- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,805

^{2}+ ... + x

^{8}= 8 - 10x

^{9}, or the real root of x + x

^{2}+ ... + x

^{10}= 8 - 10x

^{11}?

- Thread starter anemone
- Start date

- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,805

- Jan 29, 2012

- 1,151

- Jan 26, 2012

- 183

$f(x) = 1 + x + x^2+ \cdots +x^8 + 10x^9 - 8$

and

$g(x) = 1 + x + x^2+ \cdots + x^{10} + 10x^{11} - 8$

We also have $f(0.8) <0, \; f(0.9) >0, \;g(0.8) <0, \;g(0.9) >0$ meaning that both roots lie between $0.8$ and $0.9$. Now consider the difference $h(x) = f(x)-g(x) =9x^8-x^{10}-10x^{11}= -(x+1)(10x-9)x^9 $. On the interval $(0.8,0.9)$ $h(x)> 0$ meaning that $ f(x) > g(x)$ giving that the root of $g(x)$ is larger than the root of $f(x)$ as HallsofIvy pointed out.

- Thread starter
- Admin
- #4

- Feb 14, 2012

- 3,805

Thanks for participating to both of you!

$f(x) = 1 + x + x^2+ \cdots +x^8 + 10x^9 - 8$

and

$g(x) = 1 + x + x^2+ \cdots + x^{10} + 10x^{11} - 8$

We also have $f(0.8) <0, \; f(0.9) >0, \;g(0.8) <0, \;g(0.9) >0$ meaning that both roots lie between $0.8$ and $0.9$. Now consider the difference $h(x) = f(x)-g(x) =9x^8-x^{10}-10x^{11}= -(x+1)(10x-9)x^9 $. On the interval $(0.8,0.9)$ $h(x)> 0$ meaning that $ f(x) > g(x)$ giving that the root of $g(x)$ is larger than the root of $f(x)$ as HallsofIvy pointed out.

Hi

I want to also let you know that my method is so much different than you and... it's less elegant but it works!

- Thread starter
- Admin
- #5

- Feb 14, 2012

- 3,805

My solution:^{2}+ ... + x^{8}= 8 - 10x^{9}, or the real root of x + x^{2}+ ... + x^{10}= 8 - 10x^{11}?

First, notice that

$1-x^9=(1-x)(x+x^2+\cdots+x^8)$ $1-x^9=(1-x)(8-10x^9)$ $1-x^9=8-10x^9-8x+10x^{10}$ $10x^{10}-9x^9-8x+7=0$ | $1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$ $1-x^{11}=(1-x)(8-10x^{11})$ $1-x^{11}=8-10x^{11}-8x+10x^{12}$ $10x^{12}-9x^{11}-8x+7=0$ |

So I let $f(x)=10x^{10}-9x^9-8x+7$ | So I let $g(x)=10x^{12}-9x^{11}-8x+7$ |

Descartes's Rule says $f(x)$ has two positive real roots and obviously $x=1$ is one of the root of $f(x)$ and the other root lies between $(0, 1)$. | Descartes's Rule says $g(x)$ has two positive real roots and obviously $x=1$ is one of the root of $g(x)$ and the other root lies between $(0, 1)$. |

Observe also that

$g(x)=10x^{12}-9x^{11}-8x+7=x^2(10x^{10}-9x^9-8x+7)-8x+7+8x^3-7x^2=x^2f(x)+(8x-7)(x^2-1)$

If $a$ is a root of the function of $f$, then

$g(a)=a^2f(a)+(8a-7)(a^2-1)$

$g(a)=0+(8a-7)(a^2-1)$

$g(a)=(8a-7)(a^2-1)$

And here is a rough sketch of the graph $y=(8a-7)(a^2-1)$

So, if $\dfrac{7}{8}<a<1$, then $g(a)<0$ whereas if $a<\dfrac{7}{8}$, then $g(a)>0$.

Now, our objective is to find out whether the root of the function of $f$ is greater than or less than $\dfrac{7}{8}$.

$f(\dfrac{7}{8})=10(\dfrac{7}{8})^{10}-9(\dfrac{7}{8})^9-8(\dfrac{7}{8})+7=-0.075$

$f(\dfrac{6}{8})=10(\dfrac{6}{8})^{10}-9(\dfrac{6}{8})^9-8(\dfrac{6}{8})+7=0.8873$

Hence, we can say that the other positive real root of $f$, i.e. $a$, lies between $\dfrac{6}{8}$ and $\dfrac{6}{8}$, i.e. $\dfrac{6}{8}<a<\dfrac{7}{8}$, or a is less than $\dfrac{7}{8}$ and this tells us $g(a)>0$.

The graph below shows that the real root of $g(x)$ is larger than the root of $f(x)$.