Problem 1: Taylor Series Expansion of the log Function
edit
Problem Statement
edit
y
″
+
6
y
′
+
9
y
=
l
o
g
(
3
+
4
x
)
{\displaystyle y''+6y'+9y=log(3+4x)}
y
(
0
)
=
1
,
y
′
(
0
)
=
.5
{\displaystyle y(0)=1,y'(0)=.5}
Use the point
x
^
=
−
1
/
2
{\displaystyle {\widehat {x}}=-1/2}
log
x
=
log
e
x
=
ln
x
{\displaystyle \log x=\log _{e}x=\ln x}
Solution
edit
f
(
x
)
=
log
(
x
)
{\displaystyle f(x)=\log(x)}
f
(
x
^
)
=
log
(
x
^
)
{\displaystyle f({\widehat {x}})=\log({\widehat {x}})}
∑
k
=
0
∞
(
f
k
)
(
x
^
)
k
!
(
x
−
x
^
)
k
{\displaystyle \sum _{k=0}^{\infty }{\frac {(f^{k})({\widehat {x}})}{k!}}(x-{\widehat {x}})^{k}}
f
1
(
x
)
=
1
1
+
x
=
f
1
(
x
^
)
=
1
1
+
(
x
^
)
{\displaystyle f^{1}(x)={\frac {1}{1+x}}=f^{1}({\widehat {x}})={\frac {1}{1+({\widehat {x}})}}}
w
(
x
)
=
3
+
4
x
{\displaystyle w(x)=3+4x}
w
′
(
x
)
=
4
{\displaystyle w'(x)=4}
f
2
(
x
)
=
−
1
∗
w
′
∗
4
w
2
=
−
1
w
2
{\displaystyle f^{2}(x)={\frac {-1*w'*4}{w^{2}}}={\frac {-1}{w^{2}}}}
f
3
(
x
)
=
1
∗
2
∗
w
∗
w
′
∗
4
2
w
4
=
1
∗
2
w
3
{\displaystyle f^{3}(x)={\frac {1*2*w*w'*4^{2}}{w^{4}}}={\frac {1*2}{w^{3}}}}
f
4
(
x
)
=
−
1
∗
2
∗
3
∗
w
2
∗
w
′
∗
4
3
w
6
=
−
1
∗
2
∗
3
w
4
{\displaystyle f^{4}(x)={\frac {-1*2*3*w^{2}*w'*4^{3}}{w^{6}}}={\frac {-1*2*3}{w^{4}}}}
f
k
(
x
)
=
(
−
1
)
k
−
1
∗
(
k
−
1
)
!
∗
4
k
−
1
w
k
{\displaystyle f^{k}(x)={\frac {(-1)^{k-1}*(k-1)!*4^{k-1}}{w^{k}}}}
∑
k
=
0
∞
(
f
k
)
(
x
^
)
k
!
(
x
−
x
^
)
k
=
∑
k
=
0
∞
(
−
1
)
k
−
1
(
k
−
1
)
!
k
!
(
1
+
(
x
^
)
)
k
(
x
−
x
^
)
k
{\displaystyle \sum _{k=0}^{\infty }{\frac {(f^{k})({\widehat {x}})}{k!}}(x-{\widehat {x}})^{k}=\sum _{k=0}^{\infty }{\frac {(-1)^{k-1}(k-1)!}{k!(1+({\widehat {x}}))^{k}}}(x-{\widehat {x}})^{k}}
l
o
g
(
3
+
4
x
)
{\displaystyle log(3+4x)}
∑
k
=
0
∞
(
−
1
)
k
+
1
∗
(
(
2
+
4
x
)
k
k
)
{\displaystyle \sum _{k=0}^{\infty }(-1)^{k+1}*({\frac {(2+4x)^{k}}{k}})}
x
⟶
t
such that
3
+
4
x
=
1
+
t
{\displaystyle x\longrightarrow t\ {\text{ such that }}\ 3+4x=1+t}
x
=
t
−
2
4
{\displaystyle x={\frac {t-2}{4}}}
If
l
o
g
(
1
+
t
)
{\displaystyle log(1+t)}
[
−
1
,
+
1
]
{\displaystyle [-1,+1]}
l
o
g
(
3
+
4
x
)
{\displaystyle log(3+4x)}
[
−
3
4
,
−
1
4
]
{\displaystyle [{\frac {-3}{4}},{\frac {-1}{4}}]}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 2: Plots of Truncated Series
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Number 1
edit
Plot at least 3 truncated series to show convergence
∑
m
=
0
i
n
f
x
2
m
+
1
(
2
m
+
1
)
!
{\displaystyle \sum _{m=0}^{inf}{\frac {x^{2m+1}}{(2m+1)!}}}
m=0:
x
{\displaystyle x}
m=1:
x
+
x
3
3
∗
2
∗
1
{\displaystyle x+{\frac {x^{3}}{3*2*1}}}
m=2:
x
+
x
3
3
∗
2
∗
1
+
x
5
5
∗
4
∗
3
∗
2
∗
1
{\displaystyle x+{\frac {x^{3}}{3*2*1}}+{\frac {x^{5}}{5*4*3*2*1}}}
Number 2
edit
Plot at least 3 truncated series to show convergence
∑
m
=
0
i
n
f
(
2
3
)
m
x
2
m
{\displaystyle \sum _{m=0}^{inf}({\frac {2}{3}})^{m}x^{2}m}
m=0:
x
{\displaystyle x}
m=1:
x
+
2
3
x
2
{\displaystyle x+{\frac {2}{3}}x^{2}}
m=2:
x
+
2
3
x
2
+
4
9
x
4
{\displaystyle x+{\frac {2}{3}}x^{2}+{\frac {4}{9}}x^{4}}
Number 3
edit
Find the radius of convergence for the taylor series of sinx, x = 0
The Taylor series of sinx is:
∑
m
=
0
i
n
f
(
−
1
)
m
(
x
)
2
m
+
1
(
2
m
+
1
)
!
{\displaystyle \sum _{m=0}^{inf}{\frac {(-1)^{m}(x)^{2m+1}}{(2m+1)!}}}
The radius of convergence can be found by:
R
c
=
[
l
i
m
|
d
k
+
1
d
k
|
]
−
1
{\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}
R
c
=
[
l
i
m
(
−
1
)
m
+
1
(
2
m
+
3
)
!
∗
(
2
m
+
1
)
!
(
−
1
)
m
]
−
1
{\displaystyle R_{c}=[lim{\frac {(-1)^{m+1}}{(2m+3)!}}*{\frac {(2m+1)!}{(-1)^{m}}}]^{-1}}
R
c
=
∞
{\displaystyle R_{c}=\infty }
Number 4
edit
Find the radius of convergence for the taylor series of log(1+x), x = 0
The Taylor series of log(x+1) is:
−
∑
n
=
1
∞
(
−
1
)
n
(
x
)
n
n
{\displaystyle -\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x)^{n}}{n}}}
The radius of convergence can be found by:
R
c
=
[
l
i
m
|
d
k
+
1
d
k
|
]
−
1
{\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}
R
c
=
[
l
i
m
(
−
1
)
n
+
1
n
+
1
∗
n
(
−
1
)
n
]
−
1
{\displaystyle R_{c}=[lim{\frac {(-1)^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}}}]^{-1}}
R
c
=
1
{\displaystyle R_{c}=1}
Number 5
edit
Find the radius of convergence for the taylor series of log(1+x), x = 1
The Taylor series of log(x+1) is:
−
∑
n
=
1
∞
(
−
1
)
n
(
x
−
1
)
n
n
{\displaystyle -\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x-1)^{n}}{n}}}
The radius of convergence can be found by:
R
c
=
[
l
i
m
|
d
k
+
1
d
k
|
]
−
1
{\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}
R
c
=
[
l
i
m
(
−
1
)
n
+
1
n
+
1
∗
n
(
−
1
)
n
]
−
1
{\displaystyle R_{c}=[lim{\frac {(-1)^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}}}]^{-1}}
R
c
=
1
{\displaystyle R_{c}=1}
Number 6
edit
derive the expression for the radius of convergence of log(1+x) about any focus point
The taylor series of log(1+x) is:
−
∑
n
=
1
∞
(
−
1
)
n
(
x
−
x
^
)
n
n
{\displaystyle -\sum _{n=1}^{\infty }{\frac {(-1)^{n}(x-{\widehat {x}})^{n}}{n}}}
R
c
=
[
(
−
1
)
n
+
1
(
x
^
)
n
+
1
n
+
1
∗
n
(
−
1
)
n
(
x
^
)
n
]
−
1
{\displaystyle R_{c}=[{\frac {(-1)^{n+1}({\widehat {x}})^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}({\widehat {x}})^{n}}}]^{-1}}
Number 7
edit
Find the Taylor series representation of log(3+4x)
∑
k
=
0
∞
(
f
)
k
(
x
^
)
k
!
∗
(
x
−
x
^
)
k
{\displaystyle \sum _{k=0}^{\infty }{\frac {(f)^{k}({\widehat {x}})}{k!}}*(x-{\widehat {x}})^{k}}
r
(
x
)
=
0
+
4
∗
(
x
−
x
^
)
3
+
4
x
^
−
(
4
2
)
∗
(
(
x
−
x
^
)
2
(
(
3
+
4
x
^
)
2
)
∗
(
2
!
)
+
2
(
4
2
)
∗
(
(
x
−
x
^
)
3
(
(
3
+
4
x
^
)
3
)
∗
(
3
!
)
{\displaystyle r(x)=0+{\frac {4*(x-{\widehat {x}})}{3+4{\widehat {x}}}}-{\frac {(4^{2})*((x-{\widehat {x}})^{2}}{((3+4{\widehat {x}})^{2})*(2!)}}+{\frac {2(4^{2})*((x-{\widehat {x}})^{3}}{((3+4{\widehat {x}})^{3})*(3!)}}}
∑
k
=
0
∞
(
−
1
)
k
+
1
∗
(
(
2
+
4
x
)
k
k
)
{\displaystyle \sum _{k=0}^{\infty }(-1)^{k+1}*({\frac {(2+4x)^{k}}{k}})}
Number 8
edit
Radius of convergence of log(3+4x) about the point
x
^
=
−
.5
{\displaystyle {\widehat {x}}=-.5}
x
^
=
−
.5
{\displaystyle {\widehat {x}}=-.5}
∑
n
=
0
∞
(
−
1
)
n
+
1
4
n
n
(
x
+
.5
)
n
{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.5)^{n}}
R
c
=
l
i
m
n
→
∞
[
(
−
1
)
n
+
2
4
n
+
1
n
+
1
(
.25
)
n
+
1
∗
n
4
n
(
−
1
)
n
+
1
(
.25
)
n
]
−
1
{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.25)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.25)^{n}}}]^{-1}}
R
c
=
l
i
m
n
→
∞
−
(
n
+
1
)
n
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{n}}}
R
c
=
l
i
m
n
→
∞
1
1
=
1
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {1}{1}}=1}
Number 9
edit
Radius of convergence of log(3+4x) about the point
x
^
=
−
.25
{\displaystyle {\widehat {x}}=-.25}
x
^
=
−
.25
{\displaystyle {\widehat {x}}=-.25}
∑
n
=
0
∞
(
−
1
)
n
+
1
4
n
n
(
x
+
.25
)
n
{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x+.25)^{n}}
R
c
=
l
i
m
n
→
∞
[
(
−
1
)
n
+
2
4
n
+
1
n
+
1
(
.0625
)
n
+
1
∗
n
4
n
(
−
1
)
n
+
1
(
.0625
)
n
]
−
1
{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.0625)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.0625)^{n}}}]^{-1}}
R
c
=
l
i
m
n
→
∞
−
4
(
n
+
1
)
n
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-4(n+1)}{n}}}
R
c
=
l
i
m
n
→
∞
4
1
=
4
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {4}{1}}=4}
Number 10
edit
Radius of convergence of log(3+4x) about the point
x
^
=
1
{\displaystyle {\widehat {x}}=1}
x
^
=
1
{\displaystyle {\widehat {x}}=1}
∑
n
=
0
∞
(
−
1
)
n
+
1
4
n
n
(
x
−
1
)
n
{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-1)^{n}}
R
c
=
l
i
m
n
→
∞
[
(
−
1
)
n
+
2
4
n
+
1
n
+
1
(
1
)
n
+
1
∗
n
4
n
(
−
1
)
n
+
1
(
1
)
n
]
−
1
{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(1)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(1)^{n}}}]^{-1}}
R
c
=
l
i
m
n
→
∞
−
(
n
+
1
)
4
n
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {-(n+1)}{4n}}}
R
c
=
l
i
m
n
→
∞
1
4
=
1
/
4
{\displaystyle R_{c}=lim_{n\rightarrow \infty }{\frac {1}{4}}=1/4}
Number 11
edit
Radius of convergence of log(3+4x) about any given point
x
^
{\displaystyle {\widehat {x}}}
∑
n
=
0
∞
(
−
1
)
n
+
1
4
n
n
(
x
−
x
^
)
n
{\displaystyle \sum _{n=0}^{\infty }(-1)^{n+1}{\frac {4^{n}}{n}}(x-{\widehat {x}})^{n}}
R
c
=
l
i
m
n
→
∞
[
(
−
1
)
n
+
2
4
n
+
1
n
+
1
(
x
^
2
)
n
+
1
∗
n
4
n
(
−
1
)
n
+
1
(
(
x
^
2
)
n
)
]
−
1
{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}({\widehat {x}}^{2})^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(({\widehat {x}}^{2})^{n})}}]^{-1}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 3:
edit
Problem Statement
edit
Use the Determinant of the Matrix of Components and the Gramian to verify the linear independence of the two vectors
b
1
{\displaystyle b_{1}}
b
2
{\displaystyle b_{2}}
b
1
=
5
e
1
−
3
e
2
{\displaystyle \mathbf {b} _{1}=5\mathbf {e} _{1}-3\mathbf {e} _{2}}
b
2
=
−
2
e
1
+
4
e
2
{\displaystyle \mathbf {b} _{2}=-2\mathbf {e} _{1}+4\mathbf {e} _{2}}
Solution
edit
Determinant of the Matrix of Components
edit
The Matrix of components of the vectors
b
1
{\displaystyle b_{1}}
b
2
{\displaystyle b_{2}}
|
5
−
3
−
2
4
|
=
(
5
)
(
4
)
−
(
−
3
)
(
−
2
)
=
20
−
6
=
14
≠
0
{\displaystyle {\begin{vmatrix}5&-3\\-2&4\end{vmatrix}}=(5)(4)-(-3)(-2)=20-6=14\neq 0}
b
1
{\displaystyle b_{1}}
b
2
{\displaystyle b_{2}}
For vectors, the Gramian is defined as:
Γ
(
b
1
,
b
2
)
=
[
⟨
b
1
,
b
1
⟩
⟨
b
1
,
b
2
⟩
⟨
b
2
,
b
1
⟩
⟨
b
2
,
b
2
⟩
]
{\displaystyle \ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}\langle b_{1},b_{1}\rangle &\langle b_{1},b_{2}\rangle \\\langle b_{2},b_{1}\rangle &\langle b_{2},b_{2}\rangle \end{bmatrix}}}
where:
⟨
b
i
,
b
j
⟩
=
b
i
⋅
b
j
{\displaystyle \ \langle b_{i},b_{j}\rangle =b_{i}\cdot b_{j}}
⟨
b
1
,
b
1
⟩
=
(
5
e
1
−
3
e
2
)
⋅
(
5
e
1
−
3
e
2
)
=
(
5
)
(
5
)
+
(
−
3
)
(
−
3
)
=
25
+
9
=
34
{\displaystyle \ \langle b_{1},b_{1}\rangle =(5e_{1}-3e_{2})\cdot (5e_{1}-3e_{2})=(5)(5)+(-3)(-3)=25+9=34}
⟨
b
1
,
b
2
⟩
=
(
5
e
1
−
3
e
2
)
⋅
(
−
2
e
1
+
4
e
2
)
=
(
5
)
(
−
2
)
+
(
−
3
)
(
4
)
=
−
10
−
12
=
−
22
{\displaystyle \ \langle b_{1},b_{2}\rangle =(5e_{1}-3e_{2})\cdot (-2e_{1}+4e_{2})=(5)(-2)+(-3)(4)=-10-12=-22}
⟨
b
2
,
b
1
⟩
=
(
−
2
e
1
+
4
e
2
)
⋅
(
5
e
1
−
3
e
2
)
=
(
−
2
)
(
5
)
+
(
4
)
(
−
3
)
=
−
10
−
12
=
−
22
{\displaystyle \ \langle b_{2},b_{1}\rangle =(-2e_{1}+4e_{2})\cdot (5e_{1}-3e_{2})=(-2)(5)+(4)(-3)=-10-12=-22}
⟨
b
2
,
b
2
⟩
=
(
−
2
e
1
+
4
e
2
)
⋅
(
−
2
e
1
+
4
e
2
)
=
(
−
2
)
(
−
2
)
+
(
4
)
(
4
)
=
4
+
16
=
20
{\displaystyle \ \langle b_{2},b_{2}\rangle =(-2e_{1}+4e_{2})\cdot (-2e_{1}+4e_{2})=(-2)(-2)+(4)(4)=4+16=20}
Γ
(
b
1
,
b
2
)
=
[
34
−
22
−
22
20
]
{\displaystyle \ {\boldsymbol {\Gamma }}(b_{1},b_{2})={\begin{bmatrix}34&-22\\-22&20\end{bmatrix}}}
Finding the determinant of the Gramian matrix gives the Gramian:
Γ
=
(
34
)
(
20
)
−
(
−
22
)
(
−
22
)
=
680
−
484
=
156
≠
0
{\displaystyle \ \Gamma =(34)(20)-(-22)(-22)=680-484=156\neq 0}
So the vectors
b
1
{\displaystyle b_{1}}
b
2
{\displaystyle b_{2}}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
Problem 4: Wronskian and Gramian
edit
Problem Statement
edit
Use both the Wronskian and the Gramain to find whether the following functions are linearly independent. Consider the domain of these functions to be [-1, +1] for the construction of the Gramian matrix.
f
(
x
)
=
x
,
g
(
x
)
=
x
5
{\displaystyle f(x)=x,g(x)=x^{5}}
f
(
x
)
=
c
o
s
2
x
,
g
(
x
)
=
s
i
n
4
x
{\displaystyle f(x)=cos2x,g(x)=sin4x}
Solution
edit
Wronskian:
W
(
f
,
g
)
:=
[
f
g
f
′
g
′
]
=
f
g
′
−
g
f
′
{\displaystyle W(f,g):={\begin{bmatrix}f&g\\f'&g'\end{bmatrix}}=fg'-gf'}
W
(
f
,
g
)
≠
0
{\displaystyle W(f,g)\neq 0}
1)
f
(
x
)
=
x
,
g
(
x
)
=
x
5
{\displaystyle f(x)=x,g(x)=x^{5}}
f
(
x
)
=
x
,
f
′
(
x
)
=
1
{\displaystyle f(x)=x,f'(x)=1}
g
(
x
)
=
x
5
,
g
′
(
x
)
=
5
x
4
{\displaystyle g(x)=x^{5},g'(x)=5x^{4}}
W
(
f
,
g
)
:=
[
x
x
5
1
5
x
4
]
=
5
x
5
−
x
5
=
4
x
5
≠
0
{\displaystyle W(f,g):={\begin{bmatrix}x&x^{5}\\1&5x^{4}\end{bmatrix}}=5x^{5}-x^{5}=4x^{5}\neq 0}
2)
f
(
x
)
=
c
o
s
2
x
,
g
(
x
)
=
s
i
n
4
x
{\displaystyle f(x)=cos2x,g(x)=sin4x}
f
(
x
)
=
c
o
s
2
x
,
f
′
(
x
)
=
−
2
s
i
n
2
x
{\displaystyle f(x)=cos2x,f'(x)=-2sin2x}
g
(
x
)
=
s
i
n
4
x
,
g
′
(
x
)
=
4
c
o
s
4
x
{\displaystyle g(x)=sin4x,g'(x)=4cos4x}
W
(
f
,
g
)
:=
[
c
o
s
2
x
s
i
n
4
x
−
2
s
i
n
2
x
4
c
o
s
4
x
]
=
c
o
s
2
x
∗
4
c
o
s
4
x
+
s
i
n
4
x
∗
2
s
i
n
2
x
≠
0
{\displaystyle W(f,g):={\begin{bmatrix}cos2x&sin4x\\-2sin2x&4cos4x\end{bmatrix}}=cos2x*4cos4x+sin4x*2sin2x\neq 0}
Gramian:
Γ
(
f
,
g
)
:=
[
<
f
,
f
>
<
f
,
g
>
<
g
,
f
>
<
g
,
g
>
]
{\displaystyle \Gamma (f,g):={\begin{bmatrix}<f,f>&<f,g>\\<g,f>&<g,g>\end{bmatrix}}}
Γ
(
f
,
g
)
≠
0
{\displaystyle \Gamma (f,g)\neq 0}
1)
f
(
x
)
=
x
,
g
(
x
)
=
x
5
{\displaystyle f(x)=x,g(x)=x^{5}}
<
f
,
f
>=
∫
−
1
1
x
2
d
x
=
2
3
{\displaystyle <f,f>=\int _{-1}^{1}x^{2}dx={\frac {2}{3}}}
<
f
,
g
>=
∫
−
1
1
x
6
d
x
=
2
7
{\displaystyle <f,g>=\int _{-1}^{1}x^{6}dx={\frac {2}{7}}}
<
g
,
f
>=
∫
−
1
1
x
6
d
x
=
2
7
{\displaystyle <g,f>=\int _{-1}^{1}x^{6}dx={\frac {2}{7}}}
<
g
,
g
>=
∫
−
1
1
x
1
0
d
x
=
2
11
{\displaystyle <g,g>=\int _{-1}^{1}x^{1}0dx={\frac {2}{11}}}
Γ
(
f
,
g
)
:=
[
2
3
2
7
2
7
2
11
]
≠
0
{\displaystyle \Gamma (f,g):={\begin{bmatrix}{\frac {2}{3}}&{\frac {2}{7}}\\{\frac {2}{7}}&{\frac {2}{11}}\end{bmatrix}}\neq 0}
2)
f
(
x
)
=
c
o
s
2
x
,
g
(
x
)
=
s
i
n
4
x
{\displaystyle f(x)=cos2x,g(x)=sin4x}
<
f
,
f
>=
∫
−
1
1
c
o
s
2
x
∗
c
o
s
2
x
d
x
=
1
4
(
4
+
s
i
n
4
)
=
.818
{\displaystyle <f,f>=\int _{-1}^{1}cos2x*cos2xdx={\frac {1}{4}}(4+sin4)=.818}
<
f
,
g
>=
∫
−
1
1
c
o
s
2
x
∗
s
i
n
4
x
d
x
=
[
.5
c
o
s
2
x
−
1
/
12
c
o
s
6
x
]
−
1
1
=
0
{\displaystyle <f,g>=\int _{-1}^{1}cos2x*sin4xdx=[.5cos^{2}x-1/12cos6x]_{-}^{1}1=0}
<
g
,
f
>=
∫
−
1
1
c
o
s
2
x
∗
s
i
n
4
x
d
x
=
0
{\displaystyle <g,f>=\int _{-1}^{1}cos2x*sin4xdx=0}
<
g
,
g
>=
∫
−
1
1
s
i
n
4
x
∗
s
i
n
4
x
d
x
=
1
+
s
i
n
8
8
=
.876
{\displaystyle <g,g>=\int _{-1}^{1}sin4x*sin4xdx=1+{\frac {sin8}{8}}=.876}
Γ
(
f
,
g
)
:=
[
.818
0
0
.876
]
≠
0
{\displaystyle \Gamma (f,g):={\begin{bmatrix}.818&0\\0&.876\end{bmatrix}}\neq 0}
Honor Pledge
edit
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
![{\displaystyle [-1,+1]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/daa72f1a806823ec94fda7922597b19cbda684f4)
![{\displaystyle [{\frac {-3}{4}},{\frac {-1}{4}}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c7963189a61117f0b3b530b0dac055300b6c4e0e)
![{\displaystyle R_{c}=[lim\left|{\frac {d_{k+1}}{d_{k}}}\right|]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/253e55d414767ffed2bb304def1691ecae2a32f4)
![{\displaystyle R_{c}=[lim{\frac {(-1)^{m+1}}{(2m+3)!}}*{\frac {(2m+1)!}{(-1)^{m}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/391d36bc79242f468a7992cee6f1990425c64054)
![{\displaystyle R_{c}=[lim{\frac {(-1)^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18af28d2aff6507b39038376762068f59a891d0d)
![{\displaystyle R_{c}=[{\frac {(-1)^{n+1}({\widehat {x}})^{n+1}}{n+1}}*{\frac {n}{(-1)^{n}({\widehat {x}})^{n}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ca84679b80455b67b749976604a1b073d4d063d)
![{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.25)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.25)^{n}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d506acc8526199c6f45e4f0d3409e7ef8d96696c)
![{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(.0625)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(.0625)^{n}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/46d601d38f89e339d6c97d2e24ff3748df3a7743)
![{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}(1)^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(1)^{n}}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/62f43e8da697a74de61704b8583665c40591e88f)
![{\displaystyle R_{c}=lim_{n\rightarrow \infty }[{(-1)^{n+2}}{\frac {4^{n+1}}{n+1}}({\widehat {x}}^{2})^{n+1}*{\frac {n}{4^{n}(-1)^{n+1}(({\widehat {x}}^{2})^{n})}}]^{-1}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b234a8da29509a9e50b4f5235db2a15cfda7701b)
![{\displaystyle <f,g>=\int _{-1}^{1}cos2x*sin4xdx=[.5cos^{2}x-1/12cos6x]_{-}^{1}1=0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bcc0b0f66c5db6963268fc427cb89653a6a5e46c)
