I have the information that: $$ x^3 − x^2 −1 =0 $$ Has a "positive real root" of: $x \approx 1.4655\ldots$
My questions are, please:
1) What is a "positive real root".
2) How one gets from the formula to $1.4655$?
3) What is the technique used to solve this in similar problems?
$\endgroup$ 32 Answers
$\begingroup$A positive real root is the point say $x=a$ ($\forall \ \ a>0$) where the curve intersects the x-axis.
In general, to find one real root of the polynomial equation $$a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots +a_1x+a_0=0$$
Apply Newton Raphson's iteration formula given as follows $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$
Where, $f(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\ldots +a_1x+a_0$
$x=x_0$ is the initial expected real root.
Now, we $f(x)=x^3-x^2-1$ $\implies f(x)=3x^2-2x$
Let, initial expected root $x_0=1$ then we have
Iteration-1: $$f(1)=1^3-1^2-1=-1$$$$ f'(1)=3(1^2)-2(1)=1$$
$$x_1=x_0-\frac{f(x_0)}{f'(x_0)}=1-\frac{f(1)}{f'(1)}=1-\frac{(-1)}{1}=2$$
Iteration-2: $$f(2)=2^3-2^2-1=3$$$$ f'(2)=3(2^2)-2(2)=8$$
$$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=2-\frac{f(2)}{f'(2)}=2-\frac{3}{8}=1.625$$
Iteration-3: $$f(1.625)=(1.625)^3-(1.625)^2-1=0.650391$$$$ f'(1.625)=3(1.625^2)-2(1.625)=4.671875$$
$$x_3=x_2-\frac{f(x_2)}{f'(x_2)}=1.625-\frac{f(1.625)}{f'(1.625)}=1.625-\frac{0.650391}{4.671875}\approx 1.485786$$
Iteration-4: $$f(1.485786)=(1.485786)^3-(1.485786)^2-1=0.072402$$$$ f'(1.485786)=3(1.485786^2)-2(1.485786)=3.651108$$
$$x_4=x_3-\frac{f(x_3)}{f'(x_3)}=1.485786-\frac{f(1.485786)}{f'(1.485786)}=1.485786-\frac{0.072402}{3.651108}\approx 1.465956 $$
Iteration-5:
$$f(1.465956)=(1.465956)^3-(1.465956)^2-1=0.001352$$ $$f'(1.465956)=3(1.465956^2)-2(1.465956)=3.515169$$
$$ x_5=x_4-\frac{f(x_4)}{f'(x_4)}=1.465956-\frac{f(1.465956)}{f'(1.465956)}=1.465956-\frac{0.001352}{3.515169}\approx 1.465571 $$
Above root can be approximated as $x\approx 1.4655$
$\endgroup$ 1 $\begingroup$As $x>0$ it is a positive root.
There is no really nice way to get to that solution, one possibilty is to use Cardano's method which gives you a formula for these equations: $$ax^3+bx^2+cx+d=0.$$ If you don't need the exact solution I'd suggest something like Newton's method.
$\endgroup$
