Fourier Series - Sarah SchottFourier Series When the French mathematician Joseph Fourier (1768–1830) was trying to solve a prob-lem in heat conduction, he needed to express a function

Fourier Series

When the French mathematician Joseph Fourier (1768–1830) was trying to solve a prob-lem in heat conduction, he needed to express a function as an infinite series of sine andcosine functions:

Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating prob-lems concerning vibrating strings and astronomy.

The series in Equation 1 is called a trigonometric series or Fourier series and it turnsout that expressing a function as a Fourier series is sometimes more advantageous thanexpanding it as a power series. In particular, astronomical phenomena are usually periodic,as are heartbeats, tides, and vibrating strings, so it makes sense to express them in termsof periodic functions.

We start by assuming that the trigonometric series converges and has a continuous func-tion as its sum on the interval , that is,

Our aim is to find formulas for the coefficients and in terms of . Recall that for apower series we found a formula for the coefficients in terms of deriv-atives: . Here we use integrals.

If we integrate both sides of Equation 2 and assume that it’s permissible to integrate theseries term-by-term, we get

But

because is an integer. Similarly, . So

y�

�� f �x� dx � 2�a0

x�

�� sin nx dx � 0n

y�

�� cos nx dx �

1

n sin nx�

��

�

�1

n �sin n� � sin��n�� 0

� 2�a0 � ��

n�1 an y

�

�� cos nx dx � �

�

n�1 bn y

�

�� sin nx dx

y�

�� f �x� dx � y

�

�� a0 dx � y

�

��

�

n�1 �an cos nx � bn sin nx� dx

cn � f �n��a��n!f �x� � � cn�x � a�n

fbnan

�� x � �f �x� � a0 � ��

n�1 �an cos nx � bn sin nx�2

��, ��f �x�

� b1 sin x � b2 sin 2x � b3 sin 3x � � � �

� a0 � a1 cos x � a2 cos 2x � a3 cos 3x � � � �

f �x� � a0 � ��

n�1 �an cos nx � bn sin nx�1

f

1

and solving for gives

To determine for we multiply both sides of Equation 2 by (where isan integer and ) and integrate term-by-term from to :

We’ve seen that the first integral is 0. With the help of Formulas 81, 80, and 64 in the Tableof Integrals, it’s not hard to show that

for all and

So the only nonzero term in (4) is and we get

Solving for , and then replacing by , we have

Similarly, if we multiply both sides of Equation 2 by and integrate from to ,

we get

We have derived Formulas 3, 5, and 6 assuming is a continuous function such thatEquation 2 holds and for which the term-by-term integration is legitimate. But we can stillconsider the Fourier series of a wider class of functions: A piecewise continuous functionon is continuous except perhaps for a finite number of removable or jump disconti-nuities. (In other words, the function has no infinite discontinuities. See Section 2.4 for adiscussion of the different types of discontinuities.)

�a, b�

f

n � 1, 2, 3, . . .bn �1

� y

�

�� f �x� sin nx dx6

��sin mx

n � 1, 2, 3, . . .an �1

� y

�

�� f �x� cos nx dx5

nmam

y�

�� f �x� cos mx dx � am�

am�

y�

�� cos nx cos mx dx � 0

�

for n � m

for n � m

mny�

�� sin nx cos mx dx � 0

� a0 y�

�� cos mx dx � �

�

n�1 an y

�

�� cos nx cos mx dx � �

�

n�1

bn y�

�� sin nx cos mx dx4

y�

�� f �x� cos mx dx � y

�

�� a0 � ��

n�1 �an cos nx � bn sin nx�� cos mx dx

��m � 1mcos mxn � 1an

a0 �1

2� y

�

�� f �x� dx3

a0

2 ■ FOUR IER SER I ES

■ ■ Notice that is the average value of over the interval .��, ��

fa0

Definition Let be a piecewise continuous function on . Then theFourier series of is the series

where the coefficients and in this series are defined by

and are called the Fourier coefficients of .

Notice in Definition 7 that we are not saying is equal to its Fourier series. Later wewill discuss conditions under which that is actually true. For now we are just saying thatassociated with any piecewise continuous function on is a certain series calleda Fourier series.

EXAMPLE 1 Find the Fourier coefficients and Fourier series of the square-wave functiondefined by

and

So is periodic with period and its graph is shown in Figure 1.

SOLUTION Using the formulas for the Fourier coefficients in Definition 7, we have

a0 �1

2� y

�

�� f �x� dx �

1

2� y

0

�� 0 dx �

1

2� y

�

0 1 dx � 0 �

1

2� ��

1

2

0

FIGURE 1Square-wave function

(a)

π 2π_π

1

y

x0

π 2π_π

1

y

x

(b)

2�f

f �x � 2�� f �x�f �x� � 01 if �� x � 0if 0 � x � �f

��, ��f

f �x�

f

bn �1

� y

�

�� f �x� sin nx dxan �

1

� y

�

�� f �x� cos nx dx

a0 �1

2� y

�

�� f �x� dx

bnan

a0 � ��

n�1 �an cos nx � bn sin nx�

f��, ��f7

FOUR IER SER I ES ■ 3

■ ■ Engineers use the square-wave function indescribing forces acting on a mechanical systemand electromotive forces in an electric circuit(when a switch is turned on and off repeatedly).Strictly speaking, the graph of is as shown in Figure 1(a), but it’s often represented as in Figure 1(b), where you can see why it’s called asquare wave.

f

and, for ,

The Fourier series of is therefore

Since odd integers can be written as , where is an integer, we can write theFourier series in sigma notation as

In Example 1 we found the Fourier series of the square-wave function, but we don’tknow yet whether this function is equal to its Fourier series. Let’s investigate this questiongraphically. Figure 2 shows the graphs of some of the partial sums

when is odd, together with the graph of the square-wave function.n

Sn�x� �1

2�

2

� sin x �

2

3� sin 3x � � � � �

2

n� sin nx

1

2� �

�

k�1

2

�2k � 1�� sin�2k � 1�x

kn � 2k � 1

�1

2�

2

� sin x �

2

3� sin 3x �

2

5� sin 5x �

2

7� sin 7x � � � �

�2

� sin x � 0 sin 2x �

2

3� sin 3x � 0 sin 4x �

2

5� sin 5x � � � �

�1

2� 0 � 0 � 0 � � � �

� b1 sin x � b2 sin 2x � b3 sin 3x � � � �

a0 � a1 cos x � a2 cos 2x � a3 cos 3x � � � �

f

� 02n�

if n is even

if n is odd

� �1

� cos nx

n �0�

� � 1

n� �cos n� � cos 0�

bn �1

� y

�

�� f �x� sin nx dx �

1

� y

0

�� 0 dx �

1

� y

�

0 sin x dx

� 0 �1

� sin nx

n �0�

� 1

n� �sin n� � sin 0� � 0

an �1

� y

�

�� f �x� cos nx dx �

1

� y

0

�� 0 dx �

1

� y

�

0 cos nx dx

n � 1


■ ■ Note that equals 1 if is even and if is odd.n�1

ncos n�

We see that, as increases, becomes a better approximation to the square-wavefunction. It appears that the graph of is approaching the graph of , except where

or is an integer multiple of . In other words, it looks as if is equal to the sumof its Fourier series except at the points where is discontinuous.

The following theorem, which we state without proof, says that this is typical of theFourier series of piecewise continuous functions. Recall that a piecewise continuous func-tion has only a finite number of jump discontinuities on . At a number where has a jump discontinuity, the one-sided limits exist and we use the notation

Fourier Convergence Theorem If is a periodic function with period and andare piecewise continuous on , then the Fourier series (7) is convergent.

The sum of the Fourier series is equal to at all numbers where is continu-ous. At the numbers where is discontinuous, the sum of the Fourier series isthe average of the right and left limits, that is

If we apply the Fourier Convergence Theorem to the square-wave function inExample 1, we get what we guessed from the graphs. Observe that

and

and similarly for the other points at which is discontinuous. The average of these left andright limits is , so for any integer the Fourier Convergence Theorem says that

(Of course, this equation is obvious for .)x � n�

1

2� �

�

k�1

2

�2k � 1�� sin�2k � 1�x � f �x�1

2

if n � n�if x � n�

n12

f

f �0�� lim x l 0�

f �x� � 0f �0�� lim x l 0�

f �x� � 1

f

12 � f �x�� f �x��

fxfxf �x�

��, ��f f2�f8

f �a�� lim x l a�

f �x�f �a�� lim x l a�

f �x�

fa��, ��

ff�xx � 0

f �x�Sn�x�Sn�x�n

FIGURE 2 Partial sums of the Fourier series for the square-wave function

xπ_π

1

y

S¡

π x_π

1

y

S£

π x_π

1

y

S∞

π x_π

S¡∞

1

y

π x_π

1

y

S¡¡

π x_π

1

y

S¶


Functions with Period 2L

If a function has period other than , we can find its Fourier series by making a changeof variable. Suppose has period , that is for all . If we let

and

then, as you can verify, has period and corresponds to . The Fourierseries of is

where

If we now use the Substitution Rule with , then , , andwe have the following

If is a piecewise continuous function on , its Fourier series is

where

and, for ,

Of course, the Fourier Convergence Theorem (8) is also valid for functions with period.

EXAMPLE 2 Find the Fourier series of the triangular wave function defined by for and for all . (The graph of is shown in Figure 3.)For which values of is equal to the sum of its Fourier series?

FIGURE 3The triangular wave function

1 x2_1

1

y

0

f �x�xfxf �x � 2� � f �x��1 � x � 1

f �x� � � x �2L

bn �1

L y

L

�L f �x� sin�n�xL dxan � 1L y�LL f �x� cos�n�xL dx

n � 1

a0 �1

2L y

L

�L f �x� dx

a0 � ��

n�1 an cos�n�xL � bn sin�n�xL �

��L, L�f9

dt � ��L� dxt � �x�Lx � Lt��

bn �1

� y

�

�� t�t� sin nt dtan �

1

� y

�

�� t�t� cos nt dt

a0 �1

2� y

�

�� t�t� dt

a0 � ��

n�1 �an cos nt � bn sin nt�

t

t � �x � L2�t

t�t� � f �x� � f �Lt��

t � �x�Lxf �x � 2L� � f �x�2Lf �x�

2�f


■ ■ Notice that when these formulasare the same as those in (7).

L � �


SOLUTION We find the Fourier coefficients by putting in (9):

and for ,

because is an even function. Here we integrate by parts with and . Thus,

Since is an odd function, we see that

We could therefore write the series as

But if is even and if is odd, so

Therefore, the Fourier series is

The triangular wave function is continuous everywhere and so, according to the FourierConvergence Theorem, we have

for all xf �x� �1

2� �

�

n�1

4

�2k � 1�2� 2cos��2k � 1��x�

�1

2� �

�

n�1

4

�2k � 1�2� 2cos��2k � 1��x�

1

2�

4

� 2cos��x� �

4

9� 2cos�3�x� �

4

25� 2cos�5�x� � � � �

an �2

n2� 2�cos n� � 1� � 0� 4

n2� 2

if n is even

if n is odd

ncos n� � �1ncos n� � 1

1

2� �

�

n�1 2�cos n� � 1�

n2� 2cos�n�x�

bn � y1

�1 � x � sin�n�x� dx � 0

y � � x � sin�n�x�

� 0 �2

n� � cos�n�x�n� �

1

0

�2

n2� 2�cos n� � 1�

an � 2 xn� sin�n�x��01

�2

n� y

1

0 sin�n�x� dx

dv � cos�n�x� dxu � xy � � x � cos�n�x�

an � y1

�1 � x � cos�n�x� dx � 2 y1

0 x cos�n�x� dx

n � 1

� �14 x2]�10 � 14 x 2]10 � 12

a0 �12 y

1

�1 � x � dx � 12 y0

�1 ��x� dx � 12 y

1

0 x dx

L � 1

■ ■ Notice that is more easily calculated asan area.

a0


In particular,

for

Fourier Series and Music

One of the main uses of Fourier series is in solving some of the differential equations thatarise in mathematical physics, such as the wave equation and the heat equation. (This iscovered in more advanced courses.) Here we explain briefly how Fourier series play a rolein the analysis and synthesis of musical sounds.

We hear a sound when our eardrums vibrate because of variations in air pressure. If aguitar string is plucked, or a bow is drawn across a violin string, or a piano string is struck,the string starts to vibrate. These vibrations are amplified and transmitted to the air. Theresulting air pressure fluctuations arrive at our eardrums and are converted into electricalimpulses that are processed by the brain. How is it, then, that we can distinguish betweena note of a given pitch produced by two different musical instruments?

The graphs in Figure 4 show these fluctuations (deviations from average air pressure)for a flute and a violin playing the same sustained note D (294 vibrations per second) asfunctions of time. Such graphs are called waveforms and we see that the variations in airpressure are quite different from each other. In particular, the violin waveform is morecomplex than that of the flute.

We gain insight into the differences between waveforms if we express them as sums ofFourier series:

In doing so, we are expressing the sound as a sum of simple pure sounds. The differencein sounds between two instruments can be attributed to the relative sizes of the Fouriercoefficients of the respective waveforms.

The th term of the Fourier series, that is,

is called the nth harmonic of . The amplitude of the th harmonic is

and its square, , is sometimes called energy of the th harmonic. (Notice thatnA2n � a2n � b2n

An � sa2n � b2n

nP

an cos�n� tL � bn�n� tL n

P�t� � a0 � a1 cos�� tL � b1 sin�� tL � a2 cos�2� tL � b2 sin�2� tL � � � �

FIGURE 4Waveforms (b) Violin(a) Flute

tt

�1 � x � 1� x � � 12 � ��

k�1

4

�2k � 1�2� 2cos��2k � 1��x�


for a Fourier series with only sine terms, as in Example 1, the amplitude is andthe energy is .) The graph of the sequence is called the energy spectrum of

and shows at a glance the relative sizes of the harmonics.Figure 5 shows the energy spectra for the flute and violin waveforms in Figure 4. Notice

that, for the flute, tends to diminish rapidly as increases whereas, for the violin, thehigher harmonics are fairly strong. This accounts for the relative simplicity of the flutewaveform in Figure 4 and the fact that the flute produces relatively pure sounds whencompared with the more complex violin tones.

In addition to analyzing the sounds of conventional musical instruments, Fourier seriesenable us to synthesize sounds. The idea behind music synthesizers is that we can combinevarious pure tones (harmonics) to create a richer sound through emphasizing certain harmonics by assigning larger Fourier coefficients (and therefore higher correspondingenergies).

FIGURE 5Energy spectra

n2 4 6 8 10

(b) Violin

0

A@n

0 n2 4 6 8 10

(a) Flute

A@n

nAn2

P�A2n�A2n � b2n

An � � bn �

7–11 Find the Fourier series of the function.

7.

8.

9.

10. ,

11. ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

12. A voltage , where represents time, is passed through aso-called half-wave rectifier that clips the negative part of thewave. Find the Fourier series of the resulting periodic function

f �t � 2�� f �t�f �t� � 0E sin �t

if ��

� t � 0

if 0 � t ��

tE sin �t

�1 � t � 1f �t� � sin�3�t�

f �x � 2� � f �x��1 � x � 1f �x� � 1 � x

f �x � 8� � f �x�f �x� � �x0 if �4 � x � 0if 0 � x � 4

f �x � 4� � f �x�f �x� � 010

if �2 � x � 0 if 0 � x � 1 if 1 � x � 2

f �x � 4� � f �x�f �x� � 10 if � x � � 1if 1 � � x � � 21–6 A function is given on the interval and is periodic with period .(a) Find the Fourier coefficients of .(b) Find the Fourier series of . For what values of is equal

to its Fourier series?

; (c) Graph and the partial sums , , and of the Fourier series.

1.

2.

3.

4.

5.

6.

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

f �x� � �110

if �� x � ��2 if ��2 � x � 0 if 0 � x � �

f �x� � 0cos x if �� x � 0if 0 � x � �f �x� � x 2f �x� � x

f �x� � 0x if �� x � 0if 0 � x � �f �x� � 1

�1

if �� x � 0if 0 � x � �

S6S4S2f

f �x�xff

2�f��, ��f

Exercises

Click here for solutions.S

18. Use the result of Example 2 to show that

19. Use the result of Example 1 to show that

20. Use the given graph of and Simpson’s Rule with toestimate the Fourier coefficients . Then usethem to graph the second partial sum of the Fourier series andcompare with the graph of .

x

y

1

0.25

f

a0, a1, a2, b1, and b2n � 8f

1 �1

3�

1

5�

1

7� � � � �

�

4

1 �1

32�

1

52�

1

72� � � � �

� 2

8

13–16 Sketch the graph of the sum of the Fourier series of without actually calculating the Fourier series.

13.

14.

15. ,

16. ,

■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■ ■

17. (a) Show that, if , then

(b) By substituting a specific value of , show that

��

n�1

1

n2�

� 2

6

x

x 2 �1

3� �

�

n�1��1�n

4

n2� 2cos�n�x�

�1 � x � 1

�2 � x � 2f �x� � e x

�1 � x � 1f �x� � x 3

f �x� � x1 � x if �1 � x � 0if 0 � x � 1f �x� � �13 if �4 � x � 0if 0 � x � 4

f


1. (a) a0 =1

2π

∫ π−π f(x) dx =

1

2π

(∫ 0−π dx−

∫ π0dx

)= 0.

an =1

π

∫ π−π

∫ π−π f(x) cosnxdx =

1

π

∫ 0−π cosnxdx−

1

π

∫ π0cosnxdx = 0 [since cosnx is even].

bn =1

π

∫ π−π

∫ π−π f(x) sinnxdx =

1

π

∫ 0−π sinnxdx−

1

π

∫ π0sinnxdx =

2

π

∫ 0−π sinnxdx [since sinnx is odd]

= − 2nπ

[1− cos(−nπ)] =

0 if n even

− 4nπ

if n odd

(b) f(x) =∞∑k=0

− 4(2k + 1)π

sin(2k + 1)x when−π < x < 0 and 0 < x < π.

(c)

2.51.250-1.25-2.5

1

0.5

0

-0.5

-1

x

y

x

y


SOLUTIONS

3. (a) a0 =1

2π


1

2π

∫ π−π xdx = 0.

an =1

π

∫ π−π f (x) cosnxdx =

1

π

∫ π−π x cosnxdx = 0 [because x cosnx is odd]

bn =1

π

∫ π−π f (x) sinnxdx =

1

π

∫ π−π x sinnxdx =

2

π

∫ π0x sinnxdx [since x sinnx is odd]

= − 2ncosnπ [using integration by parts] =

{−(2/n) if n even(2/n) if n odd

(b) f(x) =∞∑

n=1

(−1)n+1 2nsinnx

when−π < x < π.

(c)

2.51.250-1.25-2.5

2.5

1.25

0

-1.25

-2.5

x

y

x

y


5. (a) a0 =1

2π


1

2π

∫ π0cosxdx = 0

an =1

π

∫ π−π f(x) cosnxdx =

1

π

∫ π0cosx cosnxdx =

{1

2if n = 1

0 if n �= 1 [by symmetry about x =π2]

bn =1

π

∫ π−π f(x) sinnxdx =

1

π

∫ π0cosx sinnxdx

=

2n

π(n2 − 1) if n even

0 if n odd

[using an integral table,

and simplified using the addition formula for cos(a+ b)

]

(b) f (x) = 12cosx+

∞∑k=1

4k

π (4k2 − 1) sin(2k) when−π < x < 0, 0 < x < π.

(c)

2.51.250-1.25-2.5

1

0.5

0

-0.5

-1

x

y

x

y

7. Use f (x) =

0 if − 2 ≤ x ≤ −11 if − 1 < x < 1,0 if 1 ≤ x ≤ 2

L = 2.

a0 =1

2L

∫ L−L f(x) dx =

1

4

∫ 1−1 dx =

1

2

an =1

L

∫ L−L f(x) cos

(nπxL

)dx = 1

2

∫ 1−1 cos

(nπx2

)dx =

2

nπsin

(π2n)=

0 if n even

2/nπ if n = 4n+ 1

−2/nπ if n = 4n+ 3

bn =1

L

∫ L−L f (x) sin

(nπxL

)dx = 1

2

∫ 1−1 sin

(nπx2

)dx = 0

Fourier Series: 12+

2

πcos

(πx2

)− 2

3πcos

(3πx

2

)+

2

5πcos

(5πx

2

)− · · ·

1

2+

∞∑k=1

2

(4k + 1)πsin

(π2(4k + 1)

)− 2

(4k + 3)πsin

(π2(4k + 3)

)

52.50-2.5-5

1

0.75

0.5

0.25

0

x

y

x

y


9. Use f(x) =

{−x if − 4 ≤ x < 00 if 0 ≤ x ≤ 4 , L = 4.

a0 =1

2L

∫ L−L f(x) dx =

1

8

∫ 0−4 −xdx = 1

an =1

L

∫ L−L f(x) cos

(nπxL

)dx = 1

4

∫ 0−4 −x cos

(nπx4

)dx =

4

(nπ)2(cos (nπ)− 1) =

{0 if n is even

−8/(nπ)2 if n is odd

bn =1

L

∫ L−L f(x) sin

(nπxL

)dx = 1

4

∫ 0−4 −x sin

(nπx4

)dx =

4

nπcos (nπ) =

{4/nπ if n is even

−4/nπ if n is odd

Fourier Series:

1 +∞∑k=1

− 4(2k − 1)π sin

(π4(2k − 1)x

)− 8

(2k − 1)2π2 cos(π4(2k − 1)x

)+

4

(2k)πsin

(π4(2k)x

)

52.50-2.5-5

4

3

2

1

0

x

y

x

y

11. Use f (x) = {sin(3πt) if − 1 ≤ t ≤ 1 , L = 1.Note: This can be done instantly if one observes that the period of sin(3πt) is 2

3, and the period of f(x) = 2 which

is an integer multiple of 23. Therefore f(x) is the same as sin(3πt) for all t, and its Fourier series is therefore

sin(3πt).

We can get this result using the standard coefficient formulas: a0 =1

2L

∫ L−L f(x) dx =

1

2

∫ 1−1 sin(3πx) dx = 0

an =1

L

∫ 1−1 f(x) cos

(nπxL

)dx =

∫ 1−1 sin(3πx) cos(nπx) dx

= 0 [applying change of variables to a formula in the section]

bn =1

L

∫ L−L f(x) sin

(nπxL

)dx =

∫ 1−1 sin(3πx) sin(nπx) dx

=

6sinnπ

π (−9 + n2) if n �= 3

1 if n = 3

[using integral table and addition formula =

{0 if n �= 31 if n = 3


Fourier Series: sin(3πx)

52.50-2.5-5

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0.5

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13.

3 if −5 ≤ x < −4−1 if −4 ≤ x < 03 if 0 ≤ x < 4

−1 if 4 ≤ x < 5

52.50-2.5-5

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15.

3.752.51.250-1.25

1

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17. (a) We find the Fourier series for f(x) = {x2 if −1 ≤ x ≤ 1, L = 1

a0 =1

2L

∫ L−L f(x) dx =

1

2

∫1

−1 x2 dx = 1

3

an =1

L

∫ 1−1 f(x) cos

(nπxL

)dx =

∫ 1−1 x

2 cos(nπx) dx =4

(nπ)2cosnπ =

4

(nπ)2if n even

− 4(nπ)2

if n odd

bn =1

L

∫ L−L f (x) sin

(nπxL

)dx =

∫ 1−1 x

2 sin(nπx) dx = 0 because x2 sin(nπx) is odd.

So we have x2 = 13+

∞∑n=1

(−1)n 4(nπ)2

cos(nπx) for −1 ≤ x ≤ 1.

(b) We let x = 1 in the above to obtain

1 = 13+

∞∑n=1

(−1)n 4(nπ)2

cos(nπ) 23=

∞∑n=1

4

n2π2π2

6=

∞∑n=1

1

n2


19. Example 1 says that, for 0 ≤ x < π, 1 = 12+

∑∞k=1

2

(2k − 1)π sin((2k − 1)x).Let x = π

2to obtain

1 = 12+

∞∑k=1

2

(2k − 1)π sin(π2(2k − 1)

)π

4=

∞∑k=1

1

(2k − 1) sin((2k − 1))

π

4= 1− 1

3+ 1

5− 1

7+ · · ·

a1 =∫ 1−1 f(x) cos(πx) dx ≈

1

12

−1.25(−1) + 4(0.25)

(− 1√

2

)+ 2(−2.25)(0) + 4(−1.25)

(1√2

)+ 2(3.5)(1) + 4(3.5)

(1√2

)+ 2(0)(0) + 4(−2.75)

(− 1√

2

)− 1.25(−1)

= 1924

(1 +

√2)

a2 =∫ 1−1 f(x) cos(2πx) dx ≈

1

12

[ −1.25(1) + 4(0.25)(0) + 2(−2.25)(−1) + 4(−1.25)(0)+ 2(3.5)(1) + 4(3.5)(0) + 2(0)(−1) + 4(−2.75)(0)− 1.25(1)

]

= 34

b1 =∫ 1−1 f(x) sin(πx) dx

≈ 112

[−1.25(0) + 4(0.25)

(− 1√

2

)+ 2(−2.25)(−1) + 4 (−1.25)

(− 1√

2

)+ 2(3.5)(0) + 4(3.5)

(1√2

)+ 2(0)(1)

+ 4(−2.75)(

1√2

)− 1.25(0) + 4(3.5)

(1√2

)+ 2(0)(1) + 4(−2.75)

(1√2

)− 1.25(0)

]

=9 + 7

√2

24

b2 =∫ 1−1 f(x) sin(2πx) dx ≈ 112

[ −1.25(0) + 4(0.25)(1) + 2(−2.25)(0) + 4(−1.25)(−1)+ 2(3.5)(0) + 4(3.5)(1) + 2(0)(0) + 4(−2.75)(−1)− 1.25(0)

]

= 3112

f(x) ≈ − 112

+ 1924

(1 +

√2)cos(πx) +

(9 + 7

√2

24

)sin(πx) + 3

4cos(2πx) + 31

12sin(2πx)

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