Difference between revisions of "Differences of powers"

Revision as of 09:02, 8 July 2008

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

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

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

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

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

@@ Line 1: / Line 1: @@
 If p is a positive integer and x and y are real numbers,
-[[Image:Img2.gif]]
+<math>x^{p+1}-y^{p+1}=(x-y)(x^p+x^{p-1}y+\cdots +xy^{p-1}+y^p)</math>
 For example,
-[[Image:Img3.gif]]
+<math>x^2-y^2=(x-y)(x+y)</math>
+<math>x^3-y^3=(x-y)(x^2+xy+y^2)</math>
+<math>x^4-y^4=(x-y)(x^3+x^2y+xy^2+y^3)</math>
 Note that the number of terms in the ''long'' factor is equal to the exponent in the expression being factored.
@@ Line 11: / Line 15: @@
 An amazing thing happens when x and y differ by 1, say, x = y+1. Then x-y = 1 and
-[[Image:Img4.gif]]
+<math>x^{p+1}-y^{p+1}=(y+1)^{p+1}-y^{p+1}</math>
+<math>=(y+1)^p+(y+1)^{p-1}y+\cdots +(y+1)y^{p-1} +y^p</math>.
 For example,
-[[Image:Img5.gif]]
+<math>(y+1)^2-y^2=(y+1)+y</math>
+<math>(y+1)^3-y^3=(y+1)^2+(y+1)y+y^2</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
+<math>2y\leq (y+1)^2-y^2\leq 2(y+1)</math>
+<math>3y^2\leq (y+1)^3-y^3\leq 3(y+1)^2</math>
-If we also know that [[Image:Img6.gif]] then
+<math>4y^3\leq (y+1)^4-y^4\leq 4(y+1)^3</math>
-[[Image:Img7.gif]]
+<math>(p+1)y^p\leq (y+1)^{p+1}-y^{p+1}\leq (p+1)(y+1)^p</math>