Numerical methods

1. NAME –AYUSH MISHRA
ROLL NO – 21/1456
SUBJECT - MATHEMATICS FOUNDATION
FOR ELECTRONICS

2. Sequences and Series
A sequence can be defined as a function from natural
numbers (the positions of elements in the sequence) to the
elements at each position. The notion of a sequence can be
generalized to an indexed family, defined as a function from an
index set that may not be numbers to another set of elements.

3. Examples: {3, 3, 3, 3, . . .}
Null sequence {0, 0, 0, . . . , 0, . . .} A sequence is also denoted by {an}
whose ordinate y = an at the abscissa x = n. Thus in a sequence for
each positive integer n, a number an is assigned and is denoted as an
or
{an} = {a(1), a(2), a(3), . . . , a(n), . . .} = {a1, a2, a3, . . . , an, . . .}
Here a1, a2, a3, . . . an, are known as the first , second, third and nth
terms of the sequence.

4. Real sequence
A function from N to R, the set of real numbers; denoted by f : N → R.
Thus the real sequence f is set of all ordered
pairs {n, f (n)}|{n = 1, 2, 3, . . .} i.e., set of all pairs (n, f (n)) with n a
positive integer.
Notation: Since the domain of a sequence is always the
same (the set of positive integers) a sequence may be
written as {f (n)} instead of {n, f (n)}.

5. LIMIT OF A SEQUENCE
Consider a sequence {an} = 3 + 1 n . Plotting the values n: 1
2 4 5 10 50 100 1000 10000 100000. . . an: 4 3.5 3.25 3.2 3.1
3.02 3.01 3.001 3.0001 3.00001. . .
As n increases, an = 3 + 1 n becomes closer to 3. Thus the
difference (or distance) between 3 + 1 n and 3 becomes
smaller and smaller as n becomes larger and larger i.e., we
can make 3 + 1 3 and 3 as close as we please, by choosing
an appropriately (sufficiently) large value for n, i.e., the
terms of a sequence cluster around this (limit) point.
However note that 3 + 1 n = 3 for any value of n.

7. CONVERGENCE, DIVERGENCE AND
OSCiLLATION OF A SEQUENCE
Convergent A sequence {an} is said to be
convergent if it has a finite limit i.e.,
lim n→∞an = L = finite unique limit value.
Divergent A sequence {an} is said to be divergent if it has
a infinite limit i.e
If limn→∞an = infinite = ±∞

8. Oscillatory - If limit of an is not unique (oscillates
finitely) or ±∞ (oscillates infinitely)
OSCILLATION OF A SEQUENCE
Examples:
1. 1 n2 convergent since limn→∞ 1 n2 = 0=finite unique.
2. {n:} divergent since limn→∞n := ∞
3. {(−1)n} oscillates finitely, since lim n→∞(−1)n = 1, n even −1, n odd.
4. {(−1)n · n2} oscillates infinitely, since limit = ±∞.

9. Result 1: If sequence {an} converges to limit L and {bn} converges to L∗ then a. {an + bn}
converges to L + L∗
b. {can} converges to CL
c. {an · bn} converges to L · L∗
d. { an /bn } converges to L/ L∗ , provided L∗ [not =] 0.
Result 2: Every convergent sequence is bounded. Example: 1 n is convergent and is bounded an
= 1 n < 1, for every n.
Result 3: The converse is not true i.e., a bounded sequence may not be convergent. Example:
{(−1)n} is oscillatory (has more than one limit but is bounded since −1 ≤ (−1)n ≤ 1.
Result 4: A bounded monotonic sequence is convergent. Example: 1 n2 is bounded since 1 n2
≤ 1 for every n and monotonically decreasing since 1 n2 > 1 (n+1)2 for every n. Hence the
sequence is convergent because lim n→∞an = lim n→∞ 1 n2 = 0 = finite.

10. Useful Standard Limits

11. Example 1: an = n2−n 2n2+n
Solution: lim n→∞an = lim n→∞ n2 − n
2n2 + n = lim n→∞ 1 − 1 n 2 + 1 n = 1 2
sequence is convergent since the limit of
the sequence is unique and finite.

12. INFINITE SERIES
Differential Equations are frequently solved by using infinite series.
Fourier series, Fourier-Bessel series, etc. expansions involve infinite
series. Transcendental functions (trigonometric, exponential,
logarithmic, hyperpolic, etc.) can be expressed conveniently in
terms of infinite series. Many problems that cannot be solved in
terms of elementary (algebraic and transcendental) functions can
also be solved in terms of infinite series.

13. Series
Given a sequence of numbers u1, u2, u3, . . . un, . . . the
expression
u1 + u2 + u3 + · · · + un + · · · (1) which is the sum of the
terms of the sequence, is known as a numerical series or simply
“series”. The numbers u1, u2, u3, . . . un are known as the first,
second, third,. . ., nth term of the series (1)

14. Infinite Series
If the number of terms in the series(1) is infinite, then the series is
called an infinite series (otherwise finite series when the number of
terms is finite). Infinite series (1) is usually denoted as
The main aim of this chapter is to study the nature (or behaviors) of
convergence, divergence or oscillation of a given infinite series. For this
purpose, define {Sn}
the sequence of partial sums as
S1 = u1
S2 = u1 + u2
S3 = u1 + u2 + u3 . . .
Sn = u1 + u2 + u3 + · · · + un = n k=1 uk
Here Sn is known as the nth partial sum of the series, i.e., it is the sum of the first n
terms of the series (1).

15. NECESSARY CONDITION FOR
CONVERGENCE
Necessary condition for convergence of a series un is that, its nth term un
approaches zero as n becomes infinite i.e., If series converges, then limn→∞un = 0.
Important Note: This is not a test for convergence
Proof: Let s be the sum of this convergent series. Also let Sn and Sn−1 be the nth
and (n − 1) th partial sums of the given series so that
un = Sn − Sn−1 Taking limit, we have
lim n→∞ un = lim n→∞(Sn − Sn−1) = lim n→∞ Sn − lim n→∞ Sn−1 = s − s = 0.

16. Note 1: The converse of the above result is not true, i.e., the
above result is not a sufficient condition. From the fact that the
nth term un approaches zero, SEQUENCES AND SERIES 35.5 it
does not follow that the series converges, for the series may
diverge. If limn→∞un = 0, then the series may converge or may
diverge.

17. Example: 1 + 1 2 + 1 3 + 1 4 + · · · + 1 n + · · · is a divergent series
although its nth term approaches zero i.e., lim n→∞ un = lim n→∞ 1 n =
0 Note 2: Preliminary test for divergence. If the nth term of a series does
not tend to zero as n → ∞, then the series diverges i.e., if limn→∞un = 0
then series diverges.
Example: 1/2 + 2/3 + 3/4 + 4/5 + · · · + n/n + 1 + · · ·
Since lim n→∞ un = lim n→∞ n n+1 = 1 = 0 by the above preliminary
test, the given series diverges.

18. CAUCHY’S INTEGRAL TEST
Theorem:
Let un = u1 + u2 + u3 + · · · + un + · · · (1) be series with
positive and non-increasing terms i.e., u1 ≥ u2 ≥ u3 ≥ · · ·
Let f (x) be a continuous non-increasing function such that f (1) =
u1, f (2) = u2, f (3) = u3, . . . , f (n) = un (3) Then the improper
integral ∞ 1 f (x) dx (4) and the infinite series (1) are both finite
(in which case series (1) is convergent) or both infinite (in which
case series (1) is divergent).

19. Proof:
Plot the terms u1, u2, u3, . . . of the series (1) on the y- axis so that the first
escribed rectangle is of area u2 while the first inscribed rectangle is of area
u1.
Thus the area under the curve y = f (x), above the x-axis and between any two ordinates x = a and x = b
lies between the sum of the set of inscribed (solid) and escribed (dotted) rectangles formed by the
ordinates at x = 1, 2, 3, . . ., as shown in Fig. 35.2. Hence,
(Sn+1 − u1) ≤ n+1 1 f (x) dx ≤ Sn

20. D’ALEMBERT’S∗ RATIO TEST
Theorem:
Let ∑un = u1 + u2 + u3 + · · · + un + · · · (1) be a series with positive terms. The
ratio un+1 un measures the rate or growth of the terms of the series (1)
Let lim n→∞ un+1/un = m
then
a. the series (1) converges if m < 1
b. the series (1) diverges if m > 1
c. the ratio test fails when m = 1
i.e., series may converge or diverge. Use a different test.

21. Proof:

22. Example n3+a/2n+a
Solution: lim n→∞ un+1 un = lim
n→∞ (n + 1)3 + a 2n+1 + a · 2n + a n3
+ a = lim n→∞ 1 + 1 n 3 + a n3 2 + a
2n · 1 + a 2n 1 + a n3 = 1 2 < 1 series
is convergent.

23. CAUCHY’S* nth ROOT TEST
Theorem: For a positive series
∞∑n=1 un = u1 + u2 + u3 + · · · + un + · · · (1) Let lim n→∞ √n
un = f inite value = m Then
a. series converges when m < 1
b. b. series diverges when m > 1
c. c. test fails when m = 1, use a different test
Note: Cauchy’s root test is applied when un involves the nth
power of itself as a whole.

24. Example

25. ALTERNATING SERIES LEIBNITZ’S*
THEOREM
All the series considered so far contained only positive terms. However a series may
contain some positive and some negative terms.
Alternating series is a series whose terms are alternately positive and negative, i.e., series whose terms have
alternating (positive and negative) signs in the form
u1 − u2 + u3 − u4 + · · · + (−1)n un + · · ·
where u1, u2, · · · un, · · · are all positive (i.e., un > 0, for every n).
A simple test for convergence of alternating series is given by Leibnitz’s theorem (rule) which states that in the
alternating series
u1 − u2 + u3 − u4 + · · · + (−1)nun + · · · (with un > 0) (1)
if (i) the terms are such that each term is numerically greater than its succeeding term
i.e., u1 > u2 > u3 > u4 > · · · > un > un+1 > · · · (2) and
(ii) lim n→∞un = 0 (3)
Then the alternating series (1) converges. Its sum is positive, and does not exceed the first term.

26. Proof: S2m = (u1 − u2) + (u3 − u4) + · · · + (u2m−1 − u2m) (4)
which is the sum of the first n = 2m even number of terms of the series (1). The
expression in each of the parentheses in (4) is positive (i.e., (u1 − u2) > 0, (u3 − u4) > 0 . .
. , etc.) because of the condition (2). Hence S2m > 0 and increases with increasing m as
more positive values are added. Rewriting (4) as
S2m = u1 − (u2 − u3) − (u4 − u5) − · · · −(u2m−2 − u2m−1) − u2m
and again using the condition (2), note that
S2m < u1
since positive quantities (in each bracket) are subtracted from u1. Thus S2m increases with increasing m and is
bounded above, hence the sequence of even partial sums has a limit say s,
i.e., lim m→∞ S2m = s (with 0 < s < u1)

27. Now consider the sum ofthe first n = 2m + 1, odd
number of terms of the series (1) as S2m+1 = S2m +
u2m+1
Taking the limit,
lim m→∞ S2m+1 = lim m→∞ S2m + lim m→∞ u2m+1
= s + 0 = s
since lim m→∞u2m+1 = 0 follows from condition (3).
Therefore the given alternating series (1) converges
because limn→∞ = s both for even n and for odd n.
Note: Leibnitz’s theorem holds good even if the
inequalities (2) are true from some N onwards.

28. ABSOLUTE CONVERGENCE AND
CONDITIONAL CONVERGENCE
Absolute and conditional convergence
Let u1 + u2 + u3 + · · · + un + · · · =∑ un (1) be a plus- and -minus series with the
assumption that here onwards the members u1, u2, . . . , un, . . . can be either positive or
negative i.e., some terms may be positive and others negative (not necessarily alternative).
Let us form a series made up of the absolute values of the terms of the series (1) i.e., |u1| +
|u2| + |u3| + · · · + |un| + · · · = ∑ |un| (2) Each term of the series (2)is positive and
numerically equal to the corresponding term of series (1).

29. Absolute Convergence
The plus- and -minus series ∞∑n=1 un is said to be
absolutely convergent if the corresponding series with
absolute terms∑|un| is convergent.
Example: 1 − 1 2! + 1 3! − 1 4! + · · · is absolutely convergent
because the series formed with absolute values 1 + 1 2! + 1
3! + 1 4! + · · · is convergent.

30. Conditional Convergence
If un is convergent while |un| is divergent then un is said to be
conditionally convergent.
Example: 1 − 1/2 + 1/3 − 1/4 + · · · is conditionally convergent
because the given series is convergent (by Leibnitz’s rule) while 1 +
1/2 + 1/3 + 1/4 + · · · is a divergent harmonic series (with p = 1).

31. POWER SERIES
We have considered so far, series whose terms are constants. Now we
consider, series whose terms are functions of x, more specifically series in
which nth term is a constant times xn or constant times (x − b) n where b is
a constant.
A power series is a series of the form ∞∑n=0 anxn = a0 + a1x + a2x2 + · · ·
+ anxn + · · · where a0, a1, a2 · · · an · · · are all constants known as
coefficients of the series or ∞ ∑ n=0 an(x − b) n = a0 + a1(x − b) + a2(x − b)
2+ · · · + an(x − b) n + · · ·

32. Interval of convergence
of a power series is the interval of x say −L < x < L such that the
series converges for values of x in this interval (−L, L) and diverges
for values of x outside this interval.
Test for Convergence of Power Series