Unit - 4
Series of functions
A series is the sum of the terms of a sequence. Thus if is a sequence then the sum of all the terms is called an infinite series and it is denoted by
The sequence is called the sequence of partial sums of the series and the partial sums so on, is regarded as the approximation to the full infinite sum of the series.
If the sequence { of partial sums converges the series is termed as convergent and lim is said to be the sum of the series.
Note- an infinite series is said to converge, diverge if its sequence of partial sums converges, diverges respectively.
Necessary condition for convergence
Theorem- A necessary condition for convergence of an infinite series is that
Let
So that { is the sequence of partial sums.
Since the series converges, therefore the sequence also converges.
Consequently
Now
So that
Hence for a convergent series
i.e. series cannot converge if its n’th term does not tend to zero.
Example: Show that the series given below is not convergent.
Sol:
Here we have
Here is not zero, therefore the given series is not convergent.
Key takeaways
- A series is the sum of the terms of a sequence. Thus if is a sequence then the sum of all the terms is called an infinite series and it is denoted by
2. An infinite series is said to converge, diverge if its sequence of partial sums converges, diverges respectively.
Continuous function-
Suppose f is the function, then it is said to be continuous at a point c, a < c < b, if
Note- A function f is said to be continuous from the left at c if
A function f is said to be continuous from the right at c if
Hence a function is continuous at c if it is continuous from the left as well as from right.
Continuity at an end point- A function f defined on interval [a, b] is said to be continuous at the end point ‘a’ if it is continuous from the right at ‘a’
Similarly, the function is continuous at the end point b of [a, b] if
Continuity in an interval- A function f is said to be continuous in an interval [a, b] if it is continuous at every point of the interval.
Theorems on continuity-
Theorem- if f and g be two functions continuous at a point c then the function f + g, f – g, fg are also continuous at c.
Theorem- A function f defined on an interval I, is continuous at a point c , and is a sequence such that
Since f is continuous at c, so that for any given
Such that
Again since , therefore a positive integer m, such that
From equation (1), put , we get
The sequence converges to f(c).
Or
Now let f is not continuous at c.
We will now show that through a sequence converging to c yet the sequence does not converges to f(c).
Since f is not continuous at c, therefore there exists an
Such that for every, there exists an x such that
Also
By taking , we find that for each positive integer n, there is a , such that
Thus the sequence does not converges to f(c).
Example: Prove that the function defined by
Is continuous at x = 0
Sol:
Now
So that
Therefore, the function is continuous at x = 0.
Types of discontinuities-
- A function f is said to be a removable discontinuity at x = c if exists but it is not equal to the value of f(c) of the function.
Such a discontinuity can be removed by assigning a value to the function at x = c.
2. The function f is said to have a discontinuity of the first kind at x = c if and both exists but unequal.
3. The function f is said to have a discontinuity of the first kind from left at x = c if exists but not equal to f(c).
4. The function f is said to have a discontinuity of the second kind at x = c if neither nor exists.
5. The function f is said to have a discontinuity of the second kind from left at x = c if does not exists
Example: Discuss the kind of discontinuity if any, of the following function-
Sol:
The function is continuous at all points except the origin.
Let’s test at x = 0
And
Thus the function has discontinuity of the first kind from the right at x = 0.
Example: When [x] denotes the largest integer, then discuss the continuity at x = 3 for the function,
Sol:
Here
And
Thus the function has a discontinuity of the first kind from the left at x = 3.
Example: Show that the following function is continuous at origin-
Sol:
Here
Thus
Also
Therefore, the function is continuous at the origin.
Uniform continuity
A function f defined on an interval I is said to be uniformly continuous on I if to each there exists a , such that
Theorem: A function which is uniformly continuous on an interval, is continuous on that interval
Suppose the function f is uniformly continuous on an interval I, so that for a given , there exists a such that
Where are any two points of I for which |
Let , then by taking we find that for
Such that
Hence the function is continuous at every point , which means the functions f is continuous on I.
Derivability in an interval
A function defined on [a, b] is derivable at the end point a, which means f’(a) exists if
Or we can say that
Similarly the function is derivavble at the end point b, if
Or
Example: Show that the function f(x) = |x| is not derivable at x = 0.
Sol:
Left hand derivative –
Right hand derivative –
Thus
Hence the function is not derivable at x = 0.
Theorem- A function which is derivable at a point is necessarily continuous at that point.
Poof:
Suppose the function is derivable at x = c
So that
Now
Taking limits as x tends to c, we get
So that , therefore the function f is continuous t x = c.
Note-
If the functions f, g are derivable at c then the functions f + g, f – g, fg and f/g [are also derivable at c and
Key takeaways:
- Suppose f is the function, then it is said to be continuous at a point c, a < c < b, if
2. If f and g be two functions continuous at a point c then the function f + g, f – g, fg are also continuous at c.
3. A function f defined on an interval I, is continuous at a point c , and is a sequence such that
4. A function f defined on an interval I is said to be uniformly continuous on I if to each there exists a , such that
5. A function defined on [a, b] is derivable at the end point a, which means f’(a) exists if
Cauchy’s convergence theorem-
A series of real function, each defined on a set X converges uniformly on X iff for every, such that
Proof:
Let be a partial sum.
So that is a sequence of partial sums of the series . Now the series is uniformly convergent iff the sequence is uniformly convergent.
It means for the given a positive integer m such that n
By Cauchy’s criteria of uniform convergence of sequence
Which is the complete proof of Cauchy’s criterion for series.
Weierstrass M-Test
Let < be a sequence of functions defined on E and suppose where is independent of x.
Then converges uniformly as well as absolutely on E if converges.
Proof:
In order to prove uniform convergence, we see that
But since is convergent, given , there exists N(which is independent of x) such that
Hence
And so converges uniformly by the Cauchy’s criterion for uniform convergence.
Example: Prove that
Is uniformly convergent.
Sol.
We assume that x is positive, for if x is negative.
We can change the signs of all terms.
We have
And
Thus maximum value of is
Hence
Since
is convergent, Weirstrass’s M-test implies that is uniformly convergent for all
Power series
A series of the form
Where all the ‘a’ are independent of x, is known as power series in x
A power series is an infinite series of the form where are the coefficients.
Convergence of power series
For x = 0, every power series is convergent.
There are three cases about the convergence of power series
1. The series converges for only x = 0 which is trivial point of convergence, then it is called “nowhere convergent”.
2. The series converges absolutely for all values of x, then it is called “everywhere convergent”.
3. The series converges for some values of x and diverges for others.
Note- the collection of points x for which the series is convergent is called its region of convergence.
Definition: let be a power series. Then applying Cauchy’s root test we see that the power series is convergent if
Where
The series is divergent if
Taking
We will prove that the power series is absolutely convergent if |x| < R and divergent if |x| > R.
If are real and if x is real, we get an interval –R < x < R inside which the series is convergent.
If x is replaced by a complex number z then the power series converges absolutely at all points z inside the circle |z| = R and does not converge at any point outside the circle.
This circle is called the circle of convergence and R is known as the radius of convergence.
The interval (-R, R) is called interval of convergence.
Definition: For the power series the radius of convergence is defined as
Provided the limit exists.
Interval for convergence-
In this power series-
D’Almbert’s ratio test-
If , then by ratio test, the power series is convergent, when is less than 1.
Or in other way if |x|<1/ then the series converges and diverges for other values.
Thus the interval of convergence of power series is-
Example: Find the radius of convergence of the following series
Sol.
Here we have
The n’th term of the series will be
We know that
Then
So series converges for all x where |x| < 3/2.
Example: Find the radius of convergence of the following series
Sol:
Here we have
Now, the n’th term of the series will be
We know that
So the series is convergent for all x where |x| < e.
Example-1: If the series converges, then find the value of x.
Sol. Here
Then,
By D’Almbert’s ratio test the series is convergent for |x|<1 and divergent if |x|>1.
So at x = 1
The series becomes-
At x = -1
This is an alternately convergent series.
This is also convergent series, p = 2
Here, the interval of convergence is
Example-2: If the series converges, then find the value of x.
Sol. Here
Then,
By D’Almbert’s ratio test the series is convergent for ||<1 or |1-x|>1
Or
At x = 0, the series becomes- which is divergent harmonic series.
At x = 2, the series becomes-
It is an alternate series which is convergent by Leibnitz rule.
So that the series.
Cauchy-Hadamdard theorem- If R is the radius of convergence of the power series , then the series is absolutely convergent if |x| < R and divergent if |x| > R.
Proof:
Since the n’th term of the series is , for each x belongs to R, let
, then
Case-1: suppose 0 < R < +
In this case,
So if |x| < R, then and hence the power series is convergent by Cauchy root test, and
If |x| > R, then . So power series is divergent by Cauchy root test.
Case-2: suppose R = +
Then , therefore for each x belongs to R. Hence by Cauchy root test, power series is convergent for each
Case-3: Suppose R = 0
In this case, , so that for x , therefore by Cauchy root test, the power series is divergent for x
Example: Find the interval of convergence of the following power series-
Sol:
Consider the series
Here
Hence, the series converges for all . So that the interval of convergence is (-
Example: Find the interval of convergence of the following power series-
Sol:
Consider the series,
Here
Therefore
Hence, the series converges for all . So the interval of convergence is
Key takeaways
- A series of the form
Where all the ‘a’ are independent of x, is known as power series in x
2. This circle is called the circle of convergence and R is known as the radius of convergence.
3. Cauchy-Hadamdard theorem- If R is the radius of convergence of the power series , then the series is absolutely convergent if |x| < R and divergent if |x| > R.
What is power series?
A power series is a type of series with terms involving a variable. Evidently, if the variable is x, then all the terms of the series involve powers of x. So we can say that a power series can be design of as an infinite polynomial.
Let the power series have the radius of convergence R > 0.
Let
Then for |x| < R the function is continuous.
Differentiation of power series-
The differentiation of power series is given by
Integration of power series-
The power series can also be integrated term-by-term on an interval lying inside the interval of convergence.
Hence if –R < b < x < R,
When the series is integrated on the interval [0, x], it can be written as
Where C is the constant.
Evaluating the derivative and indefinite integral in this manner is called term-by-term differentiation of a power series and term-by-term integration of a power series respectively.
Abel’s theorem (first form)
If a power series converges at the point R of the interval of convergence (-R, R), then it uniformly converges in the interval [0, R].
Proof:
Let us consider the sum
Then we have
And so on.
This gives
…..(1)
Let e > 0 be given
Now the series is convergent at x = R.
The series of numbers is convergent and hence by Cauchy’s general principal of convergence, there exists an integer N such that
Now if we take , then we have
……(3)
Consider for all
Thus we have proved
Hence by Cauchy’s criterion of convergence of series, the series converges uniformly on [0, R].
Note-
1. In case, a power series with interval of convergence (-R, R) converges at x = -R , then the series is uniformly convergent in [-R, 0].
Similarly, if a series convergent at the end points -R and R, then the series is uniformly convergent on [-
R, R].
2. If a power series with interval of convergence (-R, R) diverges at end point x R , then it cannot be uniformly convergent on [0, R].
Abel’s theorem (Second form)
Let be a power series with finite radius of convergence R and let .
If the series converges at end point x = R then
Proof:
If we take R = 1, then there is no loss of generality,
This is the power series with radius R’, where
So if any series is given we can transform it in another power series with unit radius of convergence.
We can take R = 1.
Thus, now it is sufficient to prove that let be a power series with unit radius of convergence and
Let
If the series converges then
Now let’s prove the same.
Then
Now for |x| < 1;
So that
Now since , therefore for , there exists an integer N such that
Also we have
Hence for all , we have
Now for a fixed N, is a continuous function of x having zero value at x = 1.
Thus there exists such that 1 -
Hence
Abel’s test for uniform convergence-
The series converges uniformly on E if
1. is a positive decreasing sequence for all values of x
2. is uniformly convergent
3. is bounded for all x
Example: Let us consider the series
Here we note that if p > 1, then is absolutely convergent and is independent of x.
Hence by Weierstrass’s M-test, the given series is uniformly convergent for all x
If , the series is convergent but not absolutely.
Let
Then <> is monotonically decreasing sequence for |x| < 1, because
Also
Hence, by Abel’s test, the series
Is uniformly convergent for and |x| < 1.
Example: Prove that the given series
Is converges uniformly in (0, 1).
Sol.
Here we have
Since n/n+1 tends to 0 as n tends to infinity, taking n sufficient large
Hence < is monotonically decreasing and positive.
Hence by Abel’s test, the given series is uniformly convergent in (0, 1).
Key takeaways
- Abel’s theorem (first form)
If a power series converges at the point R of the interval of convergence (-R, R), then it uniformly converges in the interval [0, R].
2. Abel’s theorem (Second form)
Let be a power series with finite radius of convergence R and let .
If the series converges at end point x = R then
This theorem was given by Karl weierstrass in 1885. He proved this theorem on the approximation by polynomials of continuousreal valued functions defined on closed intervals.
Statement- if f is real function defined on [a, b] then there exists a sequence of real polynomials which converges uniformly to f(x) on [a, b], which means
Proof:
If a = b, then f(x) = f(a)
Then the theorem is true by taking to be a constant polynomial defined by
Thus we assume that a < b
We take a = 0, b = 1.
Now we know that for positive integer n and k where , the binomial coefficient , which is defined as
Now we define the polynomial where
The polynomial (1) is known as Bernstain polynomial.
We shall prove that certain Bernstain polynomial exists which uniformly converges to f on [0, 1].
Consider the identity
Diff w.r.t. x, we get
Multiply by x(1-x) both sides, we get
Diff. w.r.t. x, we get
On applying (2) we get
Multiply by x(1-x), we get
Since the maximum value of x(1-x) in [0, 1] is ¼.
So the equation (4) can be written as
Now f is continuous on [0,1]. So, f is bounded and uniformly continuous on [0, 1].
such that
And by uniform continuity for given
Now for any fixed but arbitrary x in [0,1], then n-values 0,1,2,3,….,n of k can be divided into two parts as follows
Let A be the set of values of k for which and B be the set of remaining values for which
Now for , we get by (5)
.
Now
By (2)
We split the summation on RHS into two parts accordingly as
Let
Thus we have
Thus converges uniformly to f(x) on [0, 1].
Which is the proof of the theorem.
Example: Find a continuous function f: (-1, 1) that can not be approximated by a polynomial.
Sol:
Suppose f: (-1 1) be given by f(x) = 1/(1+x). Then f will also be a continuous function.
If possible, for , let there exists a polynomial p(x) such that
Since p(x) is bounded on (-1, 1), there exists a constant M > 0 such that
Now
Therefore, f is bounded on (-1, 1) which is a contradiction.
Key takeaways
- If f is real function defined on [a, b] then there exists a sequence of real polynomials which converges uniformly to f(x) on [a, b], which means
References
1. Erwin Kreyszig, Advanced Engineering Mathematics, 9thEdition, John Wiley & Sons, 2006.
2. Advanced engineering mathematics, by HK Dass
3. Mathematical analysis by Dr. Anju panwar
4. Real analysis by Dr. Sachin Kaushal
5. Real analysis by Ajit kumar & S.Kumaresan
6. Principles of real analysis by S.C Malik