User:Jiangpx/6321 hw3

= Problem 4 - Taylor Series Expansion =

Given
$$\begin{align}f(x)=\exp (x)\end{align}$$ and $$\begin{align}{{x}_{o}}=0\end{align}$$

Find
Taylor Series expansion of f(x) at $$\begin{align}{{x}_{o}}=0\end{align}$$

Solution
Taylor Series expansion of function f: $$f(x) = f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots. $$

For $$\begin{align}f(x)=\exp (x)\end{align}$$,

$$\begin{align}{{f}^{(n)}}(x)=f(x)=\exp (x)\end{align}$$

and at $$\begin{align}{{x}_{o}}=0\end{align}$$,

$$\begin{align}{{f}^{(n)}}({{x}_{0}})=1\end{align}$$

therefore,

$$\exp (x)=\exp ({{x}_{o}})+\exp ({{x}_{o}})(x-{{x}_{o}})+{{\exp }^{(2)}}({{x}_{0}})\frac{2!}+...$$

becomes

$$\exp (x)=1+\frac{1!}+\frac{2!}+\frac{3!}+...=\sum\limits_{k=0}^{n}{\frac{k!}}$$

= Problem 5 - Genaralization of SC-L1-ODE-VC, from scalar to vector =

Given
$$\overset{\centerdot }{\mathop{x}}\,(t)=a(t)x(t)+b(t)u(t)$$

and the solution of this L1-ODE-VC of the form

$$\begin{align}x(t)=[\exp (\int_^{t}{a(\tau )d\tau })]\cdot x({{t}_{o}})+\int_^{t}{[\exp \int_{\tau }^{t}{a(s)ds}]}\cdot bu(\tau )d\tau \end{align}$$

Find
The solution in vector form for system of linear ODE with varying coefficients

Solution
$$\overset{\centerdot }{\mathop{X}}\,(t)=A(t)X(t)+B(t)U(t)$$

$$\Rightarrow X(t)=[\exp (\int_^{t}{A(\tau )d\tau })]\cdot X({{t}_{o}})+\int_^{t}{[\exp \int_{\tau }^{t}{A(s)ds}]}\cdot BU(\tau )d\tau $$

where the dimension of matrices are:

$$\begin{align} & {{X}_{n\times 1}} \\ & {{U}_{n\times 1}} \\ & {{A}_{n\times n}} \\ & {{B}_{n\times n}} \\ \end{align}$$

= Problem 9 - Roll Control of Rocket =

Given
Rocket basic info $$\begin{align} & \delta - aileron angle \\ & \phi - roll angle \\ & \omega - roll angular velocity \\ & Q - aileron effectiveness \\ & \tau - roll-time constant \\ \end{align}$$

Express
$$\begin{align} & \overset{\centerdot }{\mathop{\phi }}\,=\omega \\ & \overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta \\ & \overset{\centerdot }{\mathop{\delta }}\,=u \\ \end{align}$$

in terms of $$\overset{\centerdot }{\mathop{X}}\,=AX(t)+BU(t)$$ - SC-L1-ODE-CC.

Solution
Let's treat phi, omega and delta as the state variables and u as control variable. The equations can be expressed as follows:

$$\left\{ \begin{align} & \overset{\centerdot }{\mathop{\phi }}\, \\ & \overset{\centerdot }{\mathop{\omega }}\, \\ & \overset{\centerdot }{\mathop{\delta }}\, \\ \end{align} \right\}=\left( \begin{matrix}  0 & 1 & 0  \\   0 & -\frac{1}{\tau } & \frac{Q}{\tau }  \\   0 & 0 & 0  \\ \end{matrix} \right)\cdot \left\{ \begin{align} & \phi \\ & \omega \\ & \delta \\ \end{align} \right\}+\left\{ \begin{align} & 0 \\ & 0 \\  & 1 \\ \end{align} \right\}u$$

= Problem 12 - Finish the story =

Given
$$\begin{align} (6x{{y}^{2}})y'+2{{y}^{3}}={{h}_{y}}y'+{{h}_{x}} \end{align}$$

Find
$$\begin{align}h(x,y)\end{align}$$ and $$\begin{align} \phi (x,y,y')\end{align} $$

Solution
1) 1st choice, assume $$\begin{align} & {{h}_{x}}=2{{y}^{3}} \\  & {{h}_{y}}=6x{{y}^{2}} \\ \end{align}$$.

then

$$\begin{align} h(x,y)=2x{{y}^{3}}+{{k}_{1}} \end{align}$$, with k1 being integration constant.

Hence

$$\begin{align} & \overline{\phi }=h(x,y)+\int{f(x,y,p)dp}=2x{{y}^{3}}+{{k}_{1}}+3{{p}^{5}}\cos ({{x}^{2}})={{k}_{2}} \\ & \phi =2x{{y}^{3}}+3{{p}^{5}}\cos ({{x}^{2}})=k \\ \end{align}$$, with k1,k1 and k being integration constants.

2) 2nd choice, assume

$$\begin{align}{{h}_{y}}y'=2{{y}^{3}}\end{align}$$, and this implicitly assumes that

$$\begin{align}{{h}_{x}}=(6x{{y}^{2}})y'\end{align}$$.

We know that $$\begin{align}h=h(x,y)\end{align}$$, so it is not possible to have y' term in hx. Hence the story ends here, and 1st choice is the only choice.