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Exam (elaborations) MATH354 (MATH354)
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Fourier Series 1Chapter One
FOURIER SERIES
1. Definitions and Examples
A trigonometric polynomial of degree N is an expression of the form
N
X
pN (x) = (ak cos kx + bk sin kx),
k=0
where a0 , a1 , . . . and b0 , b1 , . . . are constants. The value of the constant b0 is obviously irrelevant
since sin 0x ≡ 0; for convenience we set b0 = 0.
A Fourier series extends in a natural way the definition of a trigonometric polynomial, just
like the McLaurin series extends ordinary polynomials:
IDefinition: We say that a function f (x) admits a Fourier series expansion between x = −π
and x = π if the equation
∞
X
f (x) = (ak cos kx + bk sin kx) (1)
k=0
holds in the interval (−π, π). J
Note that all terms in the sum on the right-hand side of (1) have period 2π, hence the sum of
the Fourier series (1) is also periodic with period 2π. Therefore, if the expansion (1) converges
in the fundamental interval (−π, π), then it must converge for all values of x, and the function
represented by (1) is—by construction, if you wish—periodic with period 2π.
Now, if the expansion (1) does hold (for certain values of x), it would be desirable to have
a formula that gives the coefficients ak and bk in terms of the function f . This can be done
by borrowing a simple procedure from linear algebra. So, before we go into that, let’s see an
example of this procedure, even if at first sight you might think it has little to do with Fourier
expansions.
T T T
IExample 1 Show that the vectors l = [ 1 2 3 ] , m = [ 1 1 −1 ] and n = [ 5 −4 1 ]
T
form an orthogonal basis for R3 ; hence expand the vector b = [ 1 1 0 ] in terms of this basis.
Solution: By inspection, we find immediately that
1 1 1 5 5 1
l . m = 2 . 1 = 0, l . n = 2 . −4 = 0, n . m = −4 . 1 = 0.
3 −1 3 1 1 −1
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