Report 7
Problem R7.1: Determining Orthogonality
edit
Determine orthogonality of:
⟨
ϕ
i
,
ϕ
j
⟩
=
0
for
i
≠
j
{\displaystyle \langle \phi _{i},\phi _{j}\rangle =0\ {\text{ for }}\ i\neq j}
⟨
ϕ
j
,
ϕ
j
⟩
=
L
2
for
i
=
j
{\displaystyle \langle \phi _{j},\phi _{j}\rangle ={\frac {L}{2}}\ {\text{ for }}\ i=j}
(1.0)
ϕ
i
(
x
)
=
sin
(
ω
i
x
)
{\displaystyle \displaystyle \phi _{i}(x)=\sin(\omega _{i}x)}
ϕ
j
(
x
)
=
sin
(
ω
j
x
)
{\displaystyle \displaystyle \phi _{j}(x)=\sin(\omega _{j}x)}
ω
i
=
i
π
L
{\displaystyle \displaystyle \omega _{i}={\frac {i\pi }{L}}}
ω
j
=
j
π
L
{\displaystyle \displaystyle \omega _{j}={\frac {j\pi }{L}}}
The definition of a scalar product is:
⟨
f
¯
,
g
¯
⟩
:=
∫
0
L
f
¯
(
x
)
g
¯
(
x
)
d
x
{\displaystyle \langle {\bar {f}},{\bar {g}}\rangle :=\int _{0}^{L}{\bar {f}}(x)\ {\bar {g}}(x)\,dx}
(1.1)
So, substituting the problem statement values into Eq. (1.0):
⟨
ϕ
i
,
ϕ
j
⟩
:=
∫
0
L
sin
(
ω
i
x
)
sin
(
ω
j
x
)
d
x
{\displaystyle \langle \phi _{i},\phi _{j}\rangle :=\int _{0}^{L}\sin(\omega _{i}x)\ \sin(\omega _{j}x)\,dx}
(1.2)
and:
⟨
ϕ
j
,
ϕ
j
⟩
:=
∫
0
L
sin
(
ω
j
x
)
sin
(
ω
j
x
)
d
x
{\displaystyle \langle \phi _{j},\phi _{j}\rangle :=\int _{0}^{L}\sin(\omega _{j}x)\ \sin(\omega _{j}x)\,dx}
(1.3)
Solving Eq. 1.2 first:
Because
i
{\displaystyle \displaystyle i}
j
{\displaystyle \displaystyle j}
π
{\displaystyle \displaystyle \pi }
sin
(
i
π
−
j
π
)
=
0
{\displaystyle \sin(i\pi -j\pi )=0}
sin
(
i
π
+
j
π
)
=
0
{\displaystyle \sin(i\pi +j\pi )=0}
⟨
ϕ
i
,
ϕ
j
⟩
=
0
for
i
≠
j
{\displaystyle \langle \phi _{i},\phi _{j}\rangle =0\ {\text{ for }}\ i\neq j}
(1.0.a)
Again, because
j
{\displaystyle \displaystyle j}
2
j
π
{\displaystyle \displaystyle 2j\pi }
sin
(
2
j
π
)
=
0
{\displaystyle \sin(2j\pi )=0}
⟨
ϕ
j
,
ϕ
j
⟩
=
L
2
for
i
=
j
{\displaystyle \langle \phi _{j},\phi _{j}\rangle ={\frac {L}{2}}\ {\text{ for }}\ i=j}
(1.0.b)
Solved and Typed By ---Egm4313.s12.team1.durrance (talk ) 16:49, 24 April 2012 (UTC)
Egm4313.s12.team1.armanious (talk ) 23:48, 24 April 2012 (UTC)
Problem R7.2: Truncated Series
edit
Plot the truncated-series (3) p19-12 with n=5, and for:
t
=
α
p
1
=
α
2
π
c
ω
1
=
α
2
L
c
{\displaystyle \displaystyle t=\alpha p_{1}=\alpha {\frac {2\pi }{c\omega _{1}}}=\alpha {\frac {2L}{c}}}
(2.0)
α
=
.5
,
1
,
1.5
,
2
{\displaystyle \displaystyle \alpha =.5,1,1.5,2}
(2.1)
Let it be stated first that (3) p.19-12 is equal to:
u
(
x
,
t
)
=
∑
j
=
1
n
a
j
c
o
s
(
c
ω
j
t
)
s
i
n
(
ω
j
x
)
{\displaystyle \displaystyle u(x,t)=\sum {j=1}{n}a_{j}cos(c\omega _{j}t)sin(\omega _{j}x)}
(2.2)
And in this series representation, c=3, L=2, and
a
j
{\displaystyle a_{j}}
a
j
=
{
0
f
o
r
j
=
2
m
(
e
v
e
n
)
−
4
/
(
π
3
j
3
)
f
o
r
j
=
2
m
+
1
(
o
d
d
)
{\displaystyle \displaystyle a_{j}=\left\{{\begin{matrix}0forj=2m(even)\\-4/(\pi ^{3}j^{3})forj=2m+1(odd)\end{matrix}}\right.}
(2.3)
With this in mind, we are prepared to create these truncated-series.For
α
=
.5
{\displaystyle \alpha =.5}
u
(
x
,
t
)
=
[
a
1
c
o
s
(
c
ω
1
α
2
L
c
)
s
i
n
(
ω
1
x
)
]
{\displaystyle \displaystyle u(x,t)=[a_{1}cos(c\omega _{1}\alpha {\frac {2L}{c}})sin(\omega _{1}x)]}
(2.4)
+
[
a
2
c
o
s
(
c
ω
2
α
2
L
c
)
s
i
n
(
ω
2
x
)
]
{\displaystyle \displaystyle +[a_{2}cos(c\omega _{2}\alpha {\frac {2L}{c}})sin(\omega _{2}x)]}
+
[
a
3
c
o
s
(
c
ω
3
α
2
L
c
)
s
i
n
(
ω
3
x
)
]
{\displaystyle \displaystyle +[a_{3}cos(c\omega _{3}\alpha {\frac {2L}{c}})sin(\omega _{3}x)]}
+
[
a
4
c
o
s
(
c
ω
4
α
2
L
c
)
s
i
n
(
ω
4
x
)
]
{\displaystyle \displaystyle +[a_{4}cos(c\omega _{4}\alpha {\frac {2L}{c}})sin(\omega _{4}x)]}
+
[
a
5
c
o
s
(
c
ω
5
α
2
L
c
)
s
i
n
(
ω
5
x
)
]
{\displaystyle \displaystyle +[a_{5}cos(c\omega _{5}\alpha {\frac {2L}{c}})sin(\omega _{5}x)]}
These simplify below as shown, after plugging in 3 for c, 10, for L, .5 for alpha, and the value of
a
j
{\displaystyle a_{j}}
u
(
x
,
t
)
=
[
−
4
π
3
c
o
s
(
ω
1
(
.5
)
(
2
∗
10
)
)
s
i
n
(
ω
1
x
)
]
{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(.5)(2*10))sin(\omega _{1}x)]}
(2.5)
+
[
−
4
π
3
∗
3
3
c
o
s
(
ω
3
(
.5
)
(
2
∗
10
)
)
s
i
n
(
ω
3
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(.5)(2*10))sin(\omega _{3}x)]}
+
[
−
4
π
3
∗
5
3
c
o
s
(
ω
5
(
.5
)
(
2
∗
10
)
)
s
i
n
(
ω
5
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(.5)(2*10))sin(\omega _{5}x)]}
Making
ω
{\displaystyle \omega }
2
π
{\displaystyle 2\pi }
Alpha = .5 For
α
=
1
{\displaystyle \alpha =1}
These simplify below as shown, after plugging in 3 for c, 10, for L, 1 for alpha, and the value of
a
j
{\displaystyle a_{j}}
u
(
x
,
t
)
=
[
−
4
π
3
c
o
s
(
ω
1
(
1
)
(
2
∗
10
)
)
s
i
n
(
ω
1
x
)
]
{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(1)(2*10))sin(\omega _{1}x)]}
(2.6)
+
[
−
4
π
3
∗
3
3
c
o
s
(
ω
3
(
1
)
(
2
∗
10
)
)
s
i
n
(
ω
3
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(1)(2*10))sin(\omega _{3}x)]}
+
[
−
4
π
3
∗
5
3
c
o
s
(
ω
5
(
1
)
(
2
∗
10
)
)
s
i
n
(
ω
5
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(1)(2*10))sin(\omega _{5}x)]}
Making
ω
{\displaystyle \omega }
2
π
{\displaystyle 2\pi }
Alpha = 1
For
α
=
1.5
{\displaystyle \alpha =1.5}
These simplify below as shown, after plugging in 3 for c, 10, for L, 1.5 for alpha, and the value of
a
j
{\displaystyle a_{j}}
u
(
x
,
t
)
=
[
−
4
π
3
c
o
s
(
ω
1
(
1.5
)
(
2
∗
10
)
)
s
i
n
(
ω
1
x
)
]
{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(1.5)(2*10))sin(\omega _{1}x)]}
(2.7)
+
[
−
4
π
3
∗
3
3
c
o
s
(
ω
3
(
1.5
)
(
2
∗
10
)
)
s
i
n
(
ω
3
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(1.5)(2*10))sin(\omega _{3}x)]}
+
[
−
4
π
3
∗
5
3
c
o
s
(
ω
5
(
1.5
)
(
2
∗
10
)
)
s
i
n
(
ω
5
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(1.5)(2*10))sin(\omega _{5}x)]}
Making
ω
{\displaystyle \omega }
2
π
{\displaystyle 2\pi }
Alpha = 1.5 For
α
=
2
{\displaystyle \alpha =2}
These simplify below as shown, after plugging in 3 for c, 10, for L, 2 for alpha, and the value of
a
j
{\displaystyle a_{j}}
u
(
x
,
t
)
=
[
−
4
π
3
c
o
s
(
ω
1
(
2
)
(
2
∗
10
)
)
s
i
n
(
ω
1
x
)
]
{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(2)(2*10))sin(\omega _{1}x)]}
(2.8)
+
[
−
4
π
3
∗
3
3
c
o
s
(
ω
3
(
2
)
(
2
∗
10
)
)
s
i
n
(
ω
3
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(2)(2*10))sin(\omega _{3}x)]}
+
[
−
4
π
3
∗
5
3
c
o
s
(
ω
5
(
2
)
(
2
∗
10
)
)
s
i
n
(
ω
5
x
)
]
{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(2)(2*10))sin(\omega _{5}x)]}
Making
ω
{\displaystyle \omega }
2
π
{\displaystyle 2\pi }
Alpha = 2
Solved and Typed By - Egm4313.s12.team1.silvestri (talk ) 19:01, 23 April 2012 (UTC)
Reviewed By - Egm4313.s12.team1.armanious (talk ) 19:40, 25 April 2012 (UTC)
edit
Find the scalar product <f,g>, the magnitude of f and g, and the angle between f and g for (see R7.3 Lect. 11-1 pg. 8):
1)
f
(
x
)
=
cos
x
,
g
(
x
)
=
x
,
for
−
2
≤
x
≤
10
{\displaystyle f(x)=\cos x,g(x)=x,{\text{for}}-2\leq x\leq 10\!}
f
(
x
)
=
1
2
(
3
x
2
−
1
)
,
g
(
x
)
=
1
2
(
5
x
3
−
3
x
)
,
for
−
1
≤
x
≤
+
1
{\displaystyle f(x)={\frac {1}{2}}(3x^{2}-1),g(x)={\frac {1}{2}}(5x^{3}-3x),{\text{for}}-1\leq x\leq +1\!}
The scalar product, the function equivalent of the vector dot product, of two functions can be found in the following way:
<
f
,
g
>:=
∫
a
b
f
(
x
)
g
(
x
)
d
x
{\displaystyle \displaystyle <f,g>:=\int _{a}^{b}f(x)g(x)dx}
(3.0)
The magnitude of a function is defined as the square root of the scalar product between the function and itself:
|
|
f
|
|
:=<
f
,
f
>
1
/
2
=
[
∫
a
b
f
2
(
x
)
d
x
]
1
/
2
{\displaystyle \displaystyle ||f||:=<f,f>^{1/2}=\left[\int _{a}^{b}f^{2}(x)dx\right]^{1/2}}
(3.1)
The cosine of the angle between two functions can be found in the following way:
cos
θ
=
<
f
,
g
>
|
|
f
|
|
|
|
g
|
|
{\displaystyle \displaystyle \cos \theta ={\frac {<f,g>}{||f||||g||}}}
(3.2)
Note that these equations are very similar to their vector counterparts.Part 1
f
(
x
)
=
cos
x
,
g
(
x
)
=
x
,
for
−
2
≤
x
≤
10
{\displaystyle \displaystyle f(x)=\cos x,g(x)=x,{\text{for}}-2\leq x\leq 10}
(3.3)
The scalar product of f and g can be found using (3.0):
<
f
,
g
>=
∫
−
2
10
x
cos
(
x
)
d
x
=
[
x
sin
(
x
)
+
c
o
s
(
x
)
]
−
2
10
{\displaystyle \displaystyle <f,g>=\int _{-2}^{10}x\cos(x)dx=\left[x\sin(x)+cos(x)\right]_{-2}^{10}}
(3.4)
Solving this yields the scalar product:
(
10
sin
10
+
cos
10
)
−
(
−
2
sin
(
−
2
)
+
cos
(
−
2
)
)
=
−
7.68
{\displaystyle \displaystyle (10\sin 10+\cos 10)-(-2\sin(-2)+\cos(-2))=-7.68}
(3.5)
To find the magnitude of f, the scalar product of f with itself must first be found:
<
f
,
f
>=
∫
−
2
10
cos
2
(
x
)
d
x
=
∫
−
2
10
(
1
2
cos
(
2
x
)
+
1
2
)
d
x
=
[
1
4
sin
(
2
x
)
+
1
2
x
]
−
2
10
{\displaystyle \displaystyle <f,f>=\int _{-2}^{10}\cos ^{2}(x)dx=\int _{-2}^{10}\left({\frac {1}{2}}\cos(2x)+{\frac {1}{2}}\right)dx=\left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-2}^{10}}
(3.6)
Solving this yields:
[
1
4
sin
(
2
x
)
+
1
2
x
]
−
2
10
=
(
1
4
sin
(
20
)
+
5
)
−
(
1
4
sin
(
−
4
)
−
1
)
=
6.04
{\displaystyle \displaystyle \left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-2}^{10}=\left({\frac {1}{4}}\sin(20)+5\right)-\left({\frac {1}{4}}\sin(-4)-1\right)=6.04}
(3.7)
Therefore the magnitude of f is found to be:
|
|
f
|
|
=<
f
,
f
>
1
/
2
=
6.04
=
2.46
{\displaystyle \displaystyle ||f||=<f,f>^{1/2}={\sqrt {6.04}}=2.46}
(3.8)
A similar approach can be taken for the magnitude of g:
<
g
,
g
>=
∫
−
2
10
x
2
d
x
=
[
x
3
3
]
−
2
10
{\displaystyle \displaystyle <g,g>=\int _{-2}^{10}x^{2}dx=\left[{\frac {x^{3}}{3}}\right]_{-2}^{10}}
(3.9)
[
x
3
3
]
−
2
10
=
1000
3
−
−
8
3
=
1008
3
=
336
{\displaystyle \displaystyle \left[{\frac {x^{3}}{3}}\right]_{-2}^{10}={\frac {1000}{3}}-{\frac {-8}{3}}={\frac {1008}{3}}=336}
(3.10)
|
|
g
|
|
=<
g
,
g
>
1
/
2
=
336
=
18.33
{\displaystyle \displaystyle ||g||=<g,g>^{1/2}={\sqrt {336}}=18.33}
(3.11)
Using the above information with (3.2), the cosine of the angle between f and g is found to be
cos
θ
=
<
f
,
g
>
|
|
f
|
|
|
|
g
|
|
=
−
7.68
(
2.46
)
(
18.33
)
=
−
0.1705
{\displaystyle \displaystyle \cos \theta ={\frac {<f,g>}{||f||||g||}}={\frac {-7.68}{(2.46)(18.33)}}=-0.1705}
(3.12)
Therefore, the angle between the functions f and g is
θ
=
cos
−
1
(
−
0.1705
)
=
1.74
r
a
d
=
99.8
∘
{\displaystyle \displaystyle \theta =\cos ^{-1}(-0.1705)=1.74rad=99.8^{\circ }}
(3.13)
Part 2
f
(
x
)
=
1
2
(
3
x
2
−
1
)
,
g
(
x
)
=
1
2
(
5
x
3
−
3
x
)
,
for
−
1
≤
x
≤
+
1
{\displaystyle \displaystyle f(x)={\frac {1}{2}}(3x^{2}-1),g(x)={\frac {1}{2}}(5x^{3}-3x),{\text{for}}-1\leq x\leq +1}
(3.14)
The scalar product of f and g can be found using (3.0):
<
f
,
g
>=
∫
−
1
1
1
4
(
3
x
2
−
1
)
(
5
x
3
−
3
x
)
d
x
=
1
4
∫
−
1
1
(
15
x
5
−
14
x
3
+
3
x
)
d
x
=
1
4
[
15
6
x
6
−
14
4
x
4
+
3
2
x
2
]
−
1
1
{\displaystyle \displaystyle <f,g>=\int _{-1}^{1}{\frac {1}{4}}\left(3x^{2}-1\right)\left(5x^{3}-3x\right)dx={\frac {1}{4}}\int _{-1}^{1}\left(15x^{5}-14x^{3}+3x\right)dx={\frac {1}{4}}\left[{\frac {15}{6}}x^{6}-{\frac {14}{4}}x^{4}+{\frac {3}{2}}x^{2}\right]_{-1}^{1}}
(3.15)
Solving this yields the scalar product:
1
4
[
5
2
x
6
−
7
2
x
4
+
3
2
x
2
]
−
1
1
=
1
4
(
1
2
−
1
2
)
=
0
{\displaystyle \displaystyle {\frac {1}{4}}\left[{\frac {5}{2}}x^{6}-{\frac {7}{2}}x^{4}+{\frac {3}{2}}x^{2}\right]_{-1}^{1}={\frac {1}{4}}\left({\frac {1}{2}}-{\frac {1}{2}}\right)=0}
(3.16)
To find the magnitude of f, the scalar product of f with itself must first be found:
<
f
,
f
>=
∫
−
1
1
(
1
2
(
3
x
2
−
1
)
)
2
d
x
=
1
4
∫
−
1
1
(
9
x
4
−
6
x
2
+
1
)
d
x
=
1
4
[
9
5
x
5
−
2
x
3
+
x
]
−
1
1
{\displaystyle \displaystyle <f,f>=\int _{-1}^{1}\left({\frac {1}{2}}\left(3x^{2}-1\right)\right)^{2}dx={\frac {1}{4}}\int _{-1}^{1}\left(9x^{4}-6x^{2}+1\right)dx={\frac {1}{4}}\left[{\frac {9}{5}}x^{5}-2x^{3}+x\right]_{-1}^{1}}
(3.17)
Solving this yields:
1
4
[
(
9
5
−
2
+
1
)
−
(
−
9
5
+
2
−
1
)
]
=
1
4
(
4
5
+
4
5
)
=
2
5
{\displaystyle \displaystyle {\frac {1}{4}}\left[\left({\frac {9}{5}}-2+1\right)-\left(-{\frac {9}{5}}+2-1\right)\right]={\frac {1}{4}}\left({\frac {4}{5}}+{\frac {4}{5}}\right)={\frac {2}{5}}}
(3.18)
Therefore the magnitude of f is found to be:
|
|
f
|
|
=<
f
,
f
>
1
/
2
=
0.4
=
0.632
{\displaystyle \displaystyle ||f||=<f,f>^{1/2}={\sqrt {0.4}}=0.632}
(3.19)
A similar approach can be taken for the magnitude of g:
<
g
,
g
>=
∫
−
1
1
(
1
2
(
5
x
3
−
3
x
)
)
2
d
x
=
1
4
∫
−
1
1
(
25
x
6
−
15
x
4
+
9
x
2
)
d
x
=
1
4
[
25
7
x
7
−
3
x
5
+
3
x
3
]
−
1
1
{\displaystyle \displaystyle <g,g>=\int _{-1}^{1}\left({\frac {1}{2}}\left(5x^{3}-3x\right)\right)^{2}dx={\frac {1}{4}}\int _{-1}^{1}\left(25x^{6}-15x^{4}+9x^{2}\right)dx={\frac {1}{4}}\left[{\frac {25}{7}}x^{7}-3x^{5}+3x^{3}\right]_{-1}^{1}}
(3.20)
1
2
(
25
7
−
3
+
3
)
=
25
14
{\displaystyle \displaystyle {\frac {1}{2}}\left({\frac {25}{7}}-3+3\right)={\frac {25}{14}}}
(3.21)
|
|
g
|
|
=<
g
,
g
>
1
/
2
=
25
14
=
1.336
{\displaystyle \displaystyle ||g||=<g,g>^{1/2}={\sqrt {\frac {25}{14}}}=1.336}
(3.22)
Using the above information with (3.2), the cosine of the angle between f and g is found to be
cos
θ
=
<
f
,
g
>
|
|
f
|
|
|
|
g
|
|
=
0
(
0.632
)
(
1.336
)
=
0
{\displaystyle \displaystyle \cos \theta ={\frac {<f,g>}{||f||||g||}}={\frac {0}{(0.632)(1.336)}}=0}
(3.23)
Because the scalar product of the two functions is zero, the two functions are orthogonal. That is:
θ
=
π
2
r
a
d
=
90
∘
{\displaystyle \displaystyle \theta ={\frac {\pi }{2}}rad=90^{\circ }}
(3.24)
Solved and Typed By - Egm4313.s12.team1.armanious (talk ) 23:48, 21 April 2012 (UTC)
Reviewed By - --Egm4313.s12.team1.stewart (talk ) 02:22, 25 April 2012 (UTC)
Problem R7.4 The Fourier Series and Plotted Functions
edit
From Lecture 11 Pg. 8
Do K 2011 p.482 pb.6,9,12,13.
Problems 6 and 12 ask to graph the function and the Fourier series of the function below (4.1).
f
(
x
)
=
|
x
|
{\displaystyle \displaystyle f(x)=\left|x\right|}
(4.1)
Problems 9 and 13 ask to graph the function and Fourier series of the function below (4.2).
f
(
x
)
=
{
x
i
f
−
π
<
x
<
0
π
−
x
i
f
0
<
x
<
π
}
{\displaystyle \displaystyle f(x)={\begin{Bmatrix}x\;\;if\;-\pi <x<0\\\pi -x\;\;if\;0<x<\pi \end{Bmatrix}}}
(4.2)
For Question number 6, plotted using Wolfram Alpha.
The Function 4.1 The Function 4.2
Question 12 asks for the Fourier series of equation 4.1,
f
(
x
)
=
|
x
|
{\displaystyle f(x)=\left|x\right|\!}
f
(
−
x
)
=
+
f
(
x
)
{\displaystyle f(-x)=+f(x)\!}
b
n
=
0
{\displaystyle b_{n}=0\!}
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
n
π
x
L
{\displaystyle \displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos{\frac {n\pi x}{L}}}
(4.3)
In which
L
=
π
{\displaystyle L=\pi \!}
a
0
=
1
2
π
∫
−
π
0
−
x
d
x
+
1
2
π
∫
0
π
x
d
x
{\displaystyle \displaystyle a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{0}-xdx+{\frac {1}{2\pi }}\int _{0}^{\pi }xdx}
(4.4)
a
n
=
1
π
∫
−
π
π
f
(
x
)
c
o
s
(
n
x
)
d
x
{\displaystyle \displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)cos(nx)dx}
(4.5)
b
n
=
1
π
∫
−
π
π
f
(
x
)
s
i
n
(
n
x
)
d
x
=
0
{\displaystyle \displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)sin(nx)dx=0}
(4.6)
These simplify to:
a
0
=
1
2
π
π
2
2
+
1
2
π
π
2
2
=
π
2
{\displaystyle \displaystyle a_{0}={\frac {1}{2\pi }}{\frac {\pi ^{2}}{2}}+{\frac {1}{2\pi }}{\frac {\pi ^{2}}{2}}={\frac {\pi }{2}}}
(4.7)
a
n
=
1
π
∫
−
π
0
−
x
c
o
s
(
n
x
)
d
x
+
1
π
∫
−
π
0
x
c
o
s
(
n
x
)
d
x
{\displaystyle \displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{0}-xcos(nx)dx+{\frac {1}{\pi }}\int _{-\pi }^{0}xcos(nx)dx}
(4.8)
a
n
=
−
4
n
2
π
{\displaystyle \displaystyle a_{n}={\frac {-4}{n^{2}\pi }}}
(4.9)
f
(
x
)
=
π
2
+
∑
n
−
1
∞
4
n
2
π
c
o
s
n
π
x
L
{\displaystyle \displaystyle f(x)={\frac {\pi }{2}}+\sum _{n-1}^{\infty }{\frac {4}{n^{2}\pi }}cos{\frac {n\pi x}{L}}}
(4.10)
The final answer:
f
(
x
)
=
π
2
−
4
π
(
c
o
s
x
+
c
o
s
3
x
9
+
c
o
s
5
x
25
+
.
.
.
)
{\displaystyle \displaystyle f(x)={\frac {\pi }{2}}-{\frac {4}{\pi }}(cosx+{\frac {cos3x}{9}}+{\frac {cos5x}{25}}+...)}
(4.11)
Plotted with Wolfram Aplpha
Fourier Series of Function 4.1
Now the Fourier series of equation 4.2, we need to divide the function into two parts. First let's evaluate
f
(
x
)
=
x
i
f
−
π
<
x
<
0
{\displaystyle f(x)=x\;if\;-\pi <x<0\!}
f
(
−
x
)
=
−
f
(
x
)
{\displaystyle f(-x)=-f(x)\!}
a
0
=
a
n
=
0
{\displaystyle a_{0}=a_{n}=0\!}
f
(
x
)
=
∑
n
=
1
∞
[
b
n
sin
(
n
π
x
L
)
]
=
∑
n
=
1
∞
[
b
n
sin
(
n
x
)
]
{\displaystyle \displaystyle f(x)=\sum \limits _{n=1}^{\infty }{\left[{{b}_{n}}\sin({\frac {n\pi x}{L}})\right]}=\sum \limits _{n=1}^{\infty }{\left[{{b}_{n}}\sin(nx)\right]}}
(4.12)
b
n
=
1
π
∫
−
π
0
x
s
i
n
(
n
x
)
=
1
π
[
s
i
n
(
n
x
)
−
n
x
c
o
s
(
n
x
)
n
2
]
=
−
c
o
s
(
−
π
n
)
n
{\displaystyle \displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{0}xsin(nx)={\frac {1}{\pi }}[{\frac {sin(nx)-nxcos(nx)}{n^{2}}}]={\frac {-cos(-\pi n)}{n}}}
(4.13)
Which simplifies to:
b
n
=
[
1
,
−
1
2
,
1
3
,
−
1
4
,
1
5
,
−
1
6
.
.
.
]
{\displaystyle \displaystyle b_{n}=[1,{\frac {-1}{2}},{\frac {1}{3}},{\frac {-1}{4}},{\frac {1}{5}},{\frac {-1}{6}}...]}
(4.14)
The Fourier Series for just the top part of equation 4.2
f
(
x
)
=
s
i
n
x
−
s
i
n
2
x
2
+
s
i
n
3
x
3
−
s
i
n
4
x
4
+
s
i
n
5
x
5
.
.
.
{\displaystyle \displaystyle f(x)=sinx-{\frac {sin2x}{2}}+{\frac {sin3x}{3}}-{\frac {sin4x}{4}}+{\frac {sin5x}{5}}...}
(4.15)
Now the bottom part of equation 4.2 is neither even nor odd so we must add all the components and at the end combine the solution with the Fourier series for the top part of equation 4.2
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
(
n
π
x
L
)
+
∑
n
=
1
∞
b
n
s
i
n
(
n
π
x
L
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
(
n
x
)
+
∑
n
=
1
∞
b
n
s
i
n
(
n
x
)
{\displaystyle \displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos({\frac {n\pi x}{L}})+\sum _{n=1}^{\infty }b_{n}sin({\frac {n\pi x}{L}})=a_{0}+\sum _{n=1}^{\infty }a_{n}cos(nx)+\sum _{n=1}^{\infty }b_{n}sin(nx)}
(4.16)
a
0
=
1
2
π
∫
0
π
(
π
−
x
)
d
x
=
1
2
π
[
π
x
−
x
2
2
]
=
π
4
{\displaystyle \displaystyle a_{0}={\frac {1}{2\pi }}\int _{0}^{\pi }(\pi -x)dx={\frac {1}{2\pi }}[\pi x-{\frac {x^{2}}{2}}]={\frac {\pi }{4}}}
(4.17)
a
n
=
1
π
∫
0
π
(
π
−
x
)
c
o
s
(
n
x
)
d
x
=
1
π
[
n
(
π
−
x
)
s
i
n
(
n
x
)
−
c
o
s
(
n
x
)
n
2
]
=
1
−
c
o
s
(
π
n
)
π
n
2
{\displaystyle \displaystyle a_{n}={\frac {1}{\pi }}\int _{0}^{\pi }(\pi -x)cos(nx)dx={\frac {1}{\pi }}[{\frac {n(\pi -x)sin(nx)-cos(nx)}{n^{2}}}]={\frac {1-cos(\pi n)}{\pi n^{2}}}}
(4.18)
a
n
=
[
2
π
,
0
,
2
6
π
,
0
,
2
25
π
,
0
,
2
49
π
.
.
.
]
{\displaystyle \displaystyle a_{n}=[{\frac {2}{\pi }},0,{\frac {2}{6\pi }},0,{\frac {2}{25\pi }},0,{\frac {2}{49\pi }}...]}
(4.19)
b
n
=
1
π
∫
0
π
(
π
−
x
)
s
i
n
(
n
x
)
d
x
=
1
π
[
s
i
n
(
n
x
)
+
n
(
π
−
x
)
c
o
s
(
n
x
)
n
2
]
=
1
n
{\displaystyle \displaystyle b_{n}={\frac {1}{\pi }}\int _{0}^{\pi }(\pi -x)sin(nx)dx={\frac {1}{\pi }}[{\frac {sin(nx)+n(\pi -x)cos(nx)}{n^{2}}}]={\frac {1}{n}}}
(4.20)
b
n
=
[
1
,
1
2
,
1
3
,
1
4
,
1
5
,
1
6
.
.
.
]
{\displaystyle \displaystyle b_{n}=[1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},{\frac {1}{6}}...]}
(4.21)
The Fourier Series of
f
(
x
)
=
π
−
x
i
f
0
<
x
<
π
{\displaystyle f(x)=\pi -x\;if\;0<x<\pi \!}
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
c
o
s
(
n
x
)
+
∑
n
=
1
∞
b
n
s
i
n
(
n
x
)
=
π
4
+
[
2
c
o
s
x
π
+
2
c
o
s
3
x
6
π
+
2
c
o
s
5
x
25
π
.
.
.
]
+
[
1
+
s
i
n
x
2
+
s
i
n
2
x
3
+
s
i
n
3
x
4
+
s
i
n
4
x
5
+
s
i
n
5
x
6
.
.
.
]
{\displaystyle \displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos(nx)+\sum _{n=1}^{\infty }b_{n}sin(nx)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[1+{\frac {sinx}{2}}+{\frac {sin2x}{3}}+{\frac {sin3x}{4}}+{\frac {sin4x}{5}}+{\frac {sin5x}{6}}...]}
(4.22)
Combining with the Fourier Series of
f
(
x
)
=
x
i
f
−
π
<
x
<
0
{\displaystyle f(x)=x\;if\;-\pi <x<0\!}
f
(
x
)
=
π
4
+
[
2
c
o
s
x
π
+
2
c
o
s
3
x
6
π
+
2
c
o
s
5
x
25
π
.
.
.
]
+
[
s
i
n
x
+
s
i
n
2
x
2
+
s
i
n
3
x
3
+
s
i
n
4
x
4
+
s
i
n
5
x
5
+
.
.
.
]
+
[
s
i
n
x
−
s
i
n
2
x
2
+
s
i
n
3
x
3
−
s
i
n
4
x
4
+
s
i
n
5
x
5
.
.
.
]
{\displaystyle \displaystyle f(x)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[sinx+{\frac {sin2x}{2}}+{\frac {sin3x}{3}}+{\frac {sin4x}{4}}+{\frac {sin5x}{5}}+...]+[sinx-{\frac {sin2x}{2}}+{\frac {sin3x}{3}}-{\frac {sin4x}{4}}+{\frac {sin5x}{5}}...]}
(4.23)
The total series and final answer:
f
(
x
)
=
π
4
+
[
2
c
o
s
x
π
+
2
c
o
s
3
x
6
π
+
2
c
o
s
5
x
25
π
.
.
.
]
+
[
2
s
i
n
x
+
2
s
i
n
3
x
3
+
2
s
i
n
5
x
5
.
.
.
]
{\displaystyle \displaystyle f(x)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[2sinx+{\frac {2sin3x}{3}}+{\frac {2sin5x}{5}}...]}
(4.24)
Plotted with Wolfram Alpha
Fourier Series of Function 4.2
Solved and Typed By - --Egm4313.s12.team1.stewart (talk ) 02:22, 25 April 2012 (UTC)
Reviewed By - Egm4313.s12.team1.essenwein (talk ) 16:50, 25 April 2012 (UTC)
Problem R7.5 Approximation by Trigonometric Series
edit
Consider equation (5.0) below:
<
ϕ
2
j
−
1
,
ϕ
2
k
−
1
>=
∫
0
P
sin
j
ω
x
⋅
sin
k
ω
x
d
x
=
0
{\displaystyle \displaystyle <\phi _{2j-1},\phi _{2k-1}>=\int _{0}^{P}\sin {j\omega x}\cdot \sin {k\omega x}dx=0}
(5.0)
1) Find the exact integration of equation (5.0) with
P
=
2
π
{\displaystyle P=2\pi \!}
j
=
2
{\displaystyle j=2\!}
k
=
3
{\displaystyle k=3\!}
1) Substituting the given values, we get equation (5.1) below:
<
ϕ
3
,
ϕ
5
>=
∫
0
2
π
sin
2
ω
x
⋅
sin
3
ω
x
d
x
=
0
{\displaystyle \displaystyle <\phi _{3},\phi _{5}>=\int _{0}^{2\pi }\sin {2\omega x}\cdot \sin {3\omega x}dx=0}
(5.1)
Using the product and sum trigonometry identities, we can evaluate the definite integral:
∫
0
2
π
sin
2
ω
x
⋅
sin
3
ω
x
d
x
=
1
2
∫
0
2
π
(
cos
ω
x
−
cos
5
ω
x
)
d
x
=
0
{\displaystyle \displaystyle \int _{0}^{2\pi }\sin {2\omega x}\cdot \sin {3\omega x}dx={\frac {1}{2}}\int _{0}^{2\pi }(\cos {\omega x}-\cos {5\omega x})dx=0}
(5.2)
→
[
1
2
ω
sin
ω
x
−
1
5
sin
5
ω
x
]
0
2
π
=
0
{\displaystyle \displaystyle \rightarrow [{\frac {1}{2\omega }}{\sin {\omega x}-{\frac {1}{5}}\sin {5\omega x}}]_{0}^{2\pi }=0}
(5.3)
Also,
ω
=
2
π
P
=
1
{\displaystyle \omega ={\frac {2\pi }{P}}=1\!}
[
1
2
sin
x
−
1
5
sin
5
x
]
0
2
π
=
0
{\displaystyle \displaystyle [{\frac {1}{2}}{\sin {x}-{\frac {1}{5}}\sin {5x}}]_{0}^{2\pi }=0}
(5.4)
2) The following is the code for evaluating the definite integral in equation (5.1) in Matlab:
EDU>> X = 0:pi/100:2*pi;
Z =
-2.2633e-17
EDU>> plot(X,Y)
The returned value of the evaluated integral in Matlab is essentially equal to zero, and it can be seen in the graph that the areas under the curve cancel each other out.
Solved and Typed By - Egm4313.s12.team1.wyattling
Reviewed By - Egm4313.s12.team1.rosenberg
Contributing Members
edit
Team Contribution Table
Problem Number Lecture Assigned To Solved By Typed By Proofread By
7.1
R7.1 Lect. 19c-1 pg. 10
Jesse Durrance
Jesse Durrance
Jesse Durrance
George Armanious
7.2
R7.2 Lect. 19c-1 pg. 13
Emotion Silvestri
Emotion Silvestri
Emotion Silvestri
George Armanious
7.3
R7.3 Lect. 11-1 pg. 8
George Armanious
George Armanious
George Armanious
Chris Stewart
7.4
R7.4 Lect. 11-1 pg.8
Chris Stewart
Chris Stewart
Chris Stewart
Eric Essenwein
7.5
R7.5 Lect. 12-1 pg. 8
Wyatt Ling
Wyatt Ling
Wyatt Ling
Steven Rosenberg
![{\displaystyle ={\frac {1}{2(\omega _{i}-\omega _{j})}}[\sin((\omega _{i}-\omega _{j})L)-\sin(0)]-{\frac {1}{2(\omega _{i}+\omega _{j})}}[\sin((\omega _{i}+\omega _{j})L)-\sin(0)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd2e0211deaf86437120a5901232014085d8b7b7)
![{\displaystyle \displaystyle u(x,t)=[a_{1}cos(c\omega _{1}\alpha {\frac {2L}{c}})sin(\omega _{1}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/61748af7c754398de51efc2ff9d804a724492052)
![{\displaystyle \displaystyle +[a_{2}cos(c\omega _{2}\alpha {\frac {2L}{c}})sin(\omega _{2}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77b2398e0e1183e59cd982468028ad7883f71a98)
![{\displaystyle \displaystyle +[a_{3}cos(c\omega _{3}\alpha {\frac {2L}{c}})sin(\omega _{3}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/824573bd2482a150fdba80f4086ba1b4b0e4af51)
![{\displaystyle \displaystyle +[a_{4}cos(c\omega _{4}\alpha {\frac {2L}{c}})sin(\omega _{4}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/75972838f83a25173891386103fd2e2eea63bba0)
![{\displaystyle \displaystyle +[a_{5}cos(c\omega _{5}\alpha {\frac {2L}{c}})sin(\omega _{5}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/89c1b54810daabd299cf94f73a6fd10235275720)
![{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(.5)(2*10))sin(\omega _{1}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d72c7a30a94fff602376e9d177082a83c767377)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(.5)(2*10))sin(\omega _{3}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/858a686847779292b8e4a7378bc5b47ab1e9c6fe)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(.5)(2*10))sin(\omega _{5}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/680291eb2741771a15f7f868d0f57bc4184e6d8a)
![{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(1)(2*10))sin(\omega _{1}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cdb98a4fbf39fda2aee6ff8b750a5890e8bd9381)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(1)(2*10))sin(\omega _{3}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/158e34254fd514744713250bb14fda7f2b815666)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(1)(2*10))sin(\omega _{5}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15580f427478efbd512adc110432f312f8829a73)
![{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(1.5)(2*10))sin(\omega _{1}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ececea697e1edabff7afed0a028067ab2cbfefbc)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(1.5)(2*10))sin(\omega _{3}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/604a78723ac5adb446f18fea605b5137db4ab6e2)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(1.5)(2*10))sin(\omega _{5}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c27ef675415602c08e90856f1a907f43ed1b1e63)
![{\displaystyle \displaystyle u(x,t)=[{\frac {-4}{\pi ^{3}}}cos(\omega _{1}(2)(2*10))sin(\omega _{1}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dcce36df7f5124ec6a537215a6fa358e0d52f7e3)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*3^{3}}}cos(\omega _{3}(2)(2*10))sin(\omega _{3}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6a9e1cfef6df33db6ef5352f4ebd028a9e1d661e)
![{\displaystyle \displaystyle +[{\frac {-4}{\pi ^{3}*5^{3}}}cos(\omega _{5}(2)(2*10))sin(\omega _{5}x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0916e5dc7bcc5809d100630e1279972e27adec7e)
![{\displaystyle \displaystyle ||f||:=<f,f>^{1/2}=\left[\int _{a}^{b}f^{2}(x)dx\right]^{1/2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/51007b8163e92af258c58b85ee916ceae5edff9f)
![{\displaystyle \displaystyle <f,g>=\int _{-2}^{10}x\cos(x)dx=\left[x\sin(x)+cos(x)\right]_{-2}^{10}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/705e6afcfb33ebaaad00d286b3d3d6d1cefb73e6)
![{\displaystyle \displaystyle <f,f>=\int _{-2}^{10}\cos ^{2}(x)dx=\int _{-2}^{10}\left({\frac {1}{2}}\cos(2x)+{\frac {1}{2}}\right)dx=\left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-2}^{10}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c0dd25661b5092fea10eb2ce250848d77d2fcef4)
![{\displaystyle \displaystyle \left[{\frac {1}{4}}\sin(2x)+{\frac {1}{2}}x\right]_{-2}^{10}=\left({\frac {1}{4}}\sin(20)+5\right)-\left({\frac {1}{4}}\sin(-4)-1\right)=6.04}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1464701f7d43fef731ab16e602f5903288e24d4)
![{\displaystyle \displaystyle <g,g>=\int _{-2}^{10}x^{2}dx=\left[{\frac {x^{3}}{3}}\right]_{-2}^{10}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1093451823cbbd91fa77b48b1e1f1a628348476f)
![{\displaystyle \displaystyle \left[{\frac {x^{3}}{3}}\right]_{-2}^{10}={\frac {1000}{3}}-{\frac {-8}{3}}={\frac {1008}{3}}=336}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3be1fdc0e60cf332e3f871cc3fddc0b7fe083099)
![{\displaystyle \displaystyle <f,g>=\int _{-1}^{1}{\frac {1}{4}}\left(3x^{2}-1\right)\left(5x^{3}-3x\right)dx={\frac {1}{4}}\int _{-1}^{1}\left(15x^{5}-14x^{3}+3x\right)dx={\frac {1}{4}}\left[{\frac {15}{6}}x^{6}-{\frac {14}{4}}x^{4}+{\frac {3}{2}}x^{2}\right]_{-1}^{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/29af0c1e971b20f340f4bee0b266ab6f4d1e9ac0)
![{\displaystyle \displaystyle {\frac {1}{4}}\left[{\frac {5}{2}}x^{6}-{\frac {7}{2}}x^{4}+{\frac {3}{2}}x^{2}\right]_{-1}^{1}={\frac {1}{4}}\left({\frac {1}{2}}-{\frac {1}{2}}\right)=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99393cdfff8eaad5586241f64297dd92b5226861)
![{\displaystyle \displaystyle <f,f>=\int _{-1}^{1}\left({\frac {1}{2}}\left(3x^{2}-1\right)\right)^{2}dx={\frac {1}{4}}\int _{-1}^{1}\left(9x^{4}-6x^{2}+1\right)dx={\frac {1}{4}}\left[{\frac {9}{5}}x^{5}-2x^{3}+x\right]_{-1}^{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee818778b925d51fbaeff9e39db691d653c06559)
![{\displaystyle \displaystyle {\frac {1}{4}}\left[\left({\frac {9}{5}}-2+1\right)-\left(-{\frac {9}{5}}+2-1\right)\right]={\frac {1}{4}}\left({\frac {4}{5}}+{\frac {4}{5}}\right)={\frac {2}{5}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e7e032dd6bc3a7244185133bd0afec6eb088a5bd)
![{\displaystyle \displaystyle <g,g>=\int _{-1}^{1}\left({\frac {1}{2}}\left(5x^{3}-3x\right)\right)^{2}dx={\frac {1}{4}}\int _{-1}^{1}\left(25x^{6}-15x^{4}+9x^{2}\right)dx={\frac {1}{4}}\left[{\frac {25}{7}}x^{7}-3x^{5}+3x^{3}\right]_{-1}^{1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d61706bbaf30f2c0f25718072457c3af86ca75c1)
![{\displaystyle \displaystyle f(x)=\sum \limits _{n=1}^{\infty }{\left[{{b}_{n}}\sin({\frac {n\pi x}{L}})\right]}=\sum \limits _{n=1}^{\infty }{\left[{{b}_{n}}\sin(nx)\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8f4107469deedc0a0b12292cfe5c908103bdad61)
![{\displaystyle \displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{0}xsin(nx)={\frac {1}{\pi }}[{\frac {sin(nx)-nxcos(nx)}{n^{2}}}]={\frac {-cos(-\pi n)}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/94410731eb9221bc5f8131076ec0b7ec453c0756)
![{\displaystyle \displaystyle b_{n}=[1,{\frac {-1}{2}},{\frac {1}{3}},{\frac {-1}{4}},{\frac {1}{5}},{\frac {-1}{6}}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/077d68aaf1093224f25a928e077dbaca1a69387d)
![{\displaystyle \displaystyle a_{0}={\frac {1}{2\pi }}\int _{0}^{\pi }(\pi -x)dx={\frac {1}{2\pi }}[\pi x-{\frac {x^{2}}{2}}]={\frac {\pi }{4}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8c21e888677ae5b672dd1a25bbecf26898261e6a)
![{\displaystyle \displaystyle a_{n}={\frac {1}{\pi }}\int _{0}^{\pi }(\pi -x)cos(nx)dx={\frac {1}{\pi }}[{\frac {n(\pi -x)sin(nx)-cos(nx)}{n^{2}}}]={\frac {1-cos(\pi n)}{\pi n^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d8ed14276ff1d5c996b25f8a3ebc4f63db5a070)
![{\displaystyle \displaystyle a_{n}=[{\frac {2}{\pi }},0,{\frac {2}{6\pi }},0,{\frac {2}{25\pi }},0,{\frac {2}{49\pi }}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7fcbe3bd7149fee041f073c3af96f22fd9861f42)
![{\displaystyle \displaystyle b_{n}={\frac {1}{\pi }}\int _{0}^{\pi }(\pi -x)sin(nx)dx={\frac {1}{\pi }}[{\frac {sin(nx)+n(\pi -x)cos(nx)}{n^{2}}}]={\frac {1}{n}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f8f5487f4d1bce05c0e3570da4d4bd16e4ea94f5)
![{\displaystyle \displaystyle b_{n}=[1,{\frac {1}{2}},{\frac {1}{3}},{\frac {1}{4}},{\frac {1}{5}},{\frac {1}{6}}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f56796cc068b43e84163e468f6808f44dc9190a)
![{\displaystyle \displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}cos(nx)+\sum _{n=1}^{\infty }b_{n}sin(nx)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[1+{\frac {sinx}{2}}+{\frac {sin2x}{3}}+{\frac {sin3x}{4}}+{\frac {sin4x}{5}}+{\frac {sin5x}{6}}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/295ee260ad09791f03e20f21c878789328462c82)
![{\displaystyle \displaystyle f(x)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[sinx+{\frac {sin2x}{2}}+{\frac {sin3x}{3}}+{\frac {sin4x}{4}}+{\frac {sin5x}{5}}+...]+[sinx-{\frac {sin2x}{2}}+{\frac {sin3x}{3}}-{\frac {sin4x}{4}}+{\frac {sin5x}{5}}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e8526f64e3f085fcf18e3a2506fa57592f04252a)
![{\displaystyle \displaystyle f(x)={\frac {\pi }{4}}+[{\frac {2cosx}{\pi }}+{\frac {2cos3x}{6\pi }}+{\frac {2cos5x}{25\pi }}...]+[2sinx+{\frac {2sin3x}{3}}+{\frac {2sin5x}{5}}...]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/591f80f96e905c62a232c4d004cdcfbfb3b1b536)
![{\displaystyle \displaystyle \rightarrow [{\frac {1}{2\omega }}{\sin {\omega x}-{\frac {1}{5}}\sin {5\omega x}}]_{0}^{2\pi }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fe6d4a809c62a25151c0df4d81fbecbc36ca31d2)
![{\displaystyle \displaystyle [{\frac {1}{2}}{\sin {x}-{\frac {1}{5}}\sin {5x}}]_{0}^{2\pi }=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c4489b1421abb4d0b80a4700d303852765b13a6c)
