User:Eml5526.s11.team4.yang/hw2

=Homework 2.3=

Refer to lecture slide [[media:fe1.s11.mtg8.djvu|mtg-8]] for the problem statement.

Given
Use { ai } (basis),let $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)$$, then equation
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$$ \displaystyle
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\underline{w}(x)\underline{P}(\underline{v})=0\begin{matrix} {} & \forall  \\ \end{matrix}\underline{w}(x)=\sum\limits_{i=1}^{n}(x)

$$     (Eq 1) becomes
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$$ \displaystyle
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\underline{w}(x)\underline{P}(\underline{v})=0\begin{matrix} {} & \forall  \\ \end{matrix}\{{{\beta }_{1}},{{\beta }_{i}}..\}\in {{\mathbb{R}}^{n}}\begin{matrix} {} & s.t.\underline{w}(x)=\sum\limits_{i=1}^{n} \\ \end{matrix}

$$     (Eq 2)
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Objectives
Show that equation (2) is equivalent to
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$$ \displaystyle
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\underline(x)\underline{P}(\underline{v})=0\begin{matrix} {} & i=1,2,...n \\ \end{matrix}

$$     (Eq 3)
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Solutions
1). First, we try to derive Eq.(3) using Eq.(2).

Eq. (1) is valid for any function of w (x), which is a linear combination of these linearly independent bases. Thus, the Eq. (1) is valid for some particular sets of $${{\beta }_{i}}$$ value to get $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)$$. For example, we choose $${{\beta }_{1}}=1,{{\beta }_{2}}={{\beta }_{3}}=....{{\beta }_{n}}=0$$, then $$\underline{w}(x)=\sum\limits_{i=1}^{n}(x)={{\underline{a}}_{1}}(x)$$. By substituting $$\underline{w}(x)$$ into Eq.(2), we obtain:
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$$ \displaystyle
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\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$ Which is the equation for Eq.(3) of i=1.
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Then, in similar way, we choose $${{\beta }_{2}}=1,{{\beta }_{1}}={{\beta }_{3}}=....{{\beta }_{n}}=0$$, then we obtain:
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$$ \displaystyle
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\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$
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Which is the equation for Eq.(3) of i=2.

Similarly, we continuing to choose $${{\beta }_{i}}=1,{{\beta }_{1}}={{\beta }_{2}}=..={{\beta }_{i-1}}={{\beta }_{i+1}}..={{\beta }_{n}}=0$$ then we obtain:
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$$ \displaystyle
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\underline{w}(x)\underline{P}(v)=\underline(x)\underline{P}(v)=0

$$ Which is the equation for Eq.(3) of any value of i. So, Eq. (3) is derived with Eq.(2).
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2). Secondly, we try to derive Eq.(3) using Eq.(2).

{ ai(x), i=1,2…n } is a family of linearly independent basis functions, and according to Eq.(2) we have:


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$$ \displaystyle
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\underline(x)\underline{P}(v)=0

$$
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$$ \displaystyle
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\underline(x)\underline{P}(v)=0

$$
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$$ \displaystyle
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\underline(x)\underline{P}(v)=0

$$
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Linearly combining the above equations, by choosing any set of $${{\beta }_{i}}$$value, we obtain:
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$$ \displaystyle
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\begin{align} & {{\beta }_{1}}[\underline(x)\underline{P}(v)]+{{\beta }_{2}}[\underline(x)\underline{P}(v)]+..+{{\beta }_{i}}[\underline(x)\underline{P}(v)]+...{{\beta }_{n}}[\underline(x)\underline{P}(v)] \\ & ={{\beta }_{1}}\centerdot 0+{{\beta }_{2}}\centerdot 0+..+{{\beta }_{i}}\centerdot 0+..+{{\beta }_{n}}\centerdot 0 \\ & =0 \\ \end{align}

$$
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Obviously,
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$$ \displaystyle
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\begin{align} & {{\beta }_{1}}[\underline(x)\underline{P}(v)]+{{\beta }_{2}}[\underline(x)\underline{P}(v)]+..+{{\beta }_{i}}[\underline(x)\underline{P}(v)]+...{{\beta }_{n}}[\underline(x)\underline{P}(v)] \\ & =[\sum\limits_{i=1}^{n}(x)]\underline{P}(v) \\ & =\underline{w}(x)\underline{P}(v) \\ & =0 \\ \end{align}

$$
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Because the value set of $${{\beta }_{i}}$$undefinedis arbitrary, Eq.(2) is valid for any w (x).

So, we obtain Eq.(2) using Eq.(3).

According the above two aspects, we can conclude that Eq.(2) is equivalent to Eq. (3).