {-1}^{n - 1} \frac {\cos n x} {n^2}$
for $x \in \openint {-\pi} \pi$.
Setting $x = 0$:
== Sources ==
* : Exercises on Chapter $\text I$: $6$.
Category:Sum of Reciprocals of Squares Alternating in Sign
Sum of Reciprocals of Squares of Odd Integers

== Theorem ==
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Proof 4 ==
== Proof 5 ==
== Proof 6 ==
== Proof 7 ==
== Also presented as ==
== Sources ==
* : $\S 19$: Series involving Reciprocals of Powers of Positive Integers: $19.25$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$
* : $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$
* : $\S 21$: Series of Constants: Series Involving Reciprocals of Powers of Positive Integers: $21.25.$
Category:Sum of Reciprocals of Squares of Odd Integers
Category:Formulas for Pi
Sum of Reciprocals of Squares of Odd Integers/Also presented as

== Sum of Reciprocals of Squares of Odd Integers: Also presented as == 
This result can also be presented as:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \dfrac {\pi^2} 8$
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers/Proof 2

== Theorem ==
== Proof ==
Let $n$ be a positive integer.
We have:
We also have:
So we can deduce:
:$\ds \sum_{n \mathop = 0}^\infty \frac 1 {\paren {2 n + 1}^2} = \frac {\pi^2} 8$ 
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers/Proof 4

== Theorem ==
== Proof ==
Applying the substitution:
:$\tuple {x, y} = \tuple {\dfrac {\sin u} {\cos v}, \dfrac {\sin v} {\cos u} }$
the Jacobian matrix is:
:$\mathbf J_{\mathbf f} := \begin{pmatrix}
\dfrac \partial {\partial u} \dfrac {\sin u} {\cos v} & \dfrac \partial {\partial v} \dfrac {\sin u} {\cos v} \\ 
\dfrac \partial {\partial u} \dfrac {\sin v} {\cos u} & \dfrac \partial {\partial v} \dfrac {\sin v} {\cos u}
\end{pmatrix}$
and the Jacobian determinant is:
We have:
and we have:
Under this substitution, the image of the region $\closedint 0 1^2$, that is the unit square, is an isosceles triangle $\bigtriangleup$ with:
:base and height $\dfrac \pi 2$
:vertices: $\tuple {0, 0}; \tuple {0, \dfrac \pi 2}; \tuple {\dfrac \pi 2, 0}$
This gives us the region:
:$\set {\tuple {u, v}: u, v \ge 0 \land v \le \dfrac \pi 2 - u}$
By Change of Variables Theorem (Multivariable Calculus):
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers/Proof 5

== Theorem == 
== Proof ==
By Fourier Series of Absolute Value of x, for $x \in \closedint {-\pi} \pi$:
:$\ds \size x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {\paren {2 n - 1} x} } {\paren {2 n - 1}^2}$
Setting $x = \pi$:
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers/Proof 6

== Theorem == 
== Proof ==
Let $\map f x$ be the real function defined on $\openint 0 {2 \pi}$ as:
:$\map f x = \begin{cases}
-\pi & : 0 < x \le \pi \\
x - \pi & : \pi < x < 2 \pi \end{cases}$
By Fourier Series: $-\pi$ over $\openint 0 \pi$, $x - \pi$ over $\openint \pi {2 \pi}$, its Fourier series can be expressed as:
:$\map f x \sim \map S x = \displaystyle -\dfrac \pi 4 + \frac 2 \pi \sum_{r \mathop = 0}^\infty \dfrac {\cos \paren {2 r + 1} x} {\paren {2 r + 1}^2} - \sum_{n \mathop = 1}^\infty \dfrac {2 - \paren {-1}^n \sin n x} n$
Consider the point $x = \pi$.
By Fourier's Theorem:
Thus:
== Sources ==
* : Exercises on Chapter $\text I$: $1$.
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers/Proof 7

== Theorem == 
== Proof ==
By Half-Range Fourier Cosine Series for Identity Function over $\openint 0 \pi$:
:$\ds x = \frac \pi 2 - \frac 4 \pi \sum_{n \mathop = 1}^\infty \frac {\map \cos {2 n - 1} x} {\paren {2 n - 1}^2}$
for $x \in \openint 0 \pi$.
We have that:
:$\map f \pi = \map f {\pi - 2 \pi} = \map f {-\pi} = \pi$
and so:
:$\map f {\pi^-} = \map f {\pi^+}$
Hence we can set $x = \pi$:
== Sources ==
* : Exercises on Chapter $\text I$: $6$.
Category:Sum of Reciprocals of Squares of Odd Integers
Sum of Reciprocals of Squares of Odd Integers as Double Integral

== Theorem ==
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {\paren {2 n - 1}^2} = \int_0^1 \int_0^1 \frac 1 {1 - x^2 y^2} \rd x \rd y$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Sum of Reciprocals of Squares of Odd Integers
* Riemann Zeta Function as a Multiple Integral
Category:Riemann Zeta Function
Category:Sum of Reciprocals of Squares of Odd Integers as Double Integral
Sum of Reciprocals of Squares plus 1

== Theorem ==
:$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^2 + 1} = \frac 1 2 \paren {\pi \coth \pi - 1}$
== Proof ==
Let $\map f x$ be the real function defined on $\hointl {-\pi} \pi$ as:
:$\map f x = e^x$
From Fourier Series: $e^x$ over $\openint {-\pi} \pi$, we have:
:$(1): \quad \ds \map f x \sim \map S x = \frac {\sinh \pi} \pi \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\cos n x} {1 + n^2} - \frac {n \sin n x} {1 + n^2} } }$
Let $x = \pi$.
By Fourier's Theorem, we have:
Thus setting $x = \pi$ in $(1)$, we have:
== Sources ==
* : Exercises on Chapter $\text I$: $4$.
Category:Hyperbolic Cotangent Function
Sum of Reciprocals of Twin Primes

== Theorem ==
The sum of the reciprocals of all the twin primes:
:$\dfrac 1 3 + \dfrac 1 5 + \dfrac 1 7 + \dfrac 1 {11} + \dfrac 1 {13} + \dfrac 1 {17} + \dfrac 1 {19} + \dfrac 1 {29} + \dfrac 1 {31} + \cdots$
is either finite or convergent.
== Proof ==
== Historical Note ==
== Sources ==
* : Chapter $\text {B}.16$: The Sequence of Primes
Category:Prime Numbers
Sum of Replicative Functions is Replicative

== Theorem ==
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions.
Let $f$ and $g$ both be replicative functions.
Then the pointwise sum of $f$ and $g$ is also a replicative function.
== Proof ==
Hence the result by definition of replicative function.
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $39 \ \text{g)}$
Category:Replicative Functions
Sum of Riemann-Stieltjes Integrals on Adjacent Intervals

== Theorem ==
Let $a, b, c \in \R$.
Let $f, \alpha$ be real functions that are bounded on some closed interval containing $a, b, c$.
Suppose that two of the following three Riemann-Stieltjes integrals exist, with the limits of integration interpreted in the general sense:
:$\ds \int_a^b f \rd \alpha$
:$\ds \int_b^c f \rd \alpha$
:$\ds \int_c^a f \rd \alpha$
Then, the third exists as well, and:
:$\ds \int_a^b f \rd \alpha + \int_b^c f \rd \alpha + \int_c^a f \rd \alpha = 0$
== Proof ==
We have the following special cases:
=== Part to Whole ===
=== Whole to Part ===
To prove the full theorem, we will consider separate cases according to how many of $a, b, c$ are equal to one another.
=== Case: All Equal ===
First, suppose $a = b = c$.
Then, all three integrals are equal to $0$ by convention, and the result is trivial.
=== Case: Two Equal ===
Next, suppose two of $a, b, c$ are equal to each other.
, let them be $a = c$.
Then:
:$\ds \int_c^a f \rd \alpha = 0$
We are given the existence of two different integrals, so at least one of the following is among them:
:$\ds \int_a^b f \rd \alpha$
:$\ds \int_b^c f \rd \alpha = \int_b^a f \rd \alpha$
Therefore, the other is the negative of it by convention, and the result once again follows.
=== Case: All Distinct ===
Finally, suppose $a, b, c$ are all distinct.
, suppose:
:$\ds \int_a^b f \rd \alpha$
:$\ds \int_b^c f \rd \alpha$
are the two integrals known to exist.
If necessary, we can reverse the limits of integration on all three integrals and swap the roles of $a$ and $c$ to ensure:
:$a < c$
There are now three cases:
:$\paren 1 \quad a < b < c$
:$\paren 2 \quad b < a < c$
:$\paren 3 \quad a < c < b$
Suppose $\paren 1$ holds.
Then, by Part to Whole above:
:$\ds \int_a^c f \rd \alpha = \int_a^b f \rd \alpha + \int_b^c f \rd \alpha$
Therefore, by convention:
:$\ds \int_c^a f \rd \alpha = - \int_a^b f \rd \alpha - \int_b^c f \rd \alpha$
from which the result follows.
Suppose $\paren 2$ holds.
Then, by convention:
:$\ds \int_a^b f \rd \alpha = - \int_b^a f \rd \alpha$
By Whole to Part above, we know that $\ds \int_a^c f \rd \alpha$ exists and:
:$\ds \int_b^a f \rd \alpha + \int_a^c f \rd \alpha = \int_b^c f \rd \alpha$
Again by convention:
:$\ds \int_c^a f \rd \alpha = - \int_a^c f \rd \alpha$
Therefore:
:$\ds - \int_a^b f \rd \alpha - \int_c^a f \rd \alpha = \int_b^c f \rd \alpha$
from which the result follows.
Finally, suppose that $\paren 3$ holds.
Then, by convention:
:$\ds \int_b^c f \rd \alpha = - \int_c^b f \rd \alpha$
By lemma 2, we know that $\ds \int_a^c f \rd \alpha$ exists and:
:$\ds \int_a^c f \rd \alpha + \int_c^b f \rd \alpha = \int_a^b f \rd \alpha$
Again by convention:
:$\ds \int_c^a f \rd \alpha = - \int_a^c f \rd \alpha$
Therefore:
:$\ds - \int_c^a f \rd \alpha + - \int_b^c f \rd \alpha = \int_a^b f \rd \alpha$
from which the result follows.
Since the result holds in every case, it follows from Proof by Cases.
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.4$: Definition $7.5$
Category:Riemann-Stieltjes Integral
Sum of Riemann-Stieltjes Integrals on Adjacent Intervals/Part to Whole

== Theorem ==
Let $a < c < b$ be real numbers.
Let $f, \alpha$ be real functions that are bounded on $\closedint a b$
Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a c$, and also on $\closedint c b$.
Then, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$, and:
:$\ds \int_a^b f \rd \alpha = \int_a^c f \rd \alpha + \int_c^b f \rd \alpha$
== Proof ==
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, let $P'_\epsilon$ be a subdivision of $\closedint a c$ and $P' '_\epsilon$ be a subdivision of $\closedint c b$ such that:
:For every $P'$ finer than $P'_\epsilon$, $\ds \size {\map S {P', f, \alpha} - \int_a^c f \rd \alpha} < \frac \epsilon 2$
:For every $P' '$ finer than $P' '_\epsilon$, $\ds \size {\map S {P' ', f, \alpha} - \int_c^b f \rd \alpha} < \frac \epsilon 2$
Define $P_\epsilon := P'_\epsilon \cup P' '_\epsilon$.
Since:
* $P_\epsilon \subseteq \closedint a c \cup \closedint c b = \closedint a b$
* $a \in P'_\epsilon \subseteq P_\epsilon$
* $b \in P' '_\epsilon \subseteq P_\epsilon$
it follows that $P_\epsilon$ is a subdivision of $\closedint a b$.
Let $P = \set {x_0, x_1, \dotsc, x_{n - 1}, x_n}$ be a subdivision of $\closedint a b$ that is finer than $P_\epsilon$.
Since $c \in P'_\epsilon \subseteq P_\epsilon \subseteq P$, there must be some $m \in {1, 2, \dotsc, n - 2, n - 1}$ such that:
:$x_m = c$
For each $\ell \in \set {0, 1, \dotsc, n - m - 1, n - m}$, define:
:$y_\ell = x_{\ell + m}$
Then, define:
:$P' := P \cap \closedint a c = \set {x_0, x_1, \dotsc, x_{m - 1}, x_m}$
:$P' ' := P \cap \closedint c b = \set {y_0, y_1, \dotsc, y_{n - m - 1}, y_{n - m}}$
It follows that:
:$P'$ is a subdivision of $\closedint a c$ that is finer than $P'_\epsilon$
:$P' '$ is a subdivision of $\closedint c b$ that is finer than $P' '_\epsilon$
We have:
Therefore:
As $P \supseteq P_\epsilon$ and $\epsilon > 0$ were arbitrary, by the definition of the Riemann-Stieltjes integral:
:$\ds \int_a^b f \rd \alpha = \int_a^c f \rd \alpha + \int_c^b f \rd \alpha$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.4$: Theorem $7.4$
Category:Riemann-Stieltjes Integral
Sum of Riemann-Stieltjes Integrals on Adjacent Intervals/Whole to Part

== Theorem ==
Let $a < c < b$ be real numbers.
Let $f, \alpha$ be real functions that are bounded on $\closedint a b$
Suppose that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint a b$, and also on one of the two intervals $\closedint a c$ and $\closedint c b$.
Then, $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on the other interval as well, and:
:$\ds \int_a^c f \rd \alpha + \int_c^b f \rd \alpha = \int_a^b f \rd \alpha$
== Proof ==
Let $\closedint p q$ be the interval from among $\closedint a c$ and $\closedint c b$ on which we know $f$ is Riemann-Stieltjes integrable with respect to $\alpha$.
Let $\closedint u v$ be the other interval.
Then, we want to prove that $f$ is Riemann-Stieltjes integrable with respect to $\alpha$ on $\closedint u v$, and:
:$\ds \int_u^v f \rd \alpha = \int_a^b f \rd \alpha - \int_p^q f \rd \alpha$
Let $\epsilon > 0$ be arbitrary.
By definition of the Riemann-Stieltjes integral, let $P'_\epsilon$ be a subdivision of $\closedint a b$ and $P' '_\epsilon$ be a subdivision of $\closedint p q$ such that:
:For every $P'$ finer than $P'_\epsilon$, $\ds \size {\map S {P', f, \alpha} - \int_a^b f \rd \alpha} < \frac \epsilon 2$
:For every $P' '$ finer than $P' '_\epsilon$, $\ds \size {\map S {P' ', f, \alpha} - \int_p^q f \rd \alpha} < \frac \epsilon 2$
Define $P_\epsilon := \paren {P'_\epsilon \cup {u, v}} \cap \closedint u v$.
Then, $P_\epsilon$ is a subdivision of $\closedint u v$.
Let $P = \set {x_0, \dotsc, x_n}$ be a subdivision of $\closedint u v$ that is finer than $P_\epsilon$.
Define:
:$P' = \set {y_0, \dotsc, y_m} := P \cup P'_\epsilon \cup P' '_\epsilon$
:$P' ' = \set {z_0, \dotsc, z_r} := P' \cap \closedint p q$
Since $P'_\epsilon \subseteq P' \subseteq \closedint a b$:
:$P'$ is a subdivision of $\closedint a b$ that is finer than $P'_\epsilon$
Since $P' '_\epsilon \subseteq P' ' \subseteq \closedint p q$:
:$P' '$ is a subdivision of $\closedint p q$ that is finer than $P' '_\epsilon$
We have:
Therefore:
As $P$ finer than $P_\epsilon$ and $\epsilon > 0$ were arbitrary, it follows from the definition of the Riemann-Stieltjes integral that:
:$\ds \int_u^v f \rd \alpha = \int_a^b f \rd \alpha - \int_p^q f \rd \alpha$
== Sources ==
* : Chapter $7$ The Riemann-Stieltjes Integral: $\S 7.4$: Theorem $7.4$
Category:Riemann-Stieltjes Integral
Sum of Ring Products is Subring of Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring.
Let $\struct {S, +, \circ}$ and $\struct {T, +, \circ}$ be subrings of $\struct {R, +, \circ}$.
Let $S T$ be defined as:
:$\ds S T = \set {\sum_{i \mathop = 1}^n s_i \circ t_i: s_1 \in S, t_i \in T, i \in \closedint 1 n}$
Then $S T$ is a subring of $\struct {R, +, \circ}$.
== Proof ==
From Sum of All Ring Products is Additive Subgroup we have that $\struct {S T, +}$ is an additive subgroup of $R$.
Let $x_1, x_2 \in S T$.
Then:
:$\ds x_1 = \sum_{i \mathop = 1}^m s_i \circ t_i, x_2 = \sum_{i \mathop = 1}^n s_j \circ t_j$
for some $s_i, t_i, s_j, t_j, m, n$, etc.
Then:
So $x_1 \circ x_2 \in S T$ and the result follows from the Subring Test.
== Sources ==
* : $\S 2.1$: Subrings: Lemma $2.3 \ \text{(iii)}$
Category:Commutative Rings
Category:Subrings
Sum of Roots of Polynomial

== Theorem ==
Let $P$ be the polynomial equation:
: $a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$
such that $a_n \ne 0$.
The sum of the roots of $P$ is $-\dfrac {a_{n - 1} } {a_n}$.
== Proof 1 ==
== Proof 2 ==
Category:Polynomial Equations
Category:Sum of Roots of Polynomial
Sum of Roots of Polynomial/Proof 1

== Theorem ==
== Proof ==
Let the roots of $P$ be $z_1, z_2, \ldots, z_n$.
Then $P$ can be written in factored form as:
:$\ds a_n \prod_{k \mathop = 1}^n \paren {z - z_k} = a_n \paren {z - z_1} \paren {z - z_2} \cdots \paren {z - z_n}$
Multiplying this out, $P$ can be expressed as:
:$a_n \paren {z^n - \paren {z_1 + z_2 + \cdots + z_n} z^{n - 1} + \cdots + \paren {-1}^n z_1 z_2 \cdots z_n} = 0$
where the coefficients of $z^{n - 2}, z^{n - 3}, \ldots$ are more complicated and irrelevant.
Equating powers of $z$, it follows that:
:$-a_n \paren {z_1 + z_2 + \cdots + z_n} = a_{n - 1}$
from which:
:$z_1 + z_2 + \cdots + z_n = - \dfrac {a_{n - 1} } {a_n}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $35$
Category:Sum of Roots of Polynomial
Sum of Roots of Polynomial/Proof 2

== Theorem ==
== Proof ==
From Viète's Formulas:
:$\ds a_{n - k} = \paren {-1}^k a_n \sum_{1 \mathop \le i_1 \mathop < \dotsb \mathop < i_k \mathop \le n} z_{i_1} \dotsm z_{i_k}$
for $k = 1, 2, \ldots, n$.
The result follows for $k = 1$.
Category:Sum of Roots of Polynomial
Sum of Roots of Quadratic Equation

== Theorem ==
Let $P$ be the quadratic equation $a x^2 + b x + c = 0$.
Let $\alpha$ and $\beta$ be the roots of $P$.
Then:
:$\alpha + \beta = -\dfrac b a$
== Proof ==
== Also see ==
* Product of Roots of Quadratic Equation
* Viète's Formulas
== Sources ==
* : $3$: Elementary Analytic Methods: $3.8$ Algebraic Equations: Solution of Quadratic Equations: $3.8.1$
* : $\S 9$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $9.2$
* : $\S 5$: Solutions of Algebraic Equations: Quadratic Equation: $a x^2 + b x + c = 0$: $5.2.$
* 
Category:Quadratic Equations
Sum of Sequence as Summation of Difference of Adjacent Terms

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Then:
:$\ds \sum_{k \mathop = 1}^n a_k = n a_n - \sum_{k \mathop = 1}^{n - 1} k \paren {a_{k + 1} - a_k}$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $42 \ \text{(a)}$
Category:Summations
Category:Sum of Sequence as Summation of Difference of Adjacent Terms
Sum of Sequence as Summation of Difference of Adjacent Terms/Proof 2

== Theorem ==
== Proof ==
From Abel's Lemma: Formulation 2, after renaming and reassigning variables:
:$\ds \sum_{k \mathop = 1}^n a_k b_k = \sum_{k \mathop = 1}^{n - 1} \map {A_k} {a_k - a_{k + 1} } + A_n a_n$
where:
:$\sequence a$ and $\sequence b$ are sequences in $\R$
:$\ds A_n = \sum_{k \mathop = 1}^n {b_k}$ be the partial sum of $\sequence b$ from $1$ to $n$.
Let $\sequence b$ be defined as:
:$\forall k: b_k = 1$
Thus:
:$\ds A_n = \sum_{k \mathop = 1}^n 1 = n$
and so:
Category:Sum of Sequence as Summation of Difference of Adjacent Terms
Sum of Sequence of Alternating Positive and Negative Factorials being Prime

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let:
The sequence of $n$ such that $m$ is prime begins:
:$3, 4, 5, 6, 7, 8, 10, 15, 19, 41, 59, 61, 105, 160, \ldots$
The sequence of those values of $m$ begins:
:$5, 19, 101, 619, 4421, 35 \, 899, 3 \, 301 \, 819, 1 \, 226 \, 280 \, 710 \, 981, \ldots$
== Proof ==
Let $\map f n$ be defined as:
:$\map f n := \ds \sum_{k \mathop = 0}^{n - 1} \paren {-1}^k \paren {n - k}!$
First we observe that for $n > 1$:
:$\map f n := n! - \map f {n - 1}$
We have:
From here on in the numbers become unwieldy.
== Sources ==
* : $19$
* 
* : $19$
Category:Factorials
Category:Prime Numbers
Sum of Sequence of Alternating Positive and Negative Factorials being Prime/Mistake

== Source Work ==
:
:The Dictionary
::$19$
== Mistake ==
:$19! - 18! + 17! - 16! + \dotsb + 1$ is prime. The only other numbers with this property are $3$, $4$, $5$, $6$, $7$, $8$, $10$ and $15$.
== Correction ==
What should have been said is:
:... The only other known numbers with this property are ...
In , this entry has been expanded to read:
: ... The only other known numbers with this property are $3$, $4$, $5$, $6$, $7$, $8$, $10$, $15$, $41$, $59$, $61$, $105$ and $160$.
It is also worth pointing out that extends this list considerably, and reports that the list has been proven to be finite.
== Sources ==
* : $19$
Category:Mistakes/Curious and Interesting Numbers
Sum of Sequence of Binomial Coefficients by Powers of 2

== Theorem ==
Let $n \in \Z$ be an integer.
Then:
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem: Problems $1.2$: $3 \ \text{(d)}$
Category:Binomial Coefficients
Category:Sum of Sequence of Binomial Coefficients by Powers of 2