Quintic Pi Polynomial

Let $f(x) = \color{#D61F06}{3}x^{5} - \color{#D61F06}{1}x^{4} - \color{#D61F06}{4}x^{3} - \color{#D61F06}{1}x^{2} - \color{#D61F06}{5}x - \color{#D61F06}{9}$

We need to find the only real root of $f(x) = 0$ .

Observe that since $f(x)$ is a odd degree polynomial and has only one real root $\alpha$ , we can say that

• $f(x) < 0$ if $x < \alpha$ ,

• $f(x) = 0$ if $x = \alpha$ and

• $f(x) > 0$ if $x > \alpha$ .

Note that $f(0) = -9 < 0$ . Hence the root is greater than 0.

• $f(1) = -17 < 0$ , therefore $\alpha > 1$

• $f(2) = 25 > 0$ , therefore $\alpha< 2$

We can conclude that $1 < \alpha < 2$ . Thus, 1 and 2 are lower and upper bounds of $\alpha$ . We can use the bisection method to make these bounds tighter and find the approximate value of the root to great precision.

The initial lower and upper bounds are 1 and 2 respectively.

The algorithm evaluates the function $f(x)$ at the arithmetic mean of the bounds, i.e. (Lower Bound + Upper Bound)/2 .

• If the $f(\frac{L + U } {2}) < 0$ , then $\alpha > \frac{L + U } {2}$ and $\frac{L + U } {2}$ is a better lower bound than $L$ , therefore the lower bound is updated to $\frac{L + U } {2}$ .

• If the $f(\frac{L + U } {2}) > 0$ , then $\alpha < \frac{L + U } {2}$ and $\frac{L + U } {2}$ is a better upper bound than $U$ , therefore the upper bound is updated to $\frac{L + U } {2}$ .

By repeatedly applying this algorithm, the upper and lower bounds get pretty close to each other, and we can find the value of $\alpha$ to as much precision as required.

The following python code is an implementation of the bisection algorithm. The while loop repeats 20 times, and the final bounds are 1.78072547913 and 1.7807264328. Thus, $\alpha \approx 1.7807$ and $\lfloor 1000 \times \alpha \rfloor = 1780$ $_\square$

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def f(x):
    return 3*x**5 - x** 4 - 4* x**3 - x**2 - 5*x - 9

lower_bound = 1.0

upper_bound = 2.0

while i < 20:
    if f((lower_bound + upper_bound)/2.0) > 0:
        upper_bound = (lower_bound + upper_bound)/2.0
    elif f((lower_bound + upper_bound)/2.0) < 0:
        lower_bound = (lower_bound + upper_bound)/2.0  

    i += 1
    print lower_bound, upper_bound

We can carry out the same calculation by hand too. It is tedious, but has less error than using computers since floats are not exact

We will now test $f(\frac{1 + 2 } {2} )$

• $f(1.5) \approx -14.53$ , therefore $\alpha > 1.5$

We will now test $f(\frac{1.5 + 2 } {2} )$

• $f(1.75) \approx -2.3896$ , therefore $\alpha > 1.75$

We will now test $f(\frac{1.75 + 2 } {2} )$

• $f(1.875) \approx 8.90542$ , therefore $\alpha < 1.875$

We will now test $f(\frac{1.75 + 1.875 } {2} )$

• $f(1.8125) \approx 2.7256$ , therefore $\alpha < 1.8125$

We will now test $f(\frac{1.75 + 1.8125 } {2} )$

• $f(1.78125) \approx 0.04281$ , therefore $\alpha < 1.78125$

We will now test $f(\frac{1.75 + 1.78125 } {2} )$

• $f(1.765625) \approx -1.20375$ , therefore $\alpha > 1.765625$

We will now test $f(\frac{1.765625 + 1.78125 } {2} )$

• $f(1.7734375) \approx -0.588817$ , therefore $\alpha > 1.7734375$

We will now test $f(\frac{1.7734375 + 1.78125 } {2} )$

• $f(1.77734375) \approx -0.2746247$ , therefore $\alpha > 1.77734375$

We will now test $f(\frac{1.77734375 + 1.78125 } {2} )$

• $f(1.779296875) \approx -0.116396$ , therefore $\alpha > 1.779296875$

We will now test $f(\frac{1.779296875 + 1.78125 } {2} )$

• $f(1.7802599375) \approx -0.0380174$ , therefore $\alpha > 1.7802599375$

We will now test $f(\frac{1.779296875 + 1.78125 } {2} )$

• $f(1.7802599375) \approx -0.0380174$ , therefore $\alpha > 1.7802599375$

We will now test $f(\frac{1.7802599375 + 1.78125 } {2} )$

• $f(1.7807546875) \approx 0.00236288$ , therefore $\alpha < 1.7807548775$

We see that $1.7802599375 < \alpha < 1.7807548775$ , therefore $1780.2599375 < 1000 \alpha < 1780.7548775$ .

Thus, the greatest integer less than equal to $1000 \times \alpha$ is $\boxed{1780}$ . $_\square$

First get a general idea about where to find the root $x_0$ . Define

$\begin{aligned} p(x)&:=-3x^5+x^4+4x^3+x^2+5x+9,&p(1)&=17>-25 = p(2)&\Rightarrow&&x_0&\in(1;\:2) \end{aligned}$

I used fix point iteration - guessing there was a reason why there was a left hand side in the assignment. Taking $\begin{aligned} \varphi(x)&:=3^{-\frac{1}{5}}\left(x^4+4x^3+x^2+5x+9\right)^{\frac{1}{5}},&x_{k+1}&:=\varphi(x_k),&&x_1:=1 \end{aligned}$

we need 10 iteration for the fourth digit to remain constant, leading to

$\begin{aligned} \begin{array}{r | l} k &1000x_k\\\hline 1&1461.4425516219254\\ 2&1639.0698520255442\\ 3& 1716.575631186962\\ 4& 1751.4572849684504\\ 5& 1767.3306998553346\\ 6& 1774.5871029154345\\ 7& 1777.910910421906\\ 8& 1779.434744225369\\ 9&1780.1336444675542\\ 10&\boxed{1780}.4542515232284\\ \end{array} && \text{Check:}\quad p(1.780) &> 0.05 > -0.02 > p(1.781) & \Rightarrow && x_0 &\in (1.780;\:1.781) \end{aligned}$

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/* maxima code */
numer : true;
x0 : 1;

/* iteration function */
p(x) := x^4 + 4*x^3 + x^2 + 5*x + 9 - 3*x^5;
f(x) := ( (x^4 + 4*x^3 + x^2 + 5*x + 9) / 3)^(1/5);

/* fix point iteration */
for i : 1 thru 20 do(
    x0 : f(x0),
    printf(true, "k: ~d~txk: ~f~%", i, x0)
);