مبرهنة ذات الحدين Binomial theorem أوثنائي نيوتن هي صيغة وضعها نيوتن لإيجاد نشر لثنائي مرفوع بقوة سليمة ما. ويطلق على هذه الصيغة صيغة ثنائي نيوتن، أوببساطة صيغة الثنائي

الصيغة

فلنعتبر ثنائيا متكونا من عنصرين x وy معهدين على مجموعة حيث xy=yx، وعددا سليما طبييعا n،

${\displaystyle (x+y)^{n =\sum _{k=0 ^{n {n \choose k x^{n-k y^{k$

حيث الأعداد ${\displaystyle {n \choose k$ (والتي تخط أحيانا ${\displaystyle C_{n ^{k$) هي الضوارب الثنائية.

هذا المجموع يعتمد على الضوارب الثنائية الموجودة على أحد سطور مثلث باسكال.

تغيير y بـ - y داخل الصيغة، يعطي الصيغة :

${\displaystyle (x-y)^{n =\sum _{k=0 ^{n (-1)^{k {n \choose k x^{n-k y^{k$

مثال :

${\displaystyle n=2~,\qquad (x+y)^{2 =x^{2 +2xy+y^{2 \,$

${\displaystyle n=3~,\qquad (x-y)^{3 =x^{3 -3x^{2 y+3xy^{2 -y^{3 \,$

${\displaystyle n=4~,\qquad (x+y)^{4 =x^{4 +4x^{3 y+6x^{2 y^{2 +4xy^{3 +y^{4 \,$

التبيان

فلتكن x، y عناصر من مجموعة حيث xy=yx وn عددا طبيعيا سليما.

${\displaystyle (x+y)^{n =\sum _{k=0 ^{n {n \choose k x^{n-k y^{k$

فلنبين هذه الصيغة بالـ "الكيفية التراجعية" :

البداية

${\displaystyle n=0~,\qquad (x+y)^{0 =1={0 \choose 0 x^{0 y^{0$

${\displaystyle n=1~,\qquad (x+y)^{1 =x+y={1 \choose 0 x^{1 y^{0 +{1 \choose 1 x^{0 y^{1$

صحة العنصر التالي

فليكن n عددا سليما طبيعيا أكبر أومساولـ 1, فلنبين حتى العلاقات سليمة لـ n + 1 إذا كانت سليمة لـ n:

حسب الافتراض الاول :

${\displaystyle (x+y)^{n+1 =(x+y)\cdot \sum _{k=0 ^{n {n \choose k x^{n-k y^{k ,$

بتوزيعية ${\displaystyle \cdot$ على ${\displaystyle +$ :

${\displaystyle (x+y)^{n+1 =x^{n+1 +x\cdot \sum _{k=1 ^{n {n \choose k x^{n-k y^{k +y\cdot \sum _{k=0 ^{n-1 {n \choose k x^{n-k y^{k +y^{n+1$

بالتفكيك إلى جذاء :

${\displaystyle (x+y)^{n+1 =x^{n+1 +\sum _{k=1 ^{n \left\lbrack {{n \choose {k +{{n \choose {k-1 \right\rbrack x^{n-k+1 y^{k +y^{n+1$

باستعمال صيغة مثلث باسكال :

${\displaystyle (x+y)^{n+1 =x^{n+1 +\sum _{k=1 ^{n {{n+1 \choose k ~x^{n-k+1 y^{k +y^{n+1$

وهوما ينهي التبيان.

الشرح الهندسي

For positive values of a and b, the binomial theorem with n = 2 is the geometrically evident fact that a square of side a + b can be cut into a square of side a, a square of side b, and two rectangles with sides a and b. With n = 3, the theorem states that a cube of side a + b can be cut into a cube of side a, a cube of side b, three a×a×b rectangular boxes, and three a×b×b rectangular boxes.

In calculus, this picture also gives a geometric proof of the derivative ${\displaystyle (x^n{)}^{\prime }=n{x}^{n-<!--\; -\; -->1}:}$a, then this picture shows the infinitesimal change in the volume of an n-dimensional hypercube, ${\displaystyle (x+\Delta x)^{n ,$ where the coefficient of the linear term (in ${\displaystyle \Delta x$) is ${\displaystyle nx^{n-1 ,$ the area of the n faces, each of dimension ${\displaystyle (n-1):$

${\displaystyle (x+\Delta x)^{n =x^{n +nx^{n-1 \Delta x+{\tbinom {n {2 x^{n-2 (\Delta x)^{2 +\cdots .$

Substituting this into the definition of the derivative via a difference quotient and taking limits means that the higher order terms, ${\displaystyle (\Delta x)^{2$ and higher, become negligible, and yields the formula ${\displaystyle (x^{n )'=nx^{n-1 ,$ interpreted as

"the infinitesimal rate of change in volume of an n-cube as side length varies is the area of n of its ${\displaystyle (n-1)$-dimensional faces".

If one integrates this picture, which corresponds to applying the fundamental theorem of calculus, one obtains Cavalieri's quadrature formula, the integral ${\displaystyle \int <!--\; \int \; -->x^{n-<!--\; -\; -->1}dx=\frac{1}{n}{x}^n}$

المعاملات الثنائية

المعاملات التي تظهر في التمديد ثنائي الحدين تسمى المعاملات الثنائية. These are usually written ${\displaystyle {\tbinom {n {k$, and pronounced “n choose k”.

الصيغ

The coefficient of x^n−ky^k is given by the formula

${\displaystyle {n \choose k ={\frac {n! {k!(n-k)!$

which is defined in terms of the factorial function n!. Equivalently, this formula can be written

${\displaystyle {n \choose k ={\frac {n(n-1)\cdots (n-k+1) {k(k-1)\cdots 1 =\prod _{\ell =1 ^{k {\frac {n-\ell +1 {\ell =\prod _{\ell =0 ^{k-1 {\frac {n-\ell {k-\ell$

with k factors in both the numerator and denominator of the fraction. Note that, although this formula involves a fraction, the binomial coefficient ${\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}$

التفسير التوافيقي

The binomial coefficient ${\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}$

${\displaystyle (x+y)(x+y)(x+y)\cdots (x+y),$

then, according to the distributive law, there will be one term in the expansion for each choice of either x or y from each of the binomials of the product. For example, there will only be one term xⁿ, corresponding to choosing x from each binomial. However, there will be several terms of the form xⁿ⁻²y², one for each way of choosing exactly two binomials to contribute a y. Therefore, after combining like terms, the coefficient of xⁿ⁻²y² will be equal to the number of ways to choose exactly 2 elements from an n-element set.

البراهين

البرهان التوافيقي

مثال

The coefficient of xy² in

${\displaystyle {\begin{aligned (x+y)^{3 &=(x+y)(x+y)(x+y)\\&=xxx+xxy+xyx+{\underline {xyy +yxx+{\underline {yxy +{\underline {yyx +yyy\\&=x^{3 +3x^{2 y+{\underline {3xy^{2 +y^{3 .\end{aligned$

equals ${\displaystyle {\tbinom {3 {2 =3$ because there are three x,y strings of length ثلاثة with exactly two y's, namely,

${\displaystyle xyy,\;yxy,\;yyx,$

corresponding to the three 2-element subsets of { 1, 2, 3  , namely,

${\displaystyle \{2,3\ ,\;\{1,3\ ,\;\{1,2\ ,$

where each subset specifies the positions of the y in a corresponding string.

الحالة العامة

Expanding (x + y)ⁿ yields the sum of the 2 ⁿ products of the form e₁e₂ ... e _n where each e _i is x or y. Rearranging factors shows that each product equals x^n−ky^k for some k between 0 and n. For a given k, the following are proved equal in succession:

• the number of copies of x^n − ky^k in the expansion
• the number of n-character x,y strings having y in exactly k positions
• the number of k-element subsets of { 1, 2, ..., n
• ${\displaystyle {n \choose k$ (this is either by definition, or by a short combinatorial argument if one is defining ${\displaystyle {n \choose k$ as ${\displaystyle {\frac {n! {k!(n-k)!$).

This proves the binomial theorem.

Inductive proof

Induction yields another proof of the binomial theorem. When n = 0, both sides equal 1, since x⁰ = 1 and ${\displaystyle {\tbinom {0 {0 =1$. Now suppose that the equality holds for a given n; we will prove it for n + 1. For j, k ≥ 0, let [ƒ(x, y)] _j,k denote the coefficient of x^jy^k in the polynomial ƒ(x, y). By the inductive hypothesis, (x + y)ⁿ is a polynomial in x and y such that [(x + y)ⁿ] _j,k is ${\displaystyle {\tbinom {n {k$ if j + k = n, and 0 otherwise. The identity

${\displaystyle (x+y)^{n+1 =x(x+y)^{n +y(x+y)^{n$

shows that (x + y)^{n + 1} also is a polynomial in x and y, and

${\displaystyle [(x+y)^{n+1 ]_{j,k =[(x+y)^{n ]_{j-1,k +[(x+y)^{n ]_{j,k-1 ,$

since if j + k = n + 1, then (j − 1) + k = n and j + (k − 1) = n. Now, the right hand side is

${\displaystyle {\binom {n {k +{\binom {n {k-1 ={\binom {n+1 {k ,$

by Pascal's identity. On the other hand, if j +k ≠ n + 1, then (j – 1) + k ≠ n and j +(k – 1) ≠ n, so we get 0 + 0 = 0. Thus

${\displaystyle (x+y)^{n+1 =\sum _{k=0 ^{n+1 {\tbinom {n+1 {k x^{n+1-k y^{k ,$

which is the inductive hypothesis with n + 1 substituted for n and so completes the inductive step.

التعميمات

مبرهنة ذات الحدين المعممة لنيوتن

Around 1665, Isaac Newton generalized the binomial theorem to allow real exponents other than nonnegative integers. (The same generalization also applies to complex exponents.) In this generalization, the finite sum is replaced by an infinite series. In order to do this, one needs to give meaning to binomial coefficients with an arbitrary upper index, which cannot be done using the usual formula with factorials. However, for an arbitrary number r, one can define

${\displaystyle {r \choose k ={\frac {r\,(r-1)\cdots (r-k+1) {k! ={\frac {(r)_{k {k! ,$