MHB Which Real Root is Larger: x^8 = 8 - 10x^9 or x^10 = 8 - 10x^11?

  • Thread starter Thread starter anemone
  • Start date Start date
  • Tags Tags
    Root
AI Thread Summary
The discussion focuses on comparing the real roots of two polynomial equations: x + x^2 + ... + x^8 = 8 - 10x^9 and x + x^2 + ... + x^10 = 8 - 10x^11. The consensus is that the second equation has a larger root, approximately 0.884, compared to the first equation's root of about 0.882. Participants used graphing techniques to identify and verify these roots. One user noted the complexity of analyzing the behavior of the functions within a specific interval. Overall, the second equation is confirmed to have the larger real root.
anemone
Gold Member
MHB
POTW Director
Messages
3,851
Reaction score
115
Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?
 
Mathematics news on Phys.org
The second one. The first has root about 0.882, the second, bout 0.884. I got that by graphing both and then "zooming" in on the zeros.
 
Here's my solution
It's not hard to show that the derivative of each is positive so both are increasing function meaning there's only one root. Let

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

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.
 
HallsofIvy said:
The second one. The first has root about 0.882, the second, bout 0.884. I got that by graphing both and then "zooming" in on the zeros.

Jester said:
Here's my solution
It's not hard to show that the derivative of each is positive so both are increasing function meaning there's only one root. Let

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

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.

Thanks for participating to both of you!:)

Hi Jester, it took me a fraction of time to decipher why $h(x)> 0$ on the interval $(0.8,0.9) in your approach!:o Well done!

I want to also let you know that my method is so much different than you and... it's less elegant but it works!:o
 
anemone said:
Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?

My solution:
First, notice that

[TABLE="class: grid, width: 500"]
[TR]
[TD]$1-x^9=(1-x)(x+x^2+\cdots+x^8)$

$1-x^9=(1-x)(8-10x^9)$

$1-x^9=8-10x^9-8x+10x^{10}$

$10x^{10}-9x^9-8x+7=0$[/TD]
[TD]$1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$

$1-x^{11}=(1-x)(8-10x^{11})$

$1-x^{11}=8-10x^{11}-8x+10x^{12}$

$10x^{12}-9x^{11}-8x+7=0$[/TD]
[/TR]
[TR]
[TD]So I let

$f(x)=10x^{10}-9x^9-8x+7$[/TD]
[TD]So I let

$g(x)=10x^{12}-9x^{11}-8x+7$[/TD]
[/TR]
[TR]
[TD]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 $(0, 1)$.[/TD]
[TD]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 $(0, 1)$.[/TD]
[/TR]
[/TABLE]

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

View attachment 1544

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 $f(x)$.

View attachment 1545
 

Attachments

  • Graph 2.JPG
    Graph 2.JPG
    25.1 KB · Views: 105
  • Graph 3.JPG
    Graph 3.JPG
    25.7 KB · Views: 105
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top