Difference between revisions of "Sum and difference of powers"

m (Factorizations of Sums of Powers)
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If we have a difference of powers of degree <math>n</math>, then
 
If we have a difference of powers of degree <math>n</math>, then
  
<cmath>x^n-y^n=(x-y)(x^n+x^ay+x^by^2...+y^n)</cmath>
+
<cmath>x^n-y^n=(x-y)(x^{n - 1}+x^ay+x^by^2...+y^{n - 1})</cmath>
  
where <math>a, b, c,...</math> are equivalent to <math>n-1, n-2, n-3,...</math> respectively.
+
where <math>a, b, c,...</math> are equivalent to <math>n-2, n-3, n-4,...</math> respectively.
  
 
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial <math>(x+y)^n</math>, except for the fact that the coefficient on each of the terms is <math>1</math>. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
 
Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial <math>(x+y)^n</math>, except for the fact that the coefficient on each of the terms is <math>1</math>. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.
  
 
- icecreamrolls8
 
- icecreamrolls8
 +
(small fix on exponents by sr_vrd)
  
 
==See Also==
 
==See Also==

Revision as of 16:45, 8 December 2022

The sum and difference of powers are powerful factoring techniques that, respectively, factor a sum or a difference of certain powers.

Sums of Odd Powers

$a^{2n+1}+b^{2n+1}=(a+b)(a^{2n}-a^{2n-1}b+a^{2n-2}b^2-\ldots-ab^{2n-1}+b^{2n})$

Differences of Powers

If $p$ is a positive integer and $x$ and $y$ are real numbers,

$x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)$

For example:

$x^2-y^2=(x-y)(x+y)$

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

Note that the number of terms in the long factor is equal to the exponent in the expression being factored.

An amazing thing happens when $x$ and $y$ differ by $1$, say, $x = y+1$. Then $x-y = 1$ and

$x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}$

$=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p$.

For example:

$(y+1)^2-y^2=(y+1)+y$

$(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2$

$(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3$

If we also know that $y\geq 0$ then:

$2y\leq (y+1)^2-y^2\leq 2(y+1)$

$3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2$

$4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3$

$(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p$

Sum of Cubes

$1^3=1^2$

$1^3+2^3 =3^2$

$1^3 +2^3+3^3=6^2$

$1^3+2^3+3^3+4^3=10^2$

Factorizations of Sums of Powers

$x^2-y^2=(x-y)(x+y)$

$x^3-y^3=(x-y)(x^2+xy+y^2)$

$x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)$

Note that all these sums of powers can be factorized as follows:

If we have a difference of powers of degree $n$, then

\[x^n-y^n=(x-y)(x^{n - 1}+x^ay+x^by^2...+y^{n - 1})\]

where $a, b, c,...$ are equivalent to $n-2, n-3, n-4,...$ respectively.

Note, of course, that some of the signs simply change when we have sum of powers instead of difference. Regardless, observe that the "longer" polynomial in the factorization is simply a binomial theorem expansion of the binomial $(x+y)^n$, except for the fact that the coefficient on each of the terms is $1$. This can be quite useful in problems that might have a sum of powers expression as well as an application of the binomial theorem.

- icecreamrolls8 (small fix on exponents by sr_vrd)

See Also