Part.I Introduction
This blog post records some formulas commonly used in mathematics related to squares or higher powers.
Chap.I some conclusions
The following section summarizes some important conclusions
- Perfect square formula: ( a ± b ) 2 = a 2 ± 2 ab + b 2 (a±b)^2=a^2±2ab+b^2(a±b)2=a2±2 ab+b2
- Square difference formula: a 2 − b 2 = ( a + b ) ( a − b ) a^2-b^2=(a+b)(ab)a2−b2=(a+b)(a−b)
- Cubic formula: ( a ± b ) 3 = a 3 ± 3 a 2 b + 3 ab 2 ± b 3 (a±b)^3=a^3±3a^2b+3ab^2±b^3(a±b)3=a3±3 a2 b+3 a b2±b3
- Cubic formula: a 3 + b 3 = ( a + b ) ( a 2 − ab + b 2 ) a^3+b^3=(a+b)(a^2-ab+b^2)a3+b3=(a+b)(a2−ab+b2)
- The formula for cubic variance: a 3 − b 3 = ( a − b ) ( a 2 + ab + b 2 ) a^3-b^3=(ab)(a^2+ab+b^2)a3−b3=(a−b)(a2+ab+b2)
- Sum of cubes minus product of three numbers: a 3 + b 3 + c 3 − 3 abc = ( a + b + c ) ( a 2 + b 2 + c 2 − ab − bc − ac ) a^3+b^3 +c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)a3+b3+c3−3 ab c=(a+b+c)(a2+b2+c2−ab−bc−ac)
- Binomial Theorem: ( a + b ) n = C n 0 an + C n 1 a ( n − 1 ) b + ⋯ + C nka ( n − k ) bk + ⋯ + C nnbn (a+b)^n =C^0_na^n+C^1_na^{(n-1)}b+\cdots+C^k_na^{(nk)}b^k+\cdots+C^n_nb^n(a+b)n=Cn0an+Cn1a(n−1)b+⋯+Cnka(n−k)bk+⋯+Cnnbn
Part.II The nth power of two terms
Chap.I and difference to the nth power (binomial theorem)
( a + b ) 2 = a 2 + a b + b a + b 2 = a 2 + 2 a b + b 2 (a+b)^2=a^2+ab+ba+b^2=a^2+2ab+b^2 (a+b)2=a2+ab+ba+b2=a2+2 ab+b2 This complete square formula should be familiar to everyone. But would like to extend it:nnn terms andnnHow to represent the nth power?
Let's look at the nn of two different itemsn order:( a + b ) n (a+b)^n(a+b)n , this expansion term has a ready-made formula, namely the binomial theorem!
( a + b ) n = C n 0 a n + C n 1 a ( n − 1 ) b + ⋯ + C n k a ( n − k ) b k + ⋯ + C n n b n (a+b)^n=C^0_na^n+C^1_na^{(n-1)}b+\cdots+C^k_na^{(n-k)}b^k+\cdots+C^n_nb^n (a+b)n=Cn0an+Cn1a(n−1)b+⋯+Cnka(n−k)bk+⋯+Cnnbn
- Binomial coefficient: C nk ( k = 0 , ⋯ , n ) C^k_n\ (k=0,\cdots,n)Cnk (k=0,⋯,n)
- Binomial general formula: C nka ( n − k ) bk C^k_na^{(nk)}b^kCnka(n−k)bk is the kth + 1 k+1in the expansionk+1 term, its general formula can be written as:T k + 1 = C nka ( n − k ) bk T_{k+1}=C^k_na^{(nk)}b^kTk+1=Cnka(n−k)bk
Chap.II sum and difference of nth power
n-order difference formula:
an − bn = ( a − b ) ( an − 1 + an − 2 b + an − 3 b 3 + ⋯ + abn − 2 + bn − 1 ) a^nb^n=(ab )(a^{n-1}+a^{n-2}b+a^{n-3}b^3+\cdots+ab^{n-2}+b^{n-1})an−bn=(a−b)(an−1+an−2b+an − 3 b3+⋯+abn−2+bn−1)
The sum formula of the nth power. When n is odd,
an + bn = ( a + b ) ( an − 1 − an − 2 b + an − 3 b 3 + ⋯ − abn − 2 + bn − 1 ) a^n+b^n=( a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^3+\cdots-ab^{n-2}+b^{n-1 })an+bn=(a+b)(an−1−an−2b+an − 3 b3+⋯−abn−2+bn − 1 )
When n is an even number, there is no n-th power sum formula. In fact, when n is an even number,
an − bn = ( a + b ) ( an − 1 − an − 2 b + an − 3 b 3 + ⋯ − abn − 2 + bn − 1 ) a^nb^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^3+\ cdots-ab^{n-2}+b^{n-1})an−bn=(a+b)(an−1−an−2b+an − 3 b3+⋯−abn−2+bn−1)
That is, when n is even, an − bna^nb^nan−bThere are two expressions for n ; only when n is an odd number, there is a formula for the sum of powers n.
Part.III Square of n distinct terms
consider nSquare of n distinct terms:( a + b + c + ⋯ ) 2 = ? (a+b+c+\cdots)^2=?(a+b+c+⋯)2=?
Here we don’t care about the specific content of each item after expansion, first we care about how many items can be expanded into, such as ( a + b ) 2 (a+b)^2(a+b)2 After unfolding, there are4 44 items, after sorting, there are3 33 items. Why distinguish before and after sorting? Because under certain operation rules, multiplication does not have commutative law, such as matrix multiplication. A table is listed below.
| number of different items | Before unfolding | Expanded |
|---|---|---|
| 2 | 4 | 3 |
| 3 | 9 | 6 |
| 4 | 16 | 10 |
| 5 | 25 | 15 |
| … | ||
| n n n | n 2 n^2 n2 | C n 2 + n C^2_n+n Cn2+n |
Part.IV Solving Higher Degree Equations in One Variable
Chap.I primary and secondary
Linear equation in one variable (also known as linear equation in one variable)
a 1 x + x 0 = 0 ( a 1 ≠ 0 ) a_1x+x_0=0\ (a_1\neq 0) a1x+x0=0 (a1=0 ) Solvingx = − a 0 / a 1 x=-a_0/a_1x=−a0/a1
Unary quadratic equation
ax 2 + bx + c = 0 ( a ≠ 0 ) ax^2+bx+c=0 (a\neq 0)ax2+bx+c=0(a=0 ) the solution isx = − b ± b 2 − 4 ac 2 ax=\frac{-b±\sqrt{b^2-4ac}}{2a}x=2a _−b±b2 −4ac
Discriminant formula: Δ = b 2 − 4 ac \Delta=b^2-4acD=b2−4 BC _
- Δ > 0 \Delta>0D>0 : the equation has two unequal real roots
- Δ = 0 \Delta=0D=0 : the equation has two equal real roots
- Δ < 0 \Delta<0 D<0 : the equation has two unequal imaginary roots
Veda's theorem: Let x 1 , x 2 x_1,x_2x1,x2are the two roots of the equation
- x 1 + x 2 = − b a x_1+x_2=-\frac{b}{a} x1+x2=−ab
- x 1 ⋅ x 2 = c a x_1\cdot x_2=\frac{c}{a} x1⋅x2=ac
Chap.II Unary Cubic Equation
a x 3 + b x 2 + c x + d = 0 ( a ≠ 0 ) ax^3+bx^2+cx+d=0 (a\neq 0) ax3+bx2+cx+d=0(a=0)
The commonly used solution is the Cardan formula method published by the Italian scholar Cardan in 1545.
Special form of root finding formula x 3 + px 2 + q = 0 x^3+px^2+q=0x3+px2+q=0
The root-finding formula in general form Cardan's method
ps: It comes from Baidu Encyclopedia, and the specific derivation will be discussed later.
Chap.III Quaternary equation in one variable
a x 4 + b x 3 + c x 2 + d x + e = 0 ( a ≠ 0 ) ax^4+bx^3+cx^2+dx+e=0 (a\neq 0) ax4+bx3+cx2+dx+e=0(a=0)
The formula for finding the root of the quadratic equation in one variable was firstly proved by the Italian mathematician Ferrari. The unary cubic equation is solved by reducing the problem to a one-dimensional quadratic equation after performing a clever exchange of elements. Therefore, if one can ingeniously transform the one-dimensional quartic equation into one-dimensional cubic equation or one-dimensional quadratic equation, one can use known formulas to solve it.
ps: The formula is relatively lengthy, see Baidu Encyclopedia for details.