Real Numbers, Sequences and Series: Part 7


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.


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:


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


Hence, 1.835<\xi<1.9.

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


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.


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