User:Egm5526.s11.team4.zou/HW2

=Problem 2.4 supplement=

Another solution for part 4
As follows,

From part1 and part2 we know,

Sustitute Eqs 4.B into Eq 4.A and then arrange terms and omit constant ones it yields,

From Eq 4.A we can see that most of time we are dealing with $$\displaystyle (a+bx)$$ rather than $$\displaystyle x$$. So why not make a transform: $$\displaystyle u=a+bx$$, hopefully it reduces the notation and calculation.

Hence,

Sustitute Eq 4.D into Eq 4.A we have,

Replace $$\displaystyle u$$ back to $$\displaystyle x$$ in Eq 4.E we have,

=Problem 2.7 Verifying the orthogonality of the family=

Problem statement
Family of basis functions given:

with the same domain $$\displaystyle \Omega =[0,1]$$.

Do the following steps to determine whether $$\displaystyle \mathcal{F}$$ is a orthogonal family.

1)Construct the Gramian matrix $$\displaystyle \Gamma \left( \mathcal{F} \right)$$, and observe its property

2)Find the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$

3)Conclude on the orhogonality of $$\displaystyle \mathcal{F}$$.

Solution
1)Construct the Gramian matrix $$\displaystyle \Gamma \left( \mathcal{F} \right)$$, and observe its property

First let ‘ s assign notation to the family $$\displaystyle \mathcal{F}$$:

The Gramian matrix is defined as

Since,

then,

Then we can construct the Gramian matrix,

And we know that it is a symmetric matrix.

2)Find the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$

The determinant can be calculated via WolframAlpha:

3)Conclude on the orhogonality of $$\displaystyle \mathcal{F}$$.

Since the determinant of $$\displaystyle \Gamma \left( \mathcal{F} \right)$$ is non-zero, so we can conclude that the family of basis functions $$\displaystyle \mathcal{F}$$ is an orthogonal family.

Problem Statement
Given:

Where $$\displaystyle \underline{w}(x)\cong {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)$$, $$\displaystyle \underline{u}(x)\cong {{\underline{u}}^{h}}(x)=\sum\limits_{j=1}^{n}{{{d}_{j}}\underline}(x)$$.

Show that Eq. (8.1) is equivalent to Eq. (8.2).

==Solutions ==

1). First, we try to derive Eq.(8.2) using Eq.(8.1).

Eq. (8.1) is valid for any function of wh (x), which is a linear combination of these linearly independent basis.

Thus, the Eq. (8.1) is valid for some particular sets of Ci value to get $$\displaystyle {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)$$.

For example, we choose C1=1, C2=C3=..=Cn=0, then $$\displaystyle {{\underline{w}}^{h}}(x)=\sum\limits_{i=1}^{n}(x)={{\underline{b}}_{1}}(x)$$. By substituting $$\displaystyle {{\underline{w}}^{h}}(x)$$ into Eq.(8.1), we obtain:

Which is the equation for Eq.(8.2) of $$\displaystyle i=1$$.

Then, in similar way, we choose C2=1, C1=C3=..=Cn=0, then we obtain:

Which is the equation for Eq.(8.2) of $$\displaystyle i=2$$.

Similarly, we continuing to choose Ci=1, C1=C2=.. Ci-1 = Ci+1 =Cn=0 then we obtain:

Which is the equation for Eq.(8.2) of any value of $$\displaystyle i$$. So, Eq. (8.2) is derived with Eq.(8.1).

2). Secondly, we try to derive Eq.(8.2) using Eq.(8.1).

{ bi(x), i=1,2…n } is a family of linearly independent basis functions, and according to Eq.(8.1) we have:

Linearly combining the above equations, by choosing any set of Civalue, we obtain:

Obviously,

Because the value set of Ciis arbitrary, Eq.(8.1) is valid for any wh (x). So, we obtain Eq.(1) using Eq.(8.2).

According the above two aspects, we can conclude that Eq.(8.1) is equivalent to Eq. (8.2).