User:Eml5526.s11.team4.yang/hw3

=Homework 3.4-Show$$\displaystyle A\ne B$$, when $$\displaystyle i\ne j$$=

Refer to lecture slide for the problem statement.

Given

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$$ \displaystyle
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A={{b}_{i}}({{a}_{2}}^{'}{{b}_{j}}^{'}+{{a}_{2}}{{b}_{j}}^{''})

$$     (Eq 1)
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$$ \displaystyle
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B={{b}_{j}}({{a}_{2}}^{'}{{b}_{i}}^{'}+{{a}_{2}}{{b}_{i}}^{''})

$$     (Eq 2)
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Objectives
Show$$\displaystyle A\ne B$$, when $$\displaystyle i\ne j$$
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$$ \displaystyle
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{{b}_{i}}(x)=\cos (ix)

$$     (Eq 3)
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$$ \displaystyle
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{{b}_{j}}(x)=\cos (jx)

$$     (Eq 4)
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Solutions
Substituting Eqs. (3),(4) into Eqs. (1) and (2) yields:
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$$ \displaystyle
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\begin{align} & A={{b}_{i}}({{a}_{2}}^{'}{{b}_{j}}^{'}+{{a}_{2}}{{b}_{j}}^{''})=\cos (ix)[{{a}_{2}}^{'}\centerdot (-j)\sin (jx)+{{a}_{2}}\centerdot {{j}^{2}}\cos (jx)] \\ & \begin{matrix} {} & = \\ \end{matrix}{{a}_{2}}\centerdot {{j}^{2}}\cos (ix)\cos (jx)-{{a}_{2}}^{'}\centerdot j\sin (jx)\cos (ix) \\ \end{align}

$$     (Eq 5)
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$$ \displaystyle
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\begin{align} & B={{b}_{j}}({{a}_{2}}^{'}{{b}_{i}}^{'}+{{a}_{2}}{{b}_{i}}^{''})=\cos (jx)[{{a}_{2}}^{'}\centerdot (-i)\sin (ix)+{{a}_{2}}\centerdot {{i}^{2}}\cos (ix)] \\ & \begin{matrix} {} & = \\ \end{matrix}{{a}_{2}}\centerdot {{i}^{2}}\cos (ix)\cos (jx)-{{a}_{2}}^{'}\centerdot i\sin (ix)\cos (jx) \\ \end{align}

$$     (Eq 6) When $$\displaystyle i\ne j$$, it is obvious that:
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$$ \displaystyle
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{{j}^{2}}\cos (ix)\cos (jx)\ne {{i}^{2}}\cos (ix)\cos (jx)

$$     (Eq 7)
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$$ \displaystyle
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j\sin (jx)\cos (ix)\ne i\sin (ix)\cos (jx)

$$     (Eq 8) When $$\displaystyle {{a}_{2}}^{'}\ne 0$$ or $$\displaystyle {{a}_{2}}\ne 0$$, it shows '''$$\displaystyle A\ne B$$. '''
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Thus, if $$\displaystyle {{a}_{2}}^{'}$$ and  $$\displaystyle {{a}_{2}}$$ are not all zero, $$\displaystyle A\ne B$$ ( $$\displaystyle i\ne j$$).
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Specifically, we can take i=1 and j=2 for example, then Eqs. (5) and (6) become:
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$$ \displaystyle
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A=4{{a}_{2}}\centerdot \cos (x)\cos (2x)-2{{a}_{2}}^{'}\centerdot \sin (2x)\cos (x)

$$     (Eq 9)
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$$ \displaystyle
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B={{a}_{2}}\cos (x)\cos (2x)-{{a}_{2}}^{'}\sin (x)\cos (2x)

$$     (Eq 10) When $$\displaystyle {{a}_{2}}^{'}\ne 0$$ or $$\displaystyle {{a}_{2}}\ne 0$$, it shows '''$$\displaystyle A\ne B$$. '''
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The equation $$\displaystyle A=B$$is valid only $$\displaystyle {{a}_{2}}^{'}=0$$ and $$\displaystyle {{a}_{2}}=0$$.