Periodic Functions And Fourier Series
Periodic Functions And Fourier Series :-
Definition :
A function f is said to be periodic with period T if
(i) f(x) is defined for all x and f(x+T) = f(x) for all x for some positive number T . For example , sin x is periodic with period 2π , since sin (x+2π) = sin x . A periodic function has many periods, for if f(x) = f(x+T) then
f(x) = f(x+T) = f(x+2T) = ....=f(x+nT),
where n is any integer. Hence when T is a period of f, nT is also a period of f , but while referring to period we mean the smallest one .
Since each of the functions sin x , cos x , sin 2x , cos 2x , .... are of period 2π , we may think of representing a given function f of period 2π by an infinite series of these function as
∞
f(x) = a₀/2 + Σ (aₙ cos nx + bₙ sin nx) ......(1)
n=1
Now two questions arise :
(i) Supposing the representation (1) is possible , how are the aₙ 's and bₙ 's determined ?
(ii) If the appropriate values are assigned to the coefficients , does the series converge to f(x) ?
To answer the first question we assume that the series (1) can be integrated term by term from x= -π to x= π . Then integrating both sides of (1) from -π to π , we have
π π ∞
∫ f(x) = ∫ [a₀/2 + Σ (aₙ cos nx + bₙ sin nx)]dx
-π -π n=1
π
= a₀/2 ∫ dx
-π
∞ π π
+ Σ (aₙ∫cos nx dx+bₙ∫sin nx dx)
n=1 -π -π
π π
= πa₀ [∵∫ cos nx dx = 0, ∫ sin nx dx =0]
-π -π
π
Hence a₀ = 1/π ∫ f(x) dx ...........(2)
-π
To find aₙ , we multiply each side of (1) by cos mx , where m is any fixed positive integer and integrate from -π to π . Then
π π
∫ f(x) cos mx dx = a₀/2 ∫ cos mx dx
-π -π
∞ π
+ Σ [aₙ ∫ cos nx cos mx dx
n=1 -π
π
+bₙ∫ sin nx cos mx dx]
π
+bₙ∫ sin nx cos mx dx]
-π
= πaₙ
π
[∵ ∫ cos nx cos mx dx = 0 when m≠n
-π
=π when m=n
π
∫sin nx cos mx dx = 0 for all m,n]
-π
π
Hence aₙ= 1/π ∫ f(x) cos nx dx ..........(3)
-π
Similarly to find bₙ , we multiply (1) by sin mx and integrating term wise from -π to π we get
π π
∫ f(x) sin mx dx = a₀/2 ∫ sin mx dx
-π -π
∞ π
+ Σ [aₙ ∫ cos nx sin mx dx
n=1 -π
π
+ bₙ ∫sin nx sin mx dx ]
-π
= πbₙ
π
[∵∫ cos nx sin mx dx = 0 for all m,n
-π
π
∫sin nx sin mx = 0 when m≠n
-π
π
∫ sin nx sin mx = π when m=n ]
-π
π
Hence bₙ = 1/π ∫ f(x) sin nx dx .......(4)
-π
The equations (2) , (3) and (4) are called Euler's formulae , a₀ ,aₙ and bₙ are called the Fourier Coefficients of f(x) and the series
∞
1/2 a₀ + Σ (aₙ cos nx + bₙ sin nx )
n=1
when aₙ's and bₙ's are determined by (2) , (3) and (4) is called the Fourier Series of f whether or not the series converges .
= πaₙ
π
[∵ ∫ cos nx cos mx dx = 0 when m≠n
-π
=π when m=n
π
∫sin nx cos mx dx = 0 for all m,n]
-π
π
Hence aₙ= 1/π ∫ f(x) cos nx dx ..........(3)
-π
Similarly to find bₙ , we multiply (1) by sin mx and integrating term wise from -π to π we get
π π
∫ f(x) sin mx dx = a₀/2 ∫ sin mx dx
-π -π
∞ π
+ Σ [aₙ ∫ cos nx sin mx dx
n=1 -π
π
+ bₙ ∫sin nx sin mx dx ]
-π
= πbₙ
π
[∵∫ cos nx sin mx dx = 0 for all m,n
-π
π
∫sin nx sin mx = 0 when m≠n
-π
π
∫ sin nx sin mx = π when m=n ]
-π
π
Hence bₙ = 1/π ∫ f(x) sin nx dx .......(4)
-π
The equations (2) , (3) and (4) are called Euler's formulae , a₀ ,aₙ and bₙ are called the Fourier Coefficients of f(x) and the series
∞
1/2 a₀ + Σ (aₙ cos nx + bₙ sin nx )
n=1
when aₙ's and bₙ's are determined by (2) , (3) and (4) is called the Fourier Series of f whether or not the series converges .
Motivational Quote
" The truth, he thought, has never been of any real value to any human being - it is a symbol for mathematicians and philosophers to pursue. In human relations kindness and lies are worth a thousand truths."
by Graham Greene
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