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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.

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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:

enter image description here

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:

enter image description here

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:

enter image description here

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.

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  • $\begingroup$ Actually no, exactly because in this function we can't choose $a$ and $b$ such that the first criterion is met. I already knew that this was a requisite, or else we would not know how to shrink the interval maintaining a root inside it, so my concern is really about the situations where we can get $a$ and $b$ like that which englobe more than one root. Think of the function $f(x) = \mathrm{cos}(x)$ and $a=\frac{\pi}{4}$ and $b=\frac{7\pi}{4}$ for example. Can we get one of the two roots in $[a; b]$? $\endgroup$
    – araruna
    Jun 15, 2013 at 20:29
  • $\begingroup$ I am confused. What is $cos(\frac{pi}{4})$? What is $cos(\frac{7 pi}{4})$ Those values are equal and positive and do not match the condition. We need for the interval values to give us a negative and a positive result so we can begin to apply this method. Am I missing some fundamental part of your question? The example I gave has two roots inside of $(2,2)$, but we cannot use that range for our interval as it does not meet the criteria. We can have a multiplicative root at a point, but this method cannot handle those. $\endgroup$
    – Amzoti
    Jun 15, 2013 at 21:00
  • $\begingroup$ @araruna: Maybe this needs to be stated a different way. If we have f(a) positive and f(b) negative, what does that imply? What is this telling you about your choice for an interval? $\endgroup$
    – Amzoti
    Jun 15, 2013 at 21:22
  • $\begingroup$ Excellent presentation here! $\endgroup$
    – amWhy
    Jun 16, 2013 at 0:27
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    $\begingroup$ @araruna: No problem, I make all sorts of errors all the time. Please see my update to the question. $\endgroup$
    – Amzoti
    Jun 16, 2013 at 18:16

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