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This expression can be used to rewrite the Fourier series (11.2) using the basis functions

exp in x=L rather than sine and cosines.

Problem a: Replace n by in (11.15) to show that:

;n

ein x=L + e;in x=L

1

cos(n x=L) = 2

(11.16)

ein x=L e;in x=L :

1

sin(n x=L) = 2i ;

Problem b: Insert this relation in the Fourier series (11.2) to show that this Fourier

series can also be written as:

1

X

cn ein x=L

f(x) = (11.17)

n=;1

with the coe cients cn given by:

cn = (an ibn )=2 for n > 0

;

cn = (ajnj + ibjnj)=2 for n < 0 (11.18)

c0 = a0=2

Note that the absolute value is used for n < 0.

jnj

Problem c: Explain why the n-summation in (11.17) extends from to rather than;1 1

from 0 to 1.

Problem d: The relations (11.7) and (11.8) can be used to express the coe cients cn in

the function f(x). Treat the cases n > 0, n < 0 and n = 0 separately to show that

for all values of n the coe cient cn is given by:

1 Z L f(x)e;in x=L dx :

cn = 2L (11.19)

;L

The sum (11.17) with expression (11.19) constitutes the complex Fourier transform over

a nite interval. Again, there is a close analogy with the projections of vectors shown

in section (10.1). Before we can explore this analogy, the inner product between two

complex functions f(x) and g(x) needs to be de ned. This inner product is not given by

R

(f g) = f(x)g(x)dx. The reason for this is that the length of a vector is de ned by

2 = (v v), a straightforward generalization of this expression to functions using the

kvk

inner product given above would give for the norm of the function f(x): 2 = (f f) =

R f 2(x)dx. However, when f(x) is purely imaginary this would lead to a negative norm.

kf k

This can be avoided by de ning the inner product of two complex functions by:

ZL

(f g) f (x)g(x)dx (11.20)

;L

where the asterisk denotes the complex conjugate.

2 = (f

Problem e: 2Show that with this de nition the norm of f(x) is given by f) =

R (x)j dx. kf k

jf

CHAPTER 11. FOURIER ANALYSIS

134

With this inner product the norm of the function is guaranteed to be positive. Now that we

have an inner product the analogy with the projections in linear algebra can be explored.

In order to do this, de ne the following basis functions:

1 ein x=L :

un(x) (11.21)

p

2L

Problem f: Show that these functions are orthonormal with respect to the inner product

(11.20), i.e. show that:

(un um ) = nm : (11.22)

Pay special attention to the normalization of these functions i.e. to the case n = m.

Problem g: Expand f(x) in these basis functions, f(x) = P1 nun(x) and show

n=;1

that f(x) can be written as:

1

X

f= un (un f) : (11.23)

n=;1

Problem h: Make the comparison between this expression and the expressions for the

projections of vectors in section (10.1).

11.3 The Fourier transform on an in nite interval

In several applications, one wants to compute the Fourier transform of a function that

is de ned on an in nite interval. This amounts to taking the limit L However, a

! 1.

simple inspection of (11.19) shows that one cannot simply take the limit L of the

!1

expressions of the previous section because in that limit cn = 0. In order to de ne the

Fourier transform for an in nite interval de ne the variable k by:

k n: (11.24)

L

An increment n corresponds to an increment k given by: k = n=L. In the

summation over n in the Fourier expansion (11.17), n is incremented by unity: n = 1.

This corresponds to an increment k = =L of the variable k. In the limit L this

!1

increment goes to zero, this implies that the summation over n should be replaced by an

integration over k:

n Z 1 ( ) dk = L Z 1 ( ) dk as L

1

X

:

() (11.25)

k ;1 ;1

! !1

n=;1

Problem a: Explain the presence of the factor n= k and show the last identity.

This is not enough to generalize the Fourier transform of the previous section to an in nite

interval. As noted earlier, the coe cients cn vanish in the limit L Also note that

! 1.

the integral in the right hand side of (11.25) is multiplied by L= , this coe cient is in nite

in the limit L Both problems can be solved by de ning the following function:

! 1.

F(k) L cn (11.26)

where the relation between k and n is given by (11.24).

11.4. THE FOURIER TRANSFORM AND THE DELTA FUNCTION 135

Problem b: Show that with the replacements (11.25) and (11.26) the limit L of

!1

the complex Fourier transform (11.17) and (11.19) can be taken and that the result

Z1

can be written as:

F(k)eikx dk

f(x) = (11.27)

;1

1 Z 1 f(x)e;ikx dx :

F(k) = 2 (11.28)

;1

11.4 The Fourier transform and the delta function

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