Do the false position method really need that there exists only one root inside $[a; b]$?

Question

I'm studying the False Position Method for finding zeroes of real functions and in the book I'm reading the author says that it is required that only one root of $f$ is contained inside the initially guessed interval $[a; b]$.

Is this really the case? I'm asking because I couldn't find another book or reference that states the same, proving why the method fails otherwise.

As far as I can see, this shouldn't be a problem, given that, whatever is the number of roots in $[a; b]$, at each iteration this interval shrinks.

Could you please state the reason why this is either true or false?

OBS: Please, notice that counterexamples for this are functions and intervals that both meet all the criteria required by the method, except for the number of roots inside the interval, and for which the method can't find any of the roots in a finite number of iterations.

Answer 1

One of the criteria requires that you have $f(a) f(b) < 0$.

If we take the function:

$$f(x) = x^2 - \cos^2 x$$

A plot of this function shows two roots as:

Using a different method, there are two roots at $x = \pm 0.73908513321516064166$.

If we use the False Position Method to meet that initial criteria, we could choose:

• $a = -\dfrac{1}{2}, b = \dfrac{3}{2}$, and we converge in on the positive root of $0.739085133215160641$.
• $a = -2, b = 0$, and we converge in on the negative root of $-0.73908513321516064$.

What do you notice about the choice of the interval? For example, can you choose $(a, b) = (-2, 2)$?

Reading what the author is saying means that you choose ranges for a single root at a time as this method can only find one root at a time.

Does this answer your question?

Update

The question is asking if there is more than one root within an interval, will the method still work?

As an example, we will take $f(x) = \cos x$ and use $(a, b) = (\pi/4, (11 \pi)/4)$. A plot of the $f(x)$ over this range shows:

As can clearly be seen in the plot, there are in fact three roots over this range and we $f(a) > 0$, $f(b) < 0$.

When we apply the False Position Method, it does indeed converge (in two steps) to the root $x = 4.71238898038468988$. The reason is that this method is finding find the x-intercept of the straight line connected by two points $((a,f(a), (b, f(b))$. We can depict this graphically as:

In this analysis, you can see the x-intercept is that root found by the algorithm. So, there is no problem in finding a root. The trouble is that unless you do a similar analysis, you would not be sure apriori to which root, unless you do the algorithm. Other methods have similar problems, but they still work as advertised.

I think the author is trying to point out that if you are trying to find a root within an interval and there are multiple roots, you might not get the correct one and you should use whatever is at your disposal to narrow the range down to a single root.

You should also compare and contrast the pros and cons of this method when compared to things like the Secant Method.