Given ` x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=7 ` , we are asked to show that `x^16>15 ` :
First, note that `x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=x(x-1)(x+1)(x^4+1)(x^8+1) ` so the polynomial has real roots at -1,0, and 1.
For x<-1 the polynomial is negative.
For -1<x<0 the polynomial is positive but has a maximum a little less than 1/2.
For 0<x<1 the polynomial is negative.
So we know that x>1 for the polynomial to achieve 7.
Since x>1 we have `(x-1)^2>0 `
`==> x^2-2x+1>0...
Given ` x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=7 ` , we are asked to show that `x^16>15 ` :
First, note that `x^15-x^13+x^11-x^9+x^7-x^5+x^3-x=x(x-1)(x+1)(x^4+1)(x^8+1) ` so the polynomial has real roots at -1,0, and 1.
For x<-1 the polynomial is negative.
For -1<x<0 the polynomial is positive but has a maximum a little less than 1/2.
For 0<x<1 the polynomial is negative.
So we know that x>1 for the polynomial to achieve 7.
Since x>1 we have `(x-1)^2>0 `
`==> x^2-2x+1>0 `
`==>x^2+1>2x `
`==> (x^2+1)/x>2 `
Now multiply both sides of the equation by `x^2+1 ` :
`(x^2+1)(x^15-x^13+x^11-x^9+x^7-x^5+x^3-x)=7(x^2+1) `
Multiplying and factoring we get:
`x(x^16-1)=7(x^2+1) `
Then:
`x^16-1=(7(x^2+1))/x ` ; but `(x^2+1)/x>2 ` so
`x^16-1>7(2) ` and
`x^16>15 ` as required.
*********************************************************
`x^16-1=(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1) `
and
`x^15-x^13+x^11-x^9+x^7-x^5+x^3-x= `
`x(x-1)(x+1)(x^4+1)(x^8+1) `
so multiplying the degree 15 polynomial by x^2+1 gives x times the degree 16 polynomial.
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