Wikipedia:Reference desk/Archives/Mathematics/2011 September 28

= September 28 =

Power Spectrum Estimation
I have some measurements and I am trying to get a hold of its power spectrum. The measurements are discrete so the estimation of the power spectrum I get is also discrete. My question is, what is a good method to interpolate the values in between the frequencies for power spectrum estimation? I only need interpolation, no extrapolation. This is just an opinion question to see what others more experienced around here think. For example, I may only get 128 frequencies so should I use just linear interpolation, nearest neighbor method, cubic splines, etc? Cubic splines are very sensitive to the boundary conditions but will be very smooth. Which b.c. make sense here anyway (natural, clamped, etc.)? On the other hand linear interpolation is faster and that is what we all see when we plot it (graphing calculators just linearly interpolate between the nodes). Nearest neighbor is dumb but is the fastest and may be justified if my frequencies are very close-by (very small df). I am using Welch's method so I am trying to use a small window (to take more averages and lower the variance) but that lowers my resolution. What do you guys think? Thanks! -  Looking for Wisdom and Insight! (talk) 05:03, 28 September 2011 (UTC)
 * Why interpolate at all? What are you going to use the interpolated values for? Looie496 (talk) 05:52, 28 September 2011 (UTC)
 * Crudely, couldn't you just approximate each integral by a Riemann sum based on your sample?  Sławomir Biały  (talk) 11:37, 28 September 2011 (UTC)


 * The typical way of getting an interpolated power spectrum is a technique called zero-padding. After windowing (or whatever), take each set of data, remove the mean and extend the data set by adding a large number of zeros on the end.  Those zeroes will contribute nothing to the Fourier sum / integral, and so they don't affect the values you receive, but it will force the discrete Fourier transform to return values at intermediate points.  On platforms that implement the fast Fourier transform algorithm, there is a computational advantage to choosing the total number of points to be an exact power of 2.  The power spectrum produced in this way will be smoothed in a natural fashion, but remember that smoothing doesn't add any information beyond what you already had.  For example, one needs to know that the uncertainty in peak locations is still comparable to the spacing of frequencies in the original transform.  Dragons flight (talk) 17:15, 28 September 2011 (UTC)

Integral Truths and mein eigenes Problem
I have  a  problem  involving  an  integro differential equation. It goes :



\frac{d}{dt}y(t) + \int_{0}^t f(u)\,du = sin (t), \qquad y(0) = 1, \qquad \frac{d}{dt}y(0) = 0 $$

Now I  was  advised  that  the  assumption  is  that  $$ y = f(t) $$,  and  to  isolate  the  integral  by  rearranging  the  equation,  such  that  I  got  :



\frac{d}{dt}y(t) - sin (t) = \int_{0}^t f(u)\,du $$

Next I  was  told  to  take  the  derivative  of  both  sides,  and  was  told  that  this  meant  that since



\int_{0}^t f(u)\,du = \frac{d}{dt}y(t) - sin (t) $$

that



y = \cos (t) - \frac{d^2y}{dt^2} $$

But when  I  carry  out  the  integration  on $$ \int_{0}^t f(u)\,du $$,  according  to  the  Fundamental  Theorem  of  Arithmetic,  I  get :   $$\int_0^t f(u)\, du = f(t)-f(0).$$, or have  I  done  the  wrong  thing,  since  I  thought  that  taking  the  derivative  of  an  integral  is  where  they cancel  each  other  out. But also,  if  it  is  assumed  that
 * $$ y = f(t) $$, what  is
 * $$ f(u) $$ ?

If I  assume  that,  then  I  think  I  get  $$ y - 1  = \cos (t) - \frac{d^2y}{dt^2}  $$ since if   $$ y = f(t) $$, then  $$ f(0) = y(0) = 1 $$,   as  stated  to  begin  with. If I  do  say  that  $$  y = \cos (t) - \frac{d^2y}{dt^2}  $$,  this  seems  correct,  since  it  says  that   $$ y(0) = 1 $$,  and  indeed   $$  y(0) = \cos (0) - \frac{d^2y(0)}{dt^2}  $$, and  since  if  \qquad \frac{d}{dt}y(0) = 0,  being  the  first  derivative,  so  will  be  the  second,  since  that  will  just  be  the  derivative  of  zero,  so  this  seems  right.

To solve   this,  I  tried  LaPlace  once  more,  and  got $$ y = \cos (t) - \frac{d^2y}{dt^2}  $$  implies  that  $$ \cos (t)  = y + \frac{d^2y}{dt^2}  $$, so  that Laplace of  $$ \cos (t)$$  = Laplace of $$ y + \frac{d^2y}{dt^2}  $$, which means  after  carrying  out  all  the  steps  that


 * $$ y (s)  =  \frac{s}{(s^2 + 1)^2}  +  \frac {s}{s+1} $$,

which I  thought  was
 * $$ cos(t) + t sin(t)$$ ,

but it  does  not  seem  to work when I check it. Now I  had  two  different  books  giving  two  different  results,  such  that  in  one  of  them  my  answer  is  $$ cos(t) + t sin(t)$$,  but  in  another  it  is  given  as


 * $$ cos(t) + .5t sin(t)$$, which  does  actually  work.

I also  have  a  second  one  involving  the  inverse power method  for finding the eigenvalues  of  a  four by four matrix,  and  how  to  programme MatLab™  into  doing  what  we  want.

We programmed  a  4 × 4  into  MatLab™  for  loops  to  allow  it  to  give  us  the  array  showing  the  four  eigenvalues  of  the  matrix,  and  now  we  need  a  way  to  carry  out  the  inverse power  method  to  verify  the  middle  two. We had  :

% Powers of matrices n = 4;

% pick a starting vector and a matrix x0 = ones(n,1); B = [2 1 7 6; 2 0 5 6; 7 8 8 3; 9 6 3 4] P = transpose(B) A = B*P [V,D] = eig(A) pause

% first the dominant one x = x0; for I = 1:10 x = A*x; x = x/norm(x); end v1 = x l1 = dot(x,A*x)/dot(x,x) pause

% then the other end of the spectrum x = x0; B = A-402.2821*eye(n); for I = 1:40 x = B*x; x = x/norm(x); end v2 = x l2 = dot(x,A*x)/dot(x,x)

C = inv(A-402.2821*eye(n));

for I = 1:40 x = C*x; x = x/norm(x); end

This gave  us  :

B =

2    1     7     6     2     0     5     6     7     8     8     3     9     6     3     4

P =

2    2     7     9     1     0     8     6     7     5     8     3     6     6     3     4

A =

90   75    96    69    75    65    72    57    96    72   186   147    69    57   147   142

V =

0.6704   0.0834    0.6186    0.4012   -0.6837   -0.3143    0.5742    0.3226   -0.2244    0.6642   -0.2735    0.6585    0.1810   -0.6731   -0.4614    0.5489

D =

0.0038        0         0         0         0   15.0123         0         0         0         0   65.7018         0         0         0         0  402.2821

v1 =

0.4012   0.3226    0.6585    0.5489

l1 =

402.2821

v2 =

-0.5461   0.7183   -0.2871    0.3216

l2 =

6.9144

And this  last  bit  was  my  attempt  to  find  one  of  the  middle  values,  that  is,  to  do  it  by  hand  and  confirm  it  was  15.0123,  and  or  the  other  middle  one 65.7018. I am  not  sure  what  we  were  to  do  to  get  the  computer  to  work  these  out.Thank You. Chris the Russian Christopher Lilly  05:38, 28 September 2011 (UTC)
 * Please give your two questions separate headerlines such that they can be answered separately. Otherwise this is going to be messy. Bo Jacoby (talk) 07:31, 28 September 2011 (UTC).
 * Is it me or is there a minus missing in line 2, where the integral has been moved to the right hand side? Grandiose (me, talk, contribs) 17:23, 28 September 2011 (UTC)

Faith and Begorrah ! You are  right - I cannot understand how I got that wrong ! This is the trouble with trying  to  rearrange equations. Although in  this  case  it  was  a  typo  on  my  part only  in  writing  it  all  down  for  this  question on Wikipedia™. This means  then  that

sin (t) - \frac{d}{dt}y(t) =  \int_{0}^t f(u)\,du $$  which means

y = \cos (t) - \frac{d^2y}{dt^2} $$, and this  is  what  I  had  anyway,  so  the  only  mistake  I  made  was  a  typo  for  the  first  rearrangement,  but  other  than  that,  since  the  mistake  was  not  repeated  with  the  differentiation,  it  all  worked  out. This I  have  since  solved  to  my  satisfaction,  intending  to  use  the  solution
 * $$ cos(t) + .5t sin(t)$$, which,  as  stated,  does  actually  work. The  only problem I  have  now  is  simply understanding  why  the  integral  becomes  the  function
 * $$ y(t)$$

upon differentiation,  and  what  the  differentiation  of  an  integral  really  means.

As for  the  computer  one,  sorry  to  put  two  on  the  same  thing,  but  there  it  is. We are  trying  to  find  out  how  to  find  the  eigenvalues  of  the  matrix  with  the  MatLab™,   and  how  to  use  the  inverse power  method  to  do  so,  or  that  is,  how  to  find  all  four. Thank You.Chris the Russian Christopher Lilly  22:36, 28 September 2011 (UTC)  — Preceding unsigned comment added by Christopher1968 (talk • contribs) 22:31, 28 September 2011 (UTC)

polar range
Why do we need define a range for polar cordinate system?Exx8 (talk) —Preceding undated comment added 10:49, 28 September 2011 (UTC).


 * I think it's because a polar representation is otherwise non-unique. If the tuple $$(x,y)$$ for real $$x,y$$ represents a Cartesian coordinate, the set of such tuples maps to $$\mathbb{R}^2$$ bijectively.  If it is interpreted as a polar coordinate, for radius x and angle y, $$(x,y) = (x+n2\pi)$$ for all integer $$n$$.--Leon (talk) 16:41, 28 September 2011 (UTC)


 * I'm not sure that your notation's right there. In polar coordinates, the point $$(r,\theta)$$ is the same as the point $$(r,\theta + 2\pi k)$$ for each whole number k. You can go around another turn and get back to where you started. (Like 6 a.m. and 6 p.m. have the same representation on an analogue clock, even though the hour hand has made a full turn extra when it's 6 p.m.) We sometimes specify a range for the angle &theta; but we don't have to. It's often best not to, and it doesn't require much more work. For example, the point $$(x,y) = (1,0)$$ in cartesian coordinates is given by $$(r,\theta) = (1,2\pi k)$$, where k is any whole number. — Fly by Night  ( talk )  21:00, 28 September 2011 (UTC)

collide with earth?
earth orbits around the sun; relative to the speed conjunction of other comets and asteroids will one near moon size ever collide with earth? by mathematics calculation; the the eclipse ellipse of other large masses and earth over time how will it  project over 3 light years in very high speed animation? (despite the sun become a red giant) — Preceding unsigned comment added by 207.6.211.175 (talk) 19:32, 28 September 2011 (UTC)


 * That's unlikely unless the solar system enters a new phase of instability, similar to when Jupiter and Saturn reached a 2:1 orbital resonance, triggering the Late Heavy Bombardment A possible scenario involves Mercury, whose orbital parameters aren't far from a region of instability. Given the current state of the solar system, taking into account the fact that it is chaotic on a time scale of tens of millions of years, one can show that Mecury can collide with Venus or fall into the Sun within a billion years. When that happens, the solar system will reconfigure itself, which can e.g. involve Mars being ejected from the solar system, or possibly other catastrophic events. Count Iblis (talk) 20:09, 28 September 2011 (UTC)


 * See here for details. Count Iblis (talk) 20:44, 28 September 2011 (UTC)


 * Wait, the solar system is chaotic already within tens of millions of years? Then what's the deal with everyone trying to predict whether when the Sun turns to a red giant (much later) it will eat Earth or not?  If the orbits are so chaotic, the problem is not the Sun, we don't even know where Earth will be at that time.  &#x2013; b_jonas 10:17, 29 September 2011 (UTC)
 * The general features of the Earth's orbit are fairly stable. It's chaotic in as much as we can't really determine where in its orbit the Earth will be in 10 million years, but we can be reasonable sure its orbit won't be significantly different. The only exception is close interactions with other large bodies, like Count Iblis describes, but I don't think they are considered particularly likely. --Tango (talk) 12:00, 3 October 2011 (UTC)
 * That's what I thought, but then Count Ibis mentioned a possibility of Mars being ejected from the Solar System, which sounds rather scary. &#x2013; b_jonas 20:06, 4 October 2011 (UTC)

epsilon-delta
hey, it's me again. So today I was doing some volunteer tutoring for younger students. One student asked me a question about the derivative of $${x}^{1/3}$$ at 0: he asked since theoretically it doesn't exist, why does his graphing calculator give a value for the derivative there. I explained that his calculator doesn't differentiate the way we do, but calculates a bunch of $$\frac{\Delta{y}}{\Delta{x}}$$ for very small ∆x; I was wondering idly how I would turn this into a rigorous (ε,δ) argument; this is what I have: Suppose ε is the smallest value >0 that your calculator can handle, and also suppose this value is the same when it calculates your Δx and Δy. Then your calculator is saying that for every ε'≥ε, $$|\frac{f(x)-f(a)}{x-a}-L|<\epsilon'$$ as long as $$|x-a|<\delta$$, for some δ≥ε (I know this isn't how a calculator actually thinks), but the calculator can't choose any epsilon less than ε, so as long as it's true for ε',δ≥ε, the calculator says it exists. I know this is roughly right but it's not really clear and not really specific. Can someone please help me refine this argument (keeping to epsilon's and delta's though I'm sure there are plenty more informal you could do it)? — Preceding unsigned comment added by 24.92.85.35 (talk) 22:51, 28 September 2011 (UTC)
 * I imagine the calculator isn't doing anything of the sort but is using an estimate for the derivative using a formula from numerical analysis. A typical method to estimate f′ (a) would be to find (f(a+e)-f(a-e))/2e which would have error, assuming f is reasonably well behaved, proportional to e2. The fact that f is not well behaved is, I would guess, why the calculator is giving an answer instead of a divide by zero error. What is actually happening though is anybodies guess since the calculator is running proprietary software written by programmers with confidentiality agreements. Symbolic algebra systems exist that can do this kind of thing exactly and it probably won't be too long before calculators include such features, assuming they don't already.--RDBury (talk) 04:02, 29 September 2011 (UTC)