Verify (4)-(5) p.19-9.
<
Φ
i
,
Φ
j
>=
0
,
i
≠
j
{\displaystyle <\Phi _{i},\Phi _{j}>=0,i\neq j}
<
Φ
i
,
Φ
j
>=
L
2
,
i
=
j
{\displaystyle <\Phi _{i},\Phi _{j}>={\frac {L}{2}},i=j}
We know that:
Φ
j
(
x
)
:=
s
i
n
ω
j
x
,
j
=
1
,
2
,
.
.
.
{\displaystyle \Phi _{j}(x):=sin\omega _{j}x,j=1,2,...}
⟨
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}
<
Φ
i
,
Φ
j
>=
∫
0
L
s
i
n
(
ω
i
x
)
s
i
n
(
ω
j
x
)
d
x
{\displaystyle <\Phi _{i},\Phi _{j}>=\int _{0}^{L}sin(\omega _{i}x)sin(\omega _{j}x)dx}
π
{\displaystyle \pi }
∫
0
π
s
i
n
(
ω
i
x
)
s
i
n
(
ω
j
x
)
d
x
=
(
ω
i
−
ω
j
)
s
i
n
(
(
ω
i
+
ω
j
)
x
)
+
(
−
ω
i
−
ω
j
)
s
i
n
(
(
ω
i
−
ω
j
)
x
)
2
ω
i
2
−
2
ω
j
2
{\displaystyle \int _{0}^{\pi }sin(\omega _{i}x)sin(\omega _{j}x)dx={\frac {(\omega _{i}-\omega _{j})sin((\omega _{i}+\omega _{j})x)+(-\omega _{i}-\omega _{j})sin((\omega _{i}-\omega _{j})x)}{2\omega _{i}^{2}-2\omega _{j}^{2}}}}
π
{\displaystyle \pi }
s
i
n
(
C
π
)
=
0
{\displaystyle sin(C\pi )=0}
s
i
n
(
0
)
=
0
{\displaystyle sin(0)=0}
π
{\displaystyle \pi }
2
π
{\displaystyle 2\pi }
ω
=
2
π
p
=
2
π
2
π
=
1
{\displaystyle \omega ={\frac {2\pi }{p}}={\frac {2\pi }{2\pi }}=1}
π
{\displaystyle \pi }
<
Φ
i
,
Φ
j
>=
0
,
i
≠
j
{\displaystyle <\Phi _{i},\Phi _{j}>=0,i\neq j}
<
Φ
j
,
Φ
j
>=
∫
0
L
s
i
n
(
ω
j
x
)
2
d
x
{\displaystyle <\Phi _{j},\Phi _{j}>=\int _{0}^{L}sin(\omega _{j}x)^{2}dx}
π
{\displaystyle \pi }
∫
0
π
s
i
n
(
ω
j
x
)
2
d
x
=
−
(
s
i
n
(
2
ω
j
x
)
−
2
ω
j
x
4
ω
j
)
{\displaystyle \int _{0}^{\pi }sin(\omega _{j}x)^{2}dx=-({\frac {sin(2\omega _{j}x)-2\omega _{j}x}{4\omega _{j}}})}
π
{\displaystyle \pi }
ω
=
1
{\displaystyle \omega =1}
−
(
−
2
ω
j
π
4
ω
j
−
0
)
=
π
2
{\displaystyle -(-{\frac {2\omega _{j}\pi }{4\omega _{j}}}-0)={\frac {\pi }{2}}}
π
{\displaystyle \pi }
This problem was solved and uploaded by: William Knapper
Plot the truncated series
u
(
x
,
t
)
=
∑
j
=
1
n
a
j
c
o
s
(
c
w
j
t
)
s
i
n
(
ω
j
x
)
{\displaystyle \displaystyle u(x,t)=\sum _{j=1}^{n}a_{j}cos(cw_{j}t)sin(\omega _{j}x)}
with n=5 and for
t
=
α
2
L
c
{\displaystyle \displaystyle t=\alpha {\frac {2L}{c}}}
where
α
=
0.5
,
1
,
1.5
,
2
{\displaystyle \displaystyle \alpha =0.5,1,1.5,2}
Matlab Code:
% Report 7 Problem 2
% Plot truncated series for n=5
% Given values
L = 2;
c = 3;
% Calculate alphas for use in t
alpha = (0.5:.5:2); % four alpha terms
t = ((alpha.*2*L)/c); % four "t" terms corresponding to the four alpha terms
% Calculate omegas (w)
j = (1:1:5); % series terms 1-5
w = ((j*pi)/L); % omega correspoding to series terms 1-5
% Calculate "a" coefficients
% Check if "a" is odd or even, then assign value
a = j;
for k=1:5;
if mod(k,2) == 0 % even term
a(k) = 0;
else % odd term
a(k) = -4/(pi^3*k^3);
end
end
% Calculate truncated series (note that a2=a4=0, so those terms in the
% series have no contribution
x=(0:0.01:2);
% For t=0.5
y1=zeros(size(x));
for k=1:5
y1 = y1 + a(k)*cos(c*w(k)*t(1))*sin(w(k)*x);
end
% For t=1.0
y2=zeros(size(x));
for k=1:5
y2 = y2 + a(k)*cos(c*w(k)*t(2))*sin(w(k)*x);
end
% For t=1.5
y3=zeros(size(x));
for k=1:5
y3 = y3 + a(k)*cos(c*w(k)*t(3))*sin(w(k)*x);
end
% For t=2.0
y4=zeros(size(x));
for k=1:5
y4 = y4 + a(k)*cos(c*w(k)*t(4))*sin(w(k)*x);
end
% Compare to original function
yorg = x.*(x-2);
plot(x,y1,x,y2,x,y3,x,y4,x,yorg)
legend('t=0.5','t=1.0','t=1.5','t=2.0','org function')
Plot of 4 different t's and original function:
This problem was solved and uploaded by: Josh House
Find the a) scalar product, b) the magnitude, and c) the angle between:
1.
f
(
x
)
=
cos
(
x
)
,
g
(
x
)
=
x
{\displaystyle \displaystyle 1.f(x)=\cos(x),g(x)=x\ }
−
2
≤
x
≤
10
{\displaystyle \displaystyle -2\leq x\leq 10\ }
and
2.
f
(
x
)
=
1
2
(
3
x
2
−
1
)
,
g
(
x
)
=
1
2
(
5
x
3
−
3
x
)
{\displaystyle \displaystyle 2.f(x)={\frac {1}{2}}(3x^{2}-1),g(x)={\frac {1}{2}}(5x^{3}-3x)\ }
−
1
≤
x
≤
1
{\displaystyle \displaystyle -1\leq x\leq 1\ }
The scalar product between 2 functions on the interval [a,b] is defined by:
<
f
,
g
>=
∫
a
b
f
(
x
)
g
(
x
)
d
x
{\displaystyle \displaystyle <f,g>=\int _{a}^{b}f(x)g(x)dx\ }
The magnitude of a function is defined by:
‖
f
‖
=<
f
,
f
>
1
2
=
[
∫
a
b
f
2
(
x
)
d
x
]
1
2
{\displaystyle \displaystyle \left\|f\right\|=<f,f>^{\frac {1}{2}}=\left[\int _{a}^{b}f^{2}(x)dx\right]^{\frac {1}{2}}\ }
The angle between 2 functions is defined by:
cos
(
θ
)
=
<
f
,
g
>
‖
f
‖
‖
g
‖
{\displaystyle \displaystyle \cos(\theta )={\frac {<f,g>}{\left\|f\right\|\left\|g\right\|}}\ }
Problem 1:
<
f
,
g
>=
∫
−
2
10
x
cos
(
x
)
d
x
{\displaystyle \displaystyle <f,g>=\int _{-2}^{10}x\cos(x)dx\ }
Using integration by parts:
u
=
x
,
d
u
=
d
x
,
d
V
=
cos
(
x
)
d
x
,
V
=
sin
(
x
)
{\displaystyle \displaystyle u=x,du=dx,dV=\cos(x)dx,V=\sin(x)\ }
The integral becomes:
x
sin
(
x
)
−
∫
sin
(
x
)
d
x
=
x
sin
(
x
)
+
cos
(
x
)
∣
−
2
10
=
−
7.68
{\displaystyle \displaystyle x\sin(x)-\int \sin(x)dx=x\sin(x)+\cos(x)\mid _{-2}^{10}=-7.68\ }
‖
f
‖
=
[
∫
−
2
10
cos
2
(
x
)
d
x
]
1
2
=
[
∫
−
2
10
1
+
cos
(
2
x
)
2
d
x
]
1
2
=
[
x
2
+
sin
(
2
x
)
4
∣
−
2
10
]
1
2
=
2.46
{\displaystyle \displaystyle \left\|f\right\|=[\int _{-2}^{10}\cos ^{2}(x)dx]^{\frac {1}{2}}=[\int _{-2}^{10}{\frac {1+\cos(2x)}{2}}dx]^{\frac {1}{2}}=[{\frac {x}{2}}+{\frac {\sin(2x)}{4}}\mid _{-2}^{10}]^{\frac {1}{2}}=2.46\ }
‖
g
‖
=
[
∫
−
2
10
x
2
d
x
]
1
2
=
[
x
3
3
∣
−
2
10
]
1
2
=
18.33
{\displaystyle \displaystyle \left\|g\right\|=[\int _{-2}^{10}x^{2}dx]^{\frac {1}{2}}=[{\frac {x^{3}}{3}}\mid _{-2}^{10}]^{\frac {1}{2}}=18.33\ }
cos
(
θ
)
=
−
7.68
18.33
∗
2.46
⇒
θ
=
1.742
{\displaystyle \displaystyle \cos(\theta )={\frac {-7.68}{18.33*2.46}}\Rightarrow \theta =1.742\ }
Problem 2:
<
f
,
g
>=
∫
−
1
1
1
2
(
3
x
2
−
1
)
∗
1
2
(
5
x
3
−
3
x
)
d
x
=
∫
−
1
1
(
15
4
x
5
−
7
2
x
3
+
3
4
x
)
d
x
{\displaystyle \displaystyle <f,g>=\int _{-1}^{1}{\frac {1}{2}}(3x^{2}-1)*{\frac {1}{2}}(5x^{3}-3x)dx=\int _{-1}^{1}({\frac {15}{4}}x^{5}-{\frac {7}{2}}x^{3}+{\frac {3}{4}}x)dx\ }
The integral becomes:
5
8
x
6
−
7
8
x
4
+
3
8
x
2
∣
−
1
1
=
0
{\displaystyle \displaystyle {\frac {5}{8}}x^{6}-{\frac {7}{8}}x^{4}+{\frac {3}{8}}x^{2}\mid _{-1}^{1}=0\ }
‖
f
‖
=
[
∫
−
1
1
[
1
2
(
3
x
2
−
1
)
]
2
d
x
]
1
2
=
[
∫
−
1
1
(
9
4
x
4
−
3
2
x
2
+
1
4
)
d
x
]
1
2
=
[
9
20
x
5
−
1
2
x
3
+
1
4
x
∣
−
1
1
]
1
2
=
0.632
{\displaystyle \displaystyle \left\|f\right\|=[\int _{-1}^{1}[{\frac {1}{2}}(3x^{2}-1)]^{2}dx]^{\frac {1}{2}}=[\int _{-1}^{1}({\frac {9}{4}}x^{4}-{\frac {3}{2}}x^{2}+{\frac {1}{4}})dx]^{\frac {1}{2}}=[{\frac {9}{20}}x^{5}-{\frac {1}{2}}x^{3}+{\frac {1}{4}}x\mid _{-1}^{1}]^{\frac {1}{2}}=0.632\ }
‖
g
‖
=
[
∫
−
1
1
[
1
2
(
5
x
3
−
3
x
)
]
2
d
x
]
1
2
=
[
∫
−
1
1
(
25
4
x
6
−
15
2
x
4
+
9
4
x
2
)
d
x
]
1
2
=
[
25
28
x
7
−
3
2
x
5
+
3
4
x
3
∣
−
1
1
]
1
2
=
0.535
{\displaystyle \displaystyle \left\|g\right\|=[\int _{-1}^{1}[{\frac {1}{2}}(5x^{3}-3x)]^{2}dx]^{\frac {1}{2}}=[\int _{-1}^{1}({\frac {25}{4}}x^{6}-{\frac {15}{2}}x^{4}+{\frac {9}{4}}x^{2})dx]^{\frac {1}{2}}=[{\frac {25}{28}}x^{7}-{\frac {3}{2}}x^{5}+{\frac {3}{4}}x^{3}\mid _{-1}^{1}]^{\frac {1}{2}}=0.535\ }
cos
(
θ
)
=
0
⇒
θ
=
π
/
2
{\displaystyle \displaystyle \cos(\theta )=0\Rightarrow \theta =\pi /2\ }
This problem was solved and uploaded by: David Herrick
K. 2011 p482 pb. 6,9,12,13:
problems 6,9: Sketch or graph f(x) for
−
π
<
x
<
π
{\displaystyle -\pi <x<\pi }
problems 12,13: Find the fourier series for the f(x) in problems 6 and 9 respectively up to n = 5.
Problem 6: f(x) = |x| for
−
π
<
x
<
π
{\displaystyle -\pi <x<\pi }
Problem 9: f(x) = x for
−
π
<
x
<
0
{\displaystyle -\pi <x<0}
and f(x) =
π
−
x
{\displaystyle \pi -x}
0
<
x
<
π
{\displaystyle 0<x<\pi }
Problem 6:
f(x) = |x|
f
(
x
)
=
x
{\displaystyle f(x)=x}
−
π
<
x
<
0
{\displaystyle -\pi <x<0}
f
(
x
)
=
π
−
x
{\displaystyle f(x)=\pi -x}
0
<
x
<
π
{\displaystyle 0<x<\pi }
f
(
x
)
=
−
x
{\displaystyle f(x)=-x}
−
π
<
x
<
0
{\displaystyle -\pi <x<0}
f
(
x
)
=
x
{\displaystyle f(x)=x}
0
<
x
<
π
{\displaystyle 0<x<\pi }
f
(
x
)
=
a
0
+
∑
n
=
1
∞
[
a
n
cos
(
n
x
)
+
b
n
sin
(
n
x
)
]
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}\cos(nx)+b_{n}\sin(nx)]}
a
0
=
1
2
π
∫
−
π
π
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{2\pi }}\int _{-\pi }^{\pi }f(x)dx}
a
n
=
1
π
∫
−
π
π
f
(
x
)
cos
(
n
x
)
d
x
{\displaystyle a_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\cos(nx)dx}
b
n
=
1
π
∫
−
π
π
f
(
x
)
sin
(
n
x
)
d
x
{\displaystyle b_{n}={\frac {1}{\pi }}\int _{-\pi }^{\pi }f(x)\sin(nx)dx}
a
0
=
1
2
π
[
∫
−
π
0
−
x
d
x
+
∫
0
π
x
d
x
]
{\displaystyle a_{0}={\frac {1}{2\pi }}[\int _{-\pi }^{0}-xdx+\int _{0}^{\pi }xdx]}
a
n
=
1
π
[
∫
−
π
0
−
x
cos
(
n
x
)
d
x
+
∫
0
π
x
cos
(
n
x
)
d
x
]
{\displaystyle a_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}-x\cos(nx)dx+\int _{0}^{\pi }x\cos(nx)dx]}
b
n
=
1
π
[
∫
−
π
0
−
x
sin
(
n
x
)
d
x
+
∫
0
π
x
sin
(
n
x
)
d
x
]
{\displaystyle b_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}-x\sin(nx)dx+\int _{0}^{\pi }x\sin(nx)dx]}
a
0
{\displaystyle a_{0}}
a
n
{\displaystyle a_{n}}
b
n
{\displaystyle b_{n}}
a
0
=
π
2
{\displaystyle a_{0}={\frac {\pi }{2}}}
a
n
=
2
π
n
2
(
cos
(
−
n
π
)
−
1
)
{\displaystyle a_{n}={\frac {2}{\pi n^{2}}}(\cos(-n\pi )-1)}
b
n
=
0
{\displaystyle b_{n}=0}
a
n
=
0
{\displaystyle a_{n}=0}
a
n
=
−
4
n
2
π
{\displaystyle a_{n}={\frac {-4}{n^{2}\pi }}}
f
5
(
x
)
=
π
2
−
4
π
cos
(
x
)
−
4
9
π
cos
(
3
x
)
−
4
25
π
cos
(
5
x
)
{\displaystyle f_{5}(x)={\frac {\pi }{2}}-{\frac {4}{\pi }}\cos(x)-{\frac {4}{9\pi }}\cos(3x)-{\frac {4}{25\pi }}\cos(5x)}
f
(
x
)
=
x
{\displaystyle f(x)=x}
−
π
<
x
<
0
{\displaystyle -\pi <x<0}
f
(
x
)
=
π
−
x
{\displaystyle f(x)=\pi -x}
0
<
x
<
π
{\displaystyle 0<x<\pi }
a
0
=
1
2
π
[
∫
−
π
0
x
d
x
+
∫
0
π
(
π
−
x
)
d
x
]
{\displaystyle a_{0}={\frac {1}{2\pi }}[\int _{-\pi }^{0}xdx+\int _{0}^{\pi }(\pi -x)dx]}
a
n
=
1
π
[
∫
−
π
0
x
cos
(
n
x
)
d
x
+
∫
0
π
(
π
−
x
)
cos
(
n
x
)
d
x
]
{\displaystyle a_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}x\cos(nx)dx+\int _{0}^{\pi }(\pi -x)\cos(nx)dx]}
b
n
=
1
π
[
∫
−
π
0
x
sin
(
n
x
)
d
x
+
∫
0
π
(
π
−
x
)
sin
(
n
x
)
d
x
]
{\displaystyle b_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}x\sin(nx)dx+\int _{0}^{\pi }(\pi -x)\sin(nx)dx]}
a
0
{\displaystyle a_{0}}
a
n
{\displaystyle a_{n}}
b
n
{\displaystyle b_{n}}
a
0
=
0
{\displaystyle a_{0}=0}
a
n
=
2
n
2
π
(
1
−
cos
(
n
π
)
)
{\displaystyle a_{n}={\frac {2}{n^{2}\pi }}(1-\cos(n\pi ))}
b
n
=
1
n
(
1
−
cos
(
n
π
)
)
{\displaystyle b_{n}={\frac {1}{n}}(1-\cos(n\pi ))}
a
n
=
4
n
2
π
{\displaystyle a_{n}={\frac {4}{n^{2}\pi }}}
b
n
=
2
n
{\displaystyle b_{n}={\frac {2}{n}}}
a
n
=
0
=
b
n
{\displaystyle a_{n}=0=b_{n}}
f
5
(
x
)
=
4
π
[
cos
(
x
)
+
1
9
cos
(
3
x
)
+
1
25
cos
(
5
x
)
]
+
2
[
sin
(
x
)
+
1
3
sin
(
3
x
)
+
1
5
sin
(
5
x
)
]
{\displaystyle f_{5}(x)={\frac {4}{\pi }}[\cos(x)+{\frac {1}{9}}\cos(3x)+{\frac {1}{25}}\cos(5x)]+2[\sin(x)+{\frac {1}{3}}\sin(3x)+{\frac {1}{5}}\sin(5x)]}
The problem was solved and uploaded by John North .
Consider (1)p.12-4:
<
Φ
2
j
−
1
Φ
2
k
−
1
>=
∫
0
p
sin
j
ω
x
⋅
sin
k
ω
x
d
x
{\displaystyle <\Phi _{2j-1}\Phi _{2k-1}>=\int _{0}^{p}\sin j\omega x\cdot \sin k\omega xdx}
p
=
2
π
,
j
=
2
,
k
=
3.
{\displaystyle p=2\pi ,j=2,k=3.}
1. Find the exact integration of (1)p.12-4 with the given data.
Using (3)p19-10 we get:
∫
0
2
π
sin
j
ω
x
⋅
sin
k
ω
x
d
x
=
∫
0
2
π
1
2
[
c
o
s
(
j
ω
−
k
ω
)
x
−
c
o
s
(
j
ω
+
k
ω
)
x
]
d
x
{\displaystyle \int _{0}^{2\pi }\sin j\omega x\cdot \sin k\omega xdx=\int _{0}^{2\pi }{\frac {1}{2}}\left[cos(j\omega -k\omega )x-cos(j\omega +k\omega )x\right]dx}
Integrating we get:
s
i
n
(
(
j
ω
−
k
ω
)
x
)
2
(
j
ω
−
k
ω
)
−
s
i
n
(
(
j
ω
+
k
ω
)
x
)
2
(
j
ω
+
k
ω
)
{\displaystyle {\frac {sin((j\omega -k\omega )x)}{2(j\omega -k\omega )}}-{\frac {sin((j\omega +k\omega )x)}{2(j\omega +k\omega )}}}
Plugging in the known values we have:
s
i
n
(
(
2
ω
−
3
ω
)
x
)
2
(
2
ω
−
3
ω
)
−
s
i
n
(
(
2
ω
+
3
ω
)
x
)
2
(
2
ω
+
3
ω
)
=
s
i
n
(
−
1
ω
x
)
−
2
ω
−
s
i
n
(
5
ω
x
)
10
ω
{\displaystyle {\frac {sin((2\omega -3\omega )x)}{2(2\omega -3\omega )}}-{\frac {sin((2\omega +3\omega )x)}{2(2\omega +3\omega )}}={\frac {sin(-1\omega x)}{-2\omega }}-{\frac {sin(5\omega x)}{10\omega }}}
Solving from 0 to
2
π
{\displaystyle 2\pi }
[
s
i
n
(
−
2
ω
π
)
−
2
ω
−
s
i
n
(
10
ω
π
)
10
ω
]
−
[
s
i
n
(
0
)
−
2
ω
−
s
i
n
(
0
)
10
ω
]
{\displaystyle \left[{\frac {sin(-2\omega \pi )}{-2\omega }}-{\frac {sin(10\omega \pi )}{10\omega }}\right]-\left[{\frac {sin(0)}{-2\omega }}-{\frac {sin(0)}{10\omega }}\right]}
We know that:
s
i
n
(
C
π
)
=
0
{\displaystyle sin(C\pi )=0}
s
i
n
(
0
)
=
0
{\displaystyle sin(0)=0}
Thus, if
ω
{\displaystyle \omega }
0
−
0
=
0
{\displaystyle 0-0=0}
Therefore:
<
Φ
3
,
Φ
5
>=
∫
0
2
π
sin
2
ω
x
⋅
sin
3
ω
x
d
x
=
0
{\displaystyle <\Phi _{3},\Phi _{5}>=\int _{0}^{2\pi }\sin 2\omega x\cdot \sin 3\omega xdx=0}
Using MATLAB's TRAPZ method, a variable X was created from 0 to
2
π
{\displaystyle 2\pi }
sin
j
ω
x
⋅
sin
k
ω
x
d
x
{\displaystyle \sin j\omega x\cdot \sin k\omega xdx}
was found from 0 to
2
π
{\displaystyle 2\pi }
ω
{\displaystyle \omega }
The following was the outcome in MATLAB:
-3.8317e-007 is used as an equivalent to 0.
Therefore through MATLAB's TRAPZ method,
∫
0
2
π
sin
2
ω
x
⋅
sin
3
ω
x
d
x
=
0
{\displaystyle \int _{0}^{2\pi }\sin 2\omega x\cdot \sin 3\omega xdx=0}
Part 1 of this problem was solved and uploaded by Radina Dikova .
Part 2 of this problem was solved and uploaded by Derik Bell .
Contribution Summary
edit
![{\displaystyle \displaystyle \left\|f\right\|=<f,f>^{\frac {1}{2}}=\left[\int _{a}^{b}f^{2}(x)dx\right]^{\frac {1}{2}}\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/f498c1b7b7dae0730b350e885c57714e3883e205)
![{\displaystyle \displaystyle \left\|f\right\|=[\int _{-2}^{10}\cos ^{2}(x)dx]^{\frac {1}{2}}=[\int _{-2}^{10}{\frac {1+\cos(2x)}{2}}dx]^{\frac {1}{2}}=[{\frac {x}{2}}+{\frac {\sin(2x)}{4}}\mid _{-2}^{10}]^{\frac {1}{2}}=2.46\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd568773e2c6babc07073d6cad3ca4501b3b5a45)
![{\displaystyle \displaystyle \left\|g\right\|=[\int _{-2}^{10}x^{2}dx]^{\frac {1}{2}}=[{\frac {x^{3}}{3}}\mid _{-2}^{10}]^{\frac {1}{2}}=18.33\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b1ae6422200b510be5c052f3439d36d336d55f0)
![{\displaystyle \displaystyle \left\|f\right\|=[\int _{-1}^{1}[{\frac {1}{2}}(3x^{2}-1)]^{2}dx]^{\frac {1}{2}}=[\int _{-1}^{1}({\frac {9}{4}}x^{4}-{\frac {3}{2}}x^{2}+{\frac {1}{4}})dx]^{\frac {1}{2}}=[{\frac {9}{20}}x^{5}-{\frac {1}{2}}x^{3}+{\frac {1}{4}}x\mid _{-1}^{1}]^{\frac {1}{2}}=0.632\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d6d25e88751304eb0fe04c9ce9c78e092506668)
![{\displaystyle \displaystyle \left\|g\right\|=[\int _{-1}^{1}[{\frac {1}{2}}(5x^{3}-3x)]^{2}dx]^{\frac {1}{2}}=[\int _{-1}^{1}({\frac {25}{4}}x^{6}-{\frac {15}{2}}x^{4}+{\frac {9}{4}}x^{2})dx]^{\frac {1}{2}}=[{\frac {25}{28}}x^{7}-{\frac {3}{2}}x^{5}+{\frac {3}{4}}x^{3}\mid _{-1}^{1}]^{\frac {1}{2}}=0.535\ }](https://wikimedia.org/api/rest_v1/media/math/render/svg/0458ad87ee6556934c00636d6735771afd86ff1f)
![{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}\cos(nx)+b_{n}\sin(nx)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b413022ca5674757b9ead07beea1a26091d3454b)
![{\displaystyle a_{0}={\frac {1}{2\pi }}[\int _{-\pi }^{0}-xdx+\int _{0}^{\pi }xdx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffe682a15b726d13d24ea3d15195dfa15185e4c9)
![{\displaystyle a_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}-x\cos(nx)dx+\int _{0}^{\pi }x\cos(nx)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ebca64dd19175e07d451b69ac57afa322aec394f)
![{\displaystyle b_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}-x\sin(nx)dx+\int _{0}^{\pi }x\sin(nx)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6ac9b9077887b439a12659a370e050215dd6e2b9)
![{\displaystyle a_{0}={\frac {1}{2\pi }}[\int _{-\pi }^{0}xdx+\int _{0}^{\pi }(\pi -x)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b35e5028c66cca145d3add06d536ff1672b636bb)
![{\displaystyle a_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}x\cos(nx)dx+\int _{0}^{\pi }(\pi -x)\cos(nx)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6b308a38ca3ef4ae8585f20b31d3fce200eae5b2)
![{\displaystyle b_{n}={\frac {1}{\pi }}[\int _{-\pi }^{0}x\sin(nx)dx+\int _{0}^{\pi }(\pi -x)\sin(nx)dx]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3ce58d9412489c7207f0be3be4257b00285286e)
![{\displaystyle f_{5}(x)={\frac {4}{\pi }}[\cos(x)+{\frac {1}{9}}\cos(3x)+{\frac {1}{25}}\cos(5x)]+2[\sin(x)+{\frac {1}{3}}\sin(3x)+{\frac {1}{5}}\sin(5x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8ac1f6996ca6719d7cbf9bb5efaaa367acfbe7b2)
![{\displaystyle \int _{0}^{2\pi }\sin j\omega x\cdot \sin k\omega xdx=\int _{0}^{2\pi }{\frac {1}{2}}\left[cos(j\omega -k\omega )x-cos(j\omega +k\omega )x\right]dx}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c9fad2b8c92cbac84f7b4980dddcffefe2e72e77)
![{\displaystyle \left[{\frac {sin(-2\omega \pi )}{-2\omega }}-{\frac {sin(10\omega \pi )}{10\omega }}\right]-\left[{\frac {sin(0)}{-2\omega }}-{\frac {sin(0)}{10\omega }}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/152c2a43883c493bd316a7195f94dad874871c16)
