How to approximately guess the roots of a function

Question

My question is : How to approximately guess the root of a function...

By root i mean is the starting point guess when used in case of Newton's method or any other root formulating methods. (Without calculators!)

---

I searched for the answers and came across the following solutions:

1. Graphs: Of course! The point where the curve intersect with the $x-$axis...We find an integer or a simple decimal number close to it and define it as the starting point!
2. Bisection method: We find out the interval $[a,b]$ where the root lies and if it follows certain conditions (link given) we can use $a$ or $b$ as starting points.

---

But suppose 1) I cannot plot the graph and 2) It is a very complex function (consists of trig., log, etc etc) and guessing the roots is not possible...not even the intervals...

So is there a better method to find the starting point?

---

For example the function is something like : $$\dfrac{\sin{x}}{x}=\log{x}$$ or $$x^8+24x^6+32x^3+12x+1=0$$ (I guess i exaggerated!)

Thanks!

---

P.S. - This may seem like a duplicate. But in all the questions i have seen in SE, none answer these questions (or i may have looked over some) concretely!

Answer 1

I don't think there is a magic answer-you have to think about the specific problem. For your first example, I would note that $\frac {\sin 1}1 \gt 0 = \log 1, \frac {\sin e}e \lt \frac 1e \lt 1=\log e$, so I have a bracket. For the second I would note that $f(0)=1$ and the only terms that can go negative are $32x^3+12x$ and I would note that $f(-1)=-18$ and again I have a bracket. Since this goes to $+\infty$ as $x \to -\infty$ there will be another root below $-1$. $x^8$ is so big that I would try $f(-2)$, find it is positive, and have a bracket on that root.

Answer 2

Here is a reasonable method that would work for finding a zero of a nice differentiable function $f$ with limits at the endpoints of its domain.

Step 1

First find the sign of $f$ as it approaches the two endpoints of its domain. There are only three cases. If $f$ fluctuates in sign near an endpoint, we are done. Otherwise $f$ has stable signs near both endpoints, and the signs are either opposite or the same.

Step 2 - Case 1 (Opposite signs)

Choose $a,b$ such that $f(a),f(b)$ are of opposite signs, by moving $a,b$ sufficiently towards the respective endpoints of the domain. This can be done by binary search for finite endpoints or by repeated doubling for infinite endpoints, but usually it helps to observe the function carefully.

For $f(x) = \ln(x) - \frac{\sin(x)}{x}$, $f(x) \to -\infty$ as $x \to 0$ and $f(x) \to \infty$ as $x \to \infty$. A possible choice of $(a,b)$ would be $(\frac{1}{4},4)$.

Step 2 - Case 2 (Same sign)

If $f$ has any zero at all, its derivative $f'$ must have a zero too, by a suitable extension of Rolle's theorem. So we just repeat the procedure on $f'$. In many cases, eventually we will end up in Case 1.

For $f$ that is a polynomial with even degree, $f(x) \to \infty$ in both directions, but $f'$ will be a polynomial with odd degree, and so Case 1 applies to $f'$.

After we find sufficient approximations of the zeros of $f'$, for at least one of them $f$ will be of the opposite sign compared with the sign of the limits.

Step 3

We now have an interval on which we can do binary search for a zero of $f$.

Notes

For some not nice functions $f$, it will not work well, such as for $f(x) = \frac{1}{x^2+1}(\cos(\frac{1}{2x})-\cos(\frac{1}{3x})-\cos(\frac{1}{5x})+3)$.

Note also that Newton-Raphson fails in a lot of different ways. The conclusion is that it is extremely important to understand the function before attempting to find its zeros. Any algorithm only works for certain types of functions.