Am I Capable of Tricking Mr. Haussmann?

Calculus Level 5

$x^9+x^8+2x^7+2x^6+2x^5+2x^4+2x^3+3x^2+x+1=0$

Let $y$ be the real root of the equation above. Find $\lfloor |10000y| \rfloor$ .

Notations :

$\lfloor \cdot \rfloor$ denotes the floor function .
$| \cdot |$ denotes the absolute value function .

The answer is 11004.

2 solutions

Mark Hennings
Jun 21, 2016

The iterative scheme $x_{n+1} \; = \; g(x_n) \;= \; -1 - \frac{x_n^2}{x_n^8 + 2x_n^6 + 2x_n^4 + 2x_n^2 + 1} \qquad \qquad n \ge 0$ with $x_0 = -1.1$ is rapidly convergent (since $-\tfrac14 \le g'(x) < 0$ for all $x \le -1$ ) to the root $y \approx -1.1$ . The sequence is oscillatory, with $x_2 \,=\, -1.100413996... \qquad \qquad x_3 \,=\, -1.100443736...$ Thus $y$ lies between these two values, so $|10000y|$ has integer part $\boxed{11004}$ .

Do you think this should be level 3?

Sal Gard - 4 years, 11 months ago

As I understand things, Level is determined by success rate, but the algorithm is not clear.

Once you think of using an iterative scheme, it is fairly easy to find a convergent one. Without that idea...

Isn't the real problem that this can just be plugged into WA for an answer? Reasonably enough, the Level drops if you can get the answer just by typing.

Mark Hennings - 4 years, 11 months ago

Thanks for the response.

Sal Gard - 4 years, 11 months ago

Great approach as always. I applied Newton-Raphson method.

Sal Gard - 4 years, 11 months ago

I did it via plotting its graph on desmos. How did you get the value of $x_0$ ?

Tapas Mazumdar - 4 years, 2 months ago

Like always in iterative cases, a reasonable guess.

Mark Hennings - 4 years, 2 months ago

Niranjan Khanderia
Jul 2, 2017

I did it via plotting its graph on desmos.

Am I Capable of Tricking Mr. Haussmann?

The answer is 11004.

2 solutions

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