Binomial theorem

Binom of Newton - a formula for decomposing into separate terms a non-negative integer power of the sum of two variables, having the form

(a+b)^{n}=\sum _{k=0}^{n}{\binom {n}{k}}a^{n-k}b^{k}={n \choose 0}a^{n}+{n \choose 1}a^{n-1}b+\dots +{n \choose k}a^{n-k}b^{k}+\dots +{n \choose n}b^{n}

{\ displaystyle (a + b) ^ {n} = \ sum _ {k = 0} ^ {n} {\ binom {n} {k}} a ^ {nk} b ^ {k} = {n \ choose 0} a ^ {n} + {n \ choose 1} a ^ {n-1} b + \ dots + {n \ choose k} a ^ {nk} b ^ {k} + \ dots + {n \ choose n } b ^ {n}}

(a + b) ^ n = \ sum_ {k = 0} ^ n \ binom {n} {k} a ^ {n - k} b ^ k = {n \ choose 0} a ^ n + {n \ choose 1} a ^ {n - 1} b + \ dots + {n \ choose k} a ^ {n - k} b ^ k + \ dots + {n \ choose n} b ^ n

Where ${n \choose k}={\frac {n!}{k!(n-k)!}}=C_{n}^{k}$ ${\ displaystyle {n \ choose k} = {\ frac {n!} {k! (nk)!}} = C_ {n} ^ {k}}$ ${n \ choose k} = \ frac {n!} {k! (n - k)!} = C_n ^ k$ - binomial coefficients , $n$ ${\ displaystyle n}$ $n$ Is a non-negative integer .

In this form, this formula was still known to Indian and Islamic mathematicians; Newton derived the Newton binomial formula for the more general case when the exponent is an arbitrary real (or even complex ) number. In this case, the bin is an infinite series (see below).

Proof

Evidence

Let us prove the Newton binomial formula by induction on n :

Induction Base: $n=0$ ${\ displaystyle n = 0}$ $n=0$

(a+b)^{0}=1={\binom {0}{0}}a^{0}b^{0}

{\ displaystyle (a + b) ^ {0} = 1 = {\ binom {0} {0}} a ^ {0} b ^ {0}}

(a+b)^0=1=\binom{0}{0}a^0b^0

Induction step: Let the statement for $n$ ${\ displaystyle n}$ $n$ right:

(a+b)^{n}=\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k}

{\ displaystyle (a + b) ^ {n} = \ sum _ {k = 0} ^ {n} {n \ choose k} a ^ {nk} b ^ {k}}

(a+b)^n = \sum_{k=0}^n {n \choose k } a ^ {n-k} b ^ {k}

Then we need to prove the statement for $n+1$ ${\ displaystyle n + 1}$ $n+1$ :

(a+b)^{n+1}=\sum _{k=0}^{n+1}{{n+1} \choose k}a^{n+1-k}b^{k}

{\ displaystyle (a + b) ^ {n + 1} = \ sum _ {k = 0} ^ {n + 1} {{n + 1} \ choose k} a ^ {n + 1-k} b ^ {k}}

(a+b)^{n+1} = \sum_{k=0}^{n+1} {{n+1} \choose k } a ^ {n+1-k} b ^ {k}

Let's start the proof:

(a+b)^{n+1}=(a+b)(a+b)^{n}=(a+b)\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k}=\sum _{k=0}^{n}{n \choose k}{a^{n-k+1}b^{k}}\quad +\quad \sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k+1}

{\ displaystyle (a + b) ^ {n + 1} = (a + b) (a + b) ^ {n} = (a + b) \ sum _ {k = 0} ^ {n} {n \ choose k} a ^ {nk} b ^ {k} = \ sum _ {k = 0} ^ {n} {n \ choose k} {a ^ {n-k + 1} b ^ {k}} \ quad + \ quad \ sum _ {k = 0} ^ {n} {n \ choose k} a ^ {nk} b ^ {k + 1}}

(a+b)^{n+1} = (a+b)(a+b)^n=(a+b)\sum_{k=0}^{n} {n \choose k} a ^ {n-k} b ^ {k} = \sum_{k=0}^n {n \choose k} {a ^ {n - k + 1} b ^ {k}}\quad + \quad \sum_{k=0}^n {n \choose k} a ^ {n-k} b ^ {k+1}

We extract from the first sum the term for $k=0$ ${\ displaystyle k = 0}$ $k = 0$

\sum _{k=0}^{n}{n \choose k}{a^{n-k+1}b^{k}}=a^{n+1}+\sum _{k=1}^{n}{n \choose k}a^{n-k+1}b^{k}

{\ displaystyle \ sum _ {k = 0} ^ {n} {n \ choose k} {a ^ {n-k + 1} b ^ {k}} = a ^ {n + 1} + \ sum _ { k = 1} ^ {n} {n \ choose k} a ^ {n-k + 1} b ^ {k}}

We extract from the second sum the term for $k=n$ ${\ displaystyle k = n}$

\sum _{k=0}^{n}{n \choose k}a^{n-k}b^{k+1}=b^{n+1}+\sum _{k=0}^{n-1}{n \choose k}a^{n-k}b^{k+1}=b^{n+1}+\sum _{k=1}^{n}{n \choose {k-1}}a^{n-k+1}b^{k}

{\ displaystyle \ sum _ {k = 0} ^ {n} {n \ choose k} a ^ {nk} b ^ {k + 1} = b ^ {n + 1} + \ sum _ {k = 0} ^ {n-1} {n \ choose k} a ^ {nk} b ^ {k + 1} = b ^ {n + 1} + \ sum _ {k = 1} ^ {n} {n \ choose { k-1}} a ^ {n-k + 1} b ^ {k}}

Now add the converted amounts:

a^{n+1}+\sum _{k=1}^{n}{n \choose k}a^{n-k+1}b^{k}\quad +\quad b^{n+1}+\sum _{k=1}^{n}{n \choose {k-1}}a^{n-k+1}b^{k}=a^{n+1}+b^{n+1}+\sum _{k=1}^{n}\left({n \choose k}+{n \choose {k-1}}\right)a^{n-k+1}b^{k}=

{\ displaystyle a ^ {n + 1} + \ sum _ {k = 1} ^ {n} {n \ choose k} a ^ {n-k + 1} b ^ {k} \ quad + \ quad b ^ {n + 1} + \ sum _ {k = 1} ^ {n} {n \ choose {k-1}} a ^ {n-k + 1} b ^ {k} = a ^ {n + 1} + b ^ {n + 1} + \ sum _ {k = 1} ^ {n} \ left ({n \ choose k} + {n \ choose {k-1}} \ right) a ^ {n-k +1} b ^ {k} =}

=\sum _{k=0}^{0}{n+1 \choose k}a^{n+1-k}b^{k}\quad +\quad \sum _{k=n+1}^{n+1}{n+1 \choose k}a^{n+1-k}b^{k}\quad +\quad \sum _{k=1}^{n}{n+1 \choose k}a^{n+1-k}b^{k}=\sum _{k=0}^{n+1}{{n+1} \choose k}a^{n+1-k}b^{k}

{\ displaystyle = \ sum _ {k = 0} ^ {0} {n + 1 \ choose k} a ^ {n + 1-k} b ^ {k} \ quad + \ quad \ sum _ {k = n +1} ^ {n + 1} {n + 1 \ choose k} a ^ {n + 1-k} b ^ {k} \ quad + \ quad \ sum _ {k = 1} ^ {n} {n +1 \ choose k} a ^ {n + 1-k} b ^ {k} = \ sum _ {k = 0} ^ {n + 1} {{n + 1} \ choose k} a ^ {n + 1-k} b ^ {k}}

Q.E.D. ■

Generalizations

The Newton binomial formula is a special case of the expansion of a function $(1+x)^{r}$ ${\ displaystyle (1 + x) ^ {r}}$ in Taylor's row :

(1+x)^{r}=\sum _{k=0}^{\infty }{r \choose k}x^{k}

{\ displaystyle (1 + x) ^ {r} = \ sum _ {k = 0} ^ {\ infty} {r \ choose k} x ^ {k}}

,

where r can be a complex number (in particular, negative or real). The coefficients of this expansion are found by the formula:

{r \choose k}={1 \over k!}\prod _{n=0}^{k-1}(r-n)={\frac {r(r-1)(r-2)\cdots (r-(k-1))}{k!}}

{\ displaystyle {r \ choose k} = {1 \ over k!} \ prod _ {n = 0} ^ {k-1} (rn) = {\ frac {r (r-1) (r-2) \ cdots (r- (k-1))} {k!}}}

Moreover, a number

(1+z)^{\alpha }=1+\alpha {}z+{\frac {\alpha (\alpha -1)}{2}}z^{2}+...+{\frac {\alpha (\alpha -1)\cdots (\alpha -n+1)}{n!}}z^{n}+...

{\ displaystyle (1 + z) ^ {\ alpha} = 1 + \ alpha {} z + {\ frac {\ alpha (\ alpha -1)} {2}} z ^ {2} + ... + {\ frac {\ alpha (\ alpha -1) \ cdots (\ alpha -n + 1)} {n!}} z ^ {n} + ...}

.

converges at $|z|\leq 1$ ${\ displaystyle | z | \ leq 1}$ .

In particular, when $z={\frac {1}{m}}$ ${\ displaystyle z = {\ frac {1} {m}}}$ and $\alpha =x\cdot m$ ${\ displaystyle \ alpha = x \ cdot m}$ it turns out the identity

\left(1+{\frac {1}{m}}\right)^{xm}=1+x+{\frac {xm(xm-1)}{2\;m^{2}}}+...+{\frac {xm(xm-1)\cdots (xm-n+1)}{n!\;m^{n}}}+\dots .

{\ displaystyle \ left (1 + {\ frac {1} {m}} \ right) ^ {xm} = 1 + x + {\ frac {xm (xm-1)} {2 \; m ^ {2}} } + ... + {\ frac {xm (xm-1) \ cdots (xm-n + 1)} {n! \; m ^ {n}}} + \ dots.}

Passing to the limit at $m\to \infty$ ${\ displaystyle m \ to \ infty}$ and using the second wonderful limit $\lim _{m\to \infty }{\left(1+{\frac {1}{m}}\right)^{m}}=e$ ${\ displaystyle \ lim _ {m \ to \ infty} {\ left (1 + {\ frac {1} {m}} \ right) ^ {m}} = e}$ , we deduce the identity

e^{x}=1+x+{\frac {x^{2}}{2}}+\dots +{\frac {x^{n}}{n!}}+\dots ,

{\ displaystyle e ^ {x} = 1 + x + {\ frac {x ^ {2}} {2}} + \ dots + {\ frac {x ^ {n}} {n!}} + \ dots,}

which in this way was first obtained by Euler .

Multinomial Theorem

Newton’s binomial can be generalized to the Newton polynomial — raising to a power the sum of an arbitrary number of terms:

(x_{1}+x_{2}+\cdots +x_{m})^{n}=\sum \limits _{k_{j}\geqslant 0 \atop k_{1}+k_{2}+\cdots +k_{m}=n}{n \choose k_{1},k_{2},\ldots ,k_{m}}x_{1}^{k_{1}}\ldots x_{m}^{k_{m}},

{\ displaystyle (x_ {1} + x_ {2} + \ cdots + x_ {m}) ^ {n} = \ sum \ limits _ {k_ {j} \ geqslant 0 \ atop k_ {1} + k_ {2 } + \ cdots + k_ {m} = n} {n \ choose k_ {1}, k_ {2}, \ ldots, k_ {m}} x_ {1} ^ {k_ {1}} \ ldots x_ {m } ^ {k_ {m}},}

Where $\textstyle {\binom {n}{k_{1},k_{2},\ldots ,k_{m}}}={\frac {n!}{k_{1}!k_{2}!\cdots k_{m}!}}$ ${\ displaystyle \ textstyle {\ binom {n} {k_ {1}, k_ {2}, \ ldots, k_ {m}}} = {\ frac {n!} {k_ {1}! k_ {2}! \ cdots k_ {m}!}}}$ - multinomial coefficients . The sum is taken over all non-negative integer indices. $k_{j}$ ${\ displaystyle k_ {j}}$ whose sum is equal to n (that is, over all compositions of the number n of length m ). When using the Newton polynomial, it is considered that the expressions $x_{j}^{0}=1$ ${\ displaystyle x_ {j} ^ {0} = 1}$ , even $x_{j}=0$ ${\ displaystyle x_ {j} = 0}$ .

The multinomial theorem is easily proved either by induction on n , or from combinatorial considerations and the combinatorial meaning of the multinomial coefficient.

At $m=2$ ${\ displaystyle m = 2}$ expressing $k_{2}=n-k_{1}$ ${\ displaystyle k_ {2} = n-k_ {1}}$ , we get the Newton bin.

Complete Bell

Let be $B_{n}(a_{s})=B_{n}(a_{1},\dots ,a_{n})$ ${\ displaystyle B_ {n} (a_ {s}) = B_ {n} (a_ {1}, \ dots, a_ {n})}$ and $B_{0}=1$ ${\ displaystyle B_ {0} = 1}$ then the complete Bell polynomials have binomial decomposition:

B_{n}({{a_{s}}+{b_{s}}})=\sum _{i+j=n}{n \choose i,\ j}{B_{i}}({a_{s}}){B_{j}}({b_{s}}).

{\ displaystyle B_ {n} ({{a_ {s}} + {b_ {s}}}) = \ sum _ {i + j = n} {n \ choose i, \ j} {B_ {i}} ({a_ {s}}) {B_ {j}} ({b_ {s}}).}

History

For a long time it was believed that Blaise Pascal , who described it in the 17th century , invented this formula, as well as the triangle , which allows finding the coefficients. However, historians of science found that the formula was known even to the Chinese mathematician Yang Hui , Who lived in the XIII century , as well as Islamic mathematicians at-Tusi (XIII century) and al-Kashi (XV century). In the mid- sixteenth century, Michael Stifel described binomial coefficients and also compiled a table of them to degree 18.

Isaac Newton around 1677 generalized the formula for an arbitrary exponent (fractional, negative, etc.). From the binomial decomposition, Newton, and later Euler , derived the whole theory of infinite series.

In Fiction

In fiction, “Newton’s binomial” appears in several memorable contexts where it is about something complicated. ^[one]

In A. Conan Doyle ’s short story “Holmes’s Last Affair, ” Holmes talks about mathematics by Professor Moriarty :

When he was twenty-one, he wrote a treatise on Newton’s binomial, which won him European fame. After that, he got a chair of mathematics at one of our provincial universities, and, in all likelihood, he had a brilliant career.
Original text
The Final Problem At the age of twenty-one he wrote a treatise upon the binomial theorem, which has had a European vogue. On the strength of it he won the mathematical chair at one of our smaller universities, and had, to all appearances, a most brilliant career before him.

In the novel "The Master and Margarita " by M. A. Bulgakov :

“Think, Newton’s binomial! "He will die nine months later, in February next year, from liver cancer in the clinic of the First Moscow State University, in the fourth ward."

Later, the same expression, "Think of Newton’s binom!" mentioned in the film " Stalker " by A. A. Tarkovsky .
Roman E.N. Wilmont received the name "Transience, or Just Think, Newton’s Bin!"

Notes

↑ Uspensky V. A. Preface for readers of the “New Literary Review” to the semiotic messages of Andrei Nikolaevich Kolmogorov // New Literary Review . - 1997. - No. 24 .

Literature

Newton’s binomial // Brockhaus and Efron Encyclopedic Dictionary : 86 t. (82 t. And 4 ext.). - SPb. , 1890-1907.

Links

Newton’s binomial // Great Soviet Encyclopedia : [in 30 vol.] / Ch. ed. A.M. Prokhorov . - 3rd ed. - M .: Soviet Encyclopedia, 1969-1978.

[1] Uspensky V. A. Preface for readers of the “New Literary Review” to the semiotic messages of Andrei Nikolaevich Kolmogorov // New Literary Review . - 1997. - No. 24 .