Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**juriguen**- Replies: 0

Hi all

I am trying to understand the following problem a bit further, but I lack algebra background. Say that I have a noisy sequence of measurements (a vector) of length N (say it is a column vector, Nx1), that can be written as:

where is an ideal vector of samples, and is additive white Gaussian noise. So, the components are iid, of zero mean and have variance .

Now, we filter the noisy samples using:

where is a matrix of size (PxN). The rows of the matrix are orthogonal, but not orthonormal, and so the filtered noise becomes coloured Gaussian noise.

Moreover, suppose that now I form a Toeplitz matrix using the filtered measurements

:The matrix can of any size ((P-M+1)x(M+1)), not necessarily square. And it is also not a condition that P>2M, even though the example I wrote is like that.

The idea is that, now, I have the following:

It is easy to prove that matrix

is diagonal, but not a multiple of the identity matrix. In fact, assume that the matrix is such that the diagonal values of are symmetric and decreasing from the ends to the central value.Now, consider that doing SVD on

yields non-zero entries as singular values of . This means that the null-space of is spanned by the columns of related to the (M-K) zero entries in .But, does anyone know how the SVD of

would relate to that of ? If the noise was not coloured, it is straightforward to see that the null-eigenspace is preserved. But I don't know what happens with coloured noise.Thanks in advance

Jose

Hi!

I think there's only 1 solution, and I wrote it down, z = -W(-1)

Right?

Hi there!

I actually got help from quickmath.com, and then checked what did the result mean. The idea is to use what is called the Lambert W function, which, by definition (Wikipedia) allows to express any complex number as:

Then, for your equation you have to do:

But any number can be written using the W function, so:

By equating both expressions, then:

which actually has |z| = 1.3745, so the condition is satisfied.

The other problem I cannot get it yet. I will keep trying

Ok, so I would say you have to do this numerically. First plot both functions, to have an idea. then use Newton-Rapson's method to obtain when they intersect. For instance, to calculate the area in the interval [0, 3pi/2], first lets find where they cross:

Say we want to find

Using N-R, from a guess xn, we obtain a better guess doing:

Choose as x0 = 4.5, and a few iterations give x = 4.4934

I have calculated only the area between [0, 4.4934], which is:

Hope this helps!

Jose

Are you sure about the distances of the antinode to the sources?

I don't know exactly what does 2nd order antinode mean. If it would be the 1st antinode in the 2nd harmonic of the standing wave, the following should be true:

And it doesn't make sense.

If the 2nd order antinode means the 2nd antinode for a certain harmonic [n], then the following holds:

Or

And neither of both makes sense, since [n] should be an integer.

Hi!

Lets see if this works. We have the following relation between the wave speed [c] (m/s), the wavelength [lambda] (m) and the frequency [f] (Hz):

Then, for air you use c = 344 (m/s):

Now, for steel you can calculate:

which gives a speed in steel equal to:

We now have all the information to answer the time it takes to travel s = 84 (m):

(a) For air

(b) For steel

Hi!

For the first question, be careful, because the cosine rule is not like that... (maybe dividing all the terms in the rhs rather than only the last one) Calling the triangle sides a = AB, b = BC and c = CA, and the angle between a and c being alpha:

For the second question, I attach a diagram for what I understand it is described. The cosine rule can again be applied to the diagram, using exactly the same notation as above. Then we get:

Then, the distance is 3.32 + 3.2 + 2.5 = 9.02

Cheers

Jose

Hi!

I am sure there will be an easier way to solve it, but I will try to say how would I do it.

I am going to assume you have an original point in the XY plane, heading north, which I will consider in the negative part of the X axis. Then, the original coordinates would be (-x, 0, 0), being the distance to the origin of coordinates equal to x.

I also consider that rotation from north and clockwise is keep the point in the XY plane, so rotate in Z an angle q1 = -33. And then doing the same for a "dip"angle, I assume it is rotate with respect to X, an angle q2.

First, you have to calculate the new coordinates of the point, for example from http://www.siggraph.org/education/materials/HyperGraph/modeling/mod_tran/3drota.htm

1. Rotation in Z:

2. Likewise, rotation in X:

Yo can actually find the 2 matrices, multiply one with the other and obtain the total displacement.

Now have point (x'', y'', z''). Then, we have to move it in the direction normal to certain plane, a distance of 50. Ok, the equation characterizing a plane using a vector normal to it, (a, b, c) that passes through point (x0, y0, z0) is:

And the line passing through (x'', y'', z'') normal to the plane can be written using an independent variable t:

In total, you have to calculate what new point (x''', y''', z''') is at distance 50 from (x'', y'', z''):

So the point must satisfy the equation of the line and of the distance.

And that's what I can say with the info you provided

Jose

Hi!

I guess the only angle left to calculate with respect to the base is that formed by the edges of the pyramid. This can be done knowing that its projection into the base has length b, that can be computed from the base length a:

Then, the new angle, the height of the pyramid and b can be related:

Hope this is it!

Jose

Haha, you're welcome!

It is curious that you calculated the area underneath f(y) correctly, and not for f(x). That for f(y) is much more difficult!

Jose

Hi!

The area underneath the first curve is equal to 1:

And the domain is the mapping between x and y:

Jose

Hi!

You can use Taylor series approximation around a = 1 for example, or any other a closer to 2:

Then, at x = 2

The result given is using up to 1/10. It is still not very precise, since ln(2) = 0.6931, but you can keep on adding and substracting fractions to get it more accurate.

I guess that if you approximate it closer to 2, for example a = 1.9, it will converge faster, but then you have to use powers in your calculator (or do many more multiplications )

Actually, I think that approximating around 1 is the only solution to only use (+, -, ×, ÷), because if doing it around a = 1.9, you need to know f(1.9) = ln(1.9). However, with a = 1, I guess ln(1) = 0 is well known

Jose

Sure!

I actually hadn't realized how you had used that formula, and not I see it, so go for both

Jose

Hi!

You have to differentiate in order to obtain the pdf out of the cdf!

I think then they results match!

Jose

Hi!

60mm = 0.06m

100mm = 0.1m

area = h x w = 0.06 x 0.1 = 0.006 m^2

Jose

Hi!

If you would like to understand HOW to calculate f(y), from Wikipedia [http://en.wikipedia.org/wiki/Probability_density_function]:

The probability contained in a differential area must be invariant under change of variables y = g(x)

This was for the simple case of a monotonic function g. There is a general expression in Wikipedia as well.

If you would like to know WHAT does it mean, I can only provide an example: Random Number Generation. Rather than explaining it myself, I think this other article will be useful [http://en.wikipedia.org/wiki/Inverse_transform_sampling]

Jose

Hi!

I would say that, if for example you have:

you can write

So, changing c and b effectively moves the origin of coordinates for the function

The same happens with

And if you check the derivative:

you see that the parameter a controls the rate of change of y.

Then, for the exponentiation function, you can check the derivative as well

In this case, for 0 < a < 1, increasing a makes the function to decrease slower (lna < 0, and increasing a makes the lna to be smaller in module. But be careful, because in this case the function itself is decreasing). And for 1 < a < inf, increasing a makes the function to increase faster (lna > 0, and increasing a makes lna to grow).

Hope it helps!

Jose

Hi!

I guess you need the steps, but I cannot figure them out now

Sometimes, the result can give some clue, even though it's not the case for me now. Anyway, Mathematica says:

That's a weird result, though! Hope you get luckier than me

Jose

You're welcome

Jose

Hi

The first one I would say:

So every requirement holds.

The second:

Hope it helps!

Jose

Hi!

If you want the maximum of the modulus, first of all you should calculate the absolute value of g:

After that, you can differentiate that rather ugly expression , solve it equal to 0 and find for what k is maximum

Jose

Hi!

I think in this website they explain congruences very well http://primes.utm.edu/glossary/xpage/Congruence.html

Suppose a, b and m are any integers with m not zero, then we say a is congruent to b modulo m if m divides a-b. We write this as

a = b (mod m).

For your example, use the transitive property: If a = b (mod m) and b = c (mod m), then a = c (mod m).

Then, lambda = 0.25 (mod 23) and 0.25 = 6 (mod 23), then lambda = 6 (mod 23). Where the second is true because 23 / (0.25 - 6) = -4

Jose

Hi!

I think I got the first one:

1.

Jose

Hi!

Apparently, according to http://en.wikipedia.org/wiki/Sum

Moreover:

where Bk is the kth Berboulli number.

And in http://en.wikipedia.org/wiki/Bernoulli_number

Jose

Yes, thanks Macy!

I like the other approaches You can always learn many new things in this forum, even if you got the correct answer!

Jose