nyuuzyou/wikis huggingface dataset:
Binomial Coefficient of Prime Minus One Modulo Prime

== Theorem ==
Let $p$ be a prime number.
Then:
:$0 \le k \le p - 1 \implies \dbinom {p - 1} k \equiv \left({-1}\right)^k \pmod p$
where $\dbinom {p - 1} k$ denotes a binomial coefficient.
== Proof ==
From Binomial Coefficient of Prime, we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From Pascal's Rule:
:$\dbinom {p-1} k + \dbinom {p - 1} {k - 1} = \dbinom p k \equiv 0 \pmod p$
This certainly holds for $k = 1$, and so we have:
:$\dbinom {p - 1} 1 + \dbinom {p - 1} 0 = \dbinom p 1 \equiv 0 \pmod p$
But from Binomial Coefficient with Zero:
:$\dbinom {p - 1} 0 = 1 \equiv 1 \pmod p$
So:
:$\dbinom {p - 1} 1 \equiv -1 \pmod p$
Then:
:$\dbinom {p - 1} 2 + \dbinom {p - 1} 1 = \dbinom p 2 \equiv 0 \pmod p$
and so:
:$\dbinom {p - 1} 2 \equiv 1 \pmod p$
The result follows.
== Sources ==
* : $\S 2.1$: Exercise $44$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(c)}$
Category:Binomial Coefficients
Binomial Coefficient of Prime Plus One Modulo Prime

== Theorem ==
Let $p$ be a prime number.
Then:
:$2 \le k \le p - 1 \implies \dbinom {p + 1} k \equiv 0 \pmod p$
where $\dbinom {p + 1} k$ denotes a binomial coefficient.
== Proof ==
From Binomial Coefficient of Prime, we have:
:$\dbinom p k \equiv 0 \pmod p$
when $1 \le k \le p - 1$.
From Pascal's Rule:
:$\dbinom {p + 1} k = \dbinom p {k - 1} + \dbinom p k$
The result follows immediately.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $10 \ \text{(d)}$
Category:Binomial Coefficients
Binomial Coefficient of Real Number with Half

== Theorem ==
:$\dbinom r {1 / 2} = \dfrac {2^{2 r + 1} } {\dbinom {2 r} r \pi}$
where $\dbinom r {1 / 2}$ denotes a binomial coefficient.
== Proof ==
Legendre's Duplication Formula gives:
:$\forall z \notin \set {-\dfrac n 2: n \in \N}: \map \Gamma z \map \Gamma {z + \dfrac 1 2} = 2^{1 - 2 z} \sqrt \pi \, \map \Gamma {2 z}$
and so:
:$(1): \quad \map \Gamma {\rho + \dfrac 1 2} = \dfrac {2^{1 - 2 \rho} \sqrt \pi \, \map \Gamma {2 \rho} } {\map \Gamma \rho}$
Hence:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $44$
Category:Examples of Binomial Coefficients‎
Binomial Coefficient over Power Not Greater than Reciprocal of Factorial

== Theorem ==
Let $r > 0$.
Let $k \in \N$ such that $k \le 2r + 1$.
Then:
:$\dfrac {\dbinom r k} {r^k} \le \dfrac 1 {k!}$
== Proof ==
For each $j = 0, 1, \dotsc, k - 1$, we have:
:$0 \le j \le 2r$
and thus:
:$-r \le j - r \le r$
Hence:
:$\dfrac {\size {r - j} } r \le 1$
Therefore, by Negative of Absolute Value:
:$\dfrac {\binom r k} {r^k} \le \size {\dfrac {\binom r k} {r^k}} \le \dfrac 1 {k!}$
Category:Binomial Coefficients
Category:Factorials
Binomial Coefficient with One

== Theorem ==
:$\forall r \in \R: \dbinom r 1 = r$
where $\dbinom r 1$ denotes a binomial coefficient.
The usual presentation of this result is:
:$\forall n \in \N: \dbinom n 1 = n$
== Proof ==
From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
But when $k = 1$, we have:
:$\ds \prod_{j \mathop = 0}^0 \paren {x - j} = \paren {x - 0} = x$
So:
:$\forall r \in \R: \dbinom r 1 = r$
This is completely compatible with the result for natural numbers:
:$\forall n \in \N: \dbinom n 1 = n$
as from the definition:
:$\dbinom n 1 = \dfrac {n!} {1! \ \paren {n - 1}!}$
the result following directly, again from the definition of the factorial where $1! = 1$.
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.28$
* : $\S 1.2.6$: Binomial Coefficients: $(4)$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.5$: $(1)$
Category:Examples of Binomial Coefficients‎
Binomial Coefficient with Self

== Theorem ==
:$\forall n \in \Z: \dbinom n n = \sqbrk {n \ge 0}$
where $\sqbrk {n \ge 0}$ denotes Iverson's convention.
That is:
:$\forall n \in \Z_{\ge 0}: \dbinom n n = 1$
:$\forall n \in \Z_{< 0}: \dbinom n n = 0$
== Proof ==
From the definition of binomial coefficient:
:$\dbinom n n = \dfrac {n!} {n! \ \paren {n - n}!} = \dfrac {n!} {n! \ 0!}$
the result following directly from the definition of the factorial, where $0! = 1$.
From N Choose Negative Number is Zero:
:$\forall k \in \Z_{<0}: \dbinom n k = 0$
So for $n < 0$:
:$\dbinom n n = 0$
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.5$
* : Problems
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
* : $\S 1.18$: Sequences Defined Inductively: Exercise $3 \ \text{(b)}$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.27$
* : $\S 1.2.6$: Binomial Coefficients: $(48)$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $9$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Coefficients: $3.5$: $(1)$
Category:Examples of Binomial Coefficients‎
Binomial Coefficient with Self minus One

== Theorem ==
:$\forall n \in \N_{>0}: \dbinom n {n - 1} = n$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.28$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $1$
Category:Examples of Binomial Coefficients‎
Category:Binomial Coefficient with Self minus One
Binomial Coefficient with Self minus One/Proof 1

== Theorem ==
== Proof ==
The case where $n = 1$ can be taken separately.
From Binomial Coefficient with Zero:
:$\dbinom 1 0 = 1$
demonstrating that the result holds for $n = 1$.
Let $n \in \N: n > 1$.
From the definition of binomial coefficients:
:$\dbinom n {n - 1} = \dfrac {n!} {\paren {n - 1}! \paren {n - \paren {n - 1} }!} = \dfrac {n!} {\paren {n - 1}! \ 1!}$
the result following directly from the definition of the factorial.
Category:Binomial Coefficient with Self minus One
Binomial Coefficient with Self minus One/Proof 2

== Theorem ==
== Proof ==
From Cardinality of Set of Subsets, $\dbinom n {n - 1}$ is the number of combination of things taken $n - 1$ at a time.
Choosing $n - 1$ things from $n$ is the same thing as choosing which $1$ of the elements to be left out.
There are $n$ different choices for that $1$ element.
Therefore there are $n$ ways to choose $n - 1$ things from $n$.
Category:Binomial Coefficient with Self minus One
Binomial Coefficient with Two

== Theorem ==
:$\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$
=== Corollary ===
The usual presentation of this result is:
== Proof ==
From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k}} {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
When $k = 2$:
:$\ds \prod_{j \mathop = 0}^1 \paren {x - j} = \frac {\paren {x - 0} \paren {x - 1} } {2!}$
where $2! = 1 \times 2 = 2$.
So:
:$\forall r \in \R: \dbinom r 2 = \dfrac {r \paren {r - 1} } 2$
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: $(4)$
Category:Examples of Binomial Coefficients‎
Binomial Coefficient with Two/Corollary

== Theorem ==
:$\forall n \in \N: \dbinom n 2 = T_{n - 1} = \dfrac {n \paren {n - 1} } 2$
where $T_n$ is the $n$th triangular number.
== Proof ==
From the definition of binomial coefficient:
:$\dbinom n 2 = \dfrac {n!} {2! \paren {n - 2}!}$
The result follows directly from the definition of the factorial:
:$2! = 1 \times 2$
== Also see ==
* Closed Form for Triangular Numbers: Proof using Binomial Coefficients
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.3$ Early Number Theory: Problems $1.3$: $2$
* : $15$
* : $15$
Category:Examples of Binomial Coefficients‎
Category:Triangular Numbers
Binomial Coefficient with Zero

== Theorem ==
:$\forall r \in \R: \dbinom r 0 = 1$
The usual presentation of this result is:
:$\forall n \in \N: \dbinom n 0 = 1$
== Proof ==
From the definition of binomial coefficients:
:$\dbinom r k = \dfrac {r^{\underline k} } {k!}$ for $k \ge 0$
where $r^{\underline k}$ is the falling factorial.
In turn:
:$\ds x^{\underline k} := \prod_{j \mathop = 0}^{k - 1} \paren {x - j}$
But when $k = 0$, we have:
:$\ds \prod_{j \mathop = 0}^{-1} \paren {x - j} = 1$
as $\ds \prod_{j \mathop = 0}^{-1} \paren {x - j}$ is a vacuous product.
From the definition of the factorial we have that $0! = 1$.
Thus:
:$\forall r \in \R: \dbinom r 0 = 1$
=== Integer Coefficients ===
This is completely compatible with the result for natural numbers:
:$\forall n \in \N: \dbinom n 0 = 1$
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* : $\S 1.2.6$: Binomial Coefficients: $(4)$
Category:Examples of Binomial Coefficients‎
Binomial Coefficient with Zero/Integer Coefficients

== Theorem ==
:$\forall n \in \N: \dbinom n 0 = 1$
where $\dbinom n 0$ denotes a binomial coefficient.
== Proof ==
From the definition:
== Also see ==
* Particular Values of Binomial Coefficients for other similar results.
== Sources ==
* : $3$: Elementary Analytic Methods: $3.1$ Binomial Theorem etc.: Binomial Coefficients: $3.1.5$
* : Problems
* : Chapter $\text {III}$: The Natural Numbers: $\S 19$: Combinatorial Analysis: Theorem $19.10$
* : $\S 1.18$: Sequences Defined Inductively: Exercise $3 \ \text{(a)}$
* : $\S 1.2.4$: Factorials and binomial coefficients: $1.27$
Category:Examples of Binomial Coefficients‎
Binomial Distribution/Example/Arbitrary Example 1

== Example of Binomial Distribution ==
Let a die be cast $4$ times.
Let a score of $6$ be denoted as a success.
Then the experiment can be modelled by a binomial distribution $\Binomial n p$ such that $n = 4$ and $p = \dfrac 1 6$.
Thus the probability of $2$ successes is $\dfrac {25} {216}$.
== Proof ==
The experiment of casting a die can be modelled as a Bernoulli process.
Hence from Bernoulli Process as Binomial Distribution, the model is appropriate.
Let $\map P k$ be the probability of $k$ successes.
Then by definition of binomial distribution:
:$\map P k = \dbinom 4 k \paren {\dfrac 1 6}^k \paren {\dfrac 5 6}^{4 - k}$
Hence:
== Sources ==
* 
* 
Category:Examples of Binomial Distributions
Binomial Distribution Approximated by Normal Distribution

== Theorem ==
Let $X$ be a discrete random variable which has the binomial distribution $\Binomial n p$.
Then for large $n$ and such that both $n p$ and $n q$ are approximately $5$ or more:
:$\Binomial n p \approx \Gaussian {n p} {n p q}$
where $\Gaussian {n p} {n p q}$ denotes the normal distribution.
== Proof ==
== Sources ==
* 
* 
* 
Category:Binomial Distribution
Category:Normal Distribution
Binomial Distribution Approximated by Poisson Distribution

== Theorem ==
Let $X$ be a discrete random variable which has the binomial distribution with parameters $n$ and $p$.
Then for $\lambda = n p$, $X$ can be approximated by a Poisson distribution with parameter $\lambda$:
:$\ds \lim_{n \mathop \to \infty} \binom n k p^k \paren {1 - p}^{n - k} = \frac {\lambda^k} {k!} e^{-\lambda}$
== Proof ==
Let $X$ be as described.
Let $k \in \Z_{\ge 0}$ be fixed.
We write $p = \dfrac \lambda n$ and suppose that $n$ is large.
Then:
Hence the result.
== Sources ==
* : $\S 2.2$: Examples
* 
* 
* 
* 
Category:Binomial Distribution
Category:Poisson Distribution
Binomial Form of Relation between Riemann Zeta Function and Dirichlet Eta Function

== Theorem ==
Let $\zeta$ be the Riemann zeta function.
Let $s \in \C$ be a complex number with real part $\map \Re s > 1$.
Then:
$\ds \map \zeta s = \frac 1 {1 - 2^{1 - s} } \sum_{n \mathop = 0}^\infty \paren {\frac 1 {2^{n + 1} } \sum_{k \mathop = 0}^n \paren {-1}^k \binom n k \paren {k + 1}^{-s} }$
== Proof ==
Use Riemann Zeta Function in terms of Dirichlet Eta Function and Binomial Theorem.
Category:Riemann Zeta Function
Binomial Straight Line is Divisible into Terms Uniquely

== Theorem ==
=== Lemma ===
== Proof ==
Let $AB$ be a binomial straight line.
Let $AB$ be divided into its terms at $C$.
That is, let the terms of $AB$ be $AC$ and $CB$.
By definition of binomial, $AC$ and $CB$ are rational straight lines commensurable in square only.
Suppose there exists a point $D$ on $AB$ different from $C$, such that $AD$ and $DB$ are also rational straight lines which are commensurable in square only.
Let $AC \ne DB$, otherwise $AD = BC$ and $AB$ will be divided in the same way by $D$ as it is $C$.
Similarly, the points $C$ and $D$ are not equidistant from the point of bisection of $AB$.
From :
:$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
and so:
:$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right) = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
As:
: $AC$ and $CB$ are rational straight lines
and
: $AD$ and $DB$ are rational straight lines
it follows by definition that $AC^2$, $CB^2$, $AD^2$ and $DB^2$ are all rational areas.
Thus:
:$\left({AC^2 + CB^2}\right) - \left({AD^2 + DB^2}\right)$ is a rational area.
Thus $2 AC \cdot CB - 2 AD \cdot DB$ is also a rational area.
But by definition, $2 AC \cdot CB$ and $2 AD \cdot DB$ are both medial.
But from this cannot be the case.
From this contradiction it follows that $D$ cannot divide $AB$ into rational straight lines which are commensurable in square only.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Binomial Straight Line is Divisible into Terms Uniquely/Lemma

== Lemma to Binomial Straight Line is Divisible into Terms Uniquely ==
== Proof ==
:
Let $AB$ be a straight line.
Let $AB$ be divided into unequal parts at $C$ and $D$ such that $AC > DB$.
Let $AB$ be bisected at $E$.
Let $DC$ be subtracted from both $AC$ and $DB$.
Then as $AC > DB$ it follows that $AD > CB$.
But as $AE = EB$ it follows that $DE < EC$.
Therefore $C$ and $D$ are not equidistant from the point of bisection.
From :
:$AC \cdot CB = EB^2 - EC^2$
and:
:$AD \cdot DB = EB^2 - ED^2$
so:
:$AC \cdot CB + EC^2 = AD \cdot DB + ED^2$
We have that $ED^2 < EC^2$ and so:
:$AC \cdot CB < AD \cdot DB$
and so:
:$2 \cdot AC \cdot CB < 2 \cdot AD \cdot DB$
But from :
:$AB^2 = \left({AC + CB}\right)^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
and:
:$AB^2 = \left({AD + DB}\right)^2 = AD^2 + DB^2 + 2 \cdot AD \cdot DB$
and so it follows that:
:$ AC^2 + CB^2 > AD^2 + DB^2$
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Binomial Theorem

== Theorem ==
=== Integral Index ===
=== Ring Theory ===
=== General Binomial Theorem ===
=== Multiindices ===
=== Extended Binomial Theorem ===
=== Abel's Generalisation ===
=== Hurwitz's Generalisation ===
=== Approximations ===
== Examples ==
== Also known as ==
== Also see ==
* Definition:Binomial Expansion
* Definition:Binomial Series
* Definition:Pascal's Triangle
* Pascal's Rule
Category:Binomial Theorem
Category:Binomial Coefficients
Category:Named Theorems
Binomial Theorem/Abel's Generalisation

== Theorem ==
:$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$
for $n \in \Z_{\ge 0}$ and $x \in \R_{\ne 0}$.
=== Special Case: $x + y = 0$ ===
=== Negative $n$ ===
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Historical Note ==
== Sources ==
* 
* : $\S 1.2.6$: Binomial Coefficients: $(16)$
Category:Binomial Coefficients
Category:Binomial Theorem
Binomial Theorem/Abel's Generalisation/Negative n

== Theorem ==
Abel's Generalisation of Binomial Theorem:
does not hold for $n \in \Z_{< 0}$.
== Proof ==
Putting $n = x = -1$ and $y = z = 1$ into the 
:$\paren {-1 + 1}^{-1} = \dfrac 1 0$
which is undefined.
Putting the same values into the gives:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $52$
Category:Binomial Theorem
Binomial Theorem/Abel's Generalisation/Proof 1

== Theorem ==
== Proof ==
By admitting $y = \paren {x + y} - x$, we have that:
:$\paren {x + y}^n = \paren {x + \paren {x + y} - x}^n$
Expanding the in powers of $\paren {x + y}$:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $51$
Category:Binomial Theorem
Binomial Theorem/Abel's Generalisation/Proof 2

== Theorem ==
== Proof ==
From this formula:
:$(1): \quad \ds \sum_{k \mathop \in \Z} \binom n k x \paren {x + k z}^{k - 1} y \paren {y + \paren {n - k} z}^{n - k - 1} = \paren {x + y} \paren {x + y + n z}^{n - 1}$
The given formula:
:$\ds \paren {x + y}^n = \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {y + k z}^{n - k}$
can then be transformed into $(1)$ by
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $51$
Category:Binomial Theorem
Binomial Theorem/Abel's Generalisation/Proof 3

== Theorem ==
== Proof ==
From Hurwitz's Generalisation of Binomial Theorem:
:$(1): \quad \paren {x + y}^n = \ds \sum x \paren {x + \epsilon_1 z_1 + \cdots + \epsilon_n z_n}^{\epsilon_1 + \cdots + \epsilon_n - 1} \paren {y - \epsilon_1 z_1 - \cdots - \epsilon_n z_n}^{n - \epsilon_1 - \cdots - \epsilon_n}$
Setting $z = z_1 = z_2 = \cdots z_n$ we have:
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $51$
Category:Binomial Theorem
Binomial Theorem/Abel's Generalisation/x+y = 0

== Theorem ==
Consider Abel's Generalisation of Binomial Theorem:
This holds in the special case where $x + y = 0$.
== Proof ==
As $x + y = 0$, we can substitute $y = -x$, and so:
:$\ds \sum_k \binom n k x \paren {x - k z}^{k - 1} \paren {-x + k z}^{n - k} = 0$
is to be proved.
So:
Then Sum over $k$ of $\dbinom r k$ by $\paren {-1}^{r - k}$ by Polynomial can be applied:
:$\ds \sum_k \binom r k \paren {-1}^{r - k} \map {P_r} k = r! \, b_r$
where:
:$\map {P_r} k = b_0 + b_1 k + \cdots + b_r k^r$ is a polynomial in $k$ of degree $r$.
As the coefficient of $k^n = 0$, we have:
:$\ds \sum_k \binom n k \paren {-1}^{n - k} \paren {b_0 + b_1 k + \cdots + b_{n - 1} k^n + b_n 0} = 0$
Hence the result.
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $50$
Category:Binomial Theorem
Binomial Theorem/Also known as

== Binomial Theorem: Also known as == 
The binomial theorem is also known as the binomial formula.
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Formula for Positive Integral $n$: $3.3$
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Formula for Positive Integral $n$: $3.3$
Category:Binomial Theorem
Binomial Theorem/Approximations

== Theorem ==
Consider the General Binomial Theorem:
:$\paren {1 + x}^\alpha = 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots$
When $x$ is small it is often possible to neglect terms in $x$ higher than a certain power of $x$, and use what is left as an approximation to $\paren {1 + x}^\alpha$.
=== First Order ===
=== Second Order ===
== Examples ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Approximations
Category:Binomial Theorem
Binomial Theorem/Approximations/1st Order

== Theorem ==
Consider the General Binomial Theorem:
:$\paren {1 + x}^\alpha = 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots$
When $x$ is sufficiently small that $x^2$ can be neglected then:
:$\paren {1 + x}^\alpha \approx 1 + \alpha x$
and the error is of the order of $\dfrac {\alpha \paren {\alpha - 1} } 2 x^2$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Approximations
Category:Binomial Theorem
Binomial Theorem/Approximations/2nd Order

== Theorem ==
Consider the General Binomial Theorem:
:$\paren {1 + x}^\alpha = 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } {2!} x^2 + \dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } {3!} x^3 + \cdots$
When $x$ is sufficiently small that $x^3$ can be neglected, then:
:$\paren {1 + x}^\alpha \approx 1 + \alpha x + \dfrac {\alpha \paren {\alpha - 1} } 2 x^2$
and the error is of the order of:
:$\dfrac {\alpha \paren {\alpha - 1} \paren {\alpha - 2} } 6 x^3$
== Proof ==
== Sources ==
* : $\text I$. Algebra: The Binomial Theorem: Approximations
Category:Binomial Theorem
Binomial Theorem/Examples

== Examples of Use of Binomial Theorem ==
=== Cube of Sum ===
=== Cube of Difference ===
=== Fourth Power of Sum ===
=== Fourth Power of Difference ===
=== Fifth Power of Sum ===
=== Fifth Power of Difference ===
=== Sixth Power of Sum ===
=== Sixth Power of Difference ===
=== Power of $11$: $11^4$ ===
=== Binomial Theorem: $\paren {1 + x}^7$ ===
=== Square Root of 2 ===
=== Binomial Theorem: $\paren {a - 2 b}^4$ ===
Category:Examples of Use of Binomial Theorem
Binomial Theorem/Examples/(1 + x)^7

== Example of Use of Binomial Theorem ==
:$\paren {1 + x}^7 = 1 + 7 x + 21 x^2 + 35 x^3 + 35 x^4 + 21 x^5 + 7 x^6 + x^7$
== Proof ==
As can be seen, the digits are those from the $7$th row of Pascal's triangle:
== Sources ==
* : $35$
* : $35$
Category:Examples of Use of Binomial Theorem
Binomial Theorem/Examples/(a - 2 b)^4

== Example of Use of Binomial Theorem ==
:$\paren {a - 2 b}^4 = a^4 - 8 a^3 b + 24 a^2 b^2 - 32 a b^3 + 16 b^4$
== Proof ==
== Sources ==
* : $\S 3$: The Binomial Formula and Binomial Coefficients: Binomial Formula for Positive Integral $n$: Example
Category:Examples of Use of Binomial Theorem
Binomial Theorem/Examples/11^4

== Example of Use of Binomial Theorem ==
:$11^4 = \left({10 + 1}\right)^4 = 14 \, 641$
== Proof ==
As can be seen, the digits are those from the $4$th row of Pascal's triangle:
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $5$
Category:Examples of Use of Binomial Theorem
Binomial Theorem/Examples/4th Power of Difference

== Example of Use of Binomial Theorem ==
:$\paren {x - y}^4 = x^4 - 4 x^3 y + 6 x^2 y^2 - 4 x y^3 + y^4$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$
putting $n = 4$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.6$
* : $\S 2$: Special Products and Factors: $2.6.$
Category:Examples of Use of Binomial Theorem
Category:Fourth Powers
Binomial Theorem/Examples/4th Power of Sum

== Example of Use of Binomial Theorem ==
:$\paren {x + y}^4 = x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + y^4$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 4$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.5$
* : $\S 20$: Binomial Series: $20.7$
* : Chapter $1$: Some Preliminary Considerations: $1.2$ The Binomial Theorem
* : $\S 2$: Special Products and Factors: $2.5.$
* : $\S 22$: Taylor Series: Binomial Series: $22.7.$
Category:Examples of Use of Binomial Theorem
Category:Fourth Powers
Binomial Theorem/Examples/5th Power of Difference

== Example of Use of Binomial Theorem ==
:$\paren {x - y}^5 = x^5 - 5 x^4 y + 10 x^3 y^2 - 10 x^2 y^3 + 5 x y^4 - y^5$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + \paren {-y} }^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} \paren {-y}^k$
putting $n = 5$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.8$
* : $\S 2$: Special Products and Factors: $2.8.$
Category:Examples of Use of Binomial Theorem
Category:Fifth Powers
Binomial Theorem/Examples/5th Power of Sum

== Example of Use of Binomial Theorem ==
:$\paren {x + y}^5 = x^5 + 5 x^4 y + 10 x^3 y^2 + 10 x^2 y^3 + 5 x y^4 + y^5$
== Proof ==
Follows directly from the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 5$.
== Sources ==
* : $\S 2$: Special Products and Factors: $2.7$
* : $\S 2$: Special Products and Factors: $2.7.$
Category:Examples of Use of Binomial Theorem
Category:Fifth Powers