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Answer:
Since ((xT. A . x)^2) * x = b and ((xT. A . x)^2) is a scalar, we have ((xT. A . x)^2) * xT = bT.
Therefore,
(((xT. A . x)^2) * xT).A.(((xT. A . x)^2) * x)=bT.A.b,
which means
(xT. A . x)^5=bT.A.b,
or
(xT. A . x)^2=(bT.A.b)^(2/5)
Now, from the original equation ((xT. A . x)^2) * x...
Re: vector 3
i) you mean
...find \[\nabla XF\] and \[\nabla F\]?
ii) If G=f1(x,y) +f2(x,y)j. Then that G is conservative is equivalent to
df2/dx=df1/dy
G indeed satisfies this requirement by a trivial checking;
iii) \[ \Phi =y + x^2 y^3\]
Re: vectors 2
This question is not so hard to answer, but it's a little tedious as you have to take the derivatives of V(x,y,z). Well, you have to do this by yourself or you may appeal to Mathematica. I am sure it'll help.
Answer:
i) (-dV/dx,-dV/dy,-dV/dz);
ii) in the direction =...
cannot understand the equations. For example, the left hand side of
min(X,Y) = P(X<x) U P(Y<y)
is another random variable min(X,Y), while, in the right hand side, P(X<x) is a cummulative distribution function and so is P(Y<y). Both P(X<x) and P(Y<y) are function, but not random variables.
Re: probability question
Sorry for the late response (too busy).
Yes, you are right. Here is the right place to apply that theorem, but you also take the sum directly using a program.
Re: Riccati equation
Riccati equations are specific ODEs as the following:
dy/dx=f(x) + g(x)y + h(x)y^2.
These equations usually cannot be solved by closed forms, except for very special (f(x),g(x),h(x)).
Look at this simple equation:
dy/dx=x^2 + y^2,
which is a classical Riccati equation...
Re: probability question
This can be solved by a binomial consideration.
The whole process consists of 1000 bernoulli tests, each of which amounts to pick up a screw and check whether it's good or bad. Let's assume that the probability for a good screw is p while for a bad screw q. p+q=1. We'll...
Re: Repeated Trial question
Here is what happened.
First of all, what are the two numbers, 100 and 1000? These two numbers play only one role, that is, they serve to determine the probabilities of bad (p=0.1) and good (q=0.9) bulbs in the box. None of them has anything to do with the number of...
Re: Repeated Trial question
There is a confusion of notions. That is about P(2D). I suppose the P(2D) in "P(2D) = 99/999 = 0.099099..." means the probability about the second, while P(2D) in "P(2D) = [C(n,m)]*[p^m]*[(1-p)^(n-m)]..." means the probability of both the first and the second...
Honestly, I am not so confident about this. Here is what I got.
Set
V1=0.5
V2=2
Then deltaV=V2-V1=1.5
Now,
p(mean value)=Integrate[V*p, {V,V1,V2}]=Integrate[2000V^-13, {V,V1,V2}]
=(2000/12)*(V1^-12 - V2^-12)
=(500/3)*(0.5^-12 - 2^-12)
Then,
W=p(mean value)*deltaV=(500/3)*(0.5^-12 -...
Oh, well, "functional analysis" can hardly be an intuitive topic, but it talks about topological (normed) spaces and operators defined in them. There might be some simple examples, though, which might shed you a light when you walk in dark. There are a few fundamental theorems in functional...
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