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# GCD = 1 and harmonic numbers, what is the exact asymptotic?

I am looking for the exact asymptotic for this partial sum:

$$a(N) = \sum_{n=1}^{n=N}\sum_{k=1}_{GCD(n,k)=1}^{k=n*m} \frac{1}{k}$$

where $m$ is some integer $1,2,3,4,5,...$

My guess was that since this is closely related to the Euler totient: $$\phi(N) = \sum_{n=1}^{n=N}\sum_{k=1}_{GCD(n,k)=1}^{k=n} 1$$

it should be a combination of the average order $\zeta(2) = \frac{\pi ^2}{6}$ of the Euler totient, and the asymptotics for the partial sums of the Harmonic numbers, and therefore:

$$a(N) \sim \frac{\int (\log (m n)+\gamma ) \, dn}{\frac{\pi ^2}{6}}$$

$$a(N) \sim \frac{n \log (m n)+\gamma n-n}{\frac{\pi ^2}{6}}$$

would be a good approximation. However, it appears that:

$$a(N) \sim \frac{\int (\log (m n)+1.988\gamma ) \, dn}{\frac{\pi ^2}{6}}$$

$$a(N) \sim \frac{n \log (m n)+1.988\gamma n-n}{\frac{\pi ^2}{6}}$$

is better. But where does the factor $1.988$ come from, and should it be there at all?

Plotting them it looks like a good fit:

where the staircase is $a(N)$ for the case $m=1$ and the yellow line is the asymptotic.

Included is the overestimate (green curve):

$$\frac{n \log (m n)+(1.988+1)\gamma n-n}{\frac{\pi ^2}{6}}$$

and the underestimate (red curve):

$$\frac{n \log (m n)+(1.988-1)\gamma n-n}{\frac{\pi ^2}{6}}$$

I don't know if they really are over- and underestimates. I am just guessing. For small $N$ they are not.

Anyways, differentiating we have for the overestimate:

$$\frac{\log (m n)+(1.988+1)\gamma )}{\frac{\pi ^2}{6}}$$

and the underestimate:

$$\frac{\log (m n)+(1.988-1)\gamma )}{\frac{\pi ^2}{6}}$$

Plotting the staircase $a(N)$ minus the asymptotic $\frac{n \log (m n)+1.988\gamma n-n}{\frac{\pi ^2}{6}}$ with conjectured bounds from differentiated over- and underestimates:

Ultimately I am interested in the von Mangoldt function:

$$\Lambda(N) = \lim_{m-->\infty} \sum_{n=1}^{n=N}\sum_{k=1}^{k=n*m} \frac{b(G(n,k))}{k}$$

and the Chebyshev function. In the formula the sequence "b" is the Dirichlet inverse of the Euler totient function.

There might be an error in the $\Lambda(N)$ formula above, but expressed simpler, the von Mangoldt function is:

$$\Lambda(n) =\sum_{k=1}^{\infty} \frac{b(GCD(n,k))}{k}$$

where again the sequence "b" is the Dirichlet inverse of the Euler totient function.

Some related Mathematica code:

(*Det här programmet ger asymptoter för alla triangulära GCD matriser \
för alla "m"*)
nn = 80;
q = 10;
a1 = Table[Sum[1/(n*q), {n, 1, Floor[k]}], {k, 0, nn, 1/10}];
a2 = Table[(Log[k] + EulerGamma)/q, {k, 0, nn, 1/10}];
Show[ListLinePlot[a1], ListLinePlot[a2]]
ListLinePlot[Sign[N[a1 - a2]]]

a1 = Table[Sum[1/(n), {n, 1, Floor[k]}], {k, 0, nn, 1/10}];
a2 = Table[Log[k + EulerGamma] + EulerGamma, {k, 0, nn, 1/10}];
Show[ListLinePlot[a1], ListLinePlot[a2]]
ListLinePlot[Sign[N[a1 - a2]]]

a1 = Table[Sum[1/(n), {n, 1, Floor[k]}], {k, 0, nn, 1/10}];
a2 = Table[Log[k - EulerGamma] + EulerGamma, {k, 0, nn, 1/10}];
Show[ListLinePlot[a1], ListLinePlot[a2]]
ListLinePlot[Sign[N[a1 - a2]]]

b = Table[DivisorSum[m, # MoebiusMu[#] &], {m, nn}];
m = 1;
Monitor[a1 =
Table[Total[
Table[If[GCD[n, k] == 1, 1/(k*q), 0], {k, 1, m*n}]], {n, 1/q, nn,
1/q}], N[n]];
Monitor[a1 =
Table[Sum[
If[GCD[n, k] == 1, b[[GCD[n, k]]]/(k*q), 0], {k, 1, m*n}], {n,
1/q, nn, 1/q}], N[n]];
a1[[1 ;; 4*q]]
ListLinePlot[{Accumulate[a1] -
Table[-n + (1.988)*EulerGamma n + n Log[m n], {n, 1/q, nn, 1/q}]/
Zeta[2]/q, -Table[(1.988 - 0.8) EulerGamma + Log[n*m], {n, 1/q,
nn, 1/q}]/Zeta[2]/q,
Table[(1.988 + 0.8) EulerGamma + Log[n*m], {n, 1/q, nn, 1/q}]/
Zeta[2]/q}]
ListLinePlot[{Accumulate[a1],
Table[-n + (1.988 - 1)* EulerGamma n + n Log[m n], {n, 1/q, nn,
1/q}]/Zeta[2]/q,
Table[-n + 1.988* EulerGamma n + n Log[m n], {n, 1/q, nn, 1/q}]/
Zeta[2]/q,
Table[-n + (1.988 + 1)* EulerGamma n + n Log[m n], {n, 1/q, nn,
1/q}]/Zeta[2]/q}]
ListPlot[Accumulate[a1] -
Table[-n + 1.988* EulerGamma n + n Log[m n], {n, 1/q, nn, 1/q}]/
Zeta[2]/q]

Here is a modest contribution that can perhaps initiate a discussion.

We compute the asymptotics of $$A(N) = \sum_{n=1}^N \sum_{k=1 \atop (n,k)=1}^n \frac{1}{k}.$$

Introduce $$q(N,k) = \sum_{n=1\atop (n,k)=1}^N 1 \sim N \frac{\varphi(k)}{k}.$$

Rewrite the sum in terms of $q(N,k),$ getting $$\sum_{k=1}^N \frac{1}{k} \sum_{n=k\atop (n,k)=1}^N 1 = 1 + \sum_{k=1}^N \frac{1}{k} (q(N,k) - \varphi(k)).$$ The one term in front compensates for the fact that when $k=1$ we have $$\sum_{n=1\atop (n,k)=1}^N 1 = q(N, k)$$ and not $q(N,k)-1.$ Now switching to asymptotics we obtain $$\sum_{k=1}^N \frac{1}{k} \left(N \frac{\varphi(k)}{k} - \varphi(k)\right).$$ We will use the Mellin-Perron formula to predict the first term of the asymptotics (a rigorous proof would include a bound on the remainder integral).

Recall the Mellin-Perron summation formula: $$\sum_{k=1}^{n-1} \lambda_k + \frac{1}{2} \lambda_n = \frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} L(s) \frac{n^s}{s} ds \quad\text{where}\quad L(s) = \sum_{k\ge 1} \frac{\lambda_k}{k^s}.$$

Now the well-known $$\sum_{d|n} \varphi(d) = n \quad \text{immediately implies}\quad \sum_{k\ge 1} \frac{\varphi(k)}{k^s} = \frac{\zeta(s-1)}{\zeta(s)}.$$ Therefore $$\sum_{k\ge 1} \frac{\varphi(k)/k^2}{k^s} = \frac{\zeta(s+1)}{\zeta(s+2)} \quad\text{and}\quad \sum_{k\ge 1} \frac{\varphi(k)/k}{k^s} = \frac{\zeta(s)}{\zeta(s+1)}.$$

Thus by Mellin-Perron, $$A(N) \sim N\times\operatorname{Res}\left(\frac{\zeta(s+1)}{\zeta(s+2)} \frac{N^s}{s}; s=0\right) - \operatorname{Res}\left(\frac{\zeta(s)}{\zeta(s+1)} \frac{N^s}{s}; s=1\right).$$ Computing the residues we thus obtain $$A(N) \sim \frac{6}{\pi^2} N\log N + \frac{6}{\pi^2} \left(\gamma - \frac{6}{\pi^2} \zeta'(2)-1\right) N.$$

This approximation is quite good, for example $A(1000) \approx 4291.209545$ and the approximation gives $4288.884325$. For $A(2000) \approx 9423.976478$ we get $9420.534562.$ Even more impressive, $A(5000)\approx 26339.64158$ and the asymptotic formula gives $26336.52625$.

Observe that $$1.988\gamma-1 \approx 0.147504742 \quad\text{and}\quad \gamma-\frac{6}{\pi^2}\zeta'(2)-1 \approx 0.147176658,$$ confirming the numeric result from the original query.

We really needed Mellin-Perron because the pole of the first sum term is a double pole and Wiener-Ikehara only applies to simple poles, like the one in the second term.

format: plain


text:
### Home > A2C > Chapter 9 > Lesson 9.3.2 > Problem9-154

9-154.

Find the distance between each pair of points.

1. $\left(2, 3\right)$ and $\left(4, 7\right)$

Use the distance formula to find the distance.

$\textit{d}=\sqrt{(4-2)^{2}+(7-3)^{2}}$

$\textit{d}=2\sqrt{5}$

1. $\left(2, 3\right)$ and $\left(x, y\right)$

See part (a).


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# Simulation for CI, Reflecting Quantiles

I was reading Hogg/Craig and my course notes and encountered a subsection on simulation to find out Confidence Intervals. The method expressed is to generate a random sample, calculate the MLE, repeat a billion times, find upper and lower $\alpha/2$ quantiles for the empirical distribution of the MLE. (Call them $l$ and $u$.)

Then, it said that a crude CI for the MLE would be $\hat{\theta}\pm z_{\alpha/2} SE(\hat{\theta})$. I understood everything till her. Now, they said, that a better estimate would be :

$$\hat{\theta} \in (2\hat{\theta}-u,2\hat{\theta}-l)$$ and added that this is called "reflecting quantiles". I don't appreciate

• Why this is a CI in the first place?
• How is this better?
• What is the rationale behind this?
-


text:
Question: R package for Meta-Analysis of CHIP-Seq data
0
2.1 years ago by
rezapanahi2220 wrote:

hi  everyone

i am new in chip-seq analysis

i did a chip-seq analysis by command line(UBUNTU)

and i want to do a Meta -analysis to integrative all study that dirived from my analysis

but i do not know which package do it ????

thanks

chipseq ubuntu meta-analysis • 556 views
modified 2.1 years ago • written 2.1 years ago by rezapanahi2220
Answer: R package for Meta-Analysis of CHIP-Seq data
1
2.1 years ago by
Aaron Lun24k
Cambridge, United Kingdom
Aaron Lun24k wrote:

Your question is too vague. How did you do the "analysis"? Peak-calling? What kind of "meta-analysis" do you want to do? If you're trying to look for differences between samples, googling "differences chip-seq bioconductor" will already give a number of relevant hits.

i did a peak-calling analysis  by macs2.

i want to (meta-analysis) look for differences between samples.

What software or methods can I use?

Answer: R package for Meta-Analysis of CHIP-Seq data
1
2.1 years ago by
Rory Stark2.8k
CRUK, Cambridge, UK
Rory Stark2.8k wrote:

If you want to use your MACS2 peaks, the DiffBind package offers a straightforward path to identifying differentially bound peaks. The csaw package is another alternative, particularly if your dataset requires more sophisticated normalization.  In either case the original BAM files are required.

Cheers-

Rory

Answer: R package for Meta-Analysis of CHIP-Seq data
0
2.1 years ago by
rezapanahi2220 wrote:

i mean that i did chip-seq analysis data with command line on ubuntu

i used the MACS2 peak caller for obtain the peakset files .

now i want to finde a package that integrate my peakset data (obtained from macs2 ) in multiple study .

which tools or package that i used is good ???

i hope that it is simly

cheers


text:
1. Math
2. Statistics And Probability
3. a researcher is investigating the relationship between economic development x...

# Question: a researcher is investigating the relationship between economic development x...

###### Question details

A researcher is investigating the relationship between economic development (x) and level of religiosity (y) in ten countries. (The researcher has interval-level measurements for both variables.) The researcher theorizes that citizens of countries at the lower end of the development scale will profess higher levels of religiosity than will citizens of countries at the higher end of the development scale. As development increases, religiosity decreases. Draw and label four sets of axes, like the one below:


text:
# Slicing irregular spritesheet (automatically?)

If found a cool sprite sheet on the internet but its irregular.

Is there any way besides manually cutting sprites to extract the separate pngs? I need to pack them then again into the Texture Atlas under proper names.

Maybe theres some smart online tool that uses deep learning or something like that?

• This sheet looks like it's laid out on a regular grid. So, if you're OK having some objects placed as multiple tiles, you should be able to just split this using its grid frequency. – DMGregory Jul 30 '19 at 14:35
• That sheet isn't irregular, it's 32x32 with some objects consisting of more than one tile. Source: I happened to work with this exact tileset before. – Philipp Jul 31 '19 at 8:39

Since you found the picture online, without any json file attached or any other information, its not possible to cut this into multiple sprites automatically. You'd have to do it by hand.


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# Change in angular velocity of the following system

The pink spheres cause an acceleration of 1 cm per second squared towards themselves.

The aqua spheres cause an acceleration of 2 cm per second towards themselves.

All spheres have the same mass.

The spheres are attached to a massless, yet mysteriously solid perfectly circular green ring, equidistant from the closest spheres (pi/2 from one another).

The 10 cm diameter green ring is rotating at a velocity of 1 cm per second.

Everything except the pedestal experiences an acceleration of 10 cm per second squared downwards (in the direction from the bearing towards the pedestal). The pedestal is fixed in place (does not accelerate downwards).

The ring is held up from the pedestal by a post with a frictionless bearing.

At what rate does the velocity of the ring change over time if acceleration transmission is instantaneous?

At what rate does the velocity of the ring change over time if acceleration transmission is relativistic?

Last edited:

mfb
Mentor
What is this, a strange homework problem? An attempt to violate conservation of energy?
It is certainly not mathematics, but I'm unsure where to put the thread.
The pink spheres cause an acceleration of 1 cm per second squared towards themselves.

The aqua spheres cause an acceleration of 2 cm per second towards themselves.

All spheres have the same mass.
That doesn't make sense. Acceleration depends on distance, and for the same mass the gravitational force is the same.

What is this, a strange homework problem? An attempt to violate conservation of energy?
It is certainly not mathematics, but I'm unsure where to put the thread.
It's a math question about an exaggerated fictional scenario. I wouldn't call it physics. It will help me understand a concept.
That doesn't make sense. Acceleration depends on distance.....
I thought of adding that later, it should say (although I'm still curious about the mathematical solution with non proportional acceleration!):
The pink spheres cause an acceleration of 1 cm per second squared towards themselves at a distance of 5√2cm.

The aqua spheres cause an acceleration of 2 cm per second towards themselves at a distance of 5√2cm.

The arrows point to "where" (greatly exaggerated) the acceleration is towards in the relativistic scenario.

I'm assuming similar scenarios come up a lot, so thought there would be some ready made formulas to plug values into to determine the acceleration.

Last edited:
mfb
Mentor
I'm assuming similar scenarios come up a lot
Not really.

If you want to use the physical laws of our universe, the correct treatment is general relativity. Calculating that system in general relativity is way beyond your knowledge. There are also no perfectly rigid materials any more.
You can use Newton's laws as approximation, then you can ignore all internal forces and just calculate the net torque on the wheel.

If you want to use the physical laws of our universe, the correct treatment is general relativity. Calculating that system in general relativity is way beyond your knowledge.
Ehh, I want to calculate torque on the wheel using fictional forces to illustrate a point.

I was going to use the following approximation (more or less, pressed for time):

For the shortest acceleration distance:
1) Constrain 0< s < 5√2 cm / c
2) τ = v / 10π
3) calculate s by using 10 sin(π/4 - τs/2)- c s /100 = 0 (c in cm) // ??????????
4) Use s to determine the distance from accelerating body
5) calculate magnitude of acceleration vector
6) multiply by sin (π/4 +τs/2).... now you have a relativistic approximation
....

You can use Newton's laws as approximation, then you can ignore all internal forces and just calculate the net torque on the wheel.
You can't. Instantaneous transmission of forces results in no net torque on the wheel.

Last edited:
mfb
Mentor
Ehh, I want to calculate torque on the wheel using fictional forces to illustrate a point.
The point is pointless. It's like 5 random numbers you see in 5 different places. Sure you can add them, but how would that give any interesting result.
You can't. Instantaneous transmission of forces results in no net torque on the wheel.
As I said, you can ignore all internal forces.... net torque can still come from external forces, like gravity.

Dale
Mentor
2021 Award
At what rate does the velocity of the ring change over time if acceleration transmission is instantaneous?
0

CWatters and Matt Benesi
0
That cracked me up. Thanks for the morning happiness.

The point is pointless.
I'm assuming you know how cool that comment is.
It's like 5 random numbers you see in 5 different places. Sure you can add them, but how would that give any interesting result.
If you input non-random numbers from a similar (actual) physical problem the result is interesting. The results set constraints on what certain experimental setups can detect within their margins of error. It shows that claims that the experiments detected these things are false.

I'll post the results later, and try to explain some of the reasoning behind bringing up this problem in the first place.

mfb
Mentor
Please keep our policy on personal speculation in mind.

Dale
Mentor
2021 Award
That cracked me up. Thanks for the morning happiness.
You are welcome!

If it isn't obvious to you why 0 is correct then I would be glad to explain.

You are welcome!

If it isn't obvious to you why 0 is correct then I would be glad to explain.
lol. Leave it open for someone who doesn't know the answer to figure out.

Please keep our policy on personal speculation in mind.
It's never personal when it's math. It's just right, wrong, or undefined.

Dale
Mentor
2021 Award
lol. Leave it open for someone who doesn't know the answer to figure out.
OK.

I dropped the downward acceleration for the following, just did a ring with 4 spherical masses. You can run this in WxMaxima if you want to play with it.

function usage in WxMaxima is:

fiterall(r,v,mass,accmult,iter,skippy)$with: r= radius of ring body v= tangential velocity of ring mass= "mass" of sphere to calculate acceleration from G (used G because I wanted to check something) accmult= multiplier for acceleration for aqua colored spheres iter= # of iterations to perform the homing function** set it > 100 for decent accuracy skippy= number of iterations between displaying iteration number (you can set this really high if you don't care about tracking calculation times) **each iteration of the homing function reduces error in position of acceleration source Input the following code (between lines of "**************") into WxMaxima for function declaration: ********************************************************************************************************* fpprec:128$fpprintprec:32$fiterall(r,v,mass,accmult,iter,skippy):= [pi:bfloat(%pi),acc1:mass*6.67408b-14*(100b0/r)^2b0, acc2:acc1*accmult, c:299792458b0*100b0, t:v/(2b0*r*pi*2b0), i:1,ss:r*sqrt(2b0)/c,zz: r*2b0*sin(pi/4b0-ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/4b0-ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Short Error (from perfect time): ", float(zz))), smalltime:ss, i:1,ss:r*2b0/c,zz: r*2b0*sin(pi/2b0-ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/2b0-ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Diag. Error (from perfect time): ", float(zz))), diagtime:ss, i:1,ss:r*sqrt(2b0)/c,zz: r*2b0*sin(pi/4b0+ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/4b0+ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Long Error (from perfect time): ", float(zz))), longtime:ss, shortacc:(r^2b0*2b0/((smalltime*c)^2b0))*sin(t*2*smalltime+pi/2), diagacc:(r^2b0*4b0/((diagtime*c)^2b0))*sin(-t*2*diagtime+pi), longacc:(r^2b0*2b0/((longtime*c)^2b0))*sin(-t*2*longtime/c+pi/2), aqua1:(shortacc+diagacc)*acc1-longacc*acc2, aqua2:shortacc*acc2+(diagacc-longacc)*acc1, pink3:acc2*(shortacc+diagacc)-longacc*acc1, pink4:shortacc*acc1+acc2*(diagacc-longacc), total:aqua1+aqua2+pink3+pink4, disp(" "), disp(sconcat(" acceleration: ", float(acc1))), disp(sconcat(" total a: ", float(total))), disp(sconcat(" smalltime: ", float(smalltime))), disp(sconcat(" diagtime: ", float(diagtime))), disp(sconcat(" longtime: ", float(longtime))) ]$
*********************************************************************************************************

Then run it with the following command to get some results:

fiterall(5b0,1b0,1b0,1.000b0,580,190)$At v~.3289948*c, tangential acceleration of the ring is equal to acceleration towards the test masses (when all test masses have the same acceleration towards themselves). Dale Mentor 2021 Award In the Newtonian case the angular acceleration is 0. I guess you are treating the relativistic case. If so then it can only decelerate, not accelerate. The ring is rotating counterclockwise. Assume equal masses, equal acceleration. The point the masses are attracted to is shifted like (greatly exaggerated): We only add the tangential components to the acceleration of the ring. Dale Mentor 2021 Award It cannot happen that way. Any relativistic field will radiate away the angular momentum, not gain more out of nowhere. I wanted to see what relativistic propagation of a gravity like acceleration would do in that scenario. Basically, if the gravity like acceleration propagates at c, and there are no other forces acting on the ring, the ring will experience an angular acceleration if all 4 sources are the same "mass". Maybe frame dragging, or some other effect cancels out the acceleration in real life. Maybe not. I corrected the WxMaxima code: fpprec:128$fpprintprec:32$fiterall(r,v,mass,accmult,iter,skippy):=[ pi:bfloat(%pi), c:299792458b0*100b0, t:v/(2b0*r*pi*2b0), i:1,ss:r*sqrt(2b0)/c,zz: r*2b0*sin(pi/4b0-ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/4b0-ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Short Error (from perfect time): ", float(zz))), smalltime:ss, i:1,ss:r*2b0/c,zz: r*2b0*sin(pi/2b0-ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/2b0-ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Diag. Error (from perfect time): ", float(zz))), diagtime:ss, i:1,ss:r*sqrt(2b0)/c,zz: r*2b0*sin(pi/4b0+ss*t)-c*s, steppy:ss*.1b0, for i thru iter do[zold:abs(zz), if (zz>0b0) then [ss:ss+steppy] else [ss:ss-steppy], zz: r*2b0*sin(pi/4b0+ss*t)-c*ss, if (abs(zz)>zold) then steppy:steppy/2b0, if integerp(i/skippy) then disp(sconcat(" iterations: ",(i)))], disp(sconcat(" Long Error (from perfect time): ", float(zz))), longtime:ss, squareacc:sqrt(r*sqrt(2b0)/(r*sqrt(2b0)-2b0*mass*6.67408b-14/c^2b0)) *mass*6.67408b-14*(100b0/(r*sqrt(2b0)))^2b0, diagonalacc:sqrt(r*2b0/(r*2b0-2b0*mass*6.67408b-14/c^2b0)) *mass*6.67408b-14*(100b0/(r*2b0))^2b0, sr:smalltime*c, shortacc:(sqrt(sr/(sr-2b0*mass*6.67408b-14/c^2b0)) *mass*6.67408b-14*(100b0/sr)^2b0) *sin(t*2*smalltime+pi/2), sr:diagtime*c, diagacc:(sqrt(sr/(sr-2b0*mass*6.67408b-14/c^2b0)) *mass*6.67408b-14*(100b0/sr)^2b0) *sin(-t*2*diagtime+pi), sr:longtime*c, longacc:(sqrt(sr/(sr-2b0*mass*6.67408b-14/c^2b0)) *mass*6.67408b-14*(100b0/sr)^2b0) *sin(-t*2*longtime/c+pi/2), aqua1:(shortacc+diagacc)-longacc*accmult, aqua2:shortacc*accmult+(diagacc-longacc), pink3:accmult*(shortacc+diagacc)-longacc, pink4:shortacc+accmult*(diagacc-longacc), total:aqua1+aqua2+pink3+pink4, disp(" "), disp(sconcat(" square acc: ", float(squareacc))), disp(sconcat(" diagonal acc: ", float(diagonalacc))), disp(sconcat(" total acc: ", float(total))), disp(sconcat(" smalltime: ", float(smalltime))), disp(sconcat(" diagtime: ", float(diagtime))), disp(sconcat(" longtime: ", float(longtime))) ]$

squareacc and diagonalacc are the Schwarzschild approximation accelerations for the ring at rest... which is sort of pointless, because they cancel out at rest.

So, assume that the ring is super small, Planck scale small, and its momentum ramps up really quickly at that level. It would be like a tiny little gyroscope. Neat huh?

Dale
Mentor
2021 Award
. Basically, if the gravity like acceleration propagates at c
It isn't that simple. That is why general relativity was developed. Einstein tried to do exactly what you are describing, but found that it leads to an inconsistent theory.

In the simplest consistent relativistic theory of gravity, GR, the angular momentum of the system radiates away in the gravitational waves.

mfb
mfb
Mentor
If the four masses are equal, I'm not sure if that system has a variable quadrupole moment. It could rotate without radiating, in the same way a ring will do.

Matt Benesi and Dale
Dale
Mentor
2021 Award
If the four masses are equal, I'm not sure if that system has a variable quadrupole moment. It could rotate without radiating, in the same way a ring will do.
Good point. As you say, it isn't obvious to me that there is any GW emission, but it is obvious that any change in angular momentum must be negative.

Why?

Imagine observers in other locations.

mfb
Mentor
It does not matter where the observers are. All observers will agree on whether gravitational waves are emitted, and no observer will see a speedup of angular velocity.

Dale
Mentor
2021 Award
Nor does it matter how fast they are travelling (assuming an asymptotically flat background). They may disagree on the magnitude of the change in angular velocity, but not the sign.

Matt Benesi
Would frame dragging be a stretch to explain this at these distances/velocities/masses?

mfb
Mentor
What do you mean by "this"?

No.

lol

Dale
Mentor
2021 Award
These calculations are beyond me right now, and I don't have a well informed intuition about which effects will be larger. Both gravitational waves and frame dragging may be present.

These calculations are beyond me right now, and I don't have a well informed intuition about which effects will be larger. Both gravitational waves and frame dragging may be present.
Do gravitational waves emit from symmetric bodies rotating around an axis of symmetry?


text:
## Clouds over Stuhleck – A HDR Time Lapse

Recently we were going to ski at Stuhleck in Austria. Or, better my girlfriend went skiing and I went to time lapse and HDR shots… I took three time lapses and some HDR shots. Two of the videos where really nice, the third went to crap because of two reasons. First, before setting up the camera to record I didn’t check about the memory card. I recorded my footage and went with the other camera around the mountain, after I cam back I noticed the memory were full. And second, I tried to photograph trough a hole in a tree, covered in snow, to have the sun moving on the sky. Nice idea but even with HDR everything was white and blur. Also the tree was moving a lot… Anyway here are the first two lapses in one Video.

For the HDR processing I used Luminance HDR 2.4.0 and, as usual, After Effects for processing the Video. I also tried to make some processing on two pictures for uploading to 500px but the rating was poor. A single picture I have to say looks boring but the moving clouds and light on the ground looks nice in the video.

## Hyper Parliament- A Hyper Lapse of the Austrian Parliament

Yesterday I thought, enough with work, let’s go photographing! But the questions were: what? how? After a bit of sitting around I remembered the tutorial from gwegner.de (by the way, a really good site for tutorials and information) for hyper lapses. This video is in German, but here is also a nice one in English from digitalrev.com.

Basically, with every picture you move with the camera along an imaginary line and take a picture while aiming at the same spot on every picture. As described in the videos, you should use marks are already present, like stones on the ground or just use your step span. In any case this will produce a bit shaky footage, but this can be corrected in after effect. If you shoot in the night, and the are no clouds or stars visible, like in my video, the interval between the pictures is not important. But if you have something in the screen with a steady movement, like clouds, then use a intervalometer. In this video there is a very good description of that. Anyway, in my case i was just going a step forward, point, shoot, step forward, point, shoot…. 250 times. And done. The camera settings were 3.2 s exposure, f/4.5, ISO 100.

After that you have your footage, but this is really shaky. It is impossible to align the camera and the rest completely! But for this there is Adobe After Effects, if you have money, or Deshaker for Virtual Dub, if you don’t have money or just like Open-Source. In any case, the result looks really nice. In post processing I gave to the footage a little bit of a unnatural look, so it looks like rendered, but it is actually taken with a camera in 2 hours and another 2 hours for post processing.

In this video is also stated, that u don’t need fancy equipment like tripod, intervalometer and chalk marks on the ground for 1 km. It’s worth taking a look.

## Karawanken Mountains Night Time Lapse

The few days in Kärnten, Austria where really “time lapsy”… So here another time lapse from the balcony in Kärnten withe view to Klagenfurt an the Karawanken mountains.

## Night Sky Photography Shutter Speed Calculator

UPDATE: I’m glad to announce that I finally found the time to release the Night Fox Android App for my Shutter Speed Calculator!

This is the shutter speed calculator for night sky photographing. Basically you just insert the data of crop factor, the megapixel you want to archive or your camera has, the focal length and the tolerance of pixels you can accept. The description of all these different factors is below. The description of the whole process can be found here.

 Megapixel Focal length Pixel Tolerance
 Sensor Size Vertical FOV Vertical Pixel Vertical Sensor Size Horizontal FOV Horizontal Pixel Horizontal Sensor Size Diagonal FOV Diagonal Pixel Diagonal

 Shutter Speed Vertical Shutter Speed Horizontal Shutter Speed Diagonal

Crop Factor
The crop factor is depending on your camera. It is influencing the field of view (FOV) of your camera, like the focal length also does. This calculator includes the factors 1 (3:2), 1.5 (3:2), 1.6 (3:2) and now also 2.0 (4:3). (Thx to Livio for the comment)

Megapixel
This is the amount of megapixels you want to achieve. It can be the maximum of your camera or also a lower Value. I included this value, because somebody may what to have the pictures not in full resolution, but a maximum of exposure time. With a smaller resolution I’m recommending to reduce the pixel tolerance as well.

Focal Length
The focal length is very important for the field of view and therefore how fast the stars are moving. The rule of thumb is “the lower the better”. But be aware of the speed of your lens. This influences the needed exposure time as well.

Pixel Tolerance
Depending on what you can live with, you can adjust this value. Basically it describes the tolerance of how many pixels a star can “move”. In some cases 20 pixels are OK, but for having really sharp stars, I recommend 10 pixel.

The Milky Way Exposure Calculator can be found here and as well as the other articles of http://www.lonelyspeck.com/.

## Karawanken Mountain Chain Time Lapse

After New Years Eve we got an invitation for a short ski holiday trip near Klagenfurt. I was able to shot some nice time lapses from the Karawanken mountains. The clouds coming Italy seem to fall like water over the mountains. Also i go a nice day to night transition withe the moon as light source on the mountains. The day to night transition was made with LRTimelape 3. For this tool are a lot of tutorials available and the free version is more or less full featured, except that the footage is restricted to 400 pictures.

Karawanken Mountain Chain Time Lapse von Thomas leber ist lizenziert unter einer Creative Commons Namensnennung 4.0 International Lizenz.
Beruht auf dem Werk unter https://www.jamendo.com/en/track/654856/sombre.

## Vienna at Night

After I posted the Special Shot Section of Vienna at Night, here now the normal section for the Vienna at Night shots.

## Vienna at Night – Special Shots.

During the last year I was able to take some shots of Vienna at night. A very nice spot is the Karlskirche. With my EOS 600D I took some shots at once, but I had the problem that I had no tripod. So I took some pictures with the hand and ISO 3200. But because of the APS-C, I had a lot of noise and used the stacking technique and stacked tree pictures into one. This technique is really nice for people that have a small sensor size and also want to have low noise pictures.

Karlskirche in Vienna. Taken with EOS 600D and ISO 3200 free hand

I also took some HDRs of the Karlskirche an tried different setting to the algorithm. I’m not really satisfied with the Adobe Photoshop algorithm. It looks somehow false and not nice. Next time I will evaluate the programs Photomatrix from HDRsoft and also the open source alternative Luminance HDR. I’m really looking forward to test  both programs and their performance. Anyway, here are the results of Photoshop HDR after a really long time of trying. Most of the times I ended up with monochrome settings but this is also boring all the time.

The Karlskirche in Vienna. HDR Shot out of tree pictures with +- 2EV.

The Karlskirche in Vienna. HDR Shot out of tree pictures with +- 2EV.

I also made some HDR pictures of the Musikverein in Vienna. The main problem was that I had a bit of rain. This can be seen in the dark areas where the water drops form some halos.

Vienna Musikverein. Assembled out of three pictures with +-2EV.

I also really like long exposure photographs. They show you a surreal picture of the real world. Also they are easy to made. You just need a nice spot and a long exposure. The camera type, small sensor sizes and weak ISO performance doesn’t matter, as well as lenses with aperture numbers above the f/3. These pictures below are taken with a EOS 600D and the stock 18-55 mm IS II.

The picture below is stacked out of five pictures with ISO 100, f/22 and 10 s of exposure. The camera was the EOS 600D with the 18-55 mm IS II. I stacked the picture with Photoshop with the Bright Color mod for each layer.

Light Streaks in Vienna.

This picture is also stacked, but i used 40 pictures for stacking. The Camera settings were f/11, ISO 200 and 2.5 s exposure.

Highway A22 near Vienna.

The picture below is also stacked but for this I used the the Vivid Light settings. The camera settings were f/22, ISO 100 and 15 s of exposure.

Light Streaks in Vienna at Karlsplatz.

A normal long exposure is the picture below. This was taken with the EOS 6D and the Sigma 50 mm 1.4. I did not stack them, just used some Camera RAW Settings.

Long Exposure Capture at Liechtenwerderplatz in Vienna.

Also the following picture is not stacked, just reworked with some Camera RAW settings. This time also the camera was the EOS 6D and the Sigma 50 mm 1.4.

The Währinger Gürtel an Night.

## Shutter Speed Calculating for Night Sky Photographing

Night Sky at the Alois Günther Haus

Recently I wanted to improve my skills in night sky photographing. So I opened the PC and asked Google a lot of questions. I was reading a lot about shutter speed, focal length and exposure adjustment. A very good site I found was How to Photograph the Milky Way of the site www.lonelyspeck.com. They created a very good tutorial and answered many of my questions. They also created the Milky Way Exposure Calculator, witch lets you calculate how a lens may perform on your camera. For ranking the lenses they calculate a score, basically from the physical parameters of the lens. I agreed with the majority of the parameters, but I was wondering if the slowest possible shutter speed is maybe a bit inaccurate.

In night sky photography the shutter speed is mostly longer than 5-10 s or even more. Without a tracking device the stars will move over the sky and produce, depending on the shutter speed, streaks on the picture. But we want actually a clean image of the stars. So what we need is a lens with a short focal length and an aperture which should have less than f/2 to gather a lot of light. The basics can be checked in the links above. Anyway, the rule of thump for the shutter speed, 500/focal length, seems a bit inaccurate and, as stated there, also not completely usable for APS-C cameras. So I started to think, how can this be solved better…? I was drawing some sketches and made some calculations and concluded: there might be a better and more accurate way: I started to research about what information are available for all lenses and cameras. I concluded that the sensor size and the focal length is available in any case. The angle of view was not all the time available or I just couldn’t find it. Anyway, with the sensor size and the focal length we can calculate the angle of view with the following equation:

$alpha = 2 * \arctan {\dfrac{d}{2 * f}}$

In this equation, d is the diameter of the sensor and f is the effective focal length. Since f is depending on the focus, the focal length F of a 50 mm lens is only 50 mm when set to Infinity. Since in astro photography we use only the infinity setting we can assume f = F. So we can also calculate the horizontal and vertical angle of view for a lens. In this case I will just use the stock Canon EF-S 18-55mm 1:3.5-5.6 IS II on a EOS 600D with 18 MP as example. Why we need the horizontal and the vertical angle of view, will be pointed out later. With the pixel count on the camera we can now calculate the angle to pixel ratio. The 600D has 18 MP with a 3:2 ratio. This leads to a pixel count of 5.184 px by 3.456 px. The pixel to angle ratio is calculated by α / Σnpixel. The diagonal pixel count is just calculated by Pythagorean theorem.

• Vertical AOV: αv = 27°; Pixel to Angle Radio = 0,0185°/px
• Horizontal AOV αh = 37.4°; Pixel to Angle Radio = 0,0164°/px
• Diagonal AOV αd = 46.8°; Pixel to Angle Radio = 0,0153°/px

The night sky can be seen as a cylinder rotating around the camera.

The next assumption we make is that the sky is rotating around the world. I know that some people will now say “NO! Galileo was very clear with that!”. Yes its not true but since we focus anyway on infinity and the sky is more more less a flat area for the lens, this is ok. This lead so my next assumption. The sky is not a sphere, its like a cylinder. The picture demonstrates this assumption. The cylinder rotates with the angular speed of 360° per 24h or 0.00417 °/s. Depending on the pixel to angle ratio we can now calculate the minimum shutter speed. In case of my 18 mm focal length the numbers are very small and and seem not right.

• Vertical max. Shutter Speed: 4.4 s
• Horizontal max. Shutter Speed: 3.9 s
• Diagonal max. Shutter Speed: 3.6 s

But this is only the half truth. Since the stars are not only one pixel big, we can use a threshold. For example a threshold of five pixels will give us five times the shutter speed. In my test, five pixel threshold was mostly ok and not really visible. Somebody who wants to have more clear pictures can reduce the threshold.

• Vertical max. Shutter Speed: 22 s
• Horizontal max. Shutter Speed: 20 s
• Diagonal max. Shutter Speed: 18 s

I’ll program a calculator using javascript in the near future and upload it to this website. Stay tuned!

## WP Parabola Theme and Google Structured Data

While setting up my Google Webmaster Tools I ran into the problem that WordPress in combination with the nice Parabola Theme is not 100% conformed the Google expectations. I saw some errors on the page and tried to figure out what exactly triggers this error. On the overview page I found these entries.

The crawler of Google searches for the hatom information on the page. For my page it finds most of the tags, like the author, the title, the content and also the publish date. But it searches also specific for the field “updated”. The failure occures more often on category sites, because there are more entries, and only once on the simple post.

After some research I found a solution by adding  some simple lines to some files in the theme. But because you still want to update your theme, the best way to edit your theme is by adding a child to the normal theme. This is basically done by creating a new folder under /wp-content/themes/ with the name %theme-name%-child. In my case this was “parabola-child”. After creating a child folder, a file named sytle.css has to be created in the folder. The basic setup is in my case:

The most important line is  Template: parabola. This tells WordPress that this is a child of the Parabola theme. All steps can be found here. WordPress will now load all files from the child folder and override the corresponding files of the main theme. Basically you can copy a file like header.php to the child theme and edit it. The rest of the files will be loaded from the main theme. In our case I searched the content.php and found a link to the parabola_posted_on() in includes/theme-loop.php function. This function provides the meta information of the post and can also be used to write the additional tags to the page. Unfortunately, after i copied and edited my file, WordPress was still loading the main file. This seems to be a bug or i did simply something wrong, ten times or so… Anyway, I decided to change in this case the content.php file. I copied the  snipped from here and added the “hidden=true” to prevent that information from rendering to the screen.

After this modification, the category pages will have the hidden information and Google will crawl these pages without errors. But the single post will still have this failure. So again, after some searching, I figured out that this is related to the file single.php. So I added the same two lines there.

It seems that the theme also deliverers wrong data to the crawler when showing single posts. In the picture on the side,it is visible that it writes for the bookmark the date and not the name of the post. But i was fine with the result until now. Maybe i’ll fix this later or write a bug report to the programmer.


text:
## Filters

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Q. 19.     If A and B are any two events such that $\inline P(A)+P(B)-P(A \; and\; B)=P(A),$ then

(A) $\inline P(B\mid A)=1$

(B) $\inline P(A\mid B)=1$

(C) $\inline P(B\mid A)=0$

(D) $\inline P(A\mid B)=0$

Option B is correct.

Q. 18.     If $\inline P(A\mid B)> P(A)$,  then which of the following is correct :

(A) $\inline P(B\mid A)< P(B)$

(B) $\inline P(A\cap B)< P(A).P(B)$

(C) $\inline P(B\mid A)> P(B)$

(D)$\inline P(B\mid A)= P(B)$

Option C is  correct.

Q. 17.     If A and B are two events such that $\inline P(A\neq 0)$ and$\inline P(B\mid A)=1,$ then

Choose the correct answer of the following:

(A)  $\inline A\subset B$

(B)  $\inline B\subset A$

(C)  $\inline B=\phi$

(D)  $\inline A=\phi$

A and B are two events such that  and                                                                                Option A is correct.

Q. 16  Bag I contains 3 red and 4 black balls and Bag II contains 4 red and 5 black balls. One ball is transferred from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred ball is black.

Let E1 and E2 respectively denote the event that red ball is transfered from bag 1 to bag 2 and a black ball is transfered from bag 1 to bag2.             and       Let A be the event that ball drawn is red. When a red ball is transfered from bag 1 to bag 2. When a black ball is transfered from bag 1 to bag 2.

Q. 15  An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) = $0.2$

P(B fails alone) = $0.15$

P(A and B fail) = $0.15$

Evaluate the following probabilities

(ii) $\inline P(A\; fails \; alone)$

Let event in which A fails and B fails be

Q. 15.     An electronic assembly consists of two subsystems, say, A and B. From previous testing procedures, the following probabilities are assumed to be known:

P(A fails) =$\inline 0.2$

P(B fails alone) = $\inline 0.15$

P(A and B fail) = $\inline 0.15$

Evaluate the following probabilities

(i) $\inline P(A \; fails\mid B\; has\; failed)$

Let event in which A fails and B fails be

Q. 14.     If each element of a second order determinant is either zero or one, what is the probability that the value of the determinant is positive? (Assume that the individual entries of the determinant are chosen independently, each value being assumed with probability $\frac{1}{2}$ ).

Total number of determinant of second order with each element being 0 or 1 is  The values of determinant is positive in the following cases          Probability is

Q. 13  Assume that the chances of a patient having a heart attack is $\inline 40^{o}/_{o}.$ It is also assumed that a meditation and yoga course reduce the risk of heart attack by $\inline 30^{o}/_{o}$ and prescription of certain drug reduces its chances by $\inline 25^{o}/_{o}.$At a time a patient can choose any one of the two options with equal probabilities. It is given that after going through one of the two options the patient selected at random suffers a heart attack. Find the probability that  the patient followed a course of meditation and yoga?

Let A,E1, E2 respectively denote the event that a person has a heart break, selected person followed the course of yoga and meditation , and the person adopted      the drug prescription.      the probability that the patient followed a course of meditation and yoga is

Q. 12 Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box C?

Let R be event of drawing red marble. Let   respectivly denote event of selecting box A,B,C. Total marbles = 40 Red marbles =15      Probability of drawing red marble from box C  is

Q. 12.     Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from  box B?

Let R be event of drawing red marble. Let   respectivly denote event of selecting box A,B,C. Total marbles = 40 Red marbles =15      Probability of drawing red marble from box B  is

12.     Suppose we have four boxes A,B,C and D containing coloured marbles as given below:

One of the boxes has been selected at random and a single marble is drawn from it. If the marble is red, what is the probability that it was drawn from box A ?

' Let R be the event of drawing red marble. Let   respectively denote the event of selecting box A, B, C. Total marbles = 40 Red marbles =15      Probability of drawing red marble from box A  is

Q. 11  In a game, a man wins a rupee for a six and loses a rupee for any other number when a fair die is thrown. The man decided to throw a die thrice but to quit as and when he gets a six. Find the expected value of the amount he wins / loses.

In a throw of die, probability of getting six  = P                     probability of not getting six  = q               There are three cases : 1. Gets six in the first throw, required probability is       The amount he will receive is Re. 1 2.. Does not gets six in the first throw and gets six in the second throw, then the probability                                                           ...

Q. 10 How many times must a man toss a fair coin so that the probability of having at least one head is more than $90^{o}/_{o}$ ?

Let the man toss coin n times. Probability of getting head in first toss = P                                                                                The...

Q. 9  An experiment succeeds twice as often as it fails. Find the probability that in the next six trials, there will be atleast 4 successes.

Probability of success is twice the probability of failure. Let probability of failure be X then Probability of success = 2X     Sum of probabilities is 1.                             Let                    and      Let X be random variable that represent the number of success in six trials.                                                                                ...

Q. 8  If a leap year is selected at random, what is the chance that it will contain 53 tuesdays?

In a leap year, there are 366 days. In 52 weeks, there are 52 Tuesdays. The probability that a leap year will have 53 Tuesday is equal to the probability that the remaining 2 days are Tuesday. The remaining 2 days can be : 1. Monday and Tuesday 2. Tuesday and Wednesday 3. Wednesday and Thursday 4. Thursday and Friday 5.friday and Saturday 6.saturday and Sunday 7.sunday and Monday  Total cases =...

Q.7 A die is thrown again and again until three sixes are obtained. Find the probability of obtaining the third six in the sixth throw of the die.

Probability of 6 in a throw of die =P                                                                          Probability that 2 sixes come in first five throw of die :                                                                                Probability that third six comes in sixth throw :                                                                                ...

Q. 6  In a hurdle race, a player has to cross $10$  hurdles. The probability that he will clear each hurdle is $\frac{5}{6}$ . What is the probability that he will knock down fewer than $2$ hurdles?

Let p and q respectively be probability that the player will clear and knock down the hurdle.      Let X represent random variable that represent number of times the player will knock down the hurdle.

Q. 5   An urn contains 25 balls of which 10 balls bear a mark $\inline 'X'$ and the remaining 15 bear a mark $\inline 'Y'.$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(iv) the number of balls with $\inline 'X'$ mark and $\inline 'Y'$ mark will be equal.

Q. 5  An urn contains 25 balls of which 10 balls bear a mark  $'X'$  and the remaining 15 bear a mark $'Y'.$ A ball is drawn at random from the urn, its mark is noted down and it is replaced. If 6 balls are drawn in this way, find the probability that

(iii) at least one ball will bear $'Y'$ mark.

Q. 5  An urn contains $25$ balls of which $10$ balls bear a mark $'X'$  and the remaining $15$ bear a mark  $'Y'.$  A ball is drawn at random from the urn, its mark is noted down and it is replaced. If $6$  balls are drawn in this way, find the probability that

(ii) not more than $2$ will bear $'Y'$  mark.

Total balls in urn = 25 Balls bearing mark 'X' =10 Balls bearing mark 'Y' =15   balls are drawn with replacementt. Let Z be random variable that represents number of balls with Y mark on them in trial. Z has binomail distribution with n=6.
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# Explicit Automorphic Forms for the Rational Function Field, and Their Galois Representations

 MEMBERS SEMINAR Topic: Explicit Automorphic Forms for the Rational Function Field, and Their Galois Representations Speaker: Zhiwei Yun Affiliation: Member, School of Mathematics Date: Monday, March 22 Time/Room: 2:00pm - 3:00pm/S-101

In this talk, we will give explicit examples of Langlands correspondence for reductive groups over the rational function field $F=k(t)$ . Fixing appropriate local ramifications, it is sometimes possible to write down explicit Hecke-eigenforms using the combinatorics of the affine Weyl group. Some of these examples give interesting Galois representations, for example, those with Zariski dense image in E_7, E_8, F_4 and G_2 , and hypergeometric local systems. We will see how ideas from geometric Langlands correspondence help understand the more "classical" Langlands correspondence. This is joint work with J. Heinloth and B-C. Ngo.


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# Design of Experiments

## Overview

Classical design of experiments (DoE) methods and the more modern design and analysis of computer experiments (DACE) methods are both techniques which seek to extract as much trend data from a parameter space as possible using a limited number of sample points. Classical DoE techniques arose from technical disciplines that assumed some randomness and nonrepeatability in field experiments (e.g., agricultural yield, experimental chemistry). DoE approaches such as central composite design, Box-Behnken design, and full and fractional factorial design generally put sample points at the extremes of the parameter space, since these designs offer more reliable trend extraction in the presence of nonrepeatability. DACE methods are distinguished from DoE methods in that the nonrepeatability component can be omitted since computer simulations are involved. In these cases, space filling designs such as orthogonal array sampling and Latin hypercube sampling are more commonly employed in order to accurately extract trend information. Quasi-Monte Carlo sampling techniques which are constructed to fill the unit hypercube with good uniformity of coverage can also be used for DACE.

Dakota supports both DoE and DACE techniques. In common usage, only parameter bounds are used in selecting the samples within the parameter space. Thus, DoE and DACE can be viewed as special cases of the more general probabilistic sampling for uncertainty quantification (see following section), in which the DoE/DACE parameters are treated as having uniform probability distributions. The DoE/DACE techniques are commonly used for investigation of global response trends, identification of significant parameters (e.g., main effects), and as data generation methods for building response surface approximations.

Dakota includes several approaches sampling and design of experiments, all implemented in included third-party software libraries. LHS (Latin hypercube sampling) [SW04] is a general-purpose sampling package developed at Sandia that has been used by the DOE national labs for several decades. DDACE (distributed design and analysis for computer experiments) is a more recent package for computer experiments developed at Sandia Labs [TM]. DDACE provides the capability for generating orthogonal arrays, Box-Behnken designs, Central Composite designs, and random designs. The FSUDace (Florida State University’s Design and Analysis of Computer Experiments) package provides the following sampling techniques: quasi-Monte Carlo sampling based on Halton or Hammersley sequences, and Centroidal Voronoi Tessellation. Lawrence Livermore National Lab’s PSUADE (Problem Solving Environment for Uncertainty Analysis and Design Exploration) [Ton05] includes several methods for model exploration, but only the Morris screening method is exposed in Dakota.

This chapter describes DDACE, FSUDace, and PSUADE, with a focus on designing computer experiments. Latin Hypercube Sampling, also used in uncertainty quantification, is discussed in the section on sampling methods.

## Design of Computer Experiments

What distinguishes design of computer experiments? Computer experiments are often different from physical experiments, such as those performed in agriculture, manufacturing, or biology. In physical experiments, one often applies the same treatment or factor level in an experiment several times to get an understanding of the variability of the output when that treatment is applied. For example, in an agricultural experiment, several fields (e.g., 8) may be subject to a low level of fertilizer and the same number of fields may be subject to a high level of fertilizer to see if the amount of fertilizer has a significant effect on crop output. In addition, one is often interested in the variability of the output within a treatment group: is the variability of the crop yields in the low fertilizer group much higher than that in the high fertilizer group, or not?

In physical experiments, the process we are trying to examine is stochastic: that is, the same treatment may result in different outcomes. By contrast, in computer experiments, often we have a deterministic code. If we run the code with a particular set of input parameters, the code will always produce the same output. There certainly are stochastic codes, but the main focus of computer experimentation has been on deterministic codes. Thus, in computer experiments we often do not have the need to do replicates (running the code with the exact same input parameters several times to see differences in outputs). Instead, a major concern in computer experiments is to create an experimental design which can sample a high-dimensional space in a representative way with a minimum number of samples. The number of factors or parameters that we wish to explore in computer experiments is usually much higher than physical experiments. In physical experiments, one may be interested in varying a few parameters, usually five or less, while in computer experiments we often have dozens of parameters of interest. Choosing the levels of these parameters so that the samples adequately explore the input space is a challenging problem. There are many experimental designs and sampling methods which address the issue of adequate and representative sample selection.

There are many goals of running a computer experiment: one may want to explore the input domain or the design space and get a better understanding of the range in the outputs for a particular domain. Another objective is to determine which inputs have the most influence on the output, or how changes in the inputs change the output. This is usually called sensitivity analysis.

Another goal is to use the sampled input points and their corresponding output to create a response surface approximation for the computer code. The response surface approximation (e.g., a polynomial regression model, a Gaussian-process/Kriging model, a neural net) can then be used to emulate the computer code. Constructing a response surface approximation is particularly important for applications where running a computational model is extremely expensive: the computer model may take 10 or 20 hours to run on a high performance machine, whereas the response surface model may only take a few seconds. Thus, one often optimizes the response surface model or uses it within a framework such as surrogate-based optimization. Response surface models are also valuable in cases where the gradient (first derivative) and/or Hessian (second derivative) information required by optimization techniques are either not available, expensive to compute, or inaccurate because the derivatives are poorly approximated or the function evaluation is itself noisy due to roundoff errors. Furthermore, many optimization methods require a good initial point to ensure fast convergence or to converge to good solutions (e.g. for problems with multiple local minima). Under these circumstances, a good design of computer experiment framework coupled with response surface approximations can offer great advantages.

In addition to the sensitivity analysis and response surface modeling mentioned above, we also may want to do uncertainty quantification on a computer model. Uncertainty quantification (UQ) refers to taking a particular set of distributions on the inputs, and propagating them through the model to obtain a distribution on the outputs. For example, if input parameter A follows a normal with mean 5 and variance 1, the computer produces a random draw from that distribution. If input parameter B follows a weibull distribution with alpha = 0.5 and beta = 1, the computer produces a random draw from that distribution. When all of the uncertain variables have samples drawn from their input distributions, we run the model with the sampled values as inputs. We do this repeatedly to build up a distribution of outputs. We can then use the cumulative distribution function of the output to ask questions such as: what is the probability that the output is greater than 10? What is the 99th percentile of the output?

Note that sampling-based uncertainty quantification and design of computer experiments are very similar. There is significant overlap in the purpose and methods used for UQ and for DACE. We have attempted to delineate the differences within Dakota as follows: we use the methods DDACE, FSUDACE, and PSUADE primarily for design of experiments, where we are interested in understanding the main effects of parameters and where we want to sample over an input domain to obtain values for constructing a response surface. We use the nondeterministic sampling methods (sampling) for uncertainty quantification, where we are propagating specific input distributions and interested in obtaining (for example) a cumulative distribution function on the output. If one has a problem with no distributional information, we recommend starting with a design of experiments approach. Note that DDACE, FSUDACE, and PSUADE currently do not support distributional information: they take an upper and lower bound for each uncertain input variable and sample within that. The uncertainty quantification methods in sampling (primarily Latin Hypercube sampling) offer the capability to sample from many distributional types. The distinction between UQ and DACE is somewhat arbitrary: both approaches often can yield insight about important parameters and both can determine sample points for response surface approximations.

Three software packages are available in Dakota for design of computer experiments, DDACE (developed at Sandia Labs), FSUDACE (developed at Florida State University), and PSUADE (LLNL).

## DDACE

The Distributed Design and Analysis of Computer Experiments (DDACE) package includes both classical design of experiments methods [TM] and stochastic sampling methods. The classical design of experiments methods in DDACE are central composite design (CCD) and Box-Behnken (BB) sampling. A grid-based sampling (full-factorial) method is also available. The stochastic methods are orthogonal array sampling [KO96] (which permits main effects calculations), Monte Carlo (random) sampling, Latin hypercube sampling, and orthogonal array-Latin hypercube sampling. While DDACE LHS supports variables with normal or uniform distributions, only uniform are supported through Dakota. Also DDACE does not allow enforcement of user-specified correlation structure among the variables.

The sampling methods in DDACE can be used alone or in conjunction with other methods. For example, DDACE sampling can be used with both surrogate-based optimization and optimization under uncertainty advanced methods. See Listing 64 for an example of how the DDACE settings are used in Dakota.

The following sections provide more detail about the sampling methods available for design of experiments in DDACE.

### Central Composite Design

A Box-Wilson Central Composite Design, commonly called a central composite design (CCD), contains an embedded factorial or fractional factorial design with center points that is augmented with a group of ’star points’ that allow estimation of curvature. If the distance from the center of the design space to a factorial point is $$\pm$$1 unit for each factor, the distance from the center of the design space to a star point is $$\pm\alpha$$ with $$\mid\alpha\mid > 1$$. The precise value of $$\alpha$$ depends on certain properties desired for the design and on the number of factors involved. The CCD design is specified in Dakota with the method command dace central_composite.

As an example, with two input variables or factors, each having two levels, the factorial design is shown in Table 3:

Table 3 Simple Factorial Design

Input 1

Input 2

-1

-1

-1

+1

+1

-1

+1

+1

With a CCD, the design in Table 3 would be augmented with the points shown in Table 4, if $$\alpha$$ = 1.3.

Table 4 Additional Points to make the factorial design a CCD

Input 1

Input 2

0

+1.3

0

-1.3

1.3

0

-1.3

0

0

0

These points define a circle around the original factorial design. Note that the number of sample points specified in a CCD, samples, is a function of the number of variables in the problem:

$samples = 1 + 2*NumVar + 2^{NumVar}$

### Box-Behnken Design

The Box-Behnken design is similar to a Central Composite design, with some differences. The Box-Behnken design is a quadratic design in that it does not contain an embedded factorial or fractional factorial design. In this design the treatment combinations are at the midpoints of edges of the process space and at the center, as compared with CCD designs where the extra points are placed at ’star points’ on a circle outside of the process space. Box-Behken designs are rotatable (or near rotatable) and require 3 levels of each factor. The designs have limited capability for orthogonal blocking compared to the central composite designs. Box-Behnken requires fewer runs than CCD for 3 factors, but this advantage goes away as the number of factors increases. The Box-Behnken design is specified in Dakota with the method command dace box_behnken.

Note that the number of sample points specified in a Box-Behnken design, samples, is a function of the number of variables in the problem:

$samples = 1 + 4*NumVar + (NumVar-1)/2$

### Orthogonal Array Designs

Orthogonal array (OA) sampling is a widely used technique for running experiments and systematically testing factor effects [HSS99]. An orthogonal array sample can be described as a 4-tuple $$(m,n,s,r)$$, where $$m$$ is the number of sample points, $$n$$ is the number of input variables, $$s$$ is the number of symbols, and $$r$$ is the strength of the orthogonal array. The number of sample points, $$m$$, must be a multiple of the number of symbols, $$s$$. The number of symbols refers to the number of levels per input variable. The strength refers to the number of columns where we are guaranteed to see all the possibilities an equal number of times.

For example, Table 5 shows an orthogonal array of strength 2 for $$m$$ = 8, with 7 variables:

Table 5 Orthogonal Array for Seven Variables

Input 1

Input 2

Input 3

Input 4

Input 5

Input 6

Input 7

0

0

0

0

0

0

0

0

0

0

1

1

1

1

0

1

1

0

0

1

1

0

1

1

1

1

0

0

1

0

1

0

1

0

1

1

0

1

1

0

1

0

1

1

0

0

1

1

0

1

1

0

1

0

0

1

If one picks any two columns, say the first and the third, note that each of the four possible rows we might see there, 0 0, 0 1, 1 0, 1 1, appears exactly the same number of times, twice in this case.

DDACE creates orthogonal arrays of strength 2. Further, the OAs generated by DDACE do not treat the factor levels as one fixed value (0 or 1 in the above example). Instead, once a level for a variable is determined in the array, DDACE samples a random variable from within that level. The orthogonal array design is specified in Dakota with the method command dace oas.

The orthogonal array method in DDACE is the only method that allows for the calculation of main effects, specified with the command main_effects. Main effects is a sensitivity analysis method which identifies the input variables that have the most influence on the output. In main effects, the idea is to look at the mean of the response function when variable A (for example) is at level 1 vs. when variable A is at level 2 or level 3. If these mean responses of the output are statistically significantly different at different levels of variable A, this is an indication that variable A has a significant effect on the response. The orthogonality of the columns is critical in performing main effects analysis, since the column orthogonality means that the effects of the other variables ’cancel out’ when looking at the overall effect from one variable at its different levels. There are ways of developing orthogonal arrays to calculate higher order interactions, such as two-way interactions (what is the influence of Variable A * Variable B on the output?), but this is not available in DDACE currently. At present, one way interactions are supported in the calculation of orthogonal array main effects within DDACE. The main effects are presented as a series of ANOVA tables. For each objective function and constraint, the decomposition of variance of that objective or constraint is presented as a function of the input variables. The p-value in the ANOVA table is used to indicate if the input factor is significant. The p-value is the probability that you would have obtained samples more extreme than you did if the input factor has no effect on the response. For example, if you set a level of significance at 0.05 for your p-value, and the actual p-value is 0.03, then the input factor has a significant effect on the response.

### Grid Design

In a grid design, a grid is placed over the input variable space. This is very similar to a multi-dimensional parameter study where the samples are taken over a set of partitions on each variable. The main difference is that in grid sampling, a small random perturbation is added to each sample value so that the grid points are not on a perfect grid. This is done to help capture certain features in the output such as periodic functions. A purely structured grid, with the samples exactly on the grid points, has the disadvantage of not being able to capture important features such as periodic functions with relatively high frequency (due to aliasing). Adding a random perturbation to the grid samples helps remedy this problem.

Another disadvantage with grid sampling is that the number of sample points required depends exponentially on the input dimensions. In grid sampling, the number of samples is the number of symbols (grid partitions) raised to the number of variables. For example, if there are 2 variables, each with 5 partitions, the number of samples would be $$5^2$$. In this case, doubling the number of variables squares the sample size. The grid design is specified in Dakota with the method command dace grid.

Note

### Monte Carlo Design

Monte Carlo designs simply involve pure Monte-Carlo random sampling from uniform distributions between the lower and upper bounds on each of the input variables. Monte Carlo designs, specified by dace random, are a way to generate a set of random samples over an input domain.

### LHS Design

DDACE offers the capability to generate Latin Hypercube designs. Note that the version of LHS in DDACE generates uniform samples (uniform between the variable bounds). The version of LHS offered with nondeterministic sampling can generate LHS samples according to a number of distribution types, including normal, lognormal, weibull, beta, etc. To specify the DDACE version of LHS, use the method command dace lhs.

Note

### OA-LHS Design

DDACE offers a hybrid design which is combination of an orthogonal array and a Latin Hypercube sample. This design is specified with the method command dace oa_lhs. This design has the advantages of both orthogonality of the inputs as well as stratification of the samples (see [Owe92]).

## FSUDace

The Florida State University Design and Analysis of Computer Experiments (FSUDace) package provides quasi-Monte Carlo sampling (Halton and Hammersley) and Centroidal Voronoi Tessellation (CVT) methods. All three methods natively generate sets of uniform random variables on the interval $$[0,1]$$ (or in Dakota, on user-specified uniform intervals).

The quasi-Monte Carlo and CVT methods are designed with the goal of low discrepancy. Discrepancy refers to the nonuniformity of the sample points within the unit hypercube. Low discrepancy sequences tend to cover the unit hypercube reasonably uniformly. Quasi-Monte Carlo methods produce low discrepancy sequences, especially if one is interested in the uniformity of projections of the point sets onto lower dimensional faces of the hypercube (usually 1-D: how well do the marginal distributions approximate a uniform?) CVT does very well volumetrically: it spaces the points fairly equally throughout the space, so that the points cover the region and are isotropically distributed with no directional bias in the point placement. There are various measures of volumetric uniformity which take into account the distances between pairs of points, regularity measures, etc. Note that CVT does not produce low-discrepancy sequences in lower dimensions, however: the lower-dimension (such as 1-D) projections of CVT can have high discrepancy.

The quasi-Monte Carlo sequences of Halton and Hammersley are deterministic sequences determined by a set of prime bases. A Halton design is specified in Dakota with the method command fsu_quasi_mc halton, and the Hammersley design is specified with the command fsu_quasi_mc hammersley. For more details about the input specification, see the Reference Manual. CVT points tend to arrange themselves in a pattern of cells that are roughly the same shape. To produce CVT points, an almost arbitrary set of initial points is chosen, and then an internal set of iterations is carried out. These iterations repeatedly replace the current set of sample points by an estimate of the centroids of the corresponding Voronoi subregions [DFG99]. A CVT design is specified in Dakota with the method command fsu_cvt.

The methods in FSUDace are useful for design of experiments because they provide good coverage of the input space, thus allowing global sensitivity analysis.

PSUADE (Problem Solving Environment for Uncertainty Analysis and Design Exploration) is a Lawrence Livermore National Laboratory tool for metamodeling, sensitivity analysis, uncertainty quantification, and optimization. Its features include non-intrusive and parallel function evaluations, sampling and analysis methods, an integrated design and analysis framework, global optimization, numerical integration, response surfaces (MARS and higher order regressions), graphical output with Pgplot or Matlab, and fault tolerance [Ton05]. Dakota includes a prototype interface to its Morris One-At-A-Time (MOAT) screening method, a valuable tool for global sensitivity (including interaction) analysis.

The Morris One-At-A-Time method, originally proposed by M. D. Morris [Mor91], is a screening method, designed to explore a computational model to distinguish between input variables that have negligible, linear and additive, or nonlinear or interaction effects on the output. The computer experiments performed consist of individually randomized designs which vary one input factor at a time to create a sample of its elementary effects.

With MOAT, each dimension of a $$k-$$dimensional input space is uniformly partitioned into $$p$$ levels, creating a grid of $$p^k$$ points $${\bf x} \in \mathbb{R}^k$$ at which evaluations of the model $$y({\bf x})$$ might take place. An elementary effect corresponding to input $$i$$ is computed by a forward difference

$d_i({\bf x}) = \frac{y({\bf x} + \Delta {\bf e}_i) - y({\bf x})}{\Delta},$

where $$e_i$$ is the $$i^{\mbox{th}}$$ coordinate vector, and the step $$\Delta$$ is typically taken to be large (this is not intended to be a local derivative approximation). In the present implementation of MOAT, for an input variable scaled to $$[0,1]$$, $$\Delta = \frac{p}{2(p-1)}$$, so the step used to find elementary effects is slightly larger than half the input range.

The distribution of elementary effects $$d_i$$ over the input space characterizes the effect of input $$i$$ on the output of interest. After generating $$r$$ samples from this distribution, their mean,

$\mu_i = \frac{1}{r}\sum_{j=1}^{r}{d_i^{(j)}},$

modified mean

$\mu_i^* = \frac{1}{r}\sum_{j=1}^{r}{|d_i^{(j)}|},$

(using absolute value) and standard deviation

$\sigma_i = \sqrt{ \frac{1}{r}\sum_{j=1}^{r}{ \left(d_i^{(j)} - \mu_i \right)^2} }$

are computed for each input $$i$$. The mean and modified mean give an indication of the overall effect of an input on the output. Standard deviation indicates nonlinear effects or interactions, since it is an indicator of elementary effects varying throughout the input space.

The MOAT method is selected with method keyword psuade_moat as shown in the sample Dakota input file Listing 28.

Listing 28 Dakota input file showing the Morris One-at-a-Time method – see dakota/share/dakota/examples/users/morris_ps_moat.in
# Dakota Input File: morris_ps_moat.in

environment
tabular_data

method
samples = 84
partitions = 3
seed = 500

variables
continuous_design = 20
lower_bounds = 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0 0.0
upper_bounds = 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0
1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1.0

interface
analysis_drivers = 'morris'
fork
asynchronous evaluation_concurrency = 5

responses
objective_functions = 1
no_hessians

The number of samples must be a positive integer multiple of (number of continuous design variables $$k$$ + 1) and will be automatically adjusted if misspecified. The number of partitions applies to each variable being studied and must be odd (the number of MOAT levels $$p$$ per variable is partitions + 1, similar to Dakota multidimensional parameter studies). This will also be adjusted at runtime as necessary. Finite user-specified lower and upper bounds are required and will be scaled as needed by the method.

Note

For more information on the use of MOAT sampling, see this Morris example, or Saltelli, et al. [STCR04].

## Sensitivity Analysis

### Sensitivity Analysis Overview

In many engineering design applications, sensitivity analysis techniques and parameter study methods are useful in identifying which of the design parameters have the most influence on the response quantities. This information is helpful prior to an optimization study as it can be used to remove design parameters that do not strongly influence the responses. In addition, these techniques can provide assessments as to the behavior of the response functions (smooth or nonsmooth, unimodal or multimodal) which can be invaluable in algorithm selection for optimization, uncertainty quantification, and related methods. In a post-optimization role, sensitivity information is useful is determining whether or not the response functions are robust with respect to small changes in the optimum design point.

In some instances, the term sensitivity analysis is used in a local sense to denote the computation of response derivatives at a point. These derivatives are then used in a simple analysis to make design decisions. Dakota supports this type of study through numerical finite-differencing or retrieval of analytic gradients computed within the analysis code. The desired gradient data is specified in the responses section of the Dakota input file and the collection of this data at a single point is accomplished through a parameter study method with no steps. This approach to sensitivity analysis should be distinguished from the activity of augmenting analysis codes to internally compute derivatives using techniques such as direct or adjoint differentiation, automatic differentiation (e.g., ADIFOR), or complex step modifications. These sensitivity augmentation activities are completely separate from Dakota and are outside the scope of this manual. However, once completed, Dakota can utilize these analytic gradients to perform optimization, uncertainty quantification, and related studies more reliably and efficiently.

In other instances, the term sensitivity analysis is used in a more global sense to denote the investigation of variability in the response functions. Dakota supports this type of study through computation of response data sets (typically function values only, but all data sets are supported) at a series of points in the parameter space. The series of points is defined using either a vector, list, centered, or multidimensional parameter study method. For example, a set of closely-spaced points in a vector parameter study could be used to assess the smoothness of the response functions in order to select a finite difference step size, and a set of more widely-spaced points in a centered or multidimensional parameter study could be used to determine whether the response function variation is likely to be unimodal or multimodal. See Parameter Studies Capabilities for additional information on these methods. These more global approaches to sensitivity analysis can be used to obtain trend data even in situations when gradients are unavailable or unreliable, and they are conceptually similar to the design of experiments methods and sampling approaches to uncertainty quantification described in the following sections.

### Assessing Sensitivity with DACE

Like parameter studies, the DACE techniques are useful for characterizing the behavior of the response functions of interest through the parameter ranges of interest. In addition to direct interrogation and visualization of the sampling results, a number of techniques have been developed for assessing the parameters which are most influential in the observed variability in the response functions. One example of this is the well-known technique of scatter plots, in which the set of samples is projected down and plotted against one parameter dimension, for each parameter in turn. Scatter plots with a uniformly distributed cloud of points indicate parameters with little influence on the results, whereas scatter plots with a defined shape to the cloud indicate parameters which are more significant. Related techniques include analysis of variance (ANOVA) [MM95] and main effects analysis, in which the parameters which have the greatest influence on the results are identified from sampling results. Scatter plots and ANOVA may be accessed through import of Dakota tabular results into external statistical analysis programs such as S-plus, Minitab, etc.

Running any of the design of experiments or sampling methods allows the user to save the results in a tabular data file, which then can be read into a spreadsheet or statistical package for further analysis. In addition, we have provided some functions to help determine the most important variables.

We take the definition of uncertainty analysis from [STCR04]: “The study of how uncertainty in the output of a model can be apportioned to different sources of uncertainty in the model input.”

As a default, Dakota provides correlation analyses when running LHS. Correlation tables are printed with the simple, partial, and rank correlations between inputs and outputs. These can be useful to get a quick sense of how correlated the inputs are to each other, and how correlated various outputs are to inputs. The correlation analyses are explained further in the sampling section.

We also have the capability to calculate sensitivity indices through Variance-based Decomposition (VBD). Variance-based decomposition is a global sensitivity method that summarizes how the uncertainty in model output can be apportioned to uncertainty in individual input variables. VBD uses two primary measures, the main effect sensitivity index $$S_{i}$$ and the total effect index $$T_{i}$$. The main effect sensitivity index corresponds to the fraction of the uncertainty in the output, $$Y$$, that can be attributed to input $$x_{i}$$ alone. The total effects index corresponds to the fraction of the uncertainty in the output, $$Y$$, that can be attributed to input $$x_{i}$$ and its interactions with other variables. The main effect sensitivity index compares the variance of the conditional expectation $$Var_{x_{i}}[E(Y|x_{i})]$$ against the total variance $$Var(Y)$$. Formulas for the indices are:

$S_{i}=\frac{Var_{x_{i}}[E(Y|x_{i})]}{Var(Y)} \label{eq:VBD_Si}$

and

$T_{i}=\frac{E(Var(Y|x_{-i}))}{Var(Y)}=\frac{Var(Y)-Var(E[Y|x_{-i}])}{Var(Y)} \label{eq:VBD_Ti}$

where $$Y=f({\bf x})$$ and $${x_{-i}=(x_{1},...,x_{i-1},x_{i+1},...,x_{m})}$$.

The calculation of $$S_{i}$$ and $$T_{i}$$ requires the evaluation of m-dimensional integrals which are typically approximated by Monte-Carlo sampling. More details on the calculations and interpretation of the sensitivity indices can be found in [STCR04]. In Dakota version 5.1, we have improved calculations for the calculation of the $$S_{i}$$ and $$T_{i}$$ indices when using sampling. The implementation details of these calculatiosn are provided in [WKS+12]. VBD can be specified for any of the sampling or DACE methods using the command variance_based_decomp. Note that VBD is extremely computationally intensive when using sampling since replicated sets of sample values are evaluated. If the user specified a number of samples, $$N$$, and a number of nondeterministic variables, $$M$$, variance-based decomposition requires the evaluation of $$N(M+2)$$ samples. To obtain sensitivity indices that are reasonably accurate, we recommend that $$N$$, the number of samples, be at least one hundred and preferably several hundred or thousands. Because of the computational cost, variance-based decomposition is turned off as a default for sampling or DACE. Another alternative, however, is to obtain these indices using one of the stochastic expansion methods. The calculation of the indices using expansion methods is much more efficient since the VBD indices are analytic functions of the coefficients in the stochastic expansion. The paper by Weirs et al. [WKS+12] compares different methods for calculating the sensitivity indices for nonlinear problems with significant interaction effects.

In terms of interpretation of the sensitivity indices, a larger value of the sensitivity index, $$S_{i}$$, means that the uncertainty in the input variable $$i$$ has a larger effect on the variance of the output. Note that the sum of the main effect indices will be less than or equal to one. If the sum of the main effect indices is much less than one, it indicates that there are significant two-way, three-way, or higher order interactions that contribute significantly to the variance. There is no requirement that the sum of the total effect indices is one: in most cases, the sum of the total effect indices will be greater than one. An example of the Main and Total effects indices as calculated by Dakota using sampling is shown in the following spinnet of Dakota output:

Global sensitivity indices for each response function:
response_fn_1 Sobol indices:
Main             Total
4.7508913283e-01  5.3242162037e-01 uuv_1
3.8112392892e-01  4.9912486515e-01 uuv_2

Finally, we have the capability to calculate a set of quality metrics for a particular input sample. These quality metrics measure various aspects relating to the volumetric spacing of the samples: are the points equally spaced, do they cover the region, are they isotropically distributed, do they have directional bias, etc.? The quality metrics are explained in more detail in the Reference Manual.

## DOE Usage Guidelines

Parameter studies, classical design of experiments (DOE), design/analysis of computer experiments (DACE), and sampling methods share the purpose of exploring the parameter space. When a global space-filling set of samples is desired, then the DOE, DACE, and sampling methods are recommended. These techniques are useful for scatter plot and variance analysis as well as surrogate model construction.

The distinction between DOE and DACE methods is that the former are intended for physical experiments containing an element of nonrepeatability (and therefore tend to place samples at the extreme parameter vertices), whereas the latter are intended for repeatable computer experiments and are more space-filling in nature.

The distinction between DOE/DACE and sampling is drawn based on the distributions of the parameters. DOE/DACE methods typically assume uniform distributions, whereas the sampling approaches in Dakota support a broad range of probability distributions.

To use sampling in design of experiments mode (as opposed to uncertainty quantification mode), an active view override (e.g., active all) can be included in the variables specification (see “Management of Mixed Variables by Iterator”) of the Dakota input file.

Design of experiments method selection recommendations are summarized in Table 6:

Table 6 Guidelines for selection of parameter study, DOE, DACE, and sampling methods.

Method Classification

Applications

Applicable Methods

parameter study

sensitivity analysis, directed parameter space investigations

centered_parameter_study, list_parameter_study, multidim_parameter_study, vector_parameter_study

classical design of experiments

physical experiments (parameters are uniformly distributed)

dace (box_behnken, central_composite)

design of computer experiments

variance analysis, space filling design (parameters are uniformly distributed)

dace (grid, random, oas lhs, oa_lhs), fsu_quasi_mc (halton, hammersley), fsu_cvt, psuade_moat

sampling

space filling designs (parameters have general probability distributions)

sampling (random or lhs) with optional active view override

Title


text:
# Understanding a rope around the world

This is a slight break from the norm as it is something that has been flying around my brain for the past few days/decades.

The problem

Years ago (I must have been about 12 or 13) I was at a dinner party with the family and someone mentioned this mathematical problem.

For anyone not familiar it goes something along the lines of...

Take a length of rope and form it into a circle around a basketball.
Now add around 60cm to the rope and form it into a new circle.

Now...

Take a length of rope and wrap it around the equator of the earth. (assuming earth is a perfect sphere).
Now add the same 60cm to the rope and form it into a new circle.
Amazingly, mind-blowingly and converse to anything your brain might want to believe this new circle will now float 10cm above the earth all the way around.

All you've added is 60cm of length to a piece of rope that's about 40,000km long and it adds 10cm to the radius.

Year after year my brain has wandered back to this problem and always refuses to believe it. Even when I did the maths behind it and proved it to myself I still struggled to comprehend.

The maths...

Let $r$ be the radius of initial circle.
$\displaystyle r=\frac{C}{2\pi}$

Let $r'$ be the radius of the new circle.
Let $x$ be the length added to the circumference.
$\displaystyle r'=\frac{C+x}{2\pi} \\ r'=\frac{C}{2\pi}+\frac{x}{2\pi} \\ \therefore r' = r+\frac{x}{2\pi}$

In fact, plugging in the number you get an increase in radius of around 9.5493cm.

I did this years ago and still I couldn't wrap my brain around it.

Even if the original circle wraps around the edge of the universe it will still be 9.5493cm bigger in all directions!

Making my brain understand it

So, I started thinking about how to make it understandable enough for my brain to accept this obvious fact.

I wanted to see where each part of that 60cm of additional circumference length went. If I could picture this it would make the problem easier to see.

So I decided to explore what happens to a quarter of the circle when I add 15cm (one quarter of 60cm) in various different ways.

If I concentrate on the change in $h$ in the above diagram I might be able to understand what's going on.

So, initially I added all 15cm of (straight line) length at an angle of $0^{\circ}$ to the horizontal radius of the quarter circle.

Obviously in this we can see that the new height is r (the old height) plus the 15 cm we just added.

Now, let's do this again but instead I'll do it twice.

Now, how much height did we add? Well obviously at the bottom it's 7.5 cm. But at the half way mark? Let's enlarge this on its own. I have highlighted the added section in orange.

In this instance we let $\Theta = \frac{\pi}{4}rads$.

So, what is $\delta h$?

Well, we have a right angle triangle. We know the length of the hypotenuse, we know an angle and we want to get the length of the adjacent edge. So...

$\displaystyle \cos\Theta = \frac{\delta h}{7.5}\\ 7.5\cos\Theta = \delta h$

Plugging in the values we get...

$\delta h \approx 5.3cm$

So the total increase in height is...

$\displaystyle 7.5\cos0 + 7.5\cos\left(\frac{\pi}{4}\right)\\ \to 7.5\left(\cos0 + \cos\left(\frac{\pi}{4}\right)\right)$
The first line can be through of as a triangle with angle 0.

So we get an increase of around 12.8033cm.

We can do this again...

We can see from this that, using the same geometry as we did previously, the increase in height is...

$\displaystyle 5\left(\cos0 + \cos\left(\frac{\pi}{6}\right) + \cos\left(\frac{\pi}{3}\right)\right)$

To simplify this we could write it as a sum...

$\displaystyle 5 \times \sum_{n=1}^{p=3}\cos\left((n-1)\frac{\pi}{2p}\right)$

To explain the above...

p is the number of sections added (3)
5 is the length of each section added to the line. In the is case 15 / 3 (the number of measurements).
n-1 because we want to start at zero radians.

We can now use this to work out the total change in height of around 11.83 cm.

Wolfram Alpha result

So we continue using this same method with more and more sections (triangles) and calculate the change in height each time.

$p=1 \to \delta h=15\\ p=2 \to \delta h\approx 12.8033\\ p=3 \to \delta h\approx 11.8301\\ p=4 \to \delta h\approx 11.3013\\ p=5 \to \delta h\approx 10.9706\\ p=6 \to \delta h\approx 10.7447\\ p=7 \to \delta h\approx 10.5806$

In fact, we can plot this as a graph...

From here we can clearly see that the graph asymptotes towards a value of around 9.5cm... which if we look above in the post is exactly what we were expecting. The actual value of $\frac{60}{2\pi}$ which is around 9.5493cm.

Something things to consider

Yes, I was only looking at the change in height of quarter of the circle. But the problem is symmetrical. Not only is the change in height is translated to each of the other 3 quarters. But, dues to the symmetry of the cos and sin functions the increase in height is matched at every measurement by the "opposite" increase in width. i.e. for every triangle that adds a height of x there is an opposite triangle that adds a width of x and vice versa.

I have not made any assumptions about the size of the circle I have been adding sections to. The only thing I have used is the angle at which line segment is added. Therefore this works for a circle of any size.

Working through it this way I got to a point when I realised the obviousness of this whole problem.

Take this picture. I have done the first step of adding 15cm to each quarter circle. It is fairly obvious from this that by adding 4 sections of length 15 cm like this the point that is $90^{\circ}$ around from it will be 15cm further away. This can be comprehended even if the circle is a million light years in radius.

Lastly, I am pretty sure I'm just a couple of steps away from equating my summation using the circumference to the equation $C=2\pi r$ but I'll leave that for another day.

Conclusion

Anyway, I was quite pleased with how my calculations turned out and the little formula I came up with. It was nice to get my head into some proper analytical maths and for it to work so well after so many years off.

Who knows, I might make this a regular thing.


text:
# Dressed states of a strain-driven spin in diamond

Kölbl, Johannes. Dressed states of a strain-driven spin in diamond. 2020, Doctoral Thesis, University of Basel, Faculty of Science.

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Official URL: http://edoc.unibas.ch/diss/DissB_13650

In this thesis, we address this challenge by extending the widely used approach of dynamical decoupling to protect a quantum system from decoherence. Specifically, we study three-level dressed states that emerge under continuous, closed-contour' interaction driving. To implement and investigate these dressed states, we exploit well-established methods for coherent microwave and strain manipulation of the NV center spin in a hybrid spin-mechanical system. Our results reveal that this novel continuous decoupling mechanism can overcome external magnetic fluctuations in a robust way. We demonstrate experimentally that the dressed states we created are long-lived and find coherence times nearly two orders of magnitude longer than the inhomogeneous dephasing time of the NV spin, even for moderate driving strengths.
To realize direct and efficient access to the coherence-protected dressed states under closed-contour driving, we further demonstrate the use of state transfer protocols for their initialization and readout. In addition to an adiabatic approach, we apply recently developed protocols based on shortcuts to adiabaticity' to accomplish the initialization process, which ultimately accelerates the transfer speed by a factor of $2.6$ compared to the fastest adiabatic protocol with similar fidelity. Moreover, we show bidirectionality of the accelerated state transfer, which allows us to directly read out the dressed state population and to quantify the transfer fidelity of $\approx$$\,99\,\%$.