Finding one pair roots of the equation performed through the continued fractions

With the help of the rule of continued fractions we should find one pair root of this inexact equation:

(4.1)

For example, we have an equation like this:

(4.2)

Before solving it let’s convert the fraction into a continued form. And it is clear that:

(4.3)

Now let’s find the second convergent fraction from the end of the last continued fraction (it is clear that the first convergent fraction from the end is equal to ). It certain that it is of the value . As the found convergent fraction is at the even place, its value is greater than , and the difference is in the value of . Therefore an expression

(4.4)

is performed. In order to make the equation (4.4) similar to the inexact equation, let’s increase all the members of the last equation by 13 and write it in the following form: or

(4.5)

Now let’s compare the equation (4.5) and the given equation, if we take -26 instead of x - and 39 instead of y- they are appeared to be mutually exact. Consequently, one pair root of the equation is found; it is: x=-26, y=-39. That is why now it is possible to find all the roots of the equation:.

They will be found through these chains:

As it is no conditional value for number , if we make expression as a condition of the upper chain, chains of roots appeared in a convenient form like this:

Example 2: We have equation . Let’s convert into the continued fraction. So it will be as this:

Let’s find a convergent fraction second from the end, it is . As the taken convergent is at the uneven place, its meaning is less than . That is why:

Hereof expression is performed. If we multiple members of the last equation by 5, we can find one pair roots of the equation given through expression . They are:

Consequently, the value of all roots of the given equation is found from these chains:

As the number has a conditional value, if we change for , previous chains will have the following form:

Chapter II

Дата добавления: 2015-11-14