Problem 1: Taylor Series Expansion of the log Function
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}
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}}]}
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
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}}}
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}}
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 }
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}
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}
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}}
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}})}
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}
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}
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}
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}}
On our honor, we solved this problem on our own, without aid from online solutions or solutions from previous semesters.
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}}
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}}
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
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}
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}
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)
