Real Numbers, Sequences and Series: Part 7

Exercise.

Discover (and justify) an essential difference between the decimal expansions of rational and irrational numbers.

Giving a decimal expansion of a real number means that given n \in N, we can find a_{0} \in Z and 0 \leq a_{1}, \ldots, a_{n} \leq 9 such that

|x-\sum_{k=0}^{n}\frac{a_{k}}{10^{k}}|< \frac{1}{10^{n}}

In other words if we write

x_{n}=a_{0}+\frac{a_{1}}{10}+\frac{a_{2}}{10^{2}}+\ldots +\frac{a_{n}}{10^{n}}

then x_{1}, x_{2}, x_{3}, \ldots, x_{n}, \ldots are approximate values of x correct up to the first, second, third, …, nth place of decimal respectively. So when we write a real number by a non-terminating decimal expansion, we mean that we have a scheme of approximation of the real numbers by terminating decimals in such a way that if we stop after the nth place of decimal expansion, then the maximum error committed by us is 10^{-n}.

This brings us to the question of successive approximations of a number. It is obvious that when we have some approximation we ought to have some notion of the error committed. Often we try to reach a number through its approximate values, and the context determines the maximum error admissible. Now, if the error admissible is \varepsilon >0, and x_{1}, x_{2}, x_{3}, \ldots is a scheme of successive is approximation of a number x, then we should be able to tell at which stage the desired accuracy is achieved. In fact, we should find an n such that |x-x_{n}|<\varepsilon. But this could be a chance event. If the error exceeds \varepsilon at a later stage, then the scheme cannot be a good approximation as it is not “stable”. Instead, it would be desirable that accuracy is achieved at a certain stage and it should not get worse after that stage. This can be realized by demanding that there is a natural number n_{0} such that |x-x_{n}|<\varepsilon for all n > n_{0}. It is clear that n_{0} will depend on varepsilon. This leads to the notion of convergence, which is the subject of a later blog.

More later,

Nalin Pithwa

Real root of an equation

Problem. Find the real root of the equation f(x)=x^{3}-2x^{2}+3x-5=0

with an accuracy upto 10^{-4}. Use the method of chords.

Solution:

Let us first make sure that the given equation has only one real root. This follows from the fact that the derivative

f^{'}(x)=3x^{2}-4x+3>0 (use your knowledge of theory of equations here!!!)

Then, from f(1)=-3<0 and f(2)=1>0, it follows that the given polynomial has a single positive root, which lies in the interval (1,2).

Using the method of chords, we obtain the first approximation:

x_{1}=1-(-3/4).1=1.75

Since f(1.75)=-0.5156<0 and f(2)=1>0, then 1.75<\xi<2.

The second approximation:

x_{2}=1.75+(0.5156/1.5156).o.25=1.8350 and

since f(1.835)=-0.05059<0, then 1.835<\xi<2.

The sequence of approximations converges very slowly. Let us try to narrow down the interval, taking into account that the value of the function f(x) at the point x_{2}=1.835 is considerably less in absolute value than f(2). We have

f(1.9)=0.339>0.

Hence, 1.835<\xi<1.9.

Applying the method of chords to the interval (1.835,1.9), we will get a new approximation:

x_{3}=1.835-(\frac{-0.05059}{0.339+0.05059})(0.065)=1.8434

Further calculations by the method of chords yield

x_{4}=1.8437 and x_{5}=1.8438

and since f(1.8437)<0 and f(1.8438)>0, then \xi \approx 1.8438 with required accuracy of 10^{-4}.

You can also solve this problem by the method of tangents. Try it!

More later,

Nalin Pithwa

Fifth degree polynomial equation

Question: Prove that the equation 3x^{5}+15x-8 has only one real root.

Proof:

The existence of at least one real root follows from the fact that the above polynomial is an odd power.

Let us prove the uniqueness of such a root by contradiction.

Suppose there exist two roots x_{1}<x_{2}. Then, in the interval [x_{1},x_{2}] the given polynomial function satisfies all the conditions of Rolle’s Theorem: it is continuous, it vanishes at the end points and has derivative at all points. Consequently, at some point \xi such that x_{1}<\xi<x_{2},

f^{'}(\xi) equals zero. But, f^{'}(x)=15(x^{4}+1)>0. But, this contradicts hypothesis. Hence, the proof.

More later,

Nalin Pithwa