- Nam Sun Wang
- Department of Chemical Engineering
- University of Maryland
- College Park, MD 20742-2111
- 301-405-1910 (voice)
- 301-314-9126 (FAX)
e-mail: nsw@eng.umd.edu
©2002 by Nam Sun Wang
- Department of Chemical Engineering
Solution:
Yes. If x1 and x2 are linearly dependent,
we can express x2 as a scalar multiple of x1.
x2 = a*x1
And the angle between x1 and x2 is zero.
q = cos-1[(x1,x2)/(|x1|/|x2|] = cos-1[(x1,a*x1)/a*(x1,x1)]
q = cos-1[1] = 0
Solution:
No, all three vectors may all lie in the same "plane".
fi(x)=J0(lix) where li=i for i=1,2,..,5Are these Bessel's functions J0(lix) linearly independent?
Solution:
Yes, they are linearly independent from the point of view of
pure mathematics. However, no, they are linearly dependent from
engineering mathematics. Specifically, out of the five
functions, there are only three numerically significantly
independent ones in that homework assignment. Many students
constructed the last two vectors from computational artifacts
(i.e., numerical errors.)
f0(x)=1 f1(x)=x f2(x)=x2
Solution:
First, we need to define a scalar product between two functions f(x) and g(x).
ó1
(f,g) = ô f(x)*g(x)*dx
õ0
(f,g) = integral (from -1 to 1 of f(x)*g(x)*dx)
Follow the Gram-Schmidt orthogonalization process.
First, take any one of the given function as the starting
vector.
g0 = 1
Take the second given function, and subtract from it the part that
is in the g0 direction. The remaining component is orthogonal
to g0
g2 = x - (x,g0)/(g0,g0)*g0
= x - 0/2*1
= x
Take the third given function, and subtract from it the parts
common to g1 and g2. The remaining component
is orthogonal to both g1 and g2.
g2 = x2 - (x2,g0)/(g0,g0)*g0 - (x2,g1)/(g1,g1)*g1
= x2 - (2/3)/2*1 - 0/(2/3)*x
= x2 - 1/3
f0(x)=1 f1(x)=x f2(x)=x2 f3(x)=3x+x2
Solution:
No. We need to start with a set of linearly independent set of vectors
to generate a set of orthogonal vectors (which are linearly independent).
In this case, it is not possible to construct a complete set or mutually
orthogonal functions because the last function is a linear combination
of the previous functions.
Solution:
We represent f(l) as a linear combination
of fi(l), i=1,2,...,5. We further
assume that we cannot produce each given color by mixing the rest, i.e.,
fi's are linearly independent.
f(l) = a1*f1(l) + a2*f2(l) + ... + a5*f5(l) + error
error = f(l) - a1*f1(l) - a2*f2(l) - ... - a5*f5(l)
Minimize |error|
The best approximation is the projection of f onto the
subspace spanned by (f1(l), f2(l),...,
f5(l)). The length of error, |error|, is minimized if error is orthogonal to each of the fi's.
0 = (error,f1) = (f,f1) - a1*(f1,f1) - a2*(f2,f1) - ... - a5*(f5,f1)
0 = (error,f2) = (f,f2) - a1*(f1,f2) - a2*(f2,f2) - ... - a5*(f5,f2)
0 = (error,f3) = (f,f3) - a1*(f1,f3) - a2*(f2,f3) - ... - a5*(f5,f3)
0 = (error,f4) = (f,f4) - a1*(f1,f4) - a2*(f2,f4) - ... - a5*(f5,f4)
0 = (error,f5) = (f,f5) - a1*(f1,f5) - a2*(f2,f5) - ... - a5*(f5,f5)
We solve the above 5 equations for the 5 coefficients (a1, a2, ..., a5).
The above set of equations are linear wrt the coefficients and can be compacted into,
P*a=F --->Solve---> a=P-1*F
where P and F are the various scalar products.
Pij=(fi,fj) and Fi=(f,fi)
Since spectra of inks are not necessarily mutually orthogonal,
(fi,fj)¹0
in general. Of course, when the given ink colors are
mutually orthogonal, all the off-diagonal elements of P vanishes.
Pij=(fi,fj)=0 for i¹j
and the regression equation for a is greatly simplified.
For orthogonal fi, ai=(f,fi)/(fi,fi)
We can define a suitable scalar product for l=[a,b]=[400nm,
700nm]. For example, the following is a valid one.
ób
(f,g) = ô w(l)*f(l)*g(l)*dl
õa
With the above definition, the elements of P are,
ób
Pij=(fi,fj) = ô w(l)*f(l)i*f(l)j*dl
õa
In practice, we cannot have negative quantity of ink; thus, the coefficients
ai are physically constrained to be nonnegative. (Always relate
back to physical reality.)
------------------------------------------------------------------- Final Overall HW Exam Exam Exam Class Seat Grade Score Total #1 #2 #3 Attendance Row # Sex ... ------------------------------------------------------------------- A 0.91 40 29 70 69 100 4 M ... B 0.75 45 : : : : : : : C : : : : : : : : : F : : : : : : : : : : : : : : : : : : : -------------------------------------------------------------------
Solution:
This is similar to a previous homework assignment where we
attempt to predict national ranking scores from the various
factors via linear regression in eigenvector directions. Note
that since the independent variables may be highly correlated,
plain linear regression does not work.
Of course, in ENCH620, I have already explicitly specified how I
plan to calculate students' overall scores. Had we not known the model
a priori, the above calculation may conclude that
class attendance, seating position, or even sex/race seem to contribute
to the overall score. This makes it possible for people to
distort, deliberately or unwittingly, facts and statistics to fit
their agenda.
score<i>=X*v<i>
y = a1*score<1> + a2*score<2> + ... + ai*score<i> + error<i>
where the coefficients ai are given by the projection formula,
ai = (y,score<i>)/(score<i>,score<i>) for i=1,2,...
ai = (y,score<i>) for i=1,2,... if score<i> is normalized, i.e., (score<i>,score<i>)=1 (which is usually not)
Before we include any factor:
error<0> = y
sse0=(error<0>,error<0>)
As we include each additional factor:
error<i> = error<i-1> - ai*score<i>
ssei=(error<i>,error<i>)
D2y = d2y/dx2
Solution: Yes. D transforms a continuous differentiable function y into another function. In goes a function, out comes a function. Strictly speaking, the output function also needs to be continuous differentiable. Thus, the LVS for D is all continuous differentiable functions. For example, y(x)=|x| is excluded from the LVS. D2 does the operation twice.
Solution:
Yes. The operator D satisfies both linear properties; do does D2:
1) Distributive: D(y+z) = D(y) + D(z)
L(y+z) = D2(y+z) = D2y + D2z = L(y) + L(z)
2) Associative: D(a*y) = a*D(y)
L(a*y) = D2(a*y) = a*D2 = a*L(y)
Solution:
Eigenvalue-eigenvector for the derivative operator D
Note that eigenvalue-eigenvector for the double derivative
operator L=D2 is different from that of D. Do not confuse L with D.
Dy = dy/dx = l*y
From the above equation, we see that the eigenvalues are any value and
the eigenvectors are exponential functions:
For any l', y=exp(l'*x)
If we place the restriction y(-1)=y(1)=0, we have no real solution because
a real exponential function is never zero.
L(y) = D2y = d2y/dx2 = l'*y
From the above equation, we see that the eigenvalues for the operator L (but not D) are any real and complex number, and
the eigenvectors are exponential functions:
For any l', y=exp(sqrt(l')*x) and y=exp(-sqrt(l')*x)
When we place further restriction with BC y(-1)=y(1)=0, we have sine and cosine functions.
For l=1/2*p, 3/2*p, 5/2*p,... y=cos(l*x)
For l=p, 2*p, 3*p,... y=sin(l*x)
where l2=-l'
f(x) = S aj*fjFind the coefficients aj.
Solution:
Since the eigenfunctions are mutually orthogonal, we apply the projection formula,
aj = (f,fj)/(fj,fj)
Solution:
No. f(x)=1 is even, and sine functions are odd.
Solution:
No, because cos(np*x) is orthogonal to
cos(0*x)=1 for n=1,2,.... However, with a different cosine
series, (say, cos(x), cos(2x),..., or say,
cos(1/2*p*x),
cos(3/2*p*x,),...),
yes, we can approximate f(x)=1, although we cannot represent
f(x)= 1 exactly with a finite number of terms, because the
functions are all linearly independent. (Thus, the answer
depends on what "a series of cosine functions" are and how the
scalar product is defined. Thus, I grade based on students'
explanation.)
Solution:
Yes. However, we need many terms to approximate cos(x) Again,
we cannot represent cos(x) exactly, but approximate we
can. Note that as in the last part, the answer is "no" if we try
to approximate cos(p*x) in x=[-1, 1] with
cos(np*x) for n=2,3,..., because they
are mutually orthogonal. Again, when we refer to orthogonality,
we need to define first the scalar product.
Solution: No.
dw 1 d dw -- = - -- (r --) B.C. at r=1 w=0 (no slip condition at the pipe wall) dt r dr dr at t=0 w=1-r2 (zero initial velocity)One way to solve the above PDE is by separation of variables, where we assume w(t,r) is a product of two functions: T(t) and R(r). Note that function T depends only on time t, and function R depends only on radial position r.
w(t,r)=T(t)*R(r)Substituting the above product into the given PDE yields,
T' 1 1 -- = - [- (r*R')'] T R rIn the above ODE, the LHS is a function of t only and the RHS is a function of r only. Thus, the only possibility is for both the LHS and the RHS to be constants.
T'
LHS = -- = -l2 ... Eqn (1)
T
1 1
RHS = - [- (r*R')'] = -l2 ... Eqn (2)
R r
B.C.: at r=0, R=bound
at r=1, R=0
wi(t,r) = Ti(t)*Ri(r)
w(0,r)=1-r2=S aj*J0(ljr)Find aj. Hint: Define a scalar product with an integral from 0 to 1 with a weighting factor of r.
Solution:
This is an application of the eigenvalue-eigenvector idea. In
solving PDEs and in methods of weighted residuals, we rely
heavily on the technique of separation of variables coupled with
eigenvalue-eigenvector.
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