18MAB201T-U2-part1.pdf

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18MAB201T-Transforms and Boundary Value

Problems

Prepared by

Dr. S. ATHITHAN

Assistant Professor

Department of of Mathematics

Faculty of Engineering and Technology

SRM INSTITUTE OF SCIENCE AND TECHNOLOGY

Kattankulathur-603203, Kancheepuram District.

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FOURIER SERIES

TOPICS:

? Introduction of Fourier series -Dirichlet’s conditions for existence of Fourier Series

? Fourier series -related problems

? Half Range sine series-related problems

? Half Range cosine series-related problems

? Parseval’s Identity (without proof)-related problems

? Harmonic Analysis for finding fundamental harmonic

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1

Fourier Series

1.1

Basics

To fix ideas we will work over the interval[−π, π], and deal with variations later. Fourier’s setup is that given a functionf defined over that interval, it is possible to write it as a (possibly infinite) linear combination of the ‘pure’ waves 1, cos(x), cos(2x), . . . , sin(x), sin(2x), etc, so that

f(x) = A0 2 +

∞

X

n=1

Ancos(nx) +Bnsin(nx). (1.1)

Soon it will become clear why we have divided the zeroth coefficient by 2.

For the moment let’s believe (1.1), and see how to compute the coefficientsAn, Bn. The key

here are the following formulas, calledorthogonality relations.

Exercise: 1. Show that ifm,nare non-negative integers, andm6=n, then

Z π

−π

cos(nx) cos(mx)dx= 0,

Z π

−π

sin(nx) sin(mx)dx= 0.

Also, even if we drop the restrictionm6=n,

Z π

−π

cos(nx) sin(mx)dx= 0.

Exercise: 2. Show that ifnis a positive integer, then

Z π

−π

cos2(nx)dx= Z π

−π

sin2(nx)dx= π.

Without worrying about rigor, let’s follow Fourier and obtain formulas for the coefficients. The coefficientA0 is the simplest to find: integrate (1.1) from−πtoπto get

Z π

−π

f(x)dx = Z π

−π A0

2 dx+

∞

X

n=1

An

Z π

−π

cos(nx)dx+Bn Z π

−π

sin(nx)dx

.

The series on the right vanishes, and we find that

A0 =

1

π Z π

−π

f(x)dx.

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sin(mx). We get

Z π

−π

f(x) sin(mx)dx =

Z π

−π A0

2 sin(mx)dx+

∞

X

n=1

An Z π

−π

cos(nx) sin(mx)dx+Bn Z π

−π

sin(nx) sin(mx)dx.

What is important to notice is thatallof the integrals on the right side vanish, except for the one multiplyingBm. The equation forBm becomes

Bm = 1

π Z π

−π

f(x) sin(mx)dx m = 1,2,3, . . . . (1.2) Likewise the formula forAm is

Am = 1

π Z π

−π

f(x) cos(mx)dx m= 0,1,2, . . . . (1.3) (BecauseA0appears in (1.1) divided by 2, the above formula forAm also works forA0.)

Formulas (1.2) and (1.3) allow us to compute the Fourier coefficients off. We postpone any discussion about convergence until Section 6.6.

One last important point: Even though f is defined only on [−π, π], the right-hand side of (1.1) is2π-periodic, so we could viewf as being defined over the whole line, but2π-periodic as well.

1.2

Some Examples

For our first example we considerf(x) = xover the interval[−π, π]. ClearlyA0 = 0, and

forn ≥ 1 An =

1

π Z π

−π

xcos(nx)dx = 1 π

(

xsin(nx)

n π

−π

−

Z π

−π

sin(nx)

n dx

)

= 0.

Exercise: 3. Iff is anoddfunction (f(x) = −f(−x)for allx), show thatAn = 0.

Forn ≥1we obtain

Bn = 1

π Z π

−π

xsin(nx)dx = 1 nπ

−xcos(nx)|π_−π+ Z π

−π

cos(nx)dx

= (−1)

n+1_·2

n .

So, if this Fourier series converges, we obtain, forx ∈[−π, π],

x=

∞

X

n=1

(−1)n+1·2

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For our second example let’s takef(x) = x2.

Exercise: 4. Show that iff iseven(f(x) = f(−x)for allx), thenBn = 0.

Sincex2 is even we only need to computeAn: A0 =

1

π Z π

−π

x2dx= 2 3 π

2_,

and forn≥ 1

An = 1

π Z π

−π

x2cos(nx)dx

= 1

nπ

x2sin(nx) π

−π− Z π

−π

2xsin(nx)dx

= (−1) n_·4 n2 .

As before, if we do have convergence, then, forx∈ [−π, π],

x2 = π

2

3 +

∞

X

n=1

(−1)n_·4

n2 cos(nx).

1.3

Other Intervals

The first variation on the above is to consider intervals of the form[−l, l]. The trick is to rescale the problem: iff is defined forx ∈ [−l, l], we definegby setting g(xπ/l) = f(x); theng

is defined over[−π, π], and

f(x) = g

_xπ

l

= A0 2 +

∞

X

n=1

Ancos

_nπx

l

+Bnsin

_nπx

l

,

whereAnandBn are the coefficients for the functiong.

Exercise: 5. Ifan andbn are the coefficients for the functionf, show that

an = An = 1

l Z l

−l

f(x) cos

_nπx

l

dx,

bn =Bn = 1

l Z l

−l

f(x) sin

_nπx

l

dx.

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Bn = 2

l Z l

0

f(x) sin

nπx

l

dx.

Exercise: 7. Show that iff is extended to be an even function, thenBn = 0, and

An = 2

l Z l

0

f(x) cos

nπx

l

dx.

The reason why you would want to use one type of extension over the other depends on the type of problem at hand. For Dirichlet conditions on the half-interval we want the odd extension; for Neumann conditions we want the even extension.

1.4

How the Decay Happening on the Coefficients?

In this section we want to viewf as being2π-periodic and defined over the whole line. Going back to the two examples we saw in Section 6.4, whenf(x) = xwe saw that|Bn|decayed

likeC/n, for some constantC(in fact,C = 2for that example). Likewise, whenf(x) = x2, we had|An|decaying likeC/n2. The speed with which coefficients decay is rather important

if we are interested in the convergence of the series, and the following theorem is very useful.

Theorem 1.1. Letf be a2π-periodic function withk−1continuous derivatives, and whose

k-th derivative is piece-wise continuous. Then the Fourier coefficients of f decay likeC/nk, where the constantC depends only onf.

Proof. By induction onk. Let’s computeAn,n ≥ 1. An = 1

π Z π

−π

f(x) cos(nx)dx

= 1

nπ

f(x) sin(nx)|π_−π −

Z π

−π

f0(x) sinnx dx

= − 1

nπ Z π

−π

f0(x) sinnx dx.

By induction this last integral is bounded by some constant overnk−1, so that|An| ≤C/nk.

Likewise forBn,n ≥ 1we have Bn =

1 π

Z π

−π

f(x) sin(nx)dx

= 1

nπ

−f(x) cos(nx)|π_−π+ Z π

−π

f0(x) cosnx dx

= 1

nπ Z π

−π

f0(x) cosnx dx,

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us |Bn| ≤ C/nk. We are left to do the basis of induction, k = 1. That means thatf0 is

piecewise continuous. The same formulas we have above for An and Bn are valid, and the easy estimate

Z π

−π

f0(x) sin(nx)dx

≤

Z π

−π

|f0(x)|dx =C

shows that|An|decays likeC/n. A similar estimate holds forBn.

This theorem says that the more derivatives you have in hand, the better the Fourier series will converge. Also true is a converse of this theorem: The faster the coefficients decay, the more derivatives you have forf. We will not prove this fact.

NB: The above theoremis not optimal, as shown by the examples we computed. The reader is invited to sharpen the hypotheses of this theorem.

1.5

Use of Linear Algebra by adopting

L

We now ask ourselves the important question: Why, and how, would (1.1) be true? If we look at our problem from the point of view of Linear Algebra, we are claiming that the functionf is in the linear span of the functions1,cos(x),cos(2x), . . . ,sin(x),sin(2x), etc. So what we have here is a vector space problem – but which vector space?

Definition 1.5.1. The spaceL2 =L2([−π, π])is formed by those functionsf for which

Z π

−π

|f(x)|2_dx

is finite.

Technically the integral in the above definition should be the Lebesgue integral – but we will gloss over this point and work with the Riemann integral here.

Definition 1.5.2. TheL2 norm of a functionf is given (and denoted) by

kfk = s

Z π

−π

|f(x)|2_dx.

The inner product of two functionsf andginL2 is given by

(f, g) = Z π

−π

f(x)g(x)dx.

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S(f, N)(x) = A0 2 +

N X

n=1

Ancos(nx) +Bnsin(nx).

We say thatS(f, N)converges tof inL2 if and only if

lim

N→∞kf −S(f, N)k = 0.

Convergence inL2can be very strange.

Exercise: 8. Find a sequence of functionsgn ∈ L2 such thatkgnk → 0, butgn(x)does not

converge to zero for any value ofx.

Our next result shows thatS(f, N)is, in some sense, the best approximation forf. To state the result, let

g(x) = a0 2 +

N X

n=1

ancos(nx) +bnsin(nx),

where theanandbn are any coefficients.

Theorem 1.2. Withgdefined as above, we have

kf −S(f, N)k ≤ kf −gk.

In other words, of all finite trigonometric sums of order at mostN, the one with exactly the Fourier coefficients off is the one that best approximatesf inL2.

Proof. We will show the following equality:

kf −gk2 _{=kf −S(f, N)k}2 _{+kg−S(f, N)k}2_.

(1.4) As a consequence

kf −gk ≥ kf −S(f, N)k,

with equality holding if and only if g = S(f, N). In what follows, we will write S = S(f, N). To see that (1.4) is true, we start by expanding the left hand side:

kf −gk2 _{= (f −g, f −g) = (f, f)−2(f, g) + (g, g) = kfk}2 _{−2(f, g) +kgk}2_.

Now, add and subtract the quantitieskSk2_{and2(f, S):}

kf −gk2 _=kfk2 _{−2(f, S) +kSk}2_{− kSk}2 _{−2(f, g) + 2(f, S) +kgk}2_.

Notice thatkfk2 _{−2(f, S) +kSk}2 _{=kf −Sk}2_{. The above expression becomes}

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All that is left to do is see that kg − Sk2 _{= −kSk}2 _{− 2(f, g) + 2(f, S) + kgk}2_{. We}

compute:

(f, S) = A0

2 (f,1) + N X

n=1

An(f,cos(nx)) +Bn(f,sin(nx))

= πA

2 0

2 + N X

n=1

π(A2_n+B_n2) = (S, S) = kSk2_.

Likewise we find

(f, g) = πa0A0

2 +

N X

n=1

π(a0A0+b0B0) = (S, g).

The reader is now invited to finish the proof.

Now take g = 0 in (1.4). We get kfk2 _{= kf − Sk}2 _{+ kSk}2_{, and so kfk}2 _{≥ kSk}2_.

Explicitly this reads

A2 0

2 + N X

n=1

(A2_n+B_n2)≤ 1

πkfk

2_.

SinceN is arbitrary, we obtainBessel’s inequality:

A2₀

2 +

∞

X

n=1

(A2_n+B_n2)≤ 1

πkfk

2_.

(1.5)

Bessel’s inequality is related to Pythagoras’ Theorem in_Rn. What that theorem says is that the square of the norm of a vector is the sum of the squares of the sizes of each of that vector’s components, when projected in orthogonal directions. The reason why we did not get equality in Bessel’s inequality is simply because we don’t know yet if the orthogonal directions given by the trigonometric functions span the whole ofL2. If we knew that, then Bessel’s inequality would become an equality, and vice versa: if we could prove that we have equality in (1.5), then we would know that the trigonometric functions spanL2. We will assume that we have proved that the trigonometric functions spanL2. Then we have:

Theorem 1.3. (Parseval’s identity.) Iff ∈L2, then

A2 0

2 +

∞

X

n=1

(A2_n+B_n2) = 1 πkfk

2

.

Exercise: 9. Apply Parseval’s identity to the functionf(x) = xand conclude that

∞

X

n=1

1 n2 =

π2 6 .

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State Dirichlet’s condition on the existence of Fourier series.

Hints/Solution:

A function f(x) of period 2π defined in an interval (c, c + 2π) of length 2π can be ex-panded (or) represented as an infinite trigonometric series of the form a0

2 +

∞

X

n=1

ancosnx+

∞

X

n=1

bnsinnx, where a0, an and bn are constants, if the following conditions (Dirichlet’s

conditions) are satisfied:

(i)f(x)is periodic with period2πand bounded in(c, c+ 2π)

(ii)f(x)is piecewise continuous with finite number of discontinuities in(c, c+ 2π)

(iii)f(x)has atmost finite number of maxima and minima in(c, c+ 2π)

Example : 2

Find the Root Mean Square (RMS) value off(x) = x2, 0< x < 1.

Hints/Solution:W.K.T. RMS Value is given by

v u u u t

b R

a

[f(x)]2_dx

b−a

∴

v u u u t

1

Z

0

x4_dx=

r 1

5

Example : 3

Findan forf(x) = x+x2 in the interval(−π, π). Hence deduce

∞

X

n=1

1

n2 =

π2

6 .

Hints/Solution:Fourier series forf(x)in−π ≤ x≤ π is given by

f(x) = a0 2 +

∞

X

n=1

(ancosnx+bnsinnx),

wherea0 =

1

π π Z

−π

f(x)dx = 2π

2

3

an = 1

π π Z

−π

f(x) cosnxdx = 4 n2(−1)

n

andbn = 1

π π Z

−π

f(x) sinnxdx = −2

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∴ x+x2 = π

3

3 +

∞

X

n=1

(−1)n

₄

n2 cosnx−

2

nsinnx

Deduction:f(π) +f(−π)

2 =

π3

3 +

∞

X

n=1

(−1)n ₄

n2 cosnx

=⇒ ∞

X

n=1

1

n2 =

π2

6

Example : 4

(a). Obtain the Fourier series forf(x) = |x|in−l ≤ x≤ l. (b). Expandingx(π −x)as a Fourier sine series in(0, π).

(c). Finda0 and RMS value of the functionf(x) = x−x2in−1 < x < 1.

Hints/Solution:

(a). Sincef(x) = |x|is an even function, the Fourier series forf(x)in−l ≤ x ≤ lis given by

f(x) = a0 2 +

∞

X

n=1

ancos

nπx

l

,

wherea0 =

2

l l Z

0

f(x)dxand

an = 2 l

l Z

0

f(x) cos

_nπx

l

dx.

Here we geta0 = landan = 2l

n2_π2[(−1)

n _−1].

(b). The Fourier sine series forf(x)in0 ≤ x≤ πis given by

f(x) =

∞

X

n=1

bnsinnx,

wherebn = 2

l l Z

0

f(x) sinnxdx.

Here we getbn = 4

n3_π[1−(−1)

n_].

(c). a0 =

−2

3 and the RMS value is given by ¯

y2 = 1 2

1

Z

−1

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(a). Obtain the Fourier series forf(x) = 1 +x+x2 in(−π, π)

Hints/Solution:

(a). Fourier series forf(x)in−π ≤ x≤ πis given by

f(x) = a0 2 +

∞

X

n=1

(ancosnx+bnsinnx),

wherea0 =

1

π π Z

−π

f(x)dx = 1 2π

_2π3

3 + 2π

,

an = 1

π π Z

−π

f(x) cosnxdx = 4 n2(−1)

n

and

bn = 1

π π Z

−π

f(x) sinnxdx = 2 n(−1)

n+1

Example : 6

(a). Obtain the Fourier series forf(x) = (

0 if −2 < x < 0 x if 0< x < 2 .

(b). Expandf(x) = x(l−x)in(0, l)as Fourier cosine series. (c). Findbnand RMS value of the functionf(x) = xin0< x < l.

Hints/Solution:

(a). The Fourier series forf(x)in−2≤ x ≤ 2is given by

f(x) = a0 2 +

∞

X

n=1

ancos

_nπx

2

+

∞

X

n=1

bnsin

_nπx

2

,

wherea0 =

1

2

2

Z

−2

f(x)dx= 1,

an = 1

2

2

Z

−2

f(x) cos

_nπx

2

dx = 2

n2_π2[(−1)

n_−1]and

bn = 1

2

2

Z

−2

f(x) sin

_nπx

2

dx= 2

nπ[(−1) n+1_].

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f(x) = a0 2 +

∞

X

n=1

ancos

_nπx

l

,

wherea0 =

2

l l Z

0

f(x)dx= l

3

3,

an = 2

l l Z

0

f(x) cos

_nπx

l

dx = −2l

2

n2_π2[1 + (−1)

n_].

(c). bn = 2 l

l Z

0

f(x) sin

_nπx

2

dx = 2l

nπ[(−1) n+1

]and

the RMS value is given byy¯2 = 2 l

l Z

0

x2dx= 2l

3

3l .

Example : 7

(a). Obtain the Fourier series forf(x) = (π −x)2 in(−π, π)

Hints/Solution:

(a). Here the function is neither odd nor even.

∴Fourier series forf(x)in−π ≤ x ≤π is given by

f(x) = a0 2 +

∞

X

n=1

(ancosnx+bnsinnx),

wherea0 =

1

π π Z

−π

f(x)dx = 8π

2

3 ,

an = 1

π π Z

−π

f(x) cosnxdx = 4 n2(−1)

n

and

bn = 1

π π Z

−π

f(x) sinnxdx = 4π n (−1)

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2

Exercise/Practice/Assignment Problems

1. Obtain the Fourier series for the following functions in the interval(0,2π)

(a) f(x) = x(2π −x). and deduce the sum 1

12 +

1

22 +

1

32 +. . . Ans: a0 =

4π2

3 , an = −4

n2 , bn = 0Hint: Use

f(0) +f(2π)

2 to deduce the sum and put

x= 0.

(b) f(x) = eax. Ans:a0 =

eaπ

aπ[2 sinhaπ], an =

2aeaπ

π(a2 _+n2₎[sinhaπ], bn = −naeaπ

π(a2 _+n2₎[2 sinhaπ]

(c) f(x) = x2. Ans:a0 = 8π2

3 , an = 4

n2, bn = −4π

n

(d) f(x) = x−x2. Ans:a0 =

6π2−8π3

3π , an =

−4

n2 , bn =

4π−2

n

(e) f(x) = 1

4(π−x)

2_.

Ans: f(x) = π 2 12 + ∞ X n=1 cosnx n2

(f) f(x) = (

1, 0< x < π

2, π < x <2π . Ans: f(x) =

3 2 − 2 π ∞ X n=1 sinnx n

2. Obtain the Fourier series for the following functions in the interval(−π, π)

Note 2.0.1. Before solving problems in the interval(−π, π), we need to check whether the given function isEVENorODD. Then obtain the Fourier series accordingly.

(a) f(x) = 





x+ π

2, −π < x <0 π

2 −x, 0 < x < π

. Ans: f(x) = 4

π ∞ X n=1,3,5,... cosnx n2

(b) f(x) = (π−x)2. Ans:f(x) = 4π 2

3 +4

∞

X

n=1

(−1)n_cosnx

n2 +4π

∞

X

n=1

(−1)n_sinnx

n

(c) f(x) = 1 +x +x2 and deduce the sum 1

12 +

1

22 +

1

32 + · · · =

π2

6 . Ans:

f(x) = π 2

3 + 1 + 4

∞

X

n=1

(−1)n_cosnx

n2 + 4π

∞

X

n=1

(−1)n_sinnx

n Hint: Use

f(−π) +f(π)

2 and putx= πto deduce the sum

(d) f(x) = |cosx| Ans: f(x) = 2

π + 0−4

∞

X

n=2

cosnπ₂

π(n2₋₁₎cosnx

(e) f(x) = (

π+ 2x, −π < x <0

π−2x, 0 < x < π and deduce the sum 1

12 +

1

32 +

1

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8 Ans: f(x) = ∞

X

n=1,3,5,... 8

n2_π cosnx Hint: Put x = 0to deduce the sum

(Sincexis the point of continuity)

(f) f(x) = |sinx| Ans: f(x) = 2

π + 0−4

∞

X

n=2,4,6,...

1

π(n2 ₋₁₎cosnx

(g) f(x) = xsinx Ans:f(x) = 1− 1

2cosx+ 2

∞

X

n=2,4,6,...

(−1)n

(1−n2₎cosnx

(h) f(x) = x2 and deduce the sum 1

14 +

1

24 +

1

34 +· · · =

π4

90 Ans: f(x) =

π2 3 + ∞ X n=1 4(−1)n

n2 cosnx Hint: For deduction we need to use thePARSEVAL’S

THEOREM(OR IDENTITY)

1

l

l

Z

−l

[f(x)]2dx= a

2 0 2 + ∞ X n=1

(a2_n+b2_n)

3. Obtain the Fourier series for the following functions in the interval(0,2l)and(−l, l)

Note 2.0.2. Before solving problems in the interval(−l, l), we need to check whether the given function isEVENorODD. Then obtain the Fourier series accordingly.

(a) f(x) = (l − x)2 in (0,2l) and deduce the sum 1

12 +

1

22 +

1

32 +. . . Ans: f(x) = l

2

3 + 4l2

π2

∞

X

n=1

cosnπx

n2 Hint: Use

f(0) +f(2l)

2 and put x = 0to

deduce the sum

(b) f(x) =         

0, −2 < x < −1 1−x, −1 < x < 0 1 +x, 0 < x < 1 0, 1 < x < 2

. Ans:f(x) = 1 4+ 8 π2 ∞ X n=1

sin2 nπ₄

n2 cos

nπx

2

(c) f(x) = |x|,−l < x < l Ans: f(x) = l 2 − 4l π2 ∞ X n=1,3,5,... 4l

n2_π2

cosnπx

l

Ans:f(x) = 4

π ∞ X n=1,3,5,... cosnx n2

4. Obtain the Fourier series for the following functions in the interval(0, l)and(0, π)(Half range cosine or sine series)

(a) f(x) = kx(l−x)in(0, l)(Half range sine series)

Ans:f(x) = −8kl 2 π3 ∞ X n=1 1

(2n−1)3 sin

(2n−1)πx

L

E

C

T

U

R

E

N

O

T

E

S

O

F

A

T

H

IT

H

A

N

S

(b) f(x) = (

x−1, 0< x < 1

1−x, 1< x < 2 .(Half range sine series)

Ans:f(x) =

∞

X

n=1,3,5,...

_{8 sin} nπ 2

n2_π2 − 4

nπ

sinnπx 2

(c) f(x) = xsinx,0 < x < π (Half range cosine series) Ans: f(x) = 1 − 1

2 cosx−2

∞

X

n=2

(−1)n

n2₋₁cosnx

(d) f(x) = x,0 < x < l(Half range cosine series) and deduce the sum 1

14 +

1

34 +

1

54 +· · · =

π4

96 Ans: f(x) =

l 2 − 4l π2 ∞ X n=1,3,5,...

cos nπx_l

n2 Hint: For deduction

we need to use thePARSEVAL’S THEOREM(OR IDENTITY)

1

l

l

Z

−l

[f(x)]2dx= a

2 0 2 + ∞ X n=1

(a2_n+b2_n)

(a) Obtain the Fourier series for f(x) = (

xsinπx, −1 ≤ x≤ 1

0, elsewhere and deduce the sum 1

1·3 − 1

3·5 + 1

5·7 + . . . Ans: f(x) =

1

π −

1

2π cosπx −

2 π ∞ X n=2 (−1)n+1

n2₋₁ cosnπx;

π −2

4

(b) Obtain the Fourier series for f(x) = (

cosx−sinx, −π < x <0

cosx+ sinx, 0 < x < π and

de-duce the sum 1

1·3− 1

3·5+ 1

5·7+. . . Ans:f(x) =

2 π− 4 π ∞ X n=1 1

4n2 ₋₁cos 2nx;

π−2

4

(c) Obtain the Fourier series forf(x) = (

−πx−x2, −π < x < 0

πx−x2, 0 < x < π and deduce

the sum

∞

X

n=1

1

n2 and

∞

X

n=1

(−1)n−1

n2 Ans:f(x) = π2 6 − ∞ X n=1 1

n2 cos 2nx;

π2

6 ;

π2

12

(d) Find the half-range cosine series for f(x) = (

1 +x, 0 < x < 1

0, elsewhere and de-duce the sum

∞

X

n=1

1

(2n−1)4 Ans:f(x) =

3 2 − 4 π2 ∞ X n=1 1

(2n−1)2 cos(2n−

1)πx;π 4

L

E

C

T

U

R

E

N

O

T

E

S

O

F

A

T

H

IT

H

A

N

S

(e) Find the half-range sine series for f(x) = (

2x, 0 < x < 1

4−2x, 1 < x < 2 and deduce

the sum

∞

X

n=1

1

(2n−1)4 Ans: f(x) =

16

π2

∞

X

n=1 1

n2 sin

nπ

2 sin

nπx

2 ;

π4

96

(f) Obtain half-range cosine series forf(x) = 





x, 0 < x < π 2 π −x, π

2 < x < π

and deduce

the sum

∞

X

n=1

1

(2n−1)4 Ans: f(x) = π

4 − 2

π

∞

X

n=1 1

n2 cos 2nx;

π4

96

PRACTICE MORE PROBLEMS ON SOME OF THE REFERENCE BOOKS.

Acknowledgement:

Some of the portions of this material are taken from the sources available from various sources. I thank the authors for those who prepared the calculus books and related materials.