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U nit I Real Sequences and Series

Real sequences

1.1 Definition and Examples of Sequences

Definition 1.1 A sequence is a function whose domain is the collection of all

integers greater than or equal to a given integer m (usually 0 or 1).

Example 1 f (n) = 2 n for n 0.

g (n) = n2 for n 1.

h (n) = for n 1.

k (n) = for n 0.

l (n) = for n 0.

In general if f (n) = an for n 1, then the ordered set of numbers a1, a2, a3, … , completely

determines the sequence.

Hence we can write this sequence as . Similarly f (n) = a n for n m can be written as

.

For any sequence , the number n is called an index, m the initial index and a n the nth term

of the sequence.

Exercises 1.1

Write the initial five terms of the sequence:

1. 2.

3. 4.

5.

\Prepared by Tekleyohannes Negussie

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1.2 Convergence and its Properties.

Let be any sequence. We need to study the behavior of an as n increases with out bound.

Definition 1.2 Let be a sequence. A number L is the limit of

if for every 0 there is an integer N such that if n N, then an L .

In this case we write

.

If such a number L exists, we say that converges (converges to L) or

exists and if such a number L does not exist, we say that

diverges or that does not exist.

Remark: The limit of a sequence is unique if it exists.

Example 2 Show that converges and .

Solution For any 0, let N N such that N .

If n N, then = .

Therefore, .

Note that: The sequence is known as the harmonic sequence.

Example 3 Show that diverges.


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Solution (1) n =

If L 0, then for odd values of n we’ve

and if L 0, then for even values of n we’ve

.

Hence for any , 0 1, there is no such number L.

Therefore, diverges.

Note that: The sequence is called an oscillating Sequence.

Definition 1.3 Let be a sequence. If for every number M there is an

integer N such that

if n N, then an M.

We say that diverges to , and we write

= .

Similarly, if for every number M there is an integer N such that

if n N, then an M.

we say that diverges to , and we write

= .

Theorem 1.1 Let be a sequence, L a number, and f a function defined on


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m, such that f(n) = an for n m.

If = L, then converges and = L.

If = , then diverges and = .

Consequently =

Example 4 Show that for r 0.

Solution Let f (x) = x r for x 1.

Then f (n) = n r for n 1 and = 0 for r 0.

Therefore, for r 0.

Example 5 Let r . Show that the sequence diverges for 1 and for

r = 1and converges for 1 r 1

with = .

Solution Let f (x) = r x for x 1.

Then f (n) = r n for n 1 and =

Therefore, converges for 1 r 1 with =

and diverges for 1 and for r = 1.

Note that: For any r , is called a geometric sequence.

Example 6 Determine whether the following sequence converges. If it converges find

the limit.


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a) b) .

Solutions a) Let f(x) = for x 1.

= 1 + = 1.

Therefore, converges and = 1.

b) Let f(x) = tanh x for x 1.

= = = = 1. Why?

Therefore, converges and = 1.

Example 7 Show that:

a) = 1 for c 0. b) = 1.

Solutions a) Let f (x) = = for x 1.

= = e 0 = 1.

Therefore, = 1.

b) Let f (x) = for x 1.

Since = , = = e 0 = 1.

Therefore, = 1.


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Theorem 1.2 Let and be convergence sequences and let c .

a) = + . b) = c .

c) = . d) = ,

provided that 0.

Example 8 Show that converges.

Solution Let = and = n sin .

Since = = and = =

= =

Therefore, by the product property of convergent sequences

= .

Theorem 1.3 (Squeezing Theorem)

If = and if is any sequence such that

an cn bn for n m, then converges.

Moreover = = .


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Proof Let = = L.

Thus 0 N1 N such that if n N1, then .

if n N1, then L a n L +

and 0 N2 N such that if n N2, then .

if n N2, then L b n L +

Now choose N = max N1, N2, m.

Hence 0 N N such that if n N, then L a n c n b n L +

if n N, then .

Therefore, = L.

Example 10 Show that = 0.

Hint:

Solution n 1, and = 0 =

Therefore, = 0.

Example 11 Show that:

Solution 0 sin 2 n 1.

Thus 0 and = 0 = .

Therefore,

Exercises 1.2

Evaluate each of the following limits.


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1. 2.

3. 4.

5. 6. Hint: 1 ℓn n n for n 3.

7. 8.

9. 10.

1.3 Monotonic and Bounded Sequences.

Definition 1.4 A sequence is said to be:

i) Increasing if a n a n + 1 for all n m, and strictly increasing if a n a n + 1

for all n m.

ii) Decreasing if a n a n + 1 for all n m, and strictly decreasing if a n a n + 1

for all n m.

Note that: A decreasing (or an increasing) sequence is called a monotonic sequence.


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Example 12 Show that is strictly increasing.

Solution n N and n m , 3n 3n + 1.

Hence n (2n + 3) = 2n2 + 3n 2n2 + 3n + 1 = (2n + 1) (n + 1).

a n a n + 1 n m

Therefore , is strictly increasing.

Example 13 Show that is not a monotonic sequence.

Solution n 1 and n N,

Thus a n a n + 1 for odd n and a n a n + 1 for even n.

Therefore, is not a monotonic sequence.

Definition 1.5 A sequence is said to be bounded if there is a number M

such that M for every n m. Otherwise we say the sequence is

unbounded.

Example 14 and are bounded sequences while and

are unbounded sequences.

Theorem 1.4 a) If converges, then is bounded.

b) If is unbounded, then diverges.


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Proof a) Suppose converges to L. 0 N N such that if n N, then .

Take = 1.

Thus for = 1, N N such that if n N, then 1.

if n N, then = + 1 + .

Now choose M = max , , … , , 1 + .

Then M for n m. Therefore, is bounded and b) is the contra positive of part a).

Example 15 Show that is bounded.

Solution Now a n = 1 + and a n + 1 = 1 + . Since n N, n n + 1,

a n = 1 + 1 + = a n + 1 . Thus a 1 = 2 a n n N.

Therefore, is bounded.

Example 16 Show that is bounded.

Solution n N, a n = ( 1) n =

Thus = 1 n N.

Therefore, is bounded.

Example 17 Show that is unbounded.

Solution n N , 1 (n + 1) (n + 2) 1 + .

a n = n 1 + n + = a n + 1

Hence n N, a n n 1.

Therefore, is unbounded.


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Note that: converges to L if and only if converges to L.

i.e. adding or removing a finite number of terms in a sequence do not

affect the convergence of the sequence.

Theorem 1.5 A bounded monotonic sequence is convergent.

Moreover if the bounded monotonic sequence converges to L,

then

a ) L = whenever is increasing.

b) L = whenever is decreasing.

Example 18 Show that is convergent.

Solution Since n N 1.

a n = and a n + 1 =

Thus a n + 1 a n a 1 = 2 n N.

Therefore, is convergent.

Example 19 Show that is convergent.

Solution Since n N and n 3 , n 2 2n + 1.

a n = = a n + 1

Thus a n + 1 a n n N and n 3.

Hence converges.


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Therefore, by the above remark is convergent.

Exercises 1.3

In Exercises 1-5 determine whether the sequence is bounded or unbounded, decreasing or increasing

or not.

1. 2.

3. 4.

5.

1.4 Subsequences and Accumulation Points.

Definition 1.6 Let be a sequence and let be a strictly increasing

sequence of positive integers. Then the sequence is called a

subsequence of the sequence .

Example 20 Consider the sequence .

Solution a) Let n k = 2k k N.

Now is a strictly increasing sequence in N and hence

is a subsequence of .

b) Let n k = 2k + 1 k N.

Now is a strictly increasing sequence in N and hence

is a subsequence of .


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Example 21 Every sequence is a subsequence of itself.

Solution Let n k = 2k k N.

Thus = k N.

Therefore, = .

Theorem 1.6 If a sequence converges to L, then every

subsequence of converges to L.

Corollary 1.6.1

a) If a sequence has two subsequences which converges to

different limits, then the sequence diverges.

b) If a sequence has a divergent subsequence, then

diverges.


Solution Both and are subsequences of .

Now converges to 1 and converges to 1.



Solution is a subsequences of

and sin (2k 1) = .

Thus is an oscillating sequence that diverges.


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Solution is a subsequences of and cos k = .

Thus is an oscillating sequence that diverges.


Definition 1.7 Let be a sequence. The number a is called a limit

point ( or accumulation point) of if and only if there is

a subsequence of that converges to a.

Example 25 Show that

i) 1 and 1 are accumulation points for .

ii) 1, and 0 are accumulation points for .

Example 26 Find all the accumulation points for .

Solution Any n N is of the form n = 3 k or n = 3k 1 or n = 3k 2 for some k N.

Hence , and are convergent

sub-sequences of that converge to 0, and respectively.

Therefore, 0, and are accumulation points of . Exercise 1.4

In Exercises 1-5 determine the accumulation points of the sequences.


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1. 3. 5.

2. 4.

Real Series

2.1 Infinite Series.

Let be a sequence and let s1 = a 1, s2 = a 1+ a 2, s 3 = a1 + a2 + a3, … . We can form a

sequence called the sequence of partial sum. The expressions

a1 + a2 + a3, … and are called series.

Defn 2.1 Let be a sequence. For each j N, the j th partial sum sj is the

sum of the first j terms of the sequence. If exists, then we say that the

series converges, and we call the sum of the series. Otherwise,

we say the series diverges. The numbers am, am + 1, am + 2, … are terms of

the series .

Finally if converges, then denotes the sum of the series.

Remark. Usually m = 0 or m = 1.

Example 1. Show that Converges and find the sum.

Solution. Since = n N, the partial sum

sj = ( ) + ( ) + ( ) + … + ( )

= 1


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and = 1 = 1.

Therefore Converges and = 1.

Example 2. Show that diverges.

Solution. We need to show that is unbounded.

Claim: 1 + j for any j N

Proof. Use the P.M.I. on j

For j = 1, = 1 + 1 + · 1 is true.

Assume that it is true for j = k. i.e. 1 + k.

Now we need to show for j = k + 1.

Therefore by the P.M.I. for any j N, 1 + j.

Since = , is unbounded.

Therefore diverges.


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Remark: Let m2 m1. The series converges if and only if the series

converges.

If the two series converge, then = +

Theorem 2.1 ( nth term divergence test)

a) If converges, then = 0.

b) If 0 or does not exist, then diverges.

Proof. a) For n 1. an = sn sn 1. Since converges exists.

Hence by the above remark and = .

Therefore = 0.

b) Take the contra positive of part a).

Note that: i) Part b) of theorem 2.1 is called the nth term test for divergence.

ii) The converse of part a) of theorem 6.1 may not hold true.

= 0 while diverges.

Note that: A geometric series is a series of the form , where c, r .


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Theorem 2.2 (Geometric Series theorem.)

Let r be any number and let c 0 and m 0. Then the geometric series

converges if and only if 1. For 1

= .

Proof. Where 1.

If 1, then for n m.

Hence the series diverges by theorem 6.1 part (b).

If 1, then = 1

Therefore = .

Note that: The number r is called the ratio of the geometric series.

Example 3. Show that the series converges and find the sum.

Solution. = , Since = 135 and is a geometric

series with r = and m = 1. = .

Therefore converges and = .

Example 4. Remove the middle third open subinterval of the interval I0 = [0, 1]. Successively

continue the process for each of the remaining subintervals. Show that the sum of

the lengths of the intervals removed is equal to the length of I0.


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Solution. Let be the sum of the lengths of the removed subintervals on the nth step for

each n N.

Now = , = , = and in general = .

Hence = = 1=

Therefore the sum of the lengths of the intervals removed is equal to the length of I0.

Theorem 2.3

a) If and converge, then converges, and

= + .

b) If converges and c is any number, then also converges and

= c .

Proof. Let be the j th partial sums of ,

and respectively. Then and hence =

= = + .

Therefore = + .

b) Let be the j th partial sums of and respectively.

Then and = c = c


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Therefore = c .

Remark: - Part (a) of theorem 2.3 holds for finite number of convergent series.

- If and converge, then converges and

= .

Example 5. Show that converges and find the sum of the series.

Solution. = for each n N.

Thus sj = and = = .


Example 6. Show that converges and find the sum of the series.

Solution. = for each n N.

Hence = 1 and = .


Remark: Changing index.

= .

Example 7. =


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Exercises 2.1

1. Find the sum of each of the following series.

a)

b)

c)

2. A ball is dropped from a height of 3 meters onto a smooth surface. On each bounce, the ball

rises to of the height it reached on the previous bounce. Find the total distance the ball

travels.

3. Express each of the following periodic decimals as a fraction in simplest form.

a) 0.727272 . . . b) 0.00864444 . . . c) 23.45232323 . . .

2.2 Non-negative Term Series.

A series all of whose terms are non-negative is called a Non-negative term series.

Note that: If is a non-negative term series, then the sequence of partial sum

is increasing.

Theorem 2.4 Let be a non-negative term sequence and let f be a continuous

decreasing function defined on 1, such that f (n) = an for n 1. Then the

series converges if and only if the integral converges.

Proof. Since f (n) = an n N and f is a continuous decreasing function on 1, .


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0 for n 2.

0 = and 0 =

.

Therefore the theorem holds true.

Note that: Theorem 2.4 is called the integral test.

Remark: In thm 2.4 converges if and only if converges while

Example 8. Show that converges if and only if p 1.

Solution. If p 0, then 0 and hence diverges by theorem 6.1(b).

If p = 1, then is a harmonic series which diverges.

If p 0 and p 1, then define f: 1, by f (x) = .

Since f (n) = n N and f is a continuous decreasing function on 1, , we need

to determine whether converges or not.

Now =


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= =

Since exists for p 1 and does not exist for 0 p 1, converges for

p 1 and diverges for p 1.

Therefore converges for p 1.

Note that: Series of the form for any fixed number p is called a p series.

Example 9. Show that converges.

Solution. Let f (x) = x 1, . f is a continuous decreasing function on 1, .

Hence for b 1, = = = 1.

Therefore converges.


Solution. Let f (x) = for x 2. f is a continuous decreasing function on 1, .

Hence for b 2, = ln (ln x) = = .

Therefore diverges.

Theorem 2.5 (The Comparison Tests)


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a) If converges and 0 an bn for all n 1, then converges

and

.

b) If diverges and 0 bn an for all n 1, then diverges.


Solution. 2 n 1 2 n 1 n N. n N.

But is a geometric series that converges and = 2.



Solution. n N

n N.

But is a p series with p = and hence it diverges.

Therefore, by part (b) of theorem 6.5, diverges.

Theorem 2.6 Limit Comparison Test.

Let and be non-negative term series. Suppose =

L,

where L is a positive number.


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a) If converges, then converges.

b) If diverges, then diverges.

Proof. = L 0 N N such that n N

Take = .

Now n N

If converges, then and converge.

Hence by comparison test converges.

If diverges, then diverges and hence diverges by comparison

test.


Solution. For large n, . Now compare and

Since = = 1 and =

converges, because it is a p series with p = 2.

Therefore converges by the limit comparison test.



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Solution. For large n, . Now compare and

Since = = 1 and diverges, by the limit comparison test

diverges.

Therefore diverges.

Theorem 2.7 (The ratio Test)

Let be a non-negative term series. Assume that 0 for all n

and that = r (possibly )

a) If 0 r 1, then converges. b) If r 1, then diverges.

Note that: If r = 1, then from this test alone we cannot draw any conclusion

about the convergence or divergence of .

Example 15. Show that:

i) converges. ii) converges. iii) diverges.

Solutions. i) r = = = = 0.

Since r 1, converges.


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ii) r = = =

= = . Since r 1, converges.


iii) r = = = =

Since r 1, diverges.

Therefore diverges.

Theorem 2.8 The Root Test.

Let be a non-negative term series, and assume that

= r (possibly )

a) If 0 r 1, then converges. b) If r 1, then

diverges.

Note that: If r = 1, then from this test alone we cannot draw any conclusion

about the convergence or divergence of .


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Example 16. Determine whether the series converges or diverges.

i) ii) iii)

Solutions. i) =

r = = = . Since r 1, converges.


ii) Let = .

Now claim n ! 3 n n 7. Use induction on n.

For n = 7, 7! = 5040 2187 is true.

Assume for n = k. i.e. k! 3 k we want to show (k + 1)! 3 k + 1.

(k + 1)! (K + 1) 3 k 3 k + 1

Thus = 1

Therefore diverges.

iii) Let = .

r = = = 1.


Remark: Use ratio test when factorials or powers appear in the terms of the series.


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Now consider the series

If we use: i) the ratio test we get:

and ii) The root test we get: ,

But diverges while converges.

Therefore for r = 1 ratio test and root test are not helpful.

Exercise 2.2

1. Determine the values of p for which converges.

2. Show that diverges. (Hint: ln n n, for n 2)

3. Determine whether each of the following series converges or diverges.

a) b) c)

d) e) f)

g) h) i)

4. Let and be nonnegative convergent series. Then show that both

and are convergent.

2.3 Alternating Series Test and Absolute Convergence.


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2.3.1 Alternating Series Test.

Defn 2.2 If 0 for all n N, then series of the form

is called an alternating series.

Example 17.

are examples of

alternating series.

Theorem 2.9 (Alternating series Test or Leibniz’s Test)

Let be a decreasing sequence of positive numbers such that

= 0. Then the alternating series

converge.

Further more, the sum s and the sequence of partial sums satisfies

the inequality

.

Example 18. Determine whether the series converges or diverges.


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a) b) c)

Solutions. a) Since is a decreasing sequence of positive numbers and = 0,

converges.

a) Since is an increasing sequence of positive numbers

and = 0, diverges.

b) Since is a decreasing sequence of positive numbers,

= 0 and cos n = ( 1)n n N, converges.

Example 19. Approximate the sum of the following series with an error less than 0.01.

a) b)

Solutions. A partial sum is an approximation for the sum s with an error less than 0.01

if 0.01.

a) Let = n N. Then 0.01 (n + 1) 3 100 n 3

Hence =

Therefore = with an error less than 0.01.

b) Let = n N.

Then 0.01

0.01 800 n 1


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Hence =

Therefore = with an error less than 0.01.

Example 20. Show that is convergent and approximate its sum correct to

three decimal places.

Solution. Since is a decreasing sequence of positive numbers and = 0,

converges.

Now let = n N.

Then

(n + 1 ) ! 2000 n 5

Therefore = 0.632.

2.3.2 Absolute and Conditional Convergence.

Defn 2.3 The series is said to be absolutely convergent if the

series converges.

Example 21. Show that is absolutely convergent.

Solution. Let = . = n N. Since is a convergent geometric

series converges.

Therefore is an absolutely convergent series.


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Example 22. Show that is convergent but not absolutely convergent.

Solution. Let = . = n N. Since is divergent is not

an

absolutely convergent series. But we know that by A.S.T. is

convergent.

Therefore is convergent but not absolutely convergent.

Defn2.4 A series that is convergent but not absolutely convergent is said to be

conditionally convergent.

Example 23. and are examples of conditionally convergent series.

Theorem 2.10 If is an absolutely convergent series, then the series

is convergent and .

Proof. Define:

.

Then = . Since 0 and 0 n N, and

converge. (By comparison Test), thus = is convergent.

Now we need to show that .

Let p = , q = , t = , and s = .

Now t = p q = + = p + q = s. Since = + .

Therefore .


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Example 24. Show that is convergent.

Solution. n N and is convergent.

is convergent. is absolutely convergent.

Therefore, by theorem 2. 10, is convergent.

Exercises 2.3

1. Determine whether each of the following alternating series diverges, which series converges

conditionally and converges absolutely.

a) b)

c) d)

e) f)

2. Approximate the sum of the following series with an error less than 0.01.

a) b)

2.4 The Generalized Convergence Test.

Theorem 2.11 Let be a series.

Generalized Comparison Test.

If for n 1 and if converges, then

(absolutely) converges.


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Generalized Limit Comparison Test.

If = L, where L is a positive number and if

converges, then (absolutely) converges.

Generalized Ratio Test.

Suppose that 0 for n 1and that = r (possibly )

If r 1, then (absolutely) converges. If r 1, then diverges.

Note that: If r = 1, then from this test alone we cannot draw any

Conclusion about the convergence of the series.

Generalized Root Test.

Suppose that = r (possibly )

If r 1, then (absolutely) converges. If r 1, then diverges.

Note that: If r = 1, then from this test alone we cannot draw any

conclusion about the convergence of the series.


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Example 25. Show that converges absolutely for 1, converges conditionally

for x = 1 and diverges for x [ 1, 1) .

Solution. If x = 0, then the series obviously converges.

If x 0, then =

Hence by the Generalized convergence test converges absolutely for 1.

For x = 1 the series becomes a p-series with p = and hence diverges.

For x = 1 by alternating series test converges. Since =

which diverges, we conclude that converges conditionally.

Therefore converges absolutely for 1, converges conditionally for x = 1

and diverges for x [ 1, 1) .

Example 26. Show that converges absolutely for 1 and converges

conditionally for x = 1 diverges for 1.

Solution. If x = 0, then the series converges.

If x 0, then =

Hence by the Generalized convergence test converges absolutely for 1

and for x = 1 = converges by the alternating series test.

Therefore converges absolutely for 1 and converges


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conditionally for x = 1 diverges for 1.

Corollary 2.11.1 Let be a sequence.

If = r 1 or = r 1,

then = 0.

Example 27. Show that = 0 for all x.

Solution. If x = 0, then clearly the limit is 0.

If x 0, then = = 0 for all x.

Hence converges for all x.

Therefore = 0 for all x.

Exercises 2.4

1. Determine the positive number x for which converges.

2. Find the sum of each of the following series.

a) b) c)

3. A ball is dropped from a height of 3 meters onto a smooth surface. On each bounce, the ball

rises to of the height it reached on the previous bounce. Find the total distance the ball


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travels.

4. Express each of the following periodic decimals as a fraction in simplest form.

a) 0.727272 . . . b) 0.00864444 . . .

c) 23.45232323 . . .

5. Determine the values of p for which converges.

6. Show that diverges.

(Hint: ln n n, for n 2) 7. Determine whether each of the following series converges or diverges.

a) b) c)

d) e) f)

g) h) i)

8. Let and be nonnegative convergent series. Then show that both

and are convergent.

9. Determine whether each of the following alternating series diverges, which series converges conditionally and converges absolutely.

a) b)

c) d)

e) f)

10. Approximate the sum of the following series with an error less than 0.01.

a) b)

11. Suppose converges and is not necessarily nonnegative. Give an example to show

that need not converge.

12. Determine the positive number x for which converges.


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UNIT III

Power Series.

3.1 Power Series.

Defn 3.1 A series of the form is called a power series in x and

is called a power series in x a, where a .

Now assume that x 0 = 1 x , so that

= + + + . . .

If = 0 n N, then the power series is a polynomial of degree at most N.


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Example 1. x2 3 + 2x = , where = 3, = 2, = 1 and = 0 n 3.

and are examples of power series.

Note that: Every power series in x a, converges for x = a a .

Example 2. Show that converges only for x = 0.

Solution. For x 0, = = .

Hence diverges for x 0.

Therefore converges only for x = 0.

Example 3. Show that converges for every number x.

Solution. For x 0

= = = 0.

Therefore converges for every number x.

Example 4. Show that converges for 1.

Solution. For x 0, = = .

Hence by generalized ratio test converges for 1 and x 0.

Therefore converges for 1.

Moreover; , where x 0, is a geometric series and for 1, = .


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Lemma 3.1

a) If converges, then converges absolutely for .

b) If diverges, then diverges for .

Theorem 3.1 let be a power series. Then exactly one of the following

conditions hold.

a) converges only for x = 0.

b) converges for all x .

c) There is a number R 0 such that converges for R

and diverges for R.

Note that: R is called the radius of convergence of .

If converges only for x = 0, then we let R = 0 and if converges for

x , then we let R = .

The set of all values of x for which converges is called the interval of

convergence of .

Example 5. Find the interval of convergence of .


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Solution. For x 0

= = .

Hence converges for

and for x = , diverges.

Therefore ( , ) is the interval of convergence of .

Example 6. Determine the interval of convergence of the series

a) b) c) d)

Solutions. a) For x 0,

= = .

Thus converges for 1.

For x = 1 we get and both of which diverge.

Therefore ( 1, 1) is the interval of convergence of .

b) For x 0,

= = .

Thus converges for 1.

For x = 1, converges and for x = 1, diverges.


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Therefore ( 1, 1] is the interval of convergence of .

c) For x 0,

=

= = .

Thus converges for .

For x = , = converges

and for x = , = = diverges.

Therefore ( , ] is the interval of convergence of .

d) For x 2, ,

= = .

Thus converges for 1 1 x 3.

For x = 1, = which converges

and for x = 3, = which converges.

Therefore [1, 3] is the interval of convergence of .


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Remark: The radius of convergence of is given by

R =

Differentiation of Power Series.

Theorem 3.2 (Differentiation theorem for Power Series)

Let be a power series with radius of convergence R 0.

Then has the same radius of convergence and

= = for R.

Example 7. Show that = .

Solution. For x 0, = = 0.

Thus converges for x

and = = = .

Therefore = x.

From the result of the above example we get:

f (x) = f (x) x, where f (x) = .

Since f (0) = 1, then we get f (x) = ex for all x.


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Therefore ex = x.

Moreover = , e – x = and = etc.

Note that: Theorem 7.2 states that and have the same radius

of

convergence but the interval of convergence of these series may not be the same.

Example 8. Consider .

Let f (x) = .

R = = 1 0 and diverges for x = 1 and converges for x = 1.

Hence [1, 1) is the interval of convergence of

and f (x) = converges for 1 and diverges for 1.

Thus (1, 1) is the interval of convergence of .

Therefore the two series have different interval of convergences.

Theorem 3.3 Suppose a power series has radius of convergence R 0. Let


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f (x) = for R x R.

Then f has derivatives of all orders on ( R, R) and f (n) (0) = n! cn for n 0.

Consequently

f (x) = for R x R.

Corollary 3.3.1 Let R 0 and suppose and be power series that

converge for R x R.

If = for R x R, then cn = bn for each n 0.

Integration of Power Series.

Theorem 3.4 (Integration theorem for Power Series)

Let be a power series with radius of convergence R 0. Then

has the same radius of convergence and

= = for R.

Example 9. Show that

ℓn (1 + x) = = for 1.

Solution. Note that:


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= for 1.

Replacing x by t we get

= for 1.

Hence ℓn (1 + x) = =

= for 1.

Therefore ℓn (1 + x) = for 1.

Since converges for 1 x 1,

ℓn (1 + x) = for 1 x 1

= for 1 x 1

Therefore ℓn 2 = .

Note that: the power series expansion of ℓn (1 + x) for 1 x 1 is known

as Mercator’s Series.

Example 10. Show that tan 1 x = for 1.

Solution. If 1, then t 2 1.

Hence = for 1

and tan 1 x = =


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= for 1.

Therefore tan 1 x = for 1.

Since converges for = 1,

tan 1 x = for 1.

Therefore

= .

Note that: The power series expansion of tan 1 x for 1

is known as Gregory’s Series.

Example 11. Find a power series representation of x .

Solution. Since e x = x , = .

Hence =

=

=

Therefore = x .


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Example 12. Find a power series representation of and use it to verify that

= 1.

Solution. Since = + = +

= + 1 + x + + . . . + + . . .

= 1 + + + + . . .

=

Hence = and = = 1.

Therefore = and = 1.

Example 13. Find the power series expansion of and use it to evaluate

Solution. = +

= +

= + 1 + x + + . . . + + . . .

= + + + . . .

=

Hence = and = =

.


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Therefore = and = .

Example 14. Evaluate using power series expansion of .

Solution. = for x 1.

Hence = x + + . . .

= 1 + + . . .

=

and = = 1

Therefore = and = 1.

Example 15. Find the power series representation of for 1.

Solution. Since = for 1 and = .

But = = = .

Therefore = for 1.

UNIT IV

Taylor Series.

Suppose f is a function defined on an open interval I containing 0 by

f (x) = x I.


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Then we say that we have a power series representation of f on I. The value of f at x in I can be

approximated by the partial sum of the convergent power series of f provided that

converges for each x in I.

Remark: 1. the partial sums of a power series are polynomials.

2. if f has a power series representation, then

i) f must have derivatives of all order on I.

ii) this representation is unique and written as

f (x) = x I.

Defn 4.1 Suppose that f has derivatives of all orders at a, then the Taylor series of

f about a is the power series

If a = 0, then this series is called the Maclaurin Series.

Remark: Any polynomial in x is its own Taylor series.

Defn 4.1 The nth Taylor polynomial Pn and the nth Taylor remainder Rn of f about a

are defined by

Pn (x) =

and Rn (x) =

where tk is strictly between a and x.

Moreover Rn (x) = f (x) Pn (x).


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Therefore f (x) = if and only if Rn (x) = 0.

Example 16. Show that sin x = for all x .

Solution. Let f (x) = sin x for all x .

Then f (2k) (x) = ( 1) k sin x and f (2k + 1) (x) = ( 1) k cos x for all k 0 and k Z.

Thus Pn (x) = , where m = [ (n 1)], greatest integer not

exceeding (n 1) and Rn (x) = , where strictly lies between 0 and x.

Now to show that f(x) = Pn (x) we need to show that Rn (x) = 0.

Since 1,

and hence = = 0

Thus Rn (x) = 0.

Therefore sin x = for all x .

Example 17. Show that cos x = for all x .

Solution. Let f (x) = cos x for all x .

Then g (2k) (x) = ( 1) k cos x and g (2k + 1) (x) = ( 1) k + 1 sin x for all k 0 and k Z.

Thus Pn (x) = , where m = [ n], greatest integer not exceeding n .

Since Rn (x) = , where strictly lies between 0 and x, and

1, = = 0


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and Pn (x) = = .

Therefore cos x = for all x .

Example 18. Find the fifth Taylor polynomial P5 of

a) sin x b) cos x

Solutions. a) P5 (x) = sin 0 + x cos 0 + + + +

= x +

Therefore P5 (x) = x + .

b) P5 (x) = cos 0 + x ( sin 0) + + + +

= 1 +

Therefore P5 (x) = 1 + .

Taylor Series about an Arbitrary Point.

Defn 4.2 If f has derivatives of all orders at a , then we call

the Taylor Series of f about the number a.

Example 19. Consider the function g(x) = ℓn x.

g is not defined on an open interval about 0. Hence g does not have a Taylor series of the


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form . But g(x) = ℓn (1 + x 1), and hence ℓn x

for 1 x 1 1.

Therefore ℓn x = for 0 x 2.

Example 20. Express the polynomial g(x) = 2x3 9x2 + 11x 1 as a polynomial in x a,

where a .

Solution. Since g(x) = 2x3 9x2 + 11x 1,

g(a) = 2a3 9a2 + 11a 1, g (a) = 6a2 18a + 11,

g (a) = 12a 18, g (a) = 12,

and g (n) (a) = 0 for all n 4.

Therefore g(x) = .

In particular if a = 2, then g(x) = 1 (x 2) + 3 (x 2)2 + 2(x 2)3.

Example 21. Find the Taylor series of cosh x about a, where a .

Solution. Let g(x) = cosh x.

Then g (2n) (a) = cosh a and g (2n + 1) (a) = sinh a for all n 0.

Therefore cosh x = .

Example 22. Using the Taylor series of cos x, approximate cos ( ) with an error less than

0.004.

Solution. cos x = x .

We need to find the smallest value of n for which

= 0.004.

Since 1, where f(x) = cos x and tk is strictly between and x, for all n

0.

0.004 (n + 1) ! n 6.


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Hence cos ( )

= 1 + 0.2586

Therefore cos ( ) 0.2586 with an error less than 0.004.

Example 23. Approximate e with an error less than 0.000003.

Solution. Let f(x) = e x , since e x = , e = .

We need to find the smallest value of n for which

= 0.000003.

3 0.000003, where tk (0, 1).

(n + 1) ! 1,000,000 n 9.

Hence = e.

Therefore e 2.718281526 with an error less than 0.000003.

Example Find the Taylor series expansion of f(x) = 2 x about a = 1.

Solution. f(1) = 2

f (x) = 2 x ℓn 2, f (1) = 2 ℓn 2

f (x) = 2 x (ℓn 2) 2, f (1) = 2 (ℓn 2) 2

f (x) = 2 x (ℓn 2) 3, f (1) = 2 (ℓn 2) 3

In general f (n) (x) = 2 x (ℓn 2) n and f (n) (1) = 2 (ℓn 2) n for n 0.

Therefore 2 x = for every number a.

Note that:- From the result of the above example we get:

2 x = for any x .


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UNIT V

Binomial Series. The Taylor series about 0 of the function f given by

f(x) = (1 + x)s , where s is any fixed number

is called a binomial series.

Now consider the Maclaurin series of f(x) = (1 + x)s, where s is any fixed number.

Let c0 = 1, c1 = s and cn = s (s 1) (s 2) … (s (n 1)) for all n 2.

Then f(n) (x) = cn (1 + x)s – n and f(n) (0) = cn .

Hence the Maclaurin series of f(x) = (1 + x)s is given by:

f(x) = (1 + x)s = .

In particular if s = , the Maclaurin series of

f(x) = = 1 + x +

= 1 + x +

Therefore = 1 + x + .

Defn 5.1 Let s be any number. Then we define the binomial coefficient by the formula:

= 1, = s and = for n

2.

In particular = s (s 1) and if s N, then = .

Now using this definition the above Maclaurin series of f(x) = (1 + x)s is given by:

(1 + x)s =

This series is called a binomial series.


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UNIT VI

Fourier Series Fourier series are series of cosine and sine terms. Since sine and cosine are periodic functions

we shall study Trigonometric series representation of what we call periodic functions.

6.1 Periodic Functions; Trigonometric Series.

Defn 6.1 A function f is said to be periodic if there is some positive number p such

that:

f (x + p) = f (x) for all x dom. f.

The number p is called the period of f and the smallest number p is called the primitive

(or fundamental) period of f.

Example 1. Let f (x) = sin x.

Now dom. f = , f (x + p) = f (x) x

sin (x + p) = sin x x

sin x cos p + sin p cos x = sin x x

sin x (1 cos p) = sin p cos x x

If x = , then 1 cos p = 0.

Now 1 cos p = 0 p = 2n ,where n N.

Therefore 2n, where n N is a period of f and 2 is the primitive period of f.

Example 2. Let g (x) = cos x.

Now dom. g = , g (x + p) = g (x) x

cos (x + p) = cos x x

cos x cos p sin p sin x = cos x x

cos x (cos p 1) = sin p sin x x

If x = , then cos p 1= 0.

Now cos p 1= 0 p = 2n, where n N.

Therefore 2n, where n N is a period of g and 2 is the primitive period of g.


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Example 3. Let f (x) = tan x.

Now dom. f =x x (2n + 1) , n Z.

f (x + p) = f (x) x dom. f. tan (x + p) = tan x x dom. f.

= tan x x dom. f.

tan x + tan p = tan x tan p tan 2 x x dom. f.

tan p (1 + tan 2 x) = 0 x dom. f.

tan p = 0 x dom. f.

p = n, where n N.

Therefore n, where n N is a period of f and is the primitive period of f.

Example 4. Let f (x) = sin (x), where 0.

Now dom. f = .

f (x + p) = f (x) x .

sin (x + p) = sin x x .

sin x cos p + sin p cos x = sin x x .

sin x (1 cos p) = sin p cos x x .

If x = , then 1 cos p = 0.

Now 1 cos p = 0 p = 2n, where n N.

Therefore , where n N is a period of f and is the primitive period of f.

Graphs of Periodic Functions.

The graph of a periodic function of period p is obtained by periodic representation of its graph in

any interval of length p.

Example 1. Sketch the graph of f (x) = tan x.

Solution. The primitive period of f is .


149


Example 2. Sketch the graph of g (x) = sin x .

Solution. Dom. g. = .

Now g (x + p) = g (x) x .

sin (x + p) = sin x x .

sin (x + p) = sin x x or sin (x + p) = sin x x .

p = 2n, where n N or sin x (1 + cos p) = sin p cos x x .

p = 2n, where n N or p = (2n 1), where n N

p = n, where n N.

Therefore is the primitive period of g.

Example 3. Sketch the graph of f, where

f (x) =

and f (x + 2) = f (x) x .


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Properties of Periodic Functions.

1. If f is a periodic function with period p, then for any n N np is a period of f.

Proof: (By the principle of mathematical induction)

Since f (x + n p) = f (x) it is true for n = 1.

Assume for n = k, i.e. f (x + k p) = f (x).

We need to show for n = k + 1, i.e. f (x + (k + 1) p) = f (x).

f (x + (k + 1) p) = f (x + k p + p) = f (x + p + k p) = f (x + p) = f (x).

Therefore for any n N, np is a period of f.

2. If f and g are periodic functions with period p, then for any a, b

h (x) = a f(x) + b g(x) x dom. h

is a periodic function of period p.

Proof: h (x + p) = a f(x + p) + b g(x + p)

= a f(x) + b g(x)

= h (x) x dom. h.

Therefore h is a periodic function with period p.

3. If f (x) is a periodic function with period p, then for any b and b 0


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a) f (b x) is periodic with period .

b) If ( ) is periodic with period b p.

Trigonometric Series.

Defn 6.2 The system of functions:

1, cos x, sin x, cos 2x, sin 2x, cos 3x, sin 3x, …

is called a trigonometric system.

The series

a0 +

where a0, an’s and bn’s are constants is called Trigonometric series, the

constants are called coefficients.

Defn 6.3 ( Orthogonality of the Trigonometric System)

1. Let f and g be functions defined on some interval a, b. f and g are said to be

orthogonal on a, b if and only if = 0.

2. The set of functions is said to be orthogonal on a, b if and

only if = 0 for n m.

Example 1. Show that the set of functions form an

orthogonal set on , .

Solution. Let n, m N and n m.

= sin nx.


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=

=

=

= 0.

Therefore form an orthogonal set on , .

Example 2. Show that the trigonometric system 1, cos nx, sin nx, … , where n N,

form an orthogonal set on , .

Solution. We need to show that:

a) = = = 0 for n, m N and m = n.

b) = = = 0 where n, m N

and m n.

When m = n.

= = = = 0,

and = = = 0.

When m n.

=

= = 0.


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=

= = 0.

=

= = 0.

Therefore 1, cos nx, sin nx, … for n N form an orthogonal set on , .

Defn 6.4

1. A function f (x) is said to be normal or normalized in a, b if and only if

= 1

The norm of a function f (x) denoted by on a, b is defined by:

=

2. An orthogonal set of functions n (x): n N is said to be an orthonormal

set on a, b if and only if

= 1 n N.

Example 1. The set of functions n (x): n (x) = sin nx, n Nis an orthonormal

set on , .

Solution. Let m, n N and m n.

=

= = 0.


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Therefore n (x): n (x) = sin nx, n Nis an orthogonal set on , .

For m = n.

= =

Therefore n (x): n (x) = sin nx, n Nis an orthonormal set on , .

Exercises. 6.1

1. Show that the following functions are non-periodic functions.

i) f (x) = x

ii) g ( x) = 2x2

iii) h (x) = e x

2. Sketch the graphs of the following functions, which are assumed to be periodic of

period 2 and, for x , are given by the formulas:

1. f (x) = e x 2. f (x) = x2

3. f (x) = x

4. f (x) =

5. f (x) =

3. For the trigonometric system find the corresponding orthonormal system on , .

6.2 Fourier Series of Functions of Period 2.

Let us assume that f (x) is a periodic function of period 2 that can be represented by a

trigonometric series.

f (x) = a0 + (1)

i.e. the series converges and has f (x) as its sum.

Now we need to determine the coefficients a0, an and bn .

i) To determine a0 integrate (1) from to .


155


= +

= 2 a0

a0 =

ii) To determine an, n N. Multiply (1) by cos mx and integrate from to where

m N, fixed.

= +

If m n, then

= = = 0.

If m = n, then

= = 0.

and =

=

= +

= .

Therefore an = n N.

iii) To determine bn , n N. Multiply (1) by sin mx where m N and integrate from

to .


156


= +

If m n, then

= = = 0.

If m = n, then

= 0,

= = 0.

and = = .

Therefore bn = n N.

Therefore we get what we call Euler’s Formulas.

a0 =

an =

bn = , where n N.


157


Defn 6.5 The numbers a0, an and bn are called the Fourier coefficients of f (x).

The trigonometric series

a0 +

Where a0, an and bn are called the Fourier coefficients is called the Fourier

Series of f (x) (Regardless of convergence).

Remark: In the Euler’s Formulas the interval of integration may be replaced by any interval of length 2.

Example 1. Find the Fourier coefficients of each of the following periodic functions.

i) f (x) = and f (x + 2) = f (x)

ii) f (x) = for 0 x 2 and f (x + 2) = f (x).

Solutions. i) a0 = = +

= + = 0,

an = = +

= + = 0,

and bn = = +

= +


158


= ( 1 cos n)

Thus bn = .

Therefore the Fourier series of the square wave function f is:

.

ii) a0 = = = = .

an = = =

and bn = = + = .

Therefore f (x) = + .

Convergence and Sum of Fourier series.

Suppose f is any function of period 2 for which the integrals

a0 =

an = n N.

and bn = n N.

exist. Then the Fourier series of f (x) is given by:

a0 + (1)

i) If the series in (1) converges to f (x), then we write:

f (x) = a0 + .

ii) If the series in (1) doesn’t converge to f (x), then we write:


159


f (x) a0 + .

Question. When is the Fourier series in (1) converges to f (x)?

To answer this question first we need to define piecewise continuity.

Defn 6.6 Piecewise Continuity.

A function f (x) is said to be piecewise continuous on an interval I if

i) the interval can be divided into a finite number of subintervals in each of

which f (x) is continuous.

and ii) the limits of f (x) as x approaches the end points of each subinterval are

finite.

Remark: A piecewise continuous function is one that has only a finite number of jump

discontinuities.

Examples. Consider the following functions.

i) f (x) =

ii) g (x) =

f and g are examples of piecewise continuous functions.

Example. h (x) = tan x for x ( , ).

Since h (x) is discontinuous at x = and , h is not

piecewise continuous on ( , ).

Theorem 6.1 (Representation by a Fourier series)


160


If a periodic function with period 2 is piecewise continuous in the interval

x and has a left hand derivative and a right hand derivative at each point of

that interval, then the Fourier series (1) of f (x) is convergent. Its sum is f (x) except at

a point x0 at which f (x) is discontinuous and the sum of the series is the average of the

left and the right hand limits at x0.

Example 1. Find the Fourier series of the function f where

f (x) =

and f (x + 2 ) = f (x).

Solution. The Fourier coefficients are:

a0 = an = 0 and bn = n N

Therefore the Fourier series of f (x) is

a0 + = .

Since f is piecewise continuous on ( , ) this series converges to f (x) for each

x ( , ) \ 0.

For x = 0, the Fourier series is given by

= k + k = 0

and indeed at x = 0, = 0.

Therefore f (x) = x ( , ) \ 0.

If x = , then f ( ) = k.

Hence k =

= .


161


Therefore

= .

Example 2. For each of the following, find the Fourier series expansion of the function f (x)

that has period 2.

i) f (x) = x2 , x .

ii) f (x) = x sin x , 0 x 2.

iii) f (x) = x x2 , x .

Solutions. i) a0 =

an =

.

bn =

= 0.

Therefore f (x) = for x .

For x = ,

Hence 2 =

Therefore

=


162


ii) a0 =

an =

If n 1, then an = and if n = 1, then a1 = .

bn =

If n 1, then bn = 0 and if n = 1, then b1 =

Therefore f (x) = .

If x = , then f () = 0.

Thus = .

Therefore

= .

iii) a0 = .

an =

.

bn =

.

Therefore f (x) = + 2 x ,

If x = 0, then


163


= .

Therefore

= .

6.3 Functions of any Period p = 2L.

Theorem 6.2 If a function f of period p = 2L has a Fourier series representation, then the series is:

f (x) = a0 +

where the Fourier coefficients of f (x) are given by the Euler formulas:

a0 =

an = n N,

and bn = n N.

Proof. (Refer page 578 Kreyszig)

Remark: The interval of integration in the above formulas may be replaced by any

interval of length p = 2L.

Example 1. Find the Fourier series of the function f (x) with period p = 4, where

f (x) = and f (x + 4) = f (x)

Solution. p = 4 L = 2.


164


Now a0 = = = = 1.

an =

,

=

bn =

Therefore f (x) = 1 + .

If x = 0, then f (0) = 2.

= .

Example 2. Find the Fourier series representation of the periodic function f (x) with period

p = 1, where

f (x) = sin x for 0 x 1 and f (x + 1) = f (x).

Solution. p = 1 L = .

Thus a0 = .

an = 2


165


=

= +

=

=

Thus an = n N.

bn = 2

=

= = 0.

Thus bn = 0 n N.

Therefore f (x) = 2 + 4 .

If x = 0, then

.

If x = , then

6.4 Even and Odd Functions.


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Defn 6.7 (Even and Odd Functions)

1. A function f (x) is even if

f ( x) = f (x) x dom. f.

2. A function g (x) is odd if

g ( x) = g (x) x dom. g.

Example 1. f (x) = cos nx , x , where n N is even.

2. g (x) = sin nx , x , where n N is odd.

Note that:

1. If f (x) is an even function, then

= 2

2. If g (x) is an odd function, then

= 0

3. The product of

i) even and odd functions is odd.

ii) two odd functions is even.

iii) two even functions is even.

Theorem 6.3 (Fourier series of Odd and Even Functions)

a. The Fourier series of an even function f of period 2L is a Fourier cosine

series.

f (x) = a0 + with coefficients


167


a0 = and an = n N

b. The Fourier series of an odd function f of period 2L is a Fourier sine series.

f (x) = with coefficients

bn = n N.

Theorem 6.4 (Sum of Functions)

The Fourier coefficients of a sum of two function f1 and f2 are the sum of the

corresponding Fourier coefficients of f1 and f2.

The Fourier coefficients of cf are c times the corresponding Fourier

coefficients of f.

Example 1. Let f (x) = and f (x +2) = f (x) x dom. f.

Find the Fourier series of f (x).

Solution. Let f1 (x) = and f1 (x +2 ) = f1 (x) x dom. f

and let f2 (x) = k x .

Then (f1 + f2) (x) = f (x) x dom. f.

But f1 (x) = and f2 (x) = k.


168


Therefore f (x) = k + x dom. f.

Example 2. Find the Fourier series of the function f (x) where f (x) = x + if x

and f (x + 2) = f (x) x dom. f.

Solution. Let f1 (x) = x and f2 (x) = if x and be periodic functions with period 2.

Now f1 (x) = 2 and f2 (x) = .

Therefore f (x) = + 2 x dom. f.

Example 3. Find the Fourier series representation of

f (x) = x + x2 on , .

Solution. Let f1 (x) = x and f2 (x) = x2 for x , and f1 and f2 be periodic functions with

period 2.

Now f1 (x) is odd hence it has a Fourier sine series

and bn =

= n N

Thus f1 (x) = 2 .

f2 (x) is even, hence it has a Fourier cosine series representation.

Now a0 =

and an =

= n N

Thus f2 (x) = + 4 .

Therefore f (x) = +


169


Note that:

If g (x) is defined x , then the function

P (x) = is even

and q (x) = is odd.

Hence g (x) = p (x) + q (x) and g ( x) = p (x) q (x).

Exercises 6.4

Represent the following functions as a sum of even and odd functions.

i) = +

ii) = +

iii) = cosh kx + sinh kx.

6.5 Half-range Expansion.

Sometimes it is required to extend a function f (x) in the range (0, L) in a Fourier series of

period 2L, and it is immaterial what the function may be outside the range 0 x L. We could

extend f (x) with period 2L and then represent the extended function by a Fourier series.

i) If we extend the function f (x) by reflecting it in the y-axis, then the Fourier series

expansion contains only the cosine terms.

ii) If we extend the function f (x) by reflecting it in the origin, then the Fourier series

expansion contains only the sine terms.

The cosine halt-range expansion is:

f (x) = a0 +

where a0 = and an = n N


170


The sine halt-range expansion is:

f (x) = , where bn = n N

Example. Find the two half-range expansions of the function

f (x) = where k 0.

Solution. f (x) can be extended to (L, L) into two ways.

I. Make f (x) even on (L, L). i.e. f (x) = f ( x) x (L, L), to get Fourier cosine

expansion.

II. Make f (x) odd on (L, L). i.e. f ( x) = f (x) x (L, L), to get Fourier sine

expansion.

Fourier cosine expansion.

Now a0 = = +

= +

= +

=

an = =


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= +

=

=

Thus an = n N.

Therefore f (x) = + .

II. Fourier sine expansion.

Now an = =

= +

=

=

Thus bn = n N.

Therefore f (x) = .


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Since f (x) is piecewise continuous and = = k.

= k, but

Thus = .

Therefore

= .

Exercises 6.5 1. Represent each of the following functions by a Fourier sine and cosine series extensions,

where L + .

i) f (x) = x, 0 x L.

ii) f (x) = x2, 0 x L.

iii) f (x) = sin x, 0 x L.

2. Obtain a half-range sine and cosine series for

i) ex in 0 x 1.

ii) x2 x in 0 x .


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