Question
edit
Find the (smallest) period of
c
o
s
n
ω
x
{\displaystyle cosn\omega x}
s
i
n
n
ω
x
{\displaystyle sinn\omega x}
p .
a
o
{\displaystyle a_{o}}
p .
Solution
edit
Given (1) p.9-2:
f
(
x
+
n
p
)
=
f
(
x
)
{\displaystyle f(x+np)=f(x)}
n
(
.
.
.
,
−
2
,
−
1
,
0
,
1
,
2...
)
{\displaystyle n(...,-2,-1,0,1,2...)}
p is the period. n =1 thus we can use the equation as:
f
(
x
+
p
)
=
f
(
x
)
{\displaystyle f(x+p)=f(x)}
For
f
(
x
)
=
c
o
s
(
n
ω
x
)
{\displaystyle f(x)=cos(n\omega x)}
f
(
x
+
p
)
=
c
o
s
n
ω
(
x
+
p
)
=
c
o
s
(
n
ω
x
+
n
ω
p
)
=
c
o
s
(
n
ω
x
)
c
o
s
(
n
ω
p
)
−
s
i
n
(
n
ω
x
)
s
i
n
(
n
ω
p
)
{\displaystyle f(x+p)=cosn\omega (x+p)=cos(n\omega x+n\omega p)=cos(n\omega x)cos(n\omega p)-sin(n\omega x)sin(n\omega p)}
c
o
s
(
n
ω
x
)
c
o
s
(
n
ω
p
)
−
s
i
n
(
n
ω
x
)
s
i
n
(
n
ω
p
)
=
c
o
s
n
ω
x
{\displaystyle cos(n\omega x)cos(n\omega p)-sin(n\omega x)sin(n\omega p)=cosn\omega x}
c
o
s
(
n
ω
x
)
c
o
s
(
n
ω
p
)
=
c
o
s
(
n
ω
x
)
(
1
)
{\displaystyle cos(n\omega x)cos(n\omega p)=cos(n\omega x)(1)}
c
o
s
(
n
ω
p
)
=
1
{\displaystyle cos(n\omega p)=1}
n
ω
p
=
2
π
{\displaystyle n\omega p=2\pi }
p
=
2
π
n
ω
{\displaystyle p={\frac {2\pi }{n\omega }}}
n =1 so,
p
=
2
π
ω
{\displaystyle p={\frac {2\pi }{\omega }}}
f
(
x
)
=
s
i
n
(
n
ω
x
)
{\displaystyle f(x)=sin(n\omega x)}
f
(
x
+
p
)
=
s
i
n
n
ω
(
x
+
p
)
=
s
i
n
(
n
ω
x
+
n
ω
p
)
=
s
i
n
(
n
ω
x
)
c
o
s
(
n
ω
p
)
−
c
o
s
(
n
ω
x
)
s
i
n
(
n
ω
p
)
{\displaystyle f(x+p)=sinn\omega (x+p)=sin(n\omega x+n\omega p)=sin(n\omega x)cos(n\omega p)-cos(n\omega x)sin(n\omega p)}
s
i
n
(
n
ω
x
)
c
o
s
(
n
ω
p
)
−
c
o
s
(
n
ω
x
)
s
i
n
(
n
ω
p
)
=
s
i
n
(
n
ω
x
)
{\displaystyle sin(n\omega x)cos(n\omega p)-cos(n\omega x)sin(n\omega p)=sin(n\omega x)}
s
i
n
(
n
ω
x
)
c
o
s
(
n
ω
p
)
=
s
i
n
(
n
ω
x
)
(
1
)
{\displaystyle sin(n\omega x)cos(n\omega p)=sin(n\omega x)(1)}
c
o
s
(
n
ω
p
)
=
1
{\displaystyle cos(n\omega p)=1}
n
ω
p
=
2
π
{\displaystyle n\omega p=2\pi }
p
=
2
π
n
ω
{\displaystyle p={\frac {2\pi }{n\omega }}}
n =1 so,
p
=
2
π
ω
{\displaystyle p={\frac {2\pi }{\omega }}}
ω
=
2
π
p
{\displaystyle \omega ={\frac {2\pi }{p}}}
p
=
2
π
ω
=
2
π
2
π
p
=
p
{\displaystyle p={\frac {2\pi }{\omega }}={\frac {2\pi }{\frac {2\pi }{p}}}=p}
p
{\displaystyle p}
a
o
=
1
2
L
∫
0
2
L
f
(
x
)
d
x
=
1
p
<
f
,
1
>
{\displaystyle a_{o}={\frac {1}{2L}}\int _{0}^{2L}f(x)dx={\frac {1}{p}}<f,1>}
a
o
=
1
2
π
∫
0
2
π
f
(
x
)
d
x
=
1
p
<
f
,
1
>
{\displaystyle a_{o}={\frac {1}{2\pi }}\int _{0}^{2\pi }f(x)dx={\frac {1}{p}}<f,1>}
Thus,
p
=
2
π
{\displaystyle p=2\pi }
a
o
{\displaystyle a_{o}}
p
{\displaystyle p}
This problem is solved and uploaded by Radina Dikova
Question
edit
Is the given function even or odd or either even nor odd? Find its Fourier series. Show details of your work.
11.
f
(
x
)
=
x
2
,
(
−
1
<
x
<
1
)
,
p
=
2
{\displaystyle 11.f(x)=x^{2},\;(-1<x<1),\;p=2}
12.
f
(
x
)
=
1
−
x
2
4
,
(
−
2
<
x
<
2
)
,
p
=
4
{\displaystyle 12.f(x)=1-{\frac {x^{2}}{4}},\;(-2<x<2),\;p=4}
Solved and uploaded by William Knapper
Question
edit
K 2011 p.491 pbs 15,17
Solution
edit
For problem 15:
π
{\displaystyle \pi }
f
(
x
)
=
∑
n
=
1
∞
b
n
sin
(
n
x
)
{\displaystyle f(x)=\sum _{n=1}^{\infty }b_{n}\sin(nx)}
b
n
=
2
π
∫
0
π
f
(
x
)
sin
(
n
x
)
d
x
{\displaystyle b_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\sin(nx)dx}
[
0
,
π
2
]
{\displaystyle [0,{\frac {\pi }{2}}]}
[
π
2
,
π
]
{\displaystyle [{\frac {\pi }{2}},\pi ]}
f
(
x
)
=
π
−
x
{\displaystyle f(x)=\pi -x}
b
n
=
2
π
(
∫
0
π
2
x
sin
(
n
x
)
d
x
+
∫
π
2
π
(
π
−
x
)
sin
(
n
x
)
d
x
)
{\displaystyle b_{n}={\frac {2}{\pi }}(\int _{0}^{\frac {\pi }{2}}x\sin(nx)dx+\int _{\frac {\pi }{2}}^{\pi }(\pi -x)\sin(nx)dx)}
b
n
{\displaystyle b_{n}}
b
n
=
2
π
n
2
(
2
sin
(
π
2
n
)
−
sin
(
π
n
)
)
{\displaystyle b_{n}={\frac {2}{\pi n^{2}}}(2\sin({\frac {\pi }{2}}n)-\sin(\pi n))}
u
=
x
,
d
u
=
d
x
,
d
v
=
sin
(
n
x
)
d
x
,
a
n
d
v
=
−
1
n
cos
(
n
x
)
{\displaystyle u=x,du=dx,dv=\sin(nx)dx,andv=-{\frac {1}{n}}\cos(nx)}
u
=
π
−
x
,
d
u
=
−
d
x
,
d
v
=
−
sin
(
n
x
)
d
x
,
a
n
d
v
=
1
n
cos
(
n
x
)
{\displaystyle u=\pi -x,du=-dx,dv=-\sin(nx)dx,andv={\frac {1}{n}}\cos(nx)}
b
n
=
+
o
r
−
4
π
n
2
{\displaystyle b_{n}=+or-{\frac {4}{\pi n^{2}}}}
b
n
=
0
{\displaystyle b_{n}=0}
f
(
x
)
=
4
π
sin
(
x
)
{\displaystyle f(x)={\frac {4}{\pi }}\sin(x)}
f
(
x
)
=
4
π
sin
(
x
)
+
4
π
9
sin
(
3
x
)
{\displaystyle f(x)={\frac {4}{\pi }}\sin(x)+{\frac {4}{\pi 9}}\sin(3x)}
f
(
x
)
=
4
π
sin
(
x
)
+
4
π
9
sin
(
3
x
)
+
4
π
25
sin
(
5
x
)
+
4
π
49
sin
(
7
x
)
{\displaystyle f(x)={\frac {4}{\pi }}\sin(x)+{\frac {4}{\pi 9}}\sin(3x)+{\frac {4}{\pi 25}}\sin(5x)+{\frac {4}{\pi 49}}\sin(7x)}
f
(
−
x
)
=
f
(
x
)
{\displaystyle f(-x)=f(x)}
f
(
x
)
=
a
0
+
∑
n
=
1
∞
a
n
cos
(
n
x
)
{\displaystyle f(x)=a_{0}+\sum _{n=1}^{\infty }a_{n}\cos(nx)}
a
0
=
1
π
∫
0
π
f
(
x
)
d
x
{\displaystyle a_{0}={\frac {1}{\pi }}\int _{0}^{\pi }f(x)dx}
a
n
=
2
π
∫
0
π
f
(
x
)
cos
(
n
x
)
d
x
{\displaystyle a_{n}={\frac {2}{\pi }}\int _{0}^{\pi }f(x)\cos(nx)dx}
f
(
x
)
=
1
−
x
{\displaystyle f(x)=1-x}
a
0
=
∫
0
1
(
1
−
x
)
d
x
=
1
2
{\displaystyle a_{0}=\int _{0}^{1}(1-x)dx={\frac {1}{2}}}
π
{\displaystyle \pi }
a
n
{\displaystyle a_{n}}
cos
(
n
x
)
{\displaystyle \cos(nx)}
c
o
s
(
n
π
x
)
{\displaystyle cos(n\pi x)}
π
{\displaystyle \pi }
2
π
{\displaystyle {\frac {2}{\pi }}}
2
1
{\displaystyle {\frac {2}{1}}}
a
n
=
2
∫
0
1
(
f
(
x
)
cos
(
n
π
x
)
)
d
x
=
2
∫
0
1
(
(
1
−
x
)
cos
(
n
π
x
)
)
d
x
{\displaystyle a_{n}=2\int _{0}^{1}(f(x)\cos(n\pi x))dx=2\int _{0}^{1}((1-x)\cos(n\pi x))dx}
u
=
1
−
x
,
d
u
=
−
d
x
,
d
v
=
cos
(
n
π
x
)
d
x
,
v
=
1
n
π
sin
(
n
π
x
)
{\displaystyle u=1-x,du=-dx,dv=\cos(n\pi x)dx,v={\frac {1}{n\pi }}\sin(n\pi x)}
a
n
=
2
(
−
1
n
2
π
2
(
cos
(
n
π
)
+
1
n
2
π
2
)
{\displaystyle a_{n}=2({\frac {-1}{n^{2}\pi ^{2}}}(\cos(n\pi )+{\frac {1}{n^{2}\pi ^{2}}})}
T
h
e
r
e
f
o
r
e
,
w
h
e
n
n
i
s
o
d
d
,
<
m
a
t
h
>
a
n
=
4
n
2
π
2
{\displaystyle Therefore,whennisodd,<math>a_{n}={\frac {4}{n^{2}\pi ^{2}}}}
a
n
=
0
{\displaystyle a_{n}=0}
f
(
x
)
=
1
2
+
4
π
2
cos
(
π
x
)
{\displaystyle f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)}
f
(
x
)
=
1
2
+
4
π
2
cos
(
π
x
)
+
4
9
π
2
cos
(
3
π
x
)
{\displaystyle f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)+{\frac {4}{9\pi ^{2}}}\cos(3\pi x)}
f
(
x
)
=
1
2
+
4
π
2
cos
(
π
x
)
+
4
9
π
2
cos
(
3
π
x
)
+
4
25
π
2
cos
(
5
π
x
)
+
4
49
π
2
cos
(
7
π
x
)
{\displaystyle f(x)={\frac {1}{2}}+{\frac {4}{\pi ^{2}}}\cos(\pi x)+{\frac {4}{9\pi ^{2}}}\cos(3\pi x)+{\frac {4}{25\pi ^{2}}}\cos(5\pi x)+{\frac {4}{49\pi ^{2}}}\cos(7\pi x)}
This problem was solved and uploaded by John North
Question
edit
Consider the L2-ODE-CC with the window function f(x) from p9-8 as excitation:
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle \ y''-3y'+2y=r(x)\ }
and the initial conditions
y
(
0
)
=
1
,
y
′
(
0
)
=
0
{\displaystyle \ y(0)=1,y'(0)=0\ }
1. Find
y
n
(
x
)
{\displaystyle \ y_{n}(x)\ }
y
n
″
+
a
y
n
′
+
b
y
n
=
r
n
(
x
)
{\displaystyle \ y_{n}''+ay_{n}'+by_{n}=r_{n}(x)\ }
with the same initial conditions as above.
Plot
y
n
(
x
)
{\displaystyle \ y_{n}(x)\ }
2. Use the matlab command ode45 to integrate the L2-ODE-CC and plot the numerical soln to compare with the analytical soln.
Level 1: n = 0,1
Solution
edit
λ
2
−
3
λ
+
2
=
0
=>
(
λ
−
2
)
(
λ
−
1
)
=
0
{\displaystyle \ \lambda ^{2}-3\lambda +2=0=>(\lambda -2)(\lambda -1)=0\ }
λ
1
,
2
=
2
,
1
{\displaystyle \ \lambda _{1,2}=2,1\ }
The Fourier series of a periodic function
f
(
x
)
=
a
0
+
∑
n
=
1
∞
[
a
n
{\displaystyle \ f(x)=a_{0}+\sum _{n=1}^{\infty }[a_{n}\ }
b
n
{\displaystyle \ b_{n}\ }
For an odd function the Fourier series will be the following:
f
(
x
)
=
∑
n
=
1
∞
b
n
{\displaystyle \ f(x)=\sum _{n=1}^{\infty }b_{n}\ }
=
∑
n
=
1
∞
b
n
s
i
n
(
n
∗
p
i
L
x
)
d
x
{\displaystyle \ =\sum _{n=1}^{\infty }b_{n}sin({\frac {n*pi}{L}}x)dx\ }
The independent variable t will be used to shift x to the left
t
=
x
−
1
4
{\displaystyle \ t=x-{\frac {1}{4}}\ }
The period of oscillation and frequency of oscillation will be as follows:
p
=
2
L
=
4
{\displaystyle \ p=2L=4\ }
ω = π/2
where
b
n
=
2
L
∫
0
L
f
(
x
)
s
i
n
(
n
∗
p
i
L
x
)
d
x
=
1
2
∫
0
2
f
(
x
)
s
i
n
(
n
∗
p
i
2
x
)
d
x
{\displaystyle \ b_{n}={\frac {2}{L}}\int _{0}^{L}f(x)sin({\frac {n*pi}{L}}x)dx={\frac {1}{2}}\int _{0}^{2}f(x)sin({\frac {n*pi}{2}}x)dx\ }
which comes to be:
b
n
=
A
n
∗
p
i
{\displaystyle \ b_{n}={\frac {A}{n*pi}}\ }
y
″
−
3
y
′
+
2
y
=
r
(
x
)
{\displaystyle \ y''-3y'+2y=r(x)\ }
The homogeneous equation for Y will be as follows:
Y
h
=
C
1
e
2
x
+
C
2
e
x
{\displaystyle \ Y_{h}=C_{1}e^{2x}+C_{2}e^{x}\ }
Y
=
Y
h
+
Y
p
=
C
1
e
2
x
+
C
2
e
x
+
C
3
{\displaystyle \ Y=Y_{h}+Y_{p}=C_{1}e^{2x}+C_{2}e^{x}+C_{3}\ }
This problem was solved and uploaded by Mike Wallace
Question
edit
Redo R4.2 to redisplay the particular solution, the homogenous solution, and the exact solution for n = 3,5,9 over the interval [0,20π]
Redisplay the particular solution, the homogenous solution, and the exact solution. Superpose each solution with the exact solution.
Redo R4.3 with the TA code over the interval [0.10]. Zoom in about x = -0.5, 0, and 0.5 and comment on the accuracy of different approximations.
Redo R4.4 with the TA code over the interval [0.9,10] for n = 4, 7. Zoom in about x = 1, 1.5, 2, 2.5 and comment on the accuracy of different approximations.
Solution
edit
R 4.2 redo
edit
The re-displayed functions for the homogenous solutions are:
y
3
h
(
x
)
=
2
e
x
−
0.8008
e
2
x
{\displaystyle \displaystyle y_{3h}(x)=2e^{x}-0.8008e^{2x}\ }
y
5
h
(
x
)
=
2.0001
e
x
−
0.8001
e
2
x
{\displaystyle \displaystyle y_{5h}(x)=2.0001e^{x}-0.8001e^{2x}\ }
y
9
h
(
x
)
=
2
e
x
−
0.8
e
2
x
{\displaystyle \displaystyle y_{9h}(x)=2e^{x}-0.8e^{2x}\ }
The re-displayed functions for the particular solutions are:
y
3
p
(
x
)
=
−
9.9206
∗
10
−
5
x
7
−
0.0010
x
6
−
0.0031
x
5
−
0.0078
x
4
−
0.0990
x
3
−
0.3984
x
2
−
0.3984
x
−
0.1992
{\displaystyle \displaystyle y_{3p}(x)=-9.9206*10^{-5}x^{7}-0.0010x^{6}-0.0031x^{5}-0.0078x^{4}-0.0990x^{3}-0.3984x^{2}-0.3984x-0.1992\ }
y
5
p
(
x
)
=
−
1.2526
∗
10
−
8
x
11
−
2.0668
∗
10
−
7
x
10
−
1.0334
∗
10
−
6
x
9
−
4.6503
∗
10
−
6
x
8
−
1.1781
∗
10
−
4
x
7
{\displaystyle \displaystyle y_{5p}(x)=-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^{9}-4.6503*10^{-6}x^{8}-1.1781*10^{-4}x^{7}\ }
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.0999
x
3
−
0.3999
x
2
−
0.399
x
−
0.2000
{\displaystyle \displaystyle -0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.399x-0.2000\ }
y
9
p
(
x
)
=
−
4.1103
∗
10
−
18
x
19
−
1.1714
∗
10
−
16
x
18
−
1.0543
∗
10
−
15
x
17
−
8.9615
∗
10
−
15
x
16
{\displaystyle \displaystyle y_{9p}(x)=-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\ }
−
4.5405
∗
10
−
13
x
15
−
9.1408
∗
10
−
12
x
14
−
6.3985
∗
10
−
11
x
13
−
4.1590
∗
10
−
10
x
12
−
1.5021
∗
10
−
8
x
11
−
2.2040
∗
10
−
7
x
10
{\displaystyle \displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11}-2.2040*10^{-7}x^{10}}
−
1.1020
∗
10
−
6
x
9
−
4.9591
∗
10
−
6
x
8
−
1.1904
∗
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.1000
x
3
−
0.4000
x
2
−
0.4000
x
−
0.2000
{\displaystyle \displaystyle -1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000\ }
The re-displayed functions for the general solutions are:
y
3
(
x
)
=
2
e
x
−
0.8008
e
2
x
−
9.9206
∗
10
−
5
x
7
−
0.0010
x
6
−
0.0031
x
5
−
0.0078
x
4
−
0.0990
x
3
−
0.3984
x
2
−
0.3984
x
−
0.1992
{\displaystyle \displaystyle y_{3}(x)=2e^{x}-0.8008e^{2x}-9.9206*10^{-5}x^{7}-0.0010x^{6}-0.0031x^{5}-0.0078x^{4}-0.0990x^{3}-0.3984x^{2}-0.3984x-0.1992\ }
y
5
(
x
)
=
2.0001
e
x
−
0.8001
e
2
x
−
1.2526
∗
10
−
8
x
11
−
2.0668
∗
10
−
7
x
10
−
1.0334
∗
10
−
6
x
9
−
4.6503
∗
10
−
6
x
8
{\displaystyle \displaystyle y_{5}(x)=2.0001e^{x}-0.8001e^{2x}-1.2526*10^{-8}x^{11}-2.0668*10^{-7}x^{10}-1.0334*10^{-6}x^{9}-4.6503*10^{-6}x^{8}\ }
−
1.1781
∗
10
−
4
x
7
−
0.0011
x
6
−
0.0033
x
5
−
0.0083
x
4
−
0.0999
x
3
−
0.3999
x
2
−
0.399
x
−
0.2000
{\displaystyle \displaystyle -1.1781*10^{-4}x^{7}-0.0011x^{6}-0.0033x^{5}-0.0083x^{4}-0.0999x^{3}-0.3999x^{2}-0.399x-0.2000\ }
y
9
(
x
)
=
2
e
x
−
0.8
e
2
x
−
4.1103
∗
10
−
18
x
19
−
1.1714
∗
10
−
16
x
18
−
1.0543
∗
10
−
15
x
17
−
8.9615
∗
10
−
15
x
16
{\displaystyle \displaystyle y_{9}(x)=2e^{x}-0.8e^{2x}-4.1103*10^{-18}x^{19}-1.1714*10^{-16}x^{18}-1.0543*10^{-15}x^{17}-8.9615*10^{-15}x^{16}\ }
−
4.5405
∗
10
−
13
x
15
−
9.1408
∗
10
−
12
x
14
−
6.3985
∗
10
−
11
x
13
−
4.1590
∗
10
−
10
x
12
−
1.5021
∗
10
−
8
x
11
{\displaystyle \displaystyle -4.5405*10^{-13}x^{15}-9.1408*10^{-12}x^{14}-6.3985*10^{-11}x^{13}-4.1590*10^{-10}x^{12}-1.5021*10^{-8}x^{11}\ }
−
2.2040
∗
10
−
7
x
10
−
1.1020
∗
10
−
6
x
9
−
4.9591
∗
10
−
6
x
8
−
1.1904
∗
10
−
4
x
7
−
0.0011
x
6
{\displaystyle \displaystyle -2.2040*10^{-7}x^{10}-1.1020*10^{-6}x^{9}-4.9591*10^{-6}x^{8}-1.1904*10^{-4}x^{7}-0.0011x^{6}\ }
−
0.0033
x
5
−
0.0083
x
4
−
0.1000
x
3
−
0.4000
x
2
−
0.4000
x
−
0.2000
{\displaystyle \displaystyle -0.0033x^{5}-0.0083x^{4}-0.1000x^{3}-0.4000x^{2}-0.4000x-0.2000\ }
Plot for
y
3
(
x
)
{\displaystyle \displaystyle y_{3}(x)\ }
Plot for
y
5
(
x
)
{\displaystyle \displaystyle y_{5}(x)\ }
Plot for
y
9
(
x
)
{\displaystyle \displaystyle y_{9}(x)\ }
R 4.3 Redo
edit
Matlab code:
x = 0:0.01:10;
y = log(1+x);
EDU>> x1 = 0:0.01:10;
EDU>> y1 = zeros(1,1001);
EDU>> for i = 1:4
for j = 1:1001
y1(j) = y1(j) - ((-x1(j))^i)/i;
end
end
EDU>> y2 = zeros(1,1001);
EDU>> for i = 1:7
for j = 1:1001
y2(j) = y2(j) - ((-x1(j))^i)/i;
end
end
EDU>> y3 = zeros(1,1001);
EDU>> for i = 1:11
for j = 1:1001
y3(j) = y3(j) - ((-x1(j))^i)/i;
end
end
EDU>> y4 = zeros(1,1001);
EDU>> for i = 1:16
for j = 1:1001
y4(j) = y4(j) - ((-x1(j))^i)/i;
end
end
EDU>> h = plot(x,y);
orange = [1 0.5 0.2];
EDU>> set(h,'Color',orange);
EDU>> hold on;
EDU>> plot(x1,y1,'r');
EDU>> plot(x1,y2,'g');
EDU>> plot(x1,y3,'b');
EDU>> plot(x1,y4,'c');
legend('log(1+x)','T_4','T_7','T_1_1','T_1_6');
EDU>> grid on;
EDU>> axis([0 10 -10 10])
Zoom plot about 0:
This plot makes it appear that only the 16 term approximation is in the window at point 0. This is, therefore, the most accurate of the other n solution approximations.
Zoom plot about 0.5:
This plot makes it appear that most of the approximations are very accurate and close to the exact solution. The only approximation that appears to deviate a little after 0.5 is the n = 4 approximation.
R 4.4 redo
edit
Matlab code:
syms x
EDU>> f = log(x+1);
EDU>> fT1 = taylor(f,5,1);
EDU>> fT2 = taylor(f,8,1);
X = 0.9:0.1:10;
Y(:,1) = subs(fT1,'x',X);
EDU>> Y(:,2) = subs(fT2,'x',X);
EDU>> Y(:,3) = log(1+X);
EDU>> figure
EDU>> plot(X,Y);
axis([0.9 10 -10 10])
Zoom in about 1:
This shows that neither of the approximations are very close about x = 1 to the exact solution.
Zoom in about 1.5:
This shows that the n = 7 approximation is close about x = 1.5 to the exact solution.
Zoom in about 2:
This shows that all the approximations are very close to the exact solution about x = 2.
Zoom in about 2.5:
This shows that the approximations are very close about x = 2.5, however, the n = 4 approximation is beginning to deviate and soon will not be a good approximation.
This problem was solved and uploaded by David Herrick
Question
edit
(1) Find the separated ODEs for the heat equation
∂
u
∂
t
=
κ
∂
2
u
∂
x
2
{\displaystyle \displaystyle {\frac {\partial u}{\partial t}}=\kappa {\frac {\partial ^{2}u}{\partial x^{2}}}}
Solution
edit
Assuming
u
(
x
,
t
)
=
F
(
x
)
⋅
G
(
t
)
{\displaystyle \displaystyle u(x,t)=F(x)\cdot G(t)}
Then:
∂
u
(
x
,
t
)
∂
t
=
F
(
x
)
⋅
G
˙
(
t
)
{\displaystyle \displaystyle {\frac {\partial u(x,t)}{\partial t}}=F(x)\cdot {\dot {G}}(t)}
∂
2
u
(
x
,
t
)
∂
x
2
=
F
″
(
x
)
⋅
G
(
t
)
{\displaystyle \displaystyle {\frac {\partial ^{2}u(x,t)}{\partial x^{2}}}=F''(x)\cdot {G}(t)}
Substituting partial derivatives back into original PDE:
F
(
x
)
⋅
G
˙
(
t
)
=
κ
F
″
(
x
)
⋅
G
(
t
)
{\displaystyle \displaystyle F(x)\cdot {\dot {G}}(t)=\kappa F''(x)\cdot G(t)}
G
˙
(
t
)
κ
G
(
t
)
=
F
″
(
x
)
F
(
x
)
=
c
{\displaystyle \displaystyle {\frac {{\dot {G}}(t)}{\kappa G(t)}}={\frac {F''(x)}{F(x)}}=c}
* Where "c" is a constant, because if both sides were variables then they would never be equal one another (each side would be a function of a different variable.) Kreysig 2011, pp.546 Multiplying by the denominators to get two separate ODEs:
F
″
(
x
)
−
c
F
(
x
)
=
0
{\displaystyle \displaystyle F''(x)-cF(x)=0}
G
˙
(
t
)
−
κ
c
G
(
t
)
=
0
{\displaystyle \displaystyle {\dot {G}}(t)-\kappa cG(t)=0}
Solved and uploaded by Joshua House
Contribution Summary
edit
![{\displaystyle [0,{\frac {\pi }{2}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3608f07c68994fb2cec106391992ed4f4ddb0c97)
![{\displaystyle [{\frac {\pi }{2}},\pi ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/35dd686baa24fcf70706e00c15ddd690270cee0c)
![{\displaystyle \displaystyle y_{p}(x)=xe^{-2x}[Mcos(3x)+Nsin(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/86b34035937ec7550c78754280a83eb5f23d42c2)
![{\displaystyle \displaystyle y_{p}^{'}(x)=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/77ebc7c63a14c25cbbad201fb43764ed27f5b814)
![{\displaystyle \displaystyle y_{p}^{''}(x)=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5e972bba63214aa38e73a2626ac3012a5a30528e)
![{\displaystyle \displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/67ad29b7d7f864f6335d09839c2a71bd4339cbd9)
![{\displaystyle \displaystyle y_{p}^{''}=e^{-2x}[sin(3x)[6m(2x-1)-n(5x+4)]-cos(3x)[m(5x+4)+6n(2x-1)]]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/21e0f0b7b28f13c768197886c5f9e618f872f7a8)
![{\displaystyle \displaystyle y''_{p}=e^{-2x}[12mxsin(3x)-6msin(3x)-5nxsin(3x)-4nsin(3x)-5mxcos(3x)-4mcos(3x)-12nxcos(3x)+6ncos(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2704f9e872ccd54f5f69d9bb2976eaf8edb81aa8)
![{\displaystyle \displaystyle y_{p}^{'}=e^{-2x}[sin(3x)(-3mx-2nx+n)+cos(3x)(-2mx+m+3nx)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0701357ddde7af12ea8af8298f73bc6410567f10)
![{\displaystyle \displaystyle 4y'p=e^{-2x}[-12mxsin(3x)-8nxsin(3x)+4nsin(3x)-8mxcos(3x)+4mcos(3x)+12nxcos(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f654c6a3932e51546717cf71af9354bb3af654e9)
![{\displaystyle \displaystyle y''_{p}+4y'_{p}=e^{-2x}[-6msin(3x)-13nxsin(3x)-13mxcos(3x)+6ncos(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc8745f93ef9cbc2bec0a0e180489a68347a397d)
![{\displaystyle \displaystyle 13y_{p}=e^{-2x}[13mxcos(3x)+13nxcos(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7c531c38581bfee0abd1ef313f893732e7c09027)
![{\displaystyle \displaystyle y''_{p}+4y'_{p}+13y_{p}=e^{-2x}[-6msin(3x)+6ncos(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d6cc5c7e35d65e60e6b5b96773f6889c9e4e2b11)
![{\displaystyle \displaystyle 6e^{-2x}[ncos(3x)-msin(3x)]=6e^{-2x}[ncos(3x)-msin(3x)]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b3dec4df00ffab75d1ce95b3afa88925f96af738)
