University of Florida/Egm6321/f12.Rep5hid

R5.1 Proof that exponentiation of Transverse of a Matrix equals the Transverse of the Exponentiation Expansion

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

$\displaystyle \mathbf {A} \in \mathbb {R} ^{(n\times n)}$

and

${\underset {\color {red}n\times n}{\exp[\mathbf {A} ]}}={\underset {\color {red}n\times n}{\mathbf {I} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}\mathbf {A} ^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {A} ^{k}}}$

(Eq.(2)p.20-2b)

Show That

$\displaystyle \exp \mathbf {A} ^{T}=(\exp \mathbf {A} )^{T}$

(Eq.5.1.1)

Solution

We will first expand the LHS, then the RHS of (Eq. 5.1.1) using (Eq.(2)p.20-2b) and compare the two expressions.

Expanding the LHS,

${\underset {\color {red}n\times n}{\exp[\mathbf {A^{T}} ]}}={\underset {\color {red}n\times n}{\mathbf {I^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}[\mathbf {A^{T}} ]^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {[} {A^{T}}]^{k}}}$

But we know that

$\mathbf {I} ^{T}=\mathbf {I}$

$\therefore {\underset {\color {red}n\times n}{\exp[\mathbf {A^{T}} ]}}={\underset {\color {red}n\times n}{\mathbf {I} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}[\mathbf {A^{T}} ]^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {[} {A^{T}}]^{k}}}$

(Eq.5.1.2)

Now expanding the RHS,

${\underset {\color {red}n\times n}{\exp[\mathbf {A} ]^{T}}}=\left[{\underset {\color {red}n\times n}{\mathbf {I} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}[\mathbf {A} ]^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {[} {A}]^{k}}}\right]^{T}$

Which on calculating, reduces to

${\underset {\color {red}n\times n}{\exp[\mathbf {A^{T}} ]}}={\underset {\color {red}n\times n}{\mathbf {I^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}[\mathbf {A^{T}} ]^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {[} {A^{T}}]^{k}}}$

or

${\underset {\color {red}n\times n}{\exp[\mathbf {A^{T}} ]}}={\underset {\color {red}n\times n}{\mathbf {I} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A^{T}} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}[\mathbf {A^{T}} ]^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {[} {A^{T}}]^{k}}}$

(Eq.5.1.3)

Comparing (Eq. 5.1.2) and (Eq. 5.1.3)

We conclude the LHS = RHS, Hence Proved.

R5.2. Exponentiation of a Complex Diagonal Matrix

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

A Diagonal Matrix

$\displaystyle \mathbf {D} ={\text{Diag}}[d_{1},d_{2},d_{3},....,d_{n}]$, where, $\mathbf {D} \in \mathbb {C} ^{n\times n}$.

(Eq.(2)p.20-2b)

Problem

Show that

$\displaystyle \exp[\mathbf {D} ]={\text{Diag}}[e^{d_{1}},e^{d_{2}},e^{d_{3}},....,e^{d_{n}}]$, where, $\exp[\mathbf {D} ]\in \mathbb {C} ^{n\times n}$.

(Eq.(3)p.20-2b)

Solution

We know, from Lecture Notes ,

${\underset {\color {red}n\times n}{\exp[\mathbf {A} ]}}={\underset {\color {red}n\times n}{\mathbf {I} }}+{\underset {\color {red}n\times n}{{\frac {1}{1!}}\mathbf {A} }}+{\underset {\color {red}n\times n}{{\frac {1}{2!}}\mathbf {A} ^{2}}}+....={\underset {\color {red}n\times n}{\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {A} ^{k}}}$

(Eq.(2)p.15-3)

Let us consider a Simple yet Generic 4x4 Complex Diagonal Matrix $\mathbf {D}$

$\mathbf {D} ={\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}$

(Eq.5.2.2)

where $i={\sqrt {-1}}$.

Applying (Eq.(2)p.15-3) to (Eq.5.2.2) and expanding,

$\exp[\mathbf {D} ]=\underbrace {\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}} _{\color {blue}{\text{Matrix 1}}}+{\frac {1}{1!}}\underbrace {\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}} _{\color {blue}{\text{Matrix 2}}}+{\frac {1}{2!}}\underbrace {{\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}*{\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}} _{\color {blue}{\text{Matrix 3}}}+\ldots +{\frac {1}{k!}}\underbrace {{\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}*\ldots *{\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}} _{\color {blue}{\text{multiply k times - Matrix k}}}$

(Eq.5.2.3)

Simplifying Term 2 and other higher power terms (upto Term k) in the following way,

${\text{Matrix 3}}={\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}*{\begin{bmatrix}ai&0&0&0\\0&bi&0&0\\0&0&ci&0\\0&0&0&di\end{bmatrix}}$

${\text{Matrix 3}}={\begin{bmatrix}(ai)^{2}&0&0&0\\0&(bi)^{2}&0&0\\0&0&(ci)^{2}&0\\0&0&0&(di)^{2}\end{bmatrix}}$

(Eq.5.2.4)

Similarly,

${\text{Matrix k}}={\begin{bmatrix}(ai)^{k}&0&0&0\\0&(bi)^{k}&0&0\\0&0&(ci)^{k}&0\\0&0&0&(di)^{k}\end{bmatrix}}$

(Eq.5.2.5)

Using (Eq.5.2.4) and (Eq.5.2.5) in (Eq.5.2.3) and carrying out simple matrix addition, we get,

$exp[\mathbf {D} ]={\begin{bmatrix}(1+{\frac {1}{1!}}(ai)+{\frac {1}{2!}}(ai)^{2}+\ldots +{\frac {1}{k!}}(ai)^{k}&0&0&0\\0&(1+{\frac {1}{1!}}(bi)+{\frac {1}{2!}}(bi)^{2}+\ldots +{\frac {1}{k!}}(bi)^{k})&0&0\\0&0&(1+{\frac {1}{1!}}(ci)+{\frac {1}{2!}}(ci)^{2}+\ldots +{\frac {1}{k!}}(ci)^{k})&0\\0&0&0&(1+{\frac {1}{1!}}(di)+{\frac {1}{2!}}(di)^{2}+\ldots +{\frac {1}{k!}}(di)^{k})\end{bmatrix}}$

(Eq.5.2.6)

But every diagonal term of the matrix is of the form,

$\exp(x)=1+{\frac {1}{1!}}x+{\frac {1}{2!}}x^{2}+\ldots +{\frac {1}{n!}}x^{n}$

(Eq.5.2.7)

Therefore, (Eq.5.2.6) can be rewritten as,

$exp[\mathbf {D} ]={\begin{bmatrix}exp(ai)&0&0&0\\0&exp(bi)&0&0\\0&0&exp(ci)&0\\0&0&0&exp(di)\end{bmatrix}}={\begin{bmatrix}e^{ai}&0&0&0\\0&e^{bi}&0&0\\0&0&e^{ci}&0\\0&0&0&e^{di}\end{bmatrix}}$

(Eq.5.2.8)

${\text{(ai)}},{\text{(bi)}},{\text{(ci)}},{\text{(di)}}$ are nothing but the diagonal elements of the original matrix in (Eq.5.2.2). Hence,

$\displaystyle \exp[\mathbf {D} ]={\text{Diag}}[e^{d_{1}},e^{d_{2}},e^{d_{3}},e^{d_{4}}]$

(Eq.5.2.9)

Similarly it can be easily found for an $n\times n$ complex diagonal matrix that

$\displaystyle \exp[\mathbf {D} ]={\text{Diag}}[e^{d_{1}},e^{d_{2}},e^{d_{3}},....,e^{d_{n}}]$

(Eq.5.2.10)

Hence Proved.

R5.3 Show form of Exponentiation of Matrix in terms of Eigenvalues of that matrix

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

A matrix $\mathbf {A}$ can be decomposed as

$\displaystyle \mathbf {A} =\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1}$

(Eq.(2) p.20-4)

where

$\mathbf {\Lambda }$ is the diagonal matrix of eigenvalues of matrix $\mathbf {A}$

$\displaystyle \mathbf {\Lambda } ={\text{Diag}}[\lambda _{1},\lambda _{2},\ldots ,d_{n}]\in \mathbb {C} ^{n\times n}$

(Eq.(5) p.20-3)

and

$\mathbf {\Phi }$ is the matrix established by n linearly independent eigenvectors ${\boldsymbol {\phi }}_{i}(i=1,2,3,\ldots ,n)$of matrix $\mathbf {A}$, that is,

$\displaystyle \mathbf {\Phi } =[{\boldsymbol {\phi }}_{1},{\boldsymbol {\phi }}_{2},\ldots ,{\boldsymbol {\phi }}_{n}]\in \mathbb {C} ^{n\times n}$

(Eq.(2) p.20-3)

Problem

Show that

$\displaystyle \exp[\mathbf {A} ]=\mathbf {\Phi } \,{\text{Diag}}[\,e^{\lambda _{1}},\,e^{\lambda _{2}},\ldots ,\,e^{\lambda _{n}}]\mathbf {\Phi } ^{-1}$

Solution

The power series expansion of exponentiation of matrix $\mathbf {A}$ in terms of that matrix has been given as

$\displaystyle \exp[\mathbf {A} ]=\mathbf {I} +{\frac {1}{1!}}\mathbf {A} +{\frac {1}{2!}}\mathbf {A} ^{2}+{\frac {1}{3!}}\mathbf {A} ^{3}+\cdots =\sum _{k=0}^{\infty }{\frac {1}{k!}}\mathbf {A} ^{k}$

(Eq.(2) p.15-3)

Since matrix $\mathbf {A}$ can be decomposed as,

$\displaystyle \mathbf {A} =\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1}$

(Eq.(2) p.20-4)

Expanding the $k$ power of matrix $\mathbf {A}$ yields

$\displaystyle \mathbf {A} ^{k}=\underbrace {(\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1})(\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1})\cdots (\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1})} _{{\text{number of factors}}(\mathbf {\Phi } \,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1}){\text{is k}}}$

(Eq.5.3.1)

Where the factors $(\mathbf {\Phi } )$ which are neighbors of factors $(\mathbf {\Phi } ^{-1})$ can be all cancelled in pairs, that is,

$\displaystyle \mathbf {A} ^{k}=\mathbf {\Phi } \,\mathbf {\Lambda } \,{\cancelto {\mathbf {I} }{\mathbf {\Phi } ^{-1}\mathbf {\Phi } \,}}\mathbf {\Lambda } \,{\cancelto {\mathbf {I} }{\mathbf {\Phi } ^{-1}\mathbf {\Phi } \,}}\mathbf {\Lambda } \,\cdots \mathbf {\Lambda } \,{\cancelto {\mathbf {I} }{\mathbf {\Phi } ^{-1}\mathbf {\Phi } \,}}\,\mathbf {\Lambda } \,\mathbf {\Phi } ^{-1}$

(Eq.5.3.2)

$\displaystyle \Rightarrow \mathbf {A} ^{k}=\mathbf {\Phi } \,\underbrace {\mathbf {\Lambda } \,\mathbf {\Lambda } \,\mathbf {\Lambda } \,\cdots \mathbf {\Lambda } \,\mathbf {\Lambda } \,} _{{\text{k}}\mathbf {\Lambda } {\text{s}}}\mathbf {\Phi } ^{-1}$

(Eq.5.3.3)

$\displaystyle \Rightarrow \mathbf {A} ^{k}=\mathbf {\Phi } \,\mathbf {\Lambda } ^{k}\mathbf {\Phi } ^{-1}$

(Eq.5.3.4)

Thus, the equation (Eq. 5.3.1) can be expressed as

$\displaystyle \exp[\mathbf {A} ]=\sum _{k=0}^{\infty }{\frac {1}{k!}}[\mathbf {A} ^{k}]=\sum _{k=0}^{\infty }{\frac {1}{k!}}[\mathbf {\Phi } \,\mathbf {\Lambda } ^{k}\mathbf {\Phi } ^{-1}]$

(Eq.5.3.5)

$\displaystyle \Rightarrow \exp[\mathbf {A} ]=\mathbf {\Phi } \,\sum _{k=0}^{\infty }{\frac {1}{k!}}[\mathbf {\Lambda } ^{k}]\mathbf {\Phi } ^{-1}$

(Eq.5.3.6)

According to the equation (Eq.(2) p.15-3), now we have,

$\displaystyle \exp[\mathbf {A} ]=\mathbf {\Phi } \,\exp[\mathbf {\Lambda } ]\mathbf {\Phi } ^{-1}$

(Eq.5.3.7)

Referring to the conclusion obtained in R5.2, which is

$\displaystyle \exp[\mathbf {D} ]={\text{Diag}}[\,e^{d_{1}},\,e^{d_{2}},\ldots ,\,e^{d_{n}}]$

(Eq.(3) p.20-2b)

Replacing the matrix $\mathbf {D}$ with $\mathbf {\Lambda }$, the elements $d_{i}$ with $\lambda _{i}$,where $i=1,2,3,\ldots ,n$ and then substituting into (Eq. 5.3.7) yields

$\displaystyle \exp[\mathbf {A} ]=\mathbf {\Phi } \,{\text{Diag}}[\,e^{\lambda _{1}},\,e^{\lambda _{2}},\ldots ,\,e^{\lambda _{n}}]\mathbf {\Phi } ^{-1}$

(Eq.5.3.8)

R5.4 Show Decomposed Form of Matrix and its Exponentiation

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

Exponentiation of a matrix $\mathbf {A}$ can be decomposed as

$\displaystyle \exp[\mathbf {A} ]=\mathbf {\Phi } \,{\text{Diag}}[\,e^{\lambda _{1}},\,e^{\lambda _{2}},\ldots ,\,e^{\lambda _{n}}]\mathbf {\Phi } ^{-1}$

(Eq.(3) p.20-4)

The matrix $\mathbf {B}$ is defined in lecture note

$\displaystyle \mathbf {B} ={\begin{bmatrix}0&-1\\1&0\end{bmatrix}}$

(Eq.(1) p.20-2)

Problem

Show

$\displaystyle \mathbf {B} t={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}{\begin{bmatrix}{it}&0\\0&{-it}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

(Eq.(1) p.20-5)

and

$\displaystyle \displaystyle \exp[\mathbf {B} t]={\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}\neq {\begin{bmatrix}1&e^{-t}\\e^{t}&1\end{bmatrix}}$

(Eq.(2) p.20-5)

Solution

To show the equation (Eq.(1) p.20-5), we should first find eigenvalues $\lambda$ of matrix $\mathbf {B}$ using the matrix equation as follow, indroduce $\mathbf {I}$ to represent Identity matrix.

$\displaystyle f(x)=\left|\lambda \mathbf {I} -\mathbf {B} \right|=0$

$\displaystyle \Rightarrow f(x)=\left|{\begin{bmatrix}\lambda &0\\0&\lambda \end{bmatrix}}-{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}\right|=\left|{\begin{bmatrix}\lambda &1\\-1&\lambda \end{bmatrix}}\right|=0$

$\displaystyle \Rightarrow f(x)=\lambda ^{2}+1=0$

$\displaystyle \Rightarrow \lambda _{1}=i,\lambda _{2}=-i$

Since the two eigenvalues of matrix $\mathbf {B}$ are both obtained, now solve for the corresponding two eigenvectors.

$\displaystyle (\lambda _{1}\mathbf {I} -\mathbf {B} )\mathbf {X} =\mathbf {O}$

$\displaystyle \Rightarrow ({\begin{bmatrix}\lambda _{1}&0\\0&\lambda _{1}\end{bmatrix}}-{\begin{bmatrix}0&-1\\1&0\end{bmatrix}}){\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$

$\displaystyle \Rightarrow {\begin{bmatrix}\lambda _{1}&1\\-1&\lambda _{1}\end{bmatrix}}{\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}}={\begin{bmatrix}0\\0\end{bmatrix}}$

(Eq.5.4.1)

Thus, for the first value of $\lambda$, we have

$\displaystyle \left\{{\begin{matrix}\lambda _{1}x_{1}+x_{2}=0\\-x_{1}+\lambda _{1}x_{2}=0\end{matrix}}\right.$

(Eq.5.4.2)

Substituting $\lambda _{1}=i$ into the equations above and solving yields, for the eigenvalue $\lambda _{1}=i$, that

$\displaystyle \left\{{\begin{matrix}x_{1}=1\\x_{2}=-i\end{matrix}}\right.$

(Eq.5.4.3)

Similarly, we have the equation which can be used for solving eigenvector corresponding to $\lambda _{2}=-i$,

$\displaystyle \left\{{\begin{matrix}\lambda _{2}x_{1}+x_{2}=0\\-x_{1}+\lambda _{2}x_{2}=0\end{matrix}}\right.$

(Eq.5.4.4)

Substituting $\lambda _{2}=-i$ into the equations above and solving yields, for the eigenvalue $\lambda _{2}=-i$, that

$\displaystyle \left\{{\begin{matrix}x_{1}=1\\x_{2}=i\end{matrix}}\right.$

(Eq.5.4.5)

Now we have obtained two eigenvectors ${\boldsymbol {\phi }}_{1}$ and ${\boldsymbol {\phi }}_{2}$ of matrix $\mathbf {B}$, where

$\displaystyle {\boldsymbol {\phi }}_{1}={\begin{bmatrix}1\\-i\end{bmatrix}}$, $\displaystyle {\boldsymbol {\phi }}_{2}={\begin{bmatrix}1\\i\end{bmatrix}}$

(Eq.5.4.6)

Thus we have

$\displaystyle \mathbf {\Phi } =[{\boldsymbol {\phi }}_{1},{\boldsymbol {\phi }}_{2}]={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}$

(Eq.5.4.7)

Then, calculating the inverse matrix of matrix $\mathbf {\Phi }$ yields

$\displaystyle \mathbf {\Phi } ^{-1}={\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

(Eq.5.4.8)

Therefore we reach the conclusion that,

$\displaystyle \mathbf {B} ={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}{\begin{bmatrix}{i}&0\\0&{-i}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

$\displaystyle \Rightarrow \mathbf {B} t={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}{\begin{bmatrix}{it}&0\\0&{-it}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

(Eq.(1) p.20-5)

According to the conclusion we have reached in R5.3, we have,

$\displaystyle \exp[\mathbf {B} ]=\mathbf {\Phi } \,{\text{Diag}}[\,e^{\lambda _{1}},\,e^{\lambda _{2}}]\mathbf {\Phi } ^{-1}$

$\displaystyle \Rightarrow \exp[\mathbf {B} t]={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}{\begin{bmatrix}e^{i}&0\\0&e^{-i}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {t}{2i}}$

$\displaystyle \Rightarrow \exp[\mathbf {B} t]={\begin{bmatrix}1&1\\-i&i\end{bmatrix}}{\begin{bmatrix}e^{it}&0\\0&e^{-it}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

Doing the multiplication of matrices at the right side of equation above yields

$\displaystyle \exp[\mathbf {B} t]={\begin{bmatrix}e^{it}&e^{-it}\\-ie^{it}&ie^{-it}\end{bmatrix}}{\begin{bmatrix}i&-1\\i&1\end{bmatrix}}{\frac {1}{2i}}$

$\displaystyle \Rightarrow \exp[\mathbf {B} t]={\begin{bmatrix}ie^{it}+ie^{-it}&e^{-it}-e^{it}\\e^{it}-e^{-it}&-ie^{-it}+ie^{it}\end{bmatrix}}{\frac {1}{2i}}$

$\displaystyle \Rightarrow \exp[\mathbf {B} t]={\begin{bmatrix}{\frac {1}{2}}(e^{it}+e^{-it})&{\frac {1}{2i}}(e^{-it}-e^{it})\\{\frac {1}{2i}}(e^{it}-e^{-it})&{\frac {1}{2}}(e^{it}+e^{-it})\end{bmatrix}}$

(Eq.5.4.9)

Consider Euler’s Formula,

$\displaystyle e^{it}=\cos t+i\sin t$

(Eq.5.4.10)

Replacing $\displaystyle i$ with $\displaystyle -i$ yields

$\displaystyle e^{-it}=\cos t-i\sin t$

(Eq.5.4.11)

Solve (Eq.5.4.10) together with (Eq.5.4.11), we have

$\displaystyle \left\{{\begin{matrix}e^{it}=\cos t+i\sin t\\e^{-it}=\cos t-i\sin t\end{matrix}}\right.$

$\displaystyle \Rightarrow \left\{{\begin{matrix}\cos t={\frac {1}{2}}(e^{it}+e^{-it})\\\sin t={\frac {1}{2i}}(e^{it}-e^{-it})\end{matrix}}\right.$

(Eq.5.4.12)

Substituting (Eq.5.4.12) into (Eq.5.4.9) yields

$\displaystyle \exp[\mathbf {B} t]={\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}$

(Eq.(2) p.20-5)

Obviously,

$\displaystyle {\begin{bmatrix}\cos t&-\sin t\\\sin t&\cos t\end{bmatrix}}\neq {\begin{bmatrix}1&e^{-t}\\e^{t}&1\end{bmatrix}}$

R*5.5 Generating a class of exact L2-ODE-VC

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

A L2-ODE-VC :

$\displaystyle G=R(x)y+Q(x)y^{\prime }+P(x)y^{''}$

(Eq. 5.5.1)

The first intregal $\phi (x,y,p)$ can also be expressed as:

$\displaystyle \phi _{x}(x,y,p)+\phi _{y}(x,y,p)y^{\prime }=R(x)y+Q(x)y^{\prime }$

(Eq. 5.5.2)

Problem

Show that(Eq. 5.5.1) and (Eq. 5.5.2) lead to a general class of exact L2-ODE-VC of the form:

$\displaystyle \phi (x,y,p)=P(x)p+T(x)y+k$

(Eq. 5.5.3)

Solution

Nomenclature

$\displaystyle p:=y':={\frac {dy}{dx}}$

Derivation of Eq. 5.5.3

The first exactness condition for L2-ODE-VC:

$\displaystyle G={\frac {d\phi (x,y,p)}{dx}}=\phi _{x}+\phi _{y}p+\phi _{p}y^{''}$

(Eq. 5.5.4)

From (Eq. 5.5.1) and (Eq. 5.5.4), we can infer that

$\displaystyle \phi _{p}=P(x)$

(Eq. 5.5.5)

Integrating (Eq. 5.5.5), w.r.t p, we obtain:

$\displaystyle \phi =P(x)p+k(x,y)$

(Eq. 5.5.6)

Partial derivatives of $\phi$ w.r.t to x and y can be written as:

$\displaystyle \phi _{x}=P^{\prime }(x)p+{\frac {\partial k(x,y)}{\partial x}}$

(Eq. 5.5.7)

$\displaystyle \phi _{y}={\frac {\partial k(x,y)}{\partial y}}$

(Eq. 5.5.8)

Substituting the partial derivatives of $\phi$ w.r.t x,y and p [(Eq. 5.5.7), (Eq. 5.5.8), (Eq. 5.5.6)] into (Eq. 5.5.4), we obtain:

$\displaystyle G=P^{\prime }(x)p+{\frac {\partial k(x,y)}{\partial y}}p+{\frac {\partial k(x,y)}{\partial x}}+P(x)y^{''}$

(Eq. 5.5.8)

Comparing (Eq. 5.5.8) with (Eq. 5.5.1), we can write:

$\displaystyle G=P(x)y^{''}+\underbrace {\left({\frac {\partial k(x,y)}{\partial y}}+P^{\prime }(x)\right)} _{Q(x)}p+\underbrace {\frac {\partial k(x,y)}{\partial x}} _{R(x)y}$

(Eq. 5.5.9)

Thus $\displaystyle {\frac {\partial k(x,y)}{\partial x}}=R(x)y$

Integrating w.r.t x,

$\displaystyle k(x,y)=y\underbrace {\int ^{x}R(s)ds} _{T(x)}+k_{1}(y)$

(Eq. 5.5.10)

Substituting the $k(x,y)$ obtained in (Eq. 5.5.10) back into the expression for $\phi$ obtained in (Eq. 5.5.6), we obtain:

$\displaystyle \phi =P(x)p+T(x)y+k_{1}(y)$

(Eq. 5.5.11)

The partial derivative of $\phi$ (Eq. 5.5.11) w.r.t y,

$\displaystyle \phi _{y}=T(x)+k_{1}^{\prime }(y)$

(Eq. 5.5.12)

But from (Eq. 5.5.1) and (Eq. 5.5.2), we see that $\displaystyle \phi _{y}=Q(x)$.

So, $Q(x)=T(x)+k'_{1}(y)$

Since, $Q(x)$ is only a function of $x$, so, we can now say that $T(x)=Q(x)$ and ${k_{1}}^{\prime }(y)=0$.

Thus $k_{1}(y)$ is a constant.

Hence we obtain the following expression for $\phi$:

$\displaystyle \phi (x,y,p)=P(x)p+T(x)y+k$

(Eq. 5.5.13)

which represents a general class of Exact L2-ODE-VC.

R*5.6 Solving a L2-ODE-VC

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

$\displaystyle G=(cosx)y''+(x^{2}-sinx)y'+2xy=0$

(Eq. 5.6.1)

Problem

1. Show that (Eq. 5.6.1) is exact.

2. Find $\displaystyle \phi$

3. Solve for $\displaystyle y(x)$

Solution

Nomenclature

$\displaystyle p=:y'=:{\frac {dy}{dx}}$

$\displaystyle f_{ij}={\frac {d^{2}f}{\partial i\partial j}}$

Exactness Conditions

The exactness conditions for N2-ODE (Non Linear Second Order Differential Equation) are:

First Exactness condition

For an equation to be exact, they must be of the form

$\displaystyle G(x,y,y',y'')=g(x,y,p)+f(x,y,p)y''={\frac {\mathrm {d} \phi }{\mathrm {d} x}}$

$\displaystyle g(x,y,p)=\phi _{x}+\phi _{y}.y'$

(Eq. 5.6.2)

$\displaystyle f(x,y,p)=\phi _{p}$

(Eq. 5.6.3)

Second Exactness Condition

$f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y}$

(Eq. 5.6.4)

$f_{xp}+pf_{yp}+2f_{y}=g_{pp}$

(Eq. 5.6.5)

Work

We have

$\displaystyle G=(cosx)y''+(x^{2}-sinx)y'+2xy=0$

Where we can identify

$\displaystyle g(x,y,p)=(x^{2}-sinx)y'+2xy$

and

$\displaystyle f(x,y,p)=cosx$

Thus the equation satisfies the first exactness condition.

For the second exactness condition, we first calculate the various partial derivatives of f and g.

$\displaystyle g_{p}=x^{2}-sinx$

$\displaystyle g_{pp}=0$

$\displaystyle g_{xp}=2x-cosx$

$\displaystyle g_{yp}=0$

$\displaystyle g_{y}=2x$

$\displaystyle f_{p}=0$

$\displaystyle f_{xp}=f_{yp}=0$

$\displaystyle f_{y}=f_{yy}=f_{xy}=0$

$\displaystyle f_{x}=-sinx$

$\displaystyle f_{xx}=-cosx$

Substituting the values in (Eq. 5.6.4) we get

$\displaystyle L.H.S=-cosx+0+0=-cosx$

$\displaystyle R.H.S=2x-cosx+0-2x=-cosx$

Therefore the first equation satisfies.

Substituting the values in (Eq. 5.6.5) we get

$\displaystyle L.H.S.=0+0+0=0$

$\displaystyle R.H.S=0$

Therefore the second equation satisfies as well.

Thus the second exactness condition is satisfied and the given differential equation is exact.

Now, we have $\displaystyle f(x,y,p)=\phi _{p}$

Integrating w.r.t. p, we get

$\displaystyle \phi =\int cosx.dp+h(x,y)$

where h(x,y) is a function of integration as we integrated only partially w.r.t. p.

$\displaystyle \phi =p.cosx+h(x,y)$

(Eq. 5.6.6)

Partially differentiating (Eq. 5.6.6) w.r.t x

$\displaystyle \phi _{x}=-psinx.+h_{x}$

Partially differentiating (Eq. 5.6.6) w.r.t y

$\displaystyle \phi _{y}=h_{y}$

From equation (Eq. 5.6.3), we have

$\displaystyle g(x,y,p)=\phi _{x}+\phi _{y}.y'$

$\displaystyle =-p.sinx+h_{x}+h_{y}.p$

$\displaystyle =p(h_{y}-sinx)+h_{x}$

We have established that

$\displaystyle g(x,y,p)=(x^{2}-sinx)y'+2xy$

Comparing the two equations, we get,

$\displaystyle h_{x}=2xy$

On integrating,

$\displaystyle h=x^{2}y+k_{1}(x)$

Thus,

$\displaystyle h_{x}=2xy+k_{1}'=2xy,\therefore k_{1}=const$

Thus we have

$\displaystyle \phi =cosx.y'+x^{2}y+k_{1}=k_{2}$

$\displaystyle \phi =cosx.y'+x^{2}y=k$

$\displaystyle y'+{\frac {x^{2}y}{cosx}}={\frac {k}{cosx}}$

This N1-ODE can be solved using the Integrating Factor Method that we very well know.

$\displaystyle h(x)=e^{\int {\frac {x^{2}}{cosx}}.dx}$

$\displaystyle y={\frac {1}{h(x)}}.\int h(x).{\frac {k}{cosx}}.dx+c$

$\displaystyle y={\frac {1}{e^{\int {\frac {x^{2}}{cosx}}.dx}}}\int \left[e^{\int {\frac {x^{2}}{cosx}}.dx}{\frac {k}{cosx}}dx\right]+c$

R*5.7 Show equivalence to symmetry of second partial derivatives of first integral

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

$\displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0$

(Eq.(1) p.22-3)

where

$\displaystyle g_{i}:={\frac {\partial G}{\partial y^{(i)}}}={\frac {\partial }{\partial y^{(i)}}}({\frac {d\phi }{dx}}),{\rm {{for}\ i=0,1,2}}$

(Eq.(3) p.21-7)

Problem

Show equivalence to symmetry of mixed second partial derivatives of first integral, that is

$\displaystyle \phi _{xy}=\phi _{yx},\phi _{py}=\phi _{yp},\phi _{xp}=\phi _{px}$

where

$\displaystyle p(x):=y'(x)$

Solution

$\displaystyle g_{0}:={\frac {\partial }{\partial y}}({\frac {d\phi }{dx}})$

(Eq.(3) p.21-7)

$\displaystyle {\frac {d}{dx}}g_{1}={\frac {d}{dx}}(\phi _{xp}+\phi _{yp}p+\phi _{y}+\phi _{pp}y'')$

(Eq.(2) p.22-4)

From (Eq.(2)p.22-4), we have,

$\displaystyle {\frac {d}{dx}}g_{1}={\frac {d}{dx}}[\phi _{xp}+\phi _{yp}y'+\phi _{pp}y'']+{\frac {d\phi _{y}}{dx}}$

$\displaystyle \Rightarrow {\frac {d}{dx}}g_{1}={\frac {d}{dx}}[\phi _{xp}+\phi _{yp}p+\phi _{pp}y'']+{\frac {d}{dx}}({\frac {\partial \phi }{\partial y}})$

(Eq. 5.7.1)

$\displaystyle {\frac {d^{2}}{dx^{2}}}g_{2}={\frac {d}{dx}}[\phi _{px}+\phi _{py}y'+\phi _{pp}y'']$

(Eq.(3) p.22-4)

Substituting (Eq.(3)p.21-7),(Eq. 5.7.1) and (Eq.(3)p.22-4) into (Eq.(1)p.22-3) yields

$\displaystyle {\frac {\partial }{\partial y}}({\frac {d\phi }{dx}})-{\frac {d}{dx}}[\phi _{xp}+\phi _{yp}y'+\phi _{pp}y'']-{\frac {d}{dx}}({\frac {\partial \phi }{\partial y}})+{\frac {d}{dx}}[\phi _{px}+\phi _{py}y'+\phi _{pp}y'']=0$

$\displaystyle \Rightarrow {\frac {\partial }{\partial y}}({\frac {d\phi }{dx}})-{\frac {d}{dx}}({\frac {\partial \phi }{\partial y}})+{\frac {d}{dx}}[(\phi _{px}-\phi _{xp})+(\phi _{py}-\phi _{yp})p]=0$

(Eq. 5.7.2)

Because

$\displaystyle {\frac {\partial }{\partial y}}({\frac {d\phi }{dx}})={\frac {\partial }{\partial y}}({\frac {\partial \phi }{\partial x}}+{\frac {\partial \phi }{\partial y}}{\frac {dy}{dx}}+{\frac {\partial \phi }{\partial y'}}{\frac {dy'}{dx}})=\phi _{xy}+\phi _{yy}{\frac {dy}{dx}}+\phi _{py}{\frac {dy'}{dx}}$

$\displaystyle {\frac {d}{dx}}({\frac {\partial \phi }{\partial y}})=[{\frac {\partial }{\partial x}}({\frac {\partial \phi }{\partial y}})+{\frac {\partial }{\partial y}}({\frac {\partial \phi }{\partial y}}){\frac {dy}{dx}}+{\frac {\partial }{\partial y'}}({\frac {\partial \phi }{\partial y}}){\frac {dy'}{dx}}]=\phi _{yx}+\phi _{yy}{\frac {dy}{dx}}+\phi _{yp}{\frac {dy'}{dx}}$

Thus

$\displaystyle {\frac {\partial }{\partial y}}({\frac {d\phi }{dx}})-{\frac {d}{dx}}({\frac {\partial \phi }{\partial y}})=\phi _{xy}+\phi _{yy}{\frac {dy}{dx}}+\phi _{py}{\frac {dy'}{dx}}-\phi _{yx}-\phi _{yy}{\frac {dy}{dx}}-\phi _{yp}{\frac {dy'}{dx}}=\phi _{xy}-\phi _{yx}+(\phi _{py}-\phi _{yp})y''$

(Eq. 5.7.3)

Substituting (Eq. 5.7.3) into (Eq. 5.7.2) yields

$\displaystyle (\phi _{xy}-\phi _{yx})+(\phi _{py}-\phi _{yp})y''+{\frac {d}{dx}}[(\phi _{px}-\phi _{xp})+(\phi _{py}-\phi _{yp})p]=0$

(Eq. 5.7.4)

Because

$\displaystyle {\frac {d}{dx}}(\phi _{py}-\phi _{yp})y'=(\phi _{py}y''+{\frac {d\phi _{py}}{dx}}y')-(\phi _{yp}y''+{\frac {d\phi _{yp}}{dx}}y')$

(Eq. 5.7.5)

Substitute (Eq. 5.7.5) into (Eq. 5.7.4), we have

$\displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=(\phi _{xy}-\phi _{yx})+2(\phi _{py}-\phi _{yp})y''+{\frac {d}{dx}}(\phi _{px}-\phi _{xp})+y'{\frac {d}{dx}}(\phi _{py}-\phi _{yp})=0$

$\displaystyle \Rightarrow (\phi _{xy}-\phi _{yx})+(2y''+y'{\frac {d}{dx}})(\phi _{py}-\phi _{yp})+{\frac {d}{dx}}(\phi _{px}-\phi _{xp})=0$

(Eq. 5.7.6)

Since $y''$ and $y'$ can be the second and first derivative of any solution function $y$ of any second order ODE in terms of which the equation $g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0$ is hold. That is, the factor $2y''+y'{\frac {d}{dx}}$, which consists of two derivatives of solution function and the derivative operater so that depends partly on the solution functin of ODE, can be arbitrary and thus linearly indepent of the derivative operater ${\frac {d}{dx}}$, which is a factor of the third term on left hand side of (Eq. 5.7.6).

Similarly, comparing the first and the third terms on left hand side of (Eq. 5.7.6) yields that the factor 1 (which can be treated as a unit nature number basis of function space) of the first term and the derivative operater (which is another basis of derivative function space) of the third term are linearly independent of each other.

For the left side of (Eq. 5.7.6) being zero under any circumstances, we should have,

$\displaystyle (2y''+y'{\frac {d}{dx}})(\phi _{py}-\phi _{yp})=0$

(Eq. 5.7.7)

while

$\displaystyle (\phi _{xy}-\phi _{yx})=0$

(Eq. 5.7.8)

$\displaystyle {\frac {d}{dx}}(\phi _{px}-\phi _{xp})=0$

(Eq. 5.7.9)

From (Eq. 5.7.7),since the factor $2y''+y'{\frac {d}{dx}}$ is arbitrary, we obtain,

$\displaystyle \phi _{py}-\phi _{yp}=0$

(Eq. 5.7.10)

Thus,

$\displaystyle \phi _{py}=\phi _{yp}$

(Eq. 5.7.11)

From(Eq. 5.7.9), consider $(\phi _{px}-\phi _{xp})$ to be also a function of variables x,y and p, which can be represented as $h(x,y,p)$, thus,

$\displaystyle {\frac {d}{dx}}h(x,y,p)={\frac {\partial }{\partial x}}h(x,y,p)+{\frac {\partial }{\partial y}}h(x,y,p)p+{\frac {\partial }{\partial p}}h(x,y,p)p'=0$

(Eq. 5.7.12)

Since the partial derivative opraters ${\frac {\partial }{\partial x}},{\frac {\partial }{\partial y}},{\frac {\partial }{\partial p}}$ are linearly independent, we have,

$\displaystyle {\frac {\partial }{\partial x}}h(x,y,p)=0$

(Eq. 5.7.13)

$\displaystyle {\frac {\partial }{\partial y}}h(x,y,p)p=0$

(Eq. 5.7.14)

$\displaystyle {\frac {\partial }{\partial p}}h(x,y,p)p'=0$

(Eq. 5.7.15)

Obviously the only condition by which the three equations above are all satisfied is that the function $h(x,y,p)$ is a numerical constant.

Thus, we have

$\displaystyle (\phi _{px}-\phi _{xp})=C$

(Eq. 5.7.16)

where $C$ is a constant. To find the value of constant $C$, try the process as follow.

$\displaystyle \phi _{px}=\phi _{xp}+C$

(Eq. 5.7.17)

Find integral on both sides of (Eq. 5.7.17) in terms of x,

$\displaystyle \phi _{p}=\int \phi _{xp}dx+Cx+f(y,p)$

(Eq. 5.7.18)

where the term $f(y,p)$ is an arbitrarily selected function of independent variables y and p. Then find integral on both sides of (Eq. 5.7.18) in terms of p,

$\displaystyle \phi =\int \int \phi _{xp}dxdp+Cxp+\int f(y,p)dp+g(x,y)$

(Eq. 5.7.19)

where the term $g(x,y)$ is an arbitrarily selected function of variables x and y.

The first partial derivatives of both sides of (Eq. 5.7.19) in terms of x could be

$\displaystyle \phi _{x}=\int \phi _{xp}dp+Cp+{\frac {\partial }{\partial x}}[\int f(y,p)p'dx+g(x,y)]$

(Eq. 5.7.20)

$\displaystyle \Rightarrow \phi _{x}=\int \phi _{xp}dp+Cp+f(y,p)p'+g'(x,y)$

(Eq. 5.7.21)

Then find partial derivative of both sides of (Eq. 5.7.21) in terms of p,

$\displaystyle \phi _{xp}=\phi _{xp}+C+{\frac {\partial }{\partial p}}[f(y,p)p']$

(Eq. 5.7.22)

$\displaystyle \Rightarrow C+{\frac {\partial }{\partial p}}[f(y,p)]p'=0$

(Eq. 5.7.23)

$\displaystyle \Rightarrow {\frac {C}{p'}}=-{\frac {\partial }{\partial p}}[f(y,p)]$

(Eq. 5.7.24)

Because the right hand side of (Eq. 5.7.24) is a function of two variables y and p, while the left hand side is a function of p' only, the equation (Eq. 5.7.24) could not hold if the constant $C$ has a non-zero value. Thus, the only condition by which the equation (Eq. 5.7.24) will be satisfied is that $C=0$ while ${\frac {\partial }{\partial p}}[f(y,p)]=0$ ,that is, $f(y,p)=f(y)$.

Substituting $C=0$ into (Eq. 5.7.16) yields,

$\displaystyle (\phi _{px}-\phi _{xp})=0$

(Eq. 5.7.25)

Thus we have

$\displaystyle \phi _{px}=\phi _{xp}$

(Eq. 5.7.26)

We are now left with $\displaystyle (\phi _{xy}-\phi _{yx})=0$

Thus

$\displaystyle \phi _{xy}=\phi _{yx}$

(Eq. 5.7.27)

R*5.8. Working with the coefficients in 1st exactness condition

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Given

$\displaystyle g_{0}-{\frac {dg_{1}}{dx}}+{\frac {d^{2}g_{2}}{dx^{2}}}=0$

(Eq.(1) p.22-2)

Problem

Using The Coefficients in the 1st exactness condition prove that (Eq.(1)p.22-3) can be written in the form

$\displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y}+(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0$

Solution

Nomenclature

$\displaystyle p:=y(x)'$

$\displaystyle q:=y(x)''$

For an equation to be exact, they must be of the form

$\displaystyle G(x,y,y',y'')=g(x,y,p)+f(x,y,p)y''$

$\displaystyle g_{0}={\frac {\partial G}{\partial y^{(0)}}}=f_{y}q+g_{y}$ (Eq. 5.8.2)

$\displaystyle g_{1}={\frac {\partial G}{\partial y^{(1)}}}=f_{p}q+g_{p}$

$\displaystyle g_{2}={\frac {\partial G}{\partial y^{(2)}}}={\frac {\partial G}{\partial q}}=f$

$\displaystyle {\frac {dg_{1}}{dx}}={\frac {d(f_{p}q+g_{p})}{dx}}$

using chain and product rule

$\displaystyle {\frac {dg_{1}}{dx}}=f_{xp}q+f_{yp}pq+f_{pp}q^{2}+g_{px}+g_{py}p+g_{pp}q$ (Eq. 5.8.3)

$\displaystyle {\frac {dg_{2}}{dx}}={\frac {d(f(x,y,p))}{dx}}$

$\displaystyle {\frac {dg_{2}}{dx}}=f_{x}+f_{y}y'+f_{p}p'$

$\displaystyle {\frac {dg_{2}}{dx}}=f_{x}+f_{y}p+f_{p}q$

$\displaystyle {\frac {d^{2}g_{2}}{dx^{2}}}={\frac {d(f_{x}+f_{y}p+f_{p}q)}{dx}}$

$\displaystyle {\frac {d^{2}g_{2}}{dx^{2}}}=f_{xx}+f_{xy}p+f_{yx}q+f_{yx}p+f_{yy}p^{2}+f_{yp}qp+f_{xp}q+f_{py}pq+f_{pp}q^{2}$ (Eq. 5.8.4)

plugging Eq(2),(3),&(40 into Eq(1)

$\displaystyle f_{y}q+g_{y}-[f_{xp}q+f_{yp}pq+f_{pp}q^{2}+g_{px}+g_{py}p+g_{pp}q]+f_{xx}+f_{xy}p+f_{yx}q+f_{yx}p+f_{yy}p^{2}+f_{yp}qp+f_{xp}q+f_{py}pq+f_{pp}q^{2}=0$

after cancellation of the opposite term

$\displaystyle (f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y})+(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0$

Now, we can club the terms

$\displaystyle (f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y})1={\bar {g}}$

and

$\displaystyle (f_{xp}+pf_{yp}+2f_{y}-g_{pp})q={\bar {f}}$

Since 1 and q, i.e the second derivative of y, are in general non linear, for the equation to hold true, their coefficients must both be equal to zero.

Thus we say that

$\displaystyle {\bar {g}}=(f_{xx}+2pf_{xy}+p^{2}f_{yy}-g_{px}-pg_{py}+g_{y})1=0$

and

$\displaystyle {\bar {f}}=(f_{xp}+pf_{yp}+2f_{y}-g_{pp})q=0$

Which is the required proof.

R5.9: Use of MacLaurin Series

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem

Use Taylor Series at x=0 (MacLaurin Series) to derive

$\displaystyle (1-x)^{-a}=1+ax+a(a+1){\frac {x^{2}}{2!}}+a(a+1)(a+2){\frac {x^{3}}{3!}}+.......=F(a,b;b;x)$

$\displaystyle {\frac {1}{x}}arctan(1+x)=1-{\frac {x^{2}}{3}}+{\frac {x^{4}}{5}}-{\frac {x^{6}}{7}}+....=F\left({\frac {1}{2}},1,{\frac {3}{2}},-x^{2}\right)$

Solution

The Taylor's series expansion of a function f(x) about a real or complex number c is given by the formula

$f(c)+{\frac {f'(c)}{1!}}(x-c)+{\frac {f''(c)}{2!}}(x-c)^{2}+{\frac {f^{(3)}(c)}{3!}}(x-c)^{3}+\cdots .$

(Eq. 5.9.1)

When the neighborhood for the expansion is zero, i.e c = 0, the resulting series is called the Maclaurin Series.

Part a

We have the function

$\displaystyle f(x)={\frac {1}{(1-x)^{a}}}$

Table for Maclaurin Series
$f(x)={\frac {1}{(1-x)^{a}}}$	$f(0)={\frac {1}{(1-0)^{a}}}=1$
$f'(x)={\frac {a}{(1-x)^{a+1}}}$	$f'(0)={\frac {a}{(1-0)^{a+1}}}=a$
$f''(x)={\frac {a(a+1)}{(1-x)^{a+2}}}$	$f''(0)={\frac {a(a+1)}{(1-0)^{a+2}}}=a(a+1)$
$f^{(3)}(x)={\frac {a(a+1)(a+2)}{(1-x)^{a+3}}}$	$f^{(3)}(0)={\frac {a(a+1)(a+2)}{(1-0)^{a+3}}}=a(a+1)(a+2)$
And so on..	..

Rewriting the Maclaurin series expansion,

$\displaystyle {\frac {1}{(1-x)^{a}}}=f(0)+{\frac {f'(0)}{1!}}(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f^{(3)}(0)}{3!}}(x-0)^{3}+\cdots .$

(Eq. 5.9.2)

Substituting the values from the tables in (Eq. 5.9.2) we get

$\displaystyle {\frac {1}{(1-x)^{a}}}=1+{\frac {a}{1!}}(x)+{\frac {a(a+1)}{2!}}(x)^{2}+{\frac {a(a+1)(a+2)}{3!}}(x)^{3}+\cdots .$

(Eq. 5.9.3)

$\displaystyle {\frac {1}{(1-x)^{a}}}=\sum _{n=0}^{\infty }(a)_{k}{\frac {x^{k}}{n!}}$

(Eq. 5.9.4)

Where

$(a)_{0}:=1$

$(a)_{k}:=a(a+1)(a+2)\cdots (a+k-1)$

We can represent

$\sum _{n=0}^{\infty }{\frac {(a)_{k}\,(b)_{k}}{(c)_{k}}}\,{\frac {x^{k}}{n!}}=F(a,b;c;x)$

(Eq. 5.9.4) can be written as $\sum _{n=0}^{\infty }{\frac {(a)_{k}(b)_{k}}{(b)_{k}}}{\frac {x^{k}}{n!}}=F(a,b;b;x)$ , hence proved.

Part b

We have the function

$\displaystyle {\frac {1}{x}}arctan(1+x)$

We will use a slightly different approach here when compared to part a of the solution. We will expand $\displaystyle arctan(1+x)$ and multiply the resulting expanded function with $\displaystyle {\frac {1}{x}}$

Table for Maclaurin Series
$f(x)=arctan(1+x)$	$f(0)=arctan(1+0)={\frac {\pi }{4}}$
$f'(x)={\frac {1}{1+(1+x)^{2}}}$	$f'(0)={\frac {1}{1+(1+0)^{2}}}={\frac {1}{2}}$
$f''(x)={\frac {-2(x+1)}{\left[1+(1+x)^{2}\right]^{2}}}$	$f''(0)={\frac {-2(0+1)}{\left[1+(1+0)^{2}\right]^{2}}}={\frac {-1}{2}}$
$f^{(3)}(x)={\frac {2(3x^{2}+6x+2)}{\left[1+(1+x)^{2}\right]^{3}}}$	$f^{(3)}(0)={\frac {2(3(0)^{2}+6(0)+2)}{\left[1+(1+0)^{2}\right]^{3}}}={\frac {1}{2}}$
And so on..	..

Rewriting the Maclaurin series expansion,

$\displaystyle arctan(1+x)=f(0)+{\frac {f'(0)}{1!}}(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f^{(3)}(0)}{3!}}(x-0)^{3}+\cdots .$

(Eq. 5.9.5)

Substituting the values from the tables in (Eq. 5.9.5) we get

$\displaystyle arctan(1+x)={\frac {\pi }{4}}+{\frac {1}{2*1!}}(x)-{\frac {1}{2*2!}}(x)^{2}+{\frac {1}{2*3!}}(x)^{3}+\cdots .$

(Eq. 5.9.6)

Multiplying (Eq. 5.9.6) with $\displaystyle {\frac {1}{x}}$

$\displaystyle {\frac {1}{x}}arctan(1+x)={\frac {\pi }{4x}}+{\frac {1}{2*1!}}-{\frac {1}{2*2!}}(x)+{\frac {1}{2*3!}}(x)^{2}+\cdots .$

This expression does not match the expression that we have been asked to prove. This, we believe is because there has been a misprint and the expression to be found out must be ${\frac {1}{x}}arctan(x)$

Expanding $arctan(x)$ using Maclaurin's series

Table for Maclaurin Series
$f(x)=arctan(x)$	$f(0)=arctan(0)=0$
$f'(x)={\frac {1}{1+x^{2}}}$	$f'(0)={\frac {1}{1+(0)^{2}}}=1$
$f''(x)={\frac {-2x}{\left[1+(x)^{2}\right]^{2}}}$	$f''(0)={\frac {-2(0)}{\left[1+(0)^{2}\right]^{2}}}=0$
$f^{(3)}(x)={\frac {6x^{2}-2}{\left[1+(x)^{2}\right]^{3}}}$	$f^{(3)}(0)={\frac {6(0)^{2}-2)}{\left[1+(0)^{2}\right]^{3}}}=-2$
$f^{(4)}(x)={\frac {-24x(x^{2}-1}{\left[1+(x)^{2}\right]^{4}}}$	$f^{(4)}(0)={\frac {-24(0)((0)^{2}-1}{\left[1+(0)^{2}\right]^{4}}}=0$
$f^{(5)}(x)={\frac {24(5x^{4}-10x^{2}+1}{\left[1+(x)^{2}\right]^{5}}}$	$f^{(5)}(x)={\frac {24(5(0)^{4}-10(0)^{2}+1}{\left[1+(0)^{2}\right]^{5}}}=24$
And so on..	..

Rewriting the Maclaurin series expansion,

$\displaystyle arctan(x)=f(0)+{\frac {f'(0)}{1!}}(x-0)+{\frac {f''(0)}{2!}}(x-0)^{2}+{\frac {f^{(3)}(0)}{3!}}(x-0)^{3}+\cdots .$

(Eq. 5.9.7)

Substituting the values from the tables in (Eq. 5.9.7) we get

$\displaystyle arctan(x)=0+{\frac {1}{1!}}(x)+{\frac {0}{2!}}(x)^{2}+{\frac {-2}{3!}}(x)^{3}+{\frac {0}{4!}}(x)^{4}+{\frac {24}{5!}}(x)^{5}\cdots .$

(Eq. 5.9.8)

Multiplying (Eq. 5.9.8) with $\displaystyle {\frac {1}{x}}$

$\displaystyle arctan(x)=0+{\frac {1}{1!}}+{\frac {-2}{3!}}(x)^{2}+{\frac {24}{5!}}(x)^{4}\cdots .$

$\displaystyle arctan(x)=1-{\frac {(x)^{2}}{3}}+{\frac {(x)^{4}}{5}}+{\frac {(x)^{6}}{7}}\cdots .$

(Eq. 5.9.9)

Which is the expression in the RHS.

R5.10 Gauss Hypergeometric Series

On our honor, we did this problem on our own, without looking at the solutions in previous semesters or other online solutions.

Problem

1. Use MATLAB to plot $F(5,-10;1;x)$ near x=0 to show the local maximum (or maxima) in this region.

2. Show that $F(5,-10;1;x)=(1-x)^{6}(1001x^{4}-1144x^{3}+396x^{2}-44x+1)$

((1) pg. 64-9b)

Solution

The MATLAB code, shown below, will plot the hypergeometric function $F(5,-10;1;x)$ over the interval: $0\leq x\leq 0.8$.

x = [0:0.01:0.8]';
plot(x,hypergeom([5,-10],1,x))

The plot of the hypergeometric function near x=0 reveals a local maximum of 0.1481 at x = 0.23.

The hypergeometric function $F(5,-10;1;x)$ can be expressed as $\sum _{k=0}^{\infty }{\frac {(a)_{k}(b)_{k}}{(c)_{k}}}{\frac {x^{k}}{k!}}$ using the Pochhammer Symbol

where   $\frac{(a{)}_k(b{)}_k}{(c{)}_k}=\frac{a(a+1)(a+2)...(a+k-<!--\; -\; -->1)b(b+1)(b+2)...(b+k-<!--\; -\; -->1)}{c(c+1)(}$