The average value of a function f(x) over an interval (a,b) can be approximated as: f(x) = (f(x1) + f(x2) + ... + f(xn))/n, where x1, x2, ..., xn are values in the interval. The Fourier coefficients for a periodic function f(x) are: a0 = (1/π) ∫ f(x) dx an = (2/π) ∫ f(x) cos(nx) dx bn = (2/π) ∫ f(x) sin(nx) dx The Fourier series expansion of

1. MATHEMATICAL
PHYSICS

2. ASSALAMUA’LAIKUM.WR.WB
Group 16
M.Rahmad Syalehin
Jeri Hartanto

3. AVERAGE VALUE OF A FUNCTION
 The concept of the average value of a function is
often useful. You know how to find the average of a
set of numbers. You add them and divide by the
number of numbers

4. xn x1 x2 x3 .... x n
x
n n
Problem : “ How if the
average value for a
function..???

5. solution
.This process suggest that we ought to get
an approximation to the everage value of a
function f(x) on the interval (a,b) by
averaging a number value of f(x) :
Average of f(x) on
(a,b) is approximately
equal to :f ( x ) ..... f ( x
f (x )
1 2 n
)
n

6. You can probably convince your self that the area under them is the same
For any quarter-period from 0 π/2, π/2 to π, etc
Sine wave cosine wave
2 2
sin x dx cos xdx
2 2
sin nx cos nx 1
2 2
(sin nx cos nx ) dx 2
2 2
sin nxdx cos nxdx

7.  The average value (over a period) of Sin2nx= the average
value (over a period) of cos2nx
1 2 1 2 1
sin nxdx cos nxdx
2 2 2 2

8. FOURIER COEFICIENTS :
 Formulas of fourier series :
1
f ( x) a0 a1 cos x a 2 cos 2 x a 3 cos 3 x .... b1 sin x b 2 sin 2 x b3 sin 3 x ....
2
1
a0 ( a n cos nx b n sin nx )
2 n 1
Where : 1
a0 f ( x ) dx
n
1
an f ( x ) cos nxdx
n
1
bn f ( x ) sin nxdx
n

9. EXAMPLE :
Write in a fourier series the function
f(x) :
0 , -π < x < 0
f ( x)
{ 1, 0 < x < π
Answer :
a0 1
an 0
2
bn
n
1 2 sin x sin 3 x sin 5 x
Your answer for the series is : f ( x) ....
2 1 3 5

10. PROBLEMS
Expand the periodic function in a sine-cosine Fourier series !
0, -π < x < 0
 f(x)= { 1, 0 < x < π/2
0, π/2 < x < π

11. We must find
Thus

13. Putting these values for the coefficients, you can write fourier series :
1
f ( x) a0 a 1 cos x a 2 cos 2 x a 3 cos 3 x .... b1 sin x b 2 sin 2 x b 3 sin 3 x ....
2
1
f ( x) a0 ( a n cos nx b n sin nx )
2 n 1
1 2 2 2
f ( x) sin x sin 3 x sin 5 x ....
2 3 5
1 2 2 2
f ( x) (sin x sin 3 x sin 5 x ...)
2 3 5