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Difference between revisions of "Sum and difference of powers"

(have to go, open for improvisation.)
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==Differences of Powers==
 
==Differences of Powers==
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If p is a positive integer and x and y are real numbers,
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If <math>p</math> is a positive integer and <math>x</math> and <math>y</math> are real numbers,
  
 
<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
 
<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
  
−
For example,
+
For example:
  
 
<math>x^2-y^2=(x-y)(x+y)</math>
 
<math>x^2-y^2=(x-y)(x+y)</math>
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Note that the number of terms in the ''long'' factor is equal to the exponent in the expression being factored.
 
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  
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An amazing thing happens when <math>x</math> and <math>y</math> differ by <math>1</math>, say, <math>x = y+1</math>. Then <math>x-y = 1</math> and  
  
 
<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
 
<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
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<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
 
<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
  
−
For example,
+
For example:
  
 
<math>(y+1)^2-y^2=(y+1)+y</math>
 
<math>(y+1)^2-y^2=(y+1)+y</math>
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<math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
 
<math>(y+1)^4-y^4=(y+1)^3+(y+1)^2y+(y+1)y^2+y^3</math>
  
−
If we also know that <math>y\geq 0</math> then
+
If we also know that <math>y\geq 0</math> then:
  
 
<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
 
<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>

Revision as of 17:43, 8 July 2008

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

Sums of 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:

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

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

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

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

See Also

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