Let $$f(x)$$ be a continuous function defined in an interval $$[a,b]$$.
Then, if $$f(a) \cdot f(b) <0 $$ (therefore, $$f(a)<0$$ and $$f(b)>0$$ or $$f(a)>0$$ and $$f(b)<0$$), there exists at least a point $$c$$ inside the interval $$(a,b)$$ such that $$f(c)=0$$.
This theorem can turn out to be very intuitive since if we have a continuous function that is negative in $$f (a)$$ (below the $$x$$ axis) and is positive in $$f (b)$$ (above the $$x$$ axis), or vice versa, given that the function is continuous and has to connect the points $$f (a)$$ and $$f (b)$$, the graph will cross the $$x$$ axis at least once, so a value $$c$$ will exist inside the interval so that $$f (c) =0$$.
Let's see an example of the application of the theorem:
Let's take the function $$f(x)=x-\ln^2x$$ defined in the interval $$[0.1,0.5]$$ and we wonder whether it has a zero within the interval.
We observe that: $$$\begin{array}{l} f(0.1)= -5.201898111 < 0 \\ f(0.5)=0.0195469860 >0 \end{array}$$$
As the interval is closed and the function is continuous, the hypotheses of Bolzano's theorem are satisfied and consequently it can be applied.
The theorem says that t a point $$c$$ inside the interval $$[0.1,0.5]$$ exists such that $$f(c)=0$$.
Consequently it will be satisfied that $$0=f(c)=c-\ln^2c$$, or $$c=\ln^2c$$.
Notice that we found the existence of a solution in the interval $$[0.1,0.5]$$ of the equation $$x=\ln^2x$$ which could not be previously solved.
Let's see if the equation $$x^3+\ln x=- \sqrt{x}$$ has a solution.
We will consider the function $$f(x)=x^3+\ln x+\sqrt{x}$$.
If we find two points $$a$$ and $$b$$ where $$f (a) \cdot f (b) < 0 $$, it will mean that a certain value $$c$$ exists between $$a$$ and $$b$$ such that it will be a solution to our equation.
If $$a= 0.1 \Longrightarrow f(0.1)=0.1^3+\ln 0.1+\sqrt{0.1}=-1.985357 < 0 $$
If $$b=1 \Longrightarrow f(1)=1^3+\ln 1+ \sqrt{1}=2>0$$
Therefore, in the interval $$[0.1,1]$$ a point $$c$$ exists such that $$f(c)=0$$ and we can say our equation has a solution (since $$0=f(c)=c^3+\ln c+ \sqrt{c}$$).
Application of the theorem
Now, using Bolzano’s theorem, we can define a method to bound a zero of a function or a solution in an equation:
- To find an interval where at least one solution exists by Bolzano.
- To divide the interval in $$2$$ subintervals (dividing it by half, for example).
- To evaluate the function at the median point and depending on the sign of the value, repeat the process in the new subinterval where the Bolzano conditions are satisfied ($$f (a) \cdot f (b) <0$$).
This process can be done as many times as we want, until we eventually find a very small interval where it is known with certainty that a zero of our function exists.
Let's see an example of an application of the algorithm:
Let's see if the equation $$x^2 \sin x= \ln x$$ has a solution.
Before starting we will consider the function $$f(x)=x^2\sin x- \ln x$$. Finding the zeros of this function is equivalent to finding the solutions to our equation.
- First, we will look for a closed interval where Bolzano’s theorem is satisfied, containing two values, $$a$$ and $$b$$, where the function $$f(x)$$ is positive and negative respectively:
We will try with different values: $$$\begin{array}{rcl} \mbox{if } x=2 & \Longrightarrow & f(2)=2^2\sin 2-\ln 2=2.94404 \ldots >0 \\ \mbox{if } x= 6 &\Longrightarrow & f(6)=6^2-\sin 6- \ln 6= -11.8507 \ldots < 0 \end{array}$$$ Therefore, in the interval $$[2,6]$$ a point $$c$$ exists such that $$f(c)=0$$.
We have bound our solution, but of course we can improve the interval.
- We will divide our interval in two subintervals: $$$\begin{array} {l} I_1=[2,4] \\ I_2=[4,6]\end{array}$$$ and we are going to evaluate the function at the median point: $$f(4)=-13.495 \ldots < 0$$
Since we had $$f(2)>0$$, in the interval $$I_1=[2,4]$$ the theorem will be satisfied.
- Now we are going to repeat the same process until we find the bounds that we need.
We divide again the interval: $$$\begin{array}{l} I_1=[2,3] \\ I_2=[3,4] \\ f(3)=0.171467 >0 \end{array}$$$ And we are left with the interval $$I_2=[3,4]$$, since $$f(3)>0$$ and $$f(4) < 0$$.
We divide again the interval: $$$\begin{array}{l} I_1=[3,3.5] \\I_2=[3.5,4]\\f(3.5)=-5.54 \ldots < 0\end{array}$$$ We are left with the interval $$I_1=[3,3.5]$$.
We divide again the interval: $$$\begin{array}{l} I_1=[3,3.25] \\I_2=[3.25,3.5]\\f(3.25)=-2.32 \ldots < 0 \end{array}$$$ We are left with the interval $$I_1=[3,3.25]$$.
We divide the interval for the last time: $$$\begin{array}{l} I_1=[3,3.125] \\I_2=[3.125,3.25]\\f(3.125)=-0.97 \ldots < 0\end{array}$$$ We are left with the interval $$I_1=[3,3.125]$$.
In conclusion, the equation $$x^2\sin x=\ln x$$ has a solution inside the interval $$I_1=[3,3.125]$$.
For example, we might bring the solution near the median point of the interval: $$x=3.0625$$, and we would obtain the expressions: $$$f(3.0625)=-0.378 \ldots \simeq 0$$$ or on the other hand $$$x^2\sin x= \ln x \Rightarrow 0.741029 \ldots \simeq 1.1119 \ldots$$$
We would remind you that when doing more iterations a better approximation of the exact solution may be found.