In this slide fourier series of Engineering Mathematics has been described. one Example is also added for you. Hope this will help you understand fourier series.

1. Engineering Mathematics
Presentation

2. Welcome to our
Presentation
Topic : “FOURIER SERIES”

3. INDEX
• Definition.
• Conditions.
• Formula.
• Example.
• Conclusion.

4. DEFINITION
• FOURIER SERIES : Fourier Series is an infinite series
representation of periodic function in terms of the
trigonometric sine and cosine functions.
• Most of the single valued functions which occur in
applied mathematics can be expressed in the form
of Fourier series, which is in terms of sines and
cosines.

5. DEFINITION
• Fourier series is to be expressed in terms of periodic
functions- sines and cosines. Fourier series is a very
powerful method to solve ordinary and partial
differential equations, particularly with periodic
functions appearing as non-homogeneous terms.

6. CONDITONS
Let F(x) satisfy the following conditions :
1. F(x) is defined in the interval, c < x < c+2l.
2. F(x) and F’(x) sectionally continuous in c < x < c+2l.
3. F(x+2l) = F(x) i.e. F(x) is periodic with period 2l.
If these 3 conditions remains, then we can say F(x) is
Fourier series.

7. FORMULA
The formula for a Fourier series on an interval [𝑐, 𝑐 + 2𝑙] is :
𝐹 𝑥 =
𝑎0
2
+
𝑛=1
∞
( 𝑎 𝑛 𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
+ 𝑏 𝑛 𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
)
Where,
𝑎 𝑛 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑏 𝑛 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
𝑑𝑥
𝑎0 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑑𝑥
And “𝑙” defines period, if period is specified then, period = 2𝑙
and if it is not then, the maximum limit will be the value of “𝑙” .

8. FORMULA
To do this math we need a shortcut formula, because we have trigonometric term in
this formula. And we know that trigonometric term never ends.so we have to use
this shortcut formula-
𝐹 𝑥 = 𝑢0 𝑣0 𝑑𝑥
⇒ 𝑢0 𝑣0 − 𝐷𝑢0 𝑣1 + 𝐷𝑢1 𝑣2 − ⋯ 𝑢𝑛𝑡𝑖𝑙 0

9. EXAMPLE
• Expand F(x) = 𝑥2; 0<x<2𝜋 and period = 2𝜋
𝑆𝑜𝑙 𝑛
: Here, period = 2𝜋
or, 2𝑙 = 2𝜋
or, 𝑙 = 𝜋
Now,
𝑎 𝑛 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
𝑑𝑥
⇒
1
𝜋 0
2𝜋
𝑥2
𝑐𝑜𝑠𝑛𝑥 𝑑𝑥
⇒
1
𝜋
𝑥2 sin 𝑛𝑥
𝑛
−
2𝑥
𝑛
−
cos 𝑛𝑥
𝑛
+
2
𝑛2 −
sin 𝑛𝑥
𝑛 0
2𝜋
⇒
1
𝜋
𝑥2
𝑛
sin 𝑛𝑥 +
2𝑥
𝑛2 cos 𝑛𝑥 −
2
𝑛3 sin 𝑛𝑥
0
2𝜋
⇒
1
𝜋
4𝜋2
𝑛
sin 2𝑛𝜋 +
4𝜋
𝑛2 cos 2𝑛𝜋 −
2
𝑛3 sin 2𝑛𝜋 − 0 + 0 − 0
⇒
1
𝜋
0 +
4𝜋
𝑛2 − 0
⇒
4
𝑛2

10. EXAMPLE
𝑏 𝑛 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
𝑑𝑥
⇒
1
𝜋 0
2𝜋
𝑥2
. sin 𝑛𝑥 𝑑𝑥
⇒
1
𝜋
𝑥2
−
cos 𝑛𝑥
𝑛
−
2𝑥
𝑛
−
𝑠𝑖𝑛 𝑛𝑥
𝑛
+
2
𝑛2
cos 𝑛𝑥
𝑛 0
2𝜋
⇒
1
𝜋
−
𝑥2
𝑛
𝑐𝑜𝑠 𝑛𝑥 +
2𝑥
𝑛2 sin 𝑛𝑥 +
2
𝑛3 cos 𝑛𝑥
0
2𝜋
⇒
1
𝜋
−
4𝜋2
𝑛
cos 2𝑛𝜋 +
4𝜋
𝑛2 sin 2𝑛𝜋 +
2
𝑛3 cos 2𝑛𝜋 − 0 + 0 +
2
𝑛3
⇒
1
𝜋
−
4𝜋
𝑛
+ 0 +
2
𝑛3 −
2
𝑛3
⇒ −
4𝜋
𝑛

11. EXAMPLE
𝑎0 =
1
𝑙 𝑐
𝑐+2𝑙
𝐹 𝑥 𝑑𝑥 ;
⇒
1
𝜋 0
2𝜋
𝑥2
𝑑𝑥
⇒
1
𝜋
𝑥3
3 0
2𝜋
⇒
1
𝜋
8𝜋3
3
− 0
⇒
8𝜋2
3
So. The Fourier series is :
𝐹 𝑥 =
8𝜋2
3
2
+
𝑛=1
∞
(
4
𝑛2
𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
+ −
4𝜋
𝑛
𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
)
⇒
4𝜋2
3
+
𝑛=1
∞
(
4
𝑛2
𝑐𝑜𝑠
𝑛𝜋𝑥
𝑙
−
4𝜋
𝑛
𝑠𝑖𝑛
𝑛𝜋𝑥
𝑙
)
(Answer :)

12. Conclusion
Conclusions To continue researching Fourier
Series there are a few areas and specific
problems that we would address. Fourier is
a lengthy math, So we have to be careful
about the formula while doing this math.

13. Thank you all
for having patience.