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# Noobish question on time dilation!

1. Aug 14, 2007

### Übermensch

Hm. I'll use the Twin Paradox to phrase me question. if you need a refresher, http://en.wikipedia.org/wiki/Twin_paradox

Everything I read about it seems to make a blatant claim that "the twin on Earth will be older". Are such texts missing the point, or am I?

Suppose I'm twin A (so I'm on Earth as the frame of reference). My brother, B, is orbiting the Earth near light speed. Of course, from my inertial frame, I'm stationary and he's moving. It seems to me as if time moves slower for him. So when he sends me a picture of himself each year, I'm surprised how young he looks. Great, so good so far.

But let's start from the beginning again. This time, I'm B. The Earth seems to be moving around me near the speed of light. Naturally, time seems to be going slow for them. So, when I receive a picture of my brother each year, I'm surprised to see that HE is the younger one.

Yet, I've generally read that the twin that leaves Earth is objectively younger-- which assumes that the Earth's frame of reference is the most important.

On a similar note, suppose I orbit near light speed with a clock. When I get back, I find that Earth's clocks are all behind mine, since the Earth was moving rapidly compared to my inertial frame. Yet, my boss who sent me up on the mission will look at my clock and tell me that it's behind, right? So if we put the clocks side by side and look at them, then what?

Sorry, just started with this branch of physics an hour ago or so. But, I wouldn't be able to get sleep tonight without asking.

2. Aug 14, 2007

### yenchin

You should search the forum, there have been many threads on this. After reading those, if you still have question, you are welcome to ask again.

3. Aug 14, 2007

### Wallace

There are many ways to view the resolution of this apparent paradox, as suggested do a search on 'Twin paradox' and there are some good links and explanations given.

One key thing to realise is that the classic 'twin paradox' is not a situation in which two observers move at a constant speed with respect to each other, one of the paths accelerates while the other does not, meaning that the experiences of the two observers are not the same.

If you had two people who simply move past each other very fast then they both see the others time running slower in the way you suggest, however this is not the classic twin paradox since in that case one of the observers turns around and comes back while the other does not. You could argue that in the rest frame of the twin who leaves the Earth the Earth 'turns around and comes back' however the twin on the rocket would feel the non-inertial force of the rocket firing whereas the twin on earth does not experience such a force. Hence the paths are different. By the way the Wiki page pretty much explains all of this, but have a look at some of the links given previously as well.

4. Aug 14, 2007

### jcsd

Bascically someone travelling at constant velocity has worldine in spacetime that is analgous to a straight line. Between any two staright lines there's always an axis of symmetry; for two observers travelling at constant velocity time dialation is symmetric.

An obserever that is accelrating has a worldline that is analagous to a curve (or more properly a curve that is not a straight line). For a straight line and a curve there is an axis of symmetry between the straight line and every line that is tangent to the curve, but there is no axis of symmetry between the curve and the straight line. Simlairly time dilation is not symmetric for an accelrating and non-accelerating observer.

5. Aug 14, 2007

### JesseM

But you're leaving out the part that if B is orbiting, B does not have a single inertial rest frame--only moving in a straight line at constant speed is "inertial" motion, anything else involves G-forces. The laws of SR are only symmetrical between inertial frames, there is no such symmetry between non-inertial ones.

6. Aug 14, 2007

### Übermensch

I did a search, but it didn't clear up my issue, I might actually jsut be too dumb. :tongue2:

So, I understand know *what* happens, but I'm not completely sure why. Let's pretend I never said it was orbiting. I thought that would make it easier, as I haven't taken much of a look at gravity's role in GR yet.

If B was receiving a movie from Earth of A(ignore the physical problems of this). What would he see when he accelerated? Would A just suddenly become older while B instantly accelerates? Since time always seems normal for B, doesn't this mean that Earth would have to seem to speed up?

Perhaps answering another similar question would help. I'm traveling in a single inertial frame towards planet Z(near the speed of life). They've been watching me come for hundreds of years. To them, I've shown no sign of aging whatsoever.

However, to me, planet Z is moving towards me near the speed of light, and I'm perfectly still. They show no signs of aging. Yet, since I'm in my own inertial frame, I age normally. I can't "do more i my life" than anyone else. Thus, I die of old age long before I get to planet Z. Yet, according to the people on planet Z, I couldn't have aged (thus died), as no time seemed to pass, right?

7. Aug 14, 2007

### JesseM

In SR the rate you see a moving clock ticking using light-signals is different from the rate it actually is ticking in your frame, since signals from successive ticks of the clock have a different distance to travel to reach your eyes because of the object's movement. For example, if a clock is moving at 0.6c away from me, then in my frame it's ticking at $$\sqrt{1 - 0.6c^2/c^2}$$ = 0.8 the rate of my own clock, so it only ticks once every 1/0.8 = 1.25 seconds. But in 1.25 seconds, it will have moved a distance of (1.25 s)*(0.6c) = 0.75 light-seconds, so if it's moving away from me the light from a given tick takes 0.75 s longer to reach me than the light from the previous tick, meaning I only see the ticks every 1.25 + 0.75 = 2 seconds, while if it's moving towrds me the light from a given tick takes 0.75 s less time to reach me than the light from the previous tick, meaning I see the ticks every 1.25 - 0.75 = 0.5 seconds. So if the clock is moving away from me I see it ticking twice as slow as my own, and if it's moving towards me I see it ticking twice as fast as my own, but in both cases if I factor out the signal delays I'll calculate that it's really ticking 1.25 times as slow as my own. This visual change is known as the "Doppler shift", you can see here that in general the frequency I observe is $$\sqrt{\frac{1 + v/c}{1 - v/c}}$$ times the frequency of the signal in the emitter's frame, so if a clock is sending out signals at 1 tick per second in its frame and it's moving at +0.6c (positive meaning it's moving towards me), this equation tells me I'll see it ticking at $$\sqrt{\frac{1 + 0.6}{1 - 0.6}}$$ = $$\sqrt{4}$$ = 2 ticks per second, or twice as fast as my own clock, the same result I got above.

So, in the twin paradox, each twin will see the other twin's clock ticking slow before the turnaround when they're moving apart, and fast after the turnaround when they're moving towards each other. John Baez's page on The Twin Paradox has a subsection devoted to The Doppler Shift Explanation which shows what each twin actually sees throughout the process, and there's a nice spacetime diagram of when each twin's light-signals reach the other in fig. 2 of the Too Many Explanations section, I recommend taking a look at them.
In this case, since you didn't start at a common location, you have to worry about the issue of "simultaneity", meaning that different frames disagree about whether two events at distant locations happened at the "same time" or "different times". Say you're moving towards Z at 0.6c again, and say that in your frame, when you turn 30 you're 15 light years away from them, and that "Zed" on the planet is turning 30 "at the same time" in your frame. 25 years later according to your clock, you've traveled (25 years)*(0.6c) = 15 light years, so you're finally arriving at the planet, aged 30+25 = 55 years old, and you find that Zed has only aged 0.8*25 = 20 years, so he's only 50 years old. However in the Z frame the event of Zed turning 30 and your turning 30 were not simultaneous, instead the event of his turning 30 was simultaneous with the event of your turning 39, so in 20 years according to his clock, you've only aged 0.8*20 = 16 years, explaing why you're 39+16 = 55 years old when you arrive.

Now suppose there is another fellow Zeke on the planet who is turning 30 at the same time you turn 30 in the Z frame. He will actually be 11.25 years older than Zed, at an age of 61.25 when you finally arrive; he's aged 31.25 years, while you've only aged (31.25)*(0.8) = 25 years in the Z frame. In your frame you and Zeke did not turn 30 at the same time, instead you turned 30 when Zeke was already 41.25 years old (remember that you and Zed both turned 30 at the same time in your frame, and Zeke is always 11.25 years older than Zed), so in the 25 years it took you to reach the planet, Zeke only aged 25*0.8 = 20 years, bringing his age up to 61.25.

So, we see that in all these cases everyone agrees about how old each person is when they meet at a common location--there can be no disagreements about events which happen in the same local region of spacetime in relativity--but they disagree about who was aging slower during the trip, and this is connected to the fact that they disagree about the simultaneity of distant events.

8. Aug 16, 2007

### yogi

You are not going to get a satisfying answer to this question because there is a mix of observational times and actual aging that is going on - Einstein first treated the problem of two separated clocks synchronized in one frame and then predicted the time difference between the two clocks when one was put in motion toward the other and they were compared when they came together (Part IV of his 1905 paper). The clock put in motion will read less. Different authors explain the phenomena in different ways - some using SR only, others insisting that the fact of putting one clock in motion requires General Relativity.

9. Aug 16, 2007

### pervect

Staff Emeritus
It might help you to focus on what you observe, before you start interpreting it.

If someone had a giant clock, and you moved towards them at a high velocity, if you looked through a telescope you would see the clock hands spin around rapidly. Similarly, if they transmitted a video signal, it would when watched in real time appear to be moving rapidly.

If you move away from someone, the clock appears to run slowly - the video also appears to run slowly.

In fact, the speedup / slowdown factor, k, depends only on your relative velocity. This is the basis for Bondi K-calculus, which i'm not going to explain here, but you might want to look up sometime. (If you want a specific book title, try "relativity and common sense" by Bondi. It's a cheap dover paperback, you might also find it at your local library).

The accelerating case is also fairly simple - as you left, you would be moving away, and hence see everything in slow-motion. When you started to turn around, there would never be any discontinuity in what you saw, but you'd see it gradually speed up, reaching normal speed when you reach zero velocity, and going up to a high rate as you start to move towards the planet.

The idea of "time dilation" occurs when you interpret what you actually see in terms of a coordinate system, by subtracting the estimated "trip time" of the signals to come up with a time coordinate for events.

So, for instance, if you are approaching a planet at a very high speed, you will seem them moving very quickly, but you will estimate that almost all of this quickness is due to the fact that the distance between you and the planet is decreasing rapidly. When you correct for this by imposing your own ideas of simultaneity, you find that the rate of aging on the planet after corrections is slowed down.

Last edited: Aug 16, 2007
10. Sep 15, 2007

### Übermensch

Aha, sorry for disappearing. I ended up having surgery, and after the weeks in bed, I just forgot about this.

On a brighter note, I'm feeling I understand this much better now. The Zeke and Zedd story was quite a nice way to explain it. Thanks. :)


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# Of the 1400 college teachers surveyed, 42% said that they

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Of the 1400 college teachers surveyed, 42% said that they [#permalink]  24 Sep 2008, 17:57
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Of the 1400 college teachers surveyed, 42% said that they considered enaggin in research an essential goal. How many college teachers surveyed were women?

(1) In the survey 36% of the men and 50% of the women said that they considered engaging in research an essential goal
(2) In the survey, 288 men said that they considered engaging in research an essential goal

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-1400-college-teachers-surveyed-42-said-they-consid-92076.html
[Reveal] Spoiler: OA
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Re: DS - GMAT Prep CAT [#permalink]  24 Sep 2008, 18:08
GODSPEED wrote:
Of the 1400 college teachers surveyed, 42% said that they considered enaggin in research an essential goal. How many college teachers surveyed were women?

1. In the survey 36% of the men and 50% of the women said that they considered engaging in research an essential goal
2. In the survey, 288 men said that they considered engaging in research an essential goal

Say total men surveyed =m
women surveyed =w

hence m+w=1400
1)
(50/100)w +(36/100 * m) = 42/100 * (m+w)
=>3/4*m=w also m+w=1400 => m,w can be calculated
SUFFI

2)This is not sufficient to find out men and women count
INSUFFI
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Re: DS - GMAT Prep CAT [#permalink]  25 Sep 2008, 06:17
Hi Spriya,

I agree with 1st Statement.

But I believe st2 is also sufficient. Please clarify me if i am wrong.

42/100 * 1400 = m + w ( who said they considered engagin in research an essential goal. )

588 = m + w. As m given it is 288.

so w = 300.

It is sufficient.
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Re: DS - GMAT Prep CAT [#permalink]  25 Sep 2008, 06:41
1400 – 588 considered (42%)
Men = m
Women = w

(1) 0.36m+0.50w = 0.42 (m+w)
0.36m+0.50w = 588
SUFF

(2)
588 – 288 = 300 women
SUFF

D
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Re: DS - GMAT Prep CAT [#permalink]  25 Sep 2008, 15:34
scthakur wrote:
Twoone wrote:
Hi Spriya,

I agree with 1st Statement.

But I believe st2 is also sufficient. Please clarify me if i am wrong.

42/100 * 1400 = m + w ( who said they considered engagin in research an essential goal. )

588 = m + w. As m given it is 288.

so w = 300.

It is sufficient.

Agree. D should be the answer.

I suppose it's the "total number of women surveyed" we require over here and NOT women agreed to the point.....essential goal....
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Re: Of the 1400 college teachers surveyed, 42% said that they [#permalink]  12 Oct 2013, 12:23
GODSPEED wrote:
Of the 1400 college teachers surveyed, 42% said that they considered enaggin in research an essential goal. How many college teachers surveyed were women?

1. In the survey 36% of the men and 50% of the women said that they considered engaging in research an essential goal
2. In the survey, 288 men said that they considered engaging in research an essential goal

What's the OA for this one?
I chose (E)

1st statement clearly insuff as we can't do the split
2nd statement clearly insuff as we know nothing bout women

From both statatement we can know the # of women that considerd engaging in reseach an essential goal but we don't know how many of them were surveyed in total.

Let me know if this makes sense
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Re: Of the 1400 college teachers surveyed, 42% said that they [#permalink]  12 Oct 2013, 13:54
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Expert's post
jlgdr wrote:
GODSPEED wrote:
Of the 1400 college teachers surveyed, 42% said that they considered enaggin in research an essential goal. How many college teachers surveyed were women?

1. In the survey 36% of the men and 50% of the women said that they considered engaging in research an essential goal
2. In the survey, 288 men said that they considered engaging in research an essential goal

What's the OA for this one?
I chose (E)

1st statement clearly insuff as we can't do the split
2nd statement clearly insuff as we know nothing bout women

From both statatement we can know the # of women that considerd engaging in reseach an essential goal but we don't know how many of them were surveyed in total.

Let me know if this makes sense

Of the 1400 college teachers surveyed, 42% said they considered engaging in research an essential goal. How many of the college teacher surveyed were women?

(1) In the survey 36% of men and 50% of women said that they consider engaging in research activity an essential goal.
(2) In the survey 288 men said that they consider engaging in research activity an essential goal.

(1) $$m+w=1400$$ and $$1400*0.42=0.36*m+0.5*w$$. Two unknowns two equations. Sufficient.
(2) From this we can calculate only that $$1400*0.42-288=300$$ women consider engaging in research activity an essential goal. No other info. Not sufficient.

OPEN DISCUSSION OF THIS QUESTION IS HERE: of-the-1400-college-teachers-surveyed-42-said-they-consid-92076.html
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Re: Of the 1400 college teachers surveyed, 42% said that they   [#permalink] 12 Oct 2013, 13:54
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每週問題 May 26, 2014

Let $A$ be a nonsingular real matrix. If all singular values of $A$ are equal, show that $A=cU$, where $c$ is a real number and $U$ is a real orthogonal matrix.


text:
Number system

jackson6612
I'm trying to learn some math. I was studying about decimal number system.

The place values of base 10 are: ... 10000s 1000s 100s 10s 1s (decimal point) 1/10s 1/100s 1/1000s 1/10000s ...

In positional notation system every lower valued place is used to feed the higher one. Any place value left to the other is higher one.

999 means that there are 9 groups of 1s, 9 groups of 10s, and 9 groups of 100s (from right to left).

0.999 = 9/10 + 9/100 + 9/1000 = 999/1000.
0.9 = 9/10 = 90/100 = 900/1000

0.009 = 0/10 + 0/100 + 9/1000 = It means that there are 9 groups of 1/10000 things.

When 1000 is divided by 100 it can be said 1000 objects have been divided into 10 parts. What would be said when 1/100 is divided by 10? Please guide me on this with some example. Thanks.

jackson6612
Perhaps, it means this. Suppose, a pizza is divided into 100 slices and there are also 100 persons to eat those slices. Each person gets 1 slice which 1/100 of the pizza. One person goes out and he encounters some of his 9 friends as he was just going to take first bite of that slice. Now there are total 10 persons who need an equal share of that slice. It is divided into 10 smaller pieces of equal quantity. Now the original slice, 1/100 of the pizza, has been sub-divided into 10 pieces further pieces, i.e. 1/100 divided by 10 which is equivalent to 1/100 x 1/10 or 1/1000. Do I make sense?

jackson6612

sjb-2812
Sounds a reasonable interpretation to me.

jackson6612
Thanks, Sjb.

When the person went out he had one slice, 1/100 of the original pizza. Take that once as one whole piece, let's a mini pizza. That pizza was divided into 10 pieces, one piece being 1/10 of the mini. In the context of that mini pizza it would be 1/10 but taking that mini as a slice of the original pizza, it would be 1/1000 or (1/100 x 1/10), why? There aren't 1000 persons that one piece now equals 1/1000. That's where mathematics starts sounding weird. 10/1000 = 1/100 = one piece of the original pizza.

100/100 = 100 x (10/1000) = original pizza
for 1 slice, 1/100, there are ten 1/1000 pieces
for 10 slices, 10 x 1/100, there are hundred 1/1000 pieces
for 100 slices, 100 x 1/100, there are 1000 1/1000 pieces

So in this context 1/100 = 10/1000 means the original slice which was 1/100 of the pizza is equal to 10 new slices, where one slice is 1/1000 when viewed in context of the original pizza. Physically, not all the pizza was divided into 1000 pieces rather only one slice which was 1/100 was further divided into 10 pieces. But this is how mathematics work. In mathematics you have to manipulate expressions to find answers and solve real world problems at the expense of some nonsensical steps (you could call them unreal, abstract mathematical entities) while reaching the final 'sensible' answer.

Thank you for all the help.

epkid08
Why do say nonsensical?

jackson6612
Perhaps, it's a poor choice of words. But I think it's true sometimes (most of the time) mathematics uses steps, or make use of some ideas which aren't real or not too match close to reality. We have to make sense out such 'nonsense'.

Sometimes, one may get answer where a step involves 1/2 of a human being. But in reality a human being doesn't exist in a half piece. I hope you get my point.

Mentor
When 1000 is divided by 100 it can be said 1000 objects have been divided into 10 parts. What would be said when 1/100 is divided by 10? Please guide me on this with some example. Thanks.
When 1000 is divided by 100, the 1000 items are being divided into 100 parts, not 10 parts. Each part consists of 100 items.

Mentor
Thanks, Sjb.

When the person went out he had one slice, 1/100 of the original pizza. Take that once as one whole piece, let's a mini pizza. That pizza was divided into 10 pieces, one piece being 1/10 of the mini. In the context of that mini pizza it would be 1/10 but taking that mini as a slice of the original pizza, it would be 1/1000 or (1/100 x 1/10), why?
Because 1/10 of 1/100 = 1/1000. One tenth "of: something means 1/10 "times" that something.
There aren't 1000 persons that one piece now equals 1/1000. That's where mathematics starts sounding weird. 10/1000 = 1/100 = one piece of the original pizza.
The other 99 pieces of the original pizza went to the other 99 people. The one small piece (1/100 of the original pizza) is now being shared by 10 people, so each of these people gets 1/10 of the small piece, which is 1/1000 of the original pizza.

I don't understand what problem you are having with this.
100/100 = 100 x (10/1000) = original pizza
for 1 slice, 1/100, there are ten 1/1000 pieces
for 10 slices, 10 x 1/100, there are hundred 1/1000 pieces
for 100 slices, 100 x 1/100, there are 1000 1/1000 pieces

So in this context 1/100 = 10/1000 means the original slice which was 1/100 of the pizza is equal to 10 new slices, where one slice is 1/1000 when viewed in context of the original pizza. Physically, not all the pizza was divided into 1000 pieces rather only one slice which was 1/100 was further divided into 10 pieces. But this is how mathematics work. In mathematics you have to manipulate expressions to find answers and solve real world problems at the expense of some nonsensical steps (you could call them unreal, abstract mathematical entities) while reaching the final 'sensible' answer.

Thank you for all the help.

jackson6612
I don't understand what problem you are having with this.
\

Thanks a lot, Mark. I was just explaining it to myself. It would be nice of you if could comment a bit on that bold part too. Once again, thanks.

Mentor
Presumably you mean this:
jackson6612 said:
But this is how mathematics work. In mathematics you have to manipulate expressions to find answers and solve real world problems at the expense of some nonsensical steps (you could call them unreal, abstract mathematical entities) while reaching the final 'sensible' answer.
This doesn't make much sense to me. Can you think of some example problems you've seen that entail using nonsensical steps to arrive at a sensible answer? Preferably something other than your question about the pizza divided up into 109 pieces (99 pieces each 1/100 of the pizza + 10 pieces each 1/1000 of the pizza).

jackson6612
Mark: Thanks a lot for the comment. So you found my example about pizza being divided into 109 pieces nonsensical! :-) It was an extra, extra large pizza!

Perhaps, the use of the term 'nonsensical steps' is misleading. What I was unsuccessfully trying to say was that mathematics transform the entire real world into its own world where some of the steps or answers wouldn't make that much sense to us. Those steps aren't nonsensical (allow me to use this term again) in their own right, they seem that way from human point of view. Am I again speaking nonsense here?!

Best wishes
Jack

jackson6612
I'm trying to learn some math. I was studying about decimal number system.

The place values of base 10 are: ... 10000s 1000s 100s 10s 1s (decimal point) 1/10s 1/100s 1/1000s 1/10000s ...

In positional notation system every lower valued place is used to feed the higher one. Any place value left to the other is higher one.

999 means that there are 9 groups of 1s, 9 groups of 10s, and 9 groups of 100s (from right to left).

0.999 = 9/10 + 9/100 + 9/1000 = 999/1000.
0.9 = 9/10 = 90/100 = 900/1000

0.009 = 0/10 + 0/100 + 9/1000 = It means that there are 9 groups of 1/10000 things.

When 1000 is divided by 100 it can be said 1000 objects have been divided into 10 parts. What would be said when 1/100 is divided by 10? Please guide me on this with some example. Thanks.

Hi

Is it correct to say "9 groups of 1s"? It doesn't make any sense to me. To say that in '678' there are 6 groups of 100s makes sense.

In the quoted text I said in 0.009 there are 9 groups of 1/0000. I didn't say there 9 groups of 1/000. Why? Because the logic which holds true for the digits on the left side of decimal point (as I said there 6 groups of 100s in 678) doesn't hold true for the digits on the right side of the point. 1/1000 = 10 x 10/10000. That is, 1/1000 contains in itself 10 one-thousandth pieces of something. Correct. Please guide me.

Homework Helper
Hi

Is it correct to say "9 groups of 1s"?
Would it make sense to you to say that there were 9 groups, each containing 1 thing?

It doesn't make any sense to me. To say that in '678' there are 6 groups of 100s makes sense.
Why would it make more sense to say that "there are 6 groups of 100s" but not "8 groups of 1s"? I would be inclined to say "6 groups of 100 things" and "8 groups of 1 thing".

In the quoted text I said in 0.009 there are 9 groups of 1/0000. I didn't say there 9 groups of 1/000. Why?
Because you were wrong? "0.009" means 9/1000, not 9/10000.

Because the logic which holds true for the digits on the left side of decimal point (as I said there 6 groups of 100s in 678) doesn't hold true for the digits on the right side of the point. 1/1000 = 10 x 10/10000. That is, 1/1000 contains in itself 10 one-thousandth pieces of something. Correct. Please guide me.

jackson6612
Thanks a lot, Halls.

Would it make sense to you to say that there were 9 groups, each containing 1 thing?

No, it won't. Because the definition of 'group' implies that there are two or more things involved. So, guide me on this.

Why would it make more sense to say that "there are 6 groups of 100s" but not "8 groups of 1s"? I would be inclined to say "6 groups of 100 things" and "8 groups of 1 thing".

Okay, poor choice of words. I also intended to say 6 groups 100 things, and not of 100s. Sorry.

Because you were wrong? "0.009" means 9/1000, not 9/10000.

No, I'm not completely wrong. In the original post, I said:
0.009 = 0/10 + 0/100 + 9/1000 = It means that there are 9 groups of 1/10000 things.

It wouldn't make sense, at least to me, to say that there 9 groups of 1/1000 things. Then, how would extend that idea of grouping to the digits on the right side of decimal point?

1/1000 = 10 x 10/10000. That is, 1/1000 contains in itself 10 one-thousandth pieces/things of something. Correct? Please guide me.

Best wishes
Jackson

Mentor
Thanks a lot, Halls.
HallsofIvy said:
Would it make sense to you to say that there were 9 groups, each containing 1 thing?
No, it won't. Because the definition of 'group' implies that there are two or more things involved. So, guide me on this.
That's an overly narrow definition of this term. According to one of my dictionaries, a group is an assemblage of persons or objects. It doesn't specify that there must be two or more things in a group.

HallsofIvy said:
Because you were wrong? "0.009" means 9/1000, not 9/10000.
No, I'm not completely wrong. In the original post, I said:

0.009 = 0/10 + 0/100 + 9/1000 = It means that there are 9 groups of 1/10000 things.
No, you were wrong then and now. 0.009 means 9/1000. There are not "9 groups of 1/10000 things" whatever that means. There is a group with 9 things, each of which is 1/1000, to stay consistent with your phrasing. I would talk instead about how many tenths, hundredths, thousandths, etc., there are. In this case there are no tenths, no hundredths, and 9 one-thousandths. Period.
It wouldn't make sense, at least to me, to say that there 9 groups of 1/1000 things. Then, how would extend that idea of grouping to the digits on the right side of decimal point?

1/1000 = 10 x 10/10000.
No, not at all. On the left side you have 1/1000. On the right side you have 100/10000, which is the same as (equal to) 1/100.
That is, 1/1000 contains in itself 10 one-thousandth pieces/things of something.
Correct?
Again, no. 1/1000 could be divided up into 10 one-ten-thousandths. I.e., 1/1000 = 10 * 1/10000.

Think about it in terms of coins. A dime is 1/10 of a dollar, or $.10. A dime consists of 10 cents, or 10 hundredths of a dollar, so in terms of dollars as the unit, 1/10 = 10 * 1/100. jackson6612 That's an overly narrow definition of this term. According to one of my dictionaries, a group is an assemblage of persons or objects. It doesn't specify that there must be two or more things in a group. No, you were wrong then and now. 0.009 means 9/1000. There are not "9 groups of 1/10000 things" whatever that means. There is a group with 9 things, each of which is 1/1000, to stay consistent with your phrasing. I would talk instead about how many tenths, hundredths, thousandths, etc., there are. In this case there are no tenths, no hundredths, and 9 one-thousandths. Period. No, not at all. On the left side you have 1/1000. On the right side you have 100/10000, which is the same as (equal to) 1/100. Again, no. 1/1000 could be divided up into 10 one-ten-thousandths. I.e., 1/1000 = 10 * 1/10000. Think about it in terms of coins. A dime is 1/10 of a dollar, or$.10. A dime consists of 10 cents, or 10 hundredths of a dollar, so in terms of dollars as the unit, 1/10 = 10 * 1/100.

Thanks a lot, Mark.

Mark, I think now I'm also going to learn English from you too. But I would disagree on that definition of a group. Please remember that I'm an English learner. M-W nowhere says that a group could consist of one thing: http://www.merriam-webster.com/dictionary/group. Please correct me. Thanks.

'678' would mean that there are 6 groups with each containing 100 items, and 7 groups each with 10 items... I don't really get what I say for '0.009'. Should I say something like there 9 groups with each containing 1/1000 items. But the problem is '1/1000' is not even a one complete item, let alone items. Please guide me. Thanks.

Best regards
Jackson

Mentor
In my Merriam-Webster dictionary, definition 2a says "a number of individuals assembled together or having some unifying relationship"
2b says "an assemblage of objects regarded as a unit"

"a number of individuals" doesn't specify that there have to be at least two.
Also, I interpret "an assemblage of objects" to mean one or more.

For .009 it's fine to say 9 thousandths and be done with it.

With your first example, 678, as you point out, there are 6 groups of 100 each, 7 groups of 10 each, and 8 groups, each with 1 thing. Once you get on the other side of the decimal point, phrasing things in terms of groups is not very useful, I don't believe.

Maybe it's better to completely abandon the "groups" terminology and recognize that where a numeral is in a decimal number indicates which power of 10 it multiplies.

For example, 1205.37
Going from the left end to the right, we have thousands, hundreds, tens, ones, tenths, and finally hundredths.

Our decimal number system is a kind of shorthand for saying
1 x 1000 +
2 x 100 +
0 x 10 +
5 x 1 +
3 x 1/10 +
7 x 1/100

Or, in another way
1 x 103 + 2 x 102 + 0 x 101 + 5 x 100 + 3 x 10-1 + 7 x 10-2

Mensanator
I'm trying to learn some math. I was studying about decimal number system.

The place values of base 10 are: ... 10000s 1000s 100s 10s 1s (decimal point) 1/10s 1/100s 1/1000s 1/10000s ...

In positional notation system every lower valued place is used to feed the higher one. Any place value left to the other is higher one.

999 means that there are 9 groups of 1s, 9 groups of 10s, and 9 groups of 100s (from right to left).

0.999 = 9/10 + 9/100 + 9/1000 = 999/1000.
0.9 = 9/10 = 90/100 = 900/1000

0.009 = 0/10 + 0/100 + 9/1000 = It means that there are 9 groups of 1/10000 things.

When 1000 is divided by 100 it can be said 1000 objects have been divided into 10 parts. What would be said when 1/100 is divided by 10? Please guide me on this with some example. Thanks.

To divide rayionals, invert abd multiply:

1/100 / 10 = 1/100 * 1/10 = 1/1000


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# checking if matrix is diagonalizable

• Mar 29th 2011, 11:25 AM
Jskid
checking if matrix is diagonalizable
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same (e.g. $\lambda=1$ but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?
• Mar 29th 2011, 11:30 AM
tonio
Quote:

Originally Posted by Jskid
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same http://mathhelpforum.com/advanced-algebra/176204-checking-if-matrix-diagonalizable.html#post634879\" rel=\"nofollow\">
My textbook gives two theorems that are used to determine if a matrix is diagonilable
1)An n x n matrix is diagonalizable if and only if it has n linearly independent eigenvectors
2)If the roots of the characteristic polynomial of an n x n matrix A are all distinct then A is diagonizable

So if I have characteristic polynomial with two roots that are the same (e.g. $\lambda=1$ but it has n linearly independent eigenvectors, which one takes presidents? Or should the two always agree and this means I made a mistake?

the 2nd statement is sufficent but not required. It is just saying the if you have no repeated eigenvalues and using the fact that each eigenspace must have dimention at least 1 you will always get n linearly independent eigenvectors.

The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.
• Mar 29th 2011, 11:51 AM
Jskid
Quote:

Originally Posted by TheEmptySet
The problem that can occur is if you have a repeated eigenvalue say of multiplicity 2 the eigenspace for that eigenvalue may only have 1 linearly indpendant eigenvector. So the matrix is not diagonalizable. But if you do get two linearly independant eigenvectors then the matrix would be diagonalizable.

That's what happened and confused me.


text:
# Bayesian reasoning

## Idea

Bayesian reasoning is an application of probability theory to inductive reasoning (and abductive reasoning). The perspective here is that, when done correctly, inductive reasoning is simply a generalisation of deductive reasoning; the latter simply restricts to the probabilities $0$ and $1$.

## Stuff

This was moved from inductive reasoning, and I hope to edit it further today, but now I have to go:

It can be shown by so-called “Dutch Book” arguments, that a rational agent must set their degrees of belief in such a way that they satisfy the axioms of probability theory. For example, if your degree of belief that the next toss of a coin will show heads is $p$, then you should believe it will show tails to degree $\left(1-p\right)$, otherwise it will be possible for someone to take betting positions against you making a loss certain whatever the outcome.

Using Bayes' Rule?, degrees of belief can be updated on receipt of new evidence.

$P\left(h\mid e\right)=P\left(e\mid h\right)\cdot \frac{P\left(h\right)}{P\left(e\right)},$P(h|e) = P(e|h) \cdot \frac{P(h)}{P(e)},

where $h$ is a hypothesis and $e$ is evidence.

The idea here is that when $e$ is observed, your degree of belief in $h$ should be changed from $P\left(h\right)$ to $P\left(h\mid e\right)$. This is known as conditionalizing. If $P\left(h\mid e\right)>P\left(h\right)$, we say that $e$ has provided confirmation for $h$.

Typically, situations will involve a range of possible hypotheses, ${h}_{1},{h}_{2},\dots$, and applying Bayes’ Rule will allow us to compare how these fare as new observations are made. For example, comparing the fate of two hypotheses,

$\frac{P\left({h}_{1}\mid e\right)}{P\left({h}_{2}\mid e\right)}=\frac{P\left(e\mid {h}_{1}\right)}{P\left(e\mid {h}_{2}\right)}\cdot \frac{P\left({h}_{1}\right)}{P\left({h}_{2}\right)}.$\frac{P(h_1|e)}{P(h_2|e)} = \frac{P(e|h_1)}{P(e|h_2)}\cdot \frac{P(h_1)}{P(h_2)}.

How to assign prior probabilities to hypotheses when you don’t think you have an exhaustive set of rivals is not obvious. When astronomers in the nineteenth century tried to account for the anomalies in the position of Mercury’s perihelion, they tried out all manner of explanations: maybe there was a planet inside Mercury’s orbit, maybe there was a cloud of dust surrounding the sun, maybe the power in the inverse square law ought to be (2 - $ϵ$),… Assigning priors and changing these as evidence comes in is one thing, but it would have been wise to have reserved some of the prior for ‘none of the above’.

Interestingly, one of the first people to give a qualitative sketch of how such an approach would work was George Polya in ‘Mathematics and Plausible Reasoning’ (Polya), where examples from mathematics are widely used. The idea of a Bayesian account of plausible reasoning in mathematics surprises many, it being assumed that mathematicians rely solely on deduction.

Revised on June 20, 2013 23:51:00 by Urs Schreiber (82.169.65.155)


text:
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Lesson Three | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: Basic Speller Teacher Materials Go to the latest version.

# 13.3: Lesson Three

Created by: CK-12

## Review of Twinning

1. The Rule of Simple Addition says that elements combine without change unless you know some special reason for making a change. One special reason is twinning:

Twinning Rule. You twin the final consonant of a free stem that has one vowel sound in it when you add a suffix that starts with a vowel and the stem ends in the pattern CVC. You twin the final consonant of a free stem that has more than one vowel sound in it when you add a suffix that starts with a vowel and the stem ends CVC only when there is stress on the last vowel of the stem before and after the suffix is added:

twin + ing = twin + n + ing = twinning

occur + ence = occur + r + ence = occurrence

2. Analyze each of the following words into free stem plus suffix. Show any cases of twinning in your analysis. Then answer the questions in the columns on the right. Assume that in any stems that have only one vowel sound, that vowel is stressed:

Word = Free Stem + Suffix Does the suffix start with a vowel? Does the stem end in the pattern CVC? Is there stress on the last vowel in the stem before you add the suffix? Does the stress stay on the last vowel of the stem after you add the suffix?
twinning = twin + n + ing Yes Yes Yes Yes
occurrence = occur + r + ence Yes Yes Yes Yes
kidnapper = kidnap + p + er Yes Yes Yes Yes
lucky = luck + y Yes No Yes Yes
committed = commit + t + ed Yes Yes Yes Yes
symbolic = symbol + ic Yes Yes No Yes
commitment = commit + ment No Yes Yes Yes
displayed = display + ed Yes No Yes Yes
limiting = limit + ing Yes Yes No No
exceeding = exceed + ing Yes No Yes Yes
excelled = excel + l + ed Yes Yes Yes Yes
cooking = cook + ing Yes No Yes Yes
repellant = repel + l + ant Yes Yes Yes Yes
compelling = compel + l + ing Yes Yes Yes Yes
logical = logic + al Yes Yes No No
informer = inform + er Yes No Yes Yes
submits = submit + s No Yes Yes Yes
exacting = exact + ing Yes No Yes Yes
recruiter = recruit + er Yes No Yes Yes

3. Look over the results of your work. You should find that for each word in which twinning occurred you have “Yes” in all four columns on the right. You should also find that for each word in which twinning did not occur you have at least one “No” in the columns on the right. If things did not work out that way, check over your work. If you get stuck, don't be afraid to ask for some help.

Twinning Rule. You twin the final consonant of a free stem that has one vowel sound in it when you add a suffix that starts with a vowel and the stem ends in the pattern CVC. You twin the final consonant of a free stem that has more than one vowel sound in it when you add a suffix that starts with a vowel and the stem ends in the pattern CVC only when there is stress on the last vowel of the stem before and after the suffix is added.

4. Combine the following free stems and suffixes. Show any cases of twinning:

Free stem + Suffix = Word
commit + t + ee = committee
complex + ity = complexity
remark + able = remarkable
logic + ian = logician
symbol + ism = symbolism
occur + ence = occurrence
refer + ence = reference
recruit + ing = recruiting
repel + ing = repelling
overlook + ed = overlooked
republic + an = republican
reveal + ing = revealing
compel + ing = compelling
resubmit + ed = resubmitted
kidnap + ing = kidnapping

Teaching Notes.

Item 1. The treatment of twinning is quite abbreviated here. For a more detailed and extended introduction to twinning, see Book 1, Lessons 34-37 (for twinning in stems with a single vowel sound) and Book 3, Lessons 45-47 (for twinning in stems with two or more vowel sounds). For more on twinning see AES, pp. 161-76.

Item 2. A sharp-eyed student may wonder why excellent has two <l>’s, in seeming contradiction of the Twinning Rule (since the stress shifts off of the second vowel in the stem when the suffix is added). Originally the free stem excel was often spelled excell. There were two <l>’s in the Latin source word. It was not until the $18^\mathrm{th}$ or $19^\mathrm{th}$ century that the spelling excel with one <l> became standard, part of a general tendency then to avoid ending free stems with more than one vowel sound with a double-<l>. It is possible that the spelling of excellent was standardize before the change of excell to excel. Or we could treat excellent as a case of the British tendency not to require stress on the second vowel of the stem when twinning: excel + l + ent. I would analyze excellent to excell + ent, positing excell as a bound stem in partnership with the free stem excel. For more on these sets of free and bound stems see AES, pp. 170-72.

Item 4. Re: complexity. Since final <x> always spells the complex sounds [ks], free stems that end in <x>, even though in terms of letters they seem to end CVC, in terms of sounds they end CVCC and thus do not twin. We never twin the letter <x>.

## Subjects:

1 , 2 , 3 , 4 , 5

## Date Created:

Feb 23, 2012

Apr 29, 2014
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text:
Q

# From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

12. From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.

Views

Let the height of the cable tower be (AB = $h+7$)m
Given,
Height of the building is 7 m and angle of elevation of the top of the tower $\angle ACE = 60^o$ , angle of depression of its foot $\angle BCE = 45^o$.

According to question,

In triangle $\Delta DBC$,
$\\\tan 45^o = \frac{CD}{BD} = \frac{7}{BD} = 1\\ BD =7\ m$
since DB = CE = 7 m

In triangle $\Delta ACE$,

$\\\tan 60^o = \frac{h}{CE}=\frac{h}{7}=\sqrt{3}\\ \therefore h = 7\sqrt{3}\ m$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3}\) m$

Exams
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text:
1. Mar 1, 2012

jaja1990

In a try to test my knowledge of Newton's laws, I have posed the following question to myself.

My attempt:

(The arrows indicate direction.)

I have been informed that the solution is invalid, and that the question itself is wrong. Can you tell me why the problem is wrong/invalid?
Someone tried to explain to me, but I didn't really understand.

2. Mar 1, 2012

Staff: Mentor

For one thing:


text:
# Momentum of a gas expanding into a vacuum chamber in outer space

Consider a gas expanding adiabatically from a gas chamber through an orifice into a vacuum chamber as shown in the schematic figure. In order for significant number of gas molecules (which we refer to as particles for simplicity), to transition into the vacuum chamber, the center of mass of the collection of all the gas particles must have a velocity which points rightwards towards the vacuum chamber. We assume that the container shown in the figure below is placed far away from all other physical objects, say in outer space, so that there are no external forces acting on it or it's contents.

How is linear momentum conserved during the transient phase of expansion of the gas from it's resting chamber to the vacuum chamber?

$$\qquad \qquad \qquad \qquad \qquad$$

• If you imagine the vacuum chamber to be space, and the "gas chamber" (cough) a rocket, you may get an idea. Famously, the center of mass of a "rocket" ("rocket" being the entirety of its mass, including fuel, at "standstill" before ignition) never moves, because -- how could it? Feb 9 at 12:46
• @Peter-ReinstateMonica the analogy you provided is insightful. Thank you! Feb 9 at 18:46

At any moment there will be particles moving in each direction, bouncing on the walls and changing direction. The pressure on the walls of the chamber is due to this bouncing.

Those who go through the hole are not reflected by the wall. Thus though the pressure rapidly equalizes in the entire left chamber, the force on the left-hand-side wall is the pressure times the entire surface, while on the right-hand-side wall, the pressure acts on the surface minus the size of the hole. It thus transmits a non vanishing total force on the left hand side chamber.

As long as the amount of gas in the right hand side chamber is small, during the transient phase, there is no compensating force on the right-hand-side chamber. The entire solid box, during the transient phase, receives a total momentum to the left, and if not connected to any object, (floating in the vacuum ?) it will move to the left.

When pressures are equalized in both chambers, the total force on the solid box as a whole is zero. Not only that, but since the gas has stopped moving as a whole, the right hand side chamber has finally given back to the gas all the momentum the left hand side chamber had received at the beginning, so there is no momentum left, neither in the gas nor in the solid and all macroscopic motion stops.

But the box as a whole will have moved left before stopping, by just the amount needed for the center of mass of the whole thing, box plus gas, to stay put.

If the box is on a table, for instance, friction will resist to its motion, but some momentum is transmitted to the table and from the feet of the table to the ground.

• @kbakshi314 The point is that a gas in an adiabatic situation is exactly the opposite of a system with no "viscous losses". Viscous losses are dominant rather than a small correction. More precisely, the mean free path of a particle between two collisions is typically extremely small compared to the dimension of a "man-usable" box. Even if the first original jet of particles bounces off the right-hand-side wall of the vacuum chamber, most of them will bounce on the separation wall. Very, very few will ever go through the hole ! Where do you see anything like a two spring-mass system ? Feb 9 at 19:07
• @kbakshi314 You wrote "However, it is not clear why microscopic motion will stop " I assume this is a misprint. You mean "macroscopic", right ? To go back to your question all the KE of the gas that went into the initial "vacuum" chamber will rapidly dissipate there and never induce any return into the left-hand-side one. Feb 9 at 19:09
• What if we look at it in the container's frame of reference? Does the expansion of the gas cause the outside universe to move? Is this consistent with the stipulation that the system does not interact with the rest of the universe? Feb 9 at 20:19
• @JohnBollinger What do you call "the container frame of reference" ? The container is first accelerated then decelerated. Its "frame of reference" is not inertial. Calculations in a non inertial referential are famously difficult. I am not an expert in General Relativity, I leave the calculations to you. Feb 9 at 20:30
• @kbakshi314 I cannot keep commenting on all the comments! Briefly a) the "bullet and gun" analogy is great to describe the initial acceleration of the whole container when the gas just begins to move out. But the many, many bullets do not fly away, they bounce on the far wall, then back on the separation wall (except for very few ones that happen to go through the hole), then back, etc. etc. So the full situation is more complex than an "ordinary bullet and gun". But since the gas keeps coming in, there will be no oscillations, there are always a new bullet "in" for each reflected one. Feb 9 at 20:43

Because the gas imparts a transient recoil to the enclosure.

It is no different, in principle, to a pellet being fired from an air gun and hitting a mesh welded over the opening of the barrel. The expanding air forces the pellet along the barrel causing a recoil. When the pellet hits the mesh it causes an impulse on the gun opposite to the direction of the original recoil. We end up with the centre of gravity of the gun/pellet unchanged, but the gun will be displaced backwards somewhat and the pellet displaced forwards.

What you are talking about is often refereed as Adiabatic free expansion of a gas or Joule expansion

I would like to quote the page Adiabatic process from wikipedia

For an adiabatic free expansion of an ideal gas, the gas is contained in an insulated container and then allowed to expand in a vacuum. Because there is no external pressure for the gas to expand against, the work done by or on the system is zero. Since this process does not involve any heat transfer or work, the first law of thermodynamics then implies that the net internal energy change of the system is zero. For an ideal gas, the temperature remains constant because the internal energy only depends on temperature in that case. Since at constant temperature, the entropy is proportional to the volume, the entropy increases in this case, therefore this process is irreversible.

Extensive description can be found in Joule expansion (see note number 3).

So to conclude, no work against the environment during the expansion, means no change in quantities such as momentum, energy, temperature ...


text:
# Why can't mathematics be formalised in terms of classes rather than sets? [closed]

I've always been curious about the seeming compulsion to found mathematics upon sets, be it ZF(C) or some other system. Of course, there are other approaches these days like category theory and type theory (themselves inextricably linked by the Curry–Howard–Lambek correspondence), but these rather seek an entirely different approach. What I'm really getting at is: why can't we simply deal with (improper) classes? That is, throw away sets as reified mathematical objects and just deal with the classes implicitly defined by predicates in some logical system. Does this indeed make certain areas of mathematics inaccessible? Are there other problems I might not have considered? I would appreciate if someone could enlighten me here in a general way, albeit perhaps also with some specific cases within mathematical subfields.

-

## closed as unclear what you're asking by Andreas Blass, Simon Thomas, Stefan Kohl, Monroe Eskew, Ricardo AndradeMay 25 at 23:06

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

The question seems to admit many interpretations, but the most natural one seems to be that you want to deal only with classes and that these (being "improper") will not be elements of other classes. So the elements of classes will have to be some non-class entities that haven't been specified. If, on the other hand, you want to allow some (but not all?) classes to be members of others, then just define "set" to mean "class that can be a member of another class" and you're back to set-class theories like NBG. Maybe you intend something entirely different, but what? –  Andreas Blass May 25 at 14:16
@AndreasBlass: Well, I left this aspect open somewhat on purpose. But let's be more specific, for the sake of argument. First, we could envisage a universe of discourse being something like the naturals (recursively defined), with only first-order classes. Alternatively, we could work in some higher-order logic, thus allowing classes to be members of other classes, in a stratified manner. –  Noldorin May 25 at 14:21
I voted to close because there was no clear question here. Rhetorical questions in philosophical areas, with no clear content, may indeed be part of life, but they are not (nor should they be) part of MathOverflow. –  Andreas Blass May 25 at 19:28
Noldorin, you have an interesting question which could deserve a discussion from like-minded individuals. This forum is the wrong place for it. The closest place on the SE network for such a discussion is the chat rooms. This forum is more for question and answer format: read the FAQ for what formats and content are encouraged here. If you have a specific question that avails itself of a specific answer and is in line with the intent of this forum, you are encouraged to ask that kind of question. –  The Masked Avenger May 25 at 19:52
I'm going to remove comments that started us in the direction of incivility. Noldorin has asked why the votes to close, which is perfectly reasonable. So are responses from closers like Andreas Blass, who sets a nice tone for civility. Let's keep it that way, please. –  Todd Trimble May 25 at 20:01

just deal with the classes implicitly defined by predicates in some logical system.

and

we could envisage a universe of discourse being something like the naturals (recursively defined), with only first-order classes.

seems to me to yield an informal description of full second-order arithmetic, $Z_2$.

As to whether mathematics can be formalized in such a system, the answer is yes, virtually all mainstream mathematics can even be formalized in weak subsystems of $Z_2$. See the book Subsystems of Second Order Arithmetic by Stephen Simpson.

-
Intriguing! Thank you for this answer. This is very much what I was hoping for. Do these formalisations avoid set theory altogether then, I take it? –  Noldorin May 25 at 14:41
I am reading you as wanting to "eliminate the ontology of sets" by letting every predicate function as a definable collection. $Z_2$ does this via the full comprehension scheme. –  Nik Weaver May 25 at 14:41
Yep, this is very much what I'm interested in. Thanks Nik. –  Noldorin May 25 at 14:46
I should add that full comprehension allows predicates which reference quantification over all predicates. If that kind of circularity bothers you, look at the subsystem $ACA$ (which still suffices for, arguably, all or nearly all mainstream mathematics). –  Nik Weaver May 25 at 14:48
@Noldorin: Because working with set theory is much easier than working with arithmetic; because ZFC allows you to "forget your foundation" when working in "virtually all mathematics"; because model theory is much easier to work with when you have access to actual semantics, even if you want to talk about uncountable models (e.g. $\Bbb R$) without reducing to their countable counterparts; because sets are not that hard to understand, and once you forget about them they're gone, but encoding things into integers is harder; and more. –  Asaf Karagila May 25 at 17:21


text: Section Summary. Transcription. It attaches to the end of the chromosome, and complementary bases to the RNA template are added on the 3′ end of the DNA strand. Replication in eukaryotes starts at multiple origins of replication. Mechanism of DNA Replication 3. During replication, these strands are separated. Different proteins are used, and the system is more complex than it is in prokaryotes. DNA Replication Summary Enzymes and proteins involved Process of replication in eukaryotes and … In case of eukaryotes, the organisms that contain a membrane-bound nucleus, the DNA is sequestered inside the nucleus.Hence, the nucleus is the site for DNA replication in eukaryotes. Summary of DNA replication, Trnascription and Translation. In 2010, scientists found that telomerase can reverse some age-related conditions in mice. Thus, the ends of the chromosomes are replicated. SSB proteins: Bind to the single strands of unwound DNA to prevent reformation of the DNA helix during replication. Cancer is characterized by uncontrolled cell division of abnormal cells. Determine whether the characteristics describe DNA replication in prokaryotes only, eukaryotes only, or both prokaryotes and eukaryotes. The gaps that remain are sealed by DNA ligase, which forms the phosphodiester bond. State the role of telomerase in DNA replication Eukaryotic genomes are much more complex and larger in size than prokaryotic genomes. Replication occurs in the nucleus. There are multiple origins of replication on the eukaryotic chromosome; humans can have up to 100,000 origins of replication. In eukaryotes, there are multiple origin of replication present. DNA Replication in Eukaryotes The essential steps of replication are the same as in prokaryotes. An important aspect of DNA replication in eukaryotes, specifically affecting humans. In recent years, considerable advances have been made in understanding the biochemistry of the archaeal replication proteins. DNA replicates by a semi-conservative method in which each of the two parental DNA strands act as a template for new DNA to be synthesized. Replication in eukaryotes starts at multiple origins of replication. Once the tRNA donated its amino acid, it exits the ribosome. How do the linear chromosomes in eukaryotes ensure that its ends are replicated completely? Eukaryotes. The DNA is synthesized by DNA polymerase in the 5'-3' direction on leading strand by DNA polymerase. Register or login to make commenting easier. In eukaryotic cells, polymerases alpha, delta, and epsilon are the primary polymerases involved in DNA replication. Because eukaryotes have linear chromosomes, DNA replication is unable to reach the very end of the chromosomesbut ends at the telomere region of repetitive DNA close to the end. DNA replicates by a semi-conservative method in which each of the two parental DNA strands act as a template for new DNA to be synthesized. Features of Eukaryotic DNA Replication Replication is bi-directional and originates at multiple origins of replication (Ori C) in eukaryotes. The mechanism is quite similar to prokaryotes. DNA Replication in Eukaryotes Section Summary. The mechanism is quite similar to that in prokaryotes. The ends of the chromosomes pose a problem as polymerase is unable to extend them without a primer. A sliding clamp protein known as PCNA (Proliferating Cell Nuclear Antigen) holds the DNA pol in place so that it does not slide off the DNA. During G 2, any damaged DNA or replication errors are corrected. In circular DNA of bacteria and linear DNA of eukaryotes, DNA replication proceeds bidirectionarlly starting from a fixed origin of replication. The mechanism is quite similar to prokaryotes. Practice: Replication. The number of DNA polymerases in eukaryotes is much more than in prokaryotes: 14 are known, of which five are known to have major roles during replication and have been well studied. University of Rwanda/Huye Campus College of Sciences and Technology School of Science Department of Biology Option: Biotechnology 3rd Year Module: Applied Molecular Biology Topic: DNA REPLICATION IN EUKARYOTES Group members No Names Student Number 1 NTEGEREJIMANA 213000753 Theogene 2 HAKORIMANA Jean 213001789 … When the replication fork reaches the end of the linear chromosome, there is no place for a primer to be made for the DNA fragment to be copied at the end of the chromosome. Prokaryotic vs Eukaryotic DNA Replication According to the Watson and Crick model suggested for DNA, one strand of DNA is the complement of the other strand; hence each strand acts as a template for the formation of a new strand of DNA.This process is known as DNA replication.The replication of DNA basically involves the unwinding of the parent strands and the base pairing … $$\overset{\underset{\mathrm{def}}{}}{=}$$, Introduction to DNA Structure and Function, Summarizing Historical Basis of Modern Understanding, Summarizing DNA Replication in Prokaryotes, Summarizing DNA Replication in Eukaryotes, http://cnx.org/contents/185cbf87-c72e-48f5-b51e-f14f21b5eabd@11.2. During initiation, the DNA is made accessible to the proteins and enzymes involved in the replication process. After replication, each DNA has one parental or “old” strand, and one daughter or “new” strand. The human genome has 3 billion base pairs per haploid set of chromosomes, and 6 billion base pairs are replicated during the S phase of the cell cycle. The Okazaki fragments in the lagging strand are joined together after the replacement of the RNA primers with DNA. Section Summary. Section Summary. There is one origin of replication. The RNA primers are replaced with DNA nucleotides; the DNA remains one continuous strand by linking the DNA fragments with DNA ligase. It depends on the sizes and details of the molecules. This is accomplished by the process of DNA replication.The replication of DNA occurs before the cell begins to divide into two separate cells. Your browser seems to have Javascript disabled. In eukaryotes, cell division is a comparatively complex process, and DNA replication occurs during the synthesis (S) phase of the cell cycle. The leading strand is synthesized continuously, whereas the lagging strand is synthesized in short stretches called Okazaki fragments. Eukaryotic DNA is bound to basic proteins known as histones to form structures called nucleosomes. At the origin of replication, a pre-replication complex is made with other initiator proteins. RNase H removes the RNA primer, which is then replaced with DNA nucleotides. -----... All you need to know about DNA replication in prokaryotes is here. Each primer is only several bases apart. In prokaryotes, DNA replication is the first step of cell division, which is primarily through binary fission or budding. They are known as pol α , pol β , pol γ , pol δ , and pol ε . Cells that undergo cell division continue to have their telomeres shortened because most somatic cells do not make telomerase. Replication in eukaryotes starts at multiple origins of replication. DNA replication in prokaryotes and eukaryotes happens before the division of cells. […] Before replication can start, the DNA has to be made available as a template. Helicases: Unwind the DNA helix at the start of replication. The mechanism is quite similar to prokaryotes. Eukaryotic DNA is bound to proteins known as histones to form structures called nucleosomes.  The entire process of DNA replication is complex and involves multiple enzymatic activities. Bidirectional replication may have multiple replication forks. Other proteins are then recruited to start the replication process. Models for Replication of Prokaryotic DNA. DNA replication: ¥Copying genetic information for transmission to the next generation ¥Occurs in S phase of cell cycle ¥Process of DNA duplicating itself ¥Begins with the unwinding of the double helix to expose the bases in each strand of DNA ¥Each unpaired nucleotide will attract a complementary nucleotide from the medium In yeast, which is a eukaryote, special sequences known as Autonomously Replicating Sequences (ARS) are found on the chromosomes. If the action of telomerase in these cells can be inhibited by drugs during cancer therapy, then the cancerous cells could potentially be stopped from further division. Replication in eukaryotes starts at multiple origins of replication. This may have potential in regenerative medicine.1 Telomerase-deficient mice were used in these studies; these mice have tissue atrophy, stem cell depletion, organ system failure, and impaired tissue injury responses. They are known as pol α , pol β , pol γ , pol δ , and pol ε . RNA – Polymerase attaches to the promoter. The mechanism is quite similar to that in prokaryotes. tRNA’s charged with amino acids enter the ribosomes, where their amino acid is transferred on to the growing polypeptide chain. Scientists have observed that cancerous cells have considerably shortened telomeres and that telomerase is active in these cells. A primer is required to initiate synthesis, which is then extended by DNA polymerase as it adds nucleotides one by one to the growing chain. Although, eukaryotes DNA replication needs some special consideration due to differences in DNA size, a unique linear DNA end structure known as ‘telomeres’. Average eukaryotic cell has a bigger DNA than a prokaryotic cell, which is 25 times larger. Unlike prokaryotic chromosomes, eukaryotic chromosomes are linear. Telomerase, an enzyme with an inbuilt RNA template, extends the ends by copying the RNA template and extending one end of the chromosome. In eukaryotes, the vast majority of DNA synthesis occurs during S phase of the cell cycle, and the entire genome must be unwound and duplicated to form two daughter copies. Replication in eukaryotes starts at multiple origins of replication. Interestingly, only after the telomeres were shortened in the cancer cells did the telomerase become active. Therefore, DNA replication in eukaryotes is a highly regulated process and usually requires extracellular signals to coordinate the specialized cell divisions in different tissues of multicellular organisms. J.H.Taylor confirmed semi conservative replication of DNA in Vicia faba chromosomes in 1963. A primer is required to initiate synthesis, which is then extended by DNA polymerase as it adds nucleotides one by one to the growing chain. Organizing and providing relevant educational content, resources and information for students. Unwinding of DNA double helix at replication origins(s)  The daughter cells must have the same genetic information as the parental cell  So DNA replication must possess a very high degree of fidelity. When the DNA polymerase works in the opposite direction on lagging strand it synthesizes discontinuous short DNA segments known as … Thus, telomere reactivation may have potential for treating age-related diseases in humans. Helicase opens up the DNA double helix, resulting in the formation of the replication fork. Telomerase, an enzyme with an inbuilt RNA template, extends the ends by copying the RNA template and extending one end of the chromosome. Eukaryotic genomes are much more complex and larger in size than prokaryotic genomes. Telomerase is typically active in germ cells and adult stem cells. DNA structure and replication review. DNA replicates by a semi-conservative method in which each of the two parental DNA strands act as a template for new DNA to be synthesized. They are known as pol α, pol β, pol γ, pol δ, and pol ε. The 5 ' to 3 ' direction on leading strand by adding deoxyribonucleotides to the growing chain! Dna molecule charged with amino acids enter the ribosomes, where their amino is. These cells of cell cycle and activate the synthesis of cyclins replication replication is unable to reach very. A CC BY-NC-SA 4.0 License 10.59: fVAf83sY @ 12/Preface-to-Biology making sure both! 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text:
# Project Euler 40: directly calculate digits of Champernowne's constant

I implemented Project Euler problem 40 in Java. It works correctly, producing the correct answer.

During the course of implementing this program, I wrote a method that directly calculates a digit of Champernowne's constant, as opposed to the obvious brute-force approach. While brute-force is fast, this method takes under 1ms. In terms of speed, this is good code.

However, the method has a lot of repetition that I would like to get rid of, but I am not sure how. This may be a case of different code structures having similar attributes (those inner if blocks) that are not able to be combined, or any such optimization would introduce so many other complexities that it makes the code less clear.

Notes:

1. This method is long, too long. If there is no way to simplify this, I will split it into multiple methods on the top-level if statements. I have it all in one method in the meantime while looking for a way to refactor the code to be more concise.

2. I could refactor it to have multiple return statements, and I know that everyone has their own preference on this. I implemented it this way (final return variable and exceptions) to fail fast. It cannot compile if I forget a code path, and the exceptions kill the program in case of a logic error.

3. This is written in Java 8, so pretty much any language feature is fair game (e.g. lambdas).

The code:

  private static int getDigit(int d) {
/* This algorithm does the following:
*
* 1. Based on the requested digit, find which group of digits it belongs
* to. A digit group is determined by how many digits each number has: for
* example, all two-digit numbers are part of the two-digit group.
*
* 2. Find the number that contains the requested digit. This can be easily
* calculated because the size of each group is well-defined, and since all
* numbers in the group have the same number of digits, we can easily
* calculate which number contains the requested digit.
*
* 3. Get the offset of the digit inside the number. This may actually be
* in the next number. If the offset is zero, it is at the end of the
* calculated number. If the offset (actually a remainder) is positive,
* count that many digits into the next number.
*/
final int result;
if (d < 1) {
throw new IllegalArgumentException("Argument [" + d + "] out of range [1-7,688,889]");
}

// Single digits are the first 9 values and take up 9 total digits (1 - 9).
else if (d < 10) {
result = d;
}

// Double digits are the next 90 values and take up 180 total digits (10 - 189).
else if (d < 190) {
int offset = d - 9;
int number = offset / 2 + 9;
int digit = offset % 2;
if (digit == 0) {
result = number % 10;
}
else if (digit == 1) {
result = (number + 1) / 10;
}
else {
throw new AssertionError(digit);
}
}

// Triple digits are the next 900 values and take up 2,700 total digits (190 - 2,889).
else if (d < 2_890) {
int offset = d - 189;
int number = offset / 3 + 99;
int digit = offset % 3;
if (digit == 0) {
result = number % 10;
}
else if (digit == 1) {
result = (number + 1) / 100;
}
else if (digit == 2) {
result = ((number + 1) / 10) % 10;
}
else {
throw new AssertionError(digit);
}
}

// Quad digits are the next 9,000 values and take up 36,000 total digits (2,890 - 38,889).
else if (d < 38_890) {
int offset = d - 2_889;
int number = offset / 4 + 999;
int digit = offset % 4;
if (digit == 0) {
result = number % 10;
}
else if (digit == 1) {
result = (number + 1) / 1_000;
}
else if (digit == 2) {
result = ((number + 1) / 100) % 10;
}
else if (digit == 3) {
result = ((number + 1) / 10) % 10;
}
else {
throw new AssertionError(digit);
}
}

// Five digits are the next 90,000 values and take up 450,000 total digits (38,890 - 488,889).
else if (d < 488_890) {
int offset = d - 38_889;
int number = offset / 5 + 9_999;
int digit = offset % 5;
if (digit == 0) {
result = number % 10;
}
else if (digit == 1) {
result = ((number + 1) / 10_000) % 10;
}
else if (digit == 2) {
result = ((number + 1) / 1_000) % 10;
}
else if (digit == 3) {
result = ((number + 1) / 100) % 10;
}
else if (digit == 4) {
result = ((number + 1) / 10) % 10;
}
else {
throw new AssertionError(digit);
}
}

// Six digits are the next 900,000 values and take up 7,200,000 total digits (488,890 - 7,688,889).
else if (d < 7_688_890) {
int offset = d - 488_889;
int number = offset / 6 + 99_999;
int digit = offset % 6;
if (digit == 0) {
result = number % 10;
}
else if (digit == 1) {
result = ((number + 1) / 100_000) % 10;
}
else if (digit == 2) {
result = ((number + 1) / 10_000) % 10;
}
else if (digit == 3) {
result = ((number + 1) / 1_000) % 10;
}
else if (digit == 4) {
result = ((number + 1) / 100) % 10;
}
else if (digit == 5) {
result = ((number + 1) / 10) % 10;
}
else {
throw new AssertionError(digit);
}
}

else {
throw new IllegalArgumentException("Argument [" + d + "] out of range [1-7,688,889]");
}

return result;
}

The boundaries of where the number of digits increases can be generated and looped over:

int n=0, nDigits;
for(nDigits=1; n < d; nDigits++){
n+= nDigits*(pow10(nDigits)-pow10(nDigits-1));
}

pow10(n) is a helper function to create the nth power of 10.

The big if-else can also be simplified and generalized a bit:

  if (digit == 0) {
number--;
digit = nDigits;
}

result = ((number + 1) / pow10(nDigits-digit)) % 10;

• The idea is on the right track, but I am having difficulty getting this to work. I will check back later once I have more time to work with this idea some more. – user31517 Oct 11 '15 at 18:41


text:
# VCP equation

Solve the following equation with variations, combinations and permutations:

4 V(2,x)-3 C(2,x+ 1) - x P(2) = 0

Result

x =  3

#### Solution:

$4 \ V(2,x)-3 \ { { 2 } \choose x+ 1 } - x P(2) = 0 \ \\ \ \\ 4 \cdot \ x \cdot \ (x-1)-3 \cdot \ (x+1) \cdot \ x/2- x \cdot \ 2 = 0 \ \\ 2.5x^2 -7.5x = 0 \ \\ \ \\ a = 2.5; b = -7.5; c = 0 \ \\ D = b^2 - 4ac = 7.5^2 - 4\cdot 2.5 \cdot 0 = 56.25 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 7.5 \pm \sqrt{ 56.25 } }{ 5 } \ \\ x_{1,2} = 1.5 \pm 1.5 \ \\ x_{1} = 3 \ \\ x_{2} = 0 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2.5 (x -3) x = 0x>0 \ \\ x = x_{ 1 } = 3 \ \\ C_{{ 2}}(4) = \dbinom{ 4}{ 2} = \dfrac{ 4! }{ 2!(4-2)!} = \dfrac{ 4 \cdot 3 } { 2 \cdot 1 } = 6 \ \\ \ \\ b = { { 2 } \choose x+1 } = { { 2 } \choose 3+1 } = 6 \ \\ y = 4 \cdot \ 3 \cdot \ 2 \cdot \ 1/(1)-3 \cdot \ b-x \cdot \ 2 \cdot \ 1 = 4 \cdot \ 3 \cdot \ 2 \cdot \ 1/1-3 \cdot \ 6-3 \cdot \ 2 \cdot \ 1 = 0$

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text:
# Employees Relations

Order Instructions/Description

The subject area of this assignment is Human Resource Management.

Please i want you send me a draft before you start the essay.

and you HAVE to get some references from one of these text books or both

Bray M , Waring P, & Cooper R (2014), Employment Relations, 3nd edition, (Sydney: McGraw Hill)

Bray M , Waring P, & Cooper R (2011), Employment Relations, 2nd edition, (Sydney: McGraw Hill)

Assessment 2
– Essay 40%
2500 to 3500 words (Word limit – 3500 words), excluding reference and/or bibliography list
Write an essay on the following:
What has been the impact of neoliberalism on Australian employment relations? You may refer to case studies of Australian organisations to provide examples and to display knowledge of the concepts.

Refer particularly to part 1 and Chapter 4 of the text book however wider reading is expected.

Text Book. Please make sure you should get some references from this book. The topic you have to read is (neoliberalism) I want the references with the page number like this one (Hamdani & Buckley 2011, p.33). Should be all the references are accessible.

Bray  M , Waring P,  & Cooper R (2014), Employment Relations, 3nd edition, (Sydney: McGraw Hill)
Bray  M , Waring P,  & Cooper R (2011), Employment Relations, 2nd edition, (Sydney: McGraw Hill)


text:
# Saving ts objects as csv files

#### Hyndsight

Occasionally R might not be the tool you want to use (hard to believe, but apparently that happens). Then you may need to export some data from R via a csv file. When the data is stored as a ts object, the time index can easily get lost. So I wrote a little function to make this easier, using the tsibble package to do almost all of the work in looking after the time index. (Thanks to Earo in the comments for greatly simplifying my original code.)

# Convert time series data to csv
ts2csv <- function(x) {
fname <- paste0(deparse(substitute(x)), ".csv")
ts2csv(elecdemand) # Multivariate half-hourly data


text:
## Sin3A Cos3A Tan3A formulas

In this post, we’re going to discuss Sin3a,cos3a, and tan3a formulas. This is the concept of Trigonometric function at multiplies of angle such sin2x,…etc.We’ll see sin3a ,co3a and tan3a derivation with example . But, to derive given formulas you must know other basic trigonometric formulas.                     … Read more


text:
# Customize KOMA section headings without titlesec

A quick question: Is it possible to change the name of a part/chapter/section with some sort of command?

For example, I have a chapter called "Important Things". I created this chapter with the command \chapter{Important Things}. This prints the title of the chapter and adds an entry in the ToC.

I guess that the information about the chapter name is stored and can somehow be accessed by a command like \chaptername or \chaptertitle or something similar. So let's assume that "Important Things" is the current "value" of the chapter-name. What if I now want to change this "value" into "Unimportant Things". For this I cannot use \chapter{Unimportant Things}, because if I use that, then I change the value of the chapter-name but I also print a chapter heading.

Maybe I could also ask the following: Can I create a part/chapter/section without printing a heading and without having an entry in the ToC?

I am not sure if I made myself clear.

I've been asked for a MWE. There you see that the page headers after the new part continue to bear the section title and chapter title of the previous part.

%!TEX TS-program = lualatexmk
\documentclass{scrbook}
\usepackage{fmtcount}
\usepackage[noindentafter]{titlesec}

\renewcommand*{\partformat}{\partname~\numberstring{part}}

\titleformat{\chapter}[display]{\fillast}{\Large\textsc{\chaptertitlename~\numberstring{chapter}}}{1ex minus .1ex}{\LARGE\emph}
\titlespacing{\chapter}{1pc}{*3}{*20}[1pc]

\titleformat{\section}[block]{\fillast\large}{\thesection}{.5em}{\large\textsc}
\titlespacing{\section}{1pc}{*6}{*2}[1pc]

\titleformat{\subsection}[block]{\fillast\itshape}{\thesubsection}{.5em}{}
\titlespacing{\subsection}{1pc}{*3}{*2}[1pc]

\titleformat{\subsubsection}[block]{\fillast}{}{0pt}{}
\titlespacing{\subsubsection}{1pc}{*3}{*2}[1pc]

\usepackage{titleps}
\newpagestyle{chapter_and_section}{
\setfoot[][][]{}{}{}
}
\pagestyle{chapter_and_section}
\newpagestyle{first_page_of_chapter}{   % here is the definition of the first page of a chapter
\setfoot[][\thepage][]{}{\thepage}{}
}
\assignpagestyle{\chapter}{first_page_of_chapter}

\begin{document}
\chapter{Some Stuff}
\section{More Stuff}
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
\end{document}

• \chapter[Important Things]{Unimportant Things} prints Unimportant Things as title chapter but Important Things in table of contents and headings. Does it works for you? Jan 22, 2015 at 7:53
• No, not really. I am not looking for that. I want Important Things in the ToC and as the heading of the Chapter. I just want to change afterwards the name of the chapter into Unimportant Things. Jan 22, 2015 at 7:55
• \markboth{whatever you want}{something else}? Jan 22, 2015 at 8:01
• titleps does not like \markboth, as I heard (tex.stackexchange.com/a/89911/12277) Jan 22, 2015 at 8:03
• Just looked at the example, should be fairly easy with an up-to-date KOMA. Give me a bit :-) Jan 22, 2015 at 10:25

Using the interface of scrlayer-scrpage provided by the KOMA-bundle, this should be what you are after:

\documentclass{scrbook}
\usepackage{scrlayer-scrpage}

\renewcommand{\chaptermarkformat}{}
\renewcommand{\sectionmarkformat}{}

\begin{document}
\chapter{Some Stuff}
\section{More Stuff}
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
Things}
Text.\newpage
Text.\newpage
Text.\newpage
Text.\newpage
\end{document}

# Only KOMA

Taking into account the other stuff with the ambition not to use packages like titlesec or titleps which are incompatible to KOMA-script, which has its own interface.

Here should be what you want using only KOMA-script stuff. I should mention that it uses features that have just been introduced with KOMA-script v. 3.15.

Below, i tried to put the stuff and thinking process into blocks. For example, first adjusting the space above chapters, and at last the space below chapters.

\documentclass{scrbook}
%\usepackage{fmtcount}%not needed
\usepackage{blindtext}

\renewcommand{\partpagestyle}{empty}

%dealing with chapters
\KOMAoption{chapterprefix}{true}
\setkomafont{chapter}{\LARGE\emph}

%dealing with (sub/subsub)sections
%all levels have something in common, let's save typing:
\RedeclareSectionCommands[beforeskip=-3ex,
afterskip=2ex]{section,subsection,subsubsection}
\RedeclareSectionCommand[beforeskip=-6ex]{section}

\usepackage[automark]{scrlayer-scrpage}
\renewcommand{\chaptermarkformat}{}
\renewcommand{\sectionmarkformat}{}
\ofoot*{}%Clear the outer foot
\cfoot[\pagemark]{}%pagemark in the center only on plain pages

\setkomafont{descriptionlabel}{\usekomafont{disposition}\scshape}
%descriptionlabels will now look a bit pleasing with respect to the headings

\begin{document}
\blinddocument
\blinddocument
\end{document}

Since you aren't used to the interface, it might be a bit intimidating. But i myself spent more time in the titlesec documentation to look up which space is where. You should have no trouble figuring out where to change which space/skip.

The pagehead was set to itshape but it looked almost identical before?
Everytime one can set or add to a komafont it will be preset somewhere. For the pagehead it is \normalfont\slshape.

• This only works for the formatting in my MWE, which is a minimal working example. In the full document, the formatting is more complex, because the first page of each chapter has a different formatting than the rest having an empty header and only the page number centered in the footer. This I could not achieve with KOMA (I guess it is possible though, but it was so easy with titleps and I also love titlesec). Jan 22, 2015 at 8:44
• You can specify the specs in the question and i'll show you how it workds. titlesec is another package that extends the generic standard classes, beiing not aware of the KOMA extension it breaks stuff. You get respective warnings. The interface to change the appearance of sections with KOMA-script has changed, you can open a new question for that. It will be interesting for others as well. Jan 22, 2015 at 8:57
• I'd be delighted and grateful if you want to give it a shot. Jan 22, 2015 at 9:37
• Please see the update. Jan 22, 2015 at 11:48
• Usually that what users want. Just do \addsecmark{your section title}. Might be worth a new question ;-) Jan 22, 2015 at 16:00