# 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}. 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,

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

More later,

Nalin Pithwa