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- Feb 14, 2012
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Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?
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.
Thanks for participating to both of you!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.
My solution:Which is larger, the real root of x + x2 + ... + x8 = 8 - 10x9, or the real root of x + x2 + ... + x10 = 8 - 10x11?
$1-x^9=(1-x)(x+x^2+\cdots+x^8)$ $1-x^9=(1-x)(8-10x^9)$ $1-x^9=8-10x^9-8x+10x^{10}$ $10x^{10}-9x^9-8x+7=0$ | $1-x^{11}=(1-x)(x+x^2+\cdots+x^{10})$ $1-x^{11}=(1-x)(8-10x^{11})$ $1-x^{11}=8-10x^{11}-8x+10x^{12}$ $10x^{12}-9x^{11}-8x+7=0$ |
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 $(0, 1)$. |