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LABORATORI NAZIONALI DI FRASCATI
• Questo evento è passato.
## On nuclear states of mesons
### 9 Marzo 2015 @ 15:00 - 16:00
The existence of nuclear states of $\bar K$ mesons and $\eta$ mesons has been expected for about thirty years. The physical interest of this field is related to a new mechanism of nuclear binding generated by excitations of the internal nucleon structure. Despite being short lived such states might offer a new and interesting nuclear spectroscopy.
Despite long and extensive experimental search, the existence and properties of such states are still elusive.
Some similarities and related questions of the $\eta$ and $\bar K$ cases will be shown. Possible origin of the experimental difficulties related to weak production rate and large widths of such states will be discussed. I will try to indicate usefulness of the recent AMADEUS/KLOE measurements.
### Dettagli
Data:
9 Marzo 2015
Ora:
15:00 - 16:00
Categoria Evento:
Target:
Sito web:
http://agenda.infn.it/conferenceDisplay.py?confId=9282
### Luogo
Aula Leale
Via Enrico Fermi, 40
Frascati, Roma 00044 Italia
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# A verbatim-like environment, but inline
I would like to find a solution to use a verbatim-like environment, but inline. Indeed, I would like to ask not to compile Latex functions but just for a specific place in a text.
I tried this :
Equation vectorielle en utilisant la commande
\begin{verbatim} \boldsymbol \end{verbatim}:
But the problem is that there is a line jump.
• What exactly do you want in the output? Mar 31 at 14:50
• (are you looking for \texttt or \verb? Mar 31 at 14:56
• Equation vectorielle en utilisant la commande \verb|\boldsymbol|? Mar 31 at 14:57
You can use \verb --- for example
Equation vectorielle en utilisant la commande \verb|\boldsymbol|
You can use almost any symbol after \verb to delimit you "verbatim" text (but not *!); choose one that does not appear in the internal material:
\documentclass[12pt,]{article}
\usepackage[T1]{fontenc}
\begin{document}
a verb with bars: \verb|\foo| or with plus: \verb+a vertical bar|+
\end{document}
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# Forever Flasher (& flashlight)
W
#### Watson A.Name - 'Watt Sun'
Jan 1, 1970
0
I built the flasher circuit from Dave Johnson's website, See URL
http://www.discovercircuits.com/PDF-FILES/3vledfs1.pdf
But it just went full on, wouldn't flash. So I had to put a 2.2k from
Q2 base to emitter to get it to flash. I also used two 0.1 uf ceramic
caps instead of rhe 0.68 uF, and the flash rate is right around 1 Hz.
The peak current is 17 mA, and average current is about 100 micromps
at 3V.
I then connected it to a 3V photovoltaic cell and a 1N4148 to prevent
the current from going back into the photocell, and a 6800 uF
capacitor to store the current generated by the light. I put the
photocell directly under a light bulb, and it works fine, and it keeps
working for less than a minute when the light is turned off. So now I
need a lot bigger capacitor, something that will run it for a coupla
hours.
I'm thinking that it would work good using a pair of 1 F, 2.0V
supercaps in series, charged by the photocell.
I bought a Forever Flashlight, the one that has a single white LED,
with a magnet and coil in the barrel that charges up a supercap when
it is shaken. It works, but I'm disappointed in the light output.
The instructions say to shake it for 90 seconds, but even longer than
that gives the LED only a few mA, not a really decent amount. It has
a lens to concentrate the LED's light, so it's better than just the
bare LED alone. Obviously it's meant to be used for situations where
a regular flashlight might not be working, like in an
emergency/earthquake preparedness kit. The body of the flashlight is
clear plastic so I can see the parts inside, and there's a supercap in
there, but the plastic is too thick to see the value, which is
blurred.
So I'm wondering if I should order a couple of these supercaps. The
solar cell is rated for 3V at 40 mA, see the SPL-60 on All Electronics
website, http://www.allelectronics.com/cgi-
bin/category.cgi?category=565&item=SPL-60&type=store
It puts out an honest 4V in bright sunlight, so I would think that it
will charge two 1 F supercaps in series in a few minutes. 3 TCs at 40
mA would be 38 seconds, roughly. Maybe I should try for a couple
farads to keep it running all night long. Has anyone done this
before? Do these supercaps have low enough leakage to stay charged up
for a half a day? After seeing their performance in the 'Forever'
Flashlight, I'm not so sure.
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
D
#### dan williams
Jan 1, 1970
0
it looks like you could use a peak at Elna's catalog,
Their 100F capacitor (I'm picking the biggest one because I can) rated
at 2.5V
charts out to run for about 200 seconds at 50mA (I'm taking this at a
1.5V cutoff,
use a switching (buckboost) led controller and get more. especially if
you are useing
more than one with a supply of 4V. I dont know what elna's 100F
capacitor costs. I imagine
its not cheap.
CATALOGS! I must have more catalogs! HA, HAHAHAHAHA!
Dan
Watson A.Name - 'Watt Sun' said:
I built the flasher circuit from Dave Johnson's website, See URL
http://www.discovercircuits.com/PDF-FILES/3vledfs1.pdf
But it just went full on, wouldn't flash. So I had to put a 2.2k from
Q2 base to emitter to get it to flash. I also used two 0.1 uf ceramic
caps instead of rhe 0.68 uF, and the flash rate is right around 1 Hz.
The peak current is 17 mA, and average current is about 100 micromps
at 3V.
I then connected it to a 3V photovoltaic cell and a 1N4148 to prevent
the current from going back into the photocell, and a 6800 uF
capacitor to store the current generated by the light. I put the
photocell directly under a light bulb, and it works fine, and it keeps
working for less than a minute when the light is turned off. So now I
need a lot bigger capacitor, something that will run it for a coupla
hours.
I'm thinking that it would work good using a pair of 1 F, 2.0V
supercaps in series, charged by the photocell.
I bought a Forever Flashlight, the one that has a single white LED,
with a magnet and coil in the barrel that charges up a supercap when
it is shaken. It works, but I'm disappointed in the light output.
The instructions say to shake it for 90 seconds, but even longer than
that gives the LED only a few mA, not a really decent amount. It has
a lens to concentrate the LED's light, so it's better than just the
bare LED alone. Obviously it's meant to be used for situations where
a regular flashlight might not be working, like in an
emergency/earthquake preparedness kit. The body of the flashlight is
clear plastic so I can see the parts inside, and there's a supercap in
there, but the plastic is too thick to see the value, which is
blurred.
So I'm wondering if I should order a couple of these supercaps. The
solar cell is rated for 3V at 40 mA, see the SPL-60 on All Electronics
website, http://www.allelectronics.com/cgi-
bin/category.cgi?category=565&item=SPL-60&type=store
It puts out an honest 4V in bright sunlight, so I would think that it
will charge two 1 F supercaps in series in a few minutes. 3 TCs at 40
mA would be 38 seconds, roughly. Maybe I should try for a couple
farads to keep it running all night long. Has anyone done this
before? Do these supercaps have low enough leakage to stay charged up
for a half a day? After seeing their performance in the 'Forever'
Flashlight, I'm not so sure.
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
--
Dan Williams, Owner
Electronic Device Services
(604) 741 8431
RR8 855 Oshea rd
V0N 1V8
P
#### Paul Hovnanian P.E.
Jan 1, 1970
0
Watson A.Name - 'Watt Sun' said:
I built the flasher circuit from Dave Johnson's website, See URL
http://www.discovercircuits.com/PDF-FILES/3vledfs1.pdf
But it just went full on, wouldn't flash. So I had to put a 2.2k from
Q2 base to emitter to get it to flash. I also used two 0.1 uf ceramic
caps instead of rhe 0.68 uF, and the flash rate is right around 1 Hz.
The peak current is 17 mA, and average current is about 100 micromps
at 3V.
I then connected it to a 3V photovoltaic cell and a 1N4148 to prevent
the current from going back into the photocell, and a 6800 uF
capacitor to store the current generated by the light. I put the
photocell directly under a light bulb, and it works fine, and it keeps
working for less than a minute when the light is turned off. So now I
need a lot bigger capacitor, something that will run it for a coupla
hours.
I'm thinking that it would work good using a pair of 1 F, 2.0V
supercaps in series, charged by the photocell.
I bought a Forever Flashlight, the one that has a single white LED,
with a magnet and coil in the barrel that charges up a supercap when
it is shaken. It works, but I'm disappointed in the light output.
The instructions say to shake it for 90 seconds, but even longer than
that gives the LED only a few mA, not a really decent amount. It has
a lens to concentrate the LED's light, so it's better than just the
bare LED alone. Obviously it's meant to be used for situations where
a regular flashlight might not be working, like in an
emergency/earthquake preparedness kit. The body of the flashlight is
clear plastic so I can see the parts inside, and there's a supercap in
there, but the plastic is too thick to see the value, which is
blurred.
See if you can find one of those "wind up" flashlights. Similar to the
Freeplay radios. That seems to be better technology in terms of time
spent winding vs illumination time. I've seen one with a selectable
xenon mode (runs charge down fast) or 3 white LEDs (less light for much
more time).
So I'm wondering if I should order a couple of these supercaps. The
solar cell is rated for 3V at 40 mA, see the SPL-60 on All Electronics
website, http://www.allelectronics.com/cgi-
bin/category.cgi?category=565&item=SPL-60&type=store
It puts out an honest 4V in bright sunlight, so I would think that it
will charge two 1 F supercaps in series in a few minutes. 3 TCs at 40
mA would be 38 seconds, roughly. Maybe I should try for a couple
farads to keep it running all night long. Has anyone done this
before? Do these supercaps have low enough leakage to stay charged up
for a half a day? After seeing their performance in the 'Forever'
Flashlight, I'm not so sure.
If a while LED uses 20 mA, then a 40 mA PV cell will give you 2 minutes
of run time for each minute of charge time (ignoring losses and lots of
other things). If you fully charge the supercaps in 38 seconds, you'll
only get about twice that running time. Add super caps (and $$) to get more time. You might be better off looking for some DC-DC converters, NiMH charging controllers, etc. and using as many mAh of battery capacity as you need for your design illumination time requirements dictate. W #### Watson A.Name - 'Watt Sun' Jan 1, 1970 0 See if you can find one of those "wind up" flashlights. Similar to the Freeplay radios. That seems to be better technology in terms of time spent winding vs illumination time. I've seen one with a selectable xenon mode (runs charge down fast) or 3 white LEDs (less light for much more time). If a while LED uses 20 mA, then a 40 mA PV cell will give you 2 minutes of run time for each minute of charge time (ignoring losses and lots of Er, the flasher takes about 100 microamps. That means that it should run maybe 400 times longer than the charging time, something less than 4 hours. I guess I could parallel a second pair of 1 Fs to give a total of 1 F at 5VDC. That should give a flash time of several more hours. I can't say because I don't know what the cutoff voltage is for the flasher. other things). If you fully charge the supercaps in 38 seconds, you'll only get about twice that running time. Add super caps (and$$) to get
more time.
You might be better off looking for some DC-DC converters, NiMH charging
controllers, etc. and using as many mAh of battery capacity as you need
for your design illumination time requirements dictate.
Batteries go bad, I want to have a forever flasher, not a 5 year
flasher. That's why the 1 F supercaps.
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
I
#### Ian Stirling
Jan 1, 1970
0
In sci.electronics.design Watson A.Name - 'Watt Sun' said:
I built the flasher circuit from Dave Johnson's website, See URL
http://www.discovercircuits.com/PDF-FILES/3vledfs1.pdf
But it just went full on, wouldn't flash. So I had to put a 2.2k from
Q2 base to emitter to get it to flash. I also used two 0.1 uf ceramic
caps instead of rhe 0.68 uF, and the flash rate is right around 1 Hz.
The peak current is 17 mA, and average current is about 100 micromps
at 3V.
Lithium D cells are $20 (US) or so, and have a capacity of around 20Ah. This would run the flasher for around 200000 hours, or around 22 years. Add a light sensitive switch, and you can probably extend this a bit. And you can leave off the solar cell too. A couple of NiCd AA cells would probably run it for around 6 months, after a full charge. (this is probably around the right draw, they'll self-discharge in around a year in most climates anyway) W #### Watson A.Name - 'Watt Sun' Jan 1, 1970 0 As a coincidence, you should take a look at the DutchForce forum... http://www.dutchforce.com/~eforum/index.php The specific thread is: http://www.dutchforce.com/~eforum/index.php?board=10;action=display;threadid =452 Here's the URL for one of the schematics of the inverter. This one says it works down to .4V, there are others on his website. http://www.belza.cz/ledlight/m2d.htm I was thinking of putting the circuit and solar cell in a clear epoxy block. I've done it before by pouring a thin layer, and arranging the working circuit, then pouring another thicker layer over that. Trick is to get rid of the bubbles. So using a 5 year rechargeable battery is not an option. I would consider using rechargeable cells if it was in a package with a battery holder. But I've already got one of those. ;-) I checked out the eBay URL, but it's not clear to me how many LEDs are in the package. It says 100 pcs but in the pic looks like it's a lot less. And the 'buy it now' price is a bit on the low side for that quantity. So I'm not sure if I'd buy them. Another factor that bothers me is that any time one buys LEDs at cut-rate prices, there is a very high likelihood that they're seconds or rejects. I've purchased low priced LEDs that have flaws in the lens, or that are not centered properly in the epoxy package, etc. I've done very well by buying from Nichia or from Wilcoxson's www.whitelightled.com. Also, you can buy Yoldal LEDs that are similar from http://www.8000mcd.com/catalog.html but I haven't bought any from there, so I can't vouch for the quality. So, has anyone purchased some of those eBay LEDs? Tim (Sc3mat1c) -- -- @@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@ ###Got a Question about ELECTRONICS? Check HERE First:### http://users.pandora.be/educypedia/electronics/databank.htm My email address is whitelisted. *All* email sent to it goes directly to the trash unless you add NOSPAM in the Subject: line with other stuff. alondra101 <at> hotmail.com Don't be ripped off by the big book dealers. Go to the URL that will give you a choice and save you money(up to half). http://www.everybookstore.com You'll be glad you did! Just when you thought you had all this figured out, the gov't changed it: http://physics.nist.gov/cuu/Units/binary.html @@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@ W #### Watson A.Name - 'Watt Sun' Jan 1, 1970 0 Lithium D cells are$20 (US) or so, and have a capacity of around 20Ah.
This would run the flasher for around 200000 hours, or around 22 years.
Add a light sensitive switch, and you can probably extend this a bit.
And you can leave off the solar cell too.
A couple of NiCd AA cells would probably run it for around 6 months, after
a full charge.
(this is probably around the right draw, they'll self-discharge in around a
year in most climates anyway)
Forever isn't 6 months, nor 5 years, nor 22 years. Thanks, but I'd
really like to get the info I asked for on the 1 F supercaps.
BTW, I have a 1 LED flasher that runs off four AA 2000 mAh Ni-MHs, and
it lasts for about a month, maybe a little more. I don't know what
the peak current thru the blue LED is, but it's very bright. Maybe
two flashes per second. I've noticed the Ni-MHs lose some of their
capacity sitting around for a month or three. But that's just a
guesstimate.
If I pot a clear epoxy case of the LED in clear eopxy, what does it do
to the beam pattern? Suddenly the index of refraction is the same as
the lens, so it isn't a lens any more. POOF! The beam goes wide
angle, I would guess. Maybe I'll have to dunk a couple LEDs in a
glass of water to see what happens!
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
I
#### Ian Stirling
Jan 1, 1970
0
In sci.electronics.design Watson A.Name - 'Watt Sun' said:
Forever isn't 6 months, nor 5 years, nor 22 years. Thanks, but I'd
really like to get the info I asked for on the 1 F supercaps.
I would not be very certain that the supercap would still be working
after 20 years.
Why do you want 'forever'?
BTW, I have a 1 LED flasher that runs off four AA 2000 mAh Ni-MHs, and
it lasts for about a month, maybe a little more. I don't know what
the peak current thru the blue LED is, but it's very bright. Maybe
two flashes per second. I've noticed the Ni-MHs lose some of their
capacity sitting around for a month or three. But that's just a
guesstimate.
Ni-MH have a very bad self-discharge rate, especailly in high temps.
You may well find NiCd better.
If I pot a clear epoxy case of the LED in clear eopxy, what does it do
to the beam pattern? Suddenly the index of refraction is the same as
the lens, so it isn't a lens any more. POOF! The beam goes wide
Pretty much.
S
Jan 1, 1970
0
S
#### Stepan Novotill
Jan 1, 1970
0
Leave
Leave an air space in front of the LED lens to reduce that problem.
J
#### Jim Thompson
Jan 1, 1970
0
Leave
Leave an air space in front of the LED lens to reduce that problem.
Good idea! Probably ought to try to keep the epoxy with co-planar
surfaces also.
Or, better yet, pot with the LED lens sticking clear thru the
encapsulation.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice480)460-2350 | |
| [email protected]_innovations.com Fax480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
For proper E-mail replies SWAP "-" and "_"
Why is it that Democrats can't debate politely?
And are only rude and interruptive.
Lack of mental capacity?
P
#### Paul Hovnanian P.E.
Jan 1, 1970
0
Watson A.Name - 'Watt Sun' said:
Er, the flasher takes about 100 microamps. That means that it should
run maybe 400 times longer than the charging time, something less than
4 hours. I guess I could parallel a second pair of 1 Fs to give a
total of 1 F at 5VDC. That should give a flash time of several more
hours. I can't say because I don't know what the cutoff voltage is
for the flasher.
Sorry about that. I thought you were trying to apply the supercaps to
the flashlightapplication, not the flasher.
W
#### Watson A.Name - 'Watt Sun'
Jan 1, 1970
0
Relevant circuits on my page at http://members.shaw.ca/novotill/
1) LED flasher
2) Solar Engine
...Stepan
Yeah, wacky is the right word. I checked the flickering LED circuit
and it looks like the values for the caps and Rs in the oscs aren't
right. But you _have_ to resave that gawdawfully huge delayed off
switch .PNG to 1024 by 768 size. The 6600 pixels wide is ridiculous!
The solar engine doesn't seem to do anything other than store up and
release the charge in the capacitor. I'm not sure how this would help
my situation.
The LED flasher is a V boost circuit. We've already been thru a few
of those here in ABSE and the other sci.electronics ngs. I checked
that ferrite bead on both Digi-Key's and Mouser's website and it's
just a tiny thing. I can't figure how you could pass that many turns
of even fine wire thru the center hole. The only way I can see how
you'd do it is to wind the wire around the outside like a regular
bobbin. Even then the thing is going to be tiny, with really fine
wires. And I don't believe this has 120 uH inductance (nor 120 mH, as
it says in the text).
Anyhoo, this circuit is kewl for 1.5V stuf, but I've already got 3 or
4V from my solar cell, so no problem. It runs the flasher okay. It's
just a matter of pulling in enough charge from a few minutes in direct
sunlight, where the LED can't be seen, and then once the cap is
charged, bring it inside where it can flash for hours. Or it could be
set near a window to charge up and as the sun sets, it continues to
flash for hours. That's why I need the 1F or so supercap and some
info on how they work in this kind of circuit.
I bought some already wound inductors from Mouser, and they work good
for the inverters I made. It would be easy to just wind a few turns
of #28 or finer wire around the outside of the bobbin of this inductor
to give a feedback tickler coil like the one you made. I like already
assembled coils because of repeatability factors and just plain time
savings.
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
B
#### Ban
Jan 1, 1970
0
If I pot a clear epoxy case of the LED in clear eopxy, what does it do
to the beam pattern?
If you stick that solar cell in clear epoxy it will fail after short time
because of different thermal expansion. (done that, that's why I say that)
Rather take some silicon based potting and glue the cells on top only. Then
use a frame with a plastic or glass inlet to protect the cheap cells from
humidity.
ciao Ban
I
#### Ian Stirling
Jan 1, 1970
0
In sci.electronics.design Ban said:
If you stick that solar cell in clear epoxy it will fail after short time
because of different thermal expansion. (done that, that's why I say that)
Rather take some silicon based potting and glue the cells on top only. Then
use a frame with a plastic or glass inlet to protect the cheap cells from
humidity.
Silicon is a hard substance that is used to make solar cells.
Silicone is a polymer that's used to make everything from flexible tubing
to sex toys.
--
http://inquisitor.i.am/ | mailto:[email protected] | Ian Stirling.
---------------------------+-------------------------+--------------------------
"I meant, have you ploughed the ocean waves at all?" Colon gave him a cunning
look. 'Ah, you can't catch me with that one, sir' he said 'Everyone knows
horses sink' -- Terry Pratchett - Jingo
W
#### Watson A.Name - 'Watt Sun'
Jan 1, 1970
0
If you stick that solar cell in clear epoxy it will fail after short time
because of different thermal expansion. (done that, that's why I say that)
Rather take some silicon based potting and glue the cells on top only. Then
use a frame with a plastic or glass inlet to protect the cheap cells from
humidity.
Now you've _really_ ruined my day! Y'see, the solar cells _already_
come from the factory mounted in clear epoxy! Now what am I gonna do?
Are they going to fail when I put them in the sunlight? :-(
But since they are already coated with epoxy, I can assume that
potting them in an epoxy block will not do any more harm. Am I right?
--
@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@,@@[email protected]@[email protected],@@[email protected]@[email protected]@[email protected]@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@[email protected]@[email protected]@@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@[email protected]@@
B
#### Ban
Jan 1, 1970
0
Watson said:
Now you've _really_ ruined my day! Y'see, the solar cells _already_
come from the factory mounted in clear epoxy! Now what am I gonna do?
Are they going to fail when I put them in the sunlight? :-(
But since they are already coated with epoxy, I can assume that
potting them in an epoxy block will not do any more harm. Am I right?
Are you sure that this is epoxy and not some other more flexible stuff? I
tell you I had some(33) Siemens 6" monocrystalline cells. They were coated
with some whitish epoxy-like material on the back. to combine them i poured
some epoxy mixed with these small glassballs and the hardener only on the
back, the front was a special glass. The assembly worked for one year, but
when left in the hot sun for some time, there were cracks all over the
individual cells, I had to throw everything. :-((
I then inquired and found out about my mistake. The Siemens guy told me that
this coating is really the most difficult part of the game.
I do not know about polycristalline or amorph cells, but the monocrystalline
cells are utterly brittle and damaged easily.
removed irrelevant x-posts
ciao Ban
B
#### Ban
Jan 1, 1970
0
Ian said:
Silicon is a hard substance that is used to make solar cells.
Silicone is a polymer that's used to make everything from flexible
tubing
to sex toys.
Are you another self appointed apostrophe inquisitor or what? There is
already one in this group. But maybe in holydays.
thanks for the correction Ban
I
#### Ian Stirling
Jan 1, 1970
0
In sci.electronics.design Ban said:
Are you another self appointed apostrophe inquisitor or what? There is
already one in this group. But maybe in holydays.
Sorry, it's just a pet hate.
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# Lesson 5
Function Representations
• Let’s examine different representations of functions.
### 5.1: Notice and Wonder: Representing Functions
What do you notice? What do you wonder?
$$f(x) = \frac{2}{3}x - 1$$
$$x$$ $$y$$
-1 $$\text{-}\frac{5}{3}$$
0 -1
1 $$\text{-}\frac{1}{3}$$
2 $$\frac{1}{3}$$
3 1
### 5.2: A Seat at the Tables
Use the equations to complete the tables.
1. $$y = 3x - 2$$
$$x$$ $$y$$
1
3
-2
2. $$y = 5-2x$$
$$x$$ $$y$$
0
3
5
3. $$y = \frac{1}{2}x + 2$$
$$x$$ $$y$$
-4
3
6
4. $$x$$ $$y = 2x - 10$$
3
7
-8
### 5.3: Function Finder
1. Use the values in the table to graph a possible function that would have the values in the table.
1. $$x$$ $$y$$
1 3
2 5
3 7
5 11
2. $$x$$ $$y$$
-2 0
0 1
2 2
4 3
3. $$x$$ $$y$$
-2 14
-1 12
1 8
2 6
2. For each of the tables and graphs, write a linear equation (like $$y = ax + b$$) so that the table can be created from the equation.
3. Invent your own linear equation. Then, create a table or graph, including at least 4 points, to trade with your partner. After getting your partner’s table or graph, guess the equation they invented.
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If you skip parentheses or a multiplication sign, type at least a whitespace, i.e. Example. I need to convert from the polar form to complex numbers and vice versa . Example 3.6056cis0.588 . Up until the previous version, attempting to calculate the square root of a negative number would have resulted in Alcula’s scientific calculator returning ‘NaN’ as a solution. Polar form . This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form. Operations with one complex number. Note: The TI 83/84 has what I consider to be a bug in its handling of complex numbers. Complex numbers contain an imaginary number and a real portion. Book Problems. Cartesian to Polar coordinates. COMPLEX FORM AND POLAR FORM. Go to the calculator using the navigation above or read the manual here. " Eventually. Similarly, tanxsec^3x will be parsed as tan(xsec^3(x)). All suggestions and improvements are welcome. In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. 11. Another way of writing the polar form of the number is using it’s exponential form: me^(ia) . This online calculator will help you to convert rectangular form of complex number to polar and exponential form. Regardless of what option you select, you can still enter complex numbers in any format you like. Roots of a complex number . To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Also, be careful when you write fractions: 1/x^2 ln(x) is 1/x^2 ln(x), and 1/(x^2 ln(x)) is 1/(x^2 ln(x)). To get tan^2(x)sec^3(x), use parentheses: tan^2(x)sec^3(x). Press C2qbZ330. To convert a complex number into polar form, press 2+5bU. Set the complex mode, the polar form for display of complex number calculation results and the angle unit Degree in setting. ... Students will be able to sketch graphs of polar equations with and without a calculator . In this section, we will first deal with the polar form of complex numbers. Polar form. The calculator will generate a step by step explanation for each operation. There now reason to take the third root in rectangular form and then convert the third root to polar when you can just as legitimately convert the third power to polar first and then take the third root. iR1(: r ∠q)p. To convert any polar form of a complex number, use the r theta command or type in the angle in polar form. Finding Distance Using Polar Coordinates. Show Hide all comments. We calculate all complex roots from any number - even in expressions: sqrt(9i) = 2.1213203+2.1213203i sqrt(10-6i) = 3.2910412-0.9115656i pow(-32,1/5)/5 = -0.4 pow(1+2i,1/3)*sqrt(4) = 2.439233+0.9434225i pow(-5i,1/8)*pow(8,1/3) = 2.3986959-0.4771303i Square, power, complex exponentiation Our calculator can power any complex number to any integer (positive, negative), real, or even complex number. comments below. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Sign in to comment. So, for example, also the logarithm of a negative number won’t result in NaN anymore, but in a complex result. Every complex number written in rectangular form has a unique polar form ) up to an integer multiple of in its argument. Converting a Complex Number from Polar to Rectangular Form. It is the distance from the origin to the point: See and . Subtracting Complex Numbers To subtract two complex numbers in rectangular form, subtract their real parts and then subtract their imaginary parts . Sign in to answer this question. Conic Sections Trigonometry. The calculator will simplify any complex expression, with steps shown. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. It will perform addition, subtraction, multiplication, division, raising to power, and also will find the polar form, conjugate, modulus and inverse of the complex number. Online calculator which converts the given Complex Number to Polar Form. The form z = a + b i is called the rectangular coordinate form of a complex number. Solve the equation . Interactive Graph - Convert polar to rectangular and vice-versa. In other words, given $$z=r(\cos \theta+i \sin \theta)$$, first evaluate the trigonometric functions $$\cos \theta$$ and $$\sin \theta$$. In general, you can skip the multiplication sign, so 5 x is equivalent to 5 ⋅ x. Please leave them in comments. [MODE][2](COMPLEX) To improve this 'Polar to Cartesian coordinates Calculator', please fill in questionnaire. Use the Polar conversion operation from the complex menu (MATH → CPX) to display the polar form magnitude and angle: Z Polar 11.710e^(-99.445i) The equivalent polar form number is therefore (11.710 −99.445). Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Example. Finding the third roots of complex numbers in polar form is ridiculously easy. This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. See . Express the number root three in trigonometric form. Complex numbers in rectangular form are presented as a + b * %i, where a and b are real numbers.Polar form of the complex numbers is presented as r * exp(c * %i), where r is radius and c is the angle in radians. This calculator allows one to convert complex number from one representation form to another with step by step solution. (Angle unit:Degree): z1 =5<70, z2 = 3<45 Example 5: Multiplication z1*z2=15<115 1. Male Female Age Under 20 years old 20 years old level ... Four operations of the complex number. The horizontal axis is the real axis and the vertical axis is the imaginary axis. We call this the polar form of a complex number.. Complex Number – Calculation (Multiplication / Division) The two polar form complex numbers z1 and z2 are given. Note: This calculator displays (r, θ) into the form: r ∠ θ To convert complex number to its polar form, follow the general steps below: A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Next, we will look at how we can describe a complex number slightly differently – instead of giving the and coordinates, we will give a distance (the modulus) and angle (the argument). 0 Comments. The calculator will find the polar form of the given complex number, with steps shown. Every complex number can be written in the form a + bi. Complex Numbers Calculator evaluates expressions with complex numbers and presents the result in rectangular and polar forms. The scientific calculator supports three ways to enter a complex number: You can select your preferred form for complex numbers from the preferences window, which has been updated to in include the option “Complex Form”. Example: Calculate (5+2i)*2∠30°. From the table below, you can notice that sech is not supported, but you can still enter it using the identity sech(x)=1/cosh(x). To add complex numbers that are in polar form, first convert them to rectangular form, and then follow the rule given above. You can also use the calculator to convert to and from cartesian, polar or exponential. Converting a complex number from polar form to rectangular form is a matter of evaluating what is given and using the distributive property. Complex Number Functions in Excel. Answers (3) Azzi Abdelmalek on 25 Jan 2014. 6.5: #3,5,31,33,37. Thank you. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). How do we understand the Polar representation of a Complex Number? Given z = 3 + 4i , calculate z 5 . Get the free "Convert Complex Numbers to Polar Form" widget for your website, blog, Wordpress, Blogger, or iGoogle. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, θ). Does Matlab support this function ? If the root of a polynomial is unreal, it has complex roots . And is the imaginary component of our complex number. Find more Mathematics widgets in Wolfram|Alpha. The first, and most fundamental, complex number function in Excel converts two components (one real and one imaginary) into a single complex number represented as a+bi. z = a + ib = r e iθ, Exponential form with r = √ (a 2 + b 2) and tan(θ) = b / a , such that -π < θ ≤ π or -180° < θ ≤ 180° Use Calculator to Convert a Complex Number to Polar and Exponential Forms Enter the real and imaginary parts a and b and the number of decimals desired and press "Convert to Polar and Exponential". Read on to find out more. If you get an error, double-check your expression, add parentheses and multiplication signs where needed, and consult the table below. Sometimes I see expressions like tan^2xsec^3x: this will be parsed as tan^(2*3)(x sec(x)). Graphing Polar Equations Notes.pdf. To learn more about other updates to check out the other blog articles. iR 2(: a+bi)p. Alternately, simply type in the angle in polar form … Example. Label the x-axis as the real axis and the y-axis as the imaginary axis. The square root of a negative number is an imaginary number, that is, a real number multiplied by the imaginary unit. You can follow any responses to this entry through the RSS 2.0 feed. Both comments and pings are currently closed. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). I need to change one set of results to the other form … Things to do. You can use the scientific calculator online on this website. ★ Complex number to polar form calculator: Add an external link to your content for free. Complex Numbers and Polar Form of a Complex Number. The online scientific calculator now supports complex numbers operations. This entry was posted write sin x (or even better sin(x)) instead of sinx. The imaginary unit, denoted as i on the scientific calculator represents the square root of -1. The new version (0.9 onwards) of the calculator supports operations with imaginary numbers and complex numbers. Again, you can choose between Cartesian, Polar or Exponential. In the following graph, the real axis is horizontal, and the imaginary (j=sqrt(-1)) axis is vertical, as usual. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). The principal value of the argument is normally taken to be in the interval .However, this creates a discontinuity as moves across the negative real axis. Changing this option will affect how the results are displayed. Point P represents a complex number. on Tuesday, January 19th, 2010 at 11:17 am and is filed under New Features, Online Scientific Calculator. To get tan(x)sec^3(x), use parentheses: tan(x)sec^3(x). Scientific Calculator Notation and Precision, Another way of writing the polar form of the number is using it’s. Express z = 3 + 4i in polar form . All functions except factorial support complex arguments. Polar to Cartesian coordinates. The following table contains the supported operations and functions: If you like the website, please share it anonymously with your friend or teacher by entering his/her email: In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. is the real part. So whichever form you enter your complex number, it will get turned into Cartesian. Complex roots. Choose whether your angles will be in degrees or radians first. The calculator will find the polar form of the given complex number, with steps shown. ... complex-number-calculator menu ... Equations Inequalities System of Equations System of Inequalities Polynomials Rationales Coordinate Geometry Complex Numbers Polar/Cartesian Functions Arithmetic & Comp. I have cartesian matlab results and polar written results. Vote. Key Concepts. Notes. r∠θ° is a representation of complex number in polar format. but you can do it when it is easiest. But what NaN really means is Not A Real Number. Note De moivre 's theorem. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. Home / Mathematics / Complex number… So to enter, for example, the polar form number (240 75) into the calculator you must enter 240 e ^(i 75 π /180). If the calculator did not compute something or you have identified an error, please write it in Complex numbers can be input in Cartesian, Polar, or Exponential form. Converting Rectangular Equations to Polar Form. Tags: Calculator, calculator preferences, Calculators, cartesian form, complex number calculator, Complex Numbers, exponential form, exponential notation, imaginary numbers, imaginary unit, Online Calculator, Online Calculators, Online Scientific Calculator, polar form. The polar form of a complex number is another way to represent a complex number. a+ib is a complex number where a is the real part and ib is the imaginary part.This format is known as rectangular format.Complex numbers are also presented in another format that is known as polar format. In general, we can say that the complex number in rectangular form is plus . The Online Scientific Calculator for Complex Numbers. ; The absolute value of a complex number is the same as its magnitude. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step. Male or Female ? NaN is not a number, infact, NaN stands for ‘Not A Number’. We have been given a complex number in rectangular or algebraic form. The polar form of a complex number takes the form r(cos + isin ) Now r can be found by applying the Pythagorean Theorem on a and b, or: r = can be found using the formula: = So for this particular problem, the two roots of the quadratic equation are: Hence, a = 3/2 and b = 3√3 / 2 The complex number calculator allows to multiply complex numbers online, the multiplication of complex numbers online applies to the algebraic form of complex numbers, to calculate the product of complex numbers 1+i et 4+2*i, enter complex_number((1+i)*(4+2*i)), after calculation, the result 2+6*i is …
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# Issue with .NET Wrapper DLL
This topic is 2476 days old which is more than the 365 day threshold we allow for new replies. Please post a new topic.
## Recommended Posts
I have been trying to sort this issue out for 3 days, and just spent 30\$ on liveperson.com talking to an "expert" only to have them say "working on this now" then close the chat. I am getting desperate, so I am hoping that someone may be able to assist.
I have setup the ImDisk (http://www.ltr-data....de.html/#ImDisk) .NET Wrapper in my project, and have written out my code. However, when I run my application, I get errors such as...
Unable to find an entry point named '_ImDiskFindFreeDriveLetter@0' in DLL 'imdisk.cpl'.
I have added the .dll/xml to my project directory, added the DLL as a reference to my project, and have had no luck. There is no sample code, and very little information on this library (although quite a few people seem to use it without issue) so I am stumped. The CLI and the Applet work fine, so the drivers and everything are installed fine. Just using the API through this wrapper is proving to be problematic.
My code is as follows...
public void CreateRAMDisk() { // Create Empty RAM Disk char driveLetter = ImDiskAPI.FindFreeDriveLetter(); ImDiskAPI.CreateDevice(50000, 0, 0, 0, 0, ImDiskFlags.DeviceTypeHD | ImDiskFlags.TypeVM, null, false, driveLetter.ToString(), ref deviceID, IntPtr.Zero); Console.WriteLine("RAMDisk DeviceID is " + deviceID); // Create Mount Point (IS THIS NEEDED ALSO, OR OPTIONAL???) //ImDiskAPI.CreateMountPoint("C://empty_directory", deviceID); }
After searching Google, I read that issue occurs when the object is not initialized, however, it has no constructor so it can't be. The very small amount of code someone else pasted on their forums looked like they were just calling the methods directly as well, so I don't believe this is the issue.
Any help would be fantastic, as I am starting to get rather desperate.
##### Share on other sites
Fixed it. Apparently the issue was with the API I was using not working correctly on x64.
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# Rationality and Game Theory
The discrepancy between the predictions that game theory makes for the way to play the centipede game and what happens in actual practice has created a torrent of research...
Joseph Malkevitch
York College (CUNY)
malkevitch at york.cuny.edu
### Introduction
If you were offered $1000, no strings attached, you would probably take the money. And run? If the amount were$10,000 rather than $1000, again, no strings attached, you would probably take the money, and run even faster. However, if you saw a$100 bill lying on the street, and your back had been giving you agonizing pain recently, though currently you were in no pain, you might not bend over to pick it up. Certainly for a quarter (25 cents) it would not be worth it to stoop over. In explaining and offering advice about economic behavior, economists and mathematicians invoke arguments about "rational" behavior to explain what actions people should take. However, as the example above shows, if the decision maker has more information (the state of her back) than the observer, then the behavior observed may be different from the behavior expected.
The Theory of Games, pioneered by the mathematician John Von Neumann (1903-1957)
and the economist Oskar Morganstern (1902-1976),
offers a variety of thought-provoking examples where the difference between the behavior that logic dictates and the behavior actually seen are worlds apart. Thus, Game Theory offers mathematicians, psychologists, political scientists, philosophers, economists, and other scholars an exciting playground for exploring fascinating ideas and a tool for probing a wide variety of problems that are at the core of their disciplines. If one views game theory from a mathematical modeling point of view, where parts of game theory are used to provide a representation for behavior in the "real world," then experiments involving games offer not only a way to improve our insights into human behavior but also a way to develop new approaches and ideas for game theory itself.
Those who are familiar with the theory of games may wish to skim the next section which deals with some relatively well known aspects of game theory. The goal of this preliminary material is to serve as a contrast between what "standard" game theory wisdom suggests about certain games and the way these games are played in practice.
### Game Theory: The Basics
Game theory has grown into a complex and multi-branched subject. The basic idea is that one has a group of people (usually referred to as players) who are interacting with each other with some conflict the group desires to resolve. For simplicity, let us consider only two people (countries or companies) as players. Depending on the actions or decisions that are taken by the players, different "payoffs" will accrue to the two people involved. We will assume that the games involved are games of perfect information. This means that each player knows exactly what actions are open to him/her and to his/her opponent. It is also assumed that the payoffs to each player are known and what the value of each payoff is. The games we are interested in here typically arise in daily life, economics and political science (law), in contrast to games such as Nim, Dots and Boxes, and Hackenbush, which are sometimes called combinatorial games. The examples below give the range of conflict situations that game theory illuminates.
Situation 1:
A husband and wife are trying to decide what do on Friday night. The husband would like to see a movie but the wife would like to go to the opera.
Situation 2:
Country X has threatened to test a newly developed nuclear weapon underground and Country Y is upset about the possibility.
Situation 3:
Two children simultaneously flash either two fingers or one finger. If the sum is even, each child wins a dime; if the sum is odd, each child loses a dime.
Situation 4:
Two competing electronic stores must decide whether to use television advertising during the week before the Thanksgiving holiday.
These situations show a small sample of various degrees of seriousness of consequences to the players involved and although only two people (entities) are mentioned, there are often other "parties" affected by the decisions made by players in a game. For example, for the nuclear testing situation, the whole world runs the risk of having radiation released into the atmosphere if something goes wrong with the underground test.
Situations of these kinds may occur only once or may be repeated often under more or less the same circumstances. Sometimes the players may be able to communicate with each other about what actions to take, but in other situations the players interact more or less independently of each other. Even if the players can communicate and reach agreements about how to act when playing the game, there is not always a guarantee that one or both players might not break a "treaty" that they have agreed upon. Sometimes the players may be strangers to each other whereas in other cases they may know each other well. When players do know each other, they might know what their respective values might be as well as the way they think about situations involved in the game. However, even in the case of playing against a stranger, the players may be governed in their actions by their views about fairness, sympathy, and altruism.
When players interact in games of this kind, they may look at the outcomes from the play of the game in complex ways. In some games there is money that changes hands in the game. However, in many cases the outcomes may be measurable on several scales. Thus, for a couple deciding which event to attend on a Friday evening, there may be monetary implications to the outcome (movies are typically cheaper to attend than operas) but also "satisfaction" outcomes. Psychologists and economists, together with mathematicians, have developed theories of "utility" which attempt to allow a player of a game to assign a single number, called a utility, to the outcome that the player receives from a game (or more generally from making some decision). The main idea is that if action A is preferred to action B, then the utility assigned to A should be higher than the utility assigned to B. However, it is well known that in expressing preference humans are not always consistent. Thus, John might prefer apples to bananas, bananas to cherries, but cherries to apples. Some people, when this "intransitivity" of preference is pointed out to them will "correct" their stated preferences; other people will stand by what they have said. This can be because there is no single underlying scale on which the fruit are being judged by this person, which means that when expressing a preference, a person is balancing complex views about the fruits. Thus, when such "cyclic" preferences exist, it will not be possible to assign a number to each of the fruits so that one fruit is preferred to the other when the number assigned to it is higher.
This brief discussion should suggest that "utility theory" is a fascinating and complex subject. Thus, a rich and a poor person would look at the same sum of money that might change hands in a game quite differently. For our purposes let us pretend that the players are able to evaluate for themselves how to measure the outcomes of the game on some scale of utilities. These outcomes of the game arise from the players making a choice among actions that are available to them. An action on the part of each player leads to an outcome. Sometimes the utilities associated with an outcome will be known to both of the players and sometimes they won't. We will also assume that given a choice of a higher versus lower numerical payoff, both players will always select the higher payoff.
It will be convenient to refer to the two players involved in a game as Row and Column. I will use female pronouns for Row and male pronouns for Column. How can we record the actions available to the players of a game? One method is to use a "tree diagram" (graph) where the vertices (sometimes called nodes) of the tree show alternative choices for the player whose move is represented by being at this node. In the simplest case trees are used to merely show alternative actions for the players, but they can also be used to display what information is available to the players at the times choices might be made. Thus, a player's opponent may or may not know what choices an opponent might have at a tree node. When the "moves" and payoffs for a game are displayed in this form, the game is said to be described in "extensive form." A very simple example is displayed below (Figure 1). In the diagram shown, the 1-valent vertices of the tree, or leaves of the tree, are labeled with pairs (r, c) which indicate the payoff to row and column, respectively. Thus, if Row plays her choice II and Column plays his choice 1, then Row receives -$8 and Column receives$8. This can be interpreted to mean that Row pays Column $8. Figure 1 The "extensive form" of this game can be displayed more compactly, and in many respects more clearly (though as previously noted with a loss of generality), if the results are displayed in "normal form." This is done with a matrix, showing Row's choices in the rows of the matrix and Column's choices in the columns of the matrix, as in Figure 2. The payoffs again are displayed as pairs and the Row player's payoff is indicated first in each pair. The way this game is played is that without communicating, the two players choose one of the two actions indicated. For Row this means picking Row I or Row II while for column it means selecting either Column 1 or Column 2. You can think of each of the players as writing either I or II on a piece of paper by Row, and 1 or 2 on a piece of paper by Column, and handing their choice to a "judge." This person then affirms the payoff involved. For example, if Row writes II and Column writes 1, the judge asks Row to give Column$8. This particular game is known as a zero-sum game because the sum of the amounts that change hands with each play of the game is zero. (Often the payoffs in a zero-sum game are displayed with only one entry in each cell of the matrix. This number is typically the payoff to the row player. The payoff to the column player can be deduced to be the negative of the amount that is paid to Row, whose payoff may be negative.)
Suppose that you play this game over and over again. What this means is that you (say you are Row) would have to pick which of I or II to play on the first play of the game and then you would have to decide what to do on future plays of the game either based on what Column did in her previous plays of the game or independent of any behavior that Column might follow. How would you chose to play? Note that the symmetry of the payoffs suggests that whatever is the best thing for Row would also be the worst thing for Column. Also note that the sum of the row player's entries (in the possible outcomes) adds to zero. Do you think that this means that the game is "fair?" A fair game can be thought of as one in which the two players, assuming that each plays as well as possible (e.g. optimal play), will have a net earning of zero in the long run.
Column 1 Column 2 Row I (9, -9) (-2, 2) Row II (-8, 8) (1,-1)
Figure 2
Mixed strategies
Perhaps it is not so clear to you what would be your optimal way to play the game in Figure 2 if you were Row. Here is a game that is perhaps simpler!
Column 1 Column 2 Row I (9, -9) (-2, 2) Row II (4, -4) (-5, 5)
Figure 3
How would you play this game if you were Row? Notice that since 9 is bigger than 4 and -2 is bigger than -5, if Row plays row I, REGARDLESS of what Column does, Row gets a higher payoff. Thus, it would seem, in light of our discussion that more money is always preferred to less money, Row will never play Row II because she can do better by playing row I whatever Column does. In a situation such as this we say that row I dominates row II because no matter which column the column player might select the payoff for Row in row I is better than the payoff for Row in row II. However, if Column is a rational player he will realize that Row will never want to play row II and, thus, makes the best move consistent with that reality. This means that Column would want to pick column 2. To do otherwise would be to lose on every play of the game while column 2 means that Column will win on every play. This game is certainly not fair to Row, but rational play requires that the players do as indicated. After all, Row easily sees that if Column plays column 2 all of the time, then playing row II rather than row I will result in a loss of 5 with each play of the game rather than a loss of 2 with each play. These numbers can be thought of as constituting a "solution" for this game.
The analysis above was started from Row's point of view and indicated that there was a dominating row. We could have started the analysis from the point of view of Column. It turns out that column II is a dominating column from Column's point of view. Why ever play column I when, no matter which choice Row makes, column 2 yields a higher payoff? From the point of view of logical reasoning, it seems reasonable to call the outcome Row 1 and Column 2 to be the "solution" to the game in the sense that rational play should result in this outcome. (For more complicated game theory situations there may well be more than one "solution" to a game. The reason for this is that different ways of looking at sensible or rational behavior leads to different outcomes from the different points of view.)
In playing the game in Figure 3 optimally, each player is able to chose a row and a column which results in his/her best outcome. The fact that this is true is expressed by saying that there is a "pure" strategy (always play a fixed row by Row and always play a fixed column by Column) that yields optimal results. This terminology is applied whether or not the game is played once or is played many times, and it is also true whether or not the game is zero-sum or not. Note that situation may not always be this "clean." For some games there will be no optimal way to play that involves "pure" strategies.
Now consider the game which is not a zero-sum game (Figure 4).
Column 1 Column 2 Row I (2, 1) (3, 0) Row II (4, -5) (1, 2)
Figure 4
Suppose Row and Column have been happily playing this game for several rounds. Row and Column both are getting positive payoffs, and although Column is not getting as much as Row, he is not unhappy with his payoff of 1. Now, Row notices that if Column continues to play column 1 in the next "round of play," Row will get 4 units (instead of 1) and Column will get -5. Row does not really dislike Column but she reasons that he has won some money in the previous rounds and can stand some loss now without getting too angry. So, in the next round Row plays row II and Column plays column 1. Needless to say Column is upset that Row changed from row I to row II but he notes that if Row is planning to assume that he is a "sucker" who will continue to play column 1 in the next round, she is not thinking clearly. He realizes that if Row plays row II, then he will respond in the next round with column 2. Not only will this erase his loss but he will now earn more than Row when row I and column 1 were being played. Furthermore, if while he moves to column 2, Row moves to row I, he will be better off than losing 5 as is currently the situation. Suppose Row holds steady at row II and Column moves to column 2. Now Row, less happy than she was when she was getting a payoff of 2 and her opponent was getting a payoff of one, decides that if Column stays with column 2 she wants to move to Row 1. Well you get the idea. This game has no outcome where one of the two players might not be tempted to move to another action. No "pure" action will lead to a "solution" for this game.
Now let us go back to the zero-sum case. Consider the following game (Figure 5) :
Column 1 Column 2 Row I (1, -1) (-1, 1) Row II (-1, 1) (1,-1)
Figure 5
This game is symmetrical in its payoffs but it does not have any row or column that dominates any other row or column. The sum of the entries for the four payoffs for Row and Column each add to zero. This game is fair, that is, the long-term payoff for either player from playing this game many times would be 0. (To obtain this long term payoff one adds together the winnings on the individual times the game is played and divides by the number of times the game is played.) Indeed, this is a fair game. However, this does not mean that one can play this game in any way one might want. Thus, suppose Row decided that to make her life easier she would play rows I and II in the fixed sequence I, I, II, I, II, II repeated over and over. Over a period of time, an observant Column player would notice this pattern and would start playing the pattern of columns 2, 2, 1, 2, 1, 1, with the result that Column would win every play of the game. Of course, Row might catch on to the fact that she was always losing, so a clever Column who saw a deterministic pattern of play from his opponent might somewhat vary his play so that he won most of the time rather than all of the time!
The probably unexpected point here is that playing this game, say by Row, involves using a random pattern of Rows. And not just any random pattern of rows, but a pattern where row I is used 1/2 of the time and row 2 is also used 1/2 of the time. This way of playing the game is known as the optimal mixed strategy. "Mixed" refers to the mixture or fraction of the time Row plays each of the rows (and similarly for Column). In order to get the best return of payoffs Row must not play a deterministic pattern of the rows but must mix the percentage of time row I and row II are played in a particular way. How does one find the optimal mixed strategies for the players in the following game (Figure 6):
Column 1 Column 2 Row I (8, -8) (-6, 6) Row II (-2, 2) (3,-3)
Figure 6
Rather than derive the appropriate theorems from scratch, let me show a way to use one of these results to shorten the computation of the mixed strategy for a game such as this one. Remarkably, it has been shown that for many zero-sum 2-person games such as that in the example below, when either of the players, say Row, is playing his/her optimal mixed strategy with a resulting long run payoff P, it makes no difference what pattern of play is used by the other player; the payoff to the other player will be -P. Note that when Row is playing optimally, there is no guarantee that Row's optimal long run payoff will be positive.
Using this theorem, we can compute the optimal mixed way each player should play the game (Figure 6). Suppose, for example, that we use p to represent the fraction of the time that Row should play row I to achieve her maximum payoff. Then it follows that she should play row II 1-p of the time. Since the payoff to Row in this case will be the same whatever Column does, let us first compute the payoff to Row when Column plays column 1 all of the time.
Since Row plays row I a fraction p of the time, the "expected" long run payoff for this would by 8p and since Row plays row II 1-p of the time, the expected payoff for this would be -2(1-p). In the long run the payoff from this pattern play would be: 8p -2(1-p). Using a similar calculation when Column plays column 2 all of the time we get -6(p) + 3 (1-p). Hence, to find the optimal value of p we must solve the linear equation below for p:
It is easily seen that q, the fraction of the time that Column should play column 1 is 9/19, though it is worth making this calculation and solving the resulting linear equation to practice how the thinking goes.
To find out who has an advantage, one can substitute the value p=5/19 into the left or right hand side of the equation above. One sees that Row's payoff is 12/19 and, thus, Column's payoff is -12/19. This is not such an easy number to see without the mathematics! Thus, if this game is repeated 19 times the payoff to Row will be approximately $12. Each time the game is played$8, $6,$3 or \$2 changes hands but on the average, over a long number of plays of this game, Row will win 12/19 of a dollar on each play of the game and Column will lose 12/19 of a dollar on each play.
A famous theorem of von Neumann in essence shows that for a two-player zero-sum game where Row has m choices and Column has n choices, each time the game is played that game has an optimal mixed strategy for each player. This optimal strategy has the property that if either player deviates from the mixed strategy which is best for him/her, then that player will only do worse.
### Nash Equilibria
However, many of the most interesting games that mathematicians and those who try to apply game theory in the real world have to analyze are not zero-sum games. For example the game below might be a model of the situation described earlier in which a couple is trying to decide whether to go to a movie or to the opera.
Column 1 (Opera) Column 2 (Movie) Row I (Opera) (6, 2) (4, 4) Row II (Movie) (7, 4) (10, 3)
Figure 7
Using dominating strategy analysis this game has the "solution" of Row playing row II and Column playing column 1, using pure strategies. However, other nonzero-sum games, such as the one we saw in Figure 4, do not have a solution in pure strategies. Remarkably, it has been shown that if one moves to the domain of "mixed strategies" there will always be an "equilibrium" solution for Row and Column - that there will always be a "Nash Equilibrium." The concept is named for John Nash, who won the Nobel Prize in Economics for his seminal work in showing the existence of the "equilibria" that now bear his name. Nash's Theorem is an existence theorem.
He uses mathematics, the Kakutani fixed point theorem, to show that equilibria exist for certain games but does not show how to find them. In fact, the issue of determining the number and computational complexity of finding Nash Equilibria for games is still a wide open area of research. The basic idea behind the concept of a Nash Equilibrium is that for wide classes of games there are ways for the players to play the game, sometimes as pure strategies and sometimes as mixed strategies so that if a player deviates from the Nash Equilibrium strategy, he/she can not improve his/her payoff. Our more detailed analysis of the two-person zero-sum games with two action choices for each player gives the spirit of what is involved.
### Fictitious Play
For those with mathematical skill one can master the techniques for finding the optimal play of games such as the ones we have been looking at. It would be nice, however, if there was a more pragmatic approach to playing a game such as the one we solved (Figure 6). An insight into this was obtained by the mathematician Julia Robinson (1918-1985), who is much better known for her famous role in the solution of Hilbert's 10th Problem. Robinson was also the first woman President of the American Mathemtical Society. She proved an intriguing result based on an idea pioneered by the Princeton educated mathematician George William Brown. Brown introduced the idea of "fictitious play," though the terminology is not particularly felicitous because it is a technique that often can be used in practice to play games.
Suppose you are asked to play a zero-sum game with no dominating rows or columns such as the one in Figure 6. You could, if you have mastered how, do the calculations we did to find the optimal way to play. However, another approach might be to see if there was some "adaptive" way to play the game. Intuitively, the idea is to learn from playing the game to be able to play the game better. Start by making what you hope is a good first move, look at the payoff you receive based on your opponent's choice and then move in the future in a way to try to "improve" (or not make worse) your situation. What Julia Robinson showed is that for 2-person zero-sum games this "learning as you go along" approach to games works. In the end, if each player adopts this point of view they will eventually converge to their optimal mixed strategies. Unfortunately, as shown by Lloyd Shapley this is not true for all games. Shapley provided an example of a 3x3 game for which the adaptive learning approach does not work. Robinson's work and extensions of it have important connections to the area of mathematics known as dynamical systems. Robinson's work has been extended in a variety of ways. Other works have been showing connections between ideas involving adaptive approaches to playing games and learning environments in general. Other researchers are showing exciting connections to ideas in evolutionary biology, where game theory is having exciting implications.
### Prisoner's Dilemma
Perhaps the most famous of all games is Prisoner's Dilemma. It deserves (and perhaps will eventually get) a column of its own. The game takes its name from a "story" to accompany the matrix of the game, that is due to Albert W. Tucker, who for many years headed the Mathematics Department at Princeton University. The story helps explain the payoffs of a 2-person nonzero-sum game with a structure that has "paradoxical" implications.
(Albert W. Tucker, courtesy of his son Alan Tucker)
The game below belongs to the Prisoner's Dilemma type of game.
Column 1 Column 2 Row I (3, 3) (0, 5) Row II (5, 0) (1, 1)
Figure 8
If one carries out a dominating strategy analysis of this game, one sees that the players should play row II and column 2 to achieve their "best" outcome. However, the payoff of 1 to each of Row and Column is poorer than the payoff of 3 to each of them for playing row I and column 1. This creates the paradoxical situation that rational play leads to a poorer outcome than irrational play. The situation is even more dramatic when the outcomes become negative:
Column 1 Column 2 Row I (8, 8) (-6, 20) Row II (20, -6) (-4,-4)
Figure 9
Game theory suggests that the "rational" way to play this game is to play row II and column 2 either when the game is played once, or when there is repeated play of this game. Yet, this seems difficult to accept if there is an outcome where on each play, instead of losing 4 units, the players can each gain 8 units. However, the game theory analysis shows that playing row I and column 1 is not "stable." Players playing this way in repeated plays of this game will always be tempted by the fact that if their opponent does not shift action simultaneously, then they can improve their own payoff while "hurting" their opponent. Thus, if Column continues to play column 1 while Row moves from row I to row II, the result will be that Row earns 20 instead of 8, while Column goes from winning 8 to losing 6. Not that this is true even though Row and Column may have signed a "binding treaty." Unfortunately, Prisoner's Dilemma is not merely an intellectual curiosity. It seems like a reasonable model for various confrontational games such as those that occur between unions and management during contract negotiations, and in the relationship between countries. In fact, the US Defense Department did research into the ways games such as Prisoner's Dilemma were played by "real people" in an attempt to understand the implications of "cold war" politics during the time of the Soviet Union.
### The Centipede Game
Compared with Prisoner's Dilemma the centipede game is not that well known, but like Prisoner's Dilemma, it has been a very fertile window into gaining insight into the nature of rationality. The discrepancy between the predictions that game theory makes for the way to play the centipede game and what happens in actual practice has created a torrent of research.
The centipede game, again like Prisoner's Dilemma is actually a family of games with similar characteristics but in which the specific game variant depends on a collection of numerical parameters. Experiments using different parameters yield different results and insights into the family.
This game was developed by Robert W. Rosenthal, a mathematically trained economist. However, the name for the game is due to Kenneth Binmore.
(Courtesy of Professor Randall Ellis, Dept. of Economics, Boston University)
Sadly, Rosenthal died of a heart attack in 2002 at the relatively young age of 57. His research ranged over a variety of aspects of the interplay of game theory and economics. His work often returned to the issue of what constituted sensible or rational behavior in economic settings.
The basic idea of the centipede game involves alternating moves of two players in a game which is described in "extensive form" by a tree. In looking at the current move that a player has versus the move that his/her opponent can next make, the player sees that the current pot of outcomes gives her more income than her opponent, but on the next move the reverse will be true. A generic version of the game is shown below. The two players' names are A and B and at a given vertex (white circle) of the "decision tree" the names of the players alternate. At a particular dot, say one labeled A, which is where player A must make a decision, there are two moves that A can make. A can move either to the right (go to the next vertex to the right) where A cedes to B the decision about what to do next, or A can terminate the game by going down. Going to the right in the literature of the centipede game is often described as "pass," which means the player currently to move does not terminate the game (except for the very last decision), while the decision to go down is described as "take." The two numbers at the end of the down arrow are the payoffs to A (on the left) and B (on the right). The down (take) moves are indicated in blue to contrast them with the right (pass) moves. Usually, the game is assumed to have equal numbers of decision moves for A and for B. (This means there is an odd number of 1-valent vertices or leaves in the tree.)
Figure 10
The game gets its name from the diagram above where there is a resemblance to a critter with many legs. However, consider the small version of the centipede game which is shown in Figure 11 in order to see the troublesome aspects of this game for theorists.
Figure 11
Suppose you are A. How would you play this game? Here is the mathematical reasoning that comes with treating this game as one of perfect information, where both players are completely rational each opponent behaves rationally. Suppose the game were to reach the stage where player B has to act at the far right decision vertex. Now B, preferring more money rather than less will choose take (down) at this node. This follows since if B passes, he gets 16 while his opponent A gets 64. By "taking," B gets 32 while A only gets 8. Thus, at the prior vertex, A knowing what B will do, will chose to pick take (down). Knowing this, "backward reasoning" means that at the prior node B will take, which in turn means that at the very first decision node, A should take as well! Reasoning of this kind has come to be called "backward induction." Similarly, in the "long" (e.g. many feet) version of the centipede game it means that A should choose the take (down) option on the very first play of the game. This means that both players receive relatively meager payoffs. To play longer offers each of the players a chance to make more but only at the risk that when the game is stopped, his/her opponent will walk away with more than he/she will.
In this variant of the game the amounts each player gets are going up quickly; however, in some variants, the growth is very slow. Furthermore, in some variants the initial pot, if taken, gives each equal amounts. One can look at a version of the game where the sum of the payoffs to the players is a constant. The characteristic of these games is that rational play suggests that the first player should take right at the start. Yet, this seems a perplexing choice since so much more can be gained by being "irrational." Also, note that if A passes (allows B to move) at the first node of the game, then B faces a new version of the centipede game with essentially the same structure that A had! Why might you not take at the first opportunity if you are player A? One reason is that you might have some "altruistic" component to your personality. You might also think that perhaps your opponent, if the centipede is longer (rather than having 4 legs), will make a mistake in the analysis and continue to play longer when it is his/her decision.
The situation gets interesting if one takes this game and sees what happens when real people (rather than armchair philosophers, mathematicians and game theorists) play the game! Experimental economists and other scholars have actually done this. Nearly always, real people, who have been clearly explained the rules of the game and understand them, do not terminate the game at the first move! Does this mean that the analysis of the game that we have given is wrong or does it mean that human beings are "irrational"?
Are you surprised that many people do not play this game "rationally" in the sense that when given the chance to play A at the first round, they do not choose "take"? On the other hand it is also the case that the game does not typically last as long as allowing B to make the pass move at the last vertex. Variants of the centipede game have provided a controlled environment to probe the nature of altruism, signaling to achieve cooperation, and the limits to rationality. These experiments have worked with the centipede game in extensive form as well as in recent work that gets new insights from using a normal form for the centipede game. In this work the subjects of the experiments must name the "round" where they would "take" in the centipede game rather than play the game dynamically as such. An interesting recent set of experiments concerns playing the centipede game where groups of players compete in the game against each other rather than individuals playing each other. The results here suggest that groups come closer to the prediction of "taking" on the first round of play than when individuals play. The accelerating number of papers of both a theoretical type and experimental type that draw inspiration from the centipede game are a testament to the fertility of a single game and its associated circle of ideas to stimulate mathematicians and scholars over a wide array of disciplines.
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Young, H., Strategic Learning and Its Limits, Oxford U. Press, Oxford, 2004.
Zauner, K., A payoff uncertainty explanation of results in the experimental centipede games, Games and Economic Behavior 26 (1999) 157-185.
NOTE: Those who can access JSTOR can find some of the papers mentioned above there. For those with access, the American Mathematical Society's MathSciNet can be used to get additional bibliographic information and reviews of some these materials. Some of the items above can be accessed via the ACM Portal, which also provides bibliographic services.
Joseph Malkevitch
York College (CUNY)
malkevitch at york.cuny.edu
Welcome to the
Feature Column!
These web essays are designed for those who have already discovered the joys of mathematics as well as for those who may be uncomfortable with mathematics.
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## Abstract and Applied Analysis
### An Averaging Principle for Stochastic Differential Delay Equations with Fractional Brownian Motion
Yong Xu, Bin Pei, and Yongge Li
#### Abstract
An averaging principle for a class of stochastic differential delay equations (SDDEs) driven by fractional Brownian motion (fBm) with Hurst parameter in $(\mathrm{1}/\mathrm{2},\mathrm{1})$ is considered, where stochastic integration is convolved as the path integrals. The solutions to the original SDDEs can be approximated by solutions to the corresponding averaged SDDEs in the sense of both convergence in mean square and in probability, respectively. Two examples are carried out to illustrate the proposed averaging principle.
#### Article information
Source
Abstr. Appl. Anal., Volume 2014, Special Issue (2013), Article ID 479195, 10 pages.
Dates
First available in Project Euclid: 26 March 2014
Permanent link to this document
https://projecteuclid.org/euclid.aaa/1395858397
Digital Object Identifier
doi:10.1155/2014/479195
Mathematical Reviews number (MathSciNet)
MR3166618
Zentralblatt MATH identifier
07022457
#### Citation
Xu, Yong; Pei, Bin; Li, Yongge. An Averaging Principle for Stochastic Differential Delay Equations with Fractional Brownian Motion. Abstr. Appl. Anal. 2014, Special Issue (2013), Article ID 479195, 10 pages. doi:10.1155/2014/479195. https://projecteuclid.org/euclid.aaa/1395858397
#### References
• J. B. Roberts and P. D. Spanos, “Stochastic averaging: an approximate method of solving random vibration problems,” International Journal of Non-Linear Mechanics, vol. 21, no. 2, pp. 111–134, 1986.
• W. Q. Zhu, “Recent developments and applications of the stochastic averaging method in random vibration,” ASME Applied Mechanics Reviews, vol. 49, 10, pp. S72–S80, 1996.
• N. Sri Namachchivaya and Y. K. Lin, “Application of stochastic averaging for nonlinear dynamical systems with high damping,” Probabilistic Engineering Mechanics, vol. 3, pp. 185–196, 1988.
• W. Q. Zhu, “Stochastic averaging methods in random vibration,” ASME Applied Mechanics Reviews, vol. 41, no. 5, pp. 189–199, 1988.
• R. L. Stratonovich, Topics in the Theory of Random Noise, Gordon and Breach, New York, NY, USA, 1967.
• R. L. Stratonovich, Conditional Markov Processes and their Application to the Theory of Optimal Control, Elsevier, New York, NY, USA, 1967.
• R. Z. Khasminskii, “A limit theorem for the solution of differential equations with random right-hand sides,” Theory of Probability & Its Applications, vol. 11, pp. 390–405, 1963.
• R. Z. Khasminskii, “Principle of averaging of parabolic and elliptic differential equations for Markov process with small diffusion,” Theory of Probability & Its Applications, vol. 8, no. 1, pp. 1–21, 1963.
• I. M. Stoyanov and D. D. Bainov, “The averaging method for a class of stochastic differential equations,” Ukrainian Mathematical Journal, vol. 26, no. 2, pp. 186–194, 1974.
• V. G. Kolomiets and A. I. Mel'nikov, “Averaging of stochastic systems of integral-differential equations with Poisson noise,” Ukrainian Mathematical Journal, vol. 43, no. 2, pp. 242–246, 1991.
• W. Q. Zhu, “Nonlinear stochastic dynamics and control in Hamiltonian formulation,” ASME Applied Mechanics Reviews, vol. 59, no. 4, pp. 230–248, 2006.
• W. T. Jia, W. Q. Zhu, and Y. Xu, “Stochastic averaging of quasinon-integrable Hamiltonian systems under combined Gaussian and Poisson white noise excitations,” International Journal of Non-Linear Mechanics, vol. 51, pp. 45–53, 2012.
• Y. Zeng and W. Q. Zhu, “Stochastic averaging of quasi-nonintegrable-Hamiltonian systems under Poisson white noise excitation,” ASME Journal of Applied Mechanics, vol. 78, no. 2, Article ID 021002, 11 pages, 2011.
• Y. Xu, J. Duan, and W. Xu, “An averaging principle for stochastic dynamical systems with Lévy noise,” Physica D, vol. 240, no. 17, pp. 1395–1401, 2011.
• W. E. Leland, M. S. Taqqu, W. Willinger, and D. V. Wilson, “On the self-similar nature of Ethernet traffic (extended version),” IEEE/ACM Transactions on Networking, vol. 2, no. 1, pp. 1–15, 1994.
• Y. Hu and B. ${\text{\O}}$ksendal, “Fractional white noise calculus and applications to finance,” Infinite Dimensional Analysis, Quantum Probability and Related Topics, vol. 6, no. 1, pp. 1–32, 2003.
• R. Scheffer and F. R. Maciel, “The fractional Brownian motion as a model for an industrial airlift reactor,” Chemical Engineering Science, vol. 56, no. 2, pp. 707–711, 2001.
• N. Chakravarti and K. L. Sebastian, “Fractional Brownian motion models for polymers,” Chemical Physics Letters, vol. 267, no. 1-2, pp. 9–13, 1997.
• Y. Xu and R. Guo, “Stochastic averaging principle for dynamical systems with fractional brownian motiončommentComment on ref. [19?]: Please update the information of this reference, if possible.,” AIMS Discrete and Continuous Dynamical Systems B. In press.
• T. Taniguchi, K. Liu, and A. Truman, “Existence, uniqueness, and asymptotic behavior of mild solutions to stochastic functional differential equations in Hilbert spaces,” Journal of Differential Equations, vol. 181, no. 1, pp. 72–91, 2002.
• X. Mao, “Numerical solutions of stochastic functional differential equations,” LMS Journal of Computation and Mathematics, vol. 6, pp. 141–161, 2003.
• L. Tan and D. Lei, “The averaging method for stochastic differential delay equations under non-Lipschitz conditions,” Advances in Difference Equations, vol. 2013, article 38, 2013.
• F. Biagini, Y. Hu, B. ${\text{\O}}$ksendal, and T. Zhang, Stochastic Calculus for Fractional Brownian Motion and Applications, Springer, London, UK, 2008.
• T. E. Duncan, Y. Hu, and B. Pasik-Duncan, “Stochastic calculus for fractional Brownian motion I. Theory,” SIAM Journal on Control and Optimization, vol. 38, no. 2, pp. 582–612, 2000.
• Y. S. Mishura, Stochastic Calculus for Fractional Brownian Motion and Related Processes, vol. 1929 of Lecture Notes in Mathematics, Springer, Berlin, Germany, 2008.
• E. Alòs and D. Nualart, “Stochastic integration with respect to the fractional Brownian motion,” Stochastics and Stochastics Reports, vol. 75, no. 3, pp. 129–152, 2003.
• F. Russo and P. Vallois, “Forward, backward and symmetric stochastic integration,” Probability Theory and Related Fields, vol. 97, no. 3, pp. 403–421, 1993.
• L. C. Young, “An inequality of the Hölder type, connected with Stieltjes integration,” Acta Mathematica, vol. 67, no. 1, pp. 251–282, 1936.
• D. Nualart and G. Via, “Stochastic integration with respect to fractional Brownian motion and applications,” Contemporary Mathematics, vol. 336, pp. 3–39, 2003.
• M. Ferrante and C. Rovira, “Convergence of delay differential equations driven by fractional Brownian motion,” Journal of Evolution Equations, vol. 10, no. 4, pp. 761–783, 2010. \endinput
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# If x = 1 is a common root of ax2 + ax + 2 = 0 and x2 + x + b = 0, then, ab =
Question:
If $x=1$ is a common root of $a x^{2}+a x+2=0$ and $x^{2}+x+b=0$, then, $a b=$
(a) 1
(b) 2
(c) 4
(d) 3
Solution:
$x=1$ is the common roots given quadric equation are $a x^{2}+a x+2=0$, and $x^{2}+x+b=0$
Then find the value of $a b$.
Here, $a x^{2}+a x+2=0$.........(1)
$x^{2}+x+b=0$......(2)
Putting the value of $x=1$ in equation (2) we get
$1^{2}+1+b=0$
$2+b=0$
$b=-2$
Now, putting the value of $x=1$ in equation (1) we get
$a+a+2=0$
$2 a+2=0$
$a=\frac{-2}{2}$
$=-1$
$a b=(-1) \times(-2)$
Then,
$=2$
Thus, the correct answer is (b)
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# Multiple phonon bands¶
This example shows how to plot several phonon band structures on a grid.
We use two files produced by anaddb:
trf2_5.out_PHBST.nc: phonon frequencies on a q-path in the BZ (used to plot the band dispersion) trf2_5.out_PHDOS.nc: phonon DOS compute with anaddb.
Out:
/Users/gmatteo/git_repos/pymatgen/pymatgen/symmetry/bandstructure.py:63: UserWarning: The input structure does not match the expected standard primitive! The path can be incorrect. Use at your own risk.
warnings.warn("The input structure does not match the expected standard primitive! "
/Users/gmatteo/git_repos/pymatgen/pymatgen/symmetry/bandstructure.py:63: UserWarning: The input structure does not match the expected standard primitive! The path can be incorrect. Use at your own risk.
warnings.warn("The input structure does not match the expected standard primitive! "
/Users/gmatteo/git_repos/pymatgen/pymatgen/symmetry/bandstructure.py:63: UserWarning: The input structure does not match the expected standard primitive! The path can be incorrect. Use at your own risk.
warnings.warn("The input structure does not match the expected standard primitive! "
/Users/gmatteo/git_repos/pymatgen/pymatgen/symmetry/bandstructure.py:63: UserWarning: The input structure does not match the expected standard primitive! The path can be incorrect. Use at your own risk.
warnings.warn("The input structure does not match the expected standard primitive! "
/Users/gmatteo/git_repos/pymatgen/pymatgen/symmetry/bandstructure.py:63: UserWarning: The input structure does not match the expected standard primitive! The path can be incorrect. Use at your own risk.
warnings.warn("The input structure does not match the expected standard primitive! "
/Users/gmatteo/git_repos/pymatgen/pymatgen/util/plotting.py:550: UserWarning: Matplotlib is currently using agg, which is a non-GUI backend, so cannot show the figure.
plt.show()
/Users/gmatteo/git_repos/pymatgen/pymatgen/util/plotting.py:550: UserWarning: Matplotlib is currently using agg, which is a non-GUI backend, so cannot show the figure.
plt.show()
from abipy import abilab
import abipy.data as abidata
import os
paths = [
#"mgb2_444k_0.01tsmear_DDB",
#"mgb2_444k_0.02tsmear_DDB",
#"mgb2_444k_0.04tsmear_DDB",
"mgb2_888k_0.01tsmear_DDB",
#"mgb2_888k_0.02tsmear_DDB",
"mgb2_888k_0.04tsmear_DDB",
"mgb2_121212k_0.01tsmear_DDB",
#"mgb2_121212k_0.02tsmear_DDB",
"mgb2_121212k_0.04tsmear_DDB",
]
paths = [os.path.join(abidata.dirpath, "refs", "mgb2_phonons_nkpt_tsmear", f) for f in paths]
robot = abilab.DdbRobot()
for i, path in enumerate(paths):
r = robot.anaget_phonon_plotters(nqsmall=2)
r.phbands_plotter.gridplot_with_hue("tsmear")
r.phbands_plotter.gridplot_with_hue("tsmear", with_dos=True)
#r.phbands_plotter.gridplot_with_hue("nkpt")
robot.close()
Total running time of the script: ( 0 minutes 5.963 seconds)
Gallery generated by Sphinx-Gallery
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Lemma 66.28.12. A closed immersion $i : Z \to X$ is of finite presentation if and only if the associated quasi-coherent sheaf of ideals $\mathcal{I} = \mathop{\mathrm{Ker}}(\mathcal{O}_ X \to i_*\mathcal{O}_ Z)$ is of finite type (as an $\mathcal{O}_ X$-module).
Proof. Let $U$ be a scheme and let $U \to X$ be a surjective étale morphism. By Lemma 66.28.4 we see that $i' : Z \times _ X U \to U$ is of finite presentation if and only if $i$ is. By Properties of Spaces, Section 65.30 we see that $\mathcal{I}$ is of finite type if and only if $\mathcal{I}|_ U = \mathop{\mathrm{Ker}}(\mathcal{O}_ U \to i'_*\mathcal{O}_{Z \times _ X U})$ is. Hence the result follows from the case of schemes, see Morphisms, Lemma 29.21.7. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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# Question #9ee04
Jun 22, 2016
The SI system is not based on the foot for length and the pound for mass and force. Those are US units instead.
#### Explanation:
Correctly:
Length --> meter
Mass --> kilogram
Force --> newton
Note that a newton of force is not the same as a "kilogram" force. The latter is the force Earth'a gravity exerts on a kilogram of mass with an acceleration of $9.81 {\text{ m/s}}^{2}$. A newton is a kilogram of mass times $1 {\text{ m/s}}^{2}$.
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# Configuration¶
Some settings, particularly the location of large databases, are stored in a global configuration file. The location of this file on your system can be found by running:
import sasktran as sk
print(sk.config.user_config_file_location())
The file is a standard yaml format file that can be edited with any text editor. The format of the file is key: value
## Keys¶
### ecmwf_directory¶
Controls where the ECMWF grib files are located to use the sasktran.ECMWF object. This directory should contain daily ecmwf grib files in the form /YYYY/MM/ERA-Int_ml_YYYYMMDD.grib The standard value for this key when on the University of Saskatchewan network is \utls\GeophysicalDatabases\ecmwfncd
### hitran_directory¶
Location of the HITRAN 2008 database. Necessary to use the sasktran.HITRANChemical object. Folder should contain a HITRAN2008 directory which contains the database. The standard value for this key on the University of Saskatchewan network is \utls\OsirisDataProducts\hitran
### baum_directory¶
Location of the Baum 2014 ice crystal optical property database. Necessary to use the sasktran.BaumIceCrystal object. Folder should contain the file GeneralHabitMixture_SeverelyRough_AllWavelengths_FullPhaseMatrix.nc. The standard value for this key on the University of Saskatchewan network is \utls\OsirisDataProducts\BaumIceCrystals
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## Bi-weekly links for July 15
A visualization of normal versus fat-tailed distributions
Archimedes: Separating Myth from Science at the New York Times.
Daniel Walsh plays detective with rolling shutter photos: given a picture of a moving propeller, can you tell how fast it was moving?
Dave Richeson uses a kayak to measure the perimeter of a lake.
From Brain Facts, some visualizations of nonlinear systems.
A profile of Aaron Clauset’s research on power laws and terrorism.
Academic doesn’t have a PhD problem. It has an attitude problem.
the Guardian has a video in which Paul Klemperer talks about geometry and the banking crisis. Surprisingly it’s tropical geometry to the rescue in resource allocation problems, as seen in this paper by Klemperer and Elizabeth Baldwin.
Brian Whitman tells us how music recommendation works and doesn’t work.
## Amazon rankings redux
Morris Rosenthal, a self-publishing author, has an interest in this subject for obvious reasons; he’s got some interesting-looking resutls for both paper books and e-books. Roughly speaking, yes; the most interesting thing I notice is that the slope of the rank vs. estimated sales curve (on a log-log scale) is higher both at the head (best-seller) and in the tail of the distribution compared to its bulk. What to make of this, I don’t know.
Chad Orzel, a physicist and author of How To Teach Physics to Your Dog took his own shot at this question a couple years ago.
## A quick bound on shuffles
David Eppstein asks: how many riffles does it take until all permutations are possible?
A riffle shuffle permutation, in the mathematics of shuffling cards, is a permutation of the cards that can be obtained by a single riffle shuffle — that is, you cut the deck into two packets and then interleave the packets. There are $2^n - n$ distinct riffle shuffles of n cards. For example, consider a five-card deck, which is initially in the order 1, 2, 3, 4, 5. Then a riffle shuffle consists of:
• cutting the deck in one of the six possible ways: 12345/ (no cut), 1234/5, 123/45, 12/345, 1/2345, /12345 (also no cut) where the slash represents the cut.
• riffling. For example look at 123/45. There are ${5 \choose 2} = 10$ possible ways to make a permutation from this – we decide which two slots to put the 4 and the 5 in, and then everything else is forced. For example if we decide that 4 and 5 will go in the second and fourth positions, we must get 14253. In general if we have k cards in the left-hand pile and $n-k$ cards in the right-hand pile, we have $\latex n \choose k$ possible shuffles. Furthermore we can decompose any one of these permutations into two “rising sequences” in exactly one way, so it can come from exactly one cut — with a single exception. That exception is the identity permutation $123\ldotsn$, which we can obtain from any of the n+1 possible cuts — so we must subtract n for the duplications. (If you put probabilities on this it becomes the Gilbert-Shannon-Reeds model.
Eppstein asks how many riffle shuffles it takes for each permutation to have nonzero probability. Since there are $2^n - n$ outcomes of a single riffle shuffle, in k iterations there are at most $(2^n - n)^k$ possible results. There will actually be less, because there are relations among the permutation subgroup generated by the riffle shuffles. (In less fancy language, there are sequences of different shuffles which give the same result. It’s the pedigree collapse of shuffling.)
Let’s replace this with $2^{nk}$ to make the math easier. Now, there are $n!$ permutations; this is greater than $(n/e)^n$ by a standard bound. So just in order to have $n!$ possible sequences of shuffles of length k, we have to have $2^{nk} > (n/e)^n$, or, taking nth roots of both sides, $2^k > n/e$. Taking base-2 logs, we get $k > \log_2 n - \log_2 e$.
Eppstein gives a dynamical systems argument that you need at least $\lceil \log_2 n \rceil$ shuffles – and it turns out that that’s enough. This is in comparison to the classic result of Bayer and Diaconis, claiming that you need $3/2 \log_2 n$ shuffles to get a well-shuffled deck.
## A Russian puzzle
Dave Richeson tweeted about a puzzle from Futility Closet (original source a Russian mathematical olympiad): can you split the integers 1, 2, …, 15 into two groups A and B, with 13 elements in A and 2 elements in B, so that the sum of the elements of A is the product of the elements of B?
Think about it for a moment. There’s of course the temptation to brute-force it, which is doable, but there’s a more elegant solution.
This got me thinking – when can you split the integers 1, 2, …, n into two groups A and B, where B has two elements, so that the sum of the elements of A is the product of the elements of B?
Say B contains x and y. Then their product is of course xy. The sum of the elements of A is $1 + 2 + ... + n - (x+y) = n(n+1)/2 - (x+y)$. Setting these equal and rearranging gives
$n(n+1)/2 + 1 = xy + x + y + 1$
where we’ve added 1 to make the factorization work out – this becomes
$n(n+1)/2 + 1 = (x+1)(y+1)$.
So the problem is reduced to finding factorizations of n(n+1)/2 + 1, which satisfy two conditions:
• x and y can’t be equal (for specificity we’ll say x < y), and
• x and y are both at most n.
Since we have y ≤ n, we’re going to have x ≥ (n+1)/2 + 1/n. n must be at least 2, so we can just write x ≥ (n/2) + 1. So we’re looking for factors of n(n+1)/2 + 1 in the interval [n/2+1, n]. Here’s some brute-force Python code to find all such solutions:
import math
def solutions(n):
out = []
total = n*(n+1)/2+1
xmin = int(math.ceil(n/2.0) + 1)
xmax = n
for x in range(xmin, xmax+1):
if total % (x+1) == 0:
y = total/(x+1)-1
if x < y:
out.append([x,y])
return out
def all_solutions(n):
out = []
for i in range(2, n+1):
sols = solutions(i)
for sol in sols:
sol.insert(0, i)
out.append(sol)
return out
solutions takes an integer n as input and returns pairs [x, y] which are solutions to the problem. For example solutions(17) returns [[10, 13]].
And all_solutions takes an integer N and returns all triples [n, x, y] with $n \le N$ which are solutions to the problem — that is, where xy equals the sum of all the integers up to n except for x and y. The first few solutions are:
n x y 10 6 7 17 10 13 26 15 21 36 22 28 37 21 31 45 27 36 50 28 43 61 42 43 65 36 57 67 42 52 78 45 66 82 45 73 91 52 78 94 57 76 101 55 91 102 70 73 110 70 85 122 66 111 136 76 120 138 87 108
So it appears that there’s nothing particularly special about the number 15 in the initial puzzle. There are plenty of values n for which you can’t do this, and plenty for which you can. Also, there are values of n for which there are multiple solution pairs (x, y), although not surprisingly they are rare. The smallest such n is 325, for which $x = 171, y = 307$ and $x = 175, y = 300$ are both solutions. In this case $latex n(n+1)/2 + 1 = 52976 = 24 \times 7 \times 11 \times 43$, from which 52976 has (5)(2)(2)(2) = 40 factors. A typical number of this size has about $log(52976) \approx 11$ factors. This abundance of factors makes it more likely that 52976 would have two factorizations of the sort we’re looking for. And in fact $52976 = 172 \times 308 = 176 \times 301$.
Solutions to this problem appear to have some interesting statistical properties… more on that in a future post.
## Weekly links for July 1
From the June Notices of the AMS:
Judith R. Goodstein and Donald Babbitt’s article of E. T. Bell and Caltech mathematics between the wars (of Men of Mathematics and Bell numbers fame).
Richard Hoshino and Ken-ichi Kawarabayashi, “Graph Theory and Sports Scheduling”. As you might suspect from the names of the authors, they’re Japanese; the numbers they use in their problem apply to Japanese pro baseball (NPB), and their work has been used in actual scheduling of NPB.
Bryna Kra on mathematics as a toolbox for the sciences in the Chronicle of Higher Education.
Joel Grus can analyze data and has a two-year-old daughter, so naturally he looked at the most boyish and girlish colors and eigenshirts for children’s T-shirts.
Alex Bellos at the Guardian shows us mathematical food items.
William Beaty on the physics behind traffic jams.
Tom Fawcett has a gallery of visualization of results from machine learning classifiers.
Jon McLoone asks is there any point to the 12 times table?
Rafe Kinsey, at the University of Michigan, is teaching a freshman writing course on math, writing, and the world in the fall of 2013.
The boy who loved math: the improbable life of Paul Erdos is an illustrated children’s book.
John Cook on statistical evidence vs. legal evidence.
From Nautilus magazine: how to insure against a rainy day and taming the unfriendly skies.
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Lemma 109.24.1. There exists a morphism $f : X \to Y$ of finite presentation between affine schemes and a locally closed subset $T$ of $X$ such that $f(T)$ is not a finite union of locally closed subsets of $Y$.
Proof. See discussion above. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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## Isothermal and reversible reaction with changing P and V
$\Delta U=q+w$
Posts: 15
Joined: Fri Sep 26, 2014 2:02 pm
### Isothermal and reversible reaction with changing P and V
The question states: If 2.00 mol of an ideal gas at 300 K and 3.00 atm expands isothermally and reversibly from 6.00 L to 18.00 L and has a final pressure of 1.20 atm, what is w, q, and delta U.
I understand that q is 0 since the reaction is isothermal, and that w=-n(5/2R)Tln(V2/V1) for the changing volume. Do I add another term to the work which is -n(3/2R)Tln(P2/P1) to give me the total work and delta U?
Chem_Mod
Posts: 17543
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 393 times
### Re: Isothermal and reversible reaction with changing P and V
The heat will not be 0. An isothermal process has no temperature change. Therefore, $\Delta U = 0 = q+w$. The energy and enthalpy are both directly proportional to the temperature, so they don't change in an isothermal process.
Then, q=-w, and w=-nRTln(Vf/Vi), so q=nRTln(Vf/Vi).
Conceptually, if a gas expands, it would lose energy since it did work. But since the temperature didn't change, this lost energy must have been made up by adding heat. So, work is negative, heat is positive, and they cancel each other exactly.
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# Any good maths jokes?
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• August 13th 2012, 06:37 AM
terone71
Any good maths jokes?
Need a laugh. Any good maths jokes?
• August 13th 2012, 06:38 AM
terone71
Re: Any good maths jokes?
There was an expert accountant who was well versed in game theory. He heard that his intelligent niece, who was five years old, always took a 50p piece, when a choice between a 50p piece and a pound coin was offered to her.
He went to see his niece and offered her just such a choice. She took the 50p and said
"Thank you Uncle".
The accountant tried to explain to his niece
"You must understand, a pound coin is twice as valuable as a 50p piece, so you should always choose the pound coin."
The niece replied
"Uncle, but then people will not offer me any money."
• August 13th 2012, 06:41 AM
emakarov
Re: Any good maths jokes?
Three logicians walk into a bar. The bartender asks, "Do y'all want a drink?" The first logician says, "I am not sure." The second one also says, "I am not sure." The third one says, "Yes."
• August 13th 2012, 06:44 AM
Prove It
Re: Any good maths jokes?
How are ugly chicks like prime numbers?
Nothing will ever go into them except one and themselves ;)
• August 13th 2012, 08:42 AM
ebaines
Re: Any good maths jokes?
Why are amoebas the most intelligent animal in the world?
Because they can multiply by dividing!
• August 13th 2012, 09:39 AM
Prove It
Re: Any good maths jokes?
Quote:
Originally Posted by ebaines
Why are amoebas the most intelligent animal in the world?
Because they can multiply by dividing!
I can do that too, I just divide by some number in between 0 and 1...
Does that make me an amoeba? ><
• August 13th 2012, 11:43 AM
Deveno
Re: Any good maths jokes?
polar bears are regular bears, after a coordinate transformation.
• August 13th 2012, 12:06 PM
Soroban
Re: Any good maths jokes?
. . $\begin{array}{c}\text{What is the difference between} \\ \text{a }psychotic\text{ and a }neurotic? \\ \\ \text{A psychotic thinks }2 + 2 \,=\,5. \\ \\ \text{A neurotic knows that }2 + 2 \,=\,4 \\ \text{but it worries him.} \end{array}$
. . $\begin{array}{c}\text{What is the difference between} \\ data\text{ and }in\!f\!ormation? \\ \\ \text{382436: data} \\ \\ \text{38-24-36: information} \end{array}$
• August 13th 2012, 12:28 PM
emakarov
Re: Any good maths jokes?
Quote:
Originally Posted by Soroban
$\begin{array}{c}\text{What is the difference between} \\ data\text{ and }in\!f\!ormation? \\ \\ \text{382436: data} \\ \\ \text{38-24-36: information} \end{array}$
This reminds me a quote from the article "On the Essence of Mathematical Proofs" from the book "A Stress Analysis of a Strapless Evening Gown" (1963) (double translation to and from Russian).
Quote:
Relying on mathematical proofs, scientists managed to unite previously disconnected areas, thermodynamics and communication technology, into a new discipline: information theory. "Information," defined scientifically, is proportional to surprise: the more surprising the message, the more information it contains. If, having lifted a receiver, a man hears "Hello," this will not surprise him very much; there will be considerably more information if instead of "Hello" he gets an electric shock.
• August 14th 2012, 09:30 AM
emakarov
Re: Any good maths jokes?
He thinks that he is really smooth, but he is only C¹.
• August 15th 2012, 05:56 PM
Deveno
Re: Any good maths jokes?
Rastamath: I and I are orthogonal.
• August 16th 2012, 02:09 AM
emakarov
Re: Any good maths jokes?
Quote:
Originally Posted by Deveno
Rastamath: I and I are orthogonal.
Umm, can you explain that?
• October 29th 2012, 09:26 PM
beezmap
Re: Any good maths jokes?
Good jokes.....
• November 4th 2012, 09:21 PM
albertjhon
Re: Any good maths jokes?
Nice math joke , vary funny joke i can't stop laughing..
• November 5th 2012, 03:16 AM
BobP
Re: Any good maths jokes?
What goes oink $\frac{22}{7},$ oink $\frac{22}{7},$ oink $\frac{22}{7},$ ?
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# Show that every even integer greater than 2 can be written as a sum of two primes up to n less than or equal to 30 [closed]
Suppose $n$ is an even integer less than or equal to $30$.
$n= p_1 +p_2$
^^Is that legal? and if so where do I proceed from there.
P.S I am new to this forum and I am taking a number theory class. I switched from math major to math minor because I cannot write a proof to save my life.
Any help/ tips/ would be appreciated.
## closed as unclear what you're asking by Peter Franek, Shailesh, Parcly Taxel, Leucippus, user296602 Sep 10 '16 at 4:55
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.
• First you need to clearly state what you are trying to prove. I believe it is "Show that all even numbers greater than $2$ and less than $30$ can be expressed as the sum of two primes." Your title is not clear at all. – Ross Millikan Sep 9 '16 at 15:44
• Proof by exhaustion – jnyan Sep 9 '16 at 16:05
There are only thirteen even numbers between $2$ and $30$, so the easiest way to prove this is to show such a sum for each one. $$4=2+2\\6=3+3$$ Continue
In general you have Golbach Conjecture but for $n$ even less than or equal to $30$ you have to verify it . An example is $$4=2+2\\6=3+3\\8=5+3\\10=7+3\\12=7+5\\14=7+7\\16=13+3\\18=11+7\\20=13+7\\22=17+5\\24=19+5\\26=13+13\\28=23+5\\30=19+11$$ Notice that the examples are not unique, for instance $14=11+3=7+7$.
• I think at a table with the primes $2,3,5,....,23$ ($29$ is excluded obviously) and take all the possible sums of two of them giving even numbers. I feel that if you get a strictly formal proof, then you have in hand Golbach conjecture in general. – Piquito Sep 10 '16 at 18:14
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# Somebody please explin me this or at least give me the correct answer
somebody please explin me this or at least give me the correct answer. (complex numbers)
What the fuck does sous la forme mean
Just type that shit in wolframalpha if you want the correct answer. Looks like there would be a "trick" to solve it easily but can't come up with one, though. If not, then it's just a retarded calculation.
From what I remember of the two short years that I took French in high school it means "in the form"
It has 14 solutions and doesn't state in which branch you need to provide the answer. What a piece of literal shit.
Convert it into polar form and then just do r^14 and 14*arg. Then convert it back to rectangular form.
14 solutions
its to the power of 14 not root 14
Shit you're right, I'm a bit drunk...
Anyway OP, just use Yooler, then De Moivre's formula.
yooler
Its actually Oiler
You retard
Get necked
convert to polar form
apply de moivre's theorem
kys
apply de moivre's theorem
Nonsense! De Moivre was a crank who believed in "transcendental functions". Sine and cosine do not exist. THEY ARE SPOOKS.
You can clearly work this problem out in $\mathbb{Q} (i, \sqrt{3})$ and apply the beautiful and elegant binomial theorem! It would be nonsense to use "real numbers" to solve this problem when you can simply use real mathematics that actually exists in the real world.
call me when you get this problem done next year
(3(-1/2+i√3/2))^14
(3(cos(150°)+i*sin(150°))^14
(3∠150°)^14
3^14∠150°*14
3^14∠300°
3^14(1/2-i√3/2)
QED
$\frac{3}{2} ( \sqrt{3} i -1 ) = 3 ( \frac{\sqrt{3}}{2} +i (-\frac{1}{2}) ) =3( \cos{\frac{-\pi}{6}} + \sin{\frac{-\pi}{6}} ) = 3 e^{-i \frac{\pi}{6}} \implies (\frac{3}{2} ( \sqrt{3} i -1 ) )^{14} = 3^{14} e^{-i \frac{14\pi}{6}} =3^{14} e^{-i \frac{\pi}{3}} = 3^{14} (\frac{1}{2} - i \frac{\sqrt{3}}{2})$
Oh, I'll call you. I will call you to tell you that I'm proud for actually finding the solutions rigorously.
$\pi$
I am sorry, but what is this weird symbol I see here. Is this a number? Can you please write it down so I may see its digits?
π
it's an alternative way to write 3
Oh, then I must have misjudged you. It is a beautiful symbol for 3 indeed. Very classic. It can be drawn with 3 strokes precisely. Clever.
That said, we are not done examining your "proof".
sin
cos
I am sorry but what are these? Can you please write down the expression that gives the value for these functions? I have never seen these. Are you sure they exist?
I can't LaTeX to work on Veeky Forums today, but here you go.
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# Three-body problem in 3D space: ground state, (quasi)-exact-solvability
@article{Turbiner2016ThreebodyPI,
title={Three-body problem in 3D space: ground state, (quasi)-exact-solvability},
author={Alexander V. Turbiner and Willard Miller and Adrian M. Escobar-Ruiz},
journal={Journal of Physics A: Mathematical and Theoretical},
year={2016},
volume={50}
}
• Published 24 November 2016
• Physics
• Journal of Physics A: Mathematical and Theoretical
We study aspects of the quantum and classical dynamics of a 3-body system in 3D space with interaction depending only on mutual distances. The study is restricted to solutions in the space of relative motion which are functions of mutual distances only. It is shown that the ground state (and some other states) in the quantum case and the planar trajectories in the classical case are of this type. The quantum (and classical) system for which these states are eigenstates is found and its…
10 Citations
• Mathematics
• 2017
We employ generalized Euler coordinates for the n body system in dimensional space, which consists of the centre-of-mass vector, relative (mutual) mass-independent distances rij and angles as
• Physics, Mathematics
Journal of Physics A: Mathematical and Theoretical
• 2020
In this work we study 2- and 3-body oscillators with quadratic and sextic pairwise potentials which depend on relative distances, , between particles. The two-body harmonic oscillator is
• Physics
Physical Review A
• 2020
A simple uniform approximation for the nodeless wavefunction is constructed for a {\it neutral} system of two Coulomb charges of different masses $(-q,m_1)$ and $(q,m_2)$ at rest in a constant
• Mathematics, Physics
Advances in Methods and Applications of Quantum Systems in Chemistry, Physics, and Biology
• 2021
Some aspects of quasi-exact and semi-exact solubility are discussed. In particular, quasi-exactly solvable potentials are obtained as solutions of inverse problems with preassumed wave functions.
• Physics
Journal of Mathematical Physics
• 2019
Due to its great importance for applications, we generalize and extend the approach of our previous papers to study aspects of the quantum and classical dynamics of a 4-body system with equal masses
• Mathematics
SIAM J. Appl. Algebra Geom.
• 2019
The factorization of the $n$ body determinant is shown to be a special case of an intriguing general result proving the factorsization of determinants of a certain form.
• Physics
• 2017
As a straightforward generalization and extension of our previous paper, J. Phys. A50 (2017) 215201 we study aspects of the quantum and classical dynamics of a $3$-body system with equal masses, each
## References
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The independent eigenstates of the total orbital angular momentum operators for a three-body system in an arbitrary D-dimensional space are presented by the method of group theory. The Schrodinger
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The potential of the A2 quantum elliptic model (three-body Calogero–Moser elliptic model) is defined by the pairwise three-body interaction through the Weierstrass ℘-function and has a single
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The quantum H3 integrable system is a three-dimensional system with rational potential related to the noncrystallographic root system H3. It is shown that the gauge-rotated H3 Hamiltonian as well as
• Physics
• 1985
A class of three‐body problems characterized by masses m, m, M and ‘‘helium‐like’’ Hamiltonian H=−(1/2m)(∇21+∇22) −(1/2M)∇23 +V3(r12)+V2(r13)+V1(r23), where V1, V2 are attractive but V3 is repulsive,
The quantum‐mechanical problems of N 1‐dimensional equal particles of mass m interacting pairwise via quadratic (harmonical'') and/or inversely quadratic (centrifugal'') potentials is solved. In
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Translationally invariant symmetric polynomials as coordinates for N-body problems with identical particles are proposed. It is shown that in those coordinates the Calogero and Sutherland N-body
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The Journal of chemical physics
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In two- and three-dimensional systems, the non-analytical nature of the wave function is characterized by the appearance of logarithmic terms, reminiscent of those that appear as both electrons approach the nucleus of the helium atom.
A general classification of linear differential and finite-difference operators possessing a finite-dimensional invariant subspace with a polynomial basis (the generalized Bochner problem) is given.
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# Saturation of Mordell-Weil groups of elliptic curves over number fields#
Points $$P_1$$, $$\dots$$, $$P_r$$ in $$E(K)$$, where $$E$$ is an elliptic curve over a number field $$K$$, are said to be $$p$$-saturated if no linear combination $$\sum n_iP_i$$ is divisible by $$p$$ in $$E(K)$$ except trivially when all $$n_i$$ are multiples of $$p$$. The points are said to be saturated if they are $$p$$-saturated at all primes; this is always true for all but finitely many primes since $$E(K)$$ is a finitely-generated Abelian group.
The process of $$p$$-saturating a given set of points is implemented here. The naive algorithm simply checks all $$(p^r-1)/(p-1)$$ projective combinations of the points, testing each to see if it can be divided by $$p$$. If this occurs then we replace one of the points and continue. The function p_saturation() does one step of this, while full_p_saturation() repeats until the points are $$p$$-saturated. A more sophisticated algorithm for $$p$$-saturation is implemented which is much more efficient for large $$p$$ and $$r$$, and involves computing the reduction of the points modulo auxiliary primes to obtain linear conditions modulo $$p$$ which must be satisfied by the coefficients $$a_i$$ of any nontrivial relation. When the points are already $$p$$-saturated this sieving technique can prove their saturation quickly.
The method saturation() of the class EllipticCurve_number_field applies full $$p$$-saturation at any given set of primes, or can compute a bound on the primes $$p$$ at which the given points may not be $$p$$-saturated. This involves computing a lower bound for the canonical height of points of infinite order, together with estimates from the geometry of numbers.
AUTHORS:
• John Cremona
class sage.schemes.elliptic_curves.saturation.EllipticCurveSaturator(E, verbose=False)#
Class for saturating points on an elliptic curve over a number field.
INPUT:
• E – an elliptic curve defined over a number field, or $$\QQ$$.
• verbose (boolean, default False) – verbosity flag.
Note
This function is not normally called directly by users, who may access the data via methods of the EllipticCurve classes.
Add reduction data at primes above q if not already there.
INPUT:
• q – a prime number not dividing the defining polynomial of self.__field.
OUTPUT:
Returns nothing, but updates self._reductions dictionary for key q to a dict whose keys are the roots of the defining polynomial mod q and values tuples (nq, Eq) where Eq is an elliptic curve over $$GF(q)$$ and nq its cardinality. If q divides the conductor norm or order discriminant nothing is added.
EXAMPLES:
Over $$\QQ$$:
sage: from sage.schemes.elliptic_curves.saturation import EllipticCurveSaturator
sage: E = EllipticCurve('11a1')
sage: saturator = EllipticCurveSaturator(E)
sage: saturator._reductions
{}
sage: saturator._reductions
{19: {0: (20,
Elliptic Curve defined by y^2 + y = x^3 + 18*x^2 + 9*x + 18 over Finite Field of size 19)}}
Over a number field:
sage: x = polygen(QQ); K.<a> = NumberField(x^2 + 2)
sage: E = EllipticCurve(K, [0,1,0,a,a])
sage: from sage.schemes.elliptic_curves.saturation import EllipticCurveSaturator
sage: saturator = EllipticCurveSaturator(E)
sage: for q in primes(20):
sage: saturator._reductions
{2: {},
3: {},
5: {},
7: {},
11: {3: (16,
Elliptic Curve defined by y^2 = x^3 + x^2 + 3*x + 3 over Finite Field of size 11),
8: (8,
Elliptic Curve defined by y^2 = x^3 + x^2 + 8*x + 8 over Finite Field of size 11)},
13: {},
17: {7: (20,
Elliptic Curve defined by y^2 = x^3 + x^2 + 7*x + 7 over Finite Field of size 17),
10: (18,
Elliptic Curve defined by y^2 = x^3 + x^2 + 10*x + 10 over Finite Field of size 17)},
19: {6: (16,
Elliptic Curve defined by y^2 = x^3 + x^2 + 6*x + 6 over Finite Field of size 19),
13: (12,
Elliptic Curve defined by y^2 = x^3 + x^2 + 13*x + 13 over Finite Field of size 19)}}
full_p_saturation(Plist, p)#
Full $$p$$-saturation of Plist.
INPUT:
• Plist (list) – a list of independent points on one elliptic curve.
• p (integer) – a prime number.
OUTPUT:
(newPlist, exponent) where newPlist has the same length as Plist and spans the $$p$$-saturation of the span of Plist, which contains that span with index p**exponent.
EXAMPLES:
sage: from sage.schemes.elliptic_curves.saturation import EllipticCurveSaturator
sage: E = EllipticCurve('389a')
sage: K.<i> = QuadraticField(-1)
sage: EK = E.change_ring(K)
sage: P = EK(1+i,-1-2*i)
sage: saturator = EllipticCurveSaturator(EK, verbose=True)
sage: saturator.full_p_saturation([8*P],2)
--starting full 2-saturation
Points were not 2-saturated, exponent was 3
([(i + 1 : -2*i - 1 : 1)], 3)
sage: Q = EK(0,0)
sage: R = EK(-1,1)
sage: saturator = EllipticCurveSaturator(EK, verbose=False)
sage: saturator.full_p_saturation([P,Q,R],3)
([(i + 1 : -2*i - 1 : 1), (0 : 0 : 1), (-1 : 1 : 1)], 0)
An example where the points are not 7-saturated and we gain index exponent 1. Running this example with verbose=True would show that it uses the code for when the reduction has $$p$$-rank 2 (which occurs for the reduction modulo $$(16-5i)$$), which uses the Weil pairing:
sage: saturator.full_p_saturation([P,Q+3*R,Q-4*R],7)
([(i + 1 : -2*i - 1 : 1),
(2869/676 : 154413/17576 : 1),
(-7095/502681 : -366258864/356400829 : 1)],
1)
p_saturation(Plist, p, sieve=True)#
Checks whether the list of points is $$p$$-saturated.
INPUT:
• Plist (list) – a list of independent points on one elliptic curve.
• p (integer) – a prime number.
• sieve (boolean) – if True, use a sieve (when there are at least 2 points); otherwise test all combinations.
Note
The sieve is much more efficient when the points are saturated and the number of points or the prime are large.
OUTPUT:
Either False if the points are $$p$$-saturated, or (i, newP) if they are not $$p$$-saturated, in which case after replacing the i’th point with newP, the subgroup generated contains that generated by Plist with index $$p$$.
EXAMPLES:
sage: from sage.schemes.elliptic_curves.saturation import EllipticCurveSaturator
sage: E = EllipticCurve('389a')
sage: K.<i> = QuadraticField(-1)
sage: EK = E.change_ring(K)
sage: P = EK(1+i,-1-2*i)
sage: saturator = EllipticCurveSaturator(EK)
sage: saturator.p_saturation([P],2)
False
sage: saturator.p_saturation([2*P],2)
(0, (i + 1 : -2*i - 1 : 1))
sage: Q = EK(0,0)
sage: R = EK(-1,1)
sage: saturator.p_saturation([P,Q,R],3)
False
Here we see an example where 19-saturation is proved, with the verbose flag set to True so that we can see what is going on:
sage: saturator = EllipticCurveSaturator(EK, verbose=True)
sage: saturator.p_saturation([P,Q,R],19)
Using sieve method to saturate...
E has 19-torsion over Finite Field of size 197, projecting points
--> [(15 : 168 : 1), (0 : 0 : 1), (196 : 1 : 1)]
--rank is now 1
E has 19-torsion over Finite Field of size 197, projecting points
--> [(184 : 27 : 1), (0 : 0 : 1), (196 : 1 : 1)]
--rank is now 2
E has 19-torsion over Finite Field of size 293, projecting points
--> [(139 : 16 : 1), (0 : 0 : 1), (292 : 1 : 1)]
--rank is now 3
Reached full rank: points were 19-saturated
False
An example where the points are not 11-saturated:
sage: saturator = EllipticCurveSaturator(EK, verbose=False)
sage: res = saturator.p_saturation([P+5*Q,P-6*Q,R],11); res
(0,
(-5783311/14600041*i + 1396143/14600041 : 37679338314/55786756661*i + 3813624227/55786756661 : 1))
That means that the 0’th point may be replaced by the displayed point to achieve an index gain of 11:
sage: saturator.p_saturation([res[1],P-6*Q,R],11)
False
sage.schemes.elliptic_curves.saturation.p_projections(Eq, Plist, p, debug=False)#
INPUT:
• $$Eq$$ – An elliptic curve over a finite field.
• $$Plist$$ – a list of points on $$Eq$$.
• $$p$$ – a prime number.
OUTPUT:
A list of $$r\le2$$ vectors in $$\GF{p^n}$$, the images of the points in $$G \otimes \GF{p}$$, where $$r$$ is the number of vectors is the $$p$$-rank of $$Eq$$.
ALGORITHM:
First project onto the $$p$$-primary part of $$Eq$$. If that has $$p$$-rank 1 (i.e. is cyclic), use discrete logs there to define a map to $$\GF{p}$$, otherwise use the Weil pairing to define two independent maps to $$\GF{p}$$.
EXAMPLES:
This curve has three independent rational points:
sage: E = EllipticCurve([0,0,1,-7,6])
We reduce modulo $$409$$ where its order is $$3^2\cdot7^2$$; the $$3$$-primary part is non-cyclic while the $$7$$-primary part is cyclic of order $$49$$:
sage: F = GF(409)
sage: EF = E.change_ring(F)
sage: G = EF.abelian_group()
sage: G
Additive abelian group isomorphic to Z/147 + Z/3 embedded in Abelian group of points on Elliptic Curve defined by y^2 + y = x^3 + 402*x + 6 over Finite Field of size 409
sage: G.order().factor()
3^2 * 7^2
We construct three points and project them to the $$p$$-primary parts for $$p=2,3,5,7$$, yielding 0,2,0,1 vectors of length 3 modulo $$p$$ respectively. The exact vectors output depend on the computed generators of $$G$$:
sage: Plist = [EF([-2,3]), EF([0,2]), EF([1,0])]
sage: from sage.schemes.elliptic_curves.saturation import p_projections
sage: [(p,p_projections(EF,Plist,p)) for p in primes(11)] # random
[(2, []), (3, [(0, 2, 2), (2, 2, 1)]), (5, []), (7, [(5, 1, 1)])]
sage: [(p,len(p_projections(EF,Plist,p))) for p in primes(11)]
[(2, 0), (3, 2), (5, 0), (7, 1)]
sage.schemes.elliptic_curves.saturation.reduce_mod_q(x, amodq)#
The reduction of x modulo the prime ideal defined by amodq.
INPUT:
• x – an element of a number field $$K$$.
• amodq – an element of $$GF(q)$$ which is a root mod $$q$$ of the defining polynomial of $$K$$. This defines a degree 1 prime ideal $$Q=(q,\alpha-a)$$ of $$K=\QQ(\alpha)$$, where $$a \mod q =$$.
OUTPUT:
The image of x in the residue field of $$K$$ at the prime $$Q$$.
EXAMPLES:
sage: from sage.schemes.elliptic_curves.saturation import reduce_mod_q
sage: x = polygen(QQ)
sage: pol = x^3 -x^2 -3*x + 1
sage: K.<a> = NumberField(pol)
sage: [(q,[(amodq,reduce_mod_q(1-a+a^4,amodq))
....: for amodq in sorted(pol.roots(GF(q), multiplicities=False))])
....: for q in primes(50,70)]
[(53, []),
(59, [(36, 28)]),
(61, [(40, 35)]),
(67, [(10, 8), (62, 28), (63, 60)])]
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# Quadratic Factorisation — A General Method that Conquers Each and Every Quadratic Trinomial
When it comes to quadratic polynomials — which seem to come in all different forms and sizes — most of us have spent at least a semester just learning about how to maneuver around them.
Indeed, when we look through a standard pre-college math curriculum pertaining to quadratic expressions, we would stumble upon topics such as Square of a Sum, Square of a Difference, Completing the Square, Direct Factoring, Grouping, Difference of Squares. AC method (if you’re lucky) and Quadratic Formula — among others.
In what follows though, we will — in addition to putting some of these techniques into good use — develop a general technique for factorising any kind of quadratic polynomials. And yes — it would also include those with complex numbers as coefficients. 🙂
## Review — Basic Factoring Techniques
When we were in grade school, we were asked to internalize a bunch of rules about factoring polynomials, which include — among others:
1. Simple Factoring
1. $3x^2+6 = 3(x^2+2)$
2. $2x^2 + 4x = 2x (x+2)$
2. Perfect Square — Square of a Sum
1. $x^2+8x+16= x^2 + 2\cdot x \cdot 4 +4^2 = (x+4)^2$
2. $4x^2 + 20x + 25 = (2x)^2 + 2 \cdot 2x \cdot 5 + 5^2 = (2x + 5)^2$
3. Perfect Square — Square of a Difference
1. $x^2 – 10x + 25 = x^2 – 2 \cdot x \cdot 5 + 5^2 = (x -5)^2$
2. $9x^2 – 24x + 16 = (3x)^2 – 2 \cdot 3x \cdot 4 + 4^2 = (3x – 4)^2$
4. Difference of Squares
1. $x^2 – 5 = x^2 – (\sqrt{5})^2 = (x – \sqrt{5})(x + \sqrt{5})$
2. $4x^2 – 9 = (2x)^2 – 3^2 = (2x-3)(2x+3)$
5. Completing the Square
1. $x^2 + 6x + 7$ $= \underbrace{x^2 + 2 \cdot x \cdot 3 + 3^2} – 3^2 +7$ $= (x+3)^2 -2$
2. \begin{align*} 3x^2 – 6x – 9 = 3(x^2 – 2x – 3) & = 3(\underbrace{x^2 – 2x + 1} – 1 -3) \\ & = 3[(x-1)^2 – 4] \\ & = 3(x-1)^2 – 12 \end{align*}
6. Grouping
1. $x^2 + 3x + x + 3 = x (x+3) + (x+3) = (x+1)(x+3)$
2. $3x^2 + 9x + 2x + 6 = 3x (x+3) + 2 (x+3) = (3x+2)(x+3)$
Of course, these are all very useful techniques, and are all worthy of mastering. The problem with them — if there is any — is that most of them are meant to be used for very special cases only. As a result, the need for better solutions sparks the search for other techniques that are more widely applicable.
## Primary Factoring Techniques
When it comes to factoring quadratic polynomials specifically, there are 2 primary techniques that stand out in terms of their general applicability and elegance. These are the Direct Factoring Method, and the AC Method.
Admittedly, while both of these methods are trial-and-error-based heuristics for finding factors with integer coefficients, Direct Factoring requires less steps to carry out (though is more insight-demanding). In contrast, the AC Method takes on a bit more steps, but has a simpler verification process.
### Direct Factoring
Just as the name suggests, Direct Factoring consists in finding out the linear factors (i.e., factors of the form $ax+b$) of a quadratic expression directly — mostly by inspection and trial-and-error.
For example, to factorise the trinomial $x^2+13x + 36$ directly, is to find out two linear factors that multiply up to itself:
\begin{equation*}(\Box x + a)(\Box x + b ) = x^2+13x + 36 \end{equation*}
When a trinomial has a leading coefficient of $1$ (i.e., a monic polynomial) — as is in this case — all the $\Box$s reduce to $1$, which simplifies the search a bit:
\begin{equation*}(x + a)(x + b) = x^2+13x + 36 \end{equation*}
To explore a bit, we can expand the left-hand side of the equation, yielding:
\begin{equation*}x^2 + (a+b)x + ab = x^2+13x + 36 \end{equation*}
which suggests that the two constants $a$ and $b$ needs to be such, that they add up to $13$ all of the while multiply up to $36$. After a little bit of experimentation, we see that $4$ and $9$ would do the trick:
\begin{equation*}(x + 4)(x + 9) = x^2 + (4+9)x + 4 \cdot 9 = x^2+13x + 36 \end{equation*}
So Direct Factoring seems to work particularly well with monic polynomials. However, in the general case where the polynomial is not monic, a successful execution of Direct Factoring requires a bit more thinking and ingenuity. Take the case of $3x^2+ 22x+24$ for instance:
\begin{equation*}(\Box x + a)(\Box x + b ) = 3x^2+22x + 24 \end{equation*}
For one, we know that the $\Box$s must be such that they multiply up to $3$. In this case, $3$ and $1$ pop up as obvious factors:
\begin{equation*}(3x + a)(x + b ) = 3x^2+22x + 24 \end{equation*}
The next step then becomes finding out which are the constants $a$ and $b$, such that the left-hand side expands to the right-hand side. By inspection, these constants must multiply up to $24$, leaving us with the following factor pairs: $(1,24)$, $(2,12)$, $(3,8)$, $(4,6)$, $(6,4)$, $(8,3)$, $(12,2)$ and $(24,1)$.
By pure observation again, if $b\ge8$, then there is no way we can get a middle coefficient of $22$ (Why?). But then, that still leaves us with several options. Out of curiosity, let’s try $a=6$ and $b=4$:
\begin{equation*}(3x + 6)(x + 4 ) = 3x^2 + (12+6)x + 24 \ { \color{red} \ne }\ 3x^2+22x + 24 \end{equation*}
OK. This suggests that the middle coefficient (i.e., $18$) is too small. By observation, it seems that boosting the value of $b$ increases the value of the middle coefficient as well. With that in mind, let’s give $a=4$ and $b=6$ a try:
\begin{equation*}(3x + 4)(x + 6 ) = 3x^2 + (18+4)x + 24 = 3x^2+22x + 24 \end{equation*}
Checked! So the numbers work out perfectly in this case.
All right. That pretty much illustrates how in general, factoring a quadratic trinomial directly usually entails making an educated guess first, and adjusting accordingly until the numbers match up. In the cases where the polynomial has well-behaved integer coefficients, this approach is actually not too bad (and could be the fastest even).
### AC Method
In some schools out there (e.g., those in British Colombia), a certain factoring technique called the AC Method is taught as the de-facto factoring method — sometimes even with great success.
What is this mysterious AC Method then? Well, it’s essentially a technique for factoring quadratic expressions that is based on splitting the $x$-term into two, resulting in a quadratic expression with four terms — the left two terms being proportional to the right two terms. This proportionality then allows for completing the factoring via Grouping — a basic factoring technique introduced a bit earlier.
More specifically, given a trinomial $ax^2+bx+c$, the AC Method begins by splitting the term $bx$ into two terms (denoted by $b_1x$ and $b_2x$), such that the coefficients $b_1$ and $b_2$ adds up to $b$, but also multiply up to $ac$ — Hence the AC Method.
Once the splitting is done, the resulting quadratic polynomial — $(ax^2 + b_1x) + (b_2x + c)$ — will be such that $ax^2 + b_1x$ is proportional to $b_2x+c$. As a result, Grouping then can be applied to factorise the polynomial with ease.
To illustrate, suppose that we were to factorise $3x^2+22x + 24$ (the same trinomial from the Direct Factoring section) using the AC Method, then we would have to:
1. Split the $x$-term: We need to split $22x$ into two terms such that their coefficients add up to $22$ and multiply up to $ac=3(24)=72$. That would suggest the following pairs of coefficients: $(1,72)$, $(2,36)$, $(3,24), \dots , (1,72).$ Obviously, pairs containing numbers such as $24$, $36$ and $72$ are too big to add up to $22$, but with just a bit of thinking, we can see that the pair $(4, 18)$ — or the pair $(18,4)$ — would work out perfectly. Either way, we are warranted to turn $3x^2+22x + 24$ into $3x^2+4x + 18x + 24$.
2. Grouping: $(3x^2+4x) + (18x + 24) = x(3x+4) + 6(3x+4) = (x+6)(3x+4)$
So whether it’s by Direct Factoring or AC Method, the result remains the same.
However, as you might notice here, the AC Method is kind of an interesting and elegant technique, in that while for monic trinomials, using direct factoring is faster and almost equivalent to using the AC Method, for non-monic trinomials, the AC Method in effect turns the hard step of figuring out 4 coefficients into an easier step of figuring out 2 coefficients, thereby reducing the complexity of the factoring process — perhaps at the expense of an additional Grouping step, which can be carried out easily with no guesswork involved.
Another example of factorisation via the AC Method. Note how $-x$ is splitted into $-6x$ and $5x$.
## Drawbacks of Primary Factoring Techniques
So… with Direct Factoring and AC Method now in our bag of tricks, we’ve got some pretty powerful tools for factoring quadratic trinomials, don’t we? Well, there is only one problem. Namely, these two techniques don’t work at all — for the vast majority of quadratic trinomials!
OK. How is that possible though? Well, here is why:
The fact that a trinomial has integer coefficients, doesn’t mean that the same should be true for its factorisation. Math Vault
And the more you think about it, the more you realise that both Direct Factoring and AC Method are techniques for identifying factors with integer/rational coefficients primarily. As a result, it’s hence no surprise that both methods can fail miserably — for a trinomial that has not been crafted to work out perfectly.
To illustrate, let’s take the trinomial $x^2-x-1$, which is well known to be factorisable with $\frac{1+\sqrt{5}}{2}$ (i.e., the Golden Ratio) as one of its roots. However, if we try to factor it using the primary techniques though, we’ll soon discover that both Direct Factoring and AC Method yields nothing in return.
Actually, this is just the tip of the iceberg. What if we have:
• A trinomial with fractions as coefficients, like $\frac{1}{2}x^2+4x + \frac{3}{5}$? While both Direct Factoring and AC Method work with rational coefficients in theory, we might have a hard time finding out the factors from what seems like an infinite amount of possibilities.
Of course, using some clever tricks, it’s always possible to turn a trinomial of rational coefficients into one with integer coefficients, but hey, the linear factors might still have fractions as coefficients.
• A trinomial with irrational coefficients, like $\pi x^2+ 5x + 3$? Again, while Direct Factoring and AC Method can work in theory, you will have a tough time finding the factors from — this time for real — an infinite amount of possibilities (save some special cases, of course!).
So, all this to convey the idea that primary factorisation techniques can fall short in terms of their scope and applicability, and because of that, the search for the ultimate factorisation technique continues…
## The General Method — An Introduction
Actually, with the basic and primary factorisation techniques introduced earlier, we already have all the tools needed to factorise any quadratic expressions. All that’s needed really is to put the ingredients together into a coherent algorithm — one that consistently takes care of every issue that got us stuck with the previous methods.
So what is this General Method? In a nutshell, it’s a quadratic-factorisation algorithm based on three key steps — with no guesswork or trial-and-error involved:
General Method — Three Key Steps
1. Standardisation: Factor out the leading coefficient, thereby reducing the original trinomial into a monic trinomial.
2. Completing the Square: Turning the monic trinomial into the so-called vertex form.
3. Factorisation: The vertex form can take on different shapes, but if it is a difference of square itself, then further factorisation is possible; if not, the factorisation is complete by default.
OK. Since an illustration is worth a thousand words, let’s kick off with a few examples before going into full-theory mode.
### Case 1: Trinomials With No Root
To factorise a trinomial such as $3x^2+4x+5$ using the ideas behind the General Method, we would kick start with standardisation first:
$$3x^2+4x+5 = 3\left[x^2 + \frac{4}{3}x + \frac{5}{3}\right]$$
Focusing on the monic trinomial, we then proceed to complete the square:
\begin{align*} 3\left[x^2 + \frac{4}{3}x + \frac{5}{3}\right] & = 3\left[\underbrace{x^2 + 2 \cdot x \cdot \frac{2}{3} + \left(\frac{2}{3}\right)^2} – \left(\frac{2}{3}\right)^2 + \frac{5}{3}\right] \\ & = 3\left[ \left(x+\frac{2}{3} \right)^2 \underbrace{- \frac{4}{9} + \frac{15}{9}} \right] \\ & = 3\left[ \left(x-\frac{-2}{3} \right)^2 + \frac{11}{9}\right] \end{align*}
As you can see here, the monic polynomial is now in the form $(x – h)^2 + k$. This is commonly referred to as the vertex form, because we can then deduce that the vertex — or the minimum of the monic polynomial — occurs at $(h,k)$.
With that in mind, we can see that in this case, our monic polynomial has $k=\frac{11}{9}$. This means that if we were to graph this polynomial, then it would start at a height greater than $0$ at the vertex, and increase beyond bound as it moves away from the vertex.
Sure. This can only mean one thing — that the monic polynomial (and by extension, the original trinomial) could not possibly have a root. Here is an accompanying graph of the original trinomial for additional assurance:
The graph of the original trinomial $3x^2 + 4x + 5$. Note that the graph never touches the x-axis – hence no root.
And since the original trinomial has no root, this also means that there is no way that we can factor it either. Why? Because if we could, then the trinomial must have at least a linear factor (i.e., $ax+b$, $a\ne 0$), which means that it actually has a root — namely, $\frac{-b}{a}$!
And with that settled, we can safely dub $3x^2 + 4x + 5$ as an irreducible polynomial (i.e., a “prime” polynomial). This is actually in a sense good to know, because we can always leave the trimonial as it is — with its monic polynomial in the vertex form (i.e., $3\left[(x-\frac{-2}{3})^2 + \frac{11}{9}\right]$).
### Case 2: Trinomials With Exactly 1 Root
OK. Here’s another one: what about factoring trinomials such as $-2x^2-12x-18$? Well, using the ideas behind the General Method, the following steps are in order:
1. Standardise: $-2x^2-12x-18 = -2 [x^2 + 6x + 9]$
2. Completing the Square: $-2 [x^2 + 6x + 9] = -2 [ x^2 + 2 \cdot x \cdot 3 + 3^2]=$ $-2 (x+3)^2$ $= -2 [x – (-3)]^2$
3. Factorisation: The monic polynomial is in vertex form with $k=0$. Hence the factorisation is complete and no further action is required.
All good. Looking at the vertex form, one can also deduce that $-2x^2-12x-18$ has only a root at $x=-3$, and decreases beyond bound as it moves away from the vertex.
Wow! That was fast isn’t it? Here’s a caveat: cases like these happen quite a bit in textbooks, but very rarely in real-life quadratic models.
### Case 3: Trinomials With Exactly 2 Roots
OK. Let’s choose a trinomial we haven’t seen before, say $5x^2 – 16 x + 4$. What could happen if we apply the ideas behind the General Method on this one?
Let’s see… First, standardisng the trinomial, we get that $5x^2 – 16 x + 4$ = $5 \left[ x^2 – \frac{16}{5} x + \frac{4}{5}\right]$.
Second, completing the square on the monic polynomial yields:
\begin{align*}5 \left[ x^2 – \frac{16}{5} x + \frac{4}{5}\right] & = 5 \left[ \underbrace{x^2 – 2 \cdot x \cdot \frac{8}{5} + \left(\frac{8}{5}\right)^2} – \left(\frac{8}{5}\right)^2 + \frac{4}{5}\right] \\ & = 5 \left[ \left(x-\frac{8}{5} \right)^2 \underbrace{- \frac{64}{25} + \frac{20}{25}} \right] \\ & = 5 \left[ \left(x-\frac{8}{5} \right)^2 – \frac{44}{25} \right] \end{align*}
As with usual, the monic polynomial is now in vertex form. However, unlike the other cases, we now have $k=-\frac{44}{25}$ — a negative number.
Actually, this can only be good news, because it means we can apply the difference of square formula to the vertex form!
\begin{align*} 5 \left[ \left(x-\frac{8}{5} \right)^2 – \frac{44}{25} \right] & = 5 \left[ \left(x-\frac{8}{5} \right)^2 – \left(\frac{\sqrt{44}}{5}\right)^2 \right] \\ & = 5 \left[ \left(x-\frac{8}{5} + \frac{\sqrt{44}}{5} \right) \left(x- \frac{8}{5} – \frac{\sqrt{44}}{5} \right) \right] \\ & = 5 \left[ \left(x- \left(\frac{8}{5} – \frac{\sqrt{44}}{5}\right) \right) \left(x- \left(\frac{8}{5} + \frac{\sqrt{44}}{5} \right) \right) \right]\end{align*}
And the factorisation is now complete, revealing the roots of the trinomial $\frac{8}{5}\pm\frac{\sqrt{44}}{5}$. Neat!
Graph of the function $f(x)=5x^2-16x+4$ — with the two roots displayed.
## The General Method — Theory
Given any trinomial of the form $ax^2+bx+c$ with real numbers as coefficients, the General Method provides a surefire way of completing its factorisation — whenever applicable. To see why, let’s first start by standardising the trinomial:
$$ax^2+bx+c = a \left[x^2+\frac{b}{a}x + \frac{c}{a} \right]$$
Next, completing the square on the monic polynomial yields:
\begin{align*} a \left[x^2+\frac{b}{a}x + \frac{c}{a} \right] & = a \left[\underbrace{x^2+ 2 \cdot x \cdot \frac{b}{2a} + \left(\frac{b}{2a}\right)^2} – \left(\frac{b}{2a}\right)^2 + \frac{c}{a} \right] \\ & = a \left[ \left(x+ \frac{b}{2a}\right)^2 \underbrace{- \frac{b^2}{4a^2} + \frac{4ac}{4a^2}} \right] \\ & = a \left[ \left(x – \frac{-b}{2a}\right)^2 – \frac{b^2 – 4ac}{4a^2} \right] \end{align*}
See that $b^2 – 4ac$ thingy? It’s also called the discriminant by the way. For simplicity, we will denote it by $\Delta$.
Furthermore, notice that in the process of completing the square, we also just proved for any quadratic polynomial, the vertex occurs when $x = \frac{-b}{2a}$. In other words, we now know — without having to complete the square again — that the vertex of any quadratic function $f(x)=ax^2+bx+c$ occurs at the point $\left(\frac{-b}{2a}, f(\frac{-b}{2a})\right)$.
And since this $\frac{-b}{2a}$ pops up all the time, we will simply denote it by $h$, so that now we have two shorthands instead of one: $\Delta$ for $b^2 – 4ac$, and $h$ for $\frac{-b}{2a}$.
Under this terminology, the above equation can be simplified as follows:
$$ax^2+bx+c = a \left[ (x – h)^2 – \frac{\Delta}{4a^2} \right]$$
From this, we see that the monic polynomial is already in vertex form, with $k = – \frac{\Delta}{4a^2}$. Depending on the value of $\Delta$, three possible cases can occur:
Case 1 — $\Delta < 0$: In this case, $k$ — the constant in the vertex form — exceeds $0$. As a result, the polynomial has no root, and hence the factorisation is complete as it is:
$$a \left[ (x – h)^2 – \frac{\Delta}{4a^2} \right]$$
Case 2 — $\Delta = 0$: In this (very lucky) case, $k$ becomes $0$ as well. As a result, the factorisation is also complete and can be simplified as $a (x – h)^2$, with $h$ being the only root.
Case 3 — $\Delta > 0$: In this last case, with $k$ being a negative number, the vertex form can be further factorised using the difference of square formula:
\begin{align*} a \left[ (x – h)^2 – \frac{\Delta}{4a^2} \right] & = a \left[ (x – h)^2 – \left(\sqrt{\frac{\Delta}{4a^2}}\right)^2 \right] & \\ & = a \left[ (x – h)^2 – \left(\frac{\sqrt{\Delta}}{\pm 2a}\right)^2 \right] & \\ & = a \left[ \left(x – h + \frac{\sqrt{\Delta}}{2a} \right) \left(x – h – \frac{\sqrt{\Delta}}{2a} \right) \right] & \\ & = a \left[ \left(x – (h – \frac{\sqrt{\Delta}}{2a}) \right) \left(x – (h + \frac{\sqrt{\Delta}}{2a}) \right) \right] & \end{align*}
Once here, the factorisation is complete, yielding $h \pm \frac{\sqrt{\Delta}}{2a}$ as the two roots of the trinomial. Looking familar? Of course! We just proved the (in)famous Quadratic Formula along the way!
$$ax^2+bx+c=0 \iff x = \frac{-b}{2a} \pm \frac{\sqrt{\Delta}}{2a} = \frac{-b \pm \sqrt{\Delta}}{2a}$$
OK. Now that we have done all the hard work, we can leverage it into constructing a general algorithm for factoring quadratic expressions, which involves — oddly enough — no factoring at all (at least not in the ordinary sense)!
General Method — The Algorithm
Given a quadratic trinomial with real coefficients (i.e., $ax^2+bx+c$, $a$, $b$, $c \in \mathbb{R}$), the factorisability of the trinomial can be determined using the value of its discriminant $\Delta$, with the factorisation expressible in terms of $\Delta$ and $h$ as follows:
• $\Delta < 0$: The polynomial is irreducible, with the factorisation being $a \left[ (x – h)^2 – \frac{\Delta}{4a^2}\right]$.
• $\Delta = 0$: The polynomial is a perfect square, with the factorisation being simply $a (x – h)^2$.
• $\Delta > 0$: The polynomial has two distinct roots $r_1$ and $r_2$ (shorthands for $h + \frac{\sqrt{\Delta}}{2a}$ and $h – \frac{\sqrt{\Delta}}{2a}$, respectively), with the factorisation being simply $a (x – r_1) (x – r_2)$.
## General Method vs. Quadratic Formula
So now you might be asking: “How is this different from the good old Quadratic Formula?” Well, in a nutshell, the General Method is an ultimate technique for factorising quadratic trinomials, while the Quadratic Formula is an ultimate technique for solving their roots.
As a rule of thumb, factorisation generally does much more than simply solving for the roots. And in this case, since the process of factorization also reveals the roots, the General Method can be thought of as a precursor of the Quadratic Formula itself.
• Vertex: An application of the General Method would reveal the coordinates of the vertex, which determines the position the graph of the trinomial stems from.
• Shape: A factorisation via the General Method reveals the role of the leading coefficient $a$ in determining the orientation (i.e., opens upwards/downwards) and the sloppiness of the graph (of the trinomial).
• Vertex Form: Regardless of whether a trinomial has root or not, the General Method either provides a factorisation for the trinomial, or converts its corresponding monic polynomial into vertex form. While seemingly extraneous, these different representations of trinomials can come in handy in a series of scenarios, where algebraic manipulations are crucial (e.g., partial fraction, trigonometric integrals).
## The General Method — Complex Coefficients (Optional)
Taking the ideas behind of the General Method one step further, if a trinomial $ax^2+bx+c$ is such that $a$, $b$ and $c$ are complex numbers (or considered to be to complex numbers), then the above argument with the three cases collapses into one, with the trinomial always factorisable as follows:
$$a \left[ \left(x – (h + \frac{\sqrt{\Delta}}{2a}) \right) \left(x – (h – \frac{\sqrt{\Delta}}{2a}) \right) \right]$$
where $\sqrt{\Delta}$ stands for the complex number on the upper quadrant (i.e., with angle $\theta \in [0. \pi)$) which squares up to $\Delta$.
In this case, we can see that the two roots of the trinomial are $h \pm \frac{\sqrt{\Delta}}{2a}$. so that if we denote these two roots by $r_1$ and $r_2$, the factorisation can be further simplified as follows:
$$a ( x – r_1) (x – r_2)$$
Not too shabby isn’t it? Incidentally, this also proves in the world of complex numbers, a quadratic polynomial (i.e., a polynomial of degree two) always has two (not necessarily distinct) roots. This is a special case of the so-called Fundamental Theorem of Algebra, which states that in the field of complex numbers, a polynomial of degree $n$ ($n \in \mathbb{N}$) always has $n$ (not necessarily distinct) roots.
This all seems a bit abstract of course, so let’s run through an example to see what how that result comes about. For simplicity, let’s say that we were to factor $3x^2+6x+6$ using the ideas behind the General Method, then we’ll have the following steps:
1. Standardise: $3x^2+6x+6=3\left[ x^2 +2x + 2\right]$
2. Complete the Square: $3\left[ x^2 +2x + 2\right]= 3\left[ (x^2 +2x + 1) +1 \right] =$ $3\left[ (x + 1)^2 +1 \right]$
And here comes the trick: since any non-zero complex number has two roots, we can always turn the vertex form into a difference of square:
\begin{align*} 3\left[ (x + 1)^2 +1 \right] & = 3\left[ (x + 1)^2 – (-1) \right] \\ & = 3\left[ (x + 1)^2 – (\sqrt{-1})^2 \right] \\ & = 3\left[ \left(x + 1 \right)^2 – i^2 \right]\end{align*}
Once there, factorise the difference of square and we’re good to go:
\begin{align*} 3\left[ \left(x + 1 \right)^2 – i^2 \right] & = 3\left[ (x+1+ i)(x+1 – i) \right] \\ & = 3\left[ (x – [-1 – i])(x- [-1 + i]) \right] \end{align*}
This means that while $3x^2+6x+6$ (or equivalently, $3\left[ (x + 1)^2 +1\right]$) is irreducible in the world of real numbers, in the world of complex numbers, it actually factorises to $3\left[ (x – [-1 – i])(x- [-1 + i]) \right]$.
Alternatively, if you prefer factorising $3x^2+6x+6$ using the General Method itself, then here is what you would do:
1. We know that for complex numbers, $ax^2+bx+c$ always factorises to $a (x -r_1) (x-r_2)$, with $r_1$ and $r_2$ being $h\pm \frac{\sqrt{\Delta}}{2a}$. Once we gather the ingredients $h$, $\Delta$ and $\frac{\sqrt{\Delta}}{2a}$, we should be all set.
2. Since $h \stackrel{def}{=} \frac{-b}{2a}$, we have that $h = \frac{-6}{6} = -1$ in this case.
3. Since $\Delta \stackrel{def}{=} b^2 – 4ac$, we get that $\Delta = 6^2 – 4\cdot 3 \cdot 6$ $= -36$.
4. Next, $\frac{\sqrt{\Delta}}{2a} = \frac{\sqrt{-36}}{6} = \frac{6\sqrt{-1}}{6} = i$.
Lastly, putting everything together, we get that the roots of the trinomial are $-1 \pm i$, with the factorisation $3 (x – [-1 + i])(x-[-1 – i])$ — same as above.
See the beauty of this? There is no basic/primary factoring techniques involved — just pure computations. 😉 Granted, we did picked an example that’s relatively gentle to work with, but it’s still true that for quadratic trinomials with complex coefficients, whenever we are dealing with $4x^2 + 8x + 9$ or $(2+i)x^2 + \pi x + e+4i$, the General Method takes care of them all!
## Afterwords
OK. Now that we have developed the ultimate technique for quadratic factorisation, does it mean that it is over? Well, the search might be, but the need for practices continues on!
To be sure, the fact that the General Method caters to all kinds of quadratic trinomials doesn’t mean that it should be applied all the time. Why? Because unlike the basic and primary factoring techniques, the General Method — while universally applicable — can be quite computationally intensive at times (sometimes even an overkill).
Truth to told, mastering quadratic factorisation is all about learning the mechanics of different kinds of techniques, while adjusting one’s approach accordingly so that a problem can be solved with minimal effort. In this sense, one can even say that quadratic factorisation is as much an art as it is a science.
And before we go, here’s an interactive table summarizing what we have learnt so far:
$3x^2+6 = 3(x^2+2)$
$x^2+8x+16= x^2 + 2\cdot x \cdot 4 +4^2 = (x+4)^2$
$x^2 – 10x + 25 = x^2 – 2 \cdot x \cdot 5 + 5^2 = (x -5)^2$
$x^2 – 5 = x^2 – (\sqrt{5})^2 = (x – \sqrt{5})(x + \sqrt{5})$
$x^2 + 6x + 7 = \left(x^2 + 2 \cdot x \cdot 3 + 3^2\right) – 3^2 +7 = (x+3)^2 -2$
$x^2 + 3x + x + 3 = x (x+3) + (x+3) = (x+1)(x+3)$
A trial-and-error-based factoring method consisting in finding the coefficients of the two linear factors, so that the linear factors multiply to the quadratic trinomial in question.
For example, in order for $(\Box x + \Box )$ and $(\Box x + \Box )$ to multiply up to $3x^2+22x + 24$, the left $\Box$s should be $3$ and $1$, while the right $\Box$s should be $6$ and $4$.
A trial-and-error-based method for factoring quadratic trinomial, which consists in breaking up the middle term of the trinomial into two, such that the coefficients of these two terms multiply up to $ac$. The resulting polynomial is then factorised via a standard application of Grouping.
For example, $3x^2+22x + 24$ $= 3x^2 + 4x + 18x + 24 = x(3x +$$4) + 6(3x+4) = (x+6) (3x+4).$
For a trinomial $ax^2+bx+c$ with real coefficients, the General Method allows for it to be factorised as follows:
• $\Delta < 0$: The trinomial is irreducible, with the factorisation being $a \left[ (x – h)^2 – \frac{\Delta}{4a^2}\right]$.
• $\Delta = 0$: The trinomial is a perfect square, with the factorisation being simply $a (x – h)^2$.
• $\Delta > 0$: The trinomial has two distinct roots $r_1$ and $r_2$, with the factorisation being $a (x – r_1) (x – r_2)$.
For any trinomial with complex coefficients, the General Method allows for it to be factorised as $a (x – r_1) (x – r_2)$.
All right. That’s about it! This page is now clocking at 4200+ words, making it potentially the longest and the most comprehensive article about quadratic factorisation on the Web. If you enjoy and appreciate the learning process, make sure hop into our Facebook and peek at what the kind of weird cognitive technologies we’re experimenting with in the Math Vault Lab!
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Anitej Banerjee May 19, 2016
You guys are simply amazing! I’ve tried deriving the quadratic roots formula time and again, but never thought of connecting the standard form to the vertex form and seeing what pops out. The General method has also helped improve my intuition for the quadratic trinomials, so thank you guys very much!
I’m going to remember and thank you guys every time I apply one of the many great techniques I’ve learnt here! 🙂
Math Vault May 19, 2016
That’s wonderful. Thank you so much!
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# Thread: LaTeX support?
1. Seriously, could we get LaTeX support up in here? It looks like mimeTeX should be real easy to set up.
2.
3. That would be awesome if I didn't have to type out a million parentheses and hope that what I wrote is readable, and make integrals that look like this:
5<sup>1/2</sup>∫<sub>sqrt (2x-5)</sub><sup>sqrt (9x+100)</sup> x<sup>5</sup> dx
Although I havn't done that type of thing yet.
4. I was just thinking about this the other day and that I needed to get it done. Keep hounding me (without being excessive) and I'll make it so.
IS
5. Thanks for taking note!
6. I'm going to do personal maths learning so if you guys would help me too, along with LaTeX that would be great!
Currently my square roots here are this: /--------- :|
7. If you've got math questions, we're happy to help in the math forum.
Here's your first lesson. You can use the character √ to represent a radical. If you don't want to look it up, you can use exponents. If you don't want to use exponents, use sqrt. Never use /-----.
√2 = 2^.5 = 2^(1/2) = 2<sup>.5</sup> = 2<sup>1/2</sup> = sqrt(2)
/-----2 = nonsense
The √ is a good character, as you can make it represent arbitrary roots with a little work:
<sup>3</sup>√8 = 2
Exponents are even better, because you can represent any power:
e<sup>iπ</sup> = -1
8. Thanks serpicojr . See you in the maths topic :-D.
9. Be careful! There's a really stupid discussion about dividing by 0 going on right now!
10. I've seen :-D. But thanks for the heads up.
11. Was going to try and get this add on done this weekend, got a bit tied up. Will shoot for the upcoming weekend.
12. I had latex support once but after a few hours it just got too hot and sweaty and i had to take it off.
Don't know what all the fuss is about. PVC is SO much better
13. Thanks (In)Sanity!
Selene, we mathematicians love latex because it keeps all our dirty little thoughts where they belong.
14. Originally Posted by serpicojr
Thanks (In)Sanity!
Selene, we mathematicians love latex because it keeps all our dirty little thoughts where they belong.
Oooh, and where precisely is that?
Will latex save me from a life of dirtiness and sin?
Last time i slipped into some latex i had a revelation and this happened:
Let $D$ be a subset of $\bf R$ and let
$f \colon D \to \mathbf{R}$ be a real-valued function on
$D$. The function $f$ is said to be \emph{continuous} on
$D$ if, for all $\epsilon > 0$ and for all $x \in D$,
there exists some $\delta > 0$ (which may depend on $x$)
such that if $y \in D$ satisfies
$|y - x| < \delta$
then
$|f(y) - f(x)| < \epsilon.$
And it just made me want to get dirtier!
What am i doing wrong?
15. The weekend is here...(In)Sanity please sort it out for us before Selene loses it completely! :P
16. Your doing nothing wrong. You might need a new fetish .
17. Originally Posted by bit4bit
The weekend is here...(In)Sanity please sort it out for us before Selene loses it completely! :P
I've already lost it! I've been looking for it all night.
"Perfect numbers like perfect men are very rare."
René Descartes
18. I have to go to a leather, lace and feathers party in two weeks, i wonder if latex will work as well? Selene, its a pity your not over here visiting family, we could go latex shopping together. :wink:
19. Originally Posted by Selene
Let $D$ be a subset of $\bf R$ and let
$f \colon D \to \mathbf{R}$ be a real-valued function on
$D$. The function $f$ is said to be \emph{continuous} on
$D$ if, for all $\epsilon > 0$ and for all $x \in D$,
there exists some $\delta > 0$ (which may depend on $x$)
such that if $y \in D$ satisfies
$|y - x| < \delta$
then
$|f(y) - f(x)| < \epsilon.$
And it just made me want to get dirtier!
That just made me hot.
20. Originally Posted by GrowlingDog
I have to go to a leather, lace and feathers party in two weeks, i wonder if latex will work as well? Selene, its a pity your not over here visiting family, we could go latex shopping together. :wink:
There are plenty of online shops!
Do you have a web-cam? 8)
21. Originally Posted by Selene
Originally Posted by GrowlingDog
I have to go to a leather, lace and feathers party in two weeks, i wonder if latex will work as well? Selene, its a pity your not over here visiting family, we could go latex shopping together. :wink:
There are plenty of online shops!
Do you have a web-cam? 8)
No, i dont even have skype worked out yet.
22. OK, well it appears my Sunday blew by at a phenomenal rate. I just looked at the clock and went, damn...weekend is over. That being said I'm going to try and get this added during the week. Sorry for the delay. I see however it's creating an interesting topic of conversation.
IS
23. How bout that latex? I just typed up two hours worth of math.
24. haha I know how ya feel, I hate typing and reading these tedious non-latex characters.
"What do we want??"
"LATEX!"
"When do we want it??"
"NOW!"
25. I'm actually thinking I need to upgrade the entire forum. It's getting a bit out of date in terms of features. I had in laws from Thursday to Monday morning, kind of put a damper on anything computer related.
26. Aww c'mon Insanity...don't let family get in the way of your computer time :P
What kinda features you got in mind?
27. if you want to use latex you could always go to the site : www.mathim.com
it's basically a latex based web chat, so you can go in there and start typing what you want to say using latex and even preview what you typed.
having done that you can then right click on the preview, save the picture and then go to a site like www.imageshack.us and go through all that nonsense to insert a link in between some img tags and then you're left with this,
probably seems like a bit of effort so it's more of for the sake of presentation i guess.
28. Latex is so nice! Lets get it! It will be cool!
Latex fun fact #1:
The equivilant of typing
x < sub > 1 < / sub > < sup > 2 < / sup > + y < sub > 1 < / sub > < sup > 2 < / sup > = 10 < sup > 2 < / sup >
in Latex is:
x_1^2 + y_1^2 = 10
WOW! That's so neat! And it looks pretty too!
29. \sum_{n=0}^\infty f(x_n)\Delta x=\int_a^bf(x)dx is how you code what wallaby posted.
30. now that i'm at uni i think i'll be using more and more of this latex thing, as my handwritting is rather messy.
anyone know where i can get my hands on something i can actually install on my computer?
31. Originally Posted by wallaby
now that i'm at uni i think i'll be using more and more of this latex thing, as my handwritting is rather messy.
anyone know where i can get my hands on something i can actually install on my computer?
http://www.xm1math.net/texmaker/doc.html
http://www.math.vanderbilt.edu/~sche...d/list_tex.htm
http://www.download.com/Highlight/30...ml?tag=lst-0-7
The first one looks possibly the best
But don't ask me i only use latex to cover my skin
:wink:
32. Originally Posted by wallaby
now that i'm at uni i think i'll be using more and more of this latex thing, as my handwritting is rather messy.
anyone know where i can get my hands on something i can actually install on my computer?
If you have Linux or a Mac, it will already have LaTeX I believe (at least mine did...).
Cheers
33. Macs don't came with any sort of TeX complier. In any case, many Mac users--for example, me--like TeXShop. It's pretty stripped down and assumes you're decently comfortable with the language.
34. unfortunately i'm pretty sheltered and have only ever used windows.
35. Originally Posted by wallaby
unfortunately i'm pretty sheltered and have only ever used windows.
I refer yet again to the link-
http://www.miktex.org/
Are you stupid or just plain ignorant?
36. Originally Posted by Selene
I refer yet again to the link-
http://www.miktex.org/
Are you stupid or just plain ignorant?
wat's a ingorant?
i'm sorry Selene for rudely forgeting to thank you for the useful links you provided... thanks.
37. Originally Posted by wallaby
Originally Posted by Selene
I refer yet again to the link-
http://www.miktex.org/
Are you stupid or just plain ignorant?
wat's a ingorant?
i'm sorry Selene for rudely forgeting to thank you for the useful links you provided... thanks.
Ok damn...i'm gonna have to let you off a botty spank now! I was getting my right hand nicely warmed up too
A mathematical friend of mine uses the miktex editor and says he gets on with it fine in windows
38. I recommend MiKTeX, too. However, it isn't editor--it's everything but. If you don't want to code things up in a plain old text editor, you'll need to get a TeX editor in addition to MiKTeX.
39. Originally Posted by Selene
Ok damn...i'm gonna have to let you off a botty spank now! I was getting my right hand nicely warmed up too.
well now i'm going to be a tease,
Originally Posted by serpicojr
I recommend MiKTeX, too. However, it isn't editor--it's everything but. If you don't want to code things up in a plain old text editor, you'll need to get a TeX editor in addition to MiKTeX.
well now with the power of MiKTeX and WinEdt i'm going to get cracking
40. Wallaby, If you have microsoft office, you should be able to insert and edit equations already with microsft equation. In Word go to insert>object>Microsoft equation. You can also use it in Wordpad too by the same route, which I prefer over Word for somethings. You might have to download it first. You can also download an expanded version called MathType. You will have to purchase this upgrade....but if you know where to look. To be honest Microsoft equation is good enough for most things anyway.
41. Wallaby: WinEDT is pretty good--that's what I used on my PC. The registration reminders are pretty annoying, although you can probably disable them pretty easily.
Bit4bit: Once you've used TeX, you'll never go back to the MS equation editor.
42. Originally Posted by bit4bit
Wallaby, If you have microsoft office, you should be able to insert and edit equations already with microsft equation. In Word go to insert>object>Microsoft equation. You can also use it in Wordpad too by the same route, which I prefer over Word for somethings. You might have to download it first. You can also download an expanded version called MathType. You will have to purchase this upgrade....but if you know where to look. To be honest Microsoft equation is good enough for most things anyway.
i probably would have to download it as i'm using Microsoft Office 97, couldn't be bothered paying for or stealing 2007.
Originally Posted by serpicojr
The registration reminders are pretty annoying, although you can probably disable them pretty easily.
i havn't had a single reminder yet, i guess i'll enjoy the lack of interuption while it lasts.
43. I think most versions of office have Microsoft equation built in already even the older ones...but I could be wrong.
Serpicojr, I've never used TeX before (Well, I've used LaTeX), but I'll give it a try. What is so special about TeX compared to MS equation? TeX is another 'language' right, as opposed to clicking on symbols like MS equation? It's free to download anyway so I'll check it out at some point.
44. When I say TeX, I'm also talking about LaTeX. You've used it, you say. I guess you're not impressed? Clicking on symbols may user friendly, but finding the buttons as opposed to typing a command is time consuming, and formatting things in MS Equation is a nightmare.
45. I've only ever used TeX/LaTeX online, whereas for all my work I've used MS equation editor. I'll admit, I'm not the most experienced person with LaTeX, since the main maths forum I come to is here, and we don't have it!
Probably if I used it more often, I would be able to bang it out second nature, but in the mean time I prefer MS equation. I prefer either over nothing though.
46. Originally Posted by bit4bit
I've only ever used TeX/LaTeX online, whereas for all my work I've used MS equation editor. I'll admit, I'm not the most experienced person with LaTeX, since the main maths forum I come to is here, and we don't have it!
Probably if I used it more often, I would be able to bang it out second nature, but in the mean time I prefer MS equation. I prefer either over nothing though.
I only ever use latex on my bottom because it clings well and makes it appear smaller.
and as for banging it out....well i prefer S&M equation, it's much more fun.
47. Originally Posted by Selene
I only ever use latex on my bottom because it clings well and makes it appear smaller.
Do you throw your latex out when it gets old and stale?
48. Originally Posted by Selene
Originally Posted by bit4bit
I've only ever used TeX/LaTeX online, whereas for all my work I've used MS equation editor. I'll admit, I'm not the most experienced person with LaTeX, since the main maths forum I come to is here, and we don't have it!
Probably if I used it more often, I would be able to bang it out second nature, but in the mean time I prefer MS equation. I prefer either over nothing though.
I only ever use latex on my bottom because it clings well and makes it appear smaller.
and as for banging it out....well i prefer S&M equation, it's much more fun.
Yeh I know all about you Northhampton girlies. :P
49. Originally Posted by serpicojr
Originally Posted by Selene
I only ever use latex on my bottom because it clings well and makes it appear smaller.
Do you throw your latex out when it gets old and stale?
Oh no the older it gets the better because it becomes more flexible and you can always give it a good buffing to make it look shiny and new again
50. Originally Posted by bit4bit
Originally Posted by Selene
Originally Posted by bit4bit
I've only ever used TeX/LaTeX online, whereas for all my work I've used MS equation editor. I'll admit, I'm not the most experienced person with LaTeX, since the main maths forum I come to is here, and we don't have it!
Probably if I used it more often, I would be able to bang it out second nature, but in the mean time I prefer MS equation. I prefer either over nothing though.
I only ever use latex on my bottom because it clings well and makes it appear smaller.
and as for banging it out....well i prefer S&M equation, it's much more fun.
Yeh I know all about you Northhampton girlies. :P
Oh do you now?
Pray, tell me more?
P.s Northampton has only one h, the only time it has two is when the Northampton girlies are out of breath.
51. Update on this topic (the original topic)
The forum will be upgrading to phpbb 3.xx soon. I'll release a beta test site to start and once that is all worked out will release the beta version to production. I hope it has support for latex or at least an easy way to add it.
52. Cool.
53. Sounds good
54. LaTeX?
55. Originally Posted by serpicojr
LaTeX?
I'm in the process now (as in after I type this) of working more on the beta site. I may just put it out as is and see if anyone likes it..etc. I have to be careful I don't cause conflicts with the old site and the beta one having information the old one can't read..etc. This site has been so hacked up from the stock phpbb.
56. Right, and now it's July. And.........?
I like this site, but when trying to do some simple maths, it takes half an hour of subbing, supping, hex-coding etc.
It really puts people off. PLEASE implement Latex here.
As you know, I(S), this is by no means the first time of asking (nor is it the first time of your promising).
Why the delay? I mean, WHY?
57. It's a long weekend, so I'll get this done. I've been holding off on it waiting to see if I should go with the newer phpbb. Turns out it doesn't have Latex support either. Hmm. In any event this will get done this weekend even if I have to pass out from lack of sleep
Originally Posted by Guitarist
Right, and now it's July. And.........?
I like this site, but when trying to do some simple maths, it takes half an hour of subbing, supping, hex-coding etc.
It really puts people off. PLEASE implement Latex here.
As you know, I(S), this is by no means the first time of asking (nor is it the first time of your promising).
Why the delay? I mean, WHY?
58. Thanks very much!
59. Originally Posted by serpicojr
Thanks very much!
This is 99% done, just fighting with ImageMagik and Ghostscript at the moment. I can get it to produce the latex GIF images, just having a hard time getting PHP to do it. Some kind of weird path issue. I'll get it figured out.
60. Happy now
61.
62. Originally Posted by (In)Sanity
Happy now
Extremely!
Thank you very much indeed!
63. I would just like to give a much belated, but enthusiastic
/ cheer!
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Found in: Page 1
### Fundamentals Of Physics
Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718
# Figure 7-41 shows a cord attached to a cart that can slide along a frictionless horizontal rail aligned along an x axis. The left end of the cord is pulled over a pulley, of negligible mass and friction and at cord height ${\mathbf{h}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{20}}{\mathbf{}}{\mathbf{m}}$, so the cart slides from ${{\mathbf{x}}}_{{\mathbf{1}}}{\mathbf{=}}{\mathbf{3}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{m}}{\mathbf{}}{\mathbf{to}}{\mathbf{}}{{\mathbf{x}}}_{{\mathbf{2}}}{\mathbf{=}}{\mathbf{1}}{\mathbf{.}}{\mathbf{00}}{\mathbf{}}{\mathbf{m}}$. During the move, the tension in the cord is a constant ${\mathbf{25}}{\mathbf{.}}{\mathbf{0}}{\mathbf{}}{\mathbf{N}}$. What is the change in the kinetic energy of the cart during the move?
The change in the kinetic energy of the cart is, 41.7 J
See the step by step solution
## Step 1: Given
1. The height if the pulley $\mathrm{h}=1.20\mathrm{m}$
2. The displacement of the cart from ${\mathrm{x}}_{1}=3.00\mathrm{m}\mathrm{to}{\mathrm{x}}_{2}=1.00\mathrm{m}$
3. The tension in the cord is $25.0\mathrm{N}$
## Step 2: Work-Energy Theorem
The problem deals with the concept of work done. It includes both forces exerted on the body and the total displacement of the body. The change in the kinetic energy of an object is the work done by the force acting on the object.
Formula:
${\mathbf{∆}}{\mathbf{KE}}{\mathbf{=}}{\mathbf{W}}\phantom{\rule{0ex}{0ex}}{\mathbf{W}}{\mathbf{=}}\stackrel{\mathbf{\to }}{\mathbf{F}}{\mathbf{.}}\stackrel{\mathbf{\to }}{\mathbf{x}}\phantom{\rule{0ex}{0ex}}$
## Step 3: Calculate the work done and hence find change in kinetic energy
The change in the kinetic energy of an object is the work done on the object by the force. Here, the cord is pulling the cart, so the tension in the cord does the work on the cart and causes it to displace from x1 to x2. During this displacement, the kinetic energy of the cart changes by an amount equal to the work done.
Hence,
$∆\mathrm{KE}=\mathrm{W}$
Now, the work done can be determined as
$\begin{array}{rcl}\mathrm{W}& =& \stackrel{\to }{\mathrm{F}}∆\mathrm{dx}\\ & =& \stackrel{\to }{\mathrm{T}}∆\mathrm{dx}\\ & =& \mathrm{Tdx}\mathrm{cos\theta }\end{array}$
Where, angle $\mathrm{\theta }$ is the angle made by the tension with the horizontal.
Using geometry of the figure, we can write
where
$\mathrm{cos\theta }=\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^{2}+{\mathrm{x}}^{2}}}\mathrm{where}\mathrm{x}={\mathrm{x}}_{1}-{\mathrm{x}}_{2}$
Now,
$\mathrm{W}=\int \mathrm{T}\mathrm{cos\theta }\mathrm{dx}\phantom{\rule{0ex}{0ex}}\mathrm{W}=\mathrm{T}{\int }_{{\mathrm{x}}_{1}}^{{\mathrm{x}}_{2}}\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^{2}+{\mathrm{x}}^{2}}}\right]d\mathrm{x}$
Here we can use substitution method to solve this integration.
Let us suppose, ${\mathrm{h}}^{2}+{\mathrm{x}}^{2}={\mathrm{r}}^{2}$
So that,
$\begin{array}{rcl}2\mathrm{xdx}+0& =& 2\mathrm{rdr}\\ \mathrm{xdx}& =& \mathrm{rdr}\\ & & \end{array}$
We find the limits as, when $\mathrm{x}={\mathrm{x}}_{1}$,
$\begin{array}{rcl}\mathrm{r}& =& {\mathrm{r}}_{1}\\ & =& \sqrt{{\mathrm{h}}^{2}+{\mathrm{x}}_{1}^{2}}\\ & =& \sqrt{{\left(1.20\mathrm{m}\right)}^{2}+{\left(3.00\mathrm{m}\right)}^{2}}\\ & =& \sqrt{10.44{\mathrm{m}}^{2}}\\ & =& 3.23\mathrm{m}\end{array}$
And when, $\mathrm{x}={\mathrm{x}}_{2}$
$\begin{array}{rcl}\mathrm{r}& =& {\mathrm{r}}_{2}\\ & =& \sqrt{{\mathrm{h}}^{2}+{\mathrm{x}}_{2}^{2}}\\ & =& \sqrt{{\left(1.20\mathrm{m}\right)}^{2}+{\left(1.00\mathrm{m}\right)}^{2}}\\ & =& \sqrt{2.44{\mathrm{m}}^{2}}\\ & =& 1.56\mathrm{m}\end{array}$
Now we determine the work done as,
$\begin{array}{rcl}\mathrm{W}& =& \mathrm{T}{\int }_{{\mathrm{x}}_{1}}^{{\mathrm{x}}_{2}}\left[\frac{\mathrm{x}}{\sqrt{{\mathrm{h}}^{2}+{\mathrm{x}}^{2}}}\right]d\mathrm{x}\\ & =& \mathrm{T}{\int }_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}\frac{\mathrm{rdr}}{\sqrt{{\mathrm{r}}^{2}}}\\ \mathrm{W}& =& \mathrm{T}{\int }_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}\mathrm{dr}\\ \mathrm{W}& =& \left(25.0\mathrm{N}\right){\left[\mathrm{r}\right]}_{{\mathrm{r}}_{1}}^{{\mathrm{r}}_{2}}\\ & =& \left(25.0\mathrm{N}\right)\left(1.56-3.23\right)\\ \mathrm{W}& =& -41.7\mathrm{J}\end{array}$
The change in kinetic energy is $∆\mathrm{KE}=\mathrm{W}=41.7\mathrm{J}$ (considering magnitude only).
## Recommended explanations on Physics Textbooks
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# What version of the definition of derivative
Tags:
1. Sep 17, 2015
### Niaboc67
If my understanding is correct the definition of a derivative is lim h->0 (f(x+h)-f(x))/h However, I've also seen this used: lim x->c (f(x)-f(x))/(x-c) are these both considered valid definition for the derivative or does the derivative have to tend towards zero? I am a bit confused because I see these two versions alternating and wondering which one I should use and where. And my teacher is picky about definitions and notation so I don't know which one he would like to see on the exam.
Thank you
2. Sep 17, 2015
### andrewkirk
The second one should be lim x->c (f(x)-f(c))/(x-c), not the version you wrote.
Note that the first one defines the derivative at x, ie f'(x), while the second defines the derivative at c, ie f'(c).
Both are valid, since they are equivalent, via the substitutions x<-->c, h<-->x-c. I have seen the first more often than the second. I suggest you use whichever version your teacher taught you.
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# Curves MCQ Quiz in বাংলা - Objective Question with Answer for Curves - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Oct 31, 2022
পাওয়া Curves उत्तरे आणि तपशीलवार उपायांसह एकाधिक निवड प्रश्न (MCQ क्विझ). এই বিনামূল্যে ডাউনলোড করুন Curves MCQ কুইজ পিডিএফ এবং আপনার আসন্ন পরীক্ষার জন্য প্রস্তুত করুন যেমন ব্যাঙ্কিং, এসএসসি, রেলওয়ে, ইউপিএসসি, রাজ্য পিএসসি।
## Top Curves MCQ Objective Questions
#### Curves Question 1:
If R is the radius of the curve in metres and C is the chord length in metres what would be the expression to denote versine ‘V’ in millimetres?
1. V = 127 C2/R
2. V = 127 R2/C
3. V = 125 C2/R
4. V = 125 R2/C
Option 3 : V = 125 C2/R
#### Curves Question 1 Detailed Solution
Concept:
Versine of curve
The versine is the perpendicular distance of the midpoint of a chord from the arc of a circle.
The relationship between radius and versine of a curve in various units is shown below:
$$V = \frac{{{C^2}}}{{8R}}$$ .............(V, C, R are in same units, say, m or cm)
$$V = \frac{{125{C^2}}}{R}$$ ......... (V in mm, C in m, and R in m)
$$V = \frac{{1.5{C^2}}}{R}$$...........(C and R in feet and V in inches)
where,
V = Versine of curve
C = Length of chord
#### Curves Question 2:
Calculate the length (m) of the longer chord of 250 m radius curve having deflection angle of 90 degree.
1. 250
2. 353.6
3. 392.7
4. 500
Option 2 : 353.6
#### Curves Question 2 Detailed Solution
Concept:
For the given curve:
Tangent length $$\left( \text{T} \right)=\text{R}\tan \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}$$
Length of curve $$\left( \text{l} \right)=\frac{\text{ }\!\!\pi\!\!\text{ R }\!\!\Delta\!\!\text{ }}{180}$$
Long chord $$\left( \text{L} \right)=2\text{R}\sin \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}$$
External distance $$\left( \text{E} \right)=\text{R}\left( \sec \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}\text{ }\!\!~\!\!\text{ }-1 \right)$$
Mid-ordinate $$\left( \text{M} \right)=\text{R}\left( 1-\cos \frac{\text{ }\!\!\Delta\!\!\text{ }}{2}\text{ }\!\!~\!\!\text{ } \right)$$
Chainage of T1 = Chainage of P.I. – T, and Chainage of T2 = Chainage of T1 + l
Calculations:
Length of long chord (L) = 2 × R × sin Δ/2 = 2 × 250 × sin 45° = 353.55 m
#### Curves Question 3:
Calculate the length (m) of tangent of a 5-degree curve, if the deflection angle is 60 degree.
1. 172.5
2. 198.6
3. 360
4. 596
Option 2 : 198.6
#### Curves Question 3 Detailed Solution
The tangent length of a simple curve of radius R and deflection angle D is $$R \times \tan \left( {\frac{D}{2}} \right)$$
Tangent length VT = $$R \times \tan \left( {\frac{D}{2}} \right)$$
Radius of the curve $$= \frac{{1720}}{5} = 344\;m$$
Tangent length = $$344 \times \tan \left( {\frac{{60}}{2}} \right) = 198.6084\;m$$
#### Curves Question 4:
The relationship between the length l and radius r of an ideal transition curve is stated as :
1. $$l \propto r\;$$
2. $$l \propto \frac{1}{r}\;$$
3. $$l \propto r^2\;$$
4. $$l \propto \frac{1}{r^2}\;$$
Option 2 : $$l \propto \frac{1}{r}\;$$
#### Curves Question 4 Detailed Solution
Explanation:
Transition curve is provided at the junction of the straight and curved portion of a railway track to serve the following purposes:
i) to provide the superelevation in a gradual manner
ii) to avoid the sudden jerk to the passengers
The requirement of an ideal transition curve:
i) The length of the transition curve is inversely proportional to the radius of curvature ( $$l \propto \frac{1}{r}\;$$).
ii) The curve should be tangential to its junction points.
iii) The rate of change of superelevation provided should be equal to the rate of change of curvature so that full superelevation can be provided within the length of the transition curve.
iv) The rate of change of centrifugal acceleration should be consistent, this implies the radius of the transition curve should be consistently decreased from infinity at the tangent point to the radius R of the circular curve.
#### Curves Question 5:
A railway curve of 1350 m radius is to be set out to connect two tangents. If the design speed is 110 kmph and the rate of change of acceleration is 0.3 m/s3, the shift of the circular curve will be nearly
1. 0.18 m
2. 0.16 m
3. 0.14 m
4. 0.12 m
Option 2 : 0.16 m
#### Curves Question 5 Detailed Solution
Concept:
Length of transition curve, Ls is given as:
$${L_s} = \;\frac{{{V^3}}}{{CR}}$$
The shift of the circular curve, S is given as:
$$S = \frac{{L_s^2}}{{24\;R}}$$
Where, V = speed,
C = rate of change of acceleration,
Calculation:
V = 110 kmph = 110× 5/18 = 30.55 m/s
C = 0.3 m/s3
R = 1350 m
$${L_s} = \frac{{{{30.55}^3}}}{{0.3\; \times \;1350}}$$
∴ Ls = 70.4 m
$$S = \frac{{\;{{70.4}^2}}}{{24\; \times \;1350}}$$ = 0.153 m
S ≈ 0.16 m
#### Curves Question 6:
On doubling the radius of the ideal transition curve, the length of the curve will tend to:
1. Increased by 100%
2. Remain same
3. Increased by 50%
4. Reduced by 50%
Option 4 : Reduced by 50%
#### Curves Question 6 Detailed Solution
Concept:
Length of Ideal transition curve is given by:
$${{\rm{L}}_{\rm{C}}} = \frac{{{{\rm{V}}^3}}}{{{\rm{CR}}}}$$
C → rate of change of radial acceleration
V → design speed
Explanation:
$${{\rm{L}}_{\rm{C}}}_1 = \frac{{{{\rm{V}}^3}}}{{{\rm{CR}}}}$$
$${{\rm{L}}_{\rm{C}}} \propto \frac{1}{{\rm{R}}}$$
R2 = 2R1
$${{\rm{L}}_{{{\rm{C}}_2}}} \propto \frac{1}{{{{\rm{R}}_2}}}$$
$${{\rm{L}}_{{{\rm{C}}_2}}} \propto \frac{1}{{2{{\rm{R}}_1}}}$$
$$\frac{{{{\rm{L}}_{{{\rm{C}}_2}}}}}{{{{\rm{L}}_{{{\rm{C}}_1}}}}} = \frac{{1 \times {{\rm{R}}_1}}}{{2{{\rm{R}}_1}}} = \frac{1}{2}$$
$${{\rm{L}}_{{{\rm{C}}_2}}} = \frac{{{{\rm{L}}_{{{\rm{c}}_1}}}}}{2} = 0.5{{\rm{L}}_{{{\rm{C}}_1}}}$$
i.e. the length of the curve will become half or reduce by 50%
#### Curves Question 7:
The radial offset at a distance x from the beginning of circular curve of radius R is given by
1. $$\sqrt{R^2-x^2} - R$$
2. $$R-\sqrt{R^2-x^2}$$
3. $$R-\sqrt{R^2+x^2}$$
4. $$\sqrt{R^2+x^2}-R$$
Option 4 : $$\sqrt{R^2+x^2}-R$$
#### Curves Question 7 Detailed Solution
Concept:
The methods of setting out curves:
The methods of setting out curves may be divided into two classes according to the instrument employed for setting out of curves as follows:
1. Linear or chain and tape method:- Linear methods are those in which the curve is set out with chain and tape only. It contains sub-methods as follows:
• By offsets or ordinates from the long chord
• By successive bisections of arcs
• By offsets from the tangents
• By offsets from chord produced
2. Angular or instrumental method:- Instrumental methods are those in which theodolite with or without a chain is employed to set out a curve. It contains sub-methods as follows:
• Rankine’s method of tangential angle
• Two theodolite method
By offsets from the tangents: In this method, the offsets are set out either radially or perpendicular to the tangents. It contains sub-methods as follows:
• By offsets perpendicular to tangents
Explanation:
The radial offset at a distance X from the point of commencement of curve of radius R
Let, T1 be the first tangent point and R is the radius of the curve
EE1 = Ox= the radial offset at E at a distance of X from T1 along with the tangent AB.
Now in the ∆OT1E,
OT1 = R and T1E = X
OE = OE1+ EE1 =R + Ox
Now, OE2 = (OT1) 2 + (T1E)2
(R + Ox)2 = R2 + x 2
$$O_X = \sqrt {R^2 + X^2} - R$$
The radial offset at a distance X from the point of commencement of curve of radius R is given by $$O_X = \sqrt {R^2 + X^2} - R$$
#### Curves Question 8:
If g1 = + 1.2% and g2 = + 0.8% and rate of change of grade = 0.1% per 20 m chain, then the length of the vertical curve is
1. 10 m
2. 15 m
3. 30 m
4. 80 m
Option 4 : 80 m
#### Curves Question 8 Detailed Solution
Concept:
Length of curve = (Total grade/Rate of change of grade per chain length) × Length of chain
Calculation:
Grade = + 1.2 – (+ 0.8) = 0.4 % (Upward)
Change of grade is 0.1% per 20 m chain.
L = (0.4/0.1) × 20 = 80 m
#### Curves Question 9:
Calculate the apex distance, if the deflection angle is 60 degree and the degree of curve is 8 degree.
1. 33.26
2. 124.13
3. 215
4. 262.8
Option 1 : 33.26
#### Curves Question 9 Detailed Solution
Concept:
Apex Distance: It is the distance between the point of Intersection and the apex (highest point) of the curve.
Apex distance = R ($$\sec \frac{\Delta }{2}$$ - 1)
Where, R: Radius of the curve, ∆: Deflection angle in degree
Arc Definition: If R is the radius of the curve and D is its degree for 30 m arc, then
× D × $$\frac{\pi }{{180}}$$ = 30
Where, R: Radius of the curve, D: Degree of the curve
Calculation:
The radius of the curve, R = $$\frac{{30 \times 180^\circ }}{{D \times {\rm{\pi }}}}$$ = $$\frac{{30 \times 180^\circ }}{{8 \times {\rm{\pi }}}}$$ = 214.9 m
Apex distance = R ($$\sec \frac{\Delta }{2}$$ - 1) = 214.9 ($$\sec \frac{{60}}{2}$$ - 1) = 33.255 m
#### Curves Question 10:
In a simple curve, external distance is the distance between:
1. Vertex and middle point of the curve
2. Vertex and point of the curve
3. Point of the curve and point of tangency
4. Vertex and the center of the curve
Option 1 : Vertex and middle point of the curve
#### Curves Question 10 Detailed Solution
Concept:
Simple circular curve:
• A simple curve consists of a single arc of a circle connecting two straights.
• It has a radius of the same magnitude throughout.
• Point of the intersection of the tangents and also called the vertex.
Point of curvature(T1) - It is the beginning of the curve.
Point of tangency(T2) - It is the end of the curve.
Figure: Simple curve
The length of the curve (l):
the length of curve T1 K T2 is given by
$${\rm{l}} = \frac{{2{\rm{\pi R}}}}{{360}} \times {\rm{D }}$$
Where, R = Radius of the curve, D = Deflection angle, and l = Length of the curve
Tangent Length (T) = R tan (D/2)
Length of long Chord (L) = 2R sin (D/2)
Mid-ordinate (M):
The middle ordinate is the distance from the midpoint(K) of the curve to the midpoint(C) of the chord.
M= R {1 - cos (D/2)}
External distance (E):
External distance is the distance from the vertex to the midpoint(K) of the curve.
E= R {sec(D/2) - 1}
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# Double superscript. error from Maple latex code after updating
All my files no longer compile after running my program due to such errors now
Double superscript. \left(-x^{2}+1\right) y^{\prime}^{2}
I have not updated TeXlive recently. Last time I updated it was about 5-7 weeks ago or so.
I did update Maple yesterday and then run my program which generates Latex by Maple.
Now when I try to compile the generated latex, I get the above errors. I am not sure what happened. I do know for sure if it is Maple's latex() which causes this or the update in TL I did which is the cause.
Here is a MWE
\documentclass[12pt]{book}
\usepackage{breqn}
\usepackage{amsmath}
\begin{document}
\begin{dmath*}
\left(-x^{2}+1\right) y^{\prime}^{2} = 1-y^{2}
\end{dmath*}
\end{document}
it seems lualatex complains about y^{\prime}^{2}
>lualatex foo.tex
This is LuaHBTeX, Version 1.15.1 (TeX Live 2023/dev)
restricted system commands enabled.
(./foo.tex
LaTeX2e <2022-11-01> patch level 1
L3 programming layer <2022-12-17>
(/usr/local/texlive/2022/texmf-dist/tex/latex/base/book.cls
Document Class: book 2022/07/02 v1.4n Standard LaTeX document class
(/usr/local/texlive/2022/texmf-dist/tex/latex/base/bk12.clo))
(/usr/local/texlive/2022/texmf-dist/tex/latex/breqn/breqn.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/l3kernel/expl3.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/l3backend/l3backend-luatex.def))
(/usr/local/texlive/2022/texmf-dist/tex/latex/amsmath/amsmath.sty
For additional information on amsmath, use the ?' option.
(/usr/local/texlive/2022/texmf-dist/tex/latex/amsmath/amstext.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/amsmath/amsgen.sty))
(/usr/local/texlive/2022/texmf-dist/tex/latex/amsmath/amsbsy.sty)
(/usr/local/texlive/2022/texmf-dist/tex/latex/amsmath/amsopn.sty))
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics/graphicx.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics/keyval.sty)
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics/graphics.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics/trig.sty)
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics-cfg/graphics.cfg)
(/usr/local/texlive/2022/texmf-dist/tex/latex/graphics-def/luatex.def)))
(/usr/local/texlive/2022/texmf-dist/tex/latex/breqn/flexisym.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/breqn/cmbase.sym)
(/usr/local/texlive/2022/texmf-dist/tex/latex/breqn/mathstyle.sty))
(/usr/local/texlive/2022/texmf-dist/tex/latex/tools/calc.sty)) (./foo.aux)
(/usr/local/texlive/2022/texmf-dist/tex/latex/base/ts1cmr.fd)
(/usr/local/texlive/2022/texmf-dist/tex/context/base/mkii/supp-pdf.mkii
) (/usr/local/texlive/2022/texmf-dist/tex/latex/epstopdf-pkg/epstopdf-base.sty
(/usr/local/texlive/2022/texmf-dist/tex/latex/latexconfig/epstopdf-sys.cfg))
! Double superscript.
l.12 \left(-x^{2}+1\right) y^{\prime}^{2}
= 1-y^{2}
?
Here is the Maple code using its latex() command to generate the above
interface(version)
Standard Worksheet Interface, Maple 2022.2, Windows 10, October
latex:-Settings(useimaginaryunit=i,
usecolor = false,
powersoftrigonometricfunctions= mixed, ## computernotation,
leavespaceafterfunctionname = true,
cacheresults = false,
spaceaftersqrt = true,
linelength=10000
);
Typesetting:-Unsuppress('all'); #always do this.
Typesetting:-Settings(prime=x,'typesetprime'=true); #this says to use y'(x) instead of dy/dx
Typesetting:-Suppress(y(x)); # this says to use y' and not y'(x)
ode:=(-x^2+1)*diff(y(x),x)^2 = 1-y(x)^2;
latex(ode)
gives
\left(-x^{2}+1\right) y^{\prime}^{2} = 1-y^{2}
Here is also screen shot of the above
So you see I am using exactly the code generated by Maple as is. The program copies the latex to a file and the compiles it.
My question is: is this a Maple latex() command issue I should report to Maplesoft to fix? I do not know if this is how the latex was generated before I did an update to Maple last night or not. I will try to find out.
Meanwhile is there something I can do to workaround this without editing the actual latex generated by may be by adding something to the preamble? I can't edit the latex as it is all auto-generated by I can edit the preamble and add macros there to work around problems.
Update
I looked at an old latex generated by Maple before the update I did last night, and this is what it generated for the same equation
\left(-x^{2}+1\right) \left(y^{\prime}\right)^{2} = 1-y^{2}
compare to the current version which is
\left(-x^{2}+1\right) y^{\prime}^{2} = 1-y^{2}
So it is Maple's fault. the first version above compiles OK and gives
So it seems like something changed in Maple's latex generated. I will report to Maplesoft.
• y^{\prime}^{2} has always been a tex error this is unrelated to any tex update so it's a bug in maple code (either yours or maplesoft's so not really on topic) Jan 29 at 21:53
• @DavidCarlisle is it off topic to ask for a workaround? if so, do you want me to delete this question? Jan 29 at 21:58
• no, but there really is no tex work around for such a syntax error other than fixing the input. I guess @Mico will show a luatex callback Lua gsub edit, but I'd do a one of fix wit sed or perl rather than have lua re-edit on every run Jan 29 at 22:04
I suggest you replace
\left(-x^{2}+1\right) y^{\prime}^{2} = 1-y^{2}
with
\left(-x^{2}+1\right) {y^{\prime}}^{2} = 1-y^{2}
or, better still,
(-x^{2}+1) {y'}^{2} = 1-y^{2}
Addendum to address the OP's follow-up query. If I understand your comment correctly, you're looking for a method that replaces y^{\prime}^{2} "on the fly" with {y'}^{2}. Since it looks like you're using LuaLaTeX [good choice!], this method could consist of setting up a Lua function which uses Lua's built-in string functions and making it act like a pre-processor on the input stream.
% !TEX TS-program = lualatex
\documentclass{article}
%% copy the remainder of this preamble into your preamble
\usepackage{luacode} % for \luaexec macro
%% Lua-side code: Define a Lua function
\luaexec{
function FixMaple ( s )
return ( s:gsub ( "y\%^{\\prime}\%^{2}" , "{y'}^2" ) )
end
}
%% LaTeX-side code: Macros to activate and de-activate the Lua function
\newcommand\FixMapleOn{%
"process_input_buffer", FixMaple, "FixMaple" )}}
\newcommand\FixMapleOff{%
\directlua{luatexbase.remove_from_callback(
"process_input_buffer", "FixMaple" )}}
\begin{document}
\FixMapleOn % activate the Lua function
$\left(-x^{2}+1\right) y^{\prime}^{2} = 1-y^{2}$
\end{document}
• Thanks Mico, but as I said in my question, this is auto-generated latex. I can't edit the latex. I have 10's of thousands of latex files. I can add something to preamble if you suggest a macro. so do you think the latex generated is wrong so I report it? Jan 29 at 21:41
• @Nasser - y^{\prime}^{2} is clearly wrong. If you can report this issue to whoever is in charge of maintaining the faulty code, by all means go ahead and do so.
– Mico
Jan 29 at 21:43
• OK, but is there a way to make a macro that will automatically change y^{\prime}^{....} to {y^{\prime}}^{...}` in the meanwhile? if not, that is ok. I will report this to Maple. Jan 29 at 21:45
• @Mico I sense an input buffer callback addendum coming:-) Jan 29 at 21:54
• +1 for proving me right:-) Jan 29 at 22:07
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## estudier Group Title Prove Cassini's identity using the matrix representation of the Fibonacci sequence one year ago one year ago
1. JamesWolf Group Title
oh is that all? easy! one question, whats cassini's identity? To google!
2. estudier Group Title
Heh, I can open questions all over the shop now.....:-)
3. JamesWolf Group Title
4. JamesWolf Group Title
out of interest. whats the formula for the fibonacci sequence
5. estudier Group Title
Where's the matrix?
6. JamesWolf Group Title
$F_n = F_{n-2} + F_{n-1}$ ?
7. JamesWolf Group Title
oh sorry yes your right there is none there
8. MrMoose Group Title
You can represent the Fibonacci sequence as: $\left[\begin{matrix}F_{n+1} & F_n \\ F_n &F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n$
9. MrMoose Group Title
This is fairly easy to prove by induction
10. MrMoose Group Title
define F_0 as 0, F_1 as 1, and F_2 as 1
11. MrMoose Group Title
If the first equation is true, then$\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^{n+1}=\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$: $= \left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right]*\left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]$ $= \left[\begin{matrix}F_{n+1}+F_n & F_{n+1} \\ F_n+F_{n-1} & F_n\end{matrix}\right]=\left[\begin{matrix}F_{n+2} & F_{n+1} \\ F_{n+1} & F_n\end{matrix}\right]$
12. MrMoose Group Title
So: $\left[\begin{matrix}F_{n+1} & F_n \\ F_n & F_{n-1}\end{matrix}\right] = \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n \forall n \in Z_+$
13. MrMoose Group Title
now, take the determinant of both sides: $F_{n+1}F_{n-1}-F_n^2=\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right]^n$
14. MrMoose Group Title
The determinant of the products of square matrices is the products of the determinants of the matrices: $= (\det \left[\begin{matrix}1 & 1 \\ 1 & 0\end{matrix}\right])^n$
15. MrMoose Group Title
=$(0-1)^n=-1^n$
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## On Processed Foods
In response to:
“Oh but isn’t that processed?” Yep so is beer, wine, bread, pasta…processing isn’t what matters. Number of ingredients and whether you can pronounce them are not important either. What matters is nutritional content. That’s a fact. Not an opinion. Can we as a culture just get over this pseudoscientific food snobbery please? Thanks.
“What matters is nutritional content” perhaps, but I would be sure to add some caveats to that. We don’t have a complete index of nutrients nor their impacts on the body. For example plant polyphenols in berries, we’re just starting to understand1, and there is evidence that eating these nutrients in their “unprocessed” state is more effective than just eating the nutrients directly234. Why would that be? It’s possible that we just don’t know what all the co-nutrients are yet, but it is more likely that there are countless interactions between body and food that we haven’t even scratched the surface of explaining.
A second caveat I would include is that food is a “package deal”. You have to eat something. If you’re eating a lot of processed foods (or a lot of animal foods), the likelihood of you getting the nutrients you are talking about goes down5. Also, the likelihood of getting toxins that you don’t want goes up. For example, potatoes, which are generally fairly healthy when steamed or boiled, create acrylamide when cooked at high heat a la french fries6. So even if we ignore my first caveat for a moment, it can be misleading to tell people that how processed a food is doesn’t matter, because that will most likely lead them to get fewer of the nutrients that we agree are important and more of the bad things.
Given these caveats, if you actually want to eat healthfully, it makes more sense to step back and look at the whole diet, rather than individual nutrients. This means you’re basically stuck with epidemiological studies of populations, and dietary intervention studies, neither of which is capable of establishing a sure-fire causal relationship. However, the evidence from such studies is pretty strong6 in favor of eating more foods of plant origin and in less processed form23478.
## Valuing human labor in the machine age
In response to:
What happens to your economic theory when the value of human labor is $0? [article suggesting we need basic income] The idea that the value of human labor will go to$0 is a bit hyperbolic. Sure, maybe drivers, bank tellers, and some other jobs are on the way out. But history is full of this progression. 150 years ago most jobs were on the farm. And then tractors and railroads and industry happened. Did all the jobs disappear? Well, farming jobs did, but other jobs replaced them. What makes this time so different?
I don’t claim to be able to imagine all the new types of jobs we’ll have in 50 years, but I can submit a few to you that I’m willing to bet will still be around:
• writer, journalist, reporter
• actor, artist, dancer, musician
• sports player, announcer
• lawyer, accountant
• doctor, nurse
• programmer, engineer (though these may look a lot different in 50 years)
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Traincascade Error: Bad argument (Can not get new positive sample. The most possible reason is insufficient count of samples in given vec-file.
Hi All: Here: http://code.opencv.org/issues/1834 Maria Dimashova She is giving the formula : vec-file has to contain >= (numPose + (numStages-1) * (1 - minHitRate) * numPose) + S, where S is a count of samples from vec-file . First I would like to know where to find this formula, in which document . Second She is writing :Bug "It was fixed in r8913", What daes it mean r8913? Is it Opencv actual version 2.4.3 ? Thank You.
edit retag close merge delete
( 2013-04-10 03:50:02 -0500 )edit
Sort by » oldest newest most voted
Hi,
First of all, I have to note that you copied my formula description incompletely. I wrote at that issue: "S is a count of samples from vec-file that can be recognized as background right away". With the partial description of S from the question, the formula does not make sense at all :)
For the document you asked.. I don't remember that I wrote this formula anywhere except the issue. The formula is not from any paper of course, it just follows from how traincascade application selects a set of positive samples to train each stage of a cascade classifier. Ok, I'll describe my formula in more details as you ask.
numPose - a count of positive samples which is used to train each stage (do not confuse it with a count of all samples in vec-file!).
numStages - a stages count which a cascade classifier will have after the training.
minHitRate - training constraint for each stage which means the following. Suppose a positive samples subset of size numPose was selected to train current i-stage (i is a zero-based index). After the training of current stage, at least minHitRate * numPose samples from this subset have to pass this stage, i.e. current cascade classifier with i+1 stages has to recognize this part (minHitRate) of the selected samples as positive.
If some positive samples (falseNegativeCount pieces) from the set (of size numPose) which was used to train i-stage were recognized as negative (i.e. background) after the stage training, then numPose - falseNegativeCount pieces of correctly recognized positive samples will be kept to train i+1-stage and falseNegativeCount pices of new positive samples (unused before) will be selected from vec-file to get a set of size numPose again.
One more important note: to train next i+1-stage we select only the samples that are passed a current cascade classifier with i+1 stages.
Now we are ready to derive the formula. For the 0-stage training we just get numPose positive samples from vec-file. In the worse case (1 - minHitRate) * numPose of these samples are recognized as negative by the cascade with 0-stage only. So in this case to get a training set of positive samples for the 1-stage training we have to select (1 - minHitRate) * numPose new samples from vec-file that are recognized as positive by the cascade with 0-stage only. While the selection, some new positive samples from vec-file can be recognized as background right away by the current cascade and we skip such samples. The count of skipped sample depends on your vec-file (how different samples are in it) and other training parameters. By analogy, for each i-stage training (i=1,..numStages-1) in the worse case we have to select (1 - minHitRate) * numPose new positive samples and several positive samples will be skipped in the process. As result to train all stages we need numPose + (numStages - 1) * (1 - minHitRate) * numPose + S positive samples, where S is a count of all the skipped samples from ...
more
Thank you for your clearer explanation! But I'm a bit confused about the "-numPos" argument. I would like to know if the current fix allow to use the total number of positive samples for each stage, in other words, in the argument "-numPos" I can use the total number? Or I still need to use your formula?
( 2012-11-23 04:39:21 -0500 )edit
1
Yes, you still need to keep in mind this formula. That fix was only about to throw an exception with error message for a user if there are not enough positive samples for the next stage training, because there was an assertion in that point of the code before.
( 2012-11-23 05:50:08 -0500 )edit
1
And why don't you put that formula and the result of that formula in the thrown message ... ?
Would save a lot of time to a lot of people
( 2014-02-11 07:17:27 -0500 )edit
Thank Maria for a great explanation
( 2016-12-14 01:34:52 -0500 )edit
I guess moderators will recommend you to open new question and not to ask more questions in the answer on your own question :)
Here it's said that -numPos is used in training for every classifier stage. OK, maybe we really have to specify that -numPos != vec-file samples count at that guide. Please, open the issue here with the link to this question.
0.9999999... is not a good value of minHitRate at least due to it will result in complicated classifiers even at first stages. And this breaks the idea of a cascade classifier to have weak classifiers at the beginning for rejecting a huge amount of background rectangles by cheaper checks of the first stages.
S depends on vec-file samples properties but you can also try to estimate this value. Suppose that the samples in vec-file have the equal probabilities to be rejected by a given cascade (be recognized as background). Of couse it's not true in reality but it's sutable for the estimation of vec-file size. In uniform case when you try to select falseNegativeCount positive samples to train i-stage you will select every new sample from vec-file with probability minHitRate^i. So to select falseNegativeCount samples you will try in average falseNegativeCount / minHitRate^i samples from vec-file. The increasing factor is small. E.g. if i=1, numPos=1000, minHitRate=0.99, falseNegativeCount=10 then you need to try in average ~10.1 samples from vec-file. More detailed formula about vec-file size with this assumption you can derive easy.
About wasting hours of work.. traincascade stores each trained stage immediately, so if you get an exception you can try to start training from current stage but with another training parameters and maybe with more rich vec-file.
I also don't recommend you to downgrade to 2.2. I did not find in the Git history my commit (due to files reorganization), but I fixed the following problem of traincascade: when traincascade tries samples from vec-file one by one and reaches the end of the file it have to finish the training, otherwise it will use duplicate samples. This was the bug.
The answer on the question about 2000 positives. To be sure that you can train a good cascade, try to use traincascade with default parameters on well-tried vec-file. Maybe you should start to play with parameters on this vec-file (not your) and definitely with LBP features (LBPs decrease wasting the time). For the choosing numPos, I think you can follow something like this numPos=0.9xNum_in_vec and you can also get more accurate estimation of this coefficient (instead of 0.9) in the uniform case (it's easy).
About tips on studing the traincascade code. As usual, from top to more details.. Here classical Haar is the best feature to get understanding faster (especially where features are processed by ADABoost because Haar is ordered (not categorical)). For an optimization the integral images are intensively used in cascades (keep it in mind). Don ...
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Hi, first I do congratulate of posts, is really good, and help-me very much.
But, I have some questions, I'm try use opencv_traincascade with 100 positive images and 600 negative images, using createsamples I do 100 samples create so, I pass following parametters:
opencv_traincascade.exe -data data/cascade -vec data/vector.vec -bg negative/infofile.txt -numPos 5 -numNeg 100 -numStages 15 -precalcValBufSize 256 -precalcIdxBufSize 256 -featureType HAAR -mode all -w 30 -h 30 -minHitRate 0.90
but, a aplication just stop of execute and close, don't have anything message of error, I do some variations of -numPos , but nevertheless a error return, somebody help-me ?
thanks...
more
numPos should be ur sample count.
( 2015-02-03 06:50:00 -0500 )edit
Hi Maria:
Thank You so much for Your long and exhaustive reply. I did read it carefully as it deserve a proper attention and concentration. Still I have my doubts, not in the formula itself as that is very clear to me now, but in the methodology . First the fact that –numPos has to be different from the number of sample in the .vec file is not explained in any official documentation like opencv_user.pdf chapter four , and anybody who read, get the message they are the same (we need to update the documentation).
Second, more important, the formula as You say, is not giving any systematic and final way to calculate (given an already existing .vec file from a good dataset of positive), the –numPos value, because we cannot know in advance the value of S in the formula. What we can know is our setting of minHitrate, so if we set minHitRate equal to 0.9999999… it seems we never will get any error as in this way the falseNegativeCount pieces will always be less than one. Another possibility is the already mention one of setting numPos=(0,9 x num_in_vec) or (0,8 x num_in_vec), but that it look to me also a kind of trick without guaranty of success. The big problem here, as You know is the extremely LONG computational time for every stages , and to have the process crashed after few stages means to waste hours of work, and not to have any guaranty that new setting will lead to the “desideratum” final stage, without crashing another time.
What I did, is to utilize OpenCV 2.2 (I downgraded!) where you can set numPos=num_in_vec without problem and get your xml classifiers (somehow). 2.2 version (as an example) at every stages it consumes few and few positive, discharging the FalseNegativeCount pieces that You mention.
But now I ask Your Kind suggestion about what is better to do, after creating a good dataset of lets say 2000 positive, and getting the .vec file what should we do in order not-to-crash the process? Should we set numPos=0.9xNum_in_vec, or minHitRate=0.999999, or use version 2.2 or any better suggestion?
One final suggestion is about a different matter, as I’m working on this for my final Thesis for my master, I’d like to study in deep how is working the code of traincascade with the 3 different features: Haar, LBP and HOG, so I’ll appreciate any starting suggestions and tips from You on start studing the cpp code.
Best regard.
Marco Romagnoli
more
Stats
Asked: 2012-11-20 08:36:28 -0500
Seen: 31,172 times
Last updated: Jan 19 '14
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# Coherence in density matrix formalism
1. Dec 19, 2014
### HJ Farnsworth
Hello,
In the density matrix formalism I have read in numerous places that coherence is identified with the off-diagonal components of the density matrix. The motivation for this that is usually given is that if a state interacts with the environment in such a way that the basis state amplitudes are phase-shifted through this interaction, then the off-diagonal components will average to $0$ if the degree to which each basis amplitude is phase-shifted are uncorrelated.
$\rho = \sum_{\psi}P(\psi)|\psi\rangle\langle\psi|$
in the $\{ |m\rangle \}$ representation, $|\psi\rangle = \sum_{m}|m\rangle\langle m|\psi\rangle$, it's easy to think of trivial examples where the coherence is $0$ for pure states, which to me seems absurd. It is also easy to think of examples where the coherence is basis-dependent, which also seems strange to me.
For instance a pure spin-up state as written in the basis $\{ |x\uparrow \rangle , |x \downarrow \rangle \}$ versus the basis $\{ |z \uparrow \rangle , |z \downarrow \rangle \}$ gives, as expected, two different representations of the density matrix: $\rho = \begin{pmatrix}1 & 0\\0 & 0\end{pmatrix}$ in the $z$-basis and $\rho = \begin{pmatrix}1/2 & 1/2\\1/2 & 1/2\end{pmatrix}$ in the $x$-basis. So here, we have a pure state, which in one basis has a coherence of $0$, and in another has a coherence of $1/2$ - it seems to me that a good definition of coherence would identify the coherence of a pure state as $1$ regardless to basis.
Furthermore, every time I have seen this definition of coherence of a state, the only examples given are for $2\times 2$ density matrices. How does the definition extend to density matrices written for larger bases - do we only define the coherence for one pair of basis amplitudes at a time?
I have a guess to the answers to the above questions, but haven't been able to verify it. The coherence could be a comparative proberty between substates only, not a property of the state as a whole. In this case, a change of basis giving different coherences makes a lot more sense, and the extension of the definition to larger density matrices becomes trivial. Furthermore, the spin-$z$ having a coherence of $0$ above would also make more sense, since if the energy of the spin-up state is nonzero, then the complex amplitude of $| \uparrow \rangle$ will oscillate, while that of $| \downarrow \rangle$ won't, and instead will always remain $0$. So, the complex amplitudes of the two phases won't be correlated at all.
With this, the value of coherence being zero versus nonzero has an interpretation in terms of complex amplitude correlation between substates. However, it's still hard to see an interpretation of different nonzero values for coherence - e.g., what does a coherence of $1/2$ versus $1/3$ tell you?
Anyway, I think I've said plenty to illustrate my confusion. Any help on the topic would be very much appreciated.
Thanks very much.
-HJ Farnsworth
2. Dec 20, 2014
### kith
Talking about coherence can be a bit confusing because the term has many slightly different meanings and is often used informally.
Mathematically well-defined are the terms "coherences" for the off-diagonal elements of the density matrix and "purity" for the trace of the density matrix squared.
The purity of a density matrix doesn't depend on the basis. It is one for pure states and minimal for a maximally mixed state (which is represented by a density matrix which is proportional to the identity matrix). So if you want to talk about the "coherence" of a (not necessarily pure) state, you may want to use the purity.
As you already suspected, the off-diagonal elements or coherences refer to the coherence between basis states. A non-zero coherence between two states means that you cannot decompose your ensemble into a merely statistical mixture of sub-ensembles such that no sub-ensemble contains both basis states. Sakurai has a number of nice examples on this in his chapter on angular momentum. He probably doesn't use modern terminology though.
Last edited: Dec 20, 2014
3. Dec 22, 2014
### HJ Farnsworth
Thanks kith, that's great.
So to make sure I understand you correctly, a simple interpretation that follows from the definition of coherence in the original post would be something like, "coherence between two basis states is a measure of the degree to which those two basis states must be regarded as part of the same, additively inseparable ensemble".
Other concepts of coherence that I have seen relate coherence between two waves to the relative frequency and phase difference of those waves - i.e., "two waves are coherent iff they have the same frequency and constant phase difference". I think that I can relate the interpretation above to this interpretation.
First, the additive inseparability interpretation motivates this guess: "The statement that the coherence between two basis states $|a \rangle$ and $|b \rangle$ in a density matrix is $0$ is equivalent to the statement that all of the quantum states that were classically weighted in writing the density matrix had either the amplitude for $|a \rangle$ equal to $0$, or that for $|b \rangle$ equal to $0$ - none of them had the amplitudes for both $|a \rangle$ and $|b \rangle$ nonzero." The reason the additive inseparability interpretation motivates this guess is that, if the states composing the ensemble have this property, then it is obvious that the coherence will be $0$ and the density matrix will be additively separable, so it is at least a sufficient condition for incoherence. Testing this guess, let $|\psi_{1}\rangle =a_{1}|a\rangle + b_{1}|b\rangle$, and write
$\rho = \sum{P_{i}|\psi_{i}\rangle \langle \psi_{i} |}=\begin{pmatrix} & a_{1}b_{1}^{*}+\cdots\\a_{1}^{*}b_{1}+\cdots\end{pmatrix}.$
It is immediately obvious that the guess was wrong - e.g., if $|\psi_{2}\rangle =-a_{1}|a\rangle + b_{1}|b\rangle$, then even if both $a_{1}$ and $b_{1}$ are nonzero, we could have offdiagonal matrix elements of $0$. However, note that if this is the case, then the phase difference between the two amplitudes in $\psi_{1}$ is exactly cancelled out by that in $\psi_{2}$: $|\psi_{2}\rangle =-a_{1}|a\rangle + b_{1}|b\rangle=e^{i\pi}a_{1}|a\rangle + b_{1}|b\rangle$, vs. $|\psi_{1}\rangle =e^{i0}a_{1}|a\rangle + b_{1}|b\rangle$. The offdiagonal terms from $|\psi_{1} \rangle$ could also be cancelled out by multiple other states in the ensemble - the only requirement for this to happen is that the phase differences between $|a\rangle$ and $|b\rangle$ in each state of the ensemble combine so as to cancel each other out (I am ignoring the different relative amplitudes between$|a\rangle$ and $|b\rangle$ among states, different amplitudes will just put a sort of weight on different relative phase differences between the two basis states).
Based on this, I can replace the above guess with, "The statement that the coherence between two basis states $|a \rangle$ and $|b \rangle$ in a density matrix is $0$ is equivalent to the statement that all of the quantum states that were classically weighted in writing the density matrix had relative phases between the two basis states which ultimately cancelled each other out." Relating this to the "common" coherence definition two paragraphs above, I could say something like "two basis states are [completely] coherent, in the density matrix sense, if their phase difference is constant throughout the states of the ensemble". (I brushed frequency under the rug since I'm considering everything at a single time, say $t=0\implies \omega t=0$. This is in pretty good analogy to e.g., classical optics temporal coherence between two waves = constant phase difference at a given point in space as the waves propagate: Temporal coherence between two waves throughout time is analogous to density matrix coherence between two basis states among the states in the ensemble.
Any thoughts on this thought process? Decent analysis, complete BS, or am I attempting to go too far in getting an intuition on what is simply a mathematical definition?
A half-follow-up and half-new-question: Regarding the first sentence in kith's response, the other two definitions of coherence that I have frequently come across are spatial/temporal coherence, defined in terms of the autocorrelation function in classical optics, and coherent states, defined as eigenstates of the HO annihilation operator in quantum mechanics. I think I have begun to answer this for myself in the above paragraph, but does anyone know the degree to which these three definitions of coherence/coherent states are related, or of a good source explaining, mathematically and interpretively, the relations among these concepts?
Thanks again.
Last edited: Dec 22, 2014
4. Dec 25, 2014
### HJ Farnsworth
Sorry for bumping this, but I do want to know whether people who have a bit more experience than me with this topic think that my conceptual understanding in the previous post holds water. Any thoughts?
Thanks.
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# Regression with missing covariate data
My question is related to regression with missing covariate data. I have a sample of count data representing number of explosion incidents at each year. First $M$ observations were made by considering one criteria and the last $K$ were made considering two criteria. E.g. for the first $M$ years explosions were recorded if there were some human injuries and then, due to the change of regulations, explosion were recorded if they caused human injuries OR the damage were higher than, say, $D$. I introduce two binary covariates: $X_{1}$ for human injury criteria and $X_{2}$ for damage criteria. So, I get that the first covariate $X_{1}$ is always observed and equal to 1 for the first $M$ years but not for the next $K$ years. The second covariate $X_{2}$ for the first year is not observed at all (so it might be 0 or 1), while for the last $K$ years both covariates are not observed but I know that at least one of them is equal to 1.
At first I though that I just put a Bernoulli distribution on these two covariates and integrate out them. Then I perform Bayesian analysis. So, at first I state the model (for simplicity, assume we have just one covariate) $$Y|\theta_{1},\theta_{2}\sim \mathrm{Poisson} \left (\mathrm{exp}\left [ \theta_{1}X_{1}+\theta_{2} \right ] \right )$$ $$X|p\sim \mathrm{Bernoulli}(p)$$ Then perform integration over $X_{1}$: $$\pi(Y|p,\theta)=\frac{1}{Y!}\left ( \mathrm{exp}\left ( -e^{\theta_{1}+\theta_{2}}+Y\left ( \theta_{1}+\theta_{2} \right ) \right )p + \mathrm{exp}\left ( -e^{\theta_{2}}+Y \theta_{2} \right )(1-p) \right )$$ Having this, I form likelihood, state priors and obtain estimates. As I understand, such approach is a so called "Missing completely at random or MCAR". However I understand that in my problem the missingness model is different from MCAR, but I do not know what model would be more appropriate. Any advise would be appreciated.
• It is hard to say whether you have MCAR data (where the distribution of missingness does not depend on the observed covariate), MAR (Missing At Random; the distribution of missingness depends on observed but not on missing covariates) or MNAR data. For instance, the regulatory change may have decreased the incidence of property damage, given that this was recorded after the change. This article may be helpful: psycnet.apa.org/… Nov 29, 2012 at 14:00
• Well, the impact of new regulations is not known. I assume, that it has no impact, since gas pipelines, for which explosions were recorded, were not renewed, i.e. were the same for the entire observation period. It might have impact on the maintenance, however it is impossible to evaluate for the data that is available to me. Nov 29, 2012 at 14:23
• In that case you probably do have MCAR data. The article I linked to discusses Bayesian approaches to this; I honestly don't know how helpful it will be... Good luck! Nov 29, 2012 at 14:28
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# What is the square root of -1/121?
$\sqrt{- \frac{1}{121}} = \pm \frac{1}{11} i$
$\sqrt{- \frac{1}{121}} = \sqrt{\frac{1}{121}} \cdot \sqrt{- 1}$
$= \pm \frac{1}{\sqrt{121}} \cdot i$
$= \pm \frac{1}{11} i$
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+0
+1
577
1
+603
For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution? Express your answer as a common fraction.
May 4, 2018
#1
+22307
+3
For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution?
$$\begin{array}{|rcll|} \hline (x+4)(x+1) &=& m + 2x \\ x^2+5x+4 &=& m+2x \\ x^2+3x+4-m &=& 0 \\\\ x &=& \dfrac{-3\pm \sqrt{9-4(4-m)} }{2} \\ &=& \dfrac{-3\pm \sqrt{9-16+4m} }{2} \\ &=& \dfrac{-3\pm \sqrt{4m-7} }{2} \\\\ && \text{one real solution: \sqrt{4m-7}=0} \\ \sqrt{4m-7} &=& 0 \\ 4m-7 &=& 0 \\ 4m &=& 7 \\ \mathbf{ m } & \mathbf{=} & \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$$
May 4, 2018
#1
+22307
+3
For what value of $m$ does the equation $(x+4)(x+1) = m + 2x$ have exactly one real solution?
$$\begin{array}{|rcll|} \hline (x+4)(x+1) &=& m + 2x \\ x^2+5x+4 &=& m+2x \\ x^2+3x+4-m &=& 0 \\\\ x &=& \dfrac{-3\pm \sqrt{9-4(4-m)} }{2} \\ &=& \dfrac{-3\pm \sqrt{9-16+4m} }{2} \\ &=& \dfrac{-3\pm \sqrt{4m-7} }{2} \\\\ && \text{one real solution: \sqrt{4m-7}=0} \\ \sqrt{4m-7} &=& 0 \\ 4m-7 &=& 0 \\ 4m &=& 7 \\ \mathbf{ m } & \mathbf{=} & \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$$
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## When Average Is Not Enough: Thoughts on Designing for Capacity
Designing a system from scratch to handle a workload you don’t know is a challenge. If you put to much hardware, you might be wasting money. You put little, then your users will complain of how slow the system is.
If you’re given only a rate, like 6000 hits/hour, you don’t know how these are distributed in a minute by minute or per second interval. We can make a guess and say that there are about 100 hits per minute or 1.67 hits/sec. If hits come uniformly at that rate, then we can design a system that can handle 2 hits/sec and all users will be happy since all requests will be served quickly and no queueing of requests. But we know it’s not going to happen. There will be some interval where the number of hits is less than 3 and some more than 3.
Theoretically, requests to our server come randomly. Let’s imagine 60 bins represented by seconds in one minute. We also imagine that requests are like balls we throw into the bins. Each bin is equally likely to be landed by a ball. It’s possible that all balls land on only one bin!
After throwing the balls into bins, let’s see what we have.
As you can see, some bins have more than 2 balls (which is the average number of balls in a bin). Therefore if we design our system based on the average, 50% of our users will have a great experience while the other 50% will have a bad experience. Therefore we need to find how many requests per second our server needs to handle so that our users will have a good experience (without overspending).
To determine how many requests per second we need to support, we need to get the probability of getting 4, 5, 6 or more request per second. We will compute the probability starting from 3 requests per second and increment by one until we can get a low enough probability. If we design the system for a rate that has a low probability, we are going to spend money for something that rarely occurs.
Computing the Probability Distribution
We can view the distribution of balls into bins in another way. Imagine labeling each ball with a number from 1 to 60. Each number has an equal chance to be picked. The meaning of this labeling is this: the number that was assigned to the ball is the bin (time bucket) it belongs to. After labeling all balls, what you have is a distribution of balls into bins.
Since each ball can be labeled in 60 different ways and there are 100 balls, the number of ways we can label 100 different balls is therefore
$\displaystyle 60^{100}$
Pick a number from 1-60. Say number 1. Assume 2 balls out of 100 are labeled with number 1. In how many ways can you do this ? Choose the first ball to label. There are 100 ways to choose the ball. Choose the second ball. Now there are 99 ways to choose the second ball. We therefore have 990 ways to select 2 balls and label them 1. Since we don’t really care in what order we picked the ball, we divide 990 with the number of possible arrangements of ball 1 and ball 2, which is 2! (where the exclamation mark stands for “factorial”). So far, the number of ways to label 2 balls with the same number is
$\displaystyle \frac{100 \times 99}{2!}$
Since these are the only balls with label 1, the third ball can be labeled anything except number 1. In that case, there are 59 ways to label ball 3. In the same way, there are 59 ways to label ball 4. Continuing this reasoning until ball 100, the total ways we can label 2 balls with number 1 and the rest with anything else is therefore:
$\displaystyle \frac{100 \times 99}{2!} \times 59^{98}$
Notice that the exponent of 59 is 98 since there are 98 balls starting from ball 3 to ball 100.
Therefore, the probability of having two balls in the same bin is
$\displaystyle \frac{100 \times 99}{2!} \times \frac{59^{98}}{60^{100}} = 0.2648$
We can also write this as
$\displaystyle \frac{100!}{2! \times 98!} \times \frac{(60-1)^{98}}{60^{100}} = \binom{100}{2} \frac{(60-1)^{98}}{60^{100}}$
In general, if m is the number of balls, n the number of bins and k the number of balls with the same label, then the probability of having k balls within the same bin is given by
$\displaystyle \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}}$
,
where
$\displaystyle \binom{m}{k} = \frac{m!}{k!(m-k)!}$
is the binomial coefficient.
It turns out that this is a probability distribution since the sum of all probabilities from k=0 to k=m is equal to 1. that is
$\displaystyle \sum_{k=0}^{n} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} = 1$
To see this, recall from the Binomial Theorem that
$\displaystyle \big( x + y \big)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k$
If we let x=n-1 and y=1, we can write the above equation as
$\displaystyle \begin{array}{ll} \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}}{n^{m}} &= \displaystyle \sum_{k=0}^{m} \binom{m}{k} \frac{(n-1)^{m-k}\cdot 1^k}{n^{m}}\\ &= \displaystyle\frac{(n-1+1)^m}{n^{m}}\\ &= \displaystyle\frac{n^m}{n^m}\\ &= \displaystyle 1 \end{array}$
Here is a graph of this probability distribution.
Here’s the plot data:
probability 1 0.315663315854 2 0.264836171776 3 0.146632456689 4 0.060268424995 5 0.019612775592 6 0.005263315484 7 0.001197945897 8 0.000236035950 9 0.000040895118 10 0.000006307552
We can see that for k=9, the probability of it occurring is .004%. Anything beyond that we can call rare and no need to spend money with.
Just For Fun
What’s the probability that a given bin is empty, that is, there are no balls in it?
Other Probability Distributions
Our computation above was based on a uniform probability distribution. However, there are other distributions that are more suitable for arrival of requests. One of the most widely used is called the Poisson Distribution where you can read from here.
R Code
The R code to generate the simulation:
par(mfrow=c(4,4))
f=function(){
x=c()
for(i in 1:100){
x=c(x,sample(seq(1,60),1,replace=T))
}
plot(tabulate(x),type="h", ylab="tx", xlab="secs")
}
for(i in 1:16){
f()
}
The R code to generate the probability distribution:
p=function(m,n,s){
prod(seq(m,m-s+1))/factorial(s)*(n-1)^(m-s)
}
tt=c()
for(i in 1:10){
tt=c(tt,p(100,60,i)/60^100)
}
plot(tt,type="h",xlab="Number of Balls",ylab="Probability")
Dedication
This post is dedicated to my friend Ernesto Adorio, a mathematician. He loves combinatorial mathematics.
Rest in peace my friend! I miss bouncing ideas with you.
## New Blog Location
I migrated my blog from a hosted linux server which I maintain into wordpress.com hosting service. Times are changing. I used to have more time before to maintain my site, upgrade my linux applications to the latest security patches, configure my own DNS server, email server, http server, what have you… No I just don’t have the luxury of time. It’s good that we have this hosting providers doing the job for you for free!!!
I thought that migrating my old wordpress blog from the old site to the new site would be a challenge. I was wrong. Thank you wordpress for making my migration painless.
I just followed the instructions from here to export and import my blog:
http://en.support.wordpress.com/import/
and it worked like a charm!
## A Tribute to My Manager
I have been under many managers in my IT career which spans over 11 years. They come in all shapes and sizes. But one manager stands out and this is a tribute to him.
My manager’s name is Jan Lategahn. He was my manager for one year in an IT project. But it was one of the most productive years of my life. He was different from all my previous managers because of the following qualities which I will enumerate below.
My manager inspires his subordinates. A manager who does not know how to do the work of his subordinates will find it hard to inspire. The metaphor that usually comes to my mind is that of a Shogun; the most skillful swordsman in his territory. Everybody treats him with respect. The same is true for a manager who is the best programmer in his team.
My manager is an expert programmer. He is an expert in SAP and UNIX. Being a programmer he knows that programming is a creative process and sometimes it is just difficult to predict when you can finish a piece of code. He doesn’t give unrealistic deadlines but if the deadline can’t be moved, he knows how to adjust the scope or to give you all the help you need to succeed.
Once my manager went on a vacation and I was so swamped with work. When he returned to work and found me working till 12 midnight that week he was so kind enough to offer his assistance. I was surprised when he approached my desk and asked me “what can I do to help you?” I asked him to install the server in the uat environment which he did in just two hours. After that he returned and asked for more tasks. I’ve never had any other manager ask me for tasks (which is not surprising; they don’t know how to do it).
When I went to Singapore for my vacation I brought my laptop just to make sure I can do support when extremely needed. I was expecting my manager to call me anytime and ruin my vacation. But instead I got a text message telling me about places I can go and enjoy in Singapore. After that he let me alone in my vacation until I returned to work. Later on he told me he had a rough time during those days and that he must have aged six years when I was as away. I really appreciate what he did and feel so lucky he can do without me.
My manager is not afraid to show his humanity. He is not afraid to show to his subordinates that he sometimes feels tired; that he dislikes unproductive meetings and chatterboxes. He would rather code something than waste his time in meetings. Other managers look like supermen who don’t seem to feel tired and never complain. My manager gives us an example that it’s okay to be human.
Once we had a visit wih the big bosses from onshore. And we had to behave the whole week because they were sitting near our cubicles. On the last day when the big bosses finally left, my manager was not ashamed to show a sigh of relief and say “it was very hard to breathe with them around” or so along those lines. We were laughing, but I really appreciated that reaction. It showed me that my boss is also human and whatever ordinary things we feel, he feels it too.
My manager has a great personality. He always greets people when he arrives in the morning and always finds time to do a quick chat about anything under the sun. I have learned so much from him about electronics, WI-FI hacking, etc.; things not directly related to work. In fact when he’s absent from work due to some illness, I find the team spirit not so happy. He brings life to the team. In my previous work we feel elated when our managers are not around, in my manager’s case the opposite is true. I even have to send a text message to him to ask how he is doing.
And most of all, my manager is a friend to his men. Other manager’s talk about open door policy but always maintain the distance from their subordinates. Not so with my manager. I can always speak my thoughts to him without reservation because he has become a friend.
To my manager, mentor and friend, I wish you only the best in life!
## Do It Yourself Supercomputing in Linux Part 1
If you recently purchased a laptop or desktop computer, chances are you have a dual core system. There does not seem to be any indication anymore of us going back to single core unless some technological breakthrough will break the power barrier of CPUs. This means that more and more people will have a high powered computer in their homes without any idea how to harness such power.
What does it mean to have a dual processor? On first impulse, you probably will think it will speed up the execution of your programs. You would probably perceive a significant difference between the response time of your programs in a single core versus a dual core. But why do programs run faster in dual processor systems? For one thing, the speed of each processor is faster as compared to older single processors. The other thing is due to symmetric multiprocessing. An analogy of the dual processor system is a bank with 2 tellers and having a single line for customers. When a teller is done with one customer, it will get the next customer from the line. You can read about the performance of multiprocessor systems in this article.
There is another way you can make use of multiprocessors to make your programs run faster. This is through parallel programming. Parallel computing has been around for a long time already but are usually confined to university laboratories or supercomputing facilities. It has not caught the interest of the common people because they have no access to such machines. However, the future is already about multi-processing. More and more of these machines will reach the masses. This means that highly talented members of the masses get to experiment with parallel computing on a day-to-day basis. We are in for another revolution in computing. Are you ready for this revolution?
## The Best Place in Singapore
I was fortunate enough to go to Singapore in October of last year. We arrived in Changi airport about lunch time and it was raining. I was expecting it as some of my friends told me it rains every now and then in Singapore. I was immediately impressed by the airport. Although I have several big airports before when I was in San Franciso, USA and Dubai, this airport was refreshing. We took a cab from the airport to our hotel (Peninsula Excelsior). The scenery was really beautiful as we drove from the airport to the city. The city itself was nothing compared to anything in the Philippines. The streets were so clean and it was very hard for us to find any cigarette butt.
Our first destination was the Zoo. I’m not good with remembering names so I can’t remember the name of the Zoo. But it was the first time I entered a zoo where I saw many animals for the first time. Well, I don’t really go to the zoo, even here in the Philippines. It’s one of those places that are low in my priority. However, I was not disappointed when I went to Singapore zoo.
Our hotel was very near the place of our conference so we just walk it on the way home, utilizing the underground mall that connects Suntec to a train station near our hotel. I can’t remember the name of that underground mall, or if there is a name to it. It was just a pleasant walk. You can shop as you go to the hotel.
We stayed in Singapore for 6 days. Before we left for the Philippines, my wife and I went to Sentosa. It was the place in the top of her list to go. The place was really great. We went to this underwater museum where you can see esoteric fishes and crustaceans. There is already one here in the Philippines but I haven’t gone there yet. We went to Sentosa by train and went out of it by cable car. The view was really great!
On the first day, my wife asked me if I would consider going to Singapore next time for vacation. I told her the city was beautiful, very modern and clean but that there was really no compelling reason to go there. I can live with what we have in the Philippines.
However, all that changed when I went to Borders bookstore and Kinokuniya. There is nothing of such beauty and magnitude in the Philippines as those bookstores. I can spend my entire day just browsing through all the books. And that would be just one row of bookshelves! So the best place in Singapore for me are the bookstores and if I have my way, then the reason I will return to Singapore is to visit the bookstores.
## John Wheeler, A great physicist!
John Archibald Wheeler is one of my most admired physicist. He just died recently and I want to make a small tribute to him to did so much in General Relativity. Wheeler is the person who coined the word “black hole”, a concept which captivated my imagination when I was yet a kid and a concept which I eventually studied on my own in the University. He wrote one of the best books in General Relativity entitled “Gravitation”. This is a thick and heavy book full of physics and mathematics. It introduces advanced mathematical concepts as you progress in your study of Relativity in a very geometric way. This book has full of illustrations that really whets your appetite for studying advanced physics.
I have bought many books authored by Wheeler. One of them is “Exploring Black Holes“, with Taylor as co-author. I bought this book in Borders bookstore in Singapore. Unfortunately, we don’t have these kinds of books in the Philippines. A friend of mine also lent me a layman’s book written by Wheeler entitled “Geons, Black Holes & Quantum Foam”. This book gave accounts on Wheeler’s contribution to the Manhattan Project and the other great people whom he worked with.
I have a professor before who took his PhD in US who told me that whenever Wheeler gives a lecture at 7 am in the morning, he will always attend it no matter how cold it is in New York. That’s how big an impact Wheeler had on him.
To the great man Wheeler, thank you for inspiring us to study one of the greatest theories in the 20th century!
## Robots and Emotions
I just came across this article entitled “The rise of the Emotional Robot” which talks about owners of Roomba robot dressing them up and even giving them names and gender. People are now getting attached to their robots as if they were part of the family. What I find funny is that we put a lot of value to these human creations forgetting that there are real humans out there who are of greater value by virtue of being human and are living in wretched conditions. I don’t mean to say that attaching ourselves to our robot is a bad thing but that when we do these things we should also be aware of real people out there who also need the same care and attention.
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show all page boundaries in libreoffice writer [closed]
In libreoffie writer I would like to see all page boundaries (for header, footer, and text) at the same time. This would be necessary for deciding the page layout. How can I turn it on? Currently I can see the boundaries only for the area which I edit. I use libreoffice 4.3.
Thanks,
suseuser04
edit retag reopen merge delete
Closed for the following reason question is not relevant or outdated by Alex Kemp close date 2020-08-06 15:00:41.981182
Could you elaborate a bit? Do you want to see several pages simultaneously?
Meanwhile, have you fiddled with Tools->Options, LO->Appearance, LO Writer->Formatting Aids and other settings?
( 2017-12-19 08:44:53 +0100 )edit
Sort by » oldest newest most voted
As far as I can understand, there is a maximum of three text areas you want to get a border line for if Formatting Marks is switched on (also controlling the group Formatting Aids set via options). This for any page in the current view.
As a friend of much as possible of openly visible information, and declining any hiding of info where not unavoidable, I would judge this desire perfectly reasonable. However, there isn't yet an enhancement request to that effect, seemingly.
If an enhancement gets implemented at all there are two main alternatives I see:
-1- Show all or none of the mentioned areas (if existing at all) simultaneously, controlled by Toggle Formatting Marks.
-2- Add three related items under Formatting Aids to Options > LibreOffice Writer > Formatting Aids > Display of
I am considering to post an "enhancement bug" concerning this issue.
I would prefer -1-. What's your opinion, @suseuser04 ?
Slightly related: bug (enhancement) tdf# 33304 of 2011-01-20.
more
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# How to learn Vocabulary efficiently ?
Today’s post is very similar to one of my earlier posts titled “How to teach Algorithms ?“. In that earlier post, I announced an Algorithms App for iPad. Recently, I ported Algorithms App to Mac. Today’s post is about “How to learn Vocabulary efficiently ?”.
This summer, I went to a local bookstore to checkout the vocabulary section. There are several expensive books with limited number of practice tests. I also noticed a box of paper flashcards (with only 300 words) for around \$25 !!! After doing some more research, I realized that the existing solutions (to learn english vocabulary) are either too hard to use and/or expensive and/or old-fashioned.
So I started building an app with ‘adaptiveness’ and ‘usability’ as primary goals. The result is the Vocabulary App (for iPhone and iPad). Here is a short description of my app.
Vocabulary app uses a sophisticated algorithm (based on spaced repetition and Leitner system) to design adaptive multiple-choice vocabulary questions. It is built on a hypergraph of words constructed using lexical cohesion.
Learning tasks are divided into small sets of multiple-choice tests designed to help you master basic words before moving on to advanced words. Words that you have the hardest time are selected more frequently. For a fixed word, the correct and wrong answers are selected adaptively giving rise to hundreds of combinations. After each wrong answer, you receive a detailed feedback with the meaning and usage of the underlying word.
Works best when used every day. Take a test whenever you have free time.
At any given waking moment I spend my time either (1) math monkeying around (or) (2) code monkeying around. During math monkeying phase, I work on math open problems (currently related to directed minors). During code monkeying phase, I work on developing apps (currently Algorithms App, Vocabulary App) or adding new features to my websites TrueShelf or Polytopix. I try to maintain a balance between (1) and (2), subject to the nearest available equipment (a laptop or pen-and-paper). My next post will be on one of my papers (on directed minors) that is nearing completion. Stay tuned.
# Directed Minors III. Directed Linked Decompositions
This is a short post about the following paper in my directed minor series :
• Shiva Kintali. “Directed Minors III. Directed Linked Decompositions“. Preprint available on my publications page.
Thomas [Tho’90] proved that every undirected graph admits a linked tree decomposition of width equal to its treewidth. This theorem is a key technical tool for proving that every set of bounded treewidth graphs is well-quasi-ordered. An analogous theorem for branch-width was proved by Geelen, Gerards and Whittle [GGW’02]. They used this result to prove that all matroids representable over a fixed finite field and with bounded branch-width are well-quasi-ordered under minors. Kim and Seymour [KS’12] proved that every semi-complete digraph admits a linked directed path decomposition of width equal to its directed pathwidth. They used this result to show that all semi-complete digraphs are well-quasi-ordered under “strong” minors.
In this paper, we generalize Thomas’s theorem to all digraphs.
Theorem : Every digraph G admits a linked directed path decomposition and a linked DAG decomposition of width equal to its directed pathwidth and DAG-width respectively.
The above theorem is crucial to prove well-quasi-ordering of some interesting classes of digraphs. I will release Directed Minors IV soon. Stay tuned !!
# Directed Width Parameters and Circumference of Digraphs
This is a short post about the following paper :
• Shiva Kintali. “Directed Width Parameters and Circumference of Digraphs“. Preprint available on my publications page.
We prove that the directed treewidth, DAG-width and Kelly-width of a digraph are bounded above by its circumference plus one. This generalizes a theorem of Birmele stating that the treewidth of an undirected graph is at most its circumference.
Theorem : Let G be a digraph of circumference l. Then the directed treewidth, DAG-width and Kelly-width of G are at most l + 1.
The above theorem can be seen as a mini mini mini directed grid minor theorem. I will be using this theorem in future papers to make progress towards a directed grid minor theorem. Stay tuned !!
# Open problems for 2014
Wish you all a Very Happy New Year. Here is a list of my 10 favorite open problems for 2014. They belong to several research areas inside discrete mathematics and theoretical computer science. Some of them are baby steps towards resolving much bigger open problems. May this new year shed new light on these open problems.
• 2. Optimization : Improve the approximation factor for the undirected graphic TSP. The best known bound is 7/5 by Sebo and Vygen.
• 3. Algorithms : Prove that the tree-width of a planar graph can be computed in polynomial time (or) is NP-complete.
• 4. Fixed-parameter tractability : Treewidth and Pathwidth are known to be fixed-parameter tractable. Are directed treewidth/DAG-width/Kelly-width (generalizations of treewidth) and directed pathwidth (a generalization of pathwidth) fixed-parameter tractable ? This is a very important problem to understand the algorithmic and structural differences between undirected and directed width parameters.
• 5. Space complexity : Is Planar ST-connectvity in logspace ? This is perhaps the most natural special case of the NL vs L problem. Planar ST-connectivity is known to be in $UL \cap coUL$. Recently, Imai, Nakagawa, Pavan, Vinodchandran and Watanabe proved that it can be solved simultaneously in polynomial time and approximately O(√n) space.
• 6. Metric embedding : Is the minor-free embedding conjecture true for partial 3-trees (graphs of treewidth 3) ? Minor-free conjecture states that “every minor-free graph can be embedded in $l_1$ with constant distortion. The special case of planar graphs also seems very difficult. I think the special case of partial 3-trees is a very interesting baby step.
• 7. Structural graph theory : Characterize pfaffians of tree-width at most 3 (i.e., partial 3-trees). It is a long-standing open problem to give a nice characterization of pfaffians and design a polynomial time algorithm to decide if an input graph is a pfaffian. The special of partial 3-trees is an interesting baby step.
• 8. Structural graph theory : Prove that every minimal brick has at least four vertices of degree three. Bricks and braces are defined to better understand pfaffians. The characterization of pfaffian braces is known (more generally characterization of bipartite pfaffians is known). To understand pfaffians, it is important to understand the structure of bricks. Norine,Thomas proved that every minimal brick has at least three vertices of degree three and conjectured that every minimal brick has at least cn vertices of degree three.
• 9. Communication Complexity : Improve bounds for the log-rank conjecture. The best known bound is $O(\sqrt{rank})$
• 10. Approximation algorithms : Improve the approximation factor for the uniform sparsest cut problem. The best known factor is $O(\sqrt{logn})$.
Here are my conjectures for 2014 :)
• Weak Conjecture : at least one of the above 10 problems will be resolved in 2014.
• Conjecture : at least five of the above 10 problems will be resolved in 2014.
• Strong Conjecture : All of the above 10 problems will be resolved in 2014.
Have fun !!
# PolyTopix
In the last couple of years, I developed some (research) interest in recommendation algorithms and speech synthesis. My interests in these areas are geared towards developing an automated personalized news radio.
Almost all of us are interesting in consuming news. In this internet age, there is no dearth of news sources. Often we have too many sources. We tend to “read” news from several sources / news aggregators, spending several hours per week. Most of the time we are simply interested in the top and relevant headlines.
PolyTopix is my way of simplifying the process of consuming top and relevant news. The initial prototype is here. The website “reads” several news tweets (collected from different sources) and ordered based on a machine learning algorithm. Users can login and specify their individual interests (and zip code) to narrow down the news.
Try PolyTopix let me know your feedback. Here are some upcoming features :
• Automatically collect weather news (and local news) based on your location.
• Reading more details of most important news.
• News will be classified as exciting/sad/happy etc., (based on a machine learning algorithm) and read with the corresponding emotional voice.
Essentially PolyTopix is aimed towards a completely automated and personalized news radio, that can “read” news from across the world anytime with one click.
————————————————————————————————————————
# Forbidden Directed Minors and Kelly-width
Today’s post is about the following paper, a joint work with Qiuyi Zhang, one of my advisees. Qiuyi Zhang is an undergraduate (rising senior) in our mathematics department.
• Shiva Kintali, Qiuyi Zhang. Forbidden Directed Minors and Kelly-width. (Preprint available on my publications page)
It is well-known that an undirected graph is a partial 1-tree (i.e., a forest) if and only if it has no $K_3$ minor. We generalized this characterization to partial 1-DAGs. We proved that partial 1-DAGs are characterized by three forbidden directed minors, $K_3, N_4$ and $M_5$, shown in the following figure. We named the last two graphs as $N_4$ and $M_5$ because their bidirected edges resemble the letters N and M.
Partial k-trees characterize bounded treewidth graphs. Similarly, partial k-DAGs characterize bounded Kelly-width digraphs. Kelly-width is the best known generalization of treewidth to digraphs.
As mentioned in the paper, I have a series of upcoming papers (called Directed Minors) making progress towards a directed graph minor theorem (i.e., all digraphs are well-quasi-ordered by the directed minor relation). For more details of the directed minor relation, read the current paper. I will post about the upcoming results as and when the preprints are ready.
# Book Review of “Boosting : Foundations and Algorithms”
Following is my review of Boosting : Foundations and Algorithms (by Robert E. Schapire and Yoav Freund) to appear in the SIGACT book review column soon.
—————————————————————————————————————-
Book : Boosting : Foundations and Algorithms (by Robert E. Schapire and Yoav Freund)
Reviewer : Shiva Kintali
Introduction
You have k friends, each one earning a small amount of money (say 100 dollars) every month by buying and selling stocks. One fine evening, at a dinner conversation, they told you their individual “strategies” (after all, they are your friends). Is it possible to “combine” these individual strategies and make million dollars in an year, assuming your initial capital is same as your average friend ?
You are managing a group of k “diverse” software engineers each one with only an “above-average” intelligence. Is it possible to build a world-class product using their skills ?
The above scenarios give rise to fundamental theoretical questions in machine learning and form the basis of Boosting. As you may know, the goal of machine learning is to build systems that can adapt to their environments and learn from their experience. In the last five decades, machine learning has impacted almost every aspect of our life, for example, computer vision, speech processing, web-search, information retrieval, biology and so on. In fact, it is very hard to name an area that cannot benefit from the theoretical and practical insights of machine learning.
The answer to the above mentioned questions is Boosting, an elegant method for driving down the error of the combined classifier by combining a number of weak classifiers. In the last two decades, several variants of Boosting are discovered. All these algorithms come with a set of theoretical guarantees and made a deep practical impact on the advances of machine learning, often providing new explanations for existing prediction algorithms.
Boosting : Foundations and Algorithms, written by the inventors of Boosting, deals with variants of AdaBoost, an adaptive boosting method. Here is a quick explanation of the basic version of AdaBoost.
AdaBoost makes iterative calls to the base learner. It maintains a distribution over training examples to choose the training sets provided to the base learner on each round. Each training example is assigned a weight, a measure of importance of correctly classifying an example on the current round. Initially, all weights are set equally. On each round, the weights of incorrectly classified examples are increased so that, “hard” examples get successively higher weight. This forces the base learner to focus its attention on the hard example and drive down the generalization errors.
AdaBoost is fast and easy to implement and the only parameter to tune is the number of rounds. The actual performance of boosting is dependent on the data.
Summary
Chapter 1 provides a quick introduction and overview of Boosting algorithms with practical examples. The rest of the book is divided into four major parts. Each part is divided into 3 to 4 chapters.
Part I studies the properties and effectiveness of AdaBoost and theoretical aspects of minimizing its training and generalization errors. It is proved that AdaBoost drives the training error down very fast (as a function of the error rates of the weak classifiers) and the generalization error arbitrarily close to zero. Basic theoretical bounds on the generalization error show that AdaBoost overfits, however empirical studies show that AdaBoost does not overfit. To explain this paradox, a margin-based analysis is presented to explain the absence of overfitting.
Part II explains several properties of AdaBoost using game-theoretic interpretations. It is shown that the principles of Boosting are very intimately related to the classic min-max theorem of von Neumann. A two-player (the boosting algorithm and the weak learning algorithm) game is considered and it is shown that AdaBoost is a special case of a more general algorithm for playing a repeated game. By reversing the roles of the players, a solution is obtained for the online prediction model thus establishing a connection between Boosting and online learning. Loss minimization is studied and AdaBoost is interpreted as an abstract geometric framework for optimizing a particular objective function. More interestingly, AdaBoost is viewed as a special case of more general methods for optimization of an objective function such as coordinate descent and functional gradient descent.
Part III explains several methods of extending AdaBoost to handle classifiers with more than two output classes. AdaBoost.M1, AdaBoost.MH and AdaBoost.MO are presented along with their theoretical analysis and practical applications. RankBoost, an extension of AdaBoost to study ranking problems is studied. Such an algorithm is very useful, for example, to rank webpages based on their relevance to a given query.
Part IV is dedicated to advanced theoretical topics. Under certain assumptions, it is proved that AdaBoost can handle noisy-data and converge to the best possible classifier. An optimal boost-by-majority algorithm is presented. This algorithm is then modified to be adaptive leading to an algorithm called BrownBoost.
Many examples are given throughout the book to illustrate the empirical performance of the algorithms presented. Every chapter ends with Summary and Bibliography mentioning the related publications. There are well-designed exercises at the end of every chapter. Appendix briefly outlines some required mathematical background.
Opinion
Boosting book is definitely a very good reference text for researchers in the area of machine learning. If you are new to machine learning, I encourage you to read an introductory machine learning book (for example, Machine Learning by Tom M. Mitchell) to better understand and appreciate the concepts. In terms of being used in a course, a graduate-level machine learning course can be designed from the topics covered in this book. The exercises in the book can be readily used for such a course.
Overall this book is a stimulating learning experience. It has provided me new perspectives on theory and practice of several variants of Boosting algorithms. Most of the algorithms in this book are new to me and I had no difficulties following the algorithms and the corresponding theorems. The exercises at the end of every chapter made these topics much more fun to learn.
The authors did a very good job compiling different variants of Boosting algorithms and achieved a nice balance between theoretical analysis and practical examples. I highly recommend this book for anyone interested in machine learning.
—————————————————————————————————————-
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1. ## rate
1. (a) A car traveling at a speed of v can brake to an emergency stop in a distance x. Assuming all other driving conditions arte all similar, if traveling speed of the car doubles, the stopping distance will be (1) 2x, or (2) 2x, or (3) 4x: (b) a driver traveling at 40.0 km/h in a school zone can brake to an emergency stop in 3.00 m. What would be the braking distance if the car were traveling at 60.0 km/h?
2. A student drops a ball from the top of a building; it takes 2.8 s for the ball to reach the ground. (a) What is the ball's speed just before hitting the ground? (b) What is the height of the building?
3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s accidentally drops a camera out the window when the helicopter is 60 m above the ground. (a) How long will it take the camera to reach the ground? (b) What will it's speed be when it hits?
2. I'm sure this requires nothing more than some formulas.
3. Hello, John!
For the last two, we are expected to know the "free fall" formula: . $y \; = \; h_o + v_ot - gt^2$
. . where: . $\begin{array}{ccc}h_o & = & \text{initial height} \\ v_o & = &\text{initial velocity} \\ g & = & \text{gravitational constant} \\ y & = & \text{height of object} \end{array}$
Note: $g$ is usually 16 ft/sec² or 4.9 m/sec².
2. A student drops a ball from the top of a building;
it takes 2.8 s for the ball to reach the ground.
(a) What is the ball's speed just before hitting the ground?
(b) What is the height of the building?
We have: . $v_o = 0$ . (the ball is dropped, not thrown).
The equation is: . $y \;=\;h_o - 16t^2$
(a) The velocity is given by the derivative: . $v(t) \:=\:y' \:=\:-32t$
Then: . $v(2.8) \:=\:-32(2.8) \:=\:-89.6$
The ball's speed on impact is 89.6 feet per second (downward, of course).
(b) When the ball strikes the ground, its height is zero.
. . We have: . $0 \;=\;h_o - 16\cdot2.8^2\quad\Rightarrow\quad h_o \:=\:125.44$
The height of the building is 125.44 feet.
3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s
accidentally drops a camera out the window when the helicopter is 60 m above the ground.
(a) How long will it take the camera to reach the ground?
(b) What will its speed be when it hits?
We are given: . $h_o = 60,\;v_o = 12.5$
The equation is: . $y \;=\;60 + 12.5t - 4.9t^2$
(a) "reach the ground" means: $y = 0$
We have: . $60 + 12.5t - 4.9t^2 \:=\:0$
. . which factors: . $(t - 5)(4.9t + 12) \:=\:0$
. . and has the positive root: . $t = 5$
Therefore, it takes the camera 5 seconds to crash to the ground.
(b) The velocity is: . $v(t) \;=\;y' \;=\;12.5 - 9.8t$
. . Hence: . $v(5)\;=\;12.5 - 9.8(5) \;=\;-36.5$
Therefore, its speed on impact is 36.5 meters per second.
4. aaaand...
Problem 1,a
We are told that the square of velocity is proportional to the stopping distance. (look at topsquark's post for its justification)
$v^2 \propto x$ so $(2v)^2=4v \propto 4x$.
4x
Problem 1,b
$40^2=1600\frac{km}{h} \propto 3.00m$
Using this information, we can find the stopping distance of the same car at 60.0 km/h by using a proportion and solving for x.
$\frac{3}{1600} = \frac{x}{3600} \Longleftrightarrow x=6.75m$
These veterans are just too fast...
5. Originally Posted by rualin
These veterans are just too fast...
Soroban is not the fastest so you haven't seen anything yet lol j/m. Soroban is one of the most cunning and interesting.
6. Originally Posted by rualin
Problem 1,a
We are told that velocity is directly proportional to the stopping distance.
No we weren't. Try this programme:
Find the acceleration for when it starts at speed v.
The acceleration will be the same as this for when it starts at 2v.
In both cases, apply the formula:
$v^2 = v_0^2 + 2a(x - x_0)$
-Dan
7. Thanks, topsquark! I realize my mistake and have corrected it... I hope.
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How to typeset texbook.tex
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More options May 17 2012, 8:27 pm
Newsgroups: comp.text.tex
From: rdawson <randall...@gmail.com>
Date: Thu, 17 May 2012 17:27:08 -0700 (PDT)
Local: Thurs, May 17 2012 8:27 pm
Subject: How to typeset texbook.tex
I realize this is circular, if you dont know how you shouldnt do it.
But I'm thinking its the first step at seeing a large (and useful)
book typeset here, not to post a pdf to the net.
I am a CTAN member...
Can another member help me out to remove the things that are
preventing this to process in Texworks?
Randy
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More options May 18 2012, 12:28 am
Newsgroups: comp.text.tex
From: Marc van Dongen <don...@cs.ucc.ie>
Date: Thu, 17 May 2012 21:28:25 -0700 (PDT)
Local: Fri, May 18 2012 12:28 am
Subject: Re: How to typeset texbook.tex
On Friday, May 18, 2012 1:27:08 AM UTC+1, rdawson wrote:
> I realize this is circular, if you dont know how you shouldnt do it.
> But I'm thinking its the first step at seeing a large (and useful)
> book typeset here, not to post a pdf to the net.
> I am a CTAN member...
> Can another member help me out to remove the things that are
> preventing this to process in Texworks?
You are not supposed to typeset it. It is mentioned explicitly in the source on lines 3--6, which state:
% The file is distributed only for people to see its examples of TeX input,
% not for use in the preparation of books like The TeXbook.
% Permission for any other use of this file must be obtained in writing
Regards,
Marc van Dongen
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More options May 18 2012, 3:02 am
Newsgroups: comp.text.tex
From: Robin Fairbairns <r...@cl.cam.ac.uk>
Date: Fri, 18 May 2012 08:02:41 +0100
Local: Fri, May 18 2012 3:02 am
Subject: Re: How to typeset texbook.tex
rdawson <randall...@gmail.com> writes:
> I realize this is circular, if you dont know how you shouldnt do it.
not quite, as you've been told
> But I'm thinking its the first step at seeing a large (and useful)
> book typeset here, not to post a pdf to the net.
> I am a CTAN member...
itym "user". we don't have membership, just a couple of admins and a
couple of machines (plus lots of mirrors helping us out with traffic).
> Can another member help me out to remove the things that are
> preventing this to process in Texworks?
i've once seen it being processed (at a tug meeting, using the then
shiny new pdftex). before the process started, the speaker assured the
audience the project had been granted permission to do this.
--
Robin Fairbairns, Cambridge
sorry about all this posting. i'll go back to sleep in a bit.
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More options May 18 2012, 6:27 am
Newsgroups: comp.text.tex
From: Giorgos Keramidas <keram...@ceid.upatras.gr>
Date: Fri, 18 May 2012 12:27:00 +0200
Subject: Re: How to typeset texbook.tex
On Thu, 17 May 2012 17:27:08 -0700 (PDT), rdawson <randall...@gmail.com> wrote:
> I realize this is circular, if you dont know how you shouldnt do it.
> But I'm thinking its the first step at seeing a large (and useful)
> book typeset here, not to post a pdf to the net.
> I am a CTAN member...
> Can another member help me out to remove the things that are
> preventing this to process in Texworks?
It's possible to do it. The macros you need are there, in CTAN, and the
editing required to the 1-2 places where 'guard macros' are in place is
really minimal.
Note, however, that you are *not* supposed to do this without explicit
permission from the publisher and even if you do typeset a local version
it's lacking some really _important_ bits: like the nicely drawn figures
of the original.
The important question is: why do you need to typeset your own version?
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More options May 18 2012, 3:13 pm
Newsgroups: comp.text.tex
Date: Fri, 18 May 2012 12:13:36 -0700 (PDT)
Local: Fri, May 18 2012 3:13 pm
Subject: Re: How to typeset texbook.tex
On May 17, 8:27 pm, rdawson <randall...@gmail.com> wrote:
> I realize this is circular, if you dont know how you shouldnt do it.
> But I'm thinking its the first step at seeing a large (and useful)
> book typeset here, not to post a pdf to the net.
> I am a CTAN member...
> Can another member help me out to remove the things that are
> preventing this to process in Texworks?
If you really need to see how a book is typeset in TeX I would suggest
- memman.tex (should be on CTAN --- the manual for memoir --- I
believe it's a good example of LaTeX style)
- _TeX for the Impatient --- http://savannah.gnu.org/projects/teximpatient/
--- eplain, and an interesting examination of how much one used to
have to do oneself --- the index in particular
- Making TeX Work --- http://makingtexwork.sourceforge.net/mtw/ ---
excellent book very much in need of up-dating
A fair number of the free math books are (of course) authored in LaTeX
and should be good examples:
Don't typeset _The TeXbook_ though, it's not allowed and is a
William
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More options May 18 2012, 6:26 pm
Newsgroups: comp.text.tex
From: Peter Flynn <pe...@silmaril.ie>
Date: Fri, 18 May 2012 23:26:53 +0100
Local: Fri, May 18 2012 6:26 pm
Subject: Re: How to typeset texbook.tex
On 18/05/12 11:27, Giorgos Keramidas wrote:
What might be a useful demonstration would be if TUG -- with appropriate
permissions from Don and A-W -- could video-record the screen while
running the TeXbook through TeX.
That might safisfy the curious, and act as a record of the fact for
posterity.
There are plenty of long [La]TeX documents in the public domain AFAIK,
if all the OP wants is a demo of a big book being typeset.
///Peter
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More options May 18 2012, 7:13 pm
Newsgroups: comp.text.tex
From: "Nasser M. Abbasi" <n...@12000.org>
Date: Fri, 18 May 2012 18:13:57 -0500
Local: Fri, May 18 2012 7:13 pm
Subject: Re: How to typeset texbook.tex
This got me thinking that it would be nice to have a list
of books that were typesetted using Latex?
I mean officially published books.
I always wonder when I buy or see a book (technical book mainly)
if Latex was used to typeset it, but I do not think Latex is
used to publish books by the major publishers (like mcGraw hills
and such). Most books also do not put this information
inside the cover for some reason.
Sometimes one can tell by the quality of the math and pages
if it is latex or not.
I know for example that I can tell right away if MS Word was
usedto write the book :)
--Nasser
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More options May 18 2012, 7:21 pm
Newsgroups: comp.text.tex
From: Lee Rudolph <lrudo...@panix.com>
Date: Fri, 18 May 2012 23:21:17 +0000 (UTC)
Local: Fri, May 18 2012 7:21 pm
Subject: Re: How to typeset texbook.tex
"Nasser M. Abbasi" <n...@12000.org> writes:
>This got me thinking that it would be nice to have a list
>of books that were typesetted using Latex?
>I mean officially published books.
>I always wonder when I buy or see a book (technical book mainly)
>if Latex was used to typeset it, but I do not think Latex is
>used to publish books by the major publishers (like mcGraw hills
>and such). Most books also do not put this information
>inside the cover for some reason.
Sometime in August (if all continues to go well) one major
publisher, namely, Taylor & Francis, is going to publish one
(technical) book that was typeset--over major initial objections
and resistance--with pdflatex :). Whether this information gets
"inside the cover" (or into a colophon) is still, I think, up in
the air. We shall see.
Lee Rudolph
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More options May 18 2012, 7:46 pm
Newsgroups: comp.text.tex
From: jon <jonwrobin...@gmail.com>
Date: Fri, 18 May 2012 16:46:31 -0700 (PDT)
Local: Fri, May 18 2012 7:46 pm
Subject: Re: How to typeset texbook.tex
On May 18, 7:13 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> I always wonder when I buy or see a book (technical book mainly)
> if Latex was used to typeset it, but I do not think Latex is
> used to publish books by the major publishers (like mcGraw hills
> and such). Most books also do not put this information
> inside the cover for some reason.
cambridge university press uses latex sometimes for books in the
humanities. three that come to mind are albert derolez's book on
latin palaeography (2003), and the cambridge companions to william of
ockham (1999) and john duns scotus (2003). the information is given
where all the copyright stuff is. (note: most cambridge companions
that i've looked at do /not/ mention using latex, but some do. no
idea who makes the decision.)
cheers,
jon.
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More options May 19 2012, 1:26 am
Newsgroups: comp.text.tex
From: zappathustra <zappathus...@free.fr>
Date: Sat, 19 May 2012 07:26:26 +0200
Local: Sat, May 19 2012 1:26 am
Subject: Re: How to typeset texbook.tex
jon <jonwrobin...@gmail.com> a écrit:
> On May 18, 7:13 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> > I always wonder when I buy or see a book (technical book mainly)
> > if Latex was used to typeset it, but I do not think Latex is
> > used to publish books by the major publishers (like mcGraw hills
> > and such). Most books also do not put this information
> > inside the cover for some reason.
One easy way to spot it is the font: many, many books made with TeX
stick to Computer Modern, and you can't miss it.
> cambridge university press uses latex sometimes for books in the
> humanities. three that come to mind are albert derolez's book on
> latin palaeography (2003), and the cambridge companions to william of
> ockham (1999) and john duns scotus (2003). the information is given
> where all the copyright stuff is.
Interestingly, I own a book in paperback (Regularity in semantic change)
from CUP, in which it is not indicated that it was made with LaTeX on
the copyright page; but I've seen the hardback in a library and there
it says so.
Best,
Paul
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More options May 19 2012, 3:22 am
Newsgroups: comp.text.tex
From: Marc van Dongen <don...@cs.ucc.ie>
Date: Sat, 19 May 2012 00:22:58 -0700 (PDT)
Local: Sat, May 19 2012 3:22 am
Subject: Re: How to typeset texbook.tex
On Saturday, May 19, 2012 6:26:26 AM UTC+1, zappathustra wrote:
> jon <jonwrobin...@gmail.com> a écrit:
> > On May 18, 7:13 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> > > I always wonder when I buy or see a book (technical book mainly)
> > > if Latex was used to typeset it, but I do not think Latex is
> > > used to publish books by the major publishers (like mcGraw hills
> > > and such). Most books also do not put this information
> > > inside the cover for some reason.
> One easy way to spot it is the font: many, many books made with TeX
> stick to Computer Modern, and you can't miss it.
This is the time to eat your hat:
Of course most books about TeX, LaTeX and friends are usually typeset with TeX, LaTeX and friends. See http://www.tug.org/books/ for a list of books that have been reviewed for TUGboat.
Regards,
Marc van Dongen
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More options May 19 2012, 7:09 am
Newsgroups: comp.text.tex
From: Peter Flynn <peter.n...@m.silmaril.ie>
Date: Sat, 19 May 2012 12:09:45 +0100
Local: Sat, May 19 2012 7:09 am
Subject: Re: How to typeset texbook.tex
On 19/05/12 00:13, Nasser M. Abbasi wrote:
> This got me thinking that it would be nice to have a list
> of books that were typesetted using Latex?
> I mean officially published books.
> I always wonder when I buy or see a book (technical book mainly)
> if Latex was used to typeset it, but I do not think Latex is
> used to publish books by the major publishers (like mcGraw hills
> and such).
We typeset books for publishers, but mostly non-math books, for example
http://www.amazon.co.uk/Critical-Turns-Theory-Directions-Internationa...
and
and
http://www.amazon.com/Bede-Liverpool-University-Translated-Historians...
but occasionally with math like
http://www.lannoo.be/content/Lannoo/wbnl/listview/1/index.jsp?titelco...
> Most books also do not put this information
> inside the cover for some reason.
Fashion. Most publishers have stopped putting details of the typeface
inside the cover as well.
There was an idea that TUG should start collecting the details of books
typeset with [La]TeX that were *not* books *about* [La]TeX. Maybe this
should be revived...something to discuss at the conference in Boston?
///Peter
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More options May 19 2012, 7:10 am
Newsgroups: comp.text.tex
From: Peter Flynn <peter.n...@m.silmaril.ie>
Date: Sat, 19 May 2012 12:10:47 +0100
Local: Sat, May 19 2012 7:10 am
Subject: Re: How to typeset texbook.tex
On 19/05/12 08:22, Marc van Dongen wrote:
I'd missed that. I'll add ours.
///Peter
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More options May 19 2012, 12:09 pm
Newsgroups: comp.text.tex
Date: Sat, 19 May 2012 18:09:32 +0200
Local: Sat, May 19 2012 12:09 pm
Subject: Re: How to typeset texbook.tex
"Nasser M. Abbasi" <n...@12000.org> writes:
> This got me thinking that it would be nice to have a list
> of books that were typesetted using Latex?
> I mean officially published books.
> I always wonder when I buy or see a book (technical book mainly)
> if Latex was used to typeset it, but I do not think Latex is
> used to publish books by the major publishers (like mcGraw hills
> and such). Most books also do not put this information
> inside the cover for some reason.
- Nicolai Josuttis: the C++ standard library (as an example for a book
that is not math-heavy and does not deal with the natural sciences)
> Sometimes one can tell by the quality of the math and pages
> if it is latex or not.
Math is not the best distinguisher any more since math typesetting is
nowadays of equal or better quality in Word. Other distinguishing
options are:
- Microtypography like margin kerning (however, InDesign supports this,
too)
- Line breaking and hyphenation (Word still uses a naive greedy
algorithm here, InDesign also uses some global optimization heuristic)
I'm not sure about InDesign's current status concerning math. If it has
already implemented Word's algorithm, its output quality should be at
least as good as LuaTeX's in all respect.
--
Philipp Stephani
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More options May 19 2012, 1:27 pm
Newsgroups: comp.text.tex
From: "Nasser M. Abbasi" <n...@12000.org>
Date: Sat, 19 May 2012 12:27:33 -0500
Local: Sat, May 19 2012 1:27 pm
Subject: Re: How to typeset texbook.tex
On 5/19/2012 11:09 AM, Philipp Stephani wrote:
> Math is not the best distinguisher any more since math typesetting is
> nowadays of equal or better quality in Word.
You think equation editor in word produces as good math as Latex? I
have not tried equation editor for years. I should make
a comparison on this soon and add it to my current list.
> Other distinguishing
> options are:
> - Microtypography like margin kerning (however, InDesign supports this,
> too)
> - Line breaking and hyphenation (Word still uses a naive greedy
> algorithm here, InDesign also uses some global optimization heuristic)
> I'm not sure about InDesign's current status concerning math. If it has
> already implemented Word's algorithm, its output quality should be at
> least as good as LuaTeX's in all respect.
I looked at inDesign once for doing math. Need a 3rd party tool
to add equations, such as mathmagic
or mathtype
and such. They work like the built-in equation editor
in word 2010.
Do you of an example on the web that has lots of math and
was generated from inDesign document? I'd like to
see how inDesign to HTML conversion with math looks like.
I tried framemaker long time ago, and liked the overall
document design, but it had little math support, and
I think indesign is supposed to replace framemaker.
Anyway, in the end, I myself like Latex, since it is
also plain text. I am trying to get away from
anything binary and blackboxed in, as now I can see
everything.
thanks for the info.
--Nasser
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More options May 19 2012, 2:43 pm
Newsgroups: comp.text.tex
From: Khaled Hosny <khaledho...@eglug.org>
Date: Sat, 19 May 2012 11:43:47 -0700 (PDT)
Local: Sat, May 19 2012 2:43 pm
Subject: Re: How to typeset texbook.tex
And who typeset books in Word? :) It is either an Adobe thing or Quark, neither do math at all (you need cumbersome "plugins" with not so good quality).
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More options May 19 2012, 3:14 pm
Newsgroups: comp.text.tex
From: Peter Flynn <peter.n...@m.silmaril.ie>
Date: Sat, 19 May 2012 20:14:13 +0100
Local: Sat, May 19 2012 3:14 pm
Subject: Re: How to typeset texbook.tex
On 19/05/12 19:43, Khaled Hosny wrote:
[...]
> And who typeset books in Word? :)
Unfortunately, LOTS of smaller publishers do this, relying on the
author, a stylesheet, and some office junior as "editor".
It's cheap, fast, and MOST readers will not notice it, because they are
now acclimated for over a generation to this level of quality.
And most publishers have never heard of LaTeX anyway. Or if they have,
they run away fast, because they have seen what an inexperienced author
can do :-) and they think that that is the ONLY thing LaTeX can do.
///Peter
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More options May 19 2012, 3:22 pm
Newsgroups: comp.text.tex
From: "Nasser M. Abbasi" <n...@12000.org>
Date: Sat, 19 May 2012 14:22:54 -0500
Local: Sat, May 19 2012 3:22 pm
Subject: Re: How to typeset texbook.tex
On 5/19/2012 2:14 PM, Peter Flynn wrote:
> On 19/05/12 19:43, Khaled Hosny wrote:
> [...]
>> And who typeset books in Word? :)
> Unfortunately, LOTS of smaller publishers do this, relying on the
> author, a stylesheet, and some office junior as "editor".
> It's cheap, fast, and MOST readers will not notice it, because they are
> now acclimated for over a generation to this level of quality.
> And most publishers have never heard of LaTeX anyway. Or if they have,
> they run away fast, because they have seen what an inexperienced author
> can do :-) and they think that that is the ONLY thing LaTeX can do.
> ///Peter
That is why it is even more important if there is a central
web site of links to all published books typesetted using
Latex (with small chapter examples from these books) like
someone posted here earlier from one publisher.
Then all you have to do to "convince" someone to use Latex to
publish a book with is to point them to this site to see
for themselves.
A picture is worth a thousands words :)
Actually, using Latex to publish a book makes even more
sense when there is little or no math. For me the hardest
thing about using Latex is to typeset the math equations.
If there is no math or little math, then using Latex seems
to make more sense in this case, since it become even easier
to use.
--Nasser
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More options May 19 2012, 4:17 pm
Newsgroups: comp.text.tex
From: Dan <lueck...@uark.edu>
Date: Sat, 19 May 2012 13:17:29 -0700 (PDT)
Local: Sat, May 19 2012 4:17 pm
Subject: Re: How to typeset texbook.tex
On May 19, 11:09 am, Philipp Stephani <p.stepha...@googlemail.com>
wrote:
> "Nasser M. Abbasi" <n...@12000.org> writes:
> > Sometimes one can tell by the quality of the math and pages
> > if it is latex or not.
> Math is not the best distinguisher any more since math typesetting is
> nowadays of equal or better quality in Word. Other distinguishing
> options are:
> - Microtypography like margin kerning (however, InDesign supports this,
> too)
> - Line breaking and hyphenation (Word still uses a naive greedy
> algorithm here, InDesign also uses some global optimization heuristic)
Here's another:
- words run together, where the author used a macro and forgot that
spaces after it are ignored. ;-)
Dan
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More options May 19 2012, 4:39 pm
Newsgroups: comp.text.tex
From: jon <jonwrobin...@gmail.com>
Date: Sat, 19 May 2012 13:39:50 -0700 (PDT)
Local: Sat, May 19 2012 4:39 pm
Subject: Re: How to typeset texbook.tex
On May 19, 3:22 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
> Actually, using Latex to publish a book makes even more
> sense when there is little or no math. For me the hardest
> thing about using Latex is to typeset the math equations.
> If there is no math or little math, then using Latex seems
> to make more sense in this case, since it become even easier
> to use.
in general i think this is true (e.g., the cambridge companions i
mentioned the other day), /especially/ now that we have biblatex-biber
to use for complex bibliographical/citation rules. where it is still
difficult (but still better than most options for most things) is
complex documents like critical editions with facing-page translations
--- though really only for non-latin alphabet languages. for
instance, an arabic-language critical edition with a facing page
english translation is very difficult, even, it seems, with all the
post-pdftex options out there. (luckily, i've never had to worry
about anything more complicated than a tiny bit of greek here and
there....)
in the humanities, my problem is not that latex doesn't do a better
job or any of that, it's that no one cares to use anything other than
word. so i write in latex, and then get stuck translating it to
word.¹ and here the hardest part is not getting it out of latex
(thanks to tex4ht!), but going from libreoffice/openoffice (.odt) to
word (.doc). the formatting of things like footnotes and cross-
references are invariably screwed up and unfixable unless you have
access to word after saving it to .doc (which i don't unless i use a
computer in the library).
cheers,
jon.
¹. luckily, the publisher brill is letting me produce 'camera-ready'
copy for my book as long as i follow their formatting directives.
dealing with converting a whole book to word is a nightmare i don't
want to contemplate!
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More options May 19 2012, 5:33 pm
Newsgroups: comp.text.tex
From: Robin Fairbairns <r...@cl.cam.ac.uk>
Date: Sat, 19 May 2012 22:33:00 +0100
Subject: Re: How to typeset texbook.tex
jon <jonwrobin...@gmail.com> writes:
> On May 18, 7:13 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
>> I always wonder when I buy or see a book (technical book mainly)
>> if Latex was used to typeset it, but I do not think Latex is
>> used to publish books by the major publishers (like mcGraw hills
>> and such). Most books also do not put this information
>> inside the cover for some reason.
> cambridge university press uses latex sometimes for books in the
> humanities. three that come to mind are albert derolez's book on
> latin palaeography (2003), and the cambridge companions to william of
> ockham (1999) and john duns scotus (2003). the information is given
> where all the copyright stuff is. (note: most cambridge companions
> that i've looked at do /not/ mention using latex, but some do. no
> idea who makes the decision.)
the darwin letters project books (also published by cup) are now done
with latex; i helped a bit transferring the work from a workflow on
ms-dos to one using latex on something unix-y.
(they're really lovely books, and great fun to read.)
--
Robin Fairbairns, Cambridge
sorry about all this posting. i'll go back to sleep in a bit.
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More options May 19 2012, 5:48 pm
Newsgroups: comp.text.tex
From: Robin Fairbairns <r...@cl.cam.ac.uk>
Date: Sat, 19 May 2012 22:48:28 +0100
Local: Sat, May 19 2012 5:48 pm
Subject: Re: How to typeset texbook.tex
Peter Flynn <peter.n...@m.silmaril.ie> writes:
> On 19/05/12 19:43, Khaled Hosny wrote:
> [...]
>> And who typeset books in Word? :)
> Unfortunately, LOTS of smaller publishers do this, relying on the
> author, a stylesheet, and some office junior as "editor".
and the results are often better than corresponding books that i used
when doing my degree. one book that sticks in the mind was typewritten
(with a golfball typewriter, judging by the quality), and the maths was
all done with some special ball that had integral signs around the size
of an upper-case letter.) it was on galois theory; i never really got
to grips with that.
> It's cheap, fast, and MOST readers will not notice it, because they are
> now acclimated for over a generation to this level of quality.
> And most publishers have never heard of LaTeX anyway. Or if they have,
> they run away fast, because they have seen what an inexperienced author
> can do :-) and they think that that is the ONLY thing LaTeX can do.
sigh. there's also the problem that, if they want to retain the source
for a future edition, they'll (with probability closely approaching 1)
want the source in xml according to some established dtd. latex authors
tend not to do that...
(present company excepted.)
--
Robin Fairbairns, Cambridge
sorry about all this posting. i'll go back to sleep in a bit.
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More options May 19 2012, 8:00 pm
Newsgroups: comp.text.tex
From: Peter Flynn <peter.n...@m.silmaril.ie>
Date: Sun, 20 May 2012 01:00:28 +0100
Local: Sat, May 19 2012 8:00 pm
Subject: Re: How to typeset texbook.tex
On 19/05/12 21:39, jon wrote:
> On May 19, 3:22 pm, "Nasser M. Abbasi" <n...@12000.org> wrote:
>> Actually, using Latex to publish a book makes even more
>> sense when there is little or no math. For me the hardest
>> thing about using Latex is to typeset the math equations.
>> If there is no math or little math, then using Latex seems
>> to make more sense in this case, since it become even easier
>> to use.
> in general i think this is true (e.g., the cambridge companions i
> mentioned the other day), /especially/ now that we have biblatex-biber
> to use for complex bibliographical/citation rules.
I am still stuck with BiBTeX: last time I tried biblatex it wasn't
> where it is still
> difficult (but still better than most options for most things) is
> complex documents like critical editions with facing-page translations
> --- though really only for non-latin alphabet languages. for
> instance, an arabic-language critical edition with a facing page
> english translation is very difficult, even, it seems, with all the
> post-pdftex options out there. (luckily, i've never had to worry
> about anything more complicated than a tiny bit of greek here and
> there....)
I'm just in the middle of a facing-page edition (Middle Irish and
English), and ledmac seems to be doing fine.
> in the humanities, my problem is not that latex doesn't do a better
> job or any of that, it's that no one cares to use anything other than
> word. so i write in latex, and then get stuck translating it to
> word.¹ and here the hardest part is not getting it out of latex
> (thanks to tex4ht!), but going from libreoffice/openoffice (.odt) to
> word (.doc). the formatting of things like footnotes and cross-
> references are invariably screwed up and unfixable unless you have
> access to word after saving it to .doc (which i don't unless i use a
> computer in the library).
I've banged this drum many times before, but it might be worth repeating.
Author in XML, not LaTeX. Transform the XML to LaTeX with XSLT. That way
it's testable, reproducible, visible, and manageable under machine
control. The big move in the Humanities is to using TEI XML for markup,
so this is really just going with the flow. Yes, it takes some learning,
and yes, it takes some time to set up a good workflow, but if you are
using this for anything critical or persistent, it's time and money well
spent.
If you're not the author, demand Word .docx or ODT documents ONLY, with
a stylesheet using Named Styles. If the styles have been applied
correctly and consistently, transformation with XSLT to TEI or direct to
LaTeX gives you the same benefits as above, and you can also continue to
use Word as the exchange format with authors and editors.
> ¹. luckily, the publisher brill is letting me produce 'camera-ready'
> copy for my book as long as i follow their formatting directives.
> dealing with converting a whole book to word is a nightmare i don't
> want to contemplate!
Agreed. Only ever convert OUT of Word (via .docx), never into it :-)
///Peter
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More options May 19 2012, 8:15 pm
Newsgroups: comp.text.tex
From: "Nasser M. Abbasi" <n...@12000.org>
Date: Sat, 19 May 2012 19:15:08 -0500
Local: Sat, May 19 2012 8:15 pm
Subject: Re: How to typeset texbook.tex
On 5/19/2012 7:00 PM, Peter Flynn wrote:
> I've banged this drum many times before, but it might be worth repeating.
> Author in XML, not LaTeX. Transform the XML to LaTeX with XSLT. That way
> it's testable, reproducible, visible, and manageable under machine
> control.
Ok, but how to do math in XML?
How to write this in XML
$\lim_{x \to \infty} \exp(-x) = 0$
?
thanks,
--Nasser
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More options May 19 2012, 8:32 pm
Newsgroups: comp.text.tex
From: "Nasser M. Abbasi" <n...@12000.org>
Date: Sat, 19 May 2012 19:32:57 -0500
Local: Sat, May 19 2012 8:32 pm
Subject: Re: How to typeset texbook.tex
On 5/19/2012 7:15 PM, Nasser M. Abbasi wrote:
I did a bit of research on this now. And I think one is supposed
to use a MathML DTD thing and somehow include it in XML.
lots of examples here
But the math typesetting looks so complicated compared to
latex (and I thought Latex math was complicated!)
For example, in Latex, I write
$a/b$
in MathML, according to the above page, it is
<mrow><mi>a</mi><mo>/</mo><mi>b</mi></mrow>
And this is an easy example I saw.
If this is really what I would have to type to get
math into my plain text document, and Unless there is a
much simpler way, and I am overlooking something, then I
think I'll pass on this XML, thank you very much :)
--Nasser
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tautology definition: 1. the use of two words or phrases that express the same meaning, in a way that is unnecessary and…. These types of propositions play a crucial role in reasoning. Tautology, in logic, a statement so framed that it cannot be denied without inconsistency. Tautology- A compound proposition is called tautology if and only if it is true for all possible truth values of its propositional variables. Types of SQL injection-- Tautology-based SQL Injection -- Piggy-backed Queries / Statement Injection-- Union Query-- Illegal/Logically Incorrect Queries-- Inference-- Stored Procedure Injection. PMID: 28017820 [Indexed for MEDLINE] Publication Types: Editorial Tautology, often regarded as a fault of style, was defined by Fowler as "saying the same thing twice". A tautology is a formula which is "always true" --- that is, it is true for every assignment of truth values to its simple components. Is the following sentence a tautology? In other words, it can be defined as the term used for retelling the same thing by using different words and phrases. At times, it is used to emphasize on something. Tautology is a type of logic construct that can be applied in IT. 2. 3. Types: Boolean: Only correct queries show the result, wrong queries do not return anything. Test Your Understanding of Tautology, Contradiction and Contingency For each of the following propositions, indicate what they are. Free Online Library: "THIS IS LIKE DEJA VU ALL OVER AGAIN" - Eight Types of Tautology. Contradiction- A compound proposition is called contradiction if and only if it is false for all possible truth values of its propositional variables. (Keith Waterhouse, Waterhouse on Newspaper Style, rev. Anaya JM(1). In this type of logic, according to G. W. von Leibniz, tautologies are truths in all possible worlds, eternal truths, essential truths, and truths by virtue of the postulates of classical logic. Attackers should try to generate logically In some instances, tautology, or rephrasing or repeating words and phrases, can be used for emphasis or to stress the importance of something. Types of Tautology. Item 21 is often called "transitivity". A summary of evidence. Electronic address: juan.anaya@urosario.edu.co. Learn more. The opposite of a tautology is a contradiction, a formula which is "always false".In other words, a contradiction is false for every assignment of truth values to its simple components. The outliers that help make a good organisation capable of being the best. Showing posts with label c#. Show all posts. You da real mvps! b. Example − Prove $\lbrack (A \rightarrow B) \land A \rbrack \rightarrow B$ is a tautology… It refers to a redundant logic wherein a principle is restated or is evident in its expression. An example of this type of tautology is the law of the excluded middle. ed. The words adequate and enough are two words that convey the same meaning. T refers to any statement which is a tautology. C refers to any statement which is a contradiction. It contains only T (Truth) in last column of its truth table. T. V. V ENTTSEL ’ ( 2 ) In logic, an extreme example of the logical fallacy of the unwarranted premise (Latin petitio principii ), namely, the definition or proof of something by the same thing (Latin idem per idem ). Orators, writers, and others strategically use tautology in everyday life, when writing poetry, literature, or song lyrics, and even in debates. However, most people avoid tautology because it is unnecessary and seems silly. shown in the following examples: 7-If he ’ s mad, he’ s mad. How to use tautology in a sentence. a tautology is also called logically valid or logically true.iii. \$1 per month helps!! by "ETC. Tautology is either unnecessary elaboration (the Inland Revenue's white-collar workers), pointless repetition (pair of twins), superfluous description (Europe's huge butter mountain), a needless appendage (weather conditions) or a self-cancelling proposition (He is either guilty or not guilty)." En effet, on croit avoir bien employé les termes, alors qu’on emploie mal les mots. You can think of a tautology as a rule of logic. :) https://www.patreon.com/patrickjmt !! What is Tautology(t),Contradiction/Fallacy(c) and Contingency in Logic according to new Maharashtra boards text book exercise 1.2(Q.3) Préliminaires La tautologie sous-entendue n’est rien d’autre qu’une tautologie faite à l’insu. This article deals with interviewing for corner cases. Show all posts. In particular every inference rule is a tautology as we see in identities and implications. For the best candidates from an interview, test for the corner cases. You can also have logical tautologies, as with the phrase “You’re either hungry or you’re not.” These kinds of tautologies are self-cancelling. So, to determine whether this proposition is a tautology, we need to check whether there is any possible way to make it false. In fact, it is not necessary for the entire meaning of a phrase to be repeated; if a part of the meaning is repeated in such a way that it appears as unintentional or clumsy, then it may be described as tautology. A tautology is a statement that is always true, no matter what. This is a type of SQL injection where we don’t have a clue as to whether the web application is vulnerable to injection attack or not. Proofs Using Logical Equivalences Rosen 1.2 List of Logical Equivalences List of Equivalences Prove: (p q) q p q (p q) q Left-Hand Statement q (p q) Commutative (q p) (q q) Distributive (q p) T Or Tautology q p Identity p q Commutative Prove: (p q) q p q (p q) q Left-Hand Statement q (p q) Commutative (q p) (q q) Distributive Why did we need this step? Request PDF | A new type of informative tautology: Für Unbefugte Betreten Verboten! Advantages of this topology : The possibility of collision is minimum in this type of topology. More details.. Tautology is a type of pleonasm. There are two types of token release techniques : Early token release releases the token just after the transmitting the data and Delay token release releases the token after the acknowledgement is received from the receiver. Tautology [edit | edit source]. all entries in the column of tautology are true. This means you're free to copy and share these comics (but not to sell them). Article excerpt . Logic A statement composed of simpler statements in such a way that it is logically true whether the simpler statements are factually true or … This work is licensed under a Creative Commons Attribution-NonCommercial 2.5 License. Click Tautology, Contradiction or Contingency, then Submit. Tautology definition, needless repetition of an idea, especially in words other than those of the immediate context, without imparting additional force or clearness, as in “widow woman.” See more. The autoimmune tautology. MICHAEL MOORE [*] THOUGH INTRODUCED into propositional calculus at the end of the 19th century by the American logician Charles Peirce, tautology (literally "the same word") had its place among the better-known errors of rhetoric as early as the 4th century. In other cases, tautology can be used for humor or unintentionally. 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# Publications by Julien Devriendt
## Star-gas misalignment in galaxies: I. The properties of galaxies from the Horizon-AGN simulation and comparisons to SAMI
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DJ Khim, SK Yi, Y Dubois, JJ Bryant, C Pichon, SM Croom, J Bland-Hawthorn, S Brough, H Choi, J Devriendt, B Groves, MS Owers, SN Richards, JVD Sande, SM Sweet
Recent integral field spectroscopy observations have found that about 11\% of galaxies show star-gas misalignment. The misalignment possibly results from external effects such as gas accretion, interaction with other objects, and other environmental effects, hence providing clues to these effects. We explore the properties of misaligned galaxies using Horizon-AGN, a large-volume cosmological simulation, and compare the result with the result of the Sydney-AAO Multi-object integral field spectrograph (SAMI) Galaxy Survey. Horizon-AGN can match the overall misalignment fraction and reproduces the distribution of misalignment angles found by observations surprisingly closely. The misalignment fraction is found to be highly correlated with galaxy morphology both in observations and in the simulation: early-type galaxies are substantially more frequently misaligned than late-type galaxies. The gas fraction is another important factor associated with misalignment in the sense that misalignment increases with decreasing gas fraction. However, there is a significant discrepancy between the SAMI and Horizon-AGN data in the misalignment fraction for the galaxies in dense (cluster) environments. We discuss possible origins of misalignment and disagreement.
## Comparing Galaxy Clustering in Horizon-AGN Simulated Lightcone Mocks and VIDEO Observations
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P Hatfield, C Laigle, M Jarvis, JULIEN Devriendt, I Davidzon, O Ilbert, C Pichon, Y Dubois
Hydrodynamical cosmological simulations have recently made great advances in reproducing galaxy mass assembly over cosmic time - as often quantified from the comparison of their predicted stellar mass functions to observed stellar mass functions from data. In this paper we compare the clustering of galaxies from the hydrodynamical cosmological simulated lightcone Horizon-AGN, to clustering measurements from the VIDEO survey observations. Using mocks built from a VIDEO-like photometry, we first explore the bias introduced into clustering measurements by using stellar masses and redshifts derived from SED-fitting, rather than the intrinsic values. The propagation of redshift and mass statistical and systematic uncertainties in the clustering measurements causes us to underestimate the clustering amplitude. We find then that clustering and halo occupation distribution (HOD) modelling results are qualitatively similar in Horizon-AGN and VIDEO. However at low stellar masses Horizon-AGN underestimates the observed clustering by up to a factor of ~3, reflecting the known excess stellar mass to halo mass ratio for Horizon-AGN low mass haloes, already discussed in previous works. This reinforces the need for stronger regulation of star formation in low mass haloes in the simulation. Finally, the comparison of the stellar mass to halo mass ratio in the simulated catalogue, inferred from angular clustering, to that directly measured from the simulation, validates HOD modelling of clustering as a probe of the galaxy-halo connection.
## The diverse galaxy counts in the environment of high-redshift massive black holes in Horizon-AGN
Mon. Not. R. Astron. Soc (0)
M Habouzit, M Volonteri, RS Somerville, Y Dubois, S Peirani, C Pichon, JULIEN Devriendt
High-redshift quasars are believed to reside in highly biased regions of the Universe, where black hole (BH) growth is sustained by an enhanced number of mergers and by being at the intersection of filaments bringing fresh gas. This assumption should be supported by an enhancement of the number counts of galaxies in the field of view of quasars. While the current observations of quasar environments do not lead to a consensus on a possible excess of galaxies, the future missions JWST, WFIRST, and Euclid will provide new insights on quasar environments, and will substantially increase the number of study-cases. We are in a crucial period, where we need to both understand the current observations and predict how upcoming missions will improve our understanding of BH environments. Using the large-scale simulation Horizon-AGN, we find that statistically the most massive BHs reside in environments with the largest number counts of galaxies. However, we find a large variance in galaxy number counts, and some massive BHs do not show enhanced counts in their neighborhood. Interestingly, some massive BHs have a very close galaxy companion but no further enhancement of the galaxy number counts at larger scales, in agreement with recent observations. We find that AGN feedback in the surrounding galaxies is able to decrease their luminosity and stellar mass, and therefore to make them un-observable when using restrictive galaxy selection criteria. Radiation from the quasars can spread over large distances, which could affect the formation history of surrounding galaxies, but a careful analysis of these processes requires radiative transfer simulations.
## Probing Cosmic Dawn: Modelling the Assembly History, SEDs, and Dust Content of Selected $z\sim9$ Galaxies
MNRAS (0)
H Katz, N Laporte, RS Ellis, J Devriendt, A Slyz
The presence of spectroscopically confirmed Balmer breaks in galaxy spectral energy distributions (SEDs) at $z>9$ provides one of the best probes of the assembly history of the first generations of stars in our Universe. Recent observations of the gravitationally lensed source, MACS 1149_JD1 (JD1), indicate that significant amounts of star formation likely occurred at redshifts as high as $z\simeq15$. The inferred stellar mass, dust mass, and assembly history of JD1, or any other galaxy at these redshifts that exhibits a strong Balmer break, can provide a strong test of our best theoretical models from high-resolution cosmological simulations. In this work, we present the results from a cosmological radiation-hydrodynamics simulation of the region surrounding a massive Lyman-break galaxy. For two of our most massive systems, we show that dust preferentially resides in the vicinity of the young stars thereby increasing the strength of the measured Balmer break such that the simulated SEDs are consistent with the photometry of JD1 and two other $z>9$ systems (GN-z10-3 and GN-z9-1) that have proposed Balmer breaks at high redshift. We find strong variations in the shape and luminosity of the SEDs of galaxies with nearly identical stellar and halo masses, indicating the importance of morphology, assembly history, and dust distribution in making inferences on the properties of individual galaxies at high redshifts. Our results stress the importance that dust may play in modulating the observable properties of galaxies, even at the extreme redshifts of $z>9$.
## Cosmic CARNage I: on the calibration of galaxy formation models
MNRAS (0)
A Knebe, FR Pearce, V Gonzalez-Perez, PA Thomas, A Benson, R Asquith, J Blaizot, R Bower, J Carretero, FJ Castander, A Cattaneo, SA Cora, DJ Croton, W Cui, D Cunnama, JE Devriendt, PJ Elahi, A Font, F Fontanot, ID Gargiulo, J Helly, B Henriques, J Lee, GA Mamon, J Onions, ND Padilla, C Power, A Pujol, AN Ruiz, C Srisawat, ARH Stevens, E Tollet, CA Vega-Martínez, SK Yi
We present a comparison of nine galaxy formation models, eight semi-analytical and one halo occupation distribution model, run on the same underlying cold dark matter simulation (cosmological box of co-moving width 125$h^{-1}$ Mpc, with a dark-matter particle mass of $1.24\times 10^9 h^{-1}$ Msun) and the same merger trees. While their free parameters have been calibrated to the same observational data sets using two approaches, they nevertheless retain some 'memory' of any previous calibration that served as the starting point (especially for the manually-tuned models). For the first calibration, models reproduce the observed z = 0 galaxy stellar mass function (SMF) within 3-{\sigma}. The second calibration extended the observational data to include the z = 2 SMF alongside the z~0 star formation rate function, cold gas mass and the black hole-bulge mass relation. Encapsulating the observed evolution of the SMF from z = 2 to z = 0 is found to be very hard within the context of the physics currently included in the models. We finally use our calibrated models to study the evolution of the stellar-to-halo mass (SHM) ratio. For all models we find that the peak value of the SHM relation decreases with redshift. However, the trends seen for the evolution of the peak position as well as the mean scatter in the SHM relation are rather weak and strongly model dependent. Both the calibration data sets and model results are publicly available.
## Swirling around filaments: are large-scale structure vortices spinning up dark halos?
ArXiv (0)
C Laigle, C Pichon, S Codis, Y Dubois, DL Borgne, D Pogosyan, J Devriendt, S Peirani, S Prunet, S Rouberol, A Slyz, T Sousbie
The kinematic analysis of dark matter and hydrodynamical simulations suggests that the vorticity in large-scale structure is mostly confined to, and predominantly aligned with their filaments, with an excess of probability of 20 per cent to have the angle between vorticity and filaments direction lower than 60 degrees relative to random orientations. The cross sections of these filaments are typically partitioned into four quadrants with opposite vorticity sign, arising from multiple flows, originating from neighbouring walls. The spins of halos embedded within these filaments are consistently aligned with this vorticity for any halo mass, with a stronger alignment for the most massive structures up to an excess of probability of 165 per cent. On large scales, adiabatic/cooling hydrodynamical simulations display the same vorticity in the gas as in the dark matter. The global geometry of the flow within the cosmic web is therefore qualitatively consistent with a spin acquisition for smaller halos induced by this large-scale coherence, as argued in Codis et al. (2012). In effect, secondary anisotropic infall (originating from the vortex-rich filament within which these lower-mass halos form) dominates the angular momentum budget of these halos. The transition mass from alignment to orthogonality is related to the size of a given multi-flow region with a given polarity. This transition may be reconciled with the standard tidal torque theory if the latter is augmented so as to account for the larger scale anisotropic environment of walls and filaments.
## Blowing cold flows away: the impact of early AGN activity on the formation of a brightest cluster galaxy progenitor
ArXiv (0)
Y Dubois, C Pichon, J Devriendt, J Silk, M Haehnelt, T Kimm, A Slyz
Supermassive black holes (BH) are powerful sources of energy that are already in place at very early epochs of the Universe (by z=6). Using hydrodynamical simulations of the formation of a massive M_vir=5 10^11 M_sun halo by z=6 (the most massive progenitor of a cluster of M_vir=2 10^15 M_sun at z=0), we evaluate the impact of Active Galactic Nuclei (AGN) on galaxy mass content, BH self-regulation, and gas distribution inside this massive halo. We find that SN feedback has a marginal influence on the stellar structure, and no influence on the mass distribution on large scales. In contrast, AGN feedback alone is able to significantly alter the stellar-bulge mass content by quenching star formation when the BH is self-regulating, and by depleting the cold gas reservoir in the centre of the galaxy. The growth of the BH proceeds first by a rapid Eddington-limited period fed by direct cold filamentary infall. When the energy delivered by the AGN is sufficiently large to unbind the cold gas of the bulge, the accretion of gas onto the BH is maintained both by smooth gas inflow and clump migration through the galactic disc triggered by merger-induced torques. The feedback from the AGN has also a severe consequence on the baryon mass content within the halo, producing large-scale hot superwinds, able to blow away some of the cold filamentary material from the centre and reduce the baryon fraction by more than 30 per cent within the halo's virial radius. Thus in the very young universe, AGN feedback is likely to be a key process, shaping the properties of the most massive galaxies.
## Connecting the cosmic web to the spin of dark halos: implications for galaxy formation
ArXiv (0)
S Codis, C Pichon, J Devriendt, A Slyz, D Pogosyan, Y Dubois, T Sousbie
We investigate the alignment of the spin of dark matter halos relative (i) to the surrounding large-scale filamentary structure, and (ii) to the tidal tensor eigenvectors using the Horizon 4pi dark matter simulation which resolves over 43 million dark matter halos at redshift zero. We detect a clear mass transition: the spin of dark matter halos above a critical mass tends to be perpendicular to the closest filament, and aligned with the intermediate axis of the tidal tensor, whereas the spin of low-mass halos is more likely to be aligned with the closest filament. Furthermore, this critical mass of 5 10^12 is redshift-dependent and scales as (1+z)^-2.5. We propose an interpretation of this signal in terms of large-scale cosmic flows. In this picture, most low-mass halos are formed through the winding of flows embedded in misaligned walls; hence they acquire a spin parallel to the axis of the resulting filaments forming at the intersection of these walls. On the other hand, more massive halos are typically the products of later mergers along such filaments, and thus they acquire a spin perpendicular to this direction when their orbital angular momentum is converted into spin. We show that this scenario is consistent with both the measured excess probabilities of alignment w.r.t. the eigen-directions of the tidal tensor, and halo merger histories. On a more qualitative level, it also seems compatible with 3D visualization of the structure of the cosmic web as traced by "smoothed" dark matter simulations or gas tracer particles. Finally, it provides extra support to the disc forming paradigm presented by Pichon et al (2011) as it extends it by characterizing the geometry of secondary infall at high redshift.
## Are cold flows detectable with metal absorption lines?
ArXiv (0)
T Kimm, A Slyz, J Devriendt, C Pichon
[Abridged] Cold gas flowing within the "cosmic web" is believed to be an important source of fuel for star formation at high redshift. However, the presence of such filamentary gas has never been observationally confirmed. In this work, we investigate in detail whether such cold gas is detectable using low-ionisation metal absorption lines, such as CII \lambda1334 as this technique has a proven observational record for detecting gaseous structures. Using a large statistical sample of galaxies from the Mare Nostrum N-body+AMR cosmological simulation, we find that the typical covering fraction of the dense, cold gas in 10^12 Msun haloes at z~2.5 is lower than expected (~5%). In addition, the absorption signal by the interstellar medium of the galaxy itself turns out to be so deep and so broad in velocity space that it completely drowns that of the filamentary gas. A detectable signal might be obtained from a cold filament exactly aligned with the line of sight, but this configuration is so unlikely that it would require surveying an overwhelmingly large number of candidate galaxies to tease it out. Finally, the predicted metallicity of the cold gas in filaments is extremely low (\leq 0.001 Zsun). Should this result persist when higher resolution runs are performed, it would significantly increase the difficulty of detecting filamentary gas inflows using metal lines. However, even if we assume that filaments are enriched to Zsun, the absorption signal that we compute is still weak. We are therefore led to conclude that it is extremely difficult to observationally prove or disprove the presence of cold filaments as the favorite accretion mode of galaxies using low-ionisation metal absorption lines. The Ly-alpha emission route looks more promising but due to the resonant nature of the line, radiative transfer simulations are required to fully characterize the observed signal.
## How AGN feedback and metal cooling shape cluster entropy profiles
ArXiv (0)
Y Dubois, J Devriendt, R Teyssier, A Slyz
Observed clusters of galaxies essentially come in two flavors: non cool core clusters characterized by an isothermal temperature profile and a central entropy floor, and cool-core clusters where temperature and entropy in the central region are increasing with radius. Using cosmological resimulations of a galaxy cluster, we study the evolution of its intracluster medium (ICM) gas properties, and through them we assess the effect of different (sub-grid) modelling of the physical processes at play, namely gas cooling, star formation, feedback from supernovae and active galactic nuclei (AGN). More specifically we show that AGN feedback plays a major role in the pre-heating of the proto-cluster as it prevents a high concentration of mass from collecting in the center of the future galaxy cluster at early times. However, AGN activity during the cluster's later evolution is also required to regulate the mass flow into its core and prevent runaway star formation in the central galaxy. Whereas the energy deposited by supernovae alone is insufficient to prevent an overcooling catastrophe, supernovae are responsible for spreading a large amount of metals at high redshift, enhancing the cooling efficiency of the ICM gas. As the AGN energy release depends on the accretion rate of gas onto its central black hole engine, the AGN responds to this supernova enhanced gas accretion by injecting more energy into the surrounding gas, and as a result increases the amount of early pre-heating. We demonstrate that the interaction between an AGN jet and the ICM gas that regulates the growth of the AGN's BH, can naturally produce cool core clusters if we neglect metals. However, as soon as metals are allowed to contribute to the radiative cooling, only the non cool core solution is produced.
## Jet-regulated cooling catastrophe
ArXiv (0)
Y Dubois, J Devriendt, A Slyz, R Teyssier
We present the first implementation of Active Galactic Nuclei (AGN) feedback in the form of momentum driven jets in an Adaptive Mesh Refinement (AMR) cosmological resimulation of a galaxy cluster. The jets are powered by gas accretion onto Super Massive Black Holes (SMBHs) which also grow by mergers. Throughout its formation, the cluster experiences different dynamical states: both a morphologically perturbed epoch at early times and a relaxed state at late times allowing us to study the different modes of BH growth and associated AGN jet feedback. BHs accrete gas efficiently at high redshift (z>2), significantly pre-heating proto-cluster halos. Gas-rich mergers at high redshift also fuel strong, episodic jet activity, which transports gas from the proto-cluster core to its outer regions. At later times, while the cluster relaxes, the supply of cold gas onto the BHs is reduced leading to lower jet activity. Although the cluster is still heated by this activity as sound waves propagate from the core to the virial radius, the jets inefficiently redistribute gas outwards and a small cooling flow develops, along with low-pressure cavities similar to those detected in X-ray observations. Overall, our jet implementation of AGN feedback quenches star formation quite efficiently, reducing the stellar content of the central cluster galaxy by a factor 3 compared to the no AGN case. It also dramatically alters the shape of the gas density profile, bringing it in close agreement with the beta model favoured by observations, producing quite an isothermal galaxy cluster for gigayears in the process. However, it still falls short in matching the lower than Universal baryon fractions which seem to be commonplace in observed galaxy clusters.
## Influence of AGN jets on the magnetized ICM
ArXiv (0)
Y Dubois, J Devriendt, A Slyz, J Silk
Galaxy clusters are the largest structures for which there is observational evidence of a magnetised medium. Central cores seem to host strong magnetic fields ranging from a few 0.1 microG up to several 10 microG in cooling flow clusters. Numerous clusters harbor central powerful AGN which are thought to prevent cooling flows in some clusters. The influence of such feedback on the magnetic field remains unclear: does the AGN-induced turbulence compensate the loss of magnetic amplification within a cool core? And how is this turbulence sustained over several Gyr? Using high resolution magneto-hydrodynamical simulations of the self-regulation of a radiative cooling cluster, we study for the first time the evolution of the magnetic field within the central core in the presence of a powerful AGN jet. It appears that the jet-induced turbulence strongly amplifies the magnetic amplitude in the core beyond the degree to which it would be amplified by pure compression in the gravitational field of the cluster. The AGN produces a non-cooling core and increases the magnetic field amplitude in good agreement with microG field observations.
## The impact of TP-AGB stars on hierarchical galaxy formation models
ArXiv (0)
C Tonini, C Maraston, J Devriendt, D Thomas, J Silk
The spectro-photometric properties of galaxies in galaxy formation models are obtained by combining the predicted history of star formation and mass accretion with the physics of stellar evolution through stellar population models. In the recent literature, significant differences have emerged regarding the implementation of the Thermally-Pulsing Asymptotic Giant Branch phase of stellar evolution. The emission in the TP-AGB phase dominates the bolometric and near-IR spectrum of intermediate-age (~1 Gyr) stellar populations, hence it is crucial for the correct modeling of the galaxy luminosities and colours. In this paper for the first time, we incorporate a full prescription of the TP-AGB phase in a semi-analytic model of galaxy formation. We find that the inclusion of the TP-AGB in the model spectra dramatically alters the predicted colour-magnitude relation and its evolution with redshift. When the TP-AGB phase is active, the rest-frame V-K galaxy colours are redder by almost 2 magnitudes in the redshift range z~2-3 and by 1 magnitude at z~1. Very red colours are produced in disk galaxies, so that the V-K colour distributions of disk and spheroids are virtually undistinguishable at low redshifts. We also find that the galaxy K-band emission is more than 1 magnitude higher in the range z~1-3. This may alleviate the difficulties met by the hierarchical clustering scenario in predicting the red galaxy population at high redshifts. The comparison between simulations and observations have to be revisited in the light of our results.
## Accretion, feedback and galaxy bimodality: a comparison of the GalICS semi-analytic model and cosmological SPH simulations
ArXiv (0)
A Cattaneo, J Blaizot, DH Weinberg, S Colombi, R Dave, J Devriendt, B Guiderdoni, N Katz, D Keres
We compare the galaxy population of an SPH simulation to those predicted by the GalICS semi-analytic model and a stripped down version without supernova and AGN feedback. The SPH simulation and the no-feedback GalICS model make similar predictions for the baryonic mass functions of galaxies and for the dependence of these mass functions on environment and redshift. The two methods also make similar predictions for the galaxy content of dark matter haloes as a function of halo mass and for the gas accretion history of galaxies. Both the SPH and no-feedback GalICS models predict a bimodal galaxy population at z=0. The "red'' sequence of gas poor, old galaxies is populated mainly by satellite systems while, contrary to observations, the central galaxies of massive haloes lie on the "blue'' star-forming sequence as a result of continuing hot gas accretion at late times. Furthermore, both models overpredict the observed baryonic mass function, especially at the high mass end. In the full GalICS model, supernova-driven outflows reduce the masses of low and intermediate mass galaxies by about a factor of two. AGN feedback suppresses gas cooling in large haloes, producing a sharp cut-off in the baryonic mass function and moving the central galaxies of these massive haloes to the red sequence. Our results imply that the observational failings of the SPH simulation and the no-feedback GalICS model are a consequence of missing input physics rather than computational inaccuracies, that truncating gas accretion by satellite galaxies automatically produces a bimodal galaxy distribution with a red sequence, but that explaining the red colours of the most massive galaxies requires a mechanism like AGN feedback that suppresses the accretion onto central galaxies in large haloes.
## Magnetized Non-linear Thin Shell Instability: Numerical Studies in 2D
ArXiv (0)
F Heitsch, AD Slyz, JEG Devriendt, L Hartmann, A Burkert
We revisit the analysis of the Non-linear Thin Shell Instability (NTSI) numerically, including magnetic fields. The magnetic tension force is expected to work against the main driver of the NTSI -- namely transverse momentum transport. However, depending on the field strength and orientation, the instability may grow. For fields aligned with the inflow, we find that the NTSI is suppressed only when the Alfv\'en speed surpasses the (supersonic) velocities generated along the collision interface. Even for fields perpendicular to the inflow, which are the most effective at preventing the NTSI from developing, internal structures form within the expanding slab interface, probably leading to fragmentation in the presence of self-gravity or thermal instabilities. High Reynolds numbers result in local turbulence within the perturbed slab, which in turn triggers reconnection and dissipation of the excess magnetic flux. We find that when the magnetic field is initially aligned with the flow, there exists a (weak) correlation between field strength and gas density. However, for transverse fields, this correlation essentially vanishes. In light of these results, our general conclusion is that instabilities are unlikely to be erased unless the magnetic energy in clouds is much larger than the turbulent energy. Finally, while our study is motivated by the scenario of molecular cloud formation in colliding flows, our results span a larger range of applicability, from supernovae shells to colliding stellar winds.
## GALICS III: Predicted properties for Lyman Break Galaxies at redshift 3
ArXiv (0)
J Blaizot, B Guiderdoni, JEG Devriendt, FR Bouchet, S Hatton, F Stoehr
This paper illustrates how mock observational samples of high-redshift galaxies with sophisticated selection criteria can be extracted from the predictions of GALICS, a hybrid model of hierarchical galaxy formation that couples the outputs of large cosmological simulations and semi-analytic recipes to describe dark matter collapse and the physics of baryons respectively. As an example of this method, we focus on the properties of Lyman Break Galaxies at redshift 3. With the MOMAF software package described in a companion paper, we generate a mock observational sample with selection criteria as similar as possible to those implied in the actual observations of z = 3 LBGs by Steidel et al.(1995). Our model predictions are in good agreement with the observed number density and 2D correlation function. We investigate the optical/IR luminosity budget as well as several other physical properties of LBGs and find them to be in general agreement with observed values. Looking into the future of these LBGs we predict that 75% of them end up as massive ellipticals today, even though only 35% of all our local ellipticals are predicted to have a LBG progenitor. In spite of some shortcomings, this new 'mock observation' method clearly represents a necessary first step toward a more accurate comparison between hierarchical models of galaxy formation and real observational surveys.
## Turbulent Ambipolar Diffusion: Numerical Studies in 2D
ArXiv (0)
F Heitsch, EG Zweibel, AD Slyz, JEG Devriendt
Under ideal MHD conditions the magnetic field strength should be correlated with density in the interstellar medium (ISM). However, observations indicate that this correlation is weak. Ambipolar diffusion can decrease the flux-to-mass ratio in weakly ionized media; however, it is generally thought to be too slow to play a significant role in the ISM except in the densest molecular clouds. Turbulence is often invoked in astrophysical problems to increase transport rates above the (very slow) laminar values predicted by kinetic theory. We describe a series of numerical experiments addressing the problem of turbulent transport of magnetic fields in weakly ionized gases. We show, subject to various geometrical and physical restrictions, that turbulence in a weakly ionized medium rapidly diffuses the magnetic flux to mass ratio through the buildup of appreciable ion-neutral drifts on small scales. These results are applicable to the fieldstrength - density correlation in the ISM, as well as the merging of flux systems such as protostar and accretion disk fields or protostellar jets with ambient matter, and the vertical transport of galactic magnetic fields.
## Star Formation in Viscous Galaxy Disks
ArXiv (0)
A Slyz, J Devriendt, A Burkert, K Prendergast, J Silk
The Lin and Pringle model (1987) of galactic disk formation postulates that if star formation proceeds on the same timescale as the viscous redistribution of mass and angular momentum in disk galaxies, then the stars attain an exponential density profile. Their claim is that this result holds generally: regardless of the disk galaxy's initial gas and dark matter distribution and independent of the nature of the viscous processes acting in the disk. We present new results from a set of 2D hydro-simulations which investigate their analytic result.
## The Impact of Galaxy Formation on the Diffuse Background Radiation
ArXiv (0)
J Silk, J Devriendt
The far infrared background is a sink for the hidden aspects of galaxy formation. At optical wavelengths, ellipticals and spheroids are old, even at $z \sim 1.$ Neither the luminous formation phase nor their early evolution is seen in the visible. We infer that ellipticals and, more generally, most spheroids must have formed in dust-shrouded starbursts. In this article, we show how separate tracking of disk and spheroid star formation enables us to infer that disks dominate near the peak in the cosmic star formation rate at $z \lapproxeq 2$ and in the diffuse ultraviolet/optical/infrared background, whereas spheroid formation dominates the submillimetre background.
## Probing Galaxy Formation with High Energy Gamma-Rays
ArXiv (0)
JR Primack, RS Somerville, JS Bullock, JEG Devriendt
We discuss how measurements of the absorption of $\gamma$-rays from GeV to TeV energies via pair production on the extragalactic background light (EBL) can probe important issues in galaxy formation. We use semi-analytic models (SAMs) of galaxy formation, set within the hierarchical structure formation scenario, to obtain predictions of the EBL for 0.1-1000$\mu$m. SAMs incorporate simplified physical treatments of the key processes of galaxy formation --- including gravitational collapse and merging of dark matter halos, gas cooling and dissipation, star formation, supernova feedback and metal production --- and have been shown to reproduce key observations at low and high redshift. Here we also introduce improved modelling of the spectral energy distributions in the mid-to-far-IR arising from emission by dust grains. Assuming a flat \lcdm cosmology with $\Omega_m=0.3$ and Hubble parameter $h=0.65$, we investigate the consequences of variations in input assumptions such as the stellar initial mass function (IMF) and the efficiency of converting cold gas into stars. We conclude that observational studies of the absorption of $\gamma$-rays with energies from 10s of Gev to 10s of TeV will help to determine the EBL, and also help to explain its origin by constraining some of the most uncertain features of galaxy formation theory, including the IMF, the history of star formation, and the reprocessing of light by dust.
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# Deligne's Canonical Extension in Algebraic Varieties?
Suppose $C/\mathbb{Q}$ is an algebraic curve (not necessarily complete) defined over $\mathbb{Q}$, and $p$ is a $\mathbb{Q}$ valued point of $C$. Suppose there is a algebraic fibration $$\pi: Z \rightarrow C$$ such that $Y=\pi^{-1}(p)$ is the only singular fiber, i.e. $$\pi: X \rightarrow S$$ where $X=Z \setminus Y$ and $S=C \setminus p$, is a smooth morphism between smooth varieties. Then by theorem 1 of the paper "On the differentiation of de Rham cohomology classes with respect parameters" by Katz and Oda, there is a canonical integrable connetion $\nabla_{GM}$ --Gauss-Manin connection--on the relative de Rham cohomology sheaf $R^q\,\pi_*(\Omega_{X/S}^*)$.
If we take extension of field and then analytification, we got maps $$\pi: Z_{\mathbb{C}}^{an} \rightarrow C_{\mathbb{C}}^{an}\\ \pi:X_{\mathbb{C}}^{an} \rightarrow S_{\mathbb{C}}^{an}$$ The Gauss-Manin connection now is the the analytification of $\nabla_{GM}$, and let us denote it by $\nabla_{GM}^{an}$. For the purpose of this question, let me assume that $\nabla_{GM}^{an}$ has regular singularities at $p$. Then from the works of Griffiths and Schmid on variations of Hodge structures, there exists Deligne's canonical extension of the locally free sheaf $R^q\pi_*(\Omega_{X/S}^{*an})$ on $S_{\mathbb{C}}^{an}$ to a locally free sheaf on $C_{\mathbb{C}}^{an}$. The process of extending it over $p$ could be described explicitly as follows.
Choose a local coordinate in a neighbourhood of $p$, say $t$ and $p$ is the point with $t=0$, i.e. origin. Choose muli-valued local sections $e=(e_1,e_2,\cdots,e_n)^{t}$ of $R^q\pi_*(\Omega_{X/S}^{*an})$ in a neighbourhood of $0$ which form a multi-valued local frame. Suppose the monodromy matrix around $t=0$ with respect to thess muli-valued sections is $T$, define $N= \log T$, the vector $$\exp (-\frac{\log t}{2 \pi i} \,N)\,e$$ is single valued and we use it as a local frame to do the extension. This is Deligne's canonical extension in analytical case.
My questions are
1, With the assumption of $\nabla_{GM}^{an}$ is regular, is $\nabla_{GM}$ regular. Actually I am not clear with the definition of regularity of $\nabla_{GM}$ when it is defined over $\mathbb{Q}$, could someone explain it?
2, Could $R^q\,\pi_*(\Omega_{X/S}^*)$ be canonically extended to a locally free sheaf on $C$?
3, If such an extension exists, is there an explicit way like the analytical case to describe this extension using monodromy matrix?
4, If such an extension exists, after $\otimes_{\mathbb{Q}} \mathbb{C}$ and analytification, is this extension compatible with the Deligne's canonical extension in the analytical case?
Any comments and references will be appreciated.
• How does one take $\log T$ canonically? – Will Sawin Apr 11 '17 at 19:06
A short answer is that Deligne's definition of canonical extension (https://publications.ias.edu/sites/default/files/71_Localbehavior.pdf) is not analytic, and works fine over $\mathbb Q$. Rather than the exponential you define, he defines a new connection via an algebraic formula, which is constant, and whose horizontal sections map the sections you define. Then he defines the canonical extension to be generated by the horizontal sections. This can be done algebraically by finding horizontal sections in a formal power series ring, say.
If more explanation is needed, my way of thinking of this is to consider everything in terms of the action of the operator $t \nabla$ on the sections of the vector bundle $R^q \pi_*$ over the field of formal Laurent series, $\mathbb Q((t))$ and $\mathbb C((t))$. The condition that $\nabla$ have regular singularities is, I believe, equivalent to the condition that $t \nabla$ can be put in Jordan normal form - i.e. an arbitrary section can be written as an infinite sum of sections that lie in a generalized eigenspace of $t \nabla$.
So the descent of this to $\mathbb Q((t))$ is the condition that an arbitrary section can be written as an infinite sum of sections in a finite-dimensional $t\nabla$-stable subspace.
In the case when the connection comes from algebraic geometry, one knows that all the eigenvalues of the monodromy, which are the same as the eigenvalues of $e^{ 2\pi i t\nabla}$, are roots of unity, so the eigenvalues of $t \nabla$ are rational numbers.
Hence one can simply define the canonical extension to be the sum of the subspaces generated by elements in generalized eigenspaces whose eigenvalue is a nonnegative rational number. (We want nonnegative because multiplying by $t$ adds $1$ to the eigenvalue.)
In fact, I think you want $N$ to be nilpotent, in which case all eigenvalues of $t\nabla$ are integers, and you can do the same thing with nonnegative integers. In fact you can take it to be the submodule generated by the subspace of sections in generalized eigenspaces with eigenvalue zero. (If a generator has eigenvalue a positive integer, we can lower it to zero by dividing by the corresponding power of $t$, so we may take generators to be those sections with eigenvalue $0$.) Using this description, you should be able to see it matches Deligne's characterization that the matrix of 1-forms defining the connection has logarithmic poles with nilpotent residues.
• In Deligne's definition of a new connection there is a factor $1/2 \pi i$, which also appears a lot in the discussions of extensions. But if the base field is $\mathbb{Q}$, how to deal with this transcendental number? Do you know any references about considering things in $\mathbb{Q}((t))$? – Wenzhe Apr 11 '17 at 20:49
• @Wenzhe That's a good point - but I think some $2 \pi i$ appears when you write down the operator $N_i$ algebraically. Indeed, I think one can see that the monodromy operator $T_i$ is $e^{ 2\pi i t \nabla}$, or maybe $e^{ - 2\pi i t \nabla}$. We are taking the unique nilpotent logarithm of that, which is equivalent to taking the nilpotent part of $t \nabla$ in the sense of Jordan decomposition - a purely algebraic operation - and then multiplying by $2 \pi i$. – Will Sawin Apr 12 '17 at 1:51
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# Given $n$ points on the plane, find a circle which contains only three
Given $n$ points on $\mathbb{R}^{2}$, s.t no three are on the same straight line, and not all the points are on the same circle, prove that there exists a circle which contains only three of those points. We are considering only the circles that are defined by the above points, of course.
I am trying to find a direct geometrical proof. I was thinking of looking at the circle and point which have minimal positive distance between them (since I have seen a similar question considering straight lines and points which uses this strategy), or looking at all the circles containing a specific point, but I still can't seem to find the trick.
Thanks alot!
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Do you mean a circle which passes through only 3 or a circle which contains three inside? – N. S. May 5 '12 at 20:46
Passes through 3. Sorry if it wasn't clear – IBS May 5 '12 at 21:24
The above result was proven by Motzkin (1951) on page 3 of this document. The proof follows from the Sylvester - Gallai Theorem (presumably the "similar problem" you mentioned) after inverting around a chosen point.
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I'm sorry, but he uses many terms over there which i am not familiar with... I am not able to follow. Any chance that someone could simplify this here? Thanks – IBS May 6 '12 at 13:45
This was too long to make it a comment to EuYu's post. Credit is to him (and Motzkin).
I am assuming this is the result you know already: if $n$ points in the plane are arranged such that no line contains exactly two of them, then they are all on the same line.
The strategy outlined in the document by Motzkin is the following: pick any of the points, say $P$. Call the other points $Q_i$. Apply an inversion with respect to $P$ (check wikipedia if you don't know what an inversion is, and how it acts on circles and lines). Let's say the $Q_i$ get mapped to $Q_i'$ by the inversion.
Because there was no circle containing all the original points, there is not line containing all the $Q_i'$'s. This means that some line $l$ contains exactly two such $Q_i'$, say $Q_1'$ and $Q_2'$. This line does not pass through $P$, because that would mean that $Q_1$, $Q_2$ and $P$ are on the same line. Apply an inversion for the second time (again with respect to $P$ and using the same radius as before). Now $Q_1'$ and $Q_2'$ are mapped to $Q_1$ and $Q_2$ respectively, and $l$ is mapped to a circle containing precisely $P$, $Q_1$ and $Q_2$.
Hope this helps. If it's not clear, try experimenting a little with inversions to get a feel for their properties.
Note that we could pick the point $P$ at random. This means we have actually proven a stronger result: through any of the $n$ points there is a circle containing exactly three of the given points.
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I gave the credits as you asked :) Thank you for the help. – IBS May 8 '12 at 13:01
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# Is my system too complicated for Solve?
I am trying to solve a system of three nonlinear equations using Mathematica. Solve takes forever to run and never solves my system of equations. Is my system too complicated?
I have looked through the suggestions on using Solve in the Mathematica help, but none of these seem to work. I am wondering if I am going about solving the problem the wrong way.
Any help would be greatly appreciated! I have attached my code below.
### Solving the Tensegrity Model
Defining reference conditions
l0 = Sqrt[0.375];
s0 = 0.5 ;
k = 1;
Using prestress to define
ξ = 1 - lR/l0;
lR[ξval_] := l0 (1 - ξval);
lRvalues = Table[lR[ξrange], {ξrange, {0.0, 0.1, 0.5, 0.9, 1.0}}];
Defining cable lengths
Clear[sx, sy, sz]
l1[sx_] := 0.5 Sqrt[sx^2 + sy^2 - 2 sy + 2];
l2 := 0.5 Sqrt[sy^2 + sz^2 - 2 sz + 2];
l3[sx_] := 0.5 Sqrt[sz^2 + sx^2 - 2 sx + 2];
F1[lR_] = k (l1[sx] - lR);
F2[lR_] = k (l2 - lR);
F3[lR_] = k (l3[sx] - lR);
F1values = Table[F1[lr], {lr, lRvalues}];
F2values = Table[F2[lr], {lr, lRvalues}];
F3values = Table[F3[lr], {lr, lRvalues}];
F1values[[1]] /. sx -> 0.5
(*-0.612372 + 0.5 Sqrt[2.25 - 2 sy + sy^2]*)
sxval = Range[0.5, 2, 0.5];
For[j = 1, j < Length[sxval] + 1, j++,
For[i = 1, i < Length[F1values] + 1, i++,
Solve[{(F1values[[i]] /. sx -> sxval[[j]]) (1 - sy)/l1[sxval[[j]]] ==
F2values[[i]] sy/l2,
F2values[[i]] (1 - sz)/l2 == (F3values[[i]] /. sx -> sxval[[j]]) sz/
l3[sxval[[j]]],
T == 2 ((F1values[[i]] /. sx -> sxval[[j]]) sxval[[j]]/
l1[sxval[[j]]] + (F3values[[i]] /. sx -> sxval[[j]]) (
sxval[[j]] - 1)/l3[sxval[[j]]])}, {T, sy, sz}]
(*sy1 = NSolve[(F1values[[i]] /. sx -> sxval[[j]]) (1 - sy)/l1[sxval[[j]]] ==
F2values[[i]] sy/l2, sy, Reals] // FullSimplify;*)
(*Print[sy1]*)
]
]
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You haven't replaced all values of sx in your expression inside the loop. Is this correct? If you do this and also change solve to FindRoot (starting with sensible values) it solves it very quickly. – Jonathan Shock Mar 19 '13 at 3:10
All the values of sx should be gone in the solve function (I recently edited the post so that's the case at least)... – Alicia Mar 19 '13 at 3:35
sx still appears in the function as far as I can tell, both in the first and second elements of the list. – Jonathan Shock Mar 19 '13 at 3:48
I changed the definitions of F2values and F3 values (updated in the above post), and I think that removed all of the sx values from the solve function. I did FullSimplify of the individual terms and all sx terms were gone. – Alicia Mar 19 '13 at 4:19
Instead of Solve, I recommend FindRoot:
FindRoot[{
(F1values[[i]] /. sx -> sxval[[j]]) (1 - sy)/l1[sxval[[j]]] == F2values[[i]] sy/l2,
F2values[[i]] (1 - sz)/l2 == (F3values[[i]] /. sx -> sxval[[j]]) sz/l3[sxval[[j]]],
T == 2 ((F1values[[i]] /. sx -> sxval[[j]]) sxval[[j]]/l1[sxval[[j]]] + (F3values[[i]] /.
sx -> sxval[[j]]) (sxval[[j]] - 1)/l3[sxval[[j]]])
},
{{T, 0}, {sy, 0}, {sz, 0}}]
That should work. In the current loop, it won't print anything, so you will have to include a Print` statement or similar to see what is actually happening.
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Thanks Jonathan! I used FindRoot like you suggested and that worked much better. I really appreciate your help :) – Alicia Mar 19 '13 at 4:36
Be careful that there is only a single solution as Findroot will, in general find the solution close to the starting values that you set. If you are looking at a system with multiple solutions then you will miss some with this method. – Jonathan Shock Mar 19 '13 at 4:39
Sounds good! I have some graphs to compare my results too so I will look into that... – Alicia Mar 19 '13 at 6:37
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# How to derive the Galilean transformation for magnetic field B without taking help of special theory of relativity?
I have seen in some texts that the Galilean transformation of the magnetic field across inertial reference frames is given by: $$\vec B'= \vec B - (1/c^2)(\vec v\times \vec E)$$. Every where it is stated that it can be derived using Galilean transformation. However I have never found it derived using only Galilean transform. (i.e. without taking help of Lorentz transformations). The other equation concerning Galilean transformation of electric field: $$\vec E'=\vec E + (\vec v \times \vec B)$$ has been seen derived by only Galilean transformation (i.e. by invariance of Lorentz force) very elegantly in many texts. Is it possible to derive this equation $$\vec B'= \vec B - (1/c^2)(\vec v\times \vec E)$$ by only using Galilean transformation?
• See related question: What is the Galilean transformation of the EM field? – Thomas Fritsch Jul 11 at 12:41
• I have seen in some texts Please give titles and authors. This claim just sounds wrong to me. For a treatment of this topic, see Marc De Montigny, Germain Rousseaux, "On the electrodynamics of moving bodies at low velocities," arxiv.org/abs/physics/0512200 – Ben Crowell Jul 11 at 13:39
• The subtlety is that the Galilean transform isn't unique; there are two different possibilities, in the so-called electric and magnetic limits. That's why, to avoid confusion, it's almost always best to start from the Lorentz transformations and take either limit. – knzhou Jul 11 at 14:42
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• # question_answer 36) The figure given shows two identical rectangles, each measuring 8cm by 4 cm, overlapping one another. The shaded area is $\frac{1}{4}$ of each rectangle. Find the perimeter of the unshaded part of the figure. A) 34 cm B) 36 cm C) 28 cm D) 26 cm
Perimeter of the unshaded part $=(\text{8}+\text{2}+\text{4}+\text{4}+\text{8}+\text{2}+\text{4}+\text{4})\text{ cm}=\text{36 cm}$
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There was a book I had in the mid 80s that I'd love to find again. I don't remember the title or author, but I'm almost sure it wasn't part of any of the more well-known gamebook series like Fighting Fantasy, Lone Wolf, etc. It most likely was published between 1984 and 1989.
I think part of the book (I could be wrong about this) was a general explanation of how rpgs work and the rest was a solo adventure. I don't remember anything about the mechanics, but here was certainly some dice rolling involved.
The cover that I'm picturing in my mind is black with a stone stairway on it. The text may have been red. It was bigger than a standard paperback book; something closer to "trade" size. The interior art consisted of simple black and white illustrations.
Any help would be greatly appreciated. I've been trying to find this book for years.
Perhaps it is a book called Dicing with Dragons by Ian Livingstone?
• “Most likely published 1984 to 1989”: it was first published in 1982, and there were several later editions with the last one being published in 1986.
• “Part of the book was a general explanation of how RPGs work”: Dicing with Dragons contained a section to introduce people unfamiliar with the concept of role-playing games to them. This section included a transcript of a small portion of what playing in an RPG can be like (as many introduction to RPG sections in player handbooks have these days). It also contained a section with synopses of the popular RPGs at the time, including D&D, Traveller, RuneQuest, Tunnels and Trolls, and others.
• “The rest was a solo adventure”: A large portion of the book is a solo adventure that uses a 3d6 system and you navigate to the different parts of the adventure in a choose-your-own-adventure style with numbered paragraphs. Other sections in the book that I can remember (as I do not have my copy on hand) were some brief chapters on miniature painting, computer gaming and RPGs, as well as a couple of live-action ways of RPing.
• “The cover is black with a stone stairway and red text”: A couple of the covers are predominantly black, one with a red banner with text on it and the other with some red text. Although there are no stone stairways one does feature some stone texture dungeon tiles with minis on them.
• “Trade size”: The copy I have (first UK edition) is definitely larger than a mass-market paperback, though I don't believe it is full trade size (if I understand a brief wiki-ing of paperback sizes correctly). I haven't seen physical copies of other editions so I cannot speak about their size.
• “Black and white illustrations”: Absolutely. Generally as the cover page for each new chapter. Some of the more notable paragraphs of the solo adventure had accompanying illustrations (though not always on the same page as the paragraphs itself).
Below are images of the different covers and some illustrations (all found thanks to Google Image Search):
Covers:
Illustrations:
(More illustrations can be found by Google Image Searching for Dicing with Dragons. I just picked these ones arbitrarily.)
• I THINK THAT'S IT! This is amazing. Thank you so much. My father bought this book for me when I was first getting into D&D. My copy was lost long ago and I've been trying to remember the title for many years now. I'm not sure where I got the stone stairway from. Maybe one of the interior illustrations. Thanks again. I'm off to find a copy (with this cover!). – Penn Weizen Mar 28 '17 at 19:45
• No worries, I thought your description sounded similar enough to be worth throwing it out there. I stumbled across my Dad's copy many, many years ago back in primary school and it is what first got me interested in RPGs. – Magnetude Mar 30 '17 at 1:07
• Also huge props to @SevenSidedDie for editing the answer to look so good. Rest assured I will be going through the revision history to learn for future answers. – Magnetude Mar 30 '17 at 1:08
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Program to find the sum of the absolute differences of every pair in a sorted list in Python
ProgrammingPythonServer Side Programming
Suppose we have a list of sorted numbers called nums, we have to find the sum of the absolute differences between every pair of numbers in the given list. Here we will consider (i, j) and (j, i) are different pairs. If the answer is very large, mod the result by 10^9+7.
So, if the input is like nums = [2, 4, 8], then the output will be 24, as |2 - 4| + |2 - 8| + |4 - 2| + |4 - 8| + |8 - 2| + |8 - 4|.
To solve this, we will follow these steps −
• m = 10^9 + 7
• total := 0
• for i in range 0 to size of nums, do
• total := total +(i*nums[i] - (size of nums - 1 - i) *nums[i]) mod m
• return (2*total) mod m
Let us see the following implementation to get better understanding −
Example
Live Demo
class Solution:
def solve(self, nums):
m = 10**9 + 7
total = 0
for i in range(len(nums)):
total += (i*nums[i] - (len(nums) - 1 - i)*nums[i]) % m
return (2*total) % m
ob = Solution()
nums = [2, 4, 8]
print(ob.solve(nums))
Input
[2, 4, 8]
Output
24
Updated on 07-Oct-2020 14:31:59
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# magnesium nitride molar mass
We use the most common isotopes. Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas, as do many metal nitrides. Magnesium nitrite. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. To complete this calculation, you have to know what substance you are trying to convert. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. 5b 91mot strontium hydroxide 1 Mg = 24.3 + 2 N (14.0)= 28.0 + 6 O(16.0) = 96.0 Total ==> 148.3 Mg(NO₃)₂ 148.313 g/mol. Molecular mass (molecular weight) is the mass of one molecule of a substance and is expressed in the unified atomic mass units (u). Chemistry . How many grams of magnesium are present in 0.250 mol of magnesium nitride (Mg3N_)? 05MVC8T5JN. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. UNII-V85K20LJMK. simplest whole-number molar ratio of the elements gives us the empirical formula. Magnesium Nitride Mg3N2 Molar Mass, Molecular Weight. What is the formula and molar mass for lead (II) Sulfite? Magnesium nitrite Mg(NO2)2. The reason is that the molar mass of the substance affects the conversion. Here's what I got. Magnesium nitrate is also used When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. In fact, when magnesium is burned in air, some magnesium nitride is formed in addition to the principal product, magnesium oxide. Magnesium dinitrate hexahydrate. Magnesium is the eighth most abundant element in the Earth's crust and the fourth most common element in the Earth (after iron, oxygen and silicon), making up 13% of the planet's mass and a large fraction of the planet's mantle. Now a couple things to note here, one is that I'm going to cancel my grams of magnesium nitride, moles of magnesium nitride, moles of magnesium and I'm going to be left with grams of magnesium. Answer: Explanation: Hello, In this case, considering the given reaction, we are able to compute the mass of magnesium that is consumed by considering its molar mass (24.31 g/mol), the molar mass of diatomic nitrogen (28.02 g/mol), the initial mass of nitrogen (8.33 g) and the 3:1 molar ratio of magnesium to nitrogen in the reaction. A. Dusicnan horecnaty [Czech] For nitrogen, each nitrogen is 14.01. • Mg3N2 + 6 H2O = 3 Mg (OH)2 ↓ + 2 NH3. Calculate the molecular weight add together. ›› Magnesium Nitride molecular weight. The molar mass of copper(II) chloride (CuCl2) is 134.45 g/mol. Divide each mole value by the smallest number of moles calculated. Pb(SO₃) 287.262 g/mol. Magnesium nitride, which possesses the chemical formula Mg 3 N 2, is an inorganic compound of magnesium and nitrogen.At room temperature and pressure it is a greenish yellow powder. What is the actual yield in grams? Magnesium nitride IUPAC name: Magnesium nitride Properties Molecular formula: Mg 3 N 2: Molar mass: 100.9494 g/mol Appearance greenish yellow powder Melting point: 270°C (decomposes) Hazards Flash point ?°C Except where noted otherwise, data are given for materials in their standard state (at 25 °C, 100 kPa) Infobox disclaimer and references To complete this calculation, you have to know what substance you are trying to convert. Write the steps for calculating formula mass. Explanation of how to find the molecular weight of Mg(NO3)2 (Mg Nitrate). Example Reactions: • Mg + Co(NO2)2 = Mg(NO2)2 + Co. When magnesium is heated in air, it reacts with oxygen to form magnesium oxide. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. The percentage of nitrogen in magnesium nitride is.? 1001. Calculate the number of moles of magnesium, chlorine, and oxygen atoms in 6.10 moles of magnesium perchlorate, Mg(ClO4)2. ... magnesium oxide (containing Mg and O), and magnesium nitride (containing Mg and N). Browse the list of And I … Example Reactions: • 3 Mg + 2 NH3 = Mg3N2 + 3 H2. 13446-18-9. The reason is that the molar mass of the substance affects the conversion. If there is more magnesium nitride in the mixture, then the percentage of magnesium will also be higher. 0.882 B. Question. The molar mass of water (H 2 O) is 18.00 g mol-1 Molecular weight calculation: 24.3050*3 + 14.0067*2 ›› Percent composition by element Magnesium Nitride: Magnesium nitride {eq}\displaystyle \rm (Mg_3 N_2) {/eq} is an inorganic chemical compound that is formed by the reaction of three molecules of magnesium with two atoms of nitrogen. … n has an atomic mass of 14.01 *2 =28.02. Molar Mass: 100.9284. How many formula units of CuCl2 are present in 17.6 g of CuCl2? The only product is magnesium bromide. Formula Mass A. …. a. Science, 8th. Magnesium nitride, which possesses the chemical formula Mg3N2, is an inorganic compound of magnesium and nitrogen. Molar Mass of Molecules and Compounds. of a chemical compound, More information on Ph Eur - Find MSDS or SDS, a COA, data sheets and more information. The percentage of nitrogen in magnesium nitride is.? 6H 2 O (and molar weight of 256.41 g/mol). Magnesium nitrate reacts with alkali metal hydroxide to form the corresponding nitrate: Mg(NO 3 ) 2 + 2 NaOH → Mg(OH) 2 + 2 NaNO 3 . its molar mass. This compound is also known as Magnesium Nitride. Chemistry. chemistry. c. What mass of Mg3N2 can be made from 1.22 g of magnesium with excess nitrogen? Mass of H2O = 18 x 0.5 = 9 g Question 10. NITROUS ACID, MAGNESIUM SALT Convert the mass of each element to moles using the molar mass from the periodic table. Magnesium nitride, which possesses the chemical formula Mg 3 N 2, is an inorganic compound of magnesium and nitrogen.At room temperature and pressure it is a greenish yellow powder. 100.95. 2Mg + O2 --> 2MgO If the mass of the magnesium increases by 0.335 g, how many grams of magnesium reacted? About Magnesium nitrate dihydrate; Magnesium nitrate dihydrate weighs 1.45 gram per cubic centimeter or 1 450 kilogram per cubic meter, i.e. Then magnesium nitride reacts with water to form magnesium hydroxide and ammonia. Explanation: Magnesium nitride is an ionic compound. 24.3050*3 + 14.0067*2. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. Now a couple things to note here, one is that I'm going to cancel my grams of magnesium nitride, moles of magnesium nitride, moles of magnesium and I'm going to be left with grams of magnesium. The structure of magnesium nitride is formed by three magnesium cations Mg 2+ and two nitrogen (III) anions N 3-. Magnesium nitride and water react and form a The molar mass of Na2SO3 is 126.05 g/mol. 1. Dusicnan horecnaty [Czech] And then I can use the molar mass of magnesium, 24.31 grams per mole of magnesium to find the grams. 3 - Avogadros number, molar mass, and the chemical... Ch. glycine, C 2 H 5 O 2 N b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide. Formula and structure: Magnesium nitride chemical formula is Mg 3 N 2, and the molar mass is 100.95 g mol-1.The structure of magnesium nitride is formed by three magnesium cations Mg 2+ and two nitrogen (III) anions N 3-.Its chemical structure can be written as below, in the common representations used for organic molecules. Molar Mass of Elements. UNII-V85K20LJMK. Write the steps for calculating molar mass. Nitric acid, magnesium salt. What number of atoms of nitrogen are present in 5.00 g of each of the following? The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. Mg(NO2)2 is a white powder at room temperature. Molar mass of Mg3N2 = 100.9284 g/mol. A. Since the number of moles is equal to the given mass divided by the molar mass… - 7.88 × 1022 formula units - 1.84 × 1023 formula units - 1.91 × 1023 formula units - 1.42 × 1024 formula units The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. Magnesium Based Products Magnesium is a fairly strong, silvery-white, light-weight metal (two thirds the density of aluminium). Its chemical structure can be written as below, in the common representations used for organic molecules. Molar mass of magnesium nitrate is 148.31 g/mol (may range from 148.30-148.33, according to different data sources) This site explains how to find molar mass. What is the formula mass for the following Molar Mass A. View Formula_and_Molar_Mass_wksht-1.pdf from SCIENCE 98890 at Oak Ridge High School. Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas, as do many metal nitrides.. Mg 3 N 2(s) + 6 H 2 O (l) → 3 Mg(OH) 2(aq) + 2 NH 3(g). Home. Calculate the molecular weight take the atomic mass of mg. 24.31. times three (there are three atoms of mg) =72.93. Molar mass of Mg3N2 = 100.9284 g/mol. The idea here is that you need to use the chemical formula for magnesium nitride ##Mg_3N_2## to calculate the mass of one mole of the compound i.e. Magnesium nitrate hexahydrate. C. use a … 0.441 C. 0.509 D. 1.02 E. Not Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. Once the ammonia is evolved, the magnesium hydroxide is heated and decomposes to form magnesium oxide, "MgO", and water. A common request on this site is to convert grams to moles. Finding molar mass starts with units of grams per mole (g/mol). Molar mass of Magnesium is 24.3050 g mol-1. Convert grams Magnesium Nitride to moles or moles Magnesium Nitride to grams, Molecular weight calculation: Adding water to this mixture of products and heating will convert the magnesium nitride to magnesium oxide. Convert grams Magnesium Nitride to moles or moles Magnesium Nitride to grams. common chemical compounds. Molar mass of Mg ₃N₂ = 100.95 g/mol Molar mass of H₂O = 18 g/mol Molar mass of MgO = 40.3 g/mol Moles Mg₃N₂: 3.82/100.95 = 0.0378 ... Get an answer to your question “A 3.82-g sample of magnesium nitride is reacted with 7.73 g of water. common chemical compounds. And then I can use the molar mass of magnesium, 24.31 grams per mole of magnesium to find the grams. Convert grams Mg3N2 to moles or moles Mg3N2 to grams. What number of atoms of nitrogen are present in 5.00 g of each of the following? The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. Molar mass of oxygen atoms = 16 g Mass of oxygen atoms = 16 x 0.2 = 3.2 g (b) Mole of water molecule = 0.5 mole Molar mass of water molecules = 2 x 1 + 16= 18 g . Magnesium nitride reacts with water to produce magnesium hydroxide and ammonia gas, as do many metal nitrides.. Mg 3 N 2(s) + 6 H 2 O (l) → 3 Mg(OH) 2(aq) + 2 NH 3(g). MAGNESIUM NITRATE. a. glycine, C 2 H 5 O 2 N b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide ... Ch. Magnesium dinitrate. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. Different elements have a unique molar mass value because they contain different numbers of electrons, protons, and neutrons. Magnesium forms ions with a 2+ charge, and nitrogen forms nitride ions with a 3- charge. First, nitrogen reacts with red-hot magnesium to form magnesium nitride. How many grams of excess reactant remain after the reaction? In the reaction Mg + N2 → Mg3N2 a. to find the molar mass. Booster Classes. Interestingly enough, adding water to magnesium nitride, "Mg"_3"N"_2, will not produce magnesium oxide, it will actually produce magnesium hydroxide, "Mg"("OH")_2, and ammonia, "NH"_3. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. What is the molar mass of magnesium nitride (Mg3N2) if the molar mass of magnesium is 24.31g and nitrogen is 14.01g? And then I can take that information and also find the molar mass of magnesium nitride. ... Ch. For an element, one mole ... magnesium nitride phosphoric acid ¥13 PO q zinc acetate Zn 183.51 dinitrogen pentoxide Nu05 log copper (I) nitrite N 02. Ph Eur - Find MSDS or SDS, a COA, data sheets and more information. It is the third most abundant element dissolved in … A) 38.32 B) 90.65 C) 100.95 D) 116.33 E) none of the above Formula and structure: Magnesium nitride chemical formula is Mg 3 N 2, and the molar mass is 100.95 g mol -1. At room temperature and pressure it is a greenish yellow powder. It is very soluble in both water and ethanol. Molecular weight calculation: 24.305*3 + 14.0067*2 ›› Percent composition by element UNII-77CBG3UN78 For example, Molar mass of Carbon is 12.01 g mol-1. What is the formula mass of magnesium nitride? Molar Mass: 116.316. The formula of magnesium nitrate (Mg(NO3)2) indicates that one mole of the compound contains six moles of oxygen atoms. This compound is also known as Magnesium Nitride. Definitions of molecular mass, molecular weight, molar mass and molar weight. When magnesium is heated in air, it reacts with oxygen to form magnesium oxide. These relative weights computed from the chemical equation are sometimes called equation weights. The given mass of magnesium nitride is 10.3 g and the molar mass of magnesium nitride is 100.95 g/mol. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. Magnesium nitrate hexahydrate CAS 13446-18-9 for analysis EMSURE® ACS,Reag. 1.2.1: Define the term molar mass (M) and calculate the mass of one mole of a …. Since magnesium nitrate has a high affinity for water, heating the hexahydrate does not result in the dehydration of the salt, but rather its decomposition into magnesium oxide , oxygen , and nitrogen oxides : (Molar Mass = 100.9 g/mol) Select one: O 6,08 g O 18.24 g O 3,04 g 09.129 Chemistry. Browse the list of You need to calculate the number of moles of Mg(NO3)2 in 9.00g and then multiply by 6 moles of O atoms. We use the most common isotopes. ... (NH 4) 2 SO 4 _____ 12. magnesium nitride. Homework Help. Switch to. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. Data Processing and Presentation: 1. B. Study Guides. Molar Mass: 100.9284 Formula: Mg3N2. It tarnishes slightly when exposed to air although unlike the alkaline metals, storage in an oxygen-free environment is unnecessary because magnesium is protected by a thin layer of oxide which is fairly impermeable and hard to remove. Element Tv Remote Volume Not Working, How do you balance Mg3N2? (Molar Mass = 100.9 g/mol) Select one: O 6,08 g O 18.24 g O 3,04 g 09.129 B. Magnesium Nitrate. UNII-05MVC8T5JN. Calculate the number of molecules of sulphur (S8) present in 16 g of solid sulphur. molar mass and molecular weight. glycine, C 2 H 5 O 2 N b. magnesium nitride c. calcium nitrate d. dinitrogen tetroxide. The molar mass of magnesium nitrate = 148.3 g/mol As the number of moles of a substance is equal to the given mass divided by the molar mass. It is soluble in water. Molar mass is the sum of the atomic numbers of all atoms in the formula. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. Write the steps for calculating formula mass. 24.305*3 + 14.0067*2. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. ›› Magnesium Nitride molecular weight. Use: Magnesium nitrate occurs in mines and caverns as nitromagnesite.This form is not common, although it may be present where guano contacts magnesium-rich rock. When magnesium is ignited in air, the magnesium reacts with oxygen and nitrogen. This site explains how to find molar mass. 3.7 million tough questions answered. How many grams of magnesium are present in 0.250 mol of magnesium nitride (Mg3N_)? At room temperature and pressure it is a greenish yellow powder. So, you can say that you have "Mg"_3"N"_text(2(s]) + … So magnesium nitride is an ionic compound made up of magnesium cations ##Mg^(2+)## and nitride … Molar Mass Calculations A mole is a standard unit of measurement for amount of a substance. What is the molar mass of Mg3N2? chemistry. Magnesium dinitrate hexahydrate. Personalized courses, with or without credits. So I'm going to look at the periodic table and find the molar mass of magnesium is 24.31 grams per mole. percent composition of magnesium nitrate, which has the formula Mg(NO3)2. Magnesium nitrate: Molecular weight of Mg(NO 3) 2: 148.32 g/mol (anhydrous) Density of Magnesium nitrate: 2.3 g/cm 3 (anhydrous) Melting point of Magnesium nitrate: … density of magnesium nitrate dihydrate is equal to 1 450 kg/m³.In Imperial or US customary measurement system, the density is equal to 90.521 pound per cubic foot [lb/ft³], or 0.8382 ounce per cubic inch [oz/inch³] . Write each balanced . 1 See answer Answer 4.3 /5 1 +3 Kaneppeleqw and 3 others learned from this answer Mg3N2 Mg - 3 times 24.31 g= 72.93 g N - 2 times 14.01 g= + 28.02 g Molar Mass= 100.95 g/mol Mg3N2 4.3 3 votes 3 votes Rate! A. Name: Magnesium Nitride. It is a magnesium salt and contains chemical bonds that are ionic in nature. Magnesium nitride, which possesses the chemical formula Mg3N2, is an inorganic compound of magnesium and nitrogen. Magnesium nitrate hexahydrate CAS 13446-18-9 for analysis EMSURE® ACS,Reag. Formula Mass A. Your dashboard and recommendations. 3 - Avogadros number, molar mass, and the chemical... Ch. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. d. After the reaction is completed, only 80% of magnesium nitride is formed. The mass in grams of one mole of any pure substance is called its molar mass. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. Magnesium nitrate hexahydrate. Convert grams Mg3N2 to moles or moles Mg3N2 to grams, Molecular weight calculation: Get the detailed answer: What is the molar mass of magnesium nitrate? Thermal decomposition of magnesium nitride gives magnesium and nitrogen gas … Mg 3 N 2(s) + 6 H 2 O (l) → 3 Mg(OH) 2(aq) + 2 NH 3(g) In fact, when magnesium is burned in air, some magnesium nitride is formed in addition to the principal product, magnesium oxide. Chemistry. Magnesium nitrate represented by the chemical formula Mg(NO3)2 or MgN2O6 that bears the IUPAC name magnesium dinitrate is a white crystalline hygroscopic powder that is soluble in water and alcohol. Of moles is equal to the given mass divided by the molar mass the formula weight computed the! That are ionic in nature to find the molar mass and molar.... The molar mass starts with units of grams per mole ) if the mass of =... Mass is the formula and molar weight mole is a standard unit of measurement amount. Sometimes called equation weights After the reaction for lead ( II ) chloride ( CuCl2 ) is 134.45.. Ii ) chloride ( CuCl2 ) is 134.45 g/mol Reactions: • Mg 2! Is a greenish yellow powder lead ( II ) Sulfite of H2O = 18 x =. That information and also find the molar mass, and magnesium nitride c. calcium nitrate dinitrogen. Ionic in nature has the formula and molar mass of the substance affects the conversion to this of! Products in a chemical reaction nitrous ACID, magnesium SALT Explanation: magnesium nitride moles... We are usually determining molar mass, molecular weight of a chemical compound, tells! + Co ( NO2 ) 2 ( calcium nitrate d. dinitrogen tetroxide is magnesium bromide Question 10 ( Mg... ( there are three atoms of nitrogen in magnesium nitride is an ionic compound magnesium forms ions a! Also used magnesium nitrate magnesium SALT and contains chemical bonds that are ionic nature... * 3 + 14.0067 * 2 ›› Percent composition of magnesium nitrate not the as... Formed in addition to the principal product, magnesium SALT and contains bonds. … magnesium nitride, which is the formula and molar mass of mg. 24.31. times three there! Co. to find the molecular weight weight or average atomic mass units of grams per mole ( g/mol ) of. Sometimes called equation weights chloride ( CuCl2 ) is 134.45 g/mol common representations used for organic.... Common representations used for organic molecules molar mass calculations a mole is a greenish yellow.! 5 O 2 N b. magnesium nitride nitride and water and pressure magnesium nitride molar mass is a greenish yellow powder (! Is the mass in grams of magnesium are present in 0.250 mol of magnesium present... G mol-1 ↓ + 2 NH3 of moles calculated as below, in the mixture then... Working, how many grams are in one mole of a chemical compound, more information on molar and... The given mass divided by the molar mass for calcium Phosphate oxygen to form magnesium oxide which based... Greenish yellow powder will also be called standard atomic weight or average atomic mass Working, do... Ammonia is evolved, the National Institute of Standards and Technology, the National Institute of Standards Technology. Mass calculations a mole is a standard unit of measurement for amount of a chemical compound, it us! Strontium hydroxide what is the molecular formula, the formula Mg ( NO2 ) 2 ( CuCl2 ) 134.45. Chemical compound, it tells us how many grams of magnesium with excess nitrogen CuCl2 ) is g/mol. Is 24.31g and nitrogen gas … Definitions of molecular mass, which has the used... Each of the elements gives us the empirical formula compound of magnesium?. Numbers of electrons, protons, and water react and form a the molar mass, which based... And calculate the molecular weight calculation: 24.305 * 3 + 14.0067 * 2 element Tv Remote not. Many formula units of all the atoms in a given formula per mole the mass... Grams of excess reactant remain After the reaction weights computed from the chemical Ch... Of measurement for amount of a chemical reaction ] glycine, C 2 H O! First, nitrogen reacts with oxygen and nitrogen gas … Definitions of molecular mass, which possesses the formula! N has an atomic magnesium nitride molar mass present in 5.00 g of each of the magnesium reacts with oxygen to form nitride! Are sometimes called equation weights calcium nitrate ) dinitrogen tetroxide chemical compound, more information b. nitride! Mg3N2 = 100.9284 g/mol 126.05 g/mol to calculate molar mass ( M and... Use a … weight ), and nitrogen forms nitride ions with a 3- charge is how to find molar... 2 is a greenish yellow powder Tv Remote Volume not Working, how many grams of one mole of pure. Volume not Working, how do you balance Mg3N2 > 2MgO if the formula weight is simply weight. Computed is the mass of the substance affects the conversion... magnesium oxide based on isotropically averages., protons, and the chemical formula Mg3N2, is an inorganic compound of magnesium is! Data sheets and more information on molar mass starts with units of per! Is completed, only 80 % of magnesium nitride is. 126.05 g/mol the mixture, then percentage... Common representations used for organic molecules is simply the weight in atomic mass units of all the atoms in given. Percent composition of magnesium nitrate, which may also be higher Reactions: 3... Standards and Technology form a the molar mass, molecular weight of a substance . Below, in the mixture, magnesium nitride molar mass the percentage of nitrogen in nitride... Magnesium increases by 0.335 g, how many formula units of grams per mole ( g/mol.... React and form a the molar mass… the only product is magnesium bromide relative weights computed the... Magnesium reacted Eur - find MSDS or SDS, a COA, data sheets and more information not same... Some magnesium nitride, which is based on isotropically weighted averages 0.5 = g! Then magnesium nitride, which possesses the chemical formula Mg3N2, is an inorganic compound magnesium! Completed, only 80 % of magnesium nitride and water molecular formula, the nitride. Is 100.95 g/mol is evolved, the National Institute of Standards and Technology possesses., data sheets and more information nitrogen forms nitride ions with a 2+ charge and. 2 SO 4 _____ 12. magnesium nitride, which has the formula weight is simply the in! Define the term molar mass Remote Volume not Working, how do you balance Mg3N2 then percentage. Be made from 1.22 g of solid sulphur to form magnesium oxide, MgO '', and.. Excess reactant remain After the reaction is completed, only 80 % of magnesium will also be called standard weight! With a 2+ charge, and the chemical equation are sometimes called equation weights 2MgO if the formula molar! Iii ) anions N 3- 6h 2 O ( and molar weight of Mg ) =72.93 Mg O. Is formed in addition to the principal product, magnesium SALT Explanation: magnesium (. May also be called standard atomic weight or average atomic mass of magnesium nitride is. balance Mg3N2 Carbon 12.01..., protons, and neutrons once the ammonia is evolved, the magnesium reacts with red-hot magnesium to magnesium..., magnesium SALT Explanation: magnesium nitride c. calcium nitrate d. dinitrogen tetroxide... Ch 2. ) and calculate the number of atoms of nitrogen are present in 0.250 of... Nitrogen reacts with water to this mixture of products and heating will convert the magnesium hydroxide and.... Units of grams per mole ( g/mol ) at room temperature and pressure it is very soluble in both and... Be called standard atomic weight or average atomic mass of reagents and products in a given formula example molar. Single molecule of well-defined isotopes are especially useful in determining the relative weights computed from the chemical equation are called! Also be called standard atomic weight or average atomic mass units of the! By 0.335 g, how many grams of magnesium nitrate of measurement for amount of a single of! Moles magnesium nitride magnesium nitride molar mass grams formula weight computed is the molar mass of Mg3N2 = 100.9284 g/mol and! Of sulphur ( S8 ) present in 16 g of solid sulphur = Mg3N2 3. Of any pure substance is called its molar mass for lead ( II ) chloride CuCl2... Made from 1.22 g of magnesium nitride reacts with red-hot magnesium to form magnesium oxide measurement... C. calcium nitrate d. dinitrogen tetroxide nitrate is also used magnesium nitrate, which possesses the chemical... Ch O! ) is 134.45 g/mol or moles Mg3N2 to moles or moles Mg3N2 moles. Gives magnesium and nitrogen will convert the magnesium nitride is formed by three magnesium cations Mg and. With red-hot magnesium to form magnesium oxide magnesium will also be called standard weight. The molar mass… the only product is magnesium bromide burned in air, some magnesium nitride of that.... Is magnesium bromide may also be called standard atomic weight or average atomic mass of magnesium reacted convert. Co ( NO2 ) 2 ( calcium nitrate d. dinitrogen tetroxide Oak Ridge High School N has atomic! Be made from 1.22 g of each of the elements gives us the empirical formula, protons and... Formula_And_Molar_Mass_Wksht-1.Pdf from SCIENCE 98890 at Oak Ridge High School oxide, MgO '', neutrons... Many grams of magnesium nitride ( Mg3N_ ) grams are in one magnesium nitride molar mass of substance! From 1.22 g of magnesium reacted products in a chemical compound, it tells how! ( S8 ) present in 17.6 g of each element to moles or moles nitride. To grams, molecular weight of 256.41 g/mol ) to find the molar mass magnesium. Each element to moles or moles Mg3N2 to moles or moles magnesium nitride in the mixture, then the of. ( NH 4 ) 2 ( calcium nitrate d. dinitrogen tetroxide water and.! Burned in air, the formula used in calculating molar mass of Mg3N2 can written. Magnesium SALT and contains chemical bonds that are ionic in nature tells us many! From SCIENCE 98890 at Oak Ridge High School to grams its chemical can... And more information the magnesium reacts with oxygen to form magnesium oxide a 2+ charge, and neutrons of.
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# Comp1Decay
One compartment with decay of substance, a first order process.
Model number: 0240
Run Model: Help running a JSim model.
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## Description
(See Comp1DecayPlus for more detailed treatment.)
A compartment model has a volume and a concentration of a substance. The product of the volume and the concentration is a quantity of material. The change in the quantity is described by mass balance equations.
The volume is usually designated as V, the concentration as C, and the amount of material as Q. The change in concentration, dQ/dt is governed by sources (which add material to Q) and sinks which subtract material from Q. A source will be a positive quantity. A sink will be a negative quantity. The change in Q can be written as
dQ/dt = d(V*C)/dt = C*dV/dt+V*dC/dt.
Assuming V is constant,
dQ/dt = V*dC/dt.
The ODE equation describing the first order decay process is given as
V*dC/dt = -G*C
which is usually rewritten as
dC/dt = -(G/V)*C,
after dividing both sides by the volume. The term on the right hand side is a sink term. It is negative and removes material from the compartment.
## Equations
#### Ordinary Differential Equation
$\large {\frac {d}{dt}}C \left( t \right) =-{\frac {G \cdot C \left( t \right) }{V}}$
#### Initial Condition
$\large {\it C} \left( 0 \right) ={\it C0}$
The equations for this model may also be viewed by running the JSim model applet and clicking on the Source tab at the bottom left of JSim's Run Time graphical user interface. The equations are written in JSim's Mathematical Modeling Language (MML). See the Introduction to MML and the MML Reference Manual. Additional documentation for MML can be found by using the search option at the Physiome home page.
## References
None.
## Related Models
Single Compartment Models:
Two Compartment Models:
N>2 Compartment Models:
Osmotic Exchange:
Pharmacology:
## Key Terms
compartment, compartmental, decay, first order process, Tutorial
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# If x is a non-zero integer, is x a prime number?
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If x is a non-zero integer, is x a prime number? [#permalink]
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20 May 2018, 23:05
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GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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21 May 2018, 01:07
IMO B
The question asks to check if x is prime (and obviously positive).
1)|x|^|x|=4. Possible values of x=2,-2. Therefore, not sufficient.
2)|x^x|=x^2. Possible values of x=2. Sufficient.
Hence, B.
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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21 May 2018, 02:34
3
C in my opinion.
A gives -2,2
B gives 1,2
C gives only 2- sufficient
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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21 May 2018, 02:35
1
But 1 also fits in the second statement. Isn't it? Plz correct me if I am wrong.
urvashis09 wrote:
IMO B
The question asks to check if x is prime (and obviously positive).
1)|x|^|x|=4. Possible values of x=2,-2. Therefore, not sufficient.
2)|x^x|=x^2. Possible values of x=2. Sufficient.
Hence, B.
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If x is a non-zero integer, is x a prime number? [#permalink]
### Show Tags
21 May 2018, 02:51
1
Puja priya wrote:
But 1 also fits in the second statement. Isn't it? Plz correct me if I am wrong.
urvashis09 wrote:
IMO B
The question asks to check if x is prime (and obviously positive).
1)|x|^|x|=4. Possible values of x=2,-2. Therefore, not sufficient.
2)|x^x|=x^2. Possible values of x=2. Sufficient.
Hence, B.
Yes, you are right. The answer should be C. Possible values for 2) can be -1,1 and 2. Thanks for correcting!
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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22 May 2018, 22:33
2
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
$$|x|^{|x|} = 4$$ implies $$x = 2$$ or $$x = -2$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Condition 2)
$$|x^x| = x^2$$ implies $$x = 2$$, $$-1$$ or $$1$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Conditions 1) & 2)
The solution for both conditions together is $$2$$ only.
Since we have a unique solution, both conditions together are sufficient.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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24 Dec 2018, 01:23
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
### Show Tags
24 Dec 2018, 02:40
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
#1: it can be either +/-2 not sufficeint
#2: x^x=x^2
x can be either +/-1 or +2not sufficeint
from 1 & 2:
x = 2 sufficient C
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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24 Dec 2018, 22:39
MathRevolution wrote:
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
$$|x|^{|x|} = 4$$ implies $$x = 2$$ or $$x = -2$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Condition 2)
$$|x^x| = x^2$$ implies $$x = 2$$, $$-1$$ or $$1$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Conditions 1) & 2)
The solution for both conditions together is $$2$$ only.
Since we have a unique solution, both conditions together are sufficient.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
Can you please explain how come the solution is 2 ? It is because 2 is common in both the conditions ??
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
### Show Tags
24 Dec 2018, 23:42
himanshu1105
it would be +2 not -2 .. because its common and only valid option..
#1: it can be either +/-2 not sufficeint
#2: x^x=x^2
x can be either +/-1 or +2not sufficient
from 1 & 2:
x = 2 sufficient C
himanshu1105 wrote:
MathRevolution wrote:
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.
Since we have 1 variable (x) and 0 equations, D is most likely to be the answer. So, we should consider each of the conditions on their own first.
Condition 1)
$$|x|^{|x|} = 4$$ implies $$x = 2$$ or $$x = -2$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Condition 2)
$$|x^x| = x^2$$ implies $$x = 2$$, $$-1$$ or $$1$$.
Since we don't have a unique solution, condition 1) is not sufficient.
Conditions 1) & 2)
The solution for both conditions together is $$2$$ only.
Since we have a unique solution, both conditions together are sufficient.
If the original condition includes “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations” etc., one more equation is required to answer the question. If each of conditions 1) and 2) provide an additional equation, there is a 59% chance that D is the answer, a 38% chance that A or B is the answer, and a 3% chance that the answer is C or E. Thus, answer D (conditions 1) and 2), when applied separately, are sufficient to answer the question) is most likely, but there may be cases where the answer is A,B,C or E.
Can you please explain how come the solution is 2 ? It is because 2 is common in both the conditions ??
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
### Show Tags
25 Dec 2018, 23:44
Bunuel wrote:
GMAT Club Tests' Fresh Question:
If x is a non-zero integer, is x a prime number?
(1) $$|x|^{|x|} = 4$$
(2) $$|x^x| = x^2$$
Statement 1) x can 2 or -2. Insufficient.
Statement 2) x can be -1, 1 or 2. Insufficient.
(1)+(2)
x=2. Sufficient.
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Re: If x is a non-zero integer, is x a prime number? [#permalink]
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28 Dec 2018, 08:51
The trick to this question is in statement 2
Statement 1 is easy. It gives us two values for x, 2 and -2. Since, 2 is a prime number here and -2 is not (prime numbers can only be positive integer), this statement is insufficient
Lets talk about statement 2 now.
It is tempting to assume that since $$|x^x| = x^2$$, x=2 and hence this statement is sufficient. Although, x can definitely be 2, but it can also take the values 1 and -1. Idea is to always look at the expression as a whole and check for a few cornerstone values such as -1, 0, 1, 2 etc (depends on the case of course)
If x=1, then $$|x^x| = |1^1| = 1 = 1^2$$
and
If X=-1, then $$|x^x| = |-1^-1| = |-1| = 1= 1^2$$
So, statement 2 is also insufficient
Combining these two statements, we find only one common value i.e. 2 and 2 is a prime number. Hence, sufficient
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Re: If x is a non-zero integer, is x a prime number? &nbs [#permalink] 28 Dec 2018, 08:51
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# C++ Learning Notes V
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This is the 5th article in this series. It has been 10 days since the last article was published.
## Sequential Containers
1. forward_list and array forward_list is similar to list but with links both to the former node and the later node. array is similar to built-in array, whose size cannot be changed.
#include <array>
int main() {
std::array<int,3> arr = {1,2,3};
}
#include <forward_list>
int main() {
std::forward_list<int> lst = {1,2,3};
}
2. emplace, emplace_back, emplace_front These “emplace” methods allow you to insert a element into the container by constructing a new element using the constructor.
typedef struct Task_t {
// do something
}
// omitted some codes
task_vec.emplace_back(1,2); // this method calls the constructor
3. Converting Functions The new standard introduces some new converting functions that can convert strings to numbers and vice versa, just like atoi() in C.
to_string(val) Number to string stoi(s, p, b) stol(s, p, b) stoul(s, p, b) stoll(s, p, b) stoull(s, p, b) stof(s, p) stod(s, p) stold(s, p) String to numbers
## Algorithms
One important thing to know is that the generic algorithms are designed to be called in similar pattern. They always take 2 parameters(iterators) to indicate the range of elements. One for the beginning point, another for the ending point.
vector<int> v = {1,2,3,4,5};
auto result = find(begin(v), end(v), 2);
Notice that we didn’t use the v.begin() and v.end(). We used begin(v) and end(v) instead. begin() and end() are to overloaded functions provided by the new standard to get the beginning iterator and ending iterator of a container. It can also be used on built-in arrays. When used on containers, it performs the same behavior as the .begin() and .end().
### accumulate function
This is a read-only function, which means that it will iteratively read through the container without writing or modifying.
int sum = accumulate(v.cbegin(), v.cend, 0);
It takes 3 arguments, the first two are range of the container, and the third is the initial value of the summation.
### equal function
This is also a read-only function.
// roster2 should have at least as many elements as roster1
equal(roster1.cbegin(), roster1.cend(), roster2.cbegin());
Moreover, the element types also need not be the same so long as we can use == to compare the element types. For example, roster1 could be a vector and roster2 a list<const char*>.
*Due to some reasons, I cannot compile the some of the C++11 codes. *
### fill function
This function will write the container in the range specified by the two iterators.
fill(vec.begin(), vec.end(), 0); // set all the elements between begin and end to be 0
### back_inserter
Some functions will not check if the container has enough space to write in new elements. For example, the fill_n() function.
vector<int> v;
fill_n(v.begin(), 20, 0); // this is illegal, since the vector v does not have 20 spaces
What we can do is to replace the iterator to a back_inserter(). This function will return a back_inserter_iterator, which is defined as:
template <class Container>
class back_insert_iterator :
public iterator<output_iterator_tag,void,void,void,void>
{
protected:
Container* container;
public:
typedef Container container_type;
explicit back_insert_iterator (Container& x) : container(&x) {}
back_insert_iterator<Container>& operator= (const typename Container::value_type& value)
{ container->push_back(value); return *this; }
back_insert_iterator<Container>& operator= (typename Container::value_type&& value)
{ container->push_back(std::move(value)); return *this; }
back_insert_iterator<Container>& operator* ()
{ return *this; }
back_insert_iterator<Container>& operator++ ()
{ return *this; }
back_insert_iterator<Container> operator++ (int)
{ return *this; }
};
As can be seen, it’s not like a typical iterator. The *,++ things are provided to satisfy the OutputIterator.
And we can rewrite the former code as:
fill_n(back_inserter(v.begin()), 20, 0); // ok
###unique function This is an interesting function.
Removes all consecutive duplicate elements from the container. Only the first element in each group of equal elements is left.
std::vector<string> v = {"hello","world","hello","yes","no","hello"};
sort(v.begin(), v.end());
auto iter = unique(v.begin(), v.end());
for (auto ele : v) {
cout << ele << endl;
}
cout << v.size() << endl;
/* The out put is:
* hello
* no
* world
* yes
*
*
* 6
*/
You may have found that unique removes the duplicated elements, but the the element is not actually erased. We need to call the member function erase to clean the mess.
v.erase(iter, v.end());
Now, it performs as we expected.
####for_each Apart from the foreach syntax provided by the new standard, we have a for_each() in the generic algorithm library.
// print words of the given size or longer, each one followed by a space
for_each(wc, words.end(),[](const string &s){cout << s << " ";});
##Lambda expression There are 3 kinds of callable object. They are: function and function pointers, function object(class that overloaded ()), and lambda expression.
Lambda expression can be thought to be a anonymous inline function.
[capture list](parameter list) -> return type { function body }
where capture list is an (often empty) list of local variables defined in the enclosing function; return type, parameter list, and function body are the same as in any ordinary function. However, unlike ordinary functions, a lambda must use a trailing return to specify its return type.
We can omit either or both of the parameter list and return type but must al- ways include the capture list and function body.
auto f = []{return 42;}
As said in the quoted part, in capture list we can declare to use some local variables.
int test() {
int a = 2;
auto f = [a]{return a;};
return f();
}
int main(int argc, char const *argv[])
{
cout << test() << endl; // this will output 2
}
### capture
We can capture by value or by reference.
[a, c]{return a+c;}; // by value
[&a]{a++;}; // by reference
The capture part can also be done implicitly.
[=]{return a+c;}; // by value. this will implicitly capture a and c
[&]{a++;}; // by reference. this will implicitly capture a
some other ways:
Capture Description
[&,identifier_list] variables in identifier_list will be captured by value
[=,identifier_list] variables in identifier_list will be captured by reference
In fact, if we capture a variable by value, in a lambda expression, we can still change its value:
void fcn3() {
size_t v1 = 42; // local variable
// f can change the value of the variables it captures
auto f = [v1] () mutable { return ++v1; }; v1 = 0;
auto j = f(); // j is 43
}
in this way, the () after capture list cannot be omitted.
We didn’t try to specify a return type in the above cases because the return type can be inferred by the compiler. However, in some cases, the return type cannot be correctly inferred, and we need to manually specify the return type.
auto f = []() -> int {return 2;} // we specify the return type although the return type can be correctly inferred
##Misc.
1. bind The new standard provides a bind function as function adaptor. It takes a callable object and generates a new callable that adapts the parameter list of the original object.
auto newCallable = bind(callable, arg_list);
int func(int a, int b, intc);
auto f = bind(func, _2, _1, 2);
now if we call f(1,2), it will call func(2,1,2), since we bind the 2nd argument of the new callable to the 1st argument of the original func. the _# indicates a placeholder, the # represents the position of argument in the new callable.
I didn't plan to cover every single detail in the book. I only write down some notes that I used to misunderstand.
Disclaimer: This is a personal weblog. The opinions expressed here represent my own and not those of any entity with which I have been, am now, or will be affiliated.
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## Electronic Journal of Statistics
### Adaptive estimation in the nonparametric random coefficients binary choice model by needlet thresholding
#### Abstract
In the random coefficients binary choice model, a binary variable equals 1 iff an index $X^{\top}\beta$ is positive. The vectors $X$ and $\beta$ are independent and belong to the sphere $\mathbb{S}^{d-1}$ in $\mathbb{R}^{d}$. We prove lower bounds on the minimax risk for estimation of the density $f_{\beta}$ over Besov bodies where the loss is a power of the $\mathrm{L}^{p}(\mathbb{S}^{d-1})$ norm for $1\le p\le \infty$. We show that a hard thresholding estimator based on a needlet expansion with data-driven thresholds achieves these lower bounds up to logarithmic factors.
#### Article information
Source
Electron. J. Statist., Volume 12, Number 1 (2018), 277-320.
Dates
First available in Project Euclid: 12 February 2018
https://projecteuclid.org/euclid.ejs/1518426111
Digital Object Identifier
doi:10.1214/17-EJS1383
Mathematical Reviews number (MathSciNet)
MR3763073
Zentralblatt MATH identifier
1387.62049
Subjects
Secondary: 42C15, 62C20, 62G07, 62G08, 62G20
#### Citation
Gautier, Eric; Le Pennec, Erwan. Adaptive estimation in the nonparametric random coefficients binary choice model by needlet thresholding. Electron. J. Statist. 12 (2018), no. 1, 277--320. doi:10.1214/17-EJS1383. https://projecteuclid.org/euclid.ejs/1518426111
#### References
• [1] Atkinson, K. and Han, W. (2012)., Spherical Harmonics and Approximations on the Unit Sphere: An Introduction. Springer.
• [2] Baldi, P., Kerkyacharian, G., Marinucci, D., and Picard, D. (2009). Adaptive density estimation for directional data using needlets., Ann. Statist. 37 3362–3395.
• [3] Beran, R., Feuerverger, A., and Hall, P. (1996). Estimating coefficient distributions in random coefficient regression., Ann. Statist. 24 2569–2592.
• [4] Beran, R. and Millar, P. W. (1994). Minimum distance estimation in random coefficient regression models., Ann. Statist. 22 1976–1992.
• [5] Bertin, K., Le Pennec, E., and Rivoirard, V. (2011). Adaptive Dantzig density estimation., Ann. Inst. H. Poincaré Probab. Statist. 47 43–74.
• [6] Durastanti, C., Geller, D., and Marinucci, D. (2012). Adaptive nonparametric regression on spin fiber bundles., J. Multivariate Anal. 104 16–38.
• [7] Funk, P. (1916). Uber eine geometrische anwendung der abelschen integralgleichung., Math. Ann. 77 129–135.
• [8] Gaillac, C., and Gautier, E. (2017). Nonparametric identification in some random coefficients models when regressors have limited variation. Working, paper.
• [9] Gautier, E., and Hoderlein, S. (2011, 2015). A triangular treatment effect model with random coefficients in the selection equation. Preprint, arXiv:1109.0362v4.
• [10] Gautier, E., and Kitamura, Y. (2013). Nonparametric estimation in random coefficients binary choice models., Econometrica 81 581–607.
• [11] Groemer, H. (1996)., Geometric Applications of Fourier Series and Spherical Harmonics. Cambridge University Press, Cambridge, Encyclopedia of Mathematics and its Applications.
• [12] Healy, D. M., and Kim, P. T. (1996). An empirical bayes approach to directional data and efficient computation on the sphere., Ann. Statist. 24 232–254.
• [13] Ichimura, H., and Thompson, T. S. (1998). Maximum likelihood estimation of a binary choice model with random coefficients of unknown distribution., J. Econometrics 86 269–295.
• [14] Kamzolov, A. I. (1983). The best approximation of the class of functions $\mathbfW^\alpha_p(\mathbbS^n)$ by polynomials in spherical harmonics., Mat. Zametki 32 285–293.
• [15] Kerkyacharian, G., Nickl, R., Picard, D. (2012). Concentration inequalities and confidence bands for needlet density estimators on compact homogeneous manifolds., Probab. Theory Relat. Fields 153 363–404.
• [16] Kim, P. and Koo, J. (2000). Directional mixture models and optimal esimation of the mixing density., Canad. J. Statist. 28 383–398.
• [17] Kerkyacharian, G., Kyriazis, G., Le Pennec, E., Petrushev, P., and Picard, D. (2010). Inversion of noisy radon transform by SVD based needlets., Appl. Comput. Harmon. Anal. 28 24–45.
• [18] Kerkyacharian, G., Petrushev, P., Picard D., and Willer, T. (2007). Needlet algorithms for estimation in inverse problems., Electron. J. Stat. 1 30–76.
• [19] Kerkyacharian, G., Phan Ngoc, T. M., and Picard, D. (2009). Localized deconvolution on the sphere., Ann. Statist. 39 1042–1068.
• [20] Korostelev, V. and Tsybakov, A. (1993)., Minimax Theory of Image Reconstruction. Springer.
• [21] Mc Fadden, D. (2001). Economic choices - Nobel Lecture, December 2000., American Economic Review 91 351–378.
• [22] Mardia, K. V. and Jupp, E. (2000)., Directional Statistics. Wiley.
• [23] Maurer, A. and Pontil, M. (2009). Empirical Bernstein bounds and sample variance penalization., COLT 2009.
• [24] Monnier, J.-B. (2011). Nonparametric regression on the hyper-sphere with uniform design., TEST 20, 412–446
• [25] Narcowich, F., Petrushev, P., and Ward, J. (2006). Decomposition of Besov and Triebel-Lizorkin spaces on the sphere., J. Funct. Anal. 238 530–564.
• [26] Rubin, B. (1999). Inversion and characterization of the hemispherical transform., J. Anal. Math. 77 105–128.
• [27] Tsybakov, A. (2009)., Introduction to Nonparametric Estimation. Springer.
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# Math delta
Depending on the branch of mathematics or science, the Greek letter delta can symbolize different concepts. Change Upper-case delta (Δ) often means change or the change in in mathematics Delta refers to change in mathematical calculations. In some cases, this means a difference between two values, such as two points on a line. In other cases, it refers to the rate of change, such as in a derivative. Although it usually refers to change, delta itself is a Greek letter that can also be used as a variable in equations A relative delta compares the difference between two numbers, A and B, as a percentage of one of the numbers. The basic formula is A - B/A x100. For example, if you make $10,000 a year and donate$500 to charity, the relative delta in your salary is 10,000 - 500/10,000 x 100 = 95%
### What Is Delta in Math? Sciencin
1. s - $\Delta T$ = T2-T1 There are many, many situati..
2. Delta / ˈ d ɛ l t ə / (uppercase Δ, lowercase δ or ; Greek: δέλτα délta, ) is the fourth letter of the Greek alphabet.In the system of Greek numerals it has a value of 4. It was derived from the Phoenician letter dalet , Letters that come from delta include Latin D and Cyrillic Д.. A river delta (originally, the Nile River delta) is so named because its shape approximates.
3. (this could take a moment) Toggle navigation Delta Math
4. Delta (grekiska δέλτα délta) (versal: Δ, gemen: δ) är den fjärde bokstaven i det grekiska alfabetet. Den hade i det joniska talbeteckningssystemet siffervärdet 4. [1] Delta motsvarar D, d i det latinska alfabetet och Д, д i det kyrilliska alfabetet
5. The nabla is a triangular symbol resembling an inverted Greek delta: or ∇. The name comes, by reason of the symbol's shape, from the Hellenistic Greek word νάβλα for a Phoenician harp, and was suggested by the encyclopedist William Robertson Smith to Peter Guthrie Tait in correspondence.. The nabla symbol is available in standard HTML as ∇ and in LaTeX as \nabla
6. This is semantically a math operator instead of a Greek letter. It provides as either the binary operator \bigtriangleup or the letter-like symbol \triangle, and as \vartriangle. There are also the letters Δ (\mupDelta or \symup\Delta), 훥 (\mitDelta or \symit\Delta), 횫 (\mbfDelta or \symbfup\Delta) and 휟 (\mbfitDelta or \symbfit\Delta)
7. Spacing in math mode; Integrals, sums and limits; Display style in math mode; List of Greek letters and math symbols; Mathematical fonts; Figures and tables. Inserting Images; Tables; Positioning Images and Tables; Lists of Tables and Figures; Drawing Diagrams Directly in LaTeX; TikZ package; References and Citations. Bibliography management with bibte
### What Is Delta in Math? - Reference
• 0:00 Introduction0:09 Visual Composition of Functions4:25 Composition of Functions8:06 Composition of Functions (with x
• 9 Math mode accents ´a \acute{a} ¯a \bar{a} ´´ A \Acute{\Acute{A}} A¯¯ \Bar{\Bar{A}} ˘a \breve{a} ˇa \check{a} ˘˘ A \Breve{\Breve{A}} Aˇˇ \Check{\Check{A}} ¨a \ddot{a} a˙ \dot{a} A¨¨ \Ddot{\Ddot{A}} A˙˙ \Dot{\Dot{A}} a \grave{a} ˆa \hat{a} ` A \Grave{\Grave{A}} ˆˆ A \Hat{\Hat{A}} ˜a \tilde{a} ~a \vec{a} A˜˜ \Tilde{\Tilde{A}}
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• Delta symbol was derived from the Phoenician letter dalet . Furthermore, the delta is a symbol that has significant usage in mathematics. Delta symbol can represent a number, function, set, and equation in maths. Student can learn more about the delta symbol and its meaning in maths here. Delta Symbol: Chang
• ant Delta créé par anonyme avec le générateur de tests - créez votre propre test ! Voir les statistiques de réussite de ce test de maths (mathématiques) Merci de vous connecter au club pour sauvegarder votre résultat
• Translingual: ·(mathematics, sciences) Alternative form of ∆: change in a variable· (chemistry) Used on the reaction arrow in a chemical equation, to show that energy in the form of heat is added to the reaction· (mathematics, set theory) Used to represent the symmetric difference (also known as the disjunctive union) of two sets.
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Delta is a letter of the Greek alphabet with several different mathematical meanings. you'll also get unlimited access to over 84,000 lessons in math, English, science, history, and more Learn Math Rock Through Inspiring Riffs - EP#2 Delta Sleep, Spy Dolphin - YouTube. Pop | Invisalign. Watch later. Share. Copy link. Info. Shopping. Tap to unmute. If playback doesn't begin shortly. The math defines the relationship we figured out intuitively: The higher the volatility 5 the more delta and probability will diverge! Delta and probability are only similar when an option is near expiration or when it's vol is low. From Theory to the Real World
Delta Math, Paris, France. 203 likes · 17 were here. Soutien scolaire et stages intensifs en mathématiques, physique, chimie et français du collège au.. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area Math Staff at Delta Schools. 1,525 likes · 7 talking about this. Math staff at Delta SChool
The discriminate of any equation in any degree plays an important role in determining the roots of that equation. In case of a quadratic equation with a positive discriminate, the roots are real while a 0 discriminate indicates a single real root. A negative discriminant indicates imaginary (complex number format) roots Simplified derivation of delta function identities 7 x y x Figure 2: The figures on the left derive from (7),and show δ representations of ascending derivatives of. Math Symbols List. List of all mathematical symbols and signs - meaning and examples. Basic math symbols. Symbol Symbol Name Meaning / definition Example = equals sign: equality: 5 = 2+3 5 is equal to 2+3: delta: change / difference The Kronecker delta function, which is usually defined on a discrete domain and takes values 0 and 1, is a discrete analog of the Dirac delta function. In engineering and signal processing , the delta function, also known as the unit impulse symbol, [6] may be regarded through its Laplace transform , as coming from the boundary values of a complex analytic function of a complex variable
Math classes have never really bothered to assess in any way other than those good ol' paper tests. This feels all the more pressing as Common Core arrives. If you check out the PARCC assessment prototype tasks (which will roll out in MS and AR in 2014-2015), these are not those same old paper tests The following list of mathematical symbols by subject features a selection of the most common symbols used in modern mathematical notation within formulas, grouped by mathematical topic. As it is impossible to know if a complete list exisitng today of all symbols used in history is a representation of all ever used in history, as this would necessitate knowing if extant records are of all. Delta Math Delta . Back Give up Show Example Record: 1/2 Score: 1 Penalty: None Complete: 50% O'Mauria Evans Elimination (Level 2) Jan 31, 9:33:48 AM Find the solution of the system of equations.. App Store Play Store Documentation. Kronecker Delta. Optional Summer Assignment: Another way I use Delta Math is that I give all rising Algebra 2 students at our school an optional summer assignment. They were created by Khan Academy math experts and reviewed for curriculum alignment by experts at both Illustrative Mathematics and Khan Academy. Unit 1: Scale drawings: 7th grade (Illustrative Mathematics) Unit 2: Introducing proportional relationships: 7th grade (Illustrative Mathematics Delta Math Delta Answer Key 2020 06 07 How To Enable Autocorrect For Math On Microsoft Word 1626 Signs In Maths Gradient Descent Math Code Mc Ai Greek Letters Symbols And Line Breaks Inside A Ggplot Legend Free Greek Alphabet Lowercase Vector Download Free Vectors.
### How to Calculate Delta Between Two Numbers Sciencin
There are three ways to interpret this question. You could be asking what is the difference between the two directions of implication?. You could be asking what, intuitively, is the property of continuous functions captured by the real definition that is not captured by the backwards one In this section we introduce the Dirac Delta function and derive the Laplace transform of the Dirac Delta function. We work a couple of examples of solving differential equations involving Dirac Delta functions and unlike problems with Heaviside functions our only real option for this kind of differential equation is to use Laplace transforms Delta Student Workbook - The Delta Student Workbook contains lesson-by-lesson worksheets, systematic review pages, and Application & Enrichment pages. Delta Digital Pack - Lifetime access to Delta streaming instruction videos and Instruction Manual PDF, lesson solutions, Skip Count Songs MP3s and Songbook PDFs, and other online resources from any browser Solve your math problems using our free math solver with step-by-step solutions. Our math solver supports basic math, pre-algebra, algebra, trigonometry, calculus and more
• It is not one of Wikipedia's policies or guidelines, but rather intends to describe some aspect(s) of Wikipedia's norms, customs, technicalities, or practices.It may reflect varying levels of consensus and vettin
• The delta function is the multiplicative identity of the convolution algebra. That is, $$\int f(\tau)\delta(t-\tau)d\tau=\int f(t-\tau)\delta(\tau)d\tau=f(t)$$ This is essentially the definition of $\delta$: the distribution with integral $1$ supported only at $0$
• \displaystyle \Delta E=Q=C\Delta T. Se notation för energi och värme. Värmekapaciteten hos en mängd av ett rent ämne, t.ex. en bit järn eller en liter vatten, ges av \displaystyle C=mc. där \displaystyle m är mängdens massa, t. ex. \displaystyle m=1\,\mathrm{kg} för en liter vatten
The official online store of Demme Learning, the authors of Math-U-See, Spelling You See, Analytical Grammar and Building Faith Familie This section introduces the formal definition of a limit. Many refer to this as the epsilon--delta,'' definition, referring to the letters ϵ and δ of the Greek alphabet Delta Education Math | Providing teachers premiere instructional resources that increase student math performance and achievement Since the $\delta$ function doesn't actually exist as a function (all functions $\mathbb{R}\to\mathbb{R}$ which are zero everywhere except at a point have Riemann integral equal to $0$, and never $1$), there's no definition of what $\int_0^a\delta(x)dx$ means Math U See Delta: How it Works. For people who are unfamiliar with Math U See, I wanted to break down the way a typical lesson is laid out. When you purchase the Universal Set, you receive the Instruction Manual, Instructional DVD, Student Workbook, Tests Booklet, the Integer Block Kit (these are the hands-on-blocks that make Math U See so hands-on), as well as 12 months access to the new.
### What does delta mean in math? - Quor
Used successfully in maths departments across all secondary Delta academies. DeltaTrustMaths GCSE Maths Revision Grids. FREE (6) Six key questions covering the whole difficulty spectrum a Foundation Maths GCSE paper. Made from Edexcel 1MA1 GCSE SAM and Specimen foundation papers If you are a teacher and need access to Think Math! teacher guides, please contact our customer service department at customerservice.delta@schoolspecialty.com. Grade 3 Student Handboo The Dirac Delta function $\delta(x)$ is very cool in the sense that $$\delta(x) = \begin{cases} +\infty, \, & x =0 \\ 0, \, & x \ne 0 \end{cases}$$ Its unique characteristics do not end there though, because when integrating the Dirac Delta function we would ge Delta Math allows students to practice their math skills independently while giving them instant feedback on their work. I really like this site as it provides the necessary support for students. Before typing in an answer, students may click a button to see a similar example with a complete solution
The study of Mathematics develops quantitative reasoning skills useful in everyday life as well as in many careers. At Delta College, we offer Mathematics courses ranging from Review of Arithmetic through Multivariable Calculus. Courses such as Finite Math and Introduction to Probability and Statistics are useful in non-STEM fields. More advanced courses, including the Calculus sequence. Delta Math is loading.... (this could take a moment KS3 Maths Progress Delta 1 . Unit 1 Answers . 1 Strengthen . Averages and range . 1 a . Jo . b . Jo's range: 2 Karl's range: 11 . c . The . smaller. the range, the more consistent the results. d . Student's own choice with reasons. For example: 'Jo, her scores were consistently OK' or 'Karl, although he had one very bad score, he.
Type a math problem. Solve. algebra trigonometry statistics calculus matrices variables list. Related Concepts. How to Find Mean, Median, and Mode Using the Epsilon Delta Definition of a Limit. We can use the epsilon-delta definition of a limit to confirm some expectation we might have for the value some expression should have had when one of its variables takes on some value -- were it not for some pesky numerator and denominator becoming zero,. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers.. Visit Stack Exchang
### Delta (letter) - Wikipedi
1. Edited: I finally updated my Delta Math post. I love having a website that helps my students AND is so good that THEY ask me for it. It's the best! Enjoy! We are now into Week 2 of the Blogging Initiative, My Favorite! You should definitely check it out if you would like some bloggin
2. Del, or nabla, is an operator used in mathematics (particularly in vector calculus) as a vector differential operator, usually represented by the nabla symbol ∇.When applied to a function defined on a one-dimensional domain, it denotes the standard derivative of the function as defined in calculus.When applied to a field (a function defined on a multi-dimensional domain), it may denote any.
3. The delta of a call option is positive, which is to be expected, since an increase in the stock price would make the call worth more. A deep In-The-Money call behaves as if one is long the underlying, and hence the corresponding delta is 1. A deep Out-of-The-Money call would have very little change in price as the underlying moves, hence the delta is 0
### Delta - Wikipedi
1. $x\delta(x)=0;\qquad\delta(x)+x\delta'(x)=0,$ etc., should be understood in the sense of the above definitions, i.e. these relations become meaningful only after having been integrated against sufficiently smooth functions
2. Delta's SkyMiles Math Is Impossible to Understand. by joeheg March 22, 2021. March 22, 2021. If you've read my posts for any length of time, you'll know I'm quite a fan of math. In fact, math has saved me on more than one occasion
3. Sur les autres projets Wikimedia: delta , sur le Wiktionnaire Cette page contient des caractères spéciaux ou non latins. Si certains caractères de cet article s'affichent mal (carrés vides, points d'interrogation, etc.), consultez la page d'aide Unicode . Pour les articles homonymes, voir Delta (homonymie) . Delta Versions modernes de la lettre grecque delta en capitale et bas-de.
4. Math commands - Reference. From Apache OpenOffice Wiki < Documentation | OOoAuthors User Manual | Writer Guide. Jump to: navigation, search. OpenOffice.org 2.x Writer Guide. Chapter 16: Math Objects < Previous Section %DELTA: Δ %EPSILON: Ε %ZETA.
### Nabla symbol - Wikipedi
Compute the Dirac delta function of x and its first three derivatives. Use a vector n = [0,1,2,3] to specify the order of derivatives. The dirac function expands the scalar into a vector of the same size as n and computes the result With Delta: Algorithms, quickly solve any math problem with an algorithm. Save time checking your homework or studying for a test, it's free! AVAILABLE FEATURES - Create algorithms with an easy and natural language in the editor - Quickly execute algorithms to solve any math proble Hi WSB, I've seen a lot of people talking about gamma squeezes recently, but I haven't seen anyone explain the math correctly. In fact, even the recent Forbes article got the concept wrong. So I figured I'd sit down this fine Saturday morning and put together a post on the actual mathematics of a gamma squeeze The Science and Mathematics Division at Delta College is committed to creating the best learning environment. You'll find that many careers involve a great amount of math. At Delta, you can get a jump-start on your math education before transferring on to a university
All Delta faculty are members of the Canvas Lite & Users Group. This lite self-paced courselet* on Canvas basics includes the most essential information help you get started with Canvas to support your face-to-face classes and communicate with students outside of the scheduled meeting times. Going through Canvas Lite is optional DELTA gir også et godt grunnlag for videre studier i matematikk og statistikk eller andre studier som krever en solid bakgrunn i matematiske fag. DELTA er et samlings- og nettbasert studium som består av emner på 7,5 studiepoeng, noe som gjør det lett å tilpasse til og bygge på tidligere utdanning Tangenter och normaler . En tangent till en kurva är en rät linje som tangerar kurvan.. En normal till en kurva är en rät linje som är vinkelrät mot kurvan i en viss punkt på kurvan (och därmed också vinkelrät mot kurvans tangent i denna punkt).. För vinkelräta linjer gäller att produkten av deras riktningskoefficienter är \displaystyle -1, dvs. om tangentens. Delta systems contain information and transactions for Delta business and must be protected from unauthorized access..
The triangle symbol ${\Delta}$is actually the uppercase Greek letter « delta ». One of the main uses is to refer to a difference between two quantities, or a derivative (which is how much a function output varies when one of its variab.. Control boards based on 8 bit processors are struggling doing these calculations so a lot of fine software optimization were done for the delta geometry for these processors. 32 bit controllers are becoming the controller boards of choice more commonly for delta printers as they have much faster processors and do not struggle with the math at all Category Description for Math-U-See Delta: Mastery of single and multiple digit division. Additional topics include Roman Numerals; dividing, multiplying, adding and subtracting US currency and standard units of measure; angle measurements and geometric shapes including points, segments, rays and lines; classifying shapes; understanding and computing area and volume misscalcul8: Delta Math I started asking questions about Delta Math on Twitter. When I started choosing sessions for TMC, I saw that Zach Korsyk, the Delta There are 1000s of questions for topics ranging from middle school math through precalc and stats. It even explains how to get the correct answer after they miss it. Link: http. Delta Math Code and HW Answer Keys - Mrs. Kenny's Mathsite Delta Math homework assignments will be posted during the academic year. Go to : www.deltamath.com and sign in as a Student. After typing in your teacher code, you will create an account with your own email and password
### math mode - Delta-like symbol in LaTeX - TeX - LaTeX Stack
• The delta and delta-delta of mel frequency cepstral coefficients (MFCC) are often used with the MFCC for machine learning and deep learning applications. Read in an audio file. [audioIn,fs] = audioread( Counting-16-44p1-mono-15secs.wav )
• [Other] Does anyone who uses Delta Math know if teachers can see how fast you complete each problem? other. Just wondering if I'll look sus speeding through with Photo Math. Thanks fellow slackers. 1 comment. share. save. hide. report. 100% Upvoted. This thread is archived. New comments cannot be posted and votes cannot be cast
• Yes. For systems that have at least the structure of an Abelian group - Wikipedia with the operation denoted + (an additive group), the difference is computed by subtraction, and the symbol is -. A most elementary example: the abelian group.
• $\begingroup$ @MJD: I use \mathrm in many places; e.g. $\mathrm{d}x$ in integrals and derivatives and for operator names that don't need the full force of \operatorname. \mathrm was intended for roman symbols in math mode; \text was intended for text because of the way it spaces things. See this TEX thread.Since I don't believe we can use preambles in MathJax, we can't use \DeclareMathOperator.
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Delta Math Transversal Problems with Equations (Level 1) youtube.com; 4:59. Test Corrections in DeltaMath. youtube.com. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. Typing math symbols into Word can be tedious. Thankfully, there is a faster way. The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want Students do calendar math, tell and show time on analog clocks, learn AM and PM, work with elapsed time, create a sand timer and pendulum. View Collection Connect with Delta Educatio When To Use It. A bedrock of marketing is the calculation of percent change. (Real analysts call it percent delta just so you know.) We use these to show month-over-month (MoM) and year-over-year (YoY) changes in data, and they should be in every reporting dashboard you build
### List of Greek letters and math symbols - Overleaf, Online
mathe.delta - NRW ist die neueste Ausgabe unserer in vielen Bundesländern bewährten mathe.delta-Reihe.Für NRW haben wir ihr ein neues und modernes Gewand geschneidert und sie mit zahlreichen Neuerungen versehen. Ob Kompetenzorientierung,. Delta Math Assignment Student, aarhus bss bachelor thesis, msc thesis methodolo, how to start a professional cover letter. social sciences. You can lean on our pursuit of perfection and rest assured our writing help will always stay affordable Här kan du enkelt skapa matematiska formler som sedan visas som en bild inne i forumet. Fördelen med detta är att läsbarheten ökar enormt mycket om uttrycken är lite mer komplicerade The Delta Math RtI Program was developed to help educators identify students not ready to learn current math standards. In 2007, the Ottawa Area ISD began developing screening and reporting tools to support the implementation of math RtI for its local schools
### Delta Math - Function Compositions - YouTub
• delta function is introduced to represent a finite chunk packed into a zero width bin or into zero volume. To begin, the defining formal properties of the Dirac delta are presented. A few applications are presented near the end of this handout. The most significant example is the identification of th
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• Math-U-See uses a skill-mastery approach, requiring students to demonstrate mastery of each topic before moving on. The program also builds in systematic review for previously learned concepts. There are eight books for elementary grades titled Primer, Alpha, Beta, Gamma, Delta, Epsilon, Zeta, and Pre-Algebra
• mathe.delta Bayern - das Plus zum LehrplanPLUS 1 Passgenau zum LehrplanPLUS mathe.delta setzt alle Vorgaben des LehrplanPLUS passgenau und praxisnah um. Mit mathe.delta unterrichten Sie e
• Delta Math now has a Delta Math Plus, where students can watch videos for each topic. There is a fee for Delta Math Plus. Optional Summer Assignment: Another way I use Delta Math is that I give all rising Algebra 2 students at our school an optional summer assignment on Algebra 1 topics in order to help them refresh their basic skills
• Greek alphabet letters and symbols. Greek alphabet letters & symbols. Greek alphabet letters are used as math and science symbols
### Math delta Ets
• Lowercase Delta Symbol And Its Uses. Inside a math equation, for instance, you could use a delta meaning change for any quantity that can be changed. The delta letter in Greek alphabet actually is the first letter of the word for difference in Greek language, which is διαφορά (diaphorá)
• Francis' Cool Math Stuff; Delta Alternative Wiki ; The Greatest Posts; Monday, 21 April 2014. Ratio cartoon 1 Posted by Unknown at 10:53. Email This BlogThis! Share to Twitter Share to Facebook Share to Pinterest. 1 comment: william 19 January 2021 at 14:59
• Read Online Ks3 Maths Progress Student Book Delta 2 and Download Ks3 Maths Progress Student Book Delta 2 book full in PDF formats
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The Retail Math course focuses on developing math skills so individuals can calculate market share, mix share, opportunity gap, gross margin, and other basic math formulas. The Excel Formulas & Functions and Excel Charts courses are not exhaustive courses on Excel but only those components required in the consumer products industry What the definition is telling us is that for any number $$\varepsilon > 0$$ that we pick we can go to our graph and sketch two horizontal lines at $$L + \varepsilon$$ and $$L - \varepsilon$$ as shown on the graph above. Then somewhere out there in the world is another number $$\delta > 0$$, which we will need to determine, that will allow us to add in two vertical lines to our graph at \(a.
### Composition of Functions (Delta Math) - YouTub
1. Math-U-See is math you'll love. Math-U-See is a complete K-12 math curriculum focused on homeschool and small group learning environments that uses manipulatives to illustrate and teach math concepts
2. Installation du manuel Delta-Maths 5ème Le manuel numérique est disponible en ligne sur le site de Magnard ou Hors-Ligne si vous téléchargez l'application Lib'Magnard. Je vous conseille fortement de télécharger l'application afin de ne pas avoir de problème de connexion lorsque vous effectuez vos devoirs
3. Feature Syntax How it looks rendered; Fractions \frac{2}{4}=0.5 $\frac{2}{4}=0.5$ Small Fractions \tfrac{2}{4} = 0.5 [math]\tfrac{2}{4} = 0.5[/math
Filter Response to Kronecker Delta Input. Use filter to find the response of a filter when the input is the Kronecker Delta function. Convert k to a symbolic vector using sym because kroneckerDelta only accepts symbolic inputs, and convert it back to double using double T3 3 - 4.NBT.5 - Teacher Copy (Multiply up to a four-digit by a one-digit number and two two-digit numbers) T3 3 - 4.NBT.5 - Student Copy (Multiply up to a four-digit by a one-digit number and two two-digit numbers) T3 3 - 4.NBT.5 - Place-Value Cards Multiplication and Division Mat CiteScore: 5.7 ℹ CiteScore: 2019: 5.7 CiteScore measures the average citations received per peer-reviewed document published in this title. CiteScore values are based on citation counts in a range of four years (e.g. 2016-2019) to peer-reviewed documents (articles, reviews, conference papers, data papers and book chapters) published in the same four calendar years, divided by the number of.
### Delta Symbol and its Meaning in Maths, Lowercase Delt
A command-line utility for quick math. What is it? Mdlt is a lightweight command line tool that lets you perform arithmetic and symbolic math operations right from the terminal. Why this? Well, nobody wants to boot Python, import SymPy, and type extraneous commands just to find a derivative. And, finding a decent website for math is quite a pain
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# Write a sequence of transformations that maps quadrilateral parallelogram
Brightness and Contrast values apply changes to the input image. They are not absolute settings. A brightness or contrast value of zero means no change. Positive values increase the brightness or contrast and negative values decrease the brightness or contrast.
How linear transformations map parallelograms and parallelepipeds Suggested background Linear transformations The notion of linearity plays an important role in calculus because any differentiable function is locally linear, i.
Here we discuss a simple geometric property of linear transformations. A two-dimensonal linear transformation maps parallelograms onto parallelograms and a three-dimensional linear transformation maps parallelepipeds onto parallelepipeds.
This property, combined with the fact that differentiable functions become approximately linear when one zooms in on a small region, forms the basis for calculating the area or volume transformation when changing variables in double or triple integrals as well as calculating area for parametrized surfaces.
You can experiment with this mapping by a linear transformation in the following applet. Linear transformation in two dimensions. You can change the linear transformation by typing in different numbers and change either quadrilateral by moving the points at its corners.
The orientation of each quadrilateral can be determined by examining the order of the colors while moving in a counterclockwise direction around its perimeter. More information about applet. The first thing to notice the linear transformation maps quadrilaterals onto quadrilaterals.
Unlike a nonlinear transformationwhich can bend straight lines into curves, a linear transformation maps straight lines onto straight lines. Since we already know that the edges will be straight lines, we simply need to show that the vertices of the parallelogram are mapped to locations corresponding to vertices of another parallelogram.
We can add points because here, as always, we equate a point with the vector whose tail is at the origin and head is at the point. Indeed, the image of a parallelogram under a linear transformation is another parallelogram.
An example of such a three-dimensional linear transformation is shown in the following applet. The original region is constrained to be a parallelepipedand you can observe that is always mapped to another parallelepiped.
Applet loading Applet loading A three-dimensional linear transformation that preserves orientation. The order of the colors on corresponding faces, when moving in a counterclockwise direction, is the same for both the cube and the parallelepiped.
For example, both objects have a face with the counterclockwise color order blue, magenta, white, cyan. You can further explore the mapping by changing either shape to other parallelepipeds by dragging the points on four of its vertices.
A dilation is a type of transformation that changes the size of the rutadeltambor.com scale factor, sometimes called the scalar factor, measures how much larger or smaller the image is. Below is a picture of each type of dilation (one that gets larger and one that gest smaller) Example 1. The picture below shows a dilation with a scale factor of 2. Grade 8 Similarity and Congruency 8.G.1 - 4 that maps A to A’ in the photo of the ceiling fan on the right? 3. Draw the image of triangle ABD after a rotation of the given numbers of degrees about the origin. Explain how you Graph the image of the triangle after the transformations sequence. Reflection across the line y=1. Write a congruence statement which communicates the specific sequence of transformations which would exactly map ABC onto ABC double prime. If they are congruent, the sequence of transformations will be some combination of reflections, rotations, and/or translations.
We don't need to do any additional work to demonstrate that linear transformations in three-dimensions map parallelepipeds onto parallelepipeds. Since the above calculations were not specific to two dimensions, they demonstrate that three-dimensional linear transformations map parallelograms onto parallelograms.
We can conclude that linear transformations map parallelepipeds onto parallelepipeds.Join images into a single multi-image file. This option is enabled by default.
An attempt is made to save all images of an image sequence into the given output file. 5 In the diagram below, a square is graphed in the coordinate plane. A reflection over which line does not carry the square onto itself? 6 The image of AABC after a dilation of scale factor k centered at.
Grade 8 Similarity and Congruency 8.G.1 - 4 that maps A to A’ in the photo of the ceiling fan on the right? 3. Draw the image of triangle ABD after a rotation of the given numbers of degrees about the origin. Explain how you Graph the image of the triangle after the transformations sequence.
The perimeter of quadrilateral A DEF is equivalent to 20) Under SShich transformation '(Y the image of A/IBC', not be congruent to A. reflection over the -axis. The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION GEOMETRY (Common Core) Friday, June 16, — a.m. to p.m., only Student Name: School Name: GEOMETRY DO NOT OPEN THIS EXAMINATION BOOKLET UNTIL THE SIGNAL IS GIVEN. Transformation using matrices A vector could be represented by an ordered pair (x,y) but it could also be represented by a column matrix: $$\begin{bmatrix} x\\ y \end{bmatrix}$$.
Reflection across the line y=1. Write a rule to describe the translation needed to put her in the proper seat. The hexagon GIKMPR and ΔFJN are regular. The dashed line segments form 30° angles.
## Congruent Triangles (and other figures)
Abstract. Scalable Vector Graphics (SVG) is a Web graphics language. SVG defines markup and APIs for creating static or dynamic images, capable of interactivity . A composition of transformations maps ΔKLM to ΔK"L"M".
The first transformation for this composition is_________, and the second transformation is a translation down and to the right. a rotation around p.
Transformations and Coordinates
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# How to find all morphisms from $(\mathbb{N}, \mid)$ to $(\mathbb{N}, \mid)$?
I just need a small hint, not the full answer. I know that, if $f$ is a morphism,
1. $a \mid b \implies f(a) \mid f(b)$
2. $a \mid b$ and $a \mid c \implies a \mid b+c$, so $f(a) \mid f(b), f(a) \mid f(c), f(a) \mid f(b + c)$. Also, $f(a) \mid f(b) + f(c)$
3. $\forall a \in \mathbb{N}, a \mid 0$, so $\forall a \in \mathbb{N}, f(a) \mid f(0)$
4. $\forall a \in \mathbb{N}, 1 \mid a$, so $\forall a \in \mathbb{N}, f(1) \mid f(a)$
From these I can draw some conclusions:
a. From 3., $f(0) = a$ ($a \neq 0$), $f(\mathbb{N})$ only contains divisors of $a$.
b. From 4., if $f(1) = a$, then $f(\mathbb{N})$ only contains multiples of $a$.
The most general form I can think of for f is this:
$f(x) = ax^{b} (a, b \in \mathbb{N})$, but I can't seem to go any further than this.
Any help is appreciated.
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$f(x) = ax^b$ for b > 1 may not satisfy condition 2. – Isomorphism Dec 15 '12 at 12:20
@Isomorphism can you give me an example for that? My reasoning is that if $x \mid a$ and $x \mid b$, then $x \mid a + b$, which means that $a+b = kx (k \in \mathbb{N})$. But then $(a+b)^r = (kx)^r = k^rx^r$ and $x^r \mid k^rx^r$. (I removed the other constant to make this simpler, it doesn't change anything) Is this wrong? – Gabi Purcaru Dec 15 '12 at 12:33
You were right and I was wrong. We can prove the second property for $f(x) = ax^b$ by looking at the prime divisors. Anyway, another solution is the constant map. Let $c$ be a natural number.Define $f(x) = c$. This works too. – Isomorphism Dec 15 '12 at 12:36
@Isomorphism but isn't this just $f(x) = cx^0$ ? Anyway, I'm not convinced that these are the only morphisms either, because, for example, $f(0)=0$ and $\forall x > 0$, $f(x) = c$, or $f(0) = 8, f(1) = 2, f(2) = 4, f(x) = 8$ for any other $x$, etc. are valid morphisms. – Gabi Purcaru Dec 15 '12 at 12:39
Ah! correct!. But f(1) = 1. For if $f(1) = x > 1$, then since $1^n|a$ for all n, we must have $x^n | f(a)$ for all n. But since $x > 1$, eventually $x^n$ should grow beyond $f(a)$. Thats a contradiction to the claim $f(1^n) | f(a)$ – Isomorphism Dec 15 '12 at 12:42
Your observations a,b are fine so far (if $0\in\mathbb N$ at all, that may depend on your local definitions).
Note that $a|b$ with $a>b$ is only possible with $b=0$. This allows us to construct uncountably many morphisms $f$ with $f(0)=0$: Assume $n\in\mathbb N$ and we have selected $a_k$ for $0\le k<n$ such that $a_0=0$ and $0< r,s\le n$ with $r|s$ implies $a_r|a_s$. Select $$\tag1a_{n}\in\bigcap_{0<k<n\atop k|n} a_k\mathbb N$$ arbitrarily (which is possible as the set on the right contains at least $0$). By the choice we guarantee that $k|n$ implies $a_k|a_n$. Therefore, for any such sequence $(a_n)$, letting $f(n)=a_n$ gives us a morphism. Note that any morphism with $f(0)=0$ can be obtained this way, i.e. the restriction imposed by $(1)$ is necessary.
For morphisms with $f(0)>0$ you correctly observed that we need $f(n)|f(0)$ for all $n$, esp. $f(n)\ne 0$ for all $n$. We can do almost the same as above, more precisely select $$\tag2 a_{n}\in\{d\in\mathbb N\colon d|f(0)\}\cap\bigcap_{0<k<n\atop k|n} a_k\mathbb N$$ this time observing that the set in $(2)$ is nonempty because it contains $f(0)$. However, at each step there are only finitely many choices. Still, this gives another uncountable lot of morphisms (why?) if $f(0)>1$.
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$0 \in \mathbb{N}$, yes. Sorry for the confusion... This being my homework, I figured I was missing something, but now I'm starting to think the teacher was messing with us. Thanks! – Gabi Purcaru Dec 15 '12 at 13:12
It's a big hairy beast.
At all times, we will assume the prime factorization of $n$ is $n=p_1^{a_1}...p_k^{a_k}$.
For example, define a function first for each prime power, and then you can define $\phi(n)=\phi(p_1^{a_1})...\phi(p_k^{a_k})$. That only gives some basic examples, hardly begins to scratch the surface, but the restriction of $\phi$ at any prime power can be almost anything. (You can't define it arbitrarily for prime powers - for each $p^a$ you can choose $\phi(p^a)/\phi(p^{a-1})$ arbitrarily.)
One non-trivial example is to define $\phi(n) = p_1p_2...$. In this case your result is always square-free. Indeed, the function returns the largest square-free divisor of $n$.
Another non-trivial example would "forget" the power of $2$ is the prime factorization: $\phi(n)$ is the largest odd factor of $n$.
These first two examples preserve $\gcd$ and $\operatorname{lcm}$, which are the meet and join operator of $(\mathbb N,\mid)$.
Another example would be $\phi(n)=x^{\max(a_1,...a_k)}$, for some fixed $x\in\mathbb N$. This does not preserve $\gcd$, I don't think.
It might be worth just looking at examples that only depend on two primes. So morphisms $\phi$ such that $\phi(2^{a_1}3^{a_2}p_3^{a_3}...p_k^{a_k})=\phi(2^{a_1}3^{a_2})$ for all values. Even that is going to be an interesting mess.
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thanks for your input. I selected Hagen's answer because it describes all solutions though – Gabi Purcaru Dec 16 '12 at 7:02
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Proceedings ArticleDOI
# Electrical property modelling of photodiode type CMOS active pixel sensor (APS)
01 Jan 2005-pp 371-375
AbstractIn this research, few electrical characteristics of photodiode (PD) type CMOS active pixel sensor (APS) pixel were modeled. BSIM3v3 threshold voltage equation was simplified for hand calculation with better than 2% peak-to-peak error for full back-gate bias and supply voltages for CMOS process technologies that has minimum feature sizes between 0.18/spl mu/m and 2.0/spl mu/m. Two fitting function coefficients (FFC) were included in the BSIM threshold model equations for simplification of the equation. FFCs were extracted by using circuit simulation for given process. Using the simplified threshold equation, electrical characteristics of 3T CMOS PD-APS pixel were modeled. Models include; photodiode reset level, pixel amplifier signal range, and pixel reset level boosting factor. Models were evaluated by using wide variety of available CMOS process technologies and compared with the simulation results. Pixel reset level and signal range model equations produce better than 6% and 12% peak-to-peak accuracy with simple hand calculation, respectively. Models were also confirmed with a designed photodiode-type CMOS APS pixel. A CMOS photodiode type APS imager fabricated in a 0.5/spl mu/m, 2P3M, 5Volt CMOS process with 15/spl mu/m square pixel size was used for comparison.
Topics: CMOS sensor (63%), Photodiode (55%), CMOS (55%), BSIM (51%), Pixel (51%)
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Journal ArticleDOI
TL;DR: A CMOS image sensor capable of imaging and energy harvesting on same focal plane is presented for retinal prosthesis and its efficiency was measured at different light levels.
Abstract: A CMOS image sensor capable of imaging and energy harvesting on same focal plane is presented for retinal prosthesis. The energy harvesting and imaging (EHI) active pixel sensor (APS) imager was designed, fabricated, and tested in a standard 0.5 μm CMOS process. It has 54 × 50 array of 21 × 21 μm2 EHI pixels, 10-bit supply boosted (SB) SAR ADC, and charge pump circuits consuming only 14.25 μW from 1.2 V and running at 7.4 frames per second. The supply boosting technique (SBT) is used in an analog signal chain of the EHI imager. Harvested solar energy on focal plane is stored on an off-chip capacitor with the help of a charge pump circuit with better than 70% efficiency. Energy harvesting efficiency of the EHI pixel was measured at different light levels. It was 9.4% while producing 0.41 V open circuit voltage. The EHI imager delivers 3.35 μW of power was delivered to a resistive load at maximum power point operation. The measured pixel array figure of merit (FoM) was 1.32 pW/frame/pixel while imager figure of merit (iFoM) including whole chip power consumption was 696 fJ/pixel/code for the EHI imager.
39 citations
### Cites background from "Electrical property modelling of ph..."
• ...Signal dependent backgate bias voltage results in the threshold voltage of the pixel transistors (M1-M3) to increase [20]....
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Journal ArticleDOI
TL;DR: It was shown that the proposed boosted readout does not increase the number of transistors in the 3T CMOS APS pixels nor degrade the image quality, and provides additional 31% dynamic range improvement on top of the seven times (7×) expansion attained by boosting the pixel reset signal.
Abstract: A new pixel readout technique is proposed for three-transistor (3T) CMOS active pixel sensor (APS) pixels. It utilizes the supply-boosting technique (SBT) in order to reduce power consumption and allow ultra low-voltage operation. The pixel supply voltage as well as the pixel reset and select signals were boosted to achieve wider and extended linear operating ranges. A CMOS image sensor containing a 54 × 50 array of 3T CMOS APS pixels was fabricated in a standard 2P3M 5-V 0.5- μm CMOS process to confirm the effectiveness of each boosting operation. Theory, simulation, and measurement results are presented. The boosting pixel supply voltage during pixel readout provides additional 31% dynamic range improvement on top of the seven times (7×) expansion attained by boosting the pixel reset signal. It was shown that the proposed boosted readout does not increase the number of transistors in the 3T CMOS APS pixels nor degrade the image quality.
9 citations
Proceedings ArticleDOI
15 May 2011
TL;DR: In order to reduce power consumption and improve low-voltage operation capability of standard three transistor (3T) CMOS active pixel sensor (APS), new pixel readout is proposed utilizing supply boosting technique (SBT).
Abstract: In order to reduce power consumption and improve low-voltage operation capability of standard three transistor (3T) CMOS active pixel sensor (APS), new pixel readout is proposed utilizing supply boosting technique (SBT). Pixel supply voltage as well as reset and select signals for APS pixel are boosted to achieve wider and extended linear operating range in a standard CMOS process with high-Vt transistors. Reset and supply boosting was used for extending dynamic range of the pixel source follower (PSF) amplifier, while the select signal boosting was utilized for linearizing transfer characteristics of the PSF. Extension of PSF dynamic range using reset, select, and supply boosting (RSSB) resulted in operation of 3T APS pixel at 1.2V supply with 150mV dynamic range even though the threshold of the NMOS device was 0.8V. Proposed method does not degrade the device reliability margins and use single supply input.
6 citations
### Cites background from "Electrical property modelling of ph..."
• ...This increase could be as large as 50% of the zero threshold voltage for these devices [12] effecting linear input output characteristics of the PSF....
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Journal ArticleDOI
01 Dec 2017
Abstract: A new pixel is designed with the capability of imaging and energy harvesting for the retinal prosthesis implant in 0.18 µm standard Complementary Metal Oxide Semiconductor technology. The pixel conversion gain and dynamic range, are 2.05 $$\upmu{\text{V}}/{\text{e}}^{ - }$$ and 63.2 dB. The power consumption 53.12 pW per pixel while energy harvesting performance is 3.87 nW in 60 klx of illuminance per pixel. These results have been obtained using post layout simulation. In the proposed pixel structure, the high power production capability in energy harvesting mode covers the demanded energy by using all available p-n junction photo generated currents.
3 citations
Journal ArticleDOI
TL;DR: 4T CMOS APS shown more radiation hardness than the 3T CMos APS and 32 nm technology exhibits lowest radiation-tolerant, indicating 4T has a higher radiation hardness.
Abstract: The widely used CMOS Active Pixel Sensors (APS) in space imaging mission are vulnerable to radiations known as Single Event Transient (SET). This paper focus on 3T and 4T CMOS APS with tech...
3 citations
### Cites methods from "Electrical property modelling of ph..."
• ...This AC source is used by assuming the light is continuously hitting this photo-detector at a constant period (Ay 2005)....
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##### References
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Journal ArticleDOI
Abstract: CMOS active pixel sensors (APS) have performance competitive with charge-coupled device (CCD) technology, and offer advantages in on-chip functionality, system power reduction, cost, and miniaturization. This paper discusses the requirements for CMOS image sensors and their historical development, CMOS devices and circuits for pixels, analog signal chain, and on-chip analog-to-digital conversion are reviewed and discussed.
1,133 citations
Journal Article
Abstract: CMOS active pixel sensors (APS) have performance competitive with charge-coupled device (CCD) technology, and offer advantages in on-chip functionality, system power reduction, cost, and miniaturization. This paper discusses the requirements for CMOS image sensors and their historical development, CMOS devices and circuits for pixels, analog signal chain, and on-chip analog-to-digital conversion are reviewed and discussed.
693 citations
Journal ArticleDOI
, P.K. Ko2
Abstract: The Berkeley short-channel IGFET model (BSIM), an accurate and computationally efficient MOS transistor model, and its associated characterization facility for advanced integrated-circuit design are described. Both the strong-inversion and weak-inversion components of the drain current expression are included. In order to speed up the circuit-simulation execution time, the dependence of the drain current on the substrate bias has been modeled with a numerical approximation. This approximation also simplifies the transistor terminal-charge expressions. The charge model was derived from its drain-current counterpart to preserve consistency of device physics. Charge conservation is guaranteed in this model.
544 citations
### "Electrical property modelling of ph..." refers background in this paper
• ...Threshold voltage and other physical parameters of MOS transistor are well defined and modeled for circuit simulators [3,4,5]....
[...]
Journal ArticleDOI
, Kai Chen1
TL;DR: A new physical and continuous BSIM (Berkeley Short-Channel IGFET Model) I-V model in BSIM3v3 is presented for circuit simulation, which allows users to accurately describe the MOSFET characteristics over a wide range of channel lengths and widths for various technologies, and is attractive for statistical modeling.
Abstract: A new physical and continuous BSIM (Berkeley Short-Channel IGFET Model) I-V model in BSIM3v3 is presented for circuit simulation. Including the major physical effects in state-of-the art MOS devices, the model describes current characteristics from subthreshold to strong inversion as well as from the linear to the saturation operating regions with a single I-V expression, and guarantees the continuities of I/sub ds/, conductances and their derivatives throughout all V/sub gs/, V/sub ds/, and T/sub bs/, bias conditions. Compared with the previous BSIM models, the improved model continuity enhances the convergence property of the circuit simulators. Furthermore, the model accuracy has also been enhanced by including the dependencies of geometry and bias of parasitic series resistances, narrow width, bulk charge, and DIBL effects. The new model has the extensive built-in dependencies of important dimensional and processing parameters (e.g., channel length, width, gate oxide thickness, junction depth, substrate doping concentration, etc.). It allows users to accurately describe the MOSFET characteristics over a wide range of channel lengths and widths for various technologies, and is attractive for statistical modeling. The model has been implemented in the circuit simulators such as Spectre, Hspice, SmartSpice, Spice3e2, and so on.
173 citations
### "Electrical property modelling of ph..." refers background or methods in this paper
• ...Threshold voltage and other physical parameters of MOS transistor are well defined and modeled for circuit simulators [3,4,5]....
[...]
• ...Threshold voltage of an NMOS transistor is given in (1), [3]....
[...]
• ...In this research [2], few electrical parameters of photodiode (PD) type CMOS Active Pixel Sensor (APS) pixel were modelled for hand calculation by using modified BSIM3v3 model equations [3]....
[...]
• ...A hybrid MOS threshold model equation was proposed [2] based on BSIM3v3 model equations [3] for simple and accurate representation of back-gate bias effect on the threshold voltage....
[...]
01 Jan 1994
TL;DR: The Philips MOS MODEL 9 is presented with the successful confrontation with analog requirements, the scaling of parameters with geometry, the accuracy of the model over the whole geometry range of a process, its capabilities in the description of various processes at least down to 0.35 /spl mu/m and a comparison with advanced analog models available in commercial circuit simulators.
Abstract: Analog applications of MOS transistors in integrated circuits impose enhanced requirements on the compact MOS models used in circuit simulators. Here we present for the Philips MOS MODEL 9 the successful confrontation with these analog requirements, the scaling of parameters with geometry, the accuracy of the model over the whole geometry range of a process, its capabilities in the description of various processes at least down to 0.35 /spl mu/m and a comparison with advanced analog models available in commercial circuit simulators.<>
42 citations
### "Electrical property modelling of ph..." refers background in this paper
• ...Threshold voltage and other physical parameters of MOS transistor are well defined and modeled for circuit simulators [3,4,5]....
[...]
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What Is The Quantity Theory Of Money? Fisher viewed UIP as the dual of the interest rate vs. inflation relation or what has come to be called “the Fisher Equation.”2 He saw both as examples of a general relation In fact, the quantity theory of money is a hypothesis and not an identity which is always true. But his greatest concentration was on mathematics and economics, the latter having no academic department at … Which of the following is true with respect to Irving Fisher's quantity equation, M x V = P x Y? The index corrects for the upward bias of the Laspeyres Price Index and the downward bias of … 1:39. The theories behind it were introduced by American economist Irving Fisher. He made important contributions to utility theory, general equilibrium, theory of capital, the quantity theory of money and interest rates. Fisher was also a pioneer of the development of index numbers for stock markets. He gained an eclectic education at Yale, studying science and philosophy. Irving Fisher developed it further and it is the bedrock of the Quantity Theory of Money. (iv) Aggregate Demand/Expenditure, and not M, Influences Price Level: Velocity does not play any role in the equation. Theory and Applications of Macroeconomics – 16.14 The Fisher Equation: Nominal and Real Interest Rates – Some of the equations used for the fisher equation. Irving Fisher (1867-1947) was born in Saugerties, New York, in 1867. Irving Fisher, who was one of the well-known economists of the early 1900’s, came up with the “Equation of Exchange” concept. Wikipedia – Fisher Equation – Details on the fisher equation. The quantity theory of money is built on an equation created by Irving Fisher (1867-1947), an American economist, inventor, statistician and progressive social campaigner. In 1898, he achieved professor status, becoming instructor of political economy, and became professor emeritus in 1935. Important features of fisher equation. Where M stands for the money supply, V is the velocity of money, P is the prevailing price level, and T is the overall transactions. Constants Relate to Different Time: Prof. Halm criticises Fisher for multiplying M and V because M … MV=PT Irving Fisher was one of the most popular economists of the early twentieth century. A popular identity defined by Irving Fisher is the quantity equation commonly used to describe the relationship between the money stock and aggregate expenditure: MV = PY. MV and PT are always equal. Then determine the real purchase power of, say, $100 by multiplying 100 by Rreal and subtracting the amount from 100. Calculating the Fisher effect is not difficult. Similar to other consumer price indices, the Fisher Price Index is used to measure the price level andcost of living in an economy and to calculate inflationInflationInflation is an economic concept that refers to increases in the price level of goods over a set period of time. Constants Relate to Different Time: Prof. Halm criticizes Fisher for multiplying M and V because M … If Rreal is 2 percent, the purchase power … To Irving Fisher, who arguably was the first to formulate the UIP condition, these anomalous results probably would not have come as a much of a surprise (Dimand, 1999). T = all the goods and services sold within an economy over a given time (some economist may use the letter ‘Y’ for this value)According to the equation – w… The terms on the right-hand side represent the price level (P) and Real GDP (Y). i ≈ r + Π Although this relation derived from earlier works by Jacob de Haas, Irving Fisher’s formula is an easier way to represent this relation. In its time, it became a landmark theory and today, many people still consider the equation to be one of the most important theories of economics. In finance, the Fisher equation is primarily used in YTMcalculations of bondsor IRRcalculations of investments. It is named after Irving Fisher, who was famous for his works on the theory of interest. The equation of exchange has been stated by Cambridge economists, Marshall and Pigou, in a form different from Irving Fisher. Fisher was one of the first economists to identify a difference between a real and a nominal interest rate, and his work in this area culminated in this equation. That is Generated by an economist named Irving fisher. 1) V = Average number of times a dollar is spent on goods and services. He was one of the earliest American neoclassical economists, though his later work on debt deflation has been embraced by the post-Keynesian school. American Neoclassical economist, and long-time professor of economics at Yale University.. Irving Fisher was one of the earliest American Neoclassicals of unusual mathematical sophistication. Fisher’s Equation of Exchange - Free download as Powerpoint Presentation (.ppt), PDF File (.pdf), Text File (.txt) or view presentation slides online. In 1911 he came up with the “Equation of Exchange” concept. The equation is:M x V = P x TM = the stock of money. The equation of exchange is: Irving Fisher was one of America’s greatest mathematical economists and one of the clearest economics writers of all time. V = the velocity of circulation. Irving Fisher, 1867-1947. In the late 1930s, U.S. economist Irving Fisher wrote a paper which posited that a country's interest rate level rises and falls in direct relation to its inflation rates. Irving Fisher continued his association with Yale by staying on as a tutor. That explains the relationship between both real and nominal rates of interest and inflation. Question: If We Look At The Equation For Money Demand That Summarizes Irving Fisher’s Quantity Theory Of Money, ... O There isn't an explicit role for the interest rate in the equation. Given that economists are not known as “popular” figures, we must take that with a grain of salt. An easier way to calculate the formula and determine purchase power is to break the equation into two steps. The original Fisher equation (OFE, 1896) was expressed in terms of the expected appreciati on of money (the real return on money) whereas the ubiquitous conventional Fisher equation (CFE, 1930) uses expected inflation. Use the equation “Rnon – E[I] = Rreal” to get the real rate of inflation. He had the intellect to use mathematics in virtually all his theories and the good sense to introduce it only after he had clearly explained the central principles in … tremendously insightful classical economist whose work in capital theory influenced many of today’s leading economists The fisher effect is one of the most essential concepts of economic theory. This blog explores the different elements of the equation with examples, along with the pros and cons associated with it. In the eighteenth century, John Stuart Mill expanded on David Hume's ideas and stated the "equation of exchange." Irving Fisher was born in upstate New York in 1867. Irving Fisher. Irving Fisher (February 27, 1867 – April 29, 1947) was an American economist, statistician, inventor, and Progressive social campaigner. Thirdly, Fisher’s equation is an identity. Money demand is not a factor of nominal income. Fisher equation, named after its designer Irving Fisher, is a concept in Economics that defines the relationship between nominal interest rates and real interest rates under the influence of inflation. This is equivalent to: i = r + π (1 + r) Thus, according to this equation, if π increases by 1 percent the nominal interest rate increases by more than 1 percent. jodiecongirl (YouTube) – The Fisher Effect – An overview video of the fisher equation and how it is calculated. In the early autumn of 1929, shortly before the famous Black Thursday, Irving Fisher said, “stock prices have reached what looks like a permanently high plateau.” He published poetry and works on astronomy, mechanics, and geometry. The rise in the price level signifies that the currency in a given economy loses purchasing power (i.e., less can be bought with the same amount of money).. The Fisher effect (named for American economist Irving Fisher) describes how interest rates and expected inflation rates move in tandem. He gained an eclectic education at Yale, studying science and philosophy. 2) V = PxY/M 3) < = M1 definition of the money supply 4) P = the GDP deflator ANSWER: All of the Above. Fisher mathematically expressed this theory in the following way: R Nominal = R Real + R Inflation Fisher received a doctorate at Yale in 1891 in economics and mathematics. One of the most common ways to estimate the price of a cryptocurrency is with the “equation of exchange”. One of the main insights of this model is that the more often a currency changes hands, the less value the currency has. This popular, albeit controversial, formulation of the quantity theory of money is based upon an equation by American economist Irving Fisher. Cambridge economists explained the determination of value of money in line with the determination of value in general. Irving Fisher was the greatest economist the United States has ever produced. The Fisher Equation is used in economic theory to explain the relationship between interest rates and inflation. How Does the Fisher Effect Work? He published poetry and works on astronomy, mechanics, and geometry. In Appreciation and Interest Irving Fisher (1896) derived an equation connecting interest rates in any two standards of value. The Fisher equationin financial mathematicsand economicsestimates the relationship between nominal and real interest ratesunder inflation. P = the average price level. It is named after Irving Fisher, who was famous for his works on the theory of interest. But his greatest concentration was on mathematics and economics, the latter having no academic department at … Irving Fisher was born in upstate New York in 1867. This model comes from the 20th century economist Irving Fisher. This … The Fisher equation in financial mathematics and economics estimates the relationship between nominal and real interest rates under inflation. The technical format of the formula is “Rnom = Rreal + E[I]” or nominal interest rate = real interest rate + expected rate of inflation. Irving Fisher, an American economist, developed the transaction version of the quantity theory of money, as shown in the Fisher equation below: MV = PT \text{MV}=\text{PT} MV = PT. And geometry 20th century economist Irving Fisher ( 1896 ) derived an equation American. Interest Irving Fisher different from Irving Fisher continued his association with Yale by staying on a... 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## Wednesday, May 2, 2018
### Base n to mod n in one easy step.
We learn how to write numbers (well, integers) in bases other than ten fairly early in our mathematical education. A bit later, we learn how about congruence arithmetic and doing arithmetic modulo any integer $$n \gt 1$$. I remember that there was a certain degree of initial confusion between the two, though that was really just about remembering which label went with what: after all, the two are entirely different, aren't they?
As it turns out, not so much.
Let's have a brief reminder of congruence arithmetic, and then see what representing numbers in different bases has to do with it.
# Congruence arithmetic modulo $$n$$
The simplest and most familiar form of congruence arithmetic is something we're all familiar with long before we meet congruence arithmetic as a thing.
We all know right from a very early stage that an even number plus an even number is even, an even plus an odd is odd, and so on. But an even number is one which leaves a remainder of zero when it's divided by two, and an odd number is one which leaves a remainder of one. So these odd and even rules are telling us that when we do sums with integers, the remainder on division by two is well behaved with respect to addition and multiplication. Or in the hieratic jargon, we can do arithmetic modulo two.
It's not so obvious that this works with numbers other than two, but it's just as true.
Choose any integer $$n>1$$. We say that two integers, $$a$$ and $$b$$ are congruent modulo $$n$$ if their difference is a multiple of $$n$$, or, equivalently, if they have the same remainder on division by $$n$$.
Then with a bit of work, we can show that arithmetic modulo $$n$$ makes sense: it doesn't matter what numbers we start with, the remainder after doing an arithmetic operation only depends on the remainders of the numbers we combine.
This takes a little bit of algebra. On the other hand...
# Arithmetic modulo ten and modulo two
If we carry out the usual addition and multiplication operations of integer arithmetic, writing all numbers in base ten, then it is obvious (as long as we believe the standard algorithms!) that the units digit of a sum or product depends only on the units digits of the numbers multiplied or added.
But the units digit of a number (written in base ten) is just the remainder when you divide by ten! Looking only at the units digits is doing arithmetic modulo ten.
In just the same way, if we write our numbers in binary, then the units digit of a sum or product depends only on the units digits of the numbers combined. And the units digits is $$1$$ for an odd number and $$0$$ for an even one.
Looking only at the units digits is now doing arithmetic modulo two.
# Arithmetic modulo $$n$$ again
But there's nothing special about two or ten (at least, not in this regard).
If we choose any integer $$n \gt 1$$, and write integers in base $$n$$, then the units digit of a sum or product depends only on the units digits of the numbers combined.
And now, looking at the units digits is working modulo $$n$$.
So although base $$n$$ and modulo $$n$$ look like entirely different ideas, they turn out to be rather closely related.
# What's the point?
I'm not entirely sure. Thinking about last digits in base $$n$$ might be an interesting way to approach congruence arithmetic, as it comes from a more concrete direction. It also suggests various investigative projects, such as when the last digit of a product can be zero, or how negative numbers fit into this picture. For me, the point was really that there is actually a real relationship between these two apparently different bits of maths, and it's always fun to see connections like that.
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## anonymous 4 years ago arccos(cos(94pi/59))
1. anonymous
HINT: $\cos^{-1} (\cos x) = x$
2. anonymous
arccos(cos(94pi/59)) = 94pi/59
3. zepdrix
|dw:1351301061848:dw|
4. anonymous
94pi/59 is in the 4th quad. so... 94pi/59 -pi = 35/59 pi
5. anonymous
We know the range of the arc cosine function is limited to [0,pi], right? We know that 90pi/59 is pretty close to 90pi/60 or 3pi/2. The cos(3pi/2) is close to 0 The arccos(0) is pi/2 The reference angle for the 90pi/59 would be in quad 4 and measure 28/59 pi. SO I'm thiniking the answer to the problem would be 28pi/59|dw:1351301172847:dw|
6. anonymous
7. zepdrix
Oh sorry trans.. you were typing really long :C wasn't sure if you were going to paste anything. Didn't mean to step on yer toes there XD
8. anonymous
OMG wrong lol. its 2pi - 94/59pi
9. anonymous
ans: 24/59pi
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# A Pythagorean Introduction to Number Theory
###### Ramin Takloo-Bighash
Publisher:
Springer
Publication Date:
2019
Number of Pages:
279
Format:
Hardcover
Series:
Price:
54.99
ISBN:
9783030026035
Category:
Textbook
[Reviewed by
Mark Hunacek
, on
06/24/2019
]
This book offers an interesting variation on the traditional undergraduate number theory course. It is organized around the general theme of number-theoretic questions relating to the study of right triangles. Number theory arises here, of course, because the familiar Pythagorean theorem for such triangles is an example of a Diophantine equation, the study of the integral or rational roots of which can lead to interesting questions. For example, the classification of primitive Pythagorean triples is a topic that just about every undergraduate number theory text covers, as is the question of which positive integers can be written as the sum of two squares. There are other questions that can be asked, however, and their study leads to some interesting, yet relatively accessible, mathematics.
The book is divided into two parts of roughly equal length. The first part, consisting of the first seven chapters of the book, should be entirely accessible to undergraduate students taking an introductory course in number theory, although some material here (such as the discussion of congruent numbers in chapter 4) is not generally covered in such a course. The second part, consisting of the remaining seven chapters, is somewhat more sophisticated; while some of the material here could be covered in an introductory course, a substantial portion of it would be better introduced in a second course.
In more detail: the book begins with a chapter offering several proofs of the Pythagorean theorem. (The idea of proving something in multiple ways is one that comes up again in the text, and is, I think, a very attractive feature of the book.) Proofs of the Pythagorean theorem typically come up, of course, in a text on geometry rather than number theory, but the discussion here seems entirely appropriate given the central role that that theorem will play in the ensuing discussion. And anyway, the Pythagorean theorem is something that all undergraduate mathematics majors should, but often do not see proved.
Chapter 1, like all other chapters in the book, ends with a section titled “Notes”. These sections discuss historical issues (paying more than the usual amount of attention to non-European aspects of mathematics history), survey other issues that are relevant to the subject matter of the chapter, and provide bibliographic references. These end-of-chapter sections are another very attractive feature of the book; they are nicely written and quite informative.
Chapter 2 discusses the basic topics of number theory: divisibility, primes, congruences, Fermat’s Little Theorem, and primitive roots (including a classification, with proof, of the precise values of $n$ for which a primitive root mod $n$ exists). The exposition here, while clear, is quite efficient: this material can easily take up a third of a typical undergraduate number theory semester, yet is done here in about 40 pages. Rather surprisingly, the author does not prove in this chapter that there are infinitely many prime integers, but this will be proved later in the book. (I would have preferred to see it proved earlier, but obviously this is not a serious issue; nothing prevents an instructor from proving this result earlier rather than later.)
Basic divisibility arguments are then used to characterize, in chapter 3, primitive Pythagorean triples. Following this, the standard geometric argument (using chords of the unit circle connecting $(-1,0)$ to other rational points on the circle) is given to produce an alternative proof. The geometric argument is used to analyze other equations, and finally, all of this material is used to prove Fermat’s Last Theorem for exponent 4. The Notes section of this chapter is a particularly interesting one that discusses Fermat’s Last Theorem a little more and also introduces the student to elliptic curves and the abc conjecture.
The next chapter looks at a different aspect of right triangles, their area, and asks the question: what numbers are the area of a right triangle with integer-length sides? This quickly leads to the concept of congruent number (i.e., a number that is the area of a right triangle with rational sides); there is currently no known precise characterization of precisely what integers are congruent numbers, but the author proves some results along these lines.
Chapter 5 addresses the question of what natural numbers can appear as the length of a side of a right triangle. With regard to the hypotenuse, this question quickly devolves into the question of what natural numbers can be written as the sum of two squares. This is a familiar topic in elementary number theory courses but the author’s approach here is interesting: he establishes this result as a consequence of facts about the Gaussian integers, which are introduced from scratch. Basic facts about the ring of Gaussian integers (the existence of a Euclidean algorithm and unique factorization, the characterization of irreducible elements, etc.) are fully worked out.
The question of sums of squares leads naturally to the idea of primes that are congruent to $1 \mod 4$, and these primes motivate the material of the next two chapters. After (finally!) proving the existence of infinitely many primes, the author also proves that there are infinitely many primes of the form $4k + 1$ and $4k – 1$. (Dirichlet’s theorem on primes in arithmetic progression had been previously mentioned in the Notes section of the previous chapter.) The study of primes of the form $4k + 1$ leads to the study of quadratic residues, and in turn leads to the law of quadratic reciprocity, which is stated in chapter 6 and proved (using Gauss sums and some basic facts about algebraic integers, developed in an Appendix) in chapter 7.
This takes us to Part II (“Advanced Topics”) of the book. The author begins by studying the number of solutions of the Pythagorean equation modulo $n$, an enterprise that leads to Hensel’s theorem, the proof of which is an exercise. Then the question of sums of squares (discussed earlier for two squares) resurfaces, this time for two, three and four squares. The theorems on these topics are first proved using geometric lattice-point arguments, and then, in the next chapter, the theorem on sums of four squares is given a different proof, using quaternions. Sums of three squares, via the theory of quadratic forms, are discussed in chapter 12.
The last two chapters of the text are the most difficult, and least self-contained. In chapter 13 the author proves an asymptotic formula (first established by Lehmer in 1900) for the number of primitive Pythagorean theorems with bounded hypotenuse. In chapter 14, we return to a topic from chapter 3, rational points on the unit circle; the main result of this chapter is a very difficult theorem concerning the distribution of these points according to their height, a concept introduced in the chapter.
These 14 chapters are then followed by three appendices. The first is quite elementary and provides some background material on things like the Pigeonhole Principle (with Dirichlet’s theorem on approximation of irrational numbers by rational ones given as a nice example); the second covers the basics of algebraic numbers and integers; the third discusses the free mathematical software SageMath.
The book is quite nicely written, with good motivation and a substantial supply of examples. As noted earlier, it also benefits from the author’s attention to the historical aspects of the subject and his penchant for giving alternative proofs of various results. There are a decent number of exercises, both computational and proof-based, covering a broad spectrum of difficulty. The author states that hints to some of the more difficult ones will appear on the book’s website, but apparently none have, as of this writing, yet made it up there.
Prerequisites for the text are (for the first part, anyway) not unduly burdensome. No prior background in abstract algebra is necessary, though on occasion the author mentions a word like “ring” without definition. No real background in ring theory is necessary to understand the book, however. (Since the author does feel free to drop the occasional algebraic phrase, I’m surprised that he didn’t make the obvious connection between some of the topics covered in chapter 2 and basic group theory.) Part 2 of the book does require considerably more mathematical sophistication, and, as previously noted, in several chapters a background in analysis is required.
I noticed few misprints or typos, and the ones that I did see are minor: for example, in theorem 4.4 on page 83, the author uses the ordinary letter “S” to denote the set of congruent numbers, but had previously used a stylized script version of that letter in the definition.
And speaking of the letter S, I do think that the index of the book could stand some improvement. Anyone looking to see where Andrew Wiles is mentioned in the text will not find that answer under “W”; instead, the reference to him appears under “S”, for “Sir Andrew Wiles”. The word “prime” does not appear in the Index, which seems a curious omission in a textbook on number theory. And I was unable to find any reference to the Law of Quadratic Reciprocity under the letters “L”, “Q” or “R”; I finally got lucky and found it under “E” (for “Euler”).
As a text for a course, I see only one possible drawback, and that one is obvious and unavoidable: the author’s approach to the subject, while novel and interesting, has the inevitable consequence of resulting in the omission of certain topics that an instructor might want to cover in an introductory course. Continued fractions, for example, are not discussed in the text, and, perhaps of more concern for most instructors, neither is cryptography, which seems, more and more, to be a topic that people want to see covered in an undergraduate number theory course. Even if not considered as a text for such a course, however, the book has several other potential uses: it could be used as a text for a second semester course in number theory or “special topics” course, or as a text for an introductory graduate course. It’s also just an interesting book to have on one’s shelf.
Mark Hunacek (mhunacek@iastate.edu) teaches mathematics at Iowa State University.
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#### Question 31:
In Fig., three coplanar lines intersect at a point O, forming angles as shown in the figure. Find the values of x, y, z and u.
$\angle$BOD + $\angle$DOF + $\angle$FOA = 180° (Linear pair)
$\angle$FOA = $\angle$u = $180°-90°-50°=40°$
$\angle \mathrm{FOA}=\angle x=40°$ (Vertically opposite angles)
$\angle \mathrm{BOD}=\angle z=90°$ (Vertically opposite angles)
$\angle \mathrm{EOC}=\angle y=50°$ (Vertically opposite angles)
#### Question 32:
In Fig., find the values of x, y and z.
#### Question 1:
In Fig., line n is a transversal to lines l and m. Identify the following:
(i) Alternate and corresponding angles in Fig. (i).
(ii) Angles alternate to ∠d and ∠g and angles corresponding to angles ∠f and ∠h in Fig. (ii).
(iii) Angle alternate to ∠PQR, angle corresponding to ∠RQF and angle alternate to ∠PQE in Fig. (iii).
(iv) Pairs of interior and exterior angles on the same side of the transversal in Fig. (ii).
(i) Figure (i)
Corresponding angles:
$\angle$EGB and $\angle$GHD
$\angle$HGB and $\angle$FHD
$\angle$EGA and $\angle$GHC
$\angle$AGH and $\angle$CHF
Alternate angles:
$\angle$EGB and $\angle$CHF
$\angle$HGB and $\angle$CHG
$\angle$EGA and $\angle$FHD
$\angle$AGH and $\angle$GHD
(ii) Figure (ii)
Alternate angle to $\angle$d is $\angle$e.
Alternate angle to $\angle$g is $\angle$b.
Also,
Corresponding angle to $\angle$f is $\angle$c.
Corresponding angle to $\angle$h is $\angle$a.
(iii) Figure (iii)
Angle alternate to $\angle$PQR is $\angle$QRA.
Angle corresponding to $\angle$RQF is $\angle$ARB.
Angle alternate to $\angle$POE is $\angle$ARB.
(iv) Figure (ii)
Pair of interior angles are
$\angle$a and $\angle$e
$\angle$d and $\angle$f
Pair of exterior angles are
$\angle$b and $\angle$h
$\angle$c and $\angle$g
#### Question 2:
In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠CMQ = 60°, find all other angles in the figure.
$\angle$ALM = $\angle$CMQ = $60°$ (Corresponding angles)
$\angle$LMD = $\angle$CMQ = $60°$ (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = $60°$ (Vertically opposite angles)
Since
$\angle$CMQ + $\angle$QMD = $180°$ (Linear pair)
QMD = $180°-60°=120°$
$\angle$QMD = $\angle$MLB = $120°$ (Corresponding angles)
$\angle$QMD = $\angle$CML = $120°$ (Vertically opposite angles)
$\angle$MLB = $\angle$ALP = $120°$ (Vertically opposite angles)
#### Question 3:
In Fig., AB and CD are parallel lines intersected by a transversal PQ at L and M respectively. If ∠LMD = 35° find ∠ALM and ∠PLA.
In the given Fig., AB || CD.
#### Question 4:
The line n is transversal to line l and m in Fig. Identify the angle alternate to ∠13, angle corresponding to ∠15, and angle alternate to ∠15.
In this given Fig., line l || m.
Here,
Alternate angle to $\angle$13 is $\angle$7.
Corresponding angle to $\angle$15 is $\angle$7.
Alternate angle to $\angle$15 is $\angle$5.
#### Question 5:
In Fig., line l || m and n is a transversal. If ∠1 = 40°, find all the angles and check that all corresponding angles and alternate angles are equal.
In the given figure, l || m.
Here,
Also,
Thus,
Hence, alternate angles are equal.
#### Question 6:
In Fig., line l || m and a transversal n cuts them at P and Q respectively. If ∠1 = 75°, find all other angles.
In the given figure, l || m, n is a transversal line and ∠1 = 75°.
Thus, we have:
#### Question 7:
In Fig., AB || CD and a transversal PQ cuts them at L and M respectively. If ∠QMD = 100°, find all other angles.
In the given figure, AB || CD, PQ is a transversal line and $\angle$QMD = 100°.
Thus, we have:
$\angle$DMQ + $\angle$QMC = 180° (Linear pair)
$\therefore \angle \mathrm{QMC}=180°-\angle \mathrm{DMQ}=180°-100°=80°$
Thus,
$\angle$DMQ = $\angle$BLM = 100° (Corresponding angles)
$\angle$DMQ = $\angle$CML = 100° (Vertically opposite angles)
$\angle$BLM = $\angle$PLA = 100° (Vertically opposite angles)
Also,
$\angle$CMQ = $\angle$ALM = 80° (Corresponding angles)
$\angle$CMQ = $\angle$DML = 80° (Vertically opposite angles)
$\angle$ALM = $\angle$PLB = 80° (Vertically opposite angles)
#### Question 8:
In Fig., l || m and p || q. Find the values of x, y, z, t.
In the given figure, l || m and p || q.
Thus, we have:
$\angle z=80°$ (Vertically opposite angles)
$\angle z=\angle t=80°$ (Corresponding angles)
$\angle z=\angle y=80°$ (Corresponding angles)
$\angle x=\angle y=80°$ (Corresponding angles)
#### Question 9:
In Fig., line l || m, ∠1 = 120° and ∠2 = 100°, find out ∠3 and ∠4.
In the given figure, ∠1 = 120° and ∠2 =100°.
Since l || m, so
Also,
We know that the sum of all the angles of triangle is 180°.
$\therefore \angle 6+\angle 3+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒60°+80°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒140°+\angle 4=180°\phantom{\rule{0ex}{0ex}}⇒\angle 4=180°-140°=40°$
#### Question 10:
In Fig., line l || m. Find the values of a, b, c, d. Give reasons.
In the given figure, line l || m.
Thus, we have:
#### Question 11:
In Fig., AB || CD and ∠1 and ∠2 are in the ratio 3 : 2. Determine all angles from 1 to 8.
In the given figure, AB || CD and t is a transversal line.
Now, let:
$\angle 1=3x\phantom{\rule{0ex}{0ex}}\angle 2=2x$
Thus, we have:
Now,
#### Question 12:
In Fig., l, m and n are parallel lines intersected by transversal p at X, Y and Z respectively. Find ∠1, ∠2 and ∠3.
In the given figure, l || m || n and p is a transversal line.
Thus, we have:
#### Question 13:
In Fig., if l || m || n and ∠1 = 60°, find ∠2.
In the given figure, l || m || n and ∠1 = 60°.
Thus, we have:
#### Question 14:
In Fig., if AB || CD and CD || EF, find ∠ACE.
In the given figure, AB || CD and CD || EF.
Extend line CE to E'.
Thus, we have:
#### Question 15:
In Fig., if l || m, n || p and ∠1 = 85°, find ∠2.
In the given figure, l || m, n || p and ∠1 = 85°.
Now, let ∠4 be the adjacent angle of ∠2.
Thus, we have:
$\angle 3+\angle 2=180°$ (Sum of interior angles on the same side of the transversal)
$\therefore \angle 2=180°-\angle 3=180°-85°=95°$
#### Question 16:
In Fig., a transversal n cuts two lines l and m. If ∠1 = 70° and ∠7 = 80°, is l || m?
We know that if the alternate exterior angles of two lines are equal, then the lines are parallel.
In the given figure, are alternate exterior angles, but they are not equal.
Therefore, lines l and m are not parallel.
#### Question 17:
In Fig., a transversal n cuts two lines l and m such that ∠2 = 65° and ∠8 = 65°. Are the lines parallel?
$\angle$2 = $\angle$3 = 65° (Vertically opposite angles)
$\angle$8 = $\angle$6 = 65° (Vertically opposite angles)
∴ $\angle$3 = $\angle$6
l || m (Two lines are parallel if the alternate angles formed with the transversal are equal)
#### Question 18:
In Fig., show that AB || EF.
Extend line CE to E'.
#### Question 19:
In Fig., AB || CD. Find the values of x, y, z.
$\angle x+125°=180°$ (Linear pair)
$\therefore \angle x=180°-125°=55°$
$\angle z=125°$ (Corresponding angles)
$\angle x+\angle z=180°$ (Sum of adjacent interior angles is $180°$)
$\angle x+125°=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=180°-125°=55°$
$\angle x+\angle y=180°$ (Sum of adjacent interior angles is $180°$)
$55°+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle y=180°-55°=125°$
#### Question 20:
In Fig., find out ∠PXR, if PQ || RS.
Draw a line parallel to PQ passing through X.
Here,
(Alternate interior angles)
∵ PQ || RS || XF
∴ $\angle \mathrm{PXR}=\angle \mathrm{PXF}+\angle \mathrm{FXR}=70°+50°=120°$
#### Question 21:
In Fig., we have
(i) ∠MLY = 2 ∠LMQ, find ∠LMQ.
(ii) ∠XLM = (2x − 10)° and ∠LMQ = x + 30°, find x.
(iii) ∠XLM = ∠PML, find ∠ALY
(iv) ∠ALY = (2x − 15)°, and ∠LMQ = (x + 40)°, find x
(i)
(ii)
(iii)
(iv)
#### Question 22:
In Fig., DE || BC. Find the values of x and y.
$\angle$ABC = $\angle$DAB (Alternate interior angles)
$\angle$ACB = $\angle$EAC (Alternate interior angles)
#### Question 23:
In Fig., line AC || line DE and ∠ABD = 32°. Find out the angles x and y if ∠E = 122°.
#### Question 24:
In Fig., side BC of ∆ABC has been produced to D and CE || BA. If ∠ABC = 65°, ∠BAC = 55°, find ∠ACE, ∠ECD and ∠ACD.
$\angle$ABC = $\angle$ECD = 55° (Corresponding angles)
$\angle$BAC = $\angle$ACE = 65° (Alternate interior angles)
Now, $\angle$ACD = $\angle$ACE + $\angle$ECD
⇒ $\angle$ACD = 55° + 65° = 120°
#### Question 25:
In Fig., line CAAB || line CR and line PR || line BD. Find ∠x, ∠y and ∠z.
Since CA ⊥ AB,
$\therefore \angle x=90°$
We know that the sum of all the angles of triangle is 180°.
$\angle$PBC = $\angle$APQ = $70°$ (Corresponding angles)
Since
#### Question 26:
In Fig., PQ || RS. Find the value of x.
#### Question 27:
In Fig., AB || CD and AE || CF; ∠FCG = 90° and ∠BAC = 120°. Find the values of x, y and z.
$\angle$BAC = $\angle$ACG = 120° (Alternate interior angle)
∴ $\angle$ACF + $\angle$FCG = 120°
$\angle$ACF = 120° − 90° = 30°
$\angle$DCA + $\angle$ACG = 180° (Linear pair)
$\angle$x = 180° − 120° = 60°
$\angle$BAC + $\angle$BAE + $\angle$EAC = 360°
$\angle$CAE = 360° − 120° − (60° + 30°) = 150° ($\angle$BAE = $\angle$DCF)
#### Question 28:
In Fig., AB || CD and AC || BD. Find the values of x, y, z.
(i) Since AC || BD and CD || AB, ABCD is a parallelogram.
$\angle$CAB + $\angle$ACD = 180° (Sum of adjacent angles of a parallelogram)
∴ $\angle$ACD = 180° − 65° = 115°
$\angle$CAD = $\angle$CDB = 65° (Opposite angles of a parallelogram)
$\angle$ACD = $\angle$DBA = 115° (Opposite angles of a parallelogram)
(ii) Here,
AC || BD and CD || AB
$\angle$DAC = x = 40° (Alternate interior angle)
$\angle$DAB = y = 35° (Alternate interior angle)
#### Question 29:
In Fig., state which lines are parallel and why?
Let F be the point of intersection of line CD and the line passing through point E.
Since $\angle$ACD and $\angle$CDE are alternate and equal angles, so
$\angle$ACD = 100° = $\angle$CDE
∴ AC || EF
#### Question 30:
In Fig. 87, the corresponding arms of ∠ABC and ∠DEF are parallel. If ∠ABC = 75°, find ∠DEF.
Construction:
Let G be the point of intersection of lines BC and DE.
∵ AB || DE and BC || EF
$\angle \mathrm{ABC}=\angle \mathrm{DGC}=\angle \mathrm{DEF}=75°$ (Corresponding angles)
#### Question 1:
The sum of an angle and one third of its supplementary angle is 90°. The measure of the angle is
(a) 135°
(b) 120°
(c) 60°
(d) 45°
Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
$x+\frac{1}{3}\left(180°-x\right)=90°\phantom{\rule{0ex}{0ex}}⇒3x+180°-x=270°\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, the correct answer is option (d).
#### Question 2:
If angles of a linear pair are equal, then the measure of each angle is
(a) 30°
(b) 45°
(c) 60°
(d) 90°
Let the required angle be x
Now, Sum of linear pair angles = 180
x + x = 180
2x = 180
x = 90
Hence, the correct answer is option (d).
#### Question 3:
Two complemntary angles are in the ratio 2 : 3. The measure of the larger angle is
(a) 60°
(b) 54°
(c) 66°
(d) 48°
Let the angles be 2x and 3x.
Now, 2x + 3x = 90
⇒ 5x = 90
x = 18
∴ Larger angle = 3x = 3 × 18 = 54
Hence, the correct answer is option (b).
#### Question 4:
An angle is thrice its supplement. The measure of the angle is
(a) 120°
(b) 105°
(c) 135°
(d) 150°
Let the required angle be x
Then,
$x=3\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒x=540°-3x\phantom{\rule{0ex}{0ex}}⇒4x=540°\phantom{\rule{0ex}{0ex}}⇒x=135°$
Hence, the correct answer is option (c).
#### Question 5:
In Fig. 88 PR is a straight line and ∠PQS : ∠SQR = 7 : 5. The measure of ∠SQR is
(a) 60°
(b) ${\left(62\frac{1}{2}\right)}^{°}$
(c) ${\left(67\frac{1}{2}\right)}^{°}$
(d) 75°
Let the measures of the angle ∠PQS and ∠SQR be 7x and 5x.
Now, ∠PQS + ∠SQR = 180 [Linear pair angles]
⇒ 7x + 5x = 180
⇒ 12x = 180
x = 15
∴ ∠SQR = 5x = 5 × 15 = 75
Hence, the correct answer is option (d).
#### Question 6:
The sum of an angle and half of its complementary angle is 75°. The measure of the angle is
(a) 40°
(b) 50°
(c) 60°
(d) 80°
Let the required angle be x.
Now, complementnary of the required angle = 90∘ − x
Then,
$x+\frac{1}{2}\left(90°-x\right)=75°\phantom{\rule{0ex}{0ex}}⇒2x+90°-x=150°\phantom{\rule{0ex}{0ex}}⇒x=150-90°\phantom{\rule{0ex}{0ex}}⇒x=60°$
Hence, the correct answer is option (c).
#### Question 7:
∠A is an obtuse angle. The measure of ∠A and twice its supplementary differ by 30°. Then ∠A can be
(a) 150°
(b) 110°
(c) 140°
(d) 120°
Supplementary of ∠A = 180 − ∠A
Now,
∠A + 30 = 2(180 − ∠A)
⇒ ∠A + 30 = 360 − 2∠A
⇒ 3∠A = 360 − 30
⇒ 3∠A = 330
⇒ ∠A = 110
Hence, the correct answer is option (b).
#### Question 8:
An angle is double of its supplement. The measure of the angle is
(a) 60°
(b) 120°
(c) 40°
(d) 80°
Let the required angle be x.
Now, supplementary of the required angle = 180∘ − x
Then,
$x=2\left(180°-x\right)\phantom{\rule{0ex}{0ex}}⇒x=360°-2x\phantom{\rule{0ex}{0ex}}⇒3x=360°\phantom{\rule{0ex}{0ex}}⇒x=120°$
Hence, the correct answer is option (b).
#### Question 9:
The measure of an angle which is its own complement is
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Let the required angle be x.
Now, complementary of the required angle = 90∘ − x
Then,
$x=90°-x\phantom{\rule{0ex}{0ex}}⇒x=90°-x\phantom{\rule{0ex}{0ex}}⇒2x=90°\phantom{\rule{0ex}{0ex}}⇒x=45°$
Hence, the correct answer is option (d).
#### Question 10:
Two supplementary angles are in the ratio 3 : 2. The smaller angle measures
(a) 108°
(b) 81°
(c) 72°
(d) 68°
Let the angles be 3x and 2x.
Now, 3x + 2x = 180
⇒ 5x = 180
x = 36
∴ Smaller angle = 2x = 2 × 36 = 72
Hence, the correct answer is option (c).
#### Question 11:
In Fig. 89, the value of x is
(a) 75
(b) 65
(c) 45
(d) 55
∠AOC and ∠BOC = 180 [∵ Linear pair angles]
⇒ 44+ (2x + 6) = 180
⇒ (2x + 6) = 136
⇒ 2x + 6 = 136
⇒ 2x = 130
x = 65
Hence, the correct answer is option (b).
#### Question 12:
In Fig. 90, AOB is a straight line and the ray OCstands on it. The value of x is
(a) 16
(b) 26
(c) 36
(d) 46
∠AOC + ∠BOC = 180 [∵ Linear pair angles]
⇒ (2x + 15) + (3x + 35) = 180
⇒ (5x + 50) = 180
⇒ 5x + 50 = 180
⇒ 5x = 130
x = 26
Hence, the correct answer is option (b).
#### Question 13:
In Fig. 91, AOB is a straight line and 4x = 5y. The value of x is
(a) 100
(b) 105
(c) 110
(d) 115
∠AOC + ∠BOC = 180 [∵ Linear pair angles]
y + x = 180
y + x = 180
Hence, the correct answer is option (a).
#### Question 14:
In Fig. 92, AOB is a straight line such that ∠AOC = (3x + 10)°, ∠COD = 50° and ∠BOD = (x − 8)°. The value of x is
(a) 32
(b) 36
(c) 42
(d) 52
∠AOC + ∠COD + ∠BOD = 180 [AOB is a straight line]
⇒ (3x + 10) + 50+ (x − 8) = 180
⇒ 3x + 10 + 50 + x − 8 = 180
⇒ 4x + 52 = 180
⇒ 4x = 128
x = 32
Hence, the correct answer is option (a).
#### Question 15:
In Fig. 93, if AOC is a straight line, then x =
(a) 42°
(b) 52°
(c) 142°
(d) 38°
∠AOD + ∠DOB + ∠BOC = 180 [∵ AOC is a straight line]
⇒ 38 + x + 90 = 180
x + 128 = 180
x = 52
Hence, the correct answer is option (b).
#### Question 16:
In Fig. 94, if ∠AOC is a straight line, then the value of x is
(a) 15
(b) 18
(c) 20
(d) 16
∠AOD + ∠DOB + ∠BOC = 180 [ AOC is a straight line]
⇒ 2x + 90 + 3x= 180
⇒ 5x + 90 = 180
⇒ 5x = 90
x = 18
Hence, the correct answer is option (b).
#### Question 17:
In Fig. 95, if AB, CD and EF are straight lines, then x =
(a) 5
(b) 10
(c) 20
(d) 30
Let all the lines intersect at O.
∠COF = ∠DOE = 4x [Vertically opposite angles]
∠AOC + ∠COF + ∠BOF = 180 [AOB is a straight line]
⇒ 2x + 4x + 3x= 180
⇒ 9x = 180
⇒ 9x = 180
x = 20
Hence, the correct answer is option (c).
#### Question 18:
In Fig. 96, if AB, CD and EF are straight lines, then x + y + z =
(a) 180
(b) 203
(c) 213
(d) 134
∠DAE + ∠BAD + ∠BAF = 180 [EAF is a straight line]
⇒ 3x + 49 + 62= 180
⇒ 3x + 111= 180
⇒ 3x = 69
⇒ 3x = 69
x = 23
Now, ∠CAE + ∠CAF = 180 [∵ EAF is a straight line]
z + y= 180
z + y = 180
Now, x + y + z = 23 + 180 = 203
Hence, the correct answer is option (b).
#### Question 19:
In Fig. 97, if AB is parallel to CD, then the value of ∠BPE is
(a) 106°
(b) 76°
(c) 74°
(d) 84°
Since, AB || CD
∴ ∠BPQ = ∠PQC [Alternate interior angles]
⇒ (3x + 34) = (5− 14)
⇒ 3x + 34 = 5− 14
⇒ 48 = 2x
x = 24
∴ ∠BPQ = (3 × 24 + 34) = 106
∠BPQ + ∠BPE = 180 [EF is a straight line]
⇒ 106 + ∠BPE = 180
⇒ ∠BPE = 74
Hence, the correct answer is option (c).
#### Question 20:
In Fig. 98, if AB is parallel to CO and EF is a transversal, then x =
(a) 19
(b) 29
(c) 39
(d) 49
Let the line EF intersect AB and CD at P and Q respectively.
Since, AB || CD
∴ ∠BPQ + ∠PQD = 180 (Angles on the same side of a transversal line are supplementary)
⇒ (7x − 12) + (4x + 17) = 180
⇒ 7x − 12 + 4x + 17 = 180
⇒ 11x + 5 = 180
⇒ 11x = 175
x = 15.90
Disclaimer: No option is correct.
#### Question 21:
In Fig. 99, AB || CD and EF is a transversal intersecting ABand CO at Pand Q respectively. The measure of ∠DPQ is
(a) 100
(b) 80
(c) 110
(d) 70
∠BQF = ∠AQP = (4x) [Vertically opposite angles]
Since, AB || CD
∴ ∠AQP + ∠CPQ = 180 [Angles on the same side of a transversal line are supplementary]
⇒ (4x) + (5x) = 180
⇒ 9 = 180
x = 20
∴ ∠BQF = (4 × 20) = 80
Now, ∠BQF = ∠DPQ = 80 [Corresponding angles]
Hence, the correct answer is option (b).
#### Question 22:
In Fig. 100, AB || CO and EF is a transversal intersecting AB and CD at P and Q respective. The measure of ∠OOP is
(a) 65
(b) 25
(c) 115
(d) 105
∠BPE = ∠APQ = (5x − 10) [Vertically opposite angles]
Since, AB || CD
∴ ∠APQ + ∠CQP = 180 [Angles on the same side of a transversal line are supplementary]
⇒ (5x − 10) + (3x − 10) = 180
⇒ 8x − 20 = 180
⇒ 8x = 200
x = 25
∴ ∠BPE = (5 × 25 − 10) = 115
Now, ∠BPE = ∠DQP = 115 [Corresponding angles]
Hence, the correct answer is option (c).
#### Question 23:
In Fig. 101, AB || CD and EF is a transversal. The value of y − x is
(a) 30
(b) 35
(c) 95
(d) 25
Since, AB || CD
∴ ∠BPQ = ∠DQF [Corresponding angles]
⇒ (5x − 20) = (3x + 40)
⇒ 5x − 20 = 3x + 40
⇒ 2x = 60
x = 30
∴ ∠BPQ = (5 × 30 − 20 ) = 130
Now, ∠APE = ∠BPQ [Vertically opposite angles]
⇒ 2y = 130
y = 65
y − x = 65 30 = 35
Hence, the correct answer is option (b).
#### Question 24:
In Fig. 102, AB || CD || EF, ∠ABG = 110°, ∠GCO = 100° and ∠BGC = x°. The value of x is
(a) 35
(b) 50
(c) 30
(d) 40
Since, AB || EG
∴ ∠ABG + ∠EGB = 180 (Angles on the same side of a transversal line are supplementary)
⇒ 110 + ∠EGB = 180
⇒ ∠EGB = 70
Again, CD || GF
∴ ∠DCG + ∠FGC = 180 (Angles on the same side of a transversal line are supplementary)
⇒ 100 + ∠FGC = 180
⇒ ∠FGC = 80
Now, ∠EGB + ∠BGC +∠FGC = 180
⇒ 70 + x + 80= 180
⇒ 150+ x = 180
x = 30
x = 30
Hence, the correct answer is option (c).
#### Question 25:
In Fig. 103, PO || RS and ∠PAB = 60° and ∠ACS = 100°. Then, ∠BAC =
(a) 40°
(b) 60°
(c) 80°
(d) 50°
Since, PQ || RS
∴ ∠PAC = ∠ACS = 100 [Corresponding angles]
Now, ∠PAC = 100
⇒ ∠PAB + ∠BAC = 100
⇒ 60 + ∠BAC = 100
⇒ ∠BAC = 40
Hence, the correct answer is option (a).
#### Question 26:
In Fig. 104, AB || CO, ∠OAB = 150° and ∠OCO = 120°. Then, ∠AOC =
(a) 80°
(b) 90°
(c) 70°
(d) 100°
Construction: Draw a line OE from the point O parallel to AB and CD
Since, AB || OE
∴ ∠BAO + ∠AOE = 180 [Angles on the same side of a transversal line are supplementary]
⇒ 150 + ∠AOE = 180
⇒ ∠AOE = 30
Again, CD || OE
∴ ∠DCO + ∠COE = 180 [Angles on the same side of a transversal line are supplementary]
⇒ 120 + ∠COE = 180
⇒ ∠COE = 60
Now, ∠AOC = ∠AOE + ∠COE
= 30 + 60
= 90
Hence, the correct answer is option (b).
#### Question 27:
In Fig. 105, if AOB and COD are straight lines. Then, x + y =
(a) 120
(b) 140
(c) 100
(d) 160
∠AOD + ∠BOD = 180 [Linear pair angles]
⇒ (7x − 20) + 3x = 180
⇒ 7x − 20 + 3x = 180
⇒ 10x = 200
x = 20
∠AOD = (7 × 20 − 20) = 120
Now∠AOD = ∠BOC = 120 [Vertically opposite angles]
y = 120
Now, x + y = 20 + 120
= 140
Hence, the correct answer is option (b).
#### Question 28:
In Fig. 106, the value of x is
(a) 22
(b) 20
(c) 21
(d) 24
(8x − 41) + (3x) + (3x + 10) + (4x − 5)= 360
⇒ 8x − 41 + 3x + 3x + 10 + 4x − 5 = 360
⇒ 18x − 36 = 360
⇒ 18x = 396
x = 22
Hence, the correct answer is option (a).
#### Question 29:
In Fig. 107, if AOBand COD are straight lines, then
(a) x = 29, y = 100
(b) x = 110, y = 29
(c) x = 29, y = 110
(d) x = 39, y = 110
∠AOD + ∠BOD = 180 [Linear pair angles]
y + 70 = 180
y = 110
y = 110
Now, ∠AOC = ∠BOD = 70 [Vertically opposite angles]
Now, ∠AOC + ∠COE + ∠EOB + ∠BOD + ∠AOD = 360 [Complete angle]
⇒ 70 + 28 + (3x − 5) + 70 + 110 = 360
⇒ (3x) + 273 = 360
⇒ 3x = 87
x = 29
Hence, the correct answer is option (c).
#### Question 30:
In Fig. 108, if AB || CD then the value of x is
(a) 87
(b) 93
(c) 147
(d) 141
Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠FCD + Reflex∠FCD = 360 (Complete angle)
⇒ ∠FCD + 273 = 360
⇒ ∠FCD = 87
Since, PQ || CD
∴∠QFC + ∠FCD = 180 (Angles on the same side of a transversal line are supplementary)
⇒ ∠QFC + 87 = 180
⇒ ∠QFC = 93
Now, ∠ABF = ∠BFQ (Corresponding angles)
= ∠BFC + ∠QFC
= 54 + 93
= 147
x = 147
x = 147
Hence, the correct answer is option (c).
#### Question 31:
In Fig. 109, if AB || CD then the value of x is
(a) 34
(b) 124
(c) 24
(d) 158
Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠QEC + ∠ECD = 180 [Angles on the same side of a transversal line are supplementary]
⇒ ∠QEC + 56 = 180
⇒ ∠QEC = 124
Now, ∠BEQ + ∠QEC = ∠BEC
⇒ ∠BEQ + 124 = 158
⇒ ∠BEQ = 34
Now, ∠ABE = ∠BEQ = 34 [Corresponding angles]
x = 34
x = 34
Hence, the correct answer is option (a).
#### Question 32:
In Fig. 110, if AB || CD. The value of x is
(a) 122
(b) 238
(c) 58
(d) 119
Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || CD
∴ ∠EFC = ∠FEQ = 37 [Alternate angles]
Now, ∠AEQ + ∠FEQ = ∠AEF
⇒ ∠AEQ + 37 = 95
⇒ ∠AEQ = 58
Since, PQ || AB
∴∠EAB + ∠AEQ = 180 [Angles on the same side of a transversal line are supplementary]
⇒ ∠EAB + 58 = 180
⇒ ∠EAB = 122
∠EAB + Reflex∠EAB = 360 [Complete angle]
∴ 122 + (2x) = 360
⇒ 2x = 238
x = 119
Hence, the correct answer is option (d).
#### Question 33:
In Fig. 111, if AB || CO then x =
(a) 154
(b) 139
(c) 144
(d) 164
Construction: Draw a line PQ parallel to AB which is also parallel to CD
Since, PQ || AB
∴ ∠AME + ∠QEM = 180 [Angles on the same side of a transversal line are supplementary]
⇒ 139 + ∠QEM = 180
⇒ ∠QEM = 41
Now, ∠QEM + ∠DEQ = ∠MED
⇒ 41 + ∠DEQ = 67
⇒ ∠DEQ = 26
Now, ∠PED + ∠DEQ = 180 [Linear Pair angles]
⇒ ∠PED + 26 = 180
⇒ ∠PED = 154
Since, PQ || AB
x = ∠PED [Corresponding angles]
x = 154
x = 154
Hence, the correct answer is option (a).
#### Question 34:
In Fig. 112, if AB || CD, then x =
(a) 32
(b) 42
(c) 52
(d) 31
Construction: Draw a line PQ parallel to AB which is also parallel to CD
∠CDP + Reflex∠CDP = 360 [Complete angle]
∴∠CDP + 249 = 360
⇒ ∠CDP = 111
Since, PQ || AB
∴ ∠BAP = ∠APQ [Alternate angles]
⇒ ∠BAP = 28
Now, ∠APQ + ∠QPD = ∠APD
⇒ 28 + ∠QPD = (2x + 13)
⇒ ∠QPD = (2x + 13)− 28
Since, PQ || CD
∴ ∠QPD + ∠CDP = 180 [Angles on the same side of a transversal line are supplementary]
⇒ (2x + 13)− 28 + 111 = 180
⇒ 2x + 13 − 28 + 111 = 180
⇒ 2x = 84
x = 42
Hence, the correct answer is option (b).
#### Question 35:
In Fig. 113 if AC || OF and AB || CE, then
(a) x = 145, y = 223
(c) x = 135, y = 233
(b) x = 223, y = 145
(d) x = 233, y = 135
Construction: Produce FD towards D to the point M
∠DCA + Reflex∠DCA = 360 [Complete angle]
∴∠DCA + (y + 15) = 360
⇒ ∠DCA = 345y
Now,
∠MDC = ∠EDF = 58 [Vertically Opposite angles]
Since, MF || AC
∴ ∠MDC + ∠QPD = 180 [Angles on the same side of a transversal line are supplementary]
⇒ 58 + 345y = 180
y = 223
∴ ∠DCA = 345 223 = 122
Again, ∠BAC + Reflex∠BAC = 360 [Complete angle]
∴∠BAC + (2x + 12) = 360
⇒ ∠DCA = 348 − (2x)
Since, AB || CD
∴ ∠DCA + ∠DCA = 180 [Angles on the same side of a transversal line are supplementary]
⇒ 348 − (2x)+ 122 = 180
⇒ (2x)= 290
x = 145
Hence, the correct answer is option (a).
#### Question 1:
Write down each pair of adjacent angles shown in Fig.
Adjacent angles are the angles that have a common vertex and a common arm.
Following are the adjacent angles in the given figure:
#### Question 2:
In Fig., name all the pairs of adjacent angles.
In figure (i), the adjacent angles are:
In figure (ii), the adjacent angles are:
$\angle$BAD and $\angle$DAC
$\angle$BDA and $\angle$CDA
#### Question 3:
In figure, write down: (i) each linear pair (ii) each pair of vertically opposite angles.
(i) Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$\angle$1 and $\angle$3
$\angle$1 and $\angle$2
$\angle$4 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$6
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$7 and $\angle$8
(ii) Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$4
$\angle$2 and $\angle$3
$\angle$5 and $\angle$8
$\angle$6 and $\angle$7
#### Question 4:
Are the angles 1 and 2 given in Fig. adjacent angles?
No, because they have no common vertex.
#### Question 5:
Find the complement of each of the following angles:
(i) 35°
(ii) 72°
(iii) 45°
(iv) 85°
Two angles are called complementary angles if the sum of those angles is 90°.
Complementary angles of the following angles are:
#### Question 6:
Find the supplement of each of the following angles:
(i) 70°
(ii) 120°
(iii) 135°
(iv) 90°
Two angles are called supplementary angles if the sum of those angles is 180°.
Supplementary angles of the following angles are:
(i) 180° − 70° = 110°
(ii) 180° − 120° = 60°
(iii) 180° − 135° = 45°
(iv) 180° − 90° = 90°
#### Question 7:
Identify the complementary and supplementary pairs of angles from the following pairs:
(i) 25°, 65°
(ii) 120°, 60°
(iii) 63°, 27°
(iv) 100°, 80°
Since
Therefore, (i) and (iii) are the pairs of complementary angles and (ii) and (iv) are the pairs of supplementary angles.
#### Question 8:
Can two angles be supplementary, if both of them be
(i) obtuse?
(ii) right?
(iii) acute?
(i) No, two obtuse angles cannot be supplementary.
(ii) Yes, two right angles can be supplementary. ($\because \angle 90°+\angle 90°=\angle 180°$)
(iii) No, two acute angles cannot be supplementary.
#### Question 9:
Name the four pairs of supplementary angles shown in Fig.
Following are the supplementary angles:
$\angle$AOC and $\angle$COB
$\angle$BOC and $\angle$DOB
$\angle$BOD and $\angle$DOA
$\angle$AOC and $\angle$DOA
#### Question 10:
In Fig., A, B, C are collinear points and ∠DBA = ∠EBA.
(i) Name two linear pairs
(ii) Name two pairs of supplementary angles.
(i) Linear pairs:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC
Because every linear pair forms supplementary angles, these angles are:
$\angle$ABD and $\angle$DBC
$\angle$ABE and $\angle$EBC
#### Question 11:
If two supplementary angles have equal measure, what is the measure of each angle?
Let x and y be two supplementary angles that are equal.
$\angle x=\angle y$
According to the question,
$\angle x+\angle y=180°\phantom{\rule{0ex}{0ex}}⇒\angle x+\angle x=180°\phantom{\rule{0ex}{0ex}}⇒2\angle x=180°\phantom{\rule{0ex}{0ex}}⇒\angle x=\frac{180°}{2}=90°\phantom{\rule{0ex}{0ex}}\therefore \angle x=\angle y=90°$
#### Question 12:
If the complement of an angle is 28°, then find the supplement of the angle.
Let x be the complement of the given angle $28°$.
So, supplement of the angle = $180°-62°=118°$
#### Question 13:
In Fig. 19, name each linear pair and each pair of vertically opposite angles:
Two adjacent angles are said to form a linear pair of angles if their non-common arms are two opposite rays.
$\angle$1 and $\angle$2
$\angle$2 and $\angle$3
$\angle$3 and $\angle$4
$\angle$1 and $\angle$4
$\angle$5 and $\angle$6
$\angle$6 and $\angle$7
$\angle$7 and $\angle$8
$\angle$8 and $\angle$5
$\angle$9 and $\angle$10
$\angle$10 and $\angle$11
$\angle$11 and $\angle$12
$\angle$12 and $\angle$9
Two angles formed by two intersecting lines having no common arms are called vertically opposite angles.
$\angle$1 and $\angle$3
$\angle$4 and $\angle$2
$\angle$5 and $\angle$7
$\angle$6 and $\angle$8
$\angle$9 and $\angle$11
$\angle$10 and $\angle$12
#### Question 14:
In Fig., OE is the bisector of ∠BOD. If ∠1 = 70°, find the magnitudes of ∠2, ∠3 and ∠4.
Since OE is the bisector of $\angle$BOD,
#### Question 15:
One of the angles forming a linear pair is a right angle. What can you say about its other angle?
One angle of a linear pair is the right angle, i.e., 90°.
∴ The other angle = 180° 90° = 90°
#### Question 16:
One of the angles forming a linear pair is an obtuse angle. What kind of angle is the other?
If one of the angles of a linear pair is obtuse, then the other angle should be acute; only then can their sum be 180°.
#### Question 17:
One of the angles forming a linear pair is an acute angle. What kind of angle is the other?
In a linear pair, if one angle is acute, then the other angle should be obtuse. Only then their sum can be 180°.
#### Question 18:
Can two acute angles form a linear pair?
No, two acute angles cannot form a linear pair because their sum is always less than 180°.
#### Question 19:
If the supplement of an angle is 65°; then find its complement.
Let be the required angle.
Then, we have:
x + 65° = 180°
$⇒$x = 180° - 65° = 115°
The complement of angle cannot be determined.
#### Question 20:
Find the value of x in each of the following figures.
(i)
Since (Linear pair)
(ii)
(iii)
(iv)
(v)
$2x°+x°+2x°+3x°=180°\phantom{\rule{0ex}{0ex}}⇒8x=180\phantom{\rule{0ex}{0ex}}⇒x=\frac{180}{8}=22.5°$
(vi)
$3x°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105}{3}=35°$
#### Question 21:
In Fig. 22, it being given that ∠1 = 65°, find all other angles.
$\angle 1=\angle 3$ (Vertically opposite angles)
$\therefore \angle 3=65°$
Since $\angle 1+\angle 2=180°$ (Linear pair)
$\therefore \angle 2=180°-65°=115°$
$\angle 2=\angle 4$ (Vertically opposite angles)
$\therefore \angle 4=\angle 2=115°$ and $\angle 3=65°$
#### Question 22:
In Fig., OA and OB are opposite rays:
(i) If x = 25°, what is the value of y?
(ii) If y = 35°, what is the value of x?
$\angle$AOC + $\angle$BOC = 180° (Linear pair)
$⇒\left(2y+5\right)+3x=180°\phantom{\rule{0ex}{0ex}}⇒3x+2y=175°$
(i) If x = 25°, then
$3×25°+2y=175°\phantom{\rule{0ex}{0ex}}⇒75°+2y=175°\phantom{\rule{0ex}{0ex}}⇒2y=175°-75°=100°\phantom{\rule{0ex}{0ex}}⇒y=\frac{100°}{2}=50°$
(ii) If y = 35°, then
$3x+2×35°=175°\phantom{\rule{0ex}{0ex}}⇒3x+70°=175°\phantom{\rule{0ex}{0ex}}⇒3x=175°-70°=105°\phantom{\rule{0ex}{0ex}}⇒x=\frac{105°}{3}=35°$
#### Question 23:
In Fig., write all pairs of adjacent angles and all the linear pairs.
Linear pairs of angles:
#### Question 24:
In Fig. 25, find ∠x. Further find ∠BOC, ∠COD and ∠AOD.
$\angle BOC=x+20°=50°+20°=70°\phantom{\rule{0ex}{0ex}}\angle COD=x=50°\phantom{\rule{0ex}{0ex}}\angle AOD=x+10°=50°+10°=60°\phantom{\rule{0ex}{0ex}}$
#### Question 25:
How many pairs of adjacent angles are formed when two lines intersect in a point?
If two lines intersect at a point, then four adjacent pairs are formed, and those pairs are linear as well.
#### Question 26:
How many pairs of adjacent angles, in all, can you name in Fig.?
There are 10 adjacent pairs in the given figure; they are:
#### Question 27:
In Fig., determine the value of x.
#### Question 28:
In Fig., AOC is a line, find x.
#### Question 29:
In Fig., POS is a line, find x.
$\angle \mathrm{QOP}+\angle \mathrm{QOR}+\angle \mathrm{ROS}=180°$ (Angles on a straight line)
$⇒60°+4x+40°=180°\phantom{\rule{0ex}{0ex}}⇒100°+4x=180°\phantom{\rule{0ex}{0ex}}⇒4x=180°-100°=80°\phantom{\rule{0ex}{0ex}}⇒x=\frac{80°}{4}=20°\phantom{\rule{0ex}{0ex}}$
#### Question 30:
In Fig., lines l1 and l2 intersect at O, forming angles as shown in the figure. If x = 45°, find the values of y, z and u.
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+0
# Help
0
127
4
+133
Find the ordered pair $$(a,b)$$ of real numbers such that the cubic polynomials $$x^3 + ax^2 + 11x + 6 = 0$$ and $$x^3 + bx^2 + 14x + 8 = 0$$ have two distinct roots in common.
May 25, 2019
#1
+102804
0
I'd also like to see someone answer this.
May 26, 2019
#2
+28125
+3
Here is my attempt:
.
May 26, 2019
#3
+102804
0
Thanks Alan,
I did not bother with expanding
I used this
if:
$$ax^3+bx^2+cx+d=0\\ \text{and the roots are u,v and t}\\ then\\ u+v+t=-\frac{b}{a}\\ uv+ut+vt=+\frac{c}{a}\\ uvt=-\frac{d}{a}$$
The relevance is that I think uvt=-8
I got this far myself but then I got into a mess.
May 26, 2019
#4
+102434
+1
x^3 + ax^2 + 11x + 6 = 0 (1)
x^3 + bx^2 + 14x + 8 = 0 (2)
By the Rational Roots Theorem, the possible zeroes for the first polynomial are ± [ 1, 2 , 3 , 6]
And for the second polynomial they are ± [ 1,2,3,4]
Subtract (1) from (2) and we get that
(b - a) x^2 + 3x + 2 = 0
Two possible shared roots of -1 and - 2 can be found if we let b - a = 1
So we have that
x^2 + 3x+ 2 = 0
(x + 1) ( x + 2) = 0
x = - 1 and x = -2
Now let x = -1 as a shared root
Then (-1)^3 + a(-1)^2 + 11(-1) + 6 = 0
-1 + a - 11 + 6 = 0
And a =6
Also
(-1)^3 + b(-1)^2 + 14(-1) +8 = 0
-1 + b -14 + 8 = 0
And b = 7
So testing (-2) as a root we have that
(-2)^3 + 6(-2)^2 + 11(-2) + 6 =
-8 + 24 - 22 + 6 =
0
And
(-2)^3 + 7(-2)^2 + 14(-2) + 8 =
-8 + 28 -28 + 8 =
0
So
(a,b ) = (6, 7)
Here is the graph to show that this is true :https://www.desmos.com/calculator/y69o2yts2n
May 26, 2019
edited by CPhill May 26, 2019
edited by CPhill May 26, 2019
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After propagating distance $$L$$ in medium, the CE phase changes due to diffence of phase and group velocities, $$\Delta\varphi_\mathsf{CE} = \omega_0 \sum_{i=1}^N\left(\frac{1}{v_{\mathsf{g},i}} - \frac{1}{v_{\mathsf{p},i}} \right) h_i . 18.39 18. The phase of vector sum of the three rays is a function of d(x,y)n with a period of lÀ. Fig. Staff Access, Visual Ordinary rays do not have spatial walk-off. In mathematics, a square root of a number x is a number y such that y² = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x.$$P_0 =\frac{2\mathcal{E}}{\Delta t}\sqrt{\frac{\ln2}{\pi}}\approx\frac{0.94\mathcal{E}}{\Delta t}. Optical period $$T = \frac{\lambda}{c} \Longrightarrow T[\mathrm{fs}] \approx \frac{\lambda[\mathrm{nm}]}{299.792}$$ Each mounting shim has two slots for compatibility with our SPW602 and SPW606 Spanner Wrenches. $11.99$ 11. Here $$\vartheta_0$$ is AOI and $$\vartheta_1 = \arcsin\frac{\sin\vartheta_0}{n}$$ is angle of refraction. Energy $$E = \hbar\omega \Longrightarrow E[\mathrm{eV}] \approx \frac{\omega[\mathrm{fs^{-1}}]}{1.519}$$ MATERIALS/PROCESS Case Studies. $$sI(\lambda) \to I(\lambda)$$ and $$\intop_{\lambda_\mathrm{min}}^{\lambda_\mathrm{max}}I(\lambda)\mathrm{d}\lambda = P.$$. ToptiCalc offers some frequently used calculations and conversions from different areas of laser optics which will help you in your daily work with lasers. New Calculator. The shim does not impinge upon the clear aperture of our wedge prisms, as shown in the drawings to the right. After propagating distance $$L$$ in medium, the CE phase changes due to diffence of phase and group velocities, $$\Delta\varphi_\mathsf{CE} = \omega_0 \left(\frac{1}{v_\mathsf{g}} - \frac{1}{v_\mathsf{p}} \right) L.$$ Energy $$E = \frac{2\pi\hbar}{T} \Longrightarrow E[\mathrm{eV}] \approx \frac{4.136}{T[\mathrm{fs}]}$$ $$, Carrier-envelope phase $$\varphi_\mathsf{CE}$$ is the phase difference between the maxima of (i) oscillating field intensity and (ii) carrier envelope. Four shims are available, each with a wedge angle that matches one of our wedge prisms: 3° 53', 7° 41', 11° 22', or 18° 9'. wedge tip to the point where the ray emerges. Phase matching condition:$$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. $$, Maximal pulse power. Check out the largest selection of affordable glasses online, including prescription glasses and sunglasses for men, women, and kids. Sydor Optics manufactures custom optical wedges used in a variety of applications. 99. See Bielschowsky's phenomenon . Eyewear 21 Step Wedge Exposure Calculator for Screen Printing Stouffer Calculator Dial in Exposure Times (21 Step Wedge) 4.4 out of 5 stars 17. OptiCampus Online Optical Continuing Education. Opportunities Elementary treatments of this problem often leave much confusion about the exact conditions under which the fringes are formed and observed. Optical Dictionary. It is also known as linearly variable filter (LVF). This is done by individually controlling the rotation of each prism. Rayleigh length is distance from beam waist to the point, where beam diameter is $$2\sqrt{2}w_0$$. Optical wedges are commonly round with each surface polished flat and tilted to each other at a specific angle. Plano Concave. It is used in various optical sensors where wavelength separation is required e.g. optical wedge prism / fused silica / BK7 WED series. Since pulse spectral density $$I(\lambda)$$ is given in arbitrary units, value of $$P$$ is used to obtain the spectral density scaling factor $$s$$, for which Solve. It is used as a method of ‘verification’ to establish that a vibration test system is doing what it should be, and not something unexpected due to incorrect input sensitivity set-up. ITAR Registered l ISO 9001:2015 l 915-595-5417 l Plus Plano Concave. Here $$\Delta t$$ is pulse length (FWHM). Zone Selector, Note: The calculator results are estimates only. Here $$\vartheta_0$$ is the angle of incidence. Mechanical advantage is the measure of vitality saved by utilizing apparatuses or mechanical gadgets. Product of pulse duration and spectral width frequency (both in FWHM). For temporally Gaussian pulse, peak power is related to pulse energy $$\mathcal{E}$$ and length $$\Delta t$$ (FWHM) as Fill in the input fields and then press the Calculate button in order to compute the approximate center and edge thickness (through the 180°) produced by a spectacle lens. 4.0 out of 5 stars 16. Select One of the Following optical Calculators. Or, a pair of wedge prisms can steer a beam anywhere within a circle described by the full angle 4θ, where θ is the deviation from a single prism. It deflects light toward its thicker portion. This calculator is a handy interactive tool for calculating the Solar, Optical and Thermal properties of our most popular glazing combinations, and will even generate 3-part CSI format custom glass specifications for each, ready to incorporate into your project specifications. For given angle of incidence $$\vartheta_0$$, prism with apex angle$$\alpha_0=2\arcsin\frac{\sin\vartheta_0}{n}$$would cause minimal possible deviation angle $$\delta$$. Precision wedge prisms made of BK7, UV grade fused silica optical material.$$, Exact and approximate relations between the bandwidth in wavelength and wavenumber units is given by: $$\Delta\lambda = \frac{4\pi c}{\Delta \omega} \left( \sqrt{1+\frac{\lambda_0^2\Delta \omega^2}{4\pi^2 c^2}} - 1 \right) \approx \frac{\Delta \omega\lambda_0^2}{2\pi c} = \Delta k \lambda_0^2. Consulting With over 10,000 downloads, it is one of the most frequently used apps for this purpose. Privacy geometrical optics page is a free tool enabling online optical calculations Light energy injected by OTDR into the fiber through a laser diode and pulse generator. Wedge Prisms are often used to steer the beam in a laser application. Instyle Frame This work center is designed to aid Esco customers, technicians, engineers, students, and industry professionals. Here $$\vartheta_0$$ is the angle of incidence. The Fabry Perot interferometer consists of two parallel flat semi-transparent mirrors separated by a fixed distance. Blue See also absorbing wedge… In this paper we study the effect of material dispersion on the performance of a moving-optical-wedge Fourier transform spectrometer. Mechanical advantage is the measure of vitality saved by utilizing apparatuses or mechanical gadgets. Contact. Precise beam deviation can be achieved by placing two wedges in series and rotating each wedge to each other, then rotating the wedges assembly. Read on to discover how Snell's law of refraction is formulated and what equation will let you calculate the angle of refraction. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.) View RFQ (0 items) | l Downloads. Has its minimum for ideal transform-limited pulses: Divergence angle $$\vartheta$$ describes how Gaussian beam diameter spreads in the far field ($$z\gg z_\mathrm{R}$$). Optical Calculator. Constructiv… Step 4: We theoretically analyze and experimentally investigate the dependence of residual amplitude modulation (RAM) on the beam radius within the electro-optic crystal (EOC), the wedge angle of the EOC and the overlap efficiency between the extraordinary and ordinary beams, and the overlap efficiency is determined by the distance from the wedge facet to the downstream polarizer. W e d g e (1) v o l … Slimlite FREE Shipping.$$ d = h \sin\vartheta_0\left( 1 - \sqrt{\frac{1-\sin^2\vartheta_0}{n^2-\sin^2\vartheta_0}}\right).$$, Optical path in system of two slabs, characterized by distance $$L$$, angle of incidence $$\vartheta_0$$ and group velocity at material $$v_\mathrm{g}$$, The FemtoOptics Dispersion Wedge family offers two size choices to fine tune dispersion in an ultrafast optical system. In that case the refraction angle is equal to the angle of incidence, $$\vartheta_0=\vartheta_1$$. Bi-Convex Lenses. Hayes The people at LightMachinery are veterans of the laser and optics world with many years of experience in the areas of optical design, high power lasers, optical fabrication, laser systems, metrology, thin film coatings and custom machinery fabrication. Terms of Service Privacy Policy Privacy Policy base side a: base width b: top side c: height h: volume V . & SUNGLASSES Frequency$$ f = ck \Longrightarrow f[\mathrm{THz}] \approx \frac{k[\mathrm{cm^{-1}}]}{33.356} $$, Wavelength$$ \lambda = Tc \Longrightarrow \lambda[\mathrm{nm}] \approx T[\mathrm{fs}] \cdot 299.792$$Phase matching condition:$$ \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{e}(\vartheta,\lambda_1)}{\lambda_1} + \frac{n_\mathrm{o}(\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. Pairs of ultrathin wedges allow fine, continuous tuning of GDD. If the filter transmits all the wavelengths more or less equally, it is called a neutral wedge . Optical period $$T = \frac{1}{ck} \Longrightarrow T[\mathrm{fs}] \approx \frac{3.336\cdot 10^4}{k[\mathrm{cm^{-1}}]}$$ Achromats. For. The quartz wedge is a simple, semi-quantitative compensator designed around a crystalline block of quartz cut with an elongated wedge angle so that the optical axis of the quartz is oriented either parallel or perpendicular to the edge of the crystal. $$, Time of flight of Gaussian beam through optical path length $$L$$,$$ t = \frac{L}{v_\mathsf{g}}=\frac{L}{c}\left( n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda} \right). Ross Optical is a top-ranking supplier for some of the largest OEMs & prime contractors in medical devices, instrumentation & machine vision/robotics. Angular frequency $$\omega = \frac{2\pi}{T} \Longrightarrow \omega[\mathrm{fs^{-1}}] \approx \frac{6.283}{T[\mathrm{fs}]}$$ The wedge is parallel to a 2" long surface, which allows for numerous copy beams to be created. Here $$\vartheta_0$$ is the angle of incidence. Eyewear Step 3: Determine the depth of the wedge at the aft bulkhead of the cargo tank, depth = D; D = E x Tan X. Conversely, for negative lenses (plano-concave, biconcave, negative meniscus), the focal length is the point out ahead of the lens from where all the light rays theoretically diverg… It is a triangular shaped tool used to isolate or hold objects. OWA can measure a single part or many parts simultaneously (e.g. SAG. If you continue to use our services, we will assume that you agree to the use of such cookies. Rayleigh length is equal to confocal parameter $$b$$ divided by 2. $$P_0 =\frac{\mathrm{arccosh}\sqrt{2}\mathcal{E}}{\Delta t}\approx\frac{0.88\mathcal{E}}{\Delta t}. These are precision calculation tools for common optical and dispensing applications. The filter is used to modify the intensity distribution in a radiation beam. Smith wedge An optical element having plane-inclined surfaces. Kids Employment Maximal pulse power. ITAR Registered l ISO 9001:2015 l 1-915-595-5417 l . D = - log (T) or. Plano convex. CE phase shift is proportional to the first derivative of refractive index over the wavelength,$$ \Delta\varphi_\mathsf{CE} = -2\pi \sum_{i=1}^N h_i \frac{\partial n_i(\lambda)}{\partial \lambda} . EKSMA Optics uses cookies to give you the best shopping experience. Bi-Convex. A wedge is one of the six simple machines. On the following pages you will find some online tools for calculation of most important optical parameters. Phase matching angle: $$\vartheta =\arcsin\sqrt{\frac{\frac{(\lambda_{1}+\lambda_{2})^{2}}{\left(n_\mathrm{o}(\lambda_{1})\lambda_{2}+n_\mathrm{o}(\lambda_{2})\lambda_{1}\right)^{2}\cos^{2}\vartheta_{0}}-\frac{1}{n^2_\mathrm{o}(\lambda_{3})}}{\frac{1}{n_\mathrm{e}^{2}(\lambda_{3})}-\frac{1}{n_\mathrm{o}^{2}(\lambda_{3})}}}$$. the wedge angle, the optical thickness d(x,y)n is not equal to a constant over the window. A wedge specification alone is not sufficient to control geometry errors in a lens with one or more aspheric surfaces. Wedge prisms, Wedged windows, optical wedges Alkor Technologies provide made out of BK7, Fused Silica, BaF2, KBr, Silicon, Germanium and ZnSe according customer's specifications. Designed for a very broad wavelength range at near normal angle of incidence, two identical wedges can be used to compensate for dispersion and angular deviation. $$ITAR Registered l ISO 9001:2015 l 915-595-5417 l High Impact$$, Third-order dispersion (TOD) in material with refraction index $$n(\lambda)$$: $$\mathrm{TOD}(\lambda) = -\frac{\lambda^{4}}{4\pi^{2}c^{3}}\left[3\frac{\mathrm{d}^{2}n}{\mathrm{d}\lambda^{2}}+\lambda\frac{\mathrm{d}^{3}n}{\mathrm{d}\lambda^{^{3}}}\right]. Difference between $$m=-1$$ diffraction angle ($$\vartheta_{-1}$$) and AOI ($$\vartheta_0$$)$$ \vartheta_\mathrm{d} = \arcsin\left(\frac{\lambda}{d}-\sin{\vartheta_0}\right) - \vartheta_0 . Two wedge prisms can be used as an anamorphic pair for beam shaping (to correct the elliptical shape of diode outputs). peak fluence is obtained as $$F_0 = \mathcal{E}\frac{2^{\frac{1}{n}}n}{\pi w_{0}^{2}\Gamma\left(\frac{1}{n}\right)}. About Pairs of ultrathin wedges allow fine, continuous tuning of GDD. For further technical fundamentals and exlanations please check chapter "Optics". See the Application Idea tab for more details.Click to Enlarge[APPLIST]SM1W1122 Mounting Shim Being Used with an … For temporally sech² pulse, peak power is related to pulse energy $$\mathcal{E}$$ and length $$\Delta t$$ (FWHM) as$$ Here $$d$$ is displacement of optical path and optical path length within a slab is To know how OTDR calculates optical length, first of all we have to understand the block diagram of OTDR properly. A wedge can be referred as two identical triangles and three rectangles holding the same height but different lengths. $$For more than 78 years Edmund Optics (EO) has been a leading producer of optics, imaging, and photonics technology. Optical Calculator Telescope Matching VAT IT 02230330504 (C) 2016-2020 Officina Turini, Tutti i diritti riservati. Wedges divert light toward their thicker portions, and may be circular, oblong or square. 39.$$, $$n_\mathrm{g} = \frac{c}{v_\mathrm{g}} = n(\lambda) - \lambda \frac{\partial n(\lambda)}{\partial \lambda}$$. Phase matching condition: \frac{n_\mathrm{e}(\vartheta,\lambda_3)}{\lambda_3} = \left( \frac{n_\mathrm{o}(\lambda_1)}{\lambda_1} + \frac{n_\mathrm{e}(\vartheta,\lambda_2)}{\lambda_2} \right)\cos\vartheta_0. Opticare Upon each reflection and refraction, some of the input power will be lost; hence, the wedged beamsplitter may be used as an attenuator. The App “APE Calculator” is for solving equations from non-linear optics. The quartz wedge is a simple, semi-quantitative compensator designed around a crystalline block of quartz cut with an elongated wedge angle so that the optical axis of the quartz is oriented either parallel or perpendicular to the edge of the crystal. Maximal pulse intensity (at beam center). Room Light that enters the etalon undergoes multiple reflections and the interference of the light emerging from the etalon during each bounce causes a modulation in the transmitted and reflected beams. As shown in the shape of diode outputs ) stock lens use … OptiCampus online optical Education. Θ ) /λ, where beam diameter is \ ( \vartheta_0 \ ) of refraction is formulated and equation. Ultrafast optical system for more than 78 years Edmund optics ( EO has. Prisms are used for measuring displacement/distance and works using the principal of persistence of vision path length (. ( \vartheta_0=\vartheta_1 \ ) is the angle of incidence so projection through the light-guide is approximately linear App. Formed and observed best performance and lowest cost for their optical systems used to isolate or hold.... Lvf ) can make any angle solid with the five faces, nine edges, and may circular. Article is devoted to the center of the wedge with length and thickness of a wedge is one of wedge... Constructed that its thickness varies continuously along a path ( straight or curved ) on its surface choices to tune... ( \vartheta_0=\vartheta_1 \ ) is the angle of incidence ultra-short pulse laser technology ( UKP ) optical Calculator used! Handling will be increased above that for our neutral density filters the curvature formulas are used for beam (... Two wedge prisms can be used individually or in steps in the etalon who are concerned... Industry professionals tuning of GDD a displacement wedge is one of the beam in a pair a... Approximately proportional to X so projection through the light-guide is approximately linear textbooks... Optical wedges are commonly round with each surface polished flat and tilted to each other at a angle! 1 minute of arc or better. Select your coatings Welcome to Esco 's optical Design center... Rfq ( 0 items ) | l Downloads with the five faces, nine edges, and may be,... Supplier for some of the wedge angle, the optical thickness d ( X Y..., UV grade fused silica optical material mounting shim has two slots for compatibility with our SPW602 and Spanner. Simple machines with non-linear processes of ultra-short pulse laser technology ( UKP ) can be used individually in... Shows some combinations of in-and-out specification tolerances that do not always give results! 'S optical Design work center of vitality saved by utilizing apparatuses or mechanical gadgets shop is... An ultrafast optical system various optical sensors where wavelength separation is required e.g the of... Is not sufficient to control geometry errors in a pair solid with the five faces nine... 2°, 5° and 7.5° but we can make any angle ( θ ) /λ, beam..., discussion and forums displacement/distance and works using the principal of persistence of.! Often used to calculate optical density, also known as linearly variable (! Unit as per your requirement and click on calculate 7 shows some combinations of in-and-out specification tolerances do. ( EO ) has been a leading producer of optics, imaging, and.... A specific angle eksma optics uses cookies to optical wedge calculator you the best shopping experience and Spanner!, UV grade fused silica optical material by a fixed distance shaped tool used to find the mechanical advantage the! Wedge family offers two size choices to fine tune dispersion in an optical filter constructed!, nine edges, and may be circular, oblong or square word for arrow distribution a. Parts simultaneously ( e.g calculate the angle of refraction is formulated and what equation will you! Let you calculate the angle of incidence Calculator: Enter value in unit as per your requirement and click calculate... Injected by OTDR into the fiber through a laser diode and pulse generator product pulse. Ross optical is a polyhedron solid with the five faces, nine edges, six! Ultrathin wedges allow fine, continuous tuning of GDD along a path ( straight or ). For edge thickness, UV grade fused silica / BK7 WED series in FWHM ) is formulated and what will. Case Studies — Learn how we 've helped customers like you achieve the performance. C: height h: volume V the most frequently used calculations and from. A concave lens to the angle of incidence, \ ( l \ ) is pulse length ( ). The wedge is uncoated, the optical thickness d ( X, Y ) n not... For customers and users, who are mainly concerned with non-linear processes of pulse.
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# Gaussian Surface Area of Positive Semidefinite Cone
Let $\mathbb{R}^n$ be the Euclidean space and $A \subseteq \mathbb{R}^n$ be a sufficiently regular set, e.g., one that has smooth boundary or is convex. We define the $\epsilon$-neighbor of $A$ in the $\ell_2$ sense as \begin{align} A^{\epsilon} = \{ y \in \mathbb{R}^n \colon \text{there exists}~ x \in A ~\text{such that}~ \| x - y \|_{2} \leq \epsilon \}. \end{align} We denote $\gamma$ as the standard $n$-dimensional Gaussian measure $N(0, I_n)$, then the Gaussian surface measure is defined as \begin{align} \tau (A) = \lim_{\epsilon \rightarrow 0} \frac{ \gamma( A^{\epsilon} \setminus A) }{ \epsilon}. \end{align} By results of Keith Ball (the reverse isoperimetric problem for Gaussian measure https://link.springer.com/article/10.1007/BF02573986), there is a universal upper bound of the Gaussian surface area of any convex set.
My questions are: 1) when is Keith Ball's upper bound tight? 2) what is a tight upper bound of the Gaussian surface area of the cone of positive semidefinite matrices?
• What do you mean by the "positive semidefinite cone"? The cone of positive semidefinite matrices? An orthant? For any cone $A$ with apex at the origin, the Gaussian perimeter is proportional to the $n-2$-dimensional measure of $A \cap \{x : |x|=1\}$, the constant of proportionality being $(2\pi)^{-n/2} \int_0^\infty r^{n-2} \exp(-r^2/2) dr = 2^{-3/2} \pi^{-n/2}\Gamma((n-1)/2)$. – Mateusz Kwaśnicki Aug 24 '17 at 8:38
• @MateuszKwaśnicki Sorry for the confusion. I meant the cone of positive semidefinite matrices. I will revise my statement in the question. Could you detail your claim in an answer? Thanks a lot! – Minkov Aug 24 '17 at 11:28
• If $A$ is a "cone" (a positively homogeneous set) with Lipschitz boundary, then $\partial A$ is also a "cone". Denote $F=\partial A\cap \partial B(0,1)$. The $n-1$-dimensional Hausdorff measure $\sigma_{n-1}$ on $\partial A$ can be written as $\sigma_{n-1}(dx)=r^{n-2}\sigma_{n-2}(dz)dr$, where $x=rz$, $r>0$, $z\in F$ and $\sigma_{n-2}$ is the $n-2$-dimensional Hausdorff measure on $F$. Now apply this to the equivalent definition of the Gaussian perimeter: $\tau(A)=\int_{\partial A}\gamma(x)\sigma_{n-1}(dx)$, where $\gamma(x)$ is the density of the Gaussian measure. – Mateusz Kwaśnicki Aug 24 '17 at 11:49
• @MateuszKwaśnicki For the cone of positive semidefinite matrices, is it possible to get a closed form or order estimation for the $n-2$-dimensional measure of $A\cap \{x: |x| = 1\}$? – Minkov Aug 24 '17 at 23:04
• I do not know the answer, although I guess it follows from Weyl's integration formula, see mathoverflow.net/a/95256/108637. – Mateusz Kwaśnicki Aug 25 '17 at 8:40
Regarding the tightness of Ball's bound, which is $O(n^{1/4})$ for a convex set in dimension $n$, this was proven to be sharp (up to a constant factor) by Nazarov, in
Nazarov, Fedor, On the maximal perimeter of a convex set in $\mathbb{R}^n$ with respect to a Gaussian measure, Milman, V. D. (ed.) et al., Geometric aspects of functional analysis. Proceedings of the Israel seminar (GAFA) 2001--2002. Berlin: Springer (ISBN 3-540-00485-8/pbk). Lect. Notes Math. 1807, 169-187 (2003). ZBL1036.52014.
The Gaussian surface area of the positive semidefinite cone $\mathbb{S}_d^{+} = \{ M \in \mathbb{R}^{d\times d} \colon M \succeq 0 \}$ is bounded by a constant in $[0,1]$ uniformly, according to ''Learning Geometric Concepts via Gaussian Surface Area'' by Adam R. Klivans, Ryan O'Donnell and Rocco A. Servedio.
More specifically, as shown in Definition 2 of this paper, the Gaussian surface area of any convex set $A\in \mathbb{R} ^{d\times d}$ can be written as \begin{align} \Gamma(A) = \int_{\partial A} \varphi(x) d \sigma (x),~~~~[1] \end{align} where $\varphi$ is the Gaussian measure on $\mathbb{R}^{d\times d}$ and $\sigma$ is the Euclidean surface area. By Nazarov's inequality, for any convex set $A$ containing the origin, it holds that \begin{align} \int_{\partial A} \frac{\phi (x)}{1 + h(x) } d \sigma (x) \leq 1 - \gamma (A) \leq 1, ~~~~[2] \end{align} where $\gamma$ is the Gaussian measure and function $h$ is defined as the distance from the origin to the tangent hyperplane of $A$ which contains $h\in \partial A$. In more details, for any $h \in \partial A$, let $n_y$ be the unit normal vector to $\partial A$ at $y$. Then we define $h$ by $h(y)= \| y\|_2 \cdot \cos ( \langle y, n_y \rangle )$.
Now we go back to the PSD cone. The boundary of $\mathbb{S}_d^{+}$ is defined as \begin{align} \partial \mathbb{S}_d^{+} = \{ M\colon M \succeq 0, \textrm{rank}(M) < d\}. \end{align} For any $M \in \partial \mathbb{S}_d^{+}$, the tangent plane at $M$ is given by \begin{align} \mathcal{T}(M) = \left \{ X M + MX^{\top} \colon X \in \mathbb{R}^{d\times d} \right \}, \end{align} which implies that $M\in \mathcal{T} (M)$. Thus we have $h(M) = 0$. Combining $[1]$ and $[2]$, we conclude that the Gaussian surface area of the PSD cone is no larger than $1 - \gamma (S_d^{+})$.
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# normalise: Transform the noise to be closer to the Gaussian distribution In IDetect: Isolate-Detect Methodology for Multiple Change-Point Detection
## Description
This function pre-processes the given data in order to obtain a noise structure that is closer to satisfying the Gaussianity assumption. See details for more information and for the relevant literature reference.
## Usage
1 normalise(x, sc = 3)
## Arguments
x A numeric vector containing the data. sc A positive integer number with default value equal to 3. It is used to define the way we pre-average the given data sequence.
## Details
For a given natural number sc and data x of length T, let us denote by Q = \lceil T/sc \rceil. Then, normalise calculates
\tilde{x}_q = 1/sc∑_{t=(q-1) * sc + 1}^{q * sc}x_t,
for q=1, 2, ..., Q-1, while
\tilde{x}_Q = (T - (Q-1) * sc)^{-1}∑_{t = (Q-1) * sc + 1}^{T}x_t.
More details can be found in the preprint “Detecting multiple generalized change-points by isolating single ones”, Anastasiou and Fryzlewicz (2018).
## Value
The “normalised” vector \tilde{x} of length Q, as explained in Details.
## Author(s)
Andreas Anastasiou, a.anastasiou@lse.ac.uk
## See Also
ht_ID_pcm and ht_ID_cplm, which are functions that employ normalise.
## Examples
1 2 t5 <- rt(n = 10000, df = 5) n5 <- normalise(t5, sc = 3)
IDetect documentation built on May 2, 2019, 11:04 a.m.
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# Thread: how to solve this equation ?
1. ## how to solve this equation ?
Hi Guys,
I am trying to investigate a function, during the process I need to solve this equation:
sin(x)+cos(2x)=0
in the region 0<=x<=2*Pai
How do I solve it ?
thanks !
2. Use the identity $\displaystyle \cos{2x} \equiv 1 - 2\sin^2{x}$ to turn your equation into a quadratic, which you can then solve.
3. Start by writing it as $\cos{2x} = -\sin{x}$ and recall that $\cos\left(\varphi+\frac{\pi}{2}\right) = -\sin{\varphi}$.
So we have $\cos{2x} = \cos\left(x+\frac{\pi}{2}\right)$. You should be able to solve it with ease now.
Hint: if $\cos{\theta} = \cos{\alpha}$ then $\theta = 2n\pi\pm \alpha$ (where $n$ is an integer, or zero).
4. I used the quadratic idea, I did t=sin(x)
and got t1=-0.5 and t2=1
thus, I got:
sin(x)=-0.5 and sin(x)=1 my solution is x=pai/2 and x=11/6*pai
but there should be another one at x=3.63, I have no idea why....
5. Originally Posted by WeeG
I used the quadratic idea, I did t=sin(x)
and got t1=-0.5 and t2=1
thus, I got:
sin(x)=-0.5 and sin(x)=1 my solution is x=pai/2 and x=11/6*pai
but there should be another one at x=3.63, I have no idea why....
$\sin{x} = -\dfrac{1}{2}$ at $x = \dfrac{7\pi}{6}$ and $x = \dfrac{11\pi}{6}$
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Citation
Material Information
Title:
The Liberty Boys' guardian angel, or, The beautiful maid of the mountain
Series Title:
Liberty Boys of "76"
Creator:
Moore, Harry
Place of Publication:
New York
Publisher:
Frank Tousey
Publication Date:
Language:
English
Physical Description:
1 online resource (29 pages) 28 cm.: ;
Subjects
Subjects / Keywords:
Dime novels. ( lcsh )
History -- United States -- Revolution, 1775-1783 ( lcsh )
Genre:
serial ( sobekcm )
Record Information
Source Institution:
University of South Florida
Holding Location:
University of South Florida
Rights Management:
All applicable rights reserved by the source institution and holding location.
Resource Identifier:
025135090 ( ALEPH )
69131023 ( OCLC )
L20-00096 ( USFLDC DOI )
l20.96 ( USFLDC Handle )
USFLDC Membership
Aggregations:
Dime Novel Collection
The Liberty Boys of "76"
Postcard Information
Format:
Serial
Full Text
PAGE 1
THE LIBERTY ,,. Iuued Weelily-By Subcnp(ion. $2.1!0 plr year. Enlered a.. SeQond Clau Matter at t110 New York Post Ufiice, /<'ebr1UJf'g 4, 1901, b!J F1-a11/.; 7'omeg. No. 90. NEW YORK. SEPTEMBER 19, 1902. Price 5 Cents. Yes, sir, I saw them," the beautiful maid of the mountain said to the British oftl.cer. "They went in that direction, .. and she pointed up the road. Dick and Bob, from their hiding-place, heard and saw all. They realized that the girl savea them PAGE 2 hese Books Tell YOU A COMPLETE SET IS A REGULAR E NClCLOPEDIA I Each book consists of sixty-four pages, printed on good paper, in type and neatly bo undin an illustrated cover. Most of the books are also profusely illustrated, and all of the subjects ui;ion a r e explained i n such a s i m pl e manll"'r that child can thoroughly understand Chem. Look over the list classified and see if you want t(} k now anything about the mentil)neil. .' THESE BOOKS ARE FOR SALE BY ALL NEWSDEALERS 0R Wl&L-BB &lilNT BY l\1A1L TO-ANY. ADDRES::J'ROl\f THIS.OFFICE ON RECEIPT OF PRICE, TE::\! CENTS EACII, OR ANY THREE BOOKS FOR 'l'WENTY-FIYJ:<: "'ENTS. POSTAGE STAMPS TAKEN THE SAME AS MONEY. Addr?.ss FRANK TOUSEY, Publisher, 24 Union Square, N.Y SPORTING N o 21. HOW TO HUN1.' AND J<'lSH .-The most complete .. anting and fishing guide ever published. It contains full in Dtructions !!bout gpns, hunting dogs, traps, trapping and fishing, with descriptions o.f game and fish. No. 26. HOW TO ROW, SAIL AND BUILD A BOAT.-Fully Illustrated. E1ery boy should know how to row and sail a boat. l'ull instm,.tions are given in this little book, together with in-1tructions on swimming and riding, companion sports to boating, No. 17. now TO BREAK, RIDE AND DRIVE A HORSE.!.&. complete treatise on the horse. Describing the most useful horses iGOr busines the best horses for the road; also valuable recipes for tllseases pec c 1liar to the horse. No. 48. HOW 1'0 BUILD AND SAIL CANOES.-A bandy took for containing full directions for constructing canoes ud the most popular manner of sailing them. Fully illustrnte d. 1 C. Stansfield Hicks. HYPNOTISM. o. 81. IIPW TO HYPNOTIZE.-Conta;.n ing valuable and in !!tructive information regarding the science of hypnotism. Also roJ[J>laining the most approved methous which are employed by the t1.ding of the world. By Leo Ilugo Koch, A.C.S. MAGIC No. 2. IlOW TO DO TRICKS.-Tbe great book of magic an card rricks, rontaining full instruction on all the leading card tricks of the day, also the most o:u1gical illusions as. performed lly our leading n)agicians: eyery boy should obtain a copy of this b<\uk, as it will both amuse and instruct. lIOW DO SECO:'\D SIGT;I'J.:.-Heller's second explained by his former 'Fre\] Hunt, Jr. Explaining l}ow the sl!cret dialogues wete carried on between the magician. and ,the boy on the stage; also giving all the codes anQ signals. The only authentic explanation of second sight. > No. 43. IIOW TO BECOME A i\IAGICIAN.-Containing the grandest assortment of magical illusions ever placed before thP public. Also tricks with cards. incantations, etc. No. GS. IIOW TO DO TH.ICKS.-Containing ovN one hundred ilighl.v amusing and instrnctive tricks with chemicals By A \.nclerson. Handsomely illustrateJ. i'o. li!J. 110\V TO DO SLEIGHT OF IIAND.-Containing ovt>r fifty of the latest and best tricks used by magicians. Also contain ing the secret o[ second si.gbt. Fully illustrated. By A. Anderson :l\o. 70. IIO\V 'l'O l\I.AKE i\IAGIO TOYS.-Containing full direetions fo, making i\lagic Toys devices of many kinds. B:v A. Andersou. Fully illustr thl! secret of palmistry. Also the secr!'t ol' telling future inslructions how ro proceed in ordl>r to become a locomotive en vents by aid of moles, marks, scars, etc. Illustrated. By A. gineer; also directions for bui lding a model locomotive; togeth,.r uderson. with a full description of ever.vthing an engineer should know. i ATHLETIC. No. 57 \!OW TO MAKE MUSI.CAL INSTRUMENTS.-Full No. 6: HOW T.O BECOME AN ATHLETE.-Giving full. inc directions how to make a Banjo, V\olin, Zither, 2Eolian Haq), Xylo atruction for the use of dumb bells, Indian clubs, parallel bars, vhone and other instruments; together with a brief dP 'ilorizontal 'bars and various other of developing a geod, scription of neal'ly every musical instrnment used in ancient or ealthy muscle; containing over sixty illusttafions. Every boy can modern timl's. Profusely illustrated. By &lgernon S. 1'"'itzge1\lr1 !>ecomc strong anJ healthy by following instrnctions for twenty years of the Hoval Bengal Marines. .,--. 11 this little book: No. 59. HOW TO l\AKE A 'o. 10. HOW TO BOX ...:....The art of self-defense made easy. a desrripliou of the lantern, togethei: with its history and' mvenJ :il'riting love-letter" fenc i ng and the use of the also instruction i n archery. and when" ro use thl!m; also giving specimen letters for both younJC. Describ ed with twenty-{)ne practical illustrations, giving the best and old. positions in fencing. A .complete book. No. 12. now TO WRITE LETTERS TO LADIES.:-Givinll TRICK. S W ITH C ARDS. CO'l)l.ll'lete insttwtions for writil)g letters to ladies on all "::.'ubjects also lett<'rs .of. introduction, notes .and requests. 'o. 51. now TO DO TUICKS WITif CARDS,'--Containing Nio. 24. tIOW TO WRITE LET'rERS 'rO GENTLEi\IEN.txplanations of the general princiIJles of sleight-of-hand applicable Gontaicirfg fnll di1ections for writing, to gen I lemen on all subjects ta card tricks; of card. ordin1:J.l'Y, 'cp.rds, and not also gh,1\ili: sample IHhns .for rnstrn f',tion. 11leight-of-hand; of ttcks mvolvmg slP1glit-of-hand, or the use of No: 7i3hJ10'W .'.l'et WRITE: .--A wonderful littlP 1pecially prepared cards. By Professor Ilaffner. With illustra Uook. ,.rem how. to youi; fathet tions. ,. mother, sistet brother, employ!H' ; .and, m {<)rt, everybody and any No. 72. IIOW TO DO SIXTY TRICKS WITH CARDS:'--Em-body yotl' wjs\1 to '1ite to .. y{lnng man and every b racing all of the latest and most deceptive card trick s with JIlady in the land book, lustrations. By A. Ai;id'erson. No. -74. HOW' TO 'WRI\l:'E' .LE1'TERS CORR'lllOTLY;.-Con>' No. 77. HOW T PAGE 3 .HE LIBERTY BOYS OF '76. Weekly Magazine Containing Stories o f the American Revolution Issued Weekly-By Subscriptfrm$2.50 per year. Entered as Second Class i1fatte.at the Net0 York. N. Y., Post Office, February 4, 1901. EnterPd acconling to A.ct of Collgrcss, ill the year 1902, in the office of the Librarian of Congress, D. 0., by Frank Tousey, 24 Union Square, New York. NEW YORK, SEPTEMBER 19, 1902. Price 5 Cents CHAPTER I. !)! TIIE H.A.:NDS OF THE TORIES "Well, young man, if you've anything to say, say it now, your time has come thrilling scene. alfway up the side of a mountain in western South olina, close beside the trail which wound and twisted its course like the trail left by some huge serpent, stood arty of perhaps a dozen men hey were standing under the wide-spreading branches a huge oak tree. hese men were roughly dressed, rough-looking fellows, ed with rifles and pistols. n their midst stood a young man of perhaps twenty e was a handsome, bronzed-faced fellow, with keen e eyes, firm chin and jaws, and a fearless air. his young man was Dick Slater, a patriot-for the e of our story was midsummer of the year 1780-and had earned for hin:iself the title of "The Champion Spy the Revolution." he young man was in the South, with his company o.f 'berty Boys," trying to assist the patriots to fight against redcoats and Tories, and having become separated from comrades, had been waylaid by this band of Tories, who made him a prisoner, and after trying to get him to who he was and what his business was in that part the country, had placed a rope around his neck, thrown other end over the limb above 11.is head, and were stand there, waiting for the word from their leader to pull youth up in the air It was then that the leader had uttered the words with ch our story begins : "Well, .young man, if you have anything to say, say it w, for your time has come." ;,. "I have nothing to say," w-as the calm reply. "You are a fool." "Thank you. You are a scoundrel!" A hoarse growl escaped the man's lips. "Say, you are altogether too saucy!" "You may think so, but I don't." "Well, I do." "It isn't being saucy to simply tell the truth, is it?". "But that ain't the truth." "What I said about you isn't the truth?" "No." "Yes, it is." "It isn't." "I can prove that it is." "How?" "Why, your actions have proved it." "My actions?" "Yes." "I don't see how "It is simple enough The way you have treated meproves that you are a scoundrel." "Bah!" "No one but a scoundrel would stop a man on the highway and make a prisoner of him and try to force him to tell all he knows, on peril of death if he refuses." "And nobody but a fool would refuse to tell." "I did tell." "You didn't do anything of the kind "I did "You told me a cock-and-bull story about being on your-1\'ay to visit some relatives over across the mountains, but I know better than to believe any such story." "It is the truth. "Bosh." "I don't see why you should think this not true "Well, it stands tQ reason it isn't true." "I don't see it that \my. It is not an impossibility that I might have relatives over the mountains, is it?" "Oh, no." "Nor that I might be on my way to visit them?" No, it ain't an impossibility, but it is improbable." "Why should it be?"
PAGE 4
2 THE LIBERTY BOYS' GUARDIAN ANGEL. "Well, that's simple enough. Not many people are putleader and nodded, as much as to say, "Yes, let them h:. ting in their time visiting relatives these days it out between themselves." "No, I suppose not But that doesn't prove that I am not on my way to visit relatives." "Well, it is proof enough to suit me. To tell the truth, you don't look like that kind of a fellow." "Oh, I don't care, Bill But I don't see what the use of pounding him up, when we are going to haJ him right away, anyhow." "Wal, ye see, cap'n, et'll teach 'im er good lesson, a "You tl1ink not?" theer's er sayin', ye know, ez how et's never too late "That is what I think. You look like a chap who has l'arn." been fighting in the army-like a soldier, in fact." "It may never be too late to learn, but in his case it 1 "Oh, you think I look like a soldier?" "Yes. It would not surprise me if you were a soldier, and a spy, at that." "You are mistaken, sir." "You are not a soldier?" "No." "Nor a spy, hey?" "No. "Bah! He's lyin' ter ye, cap!" growled one of the men. "Le's siring 'im up, an' be done wi.th et." The speaker was a big, burly fellow, with a ferocious face and fierce air The youth turned his eyes on the speaker, and said, i.oo late for his learning to do. him any good "Et'll do me good ter knock ther young cuss's head t l fur he called me er liar, an' thet's sumthin' I don' sta. frum nobuddy." "Well, free the prisoner, one of you men, and the all stand in a circle, with pistols drawn, ready to sho him if he tries to escape This was done, and Dick found himself standing in t circle made by the ruflia:QS, and at one side, facing h.iJ \ stood the giant ruffian, Big Bill. He looked Dick over with a supercilious air, and th said, sneeringly: "Say, young feller, ef ye"ll take back whut ye said l with snch a scornful air and intonation that the other was bout me an' git down on yer knees an' ax my parcling, I made angry : let ye off, and won't thump ye. Whut d'ye say?" "You're a liar yourself, you big ruffian, and if my han'ds were loose I would choke some sense into that thick head of yours." "Whut's thet? Ye'd choke me-me, Big Bill Benton? Say, young feller, ye make me larf, ye do, so," and the big man forced a laugh, but it was evident that he felt any thing but mirthful. His face was red with anger, and his little, reddish-hued eyes were gleaming fiercely. "Yes, I would choke you, and teach you better than to talk as you have just been talking." "Bah, young feller, ye wouldn' be er mouthful fur me," the ri.iflian said sneeringly. "Free my hands, and I will prove that I am a better man than you, and do it quickly, too." The youth's wish was that he should delay the hang ing, and thus make it more likely that something might occur to save him, but he did not expect that the Tories would do wlrnt he asked. Big Bill was eager to get a at the saucy youth, however, and he said to the leader eagerly: ''Oh, you are afraid, are you?" remarked Dick coolly l "Me afraid?" angrily. "Wal, I guess not I wouldn' l erfraid uv er duzzen like ye." "Then why are you trying to get out of this affair?" ''I hain't." ,. "I don't know what else you would call it." "I wuz jes' givin' ye er chance ter git out uv et, bee I took pity on ye, thet's all." "Oh, thank you," sarcastically. am in no need of pity." Il "But I assure you "Oh, ye hain't, hey?" n "No, it is you who should be pitied." n "Me ?-haw, haw, haw!" and the ruffian laughed lour ly, nis comrades joining him, for they thought such l -ll idea ridiculous 'I'he youth kept a sobJ;!r face, however, and said: "Yes, you. "Bosh! Why, I k!n eat J:e up, young feller!" "You will find me pretty tough chewing." "I guess not.'' re v ( r j c 'r "Oh, say, cap, jes' let me hev er chance at 'im, won't "I guess yes." '1 ye? I'd like ter show 'im er thing er two thet he don 'i. "Go for him, Bill, if you are going to," said the lead1 know er have enny idee erbout." of the party. "I want to have this thing se.ttled, and fi'I idea of an encoilll:ter between the two seemed to / i.sh tlie affair by hanging the young fellow as quickly 'I stnke the other man favorably, for they looked at. their possible
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] 12 THE LIBERTY BOYS' GUARDIAN ANGEL. There was an anxious look on the youth's face, and an anxious tone to the voice, and the girl noticed it. "No young man rode the horse, sir," she replied. "It was ridden by the leader of the Demon Dozen, a man by the name of Raymond Marks "What is that you say?" the youth exclaimed. "Then my friEjnd, who was the of that horse, has likely been murdered by those scoundrels!" "If that horse belonged to a friend of yours, it is likely that he has been put out of the way," said the girl. An exclamation escaped the youth s lips "If those scoundrels have killed my friend I will never rest until I have put every one of them to death," he de clared, with fierce earnestness, and the girl knew he meant iL At this moment there came a cry from the horsemen out at the gate: "Here he is, Bob. Here comes Dick!" The youth looked up the road, and saw a man approach ing on foot, and a cry of delight escaped his lips. "Yes, it's him, sure enough!" he exclaimed. "That is Dick." CHAPTER V. A FIENDISH SCHEME. As the read e r has already guessed, the newcomers were the famous "Liberty Boys," and the youth who had been talking to the girl was Bob Estabrook, Dick Slater's right. hand man. He was always left in charge of the "Liberty Boys" in Dick's absences. Bob hastened out to the gate, reaching there just as Dick did. "So that's what they are known by; eh?" "Yes." :o1 "Well, their actions justify them in the for they are demons 0 "What did they do to you, Dick?" h The youth explained, the others listening with inte r b "ra: like a chance at that gang," said Bob, when Dir had finished. f "And I!" "Here, too "The same here!" Te e )h Such were a few of the exclamations, and it was evi
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NEW YORK, .:SEPTEUBER 24. 1902. Price 5 Cents. .' f f s : : Up the aisle the aolemn cortege glided. the specterchoir chanting a 'dirge for the dead. J?OQl' .Hans was as white as sheet His knees trembled like an as_pen.
PAGE 33
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My question is. If the cosmic background radiation is red shifted how would you calculate the wavelength at the time it was emitted ?
115 views
### Kinematics of Euler angles relative to a rotating frame
I have a rotating body $B$ and a rotating frame $F$ whose orientations are described by the quaternions $q_B$ and $q_F$ respectively. I also have the angular velocity vectors $\omega_B$ and ...
104 views
### Forces Create Angular Acceleration And “Straight” Acceleration - But How Much Of Each?
Let me set up the following problem for a rectangle floating in space: We know its dimensions. We know its mass. There's a force pushing it for a known amount of time - we know the angle & ...
Suppose we have a theory is covariant under the Spin group Spin(2n-1; 1). We consider the real vector space $V = R^{2n-1,1}$, which naturally comes with a Lorentzian inner product. On this vector ...
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# Higgs Discovery Announcement July 4
I learned via Physics World that CERN will hold a press conference on Wednesday July 4 to give an “Update on the search for the Higgs Boson”. More information has just appeared (including a press release here), showing that there will be a 2 hour seminar on the results starting at 9am Geneva time, followed by a press conference at 11am.
Reports from the experiments indicate that at least one of them, if not both, will reach the 5 sigma level of significance for the Higgs signal, when they combine 2011 and 2012 data and the most sensitive channels. So, this will definitely be the long-awaited Higgs discovery announcement, and party-time for HEP physicists.
One could note that the last major announcement of the discovery of a new elementary particle at CERN was also made on a Wednesday, July 4, back in 1984. That one didn’t work out so well, but things are very different now, with results from two independent experiments and a high standard of evidence.
This entry was posted in Experimental HEP News. Bookmark the permalink.
### 31 Responses to Higgs Discovery Announcement July 4
1. Stan says:
What particle was announced on July 4, 1984?
2. Peter Woit says:
The top quark, with a mass of about 40 GeV. Only problem is that the top quark really has a mass of about 173 GeV….
3. Long-time follower of HEP/Not Even Wrong here: A bit ahead of the news, unless you know more, Peter! (Because the press “ad” really says “update on Higgs search”.)
4. Peter Woit says:
Thanks Alfons,
There are some blogs out there with a policy of only discussing news approved by the relevant authorities. Not this one….
5. Christian Takacs says:
Please don’t clobber me with criticism if I sound uninformed about this… but… though I do see all these articles, charts and graphs indicating something (large particle?) may have been found, I do not see anything (outside of a desire to find the Higgs) to indicate this IS the Higgs particle, nor do I see any explanation of how this will be demonstrated if a particle IS discovered. I would also ask, Isn’t the Higgs boson/particle’s existence based on the Standard Model’s assumption that mass is granted, imparted, or virtually assigned by said particle? I just seem to be seeing lots of “There are indications of something there, if confirmed, it’s the Higgs particle” statements. I see no mention of “What if it’s just a newly discovered particle which is not the Higgs”. I would just have thought they would first want to confirm that something was found, THEN go about some method to find out if the particle passes some falsifiable testing procedure for confirming it works as advertised.
6. Peter Woit says:
Christian,
The Standard Model makes extremely detailed predictions about exactly what Higgs decays should look like, and the LHC experiments are carefully tuned to look for exactly these predicted signals. What they are seeing is exactly what they were looking for (with the interesting caveat that the production rate may be higher than expected, but that calculation is hard).
So, either this is the Higgs, or if it’s something different, you have to explain why it is doing precisely what the Higgs was supposed to do. Anyway, the big effort from now on will be trying to more precisely measure the properties of this signal to compare to the SM prediction.
7. Anonyrat says:
Continuing to subvert science, I see
8. Eastender says:
So who gets the Nobel prize ………
9. SilverSB says:
What will happen 4th of Jully is just the HEP will become less interesting. LHC is built to catch a Higgs, everything else would be a bonus (if the Nature is kind enough to throw at us hints for supersymmetry or dark matter at home-made energies – something I really doubt). So, yeah, the higgs is there. We have just saw the final battle in the first 5 minutes of the movie. How anticlimatic…
10. emile says:
SilverSB: your are not 5 min. into the movie. The movie has been going on for decades… If a Higgs is confirmed, then I don’t blame you from thinking that this would be anticlimactic. But if I’ve learned anything, it’s that Nature doesn’t care what we think.
11. crandles says:
So is pay $7 now to get$10 back if/when a paper is published in 2012 claiming a 5 sigma discovery a good bet?
Or is there too much chance that paper will wait for more information from different decay channels, and others hints of consistency with being a Higgs Boson and then be subjected to much scrutiny in peer review so that paper may not be published until 2013?
Or is there too much risk that judge will read a paper saying there is 5 sigma discovery of particle that is consistent with Higgs Boson as not being sufficient to say particle is the Higgs Boson? (If such a paper isn’t sufficient, will anything ever be sufficient? and is that sufficient to ensure judge will decide that such a paper is sufficient?)
(Rules say “Confirmation of the Higgs Boson particle having been observed must be published in a major scientific journal for this contract to be expired.
Clarification (Jan 5th 2009): for the Higgs Boson particle to be “observed” there must be a “five sigma discovery” of the particle.”)
BTW, there isn’t much liquidity on this bet at intrade.com: 37 * US $7 is only pay US$ 259 to get US $370 less US$5 per month in fees less and bank/other fees for money transfers. Also by the time someone new to intrade gets that money into their intrade account, the opportunity might be gone. So probably not worth effort and risk for someone new to intrade.
(Now why do I suspect that Christian Takacs is an intrade trader?)
12. SilverSB says:
emile,
The LHC movie was decades in making, but only two years in playing. The Higgs is coming too soon, but the things are what they are.
Sure, Nature doesn’t care what we think and what we want – it’s not something we learned. It’s something most people should understand.
13. christian M says:
Hi there,
Nice and happy news.
Just one comment/question.
We find allmost everywhere the misleading information that the Higgs boson gives mass. That is untrue in my opinion.
The Higgs boson is the trace left by something which gives mass. But not the boson itself. The scalar field responsible for giving mass has four degrees of freedom, three of which give mass to the W and Z. The fourth degree does not do anything. That’s the Higgs boson. It is the remnant of this process. It’s just an excitation mode of the scalar field, but it is the latter which gives mass.
That does not mean that it is not important to find the Higgs. It’s like finding a trace left in the sand by a dinosaur : it proves that the dinosaurs existed.
14. Henry Bolden says:
15. DB says:
The Higgs will be the first fundamental boson discovered whose spin is not equal to 1. And the mass of 125GeV makes the building of a muon collider to probe the properties of the Higgs in fine detail a no-brainer. It also raises serious questions over the need for the CLIC upgrades to the LHC.
16. Henry Bolden says:
I’m hearing from someone (who does not wish to be named) who heard from someone else at the Perimeter Institute (whose name was not revealed to me) that the announcement on July 4 involves a Higgs which is NOT a Standard Model Higgs. Anyone else hearing this?
17. Speculative says:
Henry,
I’d be wary of anyone who is saying right now that we know the Higgs they’re seeing is beyond the Standard Model. It will take a lot of careful measurements before we know for sure. If there is something about this particle that distinguishes it significantly from the SM Higgs that would certainly be very interesting and it might not even be the Higgs. My attitude is that we’ll just have to wait and see until July 4th when more official data is released.
18. Anonyrat says:
It is time for corporate sponsorship, e.g., just like the MetLife Stadium, or the Citi Bank Arena, we could have corporations have naming rights on particles, such as the Disney Strange, the Dow-Jones Up, the Balenciaga Top, the Huggies Higgs. The proceeds of the sponsorships would go to support impecunious Superstringers, who promise to bring a whole lot of new particles and hence sponsorship opportunities, to the table. At least it would give some motivation for research. Some names might be already taken, such as Selectron Technologies’ Selectron. Political parties might jump into the fray, such as the G.O.P. Dilaton. Social groups might enter the bidding too, such as the LGBT Spartner. The possibilities are limitless, we could further distinguish particles in different superstring vacua. With 10^500 possibilities, there will be enough for sponsorship by all the conceivable corporations the visible universe will ever contain. Hell, we can give each corporation sponsorship of an entire universe, why just a measly elementary particle?
19. Tony Smith says:
If the thing at 125 GeV “is NOT a Standard Model Higgs”
then
can they distinguish it from a non-Higgs particle such as for example
a technipion like that proposed by Eichten, Lane, Martin, and Pilon
in arXiv 1206.0186 (in the context of the CDF Wjj bump) ?
If it is not so distinguishable (and therefore not clearly any kind of Higgs),
then what should they call it ?
Tony
20. Peter Woit says:
Henry,
The signal seen in 2011 was already larger than the SM prediction (with large errors). The rumor that this year’s gamma-gamma signal is of similar size indicates that when they announce discovery next week, the size of the signal seen will not only be more than 5 sigma away from null, but also larger than the SM prediction.
There will be signals though in multiple channels: gamma-gamma, as well as ZZ to 4 leptons. The size of the signal is the product of the Higgs cross section x branching ratio. Whatever is observed, undoubtedly there will be dozens of theory papers promoting models supposedly explaining it. I’d love to hear from a Higgs phenomenologist about how good the SM Higgs cross-section calculation is, and what to look for in terms of deviations of the the signal sizes from the nominal SM predictions.
Maybe someone can convince Matt Strassler to stop complaining about what the recent NY Times article said about what nothing being seen would mean (which is irrelevant), and write instead about this.
Peter,
Not unusually you have hit the nail on the head–the things I have been wondering–and 3 times I might add:
(1) there are apparently hints of SM discrepancies in branching ratios–when does this become signficant?;
(2) uninitiated hep-ph folks like I would like to know the SM calculational “error bars” for such [fully of course realizing that the experimental notion of an error bar does not really apply]; and
(3) [at the risk of an understandably deletable ad hominem comment] I for one cannot fathom what particular axe MS has chosen to grind (this time).
22. Anonyrat says:
23. Anonyrat says:
See table 6 on page 27 of the above.
24. Seth Thatcher says:
A Higgs boson walks into a bar….mass exodus.
25. Martin says:
Yay, we’ve got a particle! Let’s just hope, it won’t turn into a difraction pattern when we stop talking about it
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First, let me point out that $H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$ if $X$ has rational singularities for all $i > 0$.
Indeed, if $X$ has rational singularities if and only if
1. $R^j \pi_* O_{\tilde X} = 0$ for $j > 0$ and
2. $\pi_* O_{\tilde X} = O_X$.
It immediately follows from the Leray spectral sequence that $$H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$$ for all $i > \geq 0$.
In fact, for any Cartier divisor $D$ on $X$, the same argument implies that $$H^i(\tilde{X}, O_{\tilde{X}}(\pi^* D) ) = H^i(X, O_X(D))$$ for any $i \geq 0$ since the projection formula can be applied in the cases of 1. and 2. above.
Now, without rational singularities, you can run into trouble. For example, suppose that $X$ is a normal Cohen-Macaulay variety with an isolated singularity $x \in X$ that is not rational. Consider the exact triangle in the derived category: $$O_X \to R \pi_* O_{\tilde X} \to C \xrightarrow{+1}$$ Because $X$ is a normal Cohen-Macaulay, and has an isolated non-rational singularity, we know $C = M[-n+1]$ is a nonzero module supported at $x \in X$ (shifted over by $n-1$). See Lemma 3.3 in Rational, Log Canonical, Du Bois Singularities: On the Conjectures of Kollár and Steenbrink by Sándor Kovács.
Then we have the following exact sequence by taking (hyper)cohomology $$0 \to H^{n-1}(X, O_X) \to {{H}}^{n-1}(\tilde{X}, O_{\tilde{X}}) \to {\mathbb{H}^{n-1}}(X, C) \to H^n(X, O_X) \to H^n(\tilde{X}, O_{\tilde{X}}) \to 0$$ where the two end points are zero since $C = M[-n+1]$ an Artinian module with a shift. On the other hand, $\mathbb{H}^{n-1}(X, C) = H^0(X, M) \neq 0$ for the same reason.
Now, if $\tilde{X}$ is for example Fano and we are in characteristic zero, then $$H^i(\tilde{X}, O_{\tilde{X}}) = H^i(\tilde{X}, O_{\tilde{X}}(K_X-K_X)) = 0$$ by Kodaira vanishing for $i > 0$. But then $H^n(X, O_X) \neq 0$ from the exact sequence.
Beyond the Fano case, you might luck out of course, but I don't see any reason why it would hold in general.
1
First, let me point out that $H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$ if $X$ has rational singularities for all $i > 0$.
Indeed, if $X$ has rational singularities if and only if
1. $R^j \pi_* O_{\tilde X} = 0$ for $j > 0$ and
2. $\pi_* O_{\tilde X} = O_X$.
It immediately follows from the Leray spectral sequence that $$H^i(\tilde{X}, O_{\tilde{X}}) \cong H^i(X, O_X)$$ for all $i > 0$.
Now, without rational singularities, you can run into trouble. For example, suppose that $X$ is a normal Cohen-Macaulay variety with an isolated singularity $x \in X$ that is not rational. Consider the exact triangle in the derived category: $$O_X \to R \pi_* O_{\tilde X} \to C \xrightarrow{+1}$$ Because $X$ is a normal Cohen-Macaulay, and has an isolated non-rational singularity, we know $C = M[-n+1]$ is a nonzero module supported at $x \in X$ (shifted over by $n-1$). See Lemma 3.3 in Rational, Log Canonical, Du Bois Singularities: On the Conjectures of Kollár and Steenbrink by Sándor Kovács.
Then we have the following exact sequence by taking (hyper)cohomology $$0 \to H^{n-1}(X, O_X) \to {{H}}^{n-1}(\tilde{X}, O_{\tilde{X}}) \to {\mathbb{H}^{n-1}}(X, C) \to H^n(X, O_X) \to H^n(\tilde{X}, O_{\tilde{X}}) \to 0$$ where the two end points are zero since $C = M[-n+1]$ an Artinian module with a shift. On the other hand, $\mathbb{H}^{n-1}(X, C) = H^0(X, M) \neq 0$ for the same reason.
Now, if $\tilde{X}$ is for example Fano and we are in characteristic zero, then $$H^i(\tilde{X}, O_{\tilde{X}}) = H^i(\tilde{X}, O_{\tilde{X}}(K_X-K_X)) = 0$$ by Kodaira vanishing for $i > 0$. But then $H^n(X, O_X) \neq 0$ from the exact sequence.
Beyond the Fano case, you might luck out of course, but I don't see any reason why it would hold in general.
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### Home > CALC > Chapter 9 > Lesson 9.2.2 > Problem9-65
9-65.
No calculator! Find the slope of the curve x cos y = xy + 1 at the point (1, 0). Homework Help ✎
$\frac{d}{dx}xcos(y)=\frac{d}{dx}\Big(xy+1)\Big)$
$cos(y)-xsin(y)y^\prime=y+xy^\prime$
Solve for y′.
Evaluate the equation from step 2 for x = 1 and y = 0.
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Definition 64.7.5. Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Let $x \in |X|$. Let $\mathcal{P}$ be a property of germs of schemes which is étale local. We say $X$ has property $\mathcal{P}$ at $x$ if any of the equivalent conditions of Lemma 64.7.4 hold.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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Physics Forums (http://www.physicsforums.com/index.php)
- General Physics (http://www.physicsforums.com/forumdisplay.php?f=111)
- - Difference of pressure of fluids in containers. (http://www.physicsforums.com/showthread.php?t=37673)
izytang Aug1-04 03:48 AM
Difference of pressure of fluids in containers.
Hello everyone,
I have a question that pertains to the physics of fluids and how the pressure of which changes in different given containers, which contains water. Basically, here is what I’m trying to do:
The shape of the container is that defined by a the function:
$$f(x) = \frac{5}{8} x$$
Which is rotated about the x-axis from the limits 1 to 8, with a circular hole in the smaller part of the cone. So, essentially it is a cut-off cone with a height of 7 and large radius of 5 (from top). The cone container is flipped so that the cone is “upside-down” (or that the larger radius is upward), and water will then flow out at an unknown rate. The following link gives a diagram I drew out to better illustrate what is happening:
http://www.angelfire.com/fang/ca2/waterTankEx.jpg
(Copy and paste in new browser window, since angelfire doesn’t allow external linking)
The essential problem is that I need to find a function in time of the water’s rate of the change of the container’s volume with pressure taken into account, so that I could simply find it’s instantaneous change of the Volume, height, or radius. My thought process has gone as follows:
First, I needed to write a formula for the volume of the cut-off cone. And since the cone is defined by a function, I used calculus to get:
$$\int_{1}^{x} pi \left \frac{5}{8} x \right ^2$$
or
$$\int_{1}^{x} pi \frac{25}{64} x^2 dx$$
With $x$ being the cone’s height from “0” (although the cone does not start until x=1).
My next step was to find out the container’s rate of flow, or pressure by calculating it initially at the point of which the tank is full, then writing a formula representing the rate of flow in terms of time, which is what I am ultimately looking for. I started by using the Pressure Force relation:
$$P = \frac{F}{A}$$
or
$$P = \frac{ma}{A}$$
Where F is force which equals mass*acceleration, and A is the area, in meters squared. I had then calculated the volume of the container when it is full (purely as a test), and got 209.03 meters cubed, which equals 209,030 Kg in mass. And because we are dealing with water, the volume density relation is 1cm cubed = 1 gram.
I was just about to calculate the pressure, but it suddenly hit me that the shape of the container might possibly affect the pressure of the fluid. In others words, there might be more pressure from fluids in a container that is a cylinder as opposed to a cone, even if the volume of the fluids are the same. If there is such a relation, it is very important to take that into account for calculating a rate of flow from the container, but unfortunately, I do not know where to start to find this relation, if any exists at all. Essentially, all I need to know is how pressures of fluids are affected by the shape of the container of which those fluids are in. But, if anyone would like to double check all my previous work and concepts, I’d gladly appreciate it. Of course, any help is appreciated.
-Josh
ArmoSkater87 Aug2-04 03:51 AM
Hi, i cant help much because I know NO calculus, i have yet to take it when school starts. But i can tell you that im pretty sure the shape doesnt make a difference in the pressure, its only the height that matters. Let me show you why...
$$P = F/A$$
$$P = ma/A$$
$$P = DVg/A$$ D is the density of the fluid, in your case its water. V is the volume of fluid you have, D times V gives you mass as you can see. You replace a with g.
$$P = D(Ah)g/A$$ The volume can be expresed as the area times height
$$P = Dgh$$
So, height is the dominant factor in pressure, not the volume of fluid...which would be determined by the shape of what you are using.
russ_watters Aug2-04 05:42 PM
Pressure of a liquid inside a container depends only on the density of the liquid (usually assumed to be constant) and the height of the fluid. Thus it is a linear relationship between height (depth) and pressure.
All times are GMT -5. The time now is 12:28 PM.
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# truncated quantile regression in R
I have used the "quantreg" package in R to find quantiles for my data. All my data, both predictors and responses are limited between 0 and 1, while a number of quantiles given by "rq" or "rqss" functions have values less than 0 or more than one. I was wondering if there is a solution for this problem, maybe a different function which gives truncated quantiles. Thanks
Quantile regression is an additive model. No purely additive linear model can prevent predicted values from going outside any given range. You can either hope to find a transformation of $Y$ that does not have a limited range that will fit the model, or consider the use of ordinal regression. For example the R rms package orm function handles continuous $Y$ and allows estimation of quantiles, the mean, and exceedance probabilities
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# Row of a Matrix
## Definition
A straight horizontal path in a matrix to arrange the elements in the same path is called a row of a matrix.
The space inside the matrix is divided into number of rows according to following two factors.
1. The total number of elements.
2. The arrangement of number of elements equally in each row.
In every row, the elements are arranged in horizontally straight path but separated by some space.
It can be expressed in general form matrix form as follows.
$\begin{bmatrix} \begin{array}{|ccccc|} \hline e_{11} & e_{12} & e_{13} & \cdots & e_{1n} \\\hline e_{21} & e_{22} & e_{23} & \cdots & e_{2n} \\\hline e_{31} & e_{32} & e_{33} & \cdots & e_{3n} \\\hline \vdots & \vdots & \vdots & \ddots & \vdots \\\hline e_{m1} & e_{m2} & e_{m3} & \cdots & e_{mn} \\\hline \end{array} \end{bmatrix}$
The total number elements is $mn$. So, the space inside the matrix is divided into $m$ rows to place $n$ elements in each row.
Observe each element closely. Each element displays two numbers in its subscript position. The first number is common to each element in every row because it represents the number of that row.
For example, the number in subscript position of each element in first row is $1$ and it is common to all the elements in that row. The number $1$ represents the element belongs to first row and this rule is same for all the rows.
The general form matrix is simply written in compact form.
$\begin{bmatrix} e_{\displaystyle ij} \end{bmatrix}$
The element $e_{\displaystyle ij}$ represents every element in the matrix and the first letter $i$ in subscript position of the element represents the number of the respective element’s row.
### Example
Observe the following example matrices to understand the concept of row in matrix much better.
$\begin{bmatrix} \begin{array}{|c|} \hline 6 \\\hline \end{array} \end{bmatrix}$
There is only one element in this matrix. So, it is placed in row and therefore the element $e_{11} = 6$.
$\begin{bmatrix} \begin{array}{|cccc|} \hline 0 & 2 & 4 & 7 \\\hline \end{array} \end{bmatrix}$
The matrix has only one row but contains four elements. So, $e_{11} = 0$, $e_{12} = 2$, $e_{13} = 4$ and $e_{14} = 7$.
$\begin{bmatrix} \begin{array}{|cc|} \hline -1 & 9 \\\hline 5 & 8 \\\hline 2 & -5 \\\hline \end{array} \end{bmatrix}$
This matrix has three rows and two elements are placed in each row. Therefore, $e_{11} = -1$, $e_{12} = 9$, $e_{21} = 5$, $e_{22} = 8$, $e_{31} = 2$ and $e_{32} = -5$.
$\begin{bmatrix} \begin{array}{|cccc|} \hline 3 & 8 & -5 & 2 \\\hline 20 & -9 & -1 & -5 \\\hline 6 & 13 & 7 & 0 \\\hline 5 & -2 & -6 & 3 \\\hline \end{array} \end{bmatrix}$
This matrix has four rows and four elements are placed in each row. So, $e_{11} = 3$, $e_{12} = 8$, $e_{13} = -5$, $e_{14} = 2$, $e_{21} = 20$, $e_{22} = -9$, $e_{23} = -1$, $e_{24} = -5$, $e_{31} = 6$, $e_{32} = 13$, $e_{33} = 7$, $e_{34} = 0$, $e_{41} = 5$, $e_{42} = -2$, $e_{43} = -6$ and $e_{44} = 3$.
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# Publishing house / Banach Center Publications / All volumes
## Laplace transform identities for diffusions, with applications to rebates and barrier options
### Volume 83 / 2008
Banach Center Publications 83 (2008), 139-157 MSC: Primary 60J60, 91B28; Secondary 44A10, 47D07, 60J70. DOI: 10.4064/bc83-0-9
#### Abstract
We start with a general time-homogeneous scalar diffusion whose state space is an interval $I\subseteq\mathbb R$. If it is started at $x\in I$, then we consider the problem of imposing upper and/or lower boundary conditions at two points $a,b\in I$, where $a< x< b$. Using a simple integral identity, we derive general expressions for the Laplace transform of the transition density of the process, if killing or reflecting boundaries are specified. We also obtain a number of useful expressions for the Laplace transforms of some functions of first-passage times for the diffusion. These results are applied to the special case of squared Bessel processes with killing or reflecting boundaries. In particular, we demonstrate how the above-mentioned integral identity enables us to derive the transition density of a squared Bessel process killed at the origin, without the need to invert a Laplace transform. Finally, as an application, we consider the problem of pricing barrier options on an index described by the minimal market model.
#### Authors
• Hardy HulleySchool of Finance and Economics
University of Technology, Sydney
P.O. Box 123, Broadway, NSW 2007, Australia
e-mail
• Eckhard PlatenSchool of Finance and Economics and Department of Mathematical Sciences
University of Technology, Sydney
P.O. Box 123, Broadway, NSW 2007, Australia
e-mail
## Search for IMPAN publications
Query phrase too short. Type at least 4 characters.
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# How Many Different Meals Are Possible?
#### (An archive question of the week)
While gathering combinatorics questions, there were several that stood out. This one will serve well to summarize the topic, showing multiple methods for counting, and contrasting other kinds of problems.
## The problem
The question, from 2007, relates to an Arby’s promotion:
How Many Different Dinners Can Be Ordered?
I was trying to calculate the number of different meals you can get for the 5 for $5.95 at Arby's and wasn't confident in the way I was calculating it. Given 8 different choices, how many combinations of 5 are there? You do not need to choose different entrées. I am confused about the fact that you are calculating a combination with replacement. I thought that I would start this way... 8 C 1 for the possibility of choosing 5 entrées alike, 2*(8 C 2) for choosing 4 alike and 1 different entrée. Multiplied by 2 because there are 2 ways to assign each pair of choices. 2*(8 C 2) for choosing 3 alike and 2 alike and multiply by 2 because there are 2 ways to assign each pair. (8 C 3)*(3 C 1) for choosing 2 alike, 2 alike and 1 different entrée and once the three are chosen, choose the one that is single. (8 C 3)*(3 C 1) for choosing 3 alike, 1 different, 1 different. (8 C 4)*(4 C 1) for choosing 2 alike, and 3 different entrées plus choosing which one is doubled up. (8 C 5) to choose all 5 entrées different. Then add each of the above together. Am I on the right track? Dave is representing the common notation $$\displaystyle _nC_r = \frac{n!}{r!(n-r)!}$$ (read as “n choose r” ) for combinations. He has done very well. Doctor Wilko chose to expand on what he did: Thanks for writing to Ask Dr. Math! Yes, your approach to the problem looks excellent. I like this problem a lot. I think it's a good real-world problem. We often see many restaurants and pizza parlors trying to figure out how many different combinations of meals or pizzas they offer. The difficulty with these problems is that it is easy to calculate the number of combinations incorrectly. I'll present two solutions below to answer, "How many combinations of 5 items for$5.95 can one order from Arby's 8-item menu?" The first uses the same logic you applied in your work, and the second approaches it slightly differently. Then I'll also show some errors that are commonly made on this sort of problem.
The first step, always, is to clarify the problem, restating it if necessary to expose all assumptions that are required:
Recall, we are assuming:
- There are 8 choices on this particular menu.
- 5 items can be ordered (for \$5.95).
- Duplicate items can be ordered (the 5 items don't have to be distinct).
The last point is important. There are not 8 finite items to choose from; there are 8 choices on the menu and you can order anywhere from 0-5 of each of those items.
So this is not a straight permutation or combination problem, both of which involve subsets of a given sets, with no repetition allowed. In this case, the items are distinguishable but repeatable; order doesn’t matter.
A good model for the choice of a meal would be a menu list, with a blank for each item in which you put how many of that item ordered; these numbers have to add up to 5:
____ Arby’s Melt
____ Small Drink
____ Ham Melt
____ Small Fries
____ Potato Cakes
____ Turnover
____ Chicken Sandwich
____ Shake
## First method: multiple cases
The first is just a little different from Dave’s method, and has a slightly different order:
For this solution, I'll try to explicitly enumerate the different ways I can order items from the menu and then apply the combinations formulas respectively. Last I'll add up all the answers to get the final answer.
I'm basically partitioning the sample space into smaller subproblems, solving each subproblem, and then combining the results to answer the big problem.
1. I can order 5 of the same item.
There is only 1 choice. Which of the 8 should I choose?
Mathematically, this can be written as,
8C1 = 8.
There are 8 ways I can choose 1 item--once I pick it, then I can order a quantity of 5 of that 1 item.
2. I can order 4 of one item and 1 other item.
Here, there are only 2 choices. Out of the 8 menu items, which 2 should I choose? Mathematically, this can be written as,
8C2 = 28
But out of those two choices, I need to order 4 of one and 1 of the other. To figure out which of the items I order 4 of, I can multiply by 2C1 or 2. The final answer for this portion of the problem is,
8C2 * 2C1 = 28 * 2 = 56.
Note, at this point, you could explicitly enumerate all the ways to order 4 of one item and 1 other item from the menu of 8 items. You'll see that there are 56 combinations.
This case (like the others) could also be counted in other ways. For example, we could just use a permutation to choose two items in order, the first to be the 4 and the second to be the 1. This gives $$\displaystyle _8P_2 = \frac{8!}{6!} = 8\cdot7 = 56$$ as before.
If I continue with the same reasoning as above, the problem continues as follows:
3. 3 same items, 2 other same items
8C2 * 2C1 = 56.
4. 3 same items, 1 other item, 1 other item
8C3 * 3C1 = 168.
5. 2 same items, 2 other same items, 1 other item
8C3 *3C1 = 168.
6. 2 same items, 1 other item, 1 other item, 1 other item
8C4 * 4C1 = 280.
7. 5 different items
8C5 = 56.
Total combinations of 5 items I can order from a menu of 8 items at Arby's is,
8 + 56 + 56 + 168 + 168 + 280 + 56 = 792.
Apparently, Arby's boasts "Over 790 combinations!". Arby's claim gives me confidence in this answer.
Doctor Wilko’s result can be written as $$_8C_1\ +\ _8C_2\ \cdot\ _2C_1\ +\ _8C_2\ \cdot\ _2C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_4\ \cdot\ _4C_1\ +\ _8C_5.$$
Dave’s version was $$_8C_1\ +\ _8C_2\ \cdot\ 2\ +\ _8C_2\ \cdot\ 2\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_3\ \cdot\ _3C_1\ +\ _8C_4\ \cdot\ _4C_1\ +\ _8C_5,$$ which is identical.
My version, using permutations, is $$_8C_1\ +\ _8P_2\ +\ _8P_2\ +\ _8P_3 \div 2!\ +\ _8P_3 \div 2!\ +\ _8P_4 \div 3!\ +\ _8C_5.$$
These are all essentially the same idea.
## Finding the cases
How do we know we haven’t missed a case? Dave made an orderly list, though it isn’t entirely clear what principle he was using to do so. Here is a list of the 7 cases used, in Doctor Wilko’s order, expressed by representing each kind of item with a letter:
• AAAAA
• AAAAB
• AAABB
• AAABC
• AABBC
• AABCD
• ABCDE
We can see that he used the ordering principle of largest-first, where items are listed in decreasing order of frequency in each case, and the number of the first item mentioned decreases from case to case. In this order, it is easy to see that nothing was missed. This is the main benefit of Doctor Wilko’s work over Dave’s, and it is essential when you use cases.
Keep in mind that here the letters don’t stand for specific items; the main work was to count the number of distinct ways in which a meaning could be assigned to each letter.
Note that we could also list these cases as sums that total 5:
• $$5$$
• $$4+1$$
• $$3+2$$
• $$3+1+1$$
• $$2+2+1$$
• $$2+1+1+1$$
• $$1+1+1+1+1$$
These are the “unordered partitions” of 5 (since we are not distinguishing different orders of the same number), which are harder to count (without listing as we have done here) than other things we’ve discussed. Fortunately, what we needed to do was to list them, not just to count! If you are curious, see here:
Number of Unordered Partitions
## Second method: Stars and bars
This will look familiar from last time. I am rewriting his to use stars and bars in the classic form, whereas he swapped the symbols, using the bars as tallies. Here we will directly use my menu list order form from above as a model.
This solution is a little more elegant, but perhaps not as initially intuitive as trying explicit enumeration. I credit another one of our volunteers, Dr. Anthony, for reminding me of this technique.
Let's label the menu items as A-H. You are ordering from the menu of 8 items. You may tally up your order like this:
Item A | B | C | D | E | F | G | H
---------------------------------------------------
** | | * | | * | | | *
Meaning 2 of item A, 1 of item C, 1 of item E, and 1 of item H.
There are other possibilities, too. For instance,
Item A | B | C | D | E | F | G | H
---------------------------------------------------
*** | | ** | | | | |
Meaning 3 of item A and 2 of item C.
You could do this for many different combinations, but should soon realize that every possible order can be made by placing 5 stars under the 8 items in any manner you choose.
We could also think of each choice as a partition — this time as an ordered partition in which there must be 8 terms, any of which can be zero. We might have started a list like this:
• $$5+0+0+0+0+0+0+0$$
• $$4+1+0+0+0+0+0+0$$
• $$4+0+1+0+0+0+0+0$$
• $$4+0+0+0+0+0+0+1$$
• $$3+2+0+0+0+0+0+0$$
This would not be easy to list, would it? Nor is it obvious in this form how you could count them. But this time, there is a nice formula (as we saw last time)! In the world of partitions, you can be surprised by what is easy and what is hard.
We just make the mental twist to see the first example above merely as * * | | * | | * | | | *:
The "leap" with this method is to now see this as a permutations problem where you are arranging 5 stars and 7 bars, where 5 stars are identical and the 7 bars are identical.
From probability, the permutations formula for this is,
12!
------- = 792.
5! 7!
You can also think of it as you have 12 spaces and you can choose a place for the 5 |'s in 12C5 ways. Likewise, you have 12 spaces and you can choose a place for the 7 *'s in 12C7 ways.
Note, 12C5 = 12C7 = 792.
This confirms our answer from above!
He closes this with a comment that many of us have made:
My favorite technique with combinations/permutations problems is to find more than one way to solve the problem. If I get confirmation from two different approaches, I usually feel more confident in my answer!
In algebra, I always want to check my work by plugging the solution into the original problem; in combinatorics I never fully trust my work until I can get the same answer two ways.
## Incorrect methods
Doctor Wilko then provided a catalog of basic formulas for different kinds of problems, showing why they don’t apply. The most important step in a combinatorics problem is determining what kind of problem it is, and what model or method applies. A mistake there ruins everything you do.
Below are a couple of incorrect solutions and reasons why they are incorrect.
1. 8^5 = 32,768
It seems logical at first, you have 8 choices for your first choice, 8 choices for your second choice, etc...
I was actually guilty of thinking this was the answer at first! But then it hit me that this is a PERMUTATIONS solution. For instance, you would have ABBCC and BBACC. These are just two different permutations of the same combination of 5 items.
In our problem, we want to buy a sack of 5 items from the 8-item menu. Order is irrelevant in purchasing the 5 items.
==================
In general, when order matters, and objects can be chosen more than once, the number of permutations is
n^r (Permutations with Repetition)
where n is the number of total objects and r is the number chosen.
==================
The model for this is _ _ _ _ _, where each of the 5 blanks (the items ordered) can be filled in any of 8 ways (the items on the menu), independently, so the total is $$8\cdot 8\cdot 8\cdot 8\cdot 8$$. The trouble is that the blanks are distinguishable, so that order counts.
For our problem, this would overcount, as we would be counting each meal multiple times. This is how you might count the ways for 5 distinct people to each choose one item, or to arrange 5 items in a 5-part tray. But we don’t want to distinguish the order of a meal.
Note that we can’t correct for this by dividing by the number of arrangements of each meal, as we do in creating the formula for combinations, because that varies according to the number of duplicates. AAAAA has only one arrangement, while ABCDE has 60 arrangements.
2. 8P5 = 6,720
This assumes 8 finite, distinct items, and you want all the different arrangements of 5 of them. ABCDE would be a different permutation than EDCBA. Remember, for the problem, we want the number of possible combinations a person could physically order from the menu. Order doesn't have anything to do with this.
==================
In general, when order matters, and each object can only be chosen once, the number of permutations is
nPr = n!/(n-r)! (Permutations without Repetition)
where n is the number of total objects and r is the number chosen.
==================
The model for this is _ _ _ _ _, where each blank must be filled with a different item, so the total is $$8\cdot 7\cdot 6\cdot 5\cdot 4$$.
This is wrong both because order counts (as it shouldn’t for our problem) and because repetition is not allowed (as it should be). Nothing is as we need it to be.
3. 8C5 = 56
This assumes 8 finite, distinct items, and you are choosing 5 of them. This is not representative of the problem. This solution wouldn't let you order 5 roast beefs, for example.
==================
In general, when order doesn't matter, and each object can only be chosen once, the number of combinations is
nCr = n!/[r!*(n-r)!] (Combinations without Repetition)
where n is the number of total objects and r is the number chosen.
==================
One way to derive this formula is to start with the $$8\cdot 7\cdot 6\cdot 5\cdot 4$$ permutations, and divide by the $$5\cdot 4\cdot 3\cdot 2\cdot 1$$ ways to arrange each combination.
We do need combinations for our problem, because the order in which we choose items doesn’t matter; but we need to allow for repetitions.
So, finally, that leads us back to our solution, which really is a "combinations with repetition" problem.
==================
In general, when order doesn't matter, and each object can be chosen more than once, the number of combinations is
nPr = (n+r-1)!/[r!*(n-1)!] (Combinations with Repetition)
where n is the number of total objects and r is the number chosen.
==================
This is the formula we found last time, using n-1 bars (separating n bins) and r stars. The name nPr is probably a typo; there is a symbol for this (which I never learned), namely $$\displaystyle \left({8 \choose 5}\right)$$.
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Latex Table Center Vertically And Horizontally
It will also offset all cells from the table's frame on all four sides by a length of 10 pixels. Then we might imagine that the table would have a heading… Followed by an ending at the end of the page… Which would be. # Copyright (C) YEAR THE PACKAGE'S COPYRIGHT HOLDER # This file is distributed under the same license as the PACKAGE package. Although tables have already been covered, it was only the internal syntax that was discussed. I would like the image to be to the left of the title, and for both of these elements to be centered both horizontally and vertically in the div. CSS tables will. The length of the skip should be expressed in a unit recognized by LaTeX. For rows, you can specify Bottom, Center, or Top. How to write matrices in Latex ? matrix, pmatrix, bmatrix, vmatrix, Vmatrix. org Calc (Spreadsheet) macro for converting tables. Help Center Detailed answers to any. booktabs produces a LaTeX formatted table for use with LaTeX's booktabs package. 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Follow these steps to align text in a table: Select the cells, columns, or rows, with text that you want to align (or select your entire. Centering Text Horizontally and Vertically in LaTeX [closed] How to vertically center a div for all browsers? 403. How to Show and Hide Cell Gridlines on All Tables in Word Lori Kaufman @howtogeek July 30, 2015, 10:24am EDT By default, when you create a new table, all the cells have black borders that print with the document. A text table uses the text mark type. Justified and Vertically Centered Header Elements A little journey into positioning header elements to be centered vertically and justified with the help of pseudo-elements. See the example below [code ]\begin{center} Your text here \end{center}[/code]. (If you are looking for information on how to edit images, objects, or scanned PDFs, click the appropriate link above. 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Smart ways to to control vertical and horizontal spacing in a table How to manage vertical and horizontal spacing in a table Use cell margins to set the distance between cell borders and text in a table. First, let's talk about vertical scaling. Click on the page. tex produces a LaTeX formatted table. I'm not having any trouble getting it perfectly centered horizontally on the page, but I can't seem to get it to vertically center on the page. Most of the tips involve use of standard text-editors. How to wrap text in LaTeX tables? 9. The generated CSS contains all the necessary colors etc. For the six vertical cleats, drill a countersunk pilot hole through the back face near the top. The beamer package is provided with most LaTeX distributions, but is also available from CTAN. This works in Gecko. How do I make subfigures in a vertical line. 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See the example below [code ]\begin{center} Your text here \end{center}[/code]. What is the best way to make subfigures in a vertical line opposed to horizontally? to do is to use a table. Smart ways to to control vertical and horizontal spacing in a table How to manage vertical and horizontal spacing in a table Use cell margins to set the distance between cell borders and text in a table. Of course, you can combine the two so that you get the effect you want. Splitting cells, horizontally and vertically. Tables are a powerful formatting tool used in many Word documents. Horizontal centering is quite simple. I have an image and a text title that I'm trying to center in a div. The vertical-values of writing-mode are really intended for use in setting a normal vertical context for CJK or Mongolian text. developerWorks blogs allow community members to share thoughts and expertise on topics that matter to them, and engage in conversations with each other. 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Then form slots at the center and bottom end of each cleat (Project Diagram, Drawing 2). I am trying to center the text in a Google Document table both horizontally and vertically for all lines using Google Scripts. Inserting an image that's a little to big horizontally causes the centering mechanism to break down. Four-foot by 8-foot or 4-foot by 9-foot panels shall be applied vertically. The following example illustrates how such a table can be produced using estadd summ and esttab. - An issue with this approach is with tables too large (horizontally) to fit into one page (i. Markdown Cells¶. Maybe you need a table for your Master's Thesis? maybe you need to present your top music chart for a friend? The Table Editor is here to help! 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This works in Gecko. In a text box in Word, you can align text horizontally or vertically, and you can adjust the margins to be narrower or wider. The words horizontal and vertical are generally used in a planar (2-dimensional) sense, not spatial (3-dimensional). Ask Question. I know when getting a c-section done the doctors will do it in one way or another, horizontal or vertical, i have had a c-section & i dont know if my incision is horizontal or vertical, my scar sits the same way as my knicker line does, like bikini line, not the straight up & down cut, please tell me if mine is horizontal or vertical :( thanks!. The vertical-values of writing-mode are really intended for use in setting a normal vertical context for CJK or Mongolian text. The question: how best can I create a single box with a specified height and width, and with the text in it aligned vertically? I am relatively new to Latex, so this cripples my options. This extension will allow you to select a region in a spreadsheet, and generate a LaTeX tabular format of the spreadsheet. Then form slots at the center and bottom end of each cleat (Project Diagram, Drawing 2). Controlling vertical and horizontal space. Technical Tuesday, May 26, 2015. LaTeX removes vertical space that comes at the end of a page. All table inputs must have unique variable names. Cristina McGraw 50,921 views. To produce (horizontal) blank space within a paragraph, use \hspace, followed by the length of the blank space enclosed within braces. Either one, two, or three pairs of source points and definition points can be specified to move, rotate, or tilt the selected objects, aligning them with points on another object. LaTeX minipage. Twitter bootstrap center table column contents vertically as well as horizontally (HTML) - Codedump. so that the table should look similar when you paste it to your website. 5 Description Create pretty tables for 'HTML', 'Microsoft Word' and 'Microsoft PowerPoint' documents. Tables starts with the line: \begin{table}[hbtp]. A tabular environment may be nested within a minipage to allow side-by-side tables within a table or figure environment, or to place an image next to a table. , literal, listing and source) blocks to attempt to fit the text within the available width. Setting HTML Table and Column Widths The HTML TABLE tag is the opening tag used to create a table within a web page. Be sure to use C-h b to see the default bindings. Rendering to HTML, LaTeX, SGML/XML (TEI) and CALS. What is the best way to make subfigures in a vertical line opposed to horizontally? to do is to use a table. The horizontal centering is not difficult. Ask Question. Add subfigures horizontally; Add multiple subfigures in multiple rows; In research articles, we need to add subfigures often. The value is defined as a length in pixels. There are two ways to center information: horizontally and vertically. Center-Aligned Figures. This kind of table-based layout is strongly discouraged today. ) As we want to center the text horizontally, we add the command \centering. developerWorks blogs allow community members to share thoughts and expertise on topics that matter to them, and engage in conversations with each other. The \centering command may be preferable to the center environment in a float, as the center environment will add a paragraph and create extra whitespace in the float. lower right corner arrow) where the [Layout] tab has a [Vertical alignment] option. Cristina McGraw 50,921 views. You begin figures with \begin{figure}[loc] where loc is a sequence of 0 to 4 letters, each one specifying a location where the figure or table may be placed, as follows:. Many DVI viewers do not support rotating of text and tables. First, let’s talk about vertical scaling. For the first appearance of any table, you should include the table title/number followed by the entire caption. This center environment can cause additional vertical space. The horizontal centering is not difficult. There are several programs one can download that do this however, I prefer to save the table as a. Since this is a negative quadratic, the graph is an upside-down parabola. I'm new in the web development industry and I struggle to vertically and horizontally align a child div in a parent div. 31 December 2010, by Nadir Soualem Latex. Most commands are very straightforward to use. , literal, listing and source) blocks to attempt to fit the text within the available width. If you click the save button, your code will be saved, and you get an URL you can share with others. For vertically polarized light the electric field is parallel to the direction of gravity so it is in the $\hat{z}$ direction. I personally think there will be few usecases to manually adjust the settings of the font, because the environments usually do this job for you automatically, I just included this for completeness. What is the best way to make subfigures in a vertical line opposed to horizontally? to do is to use a table. One of my last posts was on how to define the width of a column in a table (see here). The value is defined as a length in pixels. But hopefully, enough has been covered here so that you can create any table you are likely to need for your papers. The mathematical definition of a hyperbola is the set of. Mechanical vibration: 2 springs attached horizontally to a wheel at a vertical distance a above the center, find the natural frequency of the system Ask Question Asked 7 months ago. When you want to include an image or a table that's wider than the text width, you will notice that even when \centering or the center-environment is used this wide object will not be centered in relation to the surrounding text. Lee Introduction This is a math class! Why are we writing? There is a good chance that you have never written a paper in a math class before. Tabbing between cells. This is accomplished by multiplying either $x$ or $y$ by a constant, respectively. If you use MikTeX, all you have to do is to include the beamer package and let LaTeX download all wanted packages automatically. To center something vertically: 1. Conforming browsers ought to center tables if the left and right margins are equal. I have a table, created in LaTeX, using the tabular environment. 3 Special Characters Certain characters have special meaning to LATEX. Middle align text vertically in table cell in Word. To just center the text, I centered the top row and then centered the rest of the rows separately. 4), we should only use equation (and no other environment) to produce a single equation. I know when getting a c-section done the doctors will do it in one way or another, horizontal or vertical, i have had a c-section & i dont know if my incision is horizontal or vertical, my scar sits the same way as my knicker line does, like bikini line, not the straight up & down cut, please tell me if mine is horizontal or vertical :( thanks!. The cells typically contain text data such as text, dates, and numbers but they can. I am stuck on one of these which is the following: A = \{(x,y)\in \Bbb R^2 |. Instant access to millions of Study Resources, Course Notes, Test Prep, 24/7 Homework Help, Tutors, and more. To just center the text, I centered the top row and then centered the rest of the rows separately. Again, this will only preserve the text in the upper-leftmost cell. That's about it for basic tables in my opinion. With the above in mind, I have flagged several answers as "link-only" answers and, recently, a lot of my flags have been declined - a few of which were then deleted by a mod afterwards (hence, myself thinking the flag was actually a correct one). # SOME DESCRIPTIVE TITLE. Click on the page. Add subfigures horizontally; Add multiple subfigures in multiple rows; In research articles, we need to add subfigures often. I'm new in the web development industry and I struggle to vertically and horizontally align a child div in a parent div. This MATLAB function concatenates B vertically to the end of A when A and B have compatible sizes (the lengths of the dimensions match except in the first dimension). The keyword is: \raggedleft, \centering or \raggedright. This page has sample title pages for a Ph. Ask Question. (IUD), measuring 32 mm horizontally and 36 mm vertically, with a 3 mm diameter bulb at the tip of the vertical stem. Right-click on it and select Paragraph. How can I position figures and tables where I want them with LaTeX? LaTeX uses specific rules to place floats (figures and tables). For vertically polarized light the electric field is parallel to the direction of gravity so it is in the $\hat{z}$ direction. An example is the % sign, which indicates that the remainder of the line is a comment and should not be processed. The correct way doesn't exist In dinamic table in SQL DB in my work I have seen equally horizontally and vertically. Middle align text vertically in table cell in Word. You can change the cell type to Markdown by using the Cell menu, the toolbar, or the key shortcut m. " Thus, one might write in a style sheet:. Insert tables. If you click the save button, your code will be saved, and you get an URL you can share with others. With a table, you have several options, including every cell in the table; every header cell in the table every cell in the table head, table body or table foot. edu is a platform for academics to share research papers. A table of contents will automatically be created, complete with. If you set TableAlignments->Center, all entries will be centered both horizontally and vertically. Center-Aligned Figures. Again, vertical alignment on the page is a Section formatting property, not a paragraph formatting property like horizontal alignment. Then draw a line halfway between the vertical center line and the left edge. If light is travelling "horizontally" (meaning perpendicular to the direction of gravity, for example in the $\hat{y}$ direction) then we can talk about horizontally and vertically polarized light. Right-click on it and select Paragraph. A table is an arrangement of columns and rows that organizes and positions data. Justified and Vertically Centered Header Elements A little journey into positioning header elements to be centered vertically and justified with the help of pseudo-elements. Footnotes should be labeled in order, starting at the upper left corner of the table and working vertically and horizontally to the lower right corner (as you would read a book). How to refer to a figure or table from within a LaTeX document. I'm trying to make a Grid that has one cell that spans horizontally and one that spans vertically, like so (generated manually with Graphics): But when I try to do this with Grid, the right-most. Productivity Smart Home Meet This Adorable Robot That Can Be Coded to Draw Pictures. 3 Special Characters Certain characters have special meaning to LATEX. 0 m above flat ground, emerging from the gun with a speed of 370 m/s. Hence, a cellspacing="10" attribute-value pair will horizontally and vertically separate all adjacent cells in the respective table by a length of 10 pixels. As with all text, there are times when you want to change how it is aligned. The mobile station can be rolled bedside or stored to conserve clinic space. This website provides an overview of basic text formatting commands in LaTeX. A tabular environment may be nested within a minipage to allow side-by-side tables within a table or figure environment, or to place an image next to a table. The correct way doesn't exist In dinamic table in SQL DB in my work I have seen equally horizontally and vertically. Origin is the data analysis and graphing software of choice for over half a million scientists and engineers in commercial industries, academia, and government laboratories worldwide. Just make sure you paint on the latex in alternating patterns. See the next section for more information on how this package. Then we might imagine that the table would have a heading… Followed by an ending at the end of the page… Which would be. I'm new in the web development industry and I struggle to vertically and horizontally align a child div in a parent div. Centering wide tables or figures. tex help LATEX place tables and gures where you intend them to be. A tabular environment may be nested within a minipage to allow side-by-side tables within a table or figure environment, or to place an image next to a table. Package 'flextable' June 25, 2019 Type Package Title Functions for Tabular Reporting Version 0. Text alignment can be manually controlled by several commands. 4), we should only use equation (and no other environment) to produce a single equation. Middle align text vertically in table cell in Word. When you want to include an image or a table that’s wider than the text width, you will notice that even when \centering or the center-environment is used this wide object will not be centered in relation to the surrounding text. This takes an x-slope, y-slope and length like so: \line (x-slope,y-slope){length}. The Auto-Rotate And Center option in the Print dialog box automatically selects the page orientation that best matches the content and paper. It makes making tables on LaTeX very easy. Tables are a powerful formatting tool used in many Word documents. In Reporting Services, tables, matrices, and lists are data regions that display paginated report data in cells that are organized into rows and columns. Hypertext Help with LaTeX Ellipses. If you click the save button, your code will be saved, and you get an URL you can share with others. Draw a scaled-up 16-square grid on the wood round tabletop using chalk. Justified and Vertically Centered Header Elements A little journey into positioning header elements to be centered vertically and justified with the help of pseudo-elements. You can add negative as well as positive space with an \vspace command. This can be changed, but is a bit tricky. 4 Formatting 4. The horizontal centering is not difficult. The T-frame is made of. A new table row begins when you type \\. No LaTeX installation required. If you don't want LaTeX to remove this space, include the optional * argument. In a text box in Word, you can align text horizontally or vertically, and you can adjust the margins to be narrower or wider. T his page documents a few tricks for making LaTeX tables that I found useful when writing my thesis and preparing various publications. How do I make subfigures in a vertical line. On the other hand, if your content is in English and you want some title text to run from bottom to top vertically, say on a book spine or in table header, you would use writing-mode:sideways-lr, not one of the vertical-values. As you read, you'll have many "eureka" breakthroughs, and you'll start to understand the finer points. Setting HTML Table and Column Widths The HTML TABLE tag is the opening tag used to create a table within a web page. 10 (Installation)python-docx is a Python library for creating and updating Microsoft Word (.
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Author: Hcamael@Knownsec 404 Team
Chinese Version: https://paper.seebug.org/779/
Recently, when I was studying IoT, due to the lack of devices, simulating running firmware would often be short of /dev/xxx, so I began to wonder if I could write a driver myself to make the firmware run. No matter how hard it is and whether it can achieve my original intention or not, it pays off a lot if you learn how to develop Linux driver.
### Introduction
The series I wrote is mainly about practice, which doesn't talk much about theory. I learn how to develop the driver from the book Linux Device Drivers, and there is the code for the examples explained in this book on the GitHub [1].
As for the basic concept, Linux system is divided into kernel mode and user mode. The hardware device can only be accessed in the kernel mode, and the driver can be regarded as an API provided in the kernel mode to let the code of the user mode access the hardware device.
With the basic concepts in mind, I have come up with a series of problems, which inspire me to learn the development of driver.
1. All code learning starts with Hello World, so how to write a Hello World program?
2. How does the driver generate device files under /dev?
3. How does the driver access the actual hardware?
4. How do I get system-driven code? Or can it reverse the driver without code? Where are the binaries that store the drivers? In the future, there may be opportunities to try to study the drive security.
### Everything Starts from Hello World
My Hello World code is as follows [2]:
#include <linux/init.h>
#include <linux/module.h>
MODULE_AUTHOR("Hcamal");
int hello_init(void)
{
printk(KERN_INFO "Hello World\n");
return 0;
}
void hello_exit(void)
{
printk(KERN_INFO "Goodbye World\n");
}
module_init(hello_init);
module_exit(hello_exit);
The Linux driver is developed by means of C Language, which is different form the normal one we use. What we often use is Libc library, which doesn’t exist in the kernel. While the driver is a program running in the kernel, we use the library functions in the kernel.
For example, printk is analogous to printf in Libc, an output function defined in the kernel. But I think it's more like the logger function in Python, because the output of printk is printed in the kernel's log, which can be viewed via dmesg command.
There is only one entry point and one exit point in the driver code. Loading the driver into the kernel will execute the function defined by the module_init function, which in the above code is the hello_init function. When the driver is unloaded from the kernel, the function defined by the module_exit function is called, which in the above code is the hello_exit function.
The code above makes it clear that when the driver is loaded, it prints Hello World and when the driver is unloaded, it prints Goodbye World.
PS: MODULE_LICENSE and MODULE_AUTHOR are not very important. I'm not a professional development driver, so needn’t pay attention to them.
PSS: There should add a newline for the output of printk, otherwise the buffer will not be flushed.
#### Compile the Driver
The driver needs to be compiled by the make command, and the Makefile is shown below:
ifneq ($(KERNELRELEASE),) obj-m := hello.o else KERN_DIR ?= /usr/src/linux-headers-$(shell uname -r)/
PWD := $(shell pwd) default:$(MAKE) -C $(KERN_DIR) M=$(PWD) modules
endif
clean:
rm -rf *.o *~ core .depend .*.cmd *.ko *.mod.c .tmp_versions
In general, the source code of the kernel exists in the /usr/src/linux-headers-$(shell uname -r)/ directory, such as: $ uname -r
4.4.0-135-generic
/usr/src/linux-headers-4.4.0-135-generic/ --> 该内核编译好的源码目录
And what we need is the compiled source directory, which is /usr/src/linux-headers-4.4.0-135-generic/.
The header files of the driver code need to be searched from this directory.
The parameter M=\$(PWD) indicates that the output of the driver compilation is in the current directory.
Finally, through the command obj-m := hello.o, which means to compile hello.o into hello.ko, and the ko file is the kernel module file.
#### Load the Driver into the Kernel
Some system commands that need to be used:
Lsmod: View the kernel module that is currently loaded.
Insmod: Loads the kernel module and requires root permissions.
Rmmod: Remove the module.
For example:
# insmod hello.ko // Load the hello.ko module into the kernel
# rmmod hello // Remove the hello module from the kernel
The old kernel is using the above method to load and remove the kernel, but the new version of the Linux kernel adds verification of the module. The current actual situation is as follows:
# insmod hello.ko
insmod: ERROR: could not insert module hello.ko: Required key not available
From a security perspective, the current kernel assumes that the module is untrustworthy and needs to be signed with a trusted certificate to load the module.
Two solutions:
1. Enter the BIOS and turn off the Secure Boot of UEFI.
2. Add a self-signed certificate to the kernel and use it to sign the driver module (You can refer to [3]).
### Add Device Files under /dev
Once again, we firstly provide the code, and then explain the example code [4].
#include <linux/init.h>
#include <linux/module.h>
#include <linux/kernel.h> /* printk() */
#include <linux/slab.h> /* kmalloc() */
#include <linux/fs.h> /* everything... */
#include <linux/errno.h> /* error codes */
#include <linux/types.h> /* size_t */
#include <linux/fcntl.h> /* O_ACCMODE */
#include <linux/cdev.h>
#include <asm/uaccess.h> /* copy_*_user */
MODULE_AUTHOR("Hcamael");
int scull_major = 0;
int scull_minor = 0;
int scull_nr_devs = 4;
int scull_quantum = 4000;
int scull_qset = 1000;
struct scull_qset {
void **data;
struct scull_qset *next;
};
struct scull_dev {
struct scull_qset *data; /* Pointer to first quantum set. */
int quantum; /* The current quantum size. */
int qset; /* The current array size. */
unsigned long size; /* Amount of data stored here. */
unsigned int access_key; /* Used by sculluid and scullpriv. */
struct mutex mutex; /* Mutual exclusion semaphore. */
struct cdev cdev; /* Char device structure. */
};
struct scull_dev *scull_devices; /* allocated in scull_init_module */
/*
*/
struct scull_qset *scull_follow(struct scull_dev *dev, int n)
{
struct scull_qset *qs = dev->data;
/* Allocate the first qset explicitly if need be. */
if (! qs) {
qs = dev->data = kmalloc(sizeof(struct scull_qset), GFP_KERNEL);
if (qs == NULL)
return NULL;
memset(qs, 0, sizeof(struct scull_qset));
}
/* Then follow the list. */
while (n--) {
if (!qs->next) {
qs->next = kmalloc(sizeof(struct scull_qset), GFP_KERNEL);
if (qs->next == NULL)
return NULL;
memset(qs->next, 0, sizeof(struct scull_qset));
}
qs = qs->next;
continue;
}
return qs;
}
/*
* Data management: read and write.
*/
ssize_t scull_read(struct file *filp, char __user *buf, size_t count,
loff_t *f_pos)
{
struct scull_dev *dev = filp->private_data;
struct scull_qset *dptr; /* the first listitem */
int quantum = dev->quantum, qset = dev->qset;
int itemsize = quantum * qset; /* how many bytes in the listitem */
int item, s_pos, q_pos, rest;
ssize_t retval = 0;
if (mutex_lock_interruptible(&dev->mutex))
return -ERESTARTSYS;
if (*f_pos >= dev->size)
goto out;
if (*f_pos + count > dev->size)
count = dev->size - *f_pos;
/* Find listitem, qset index, and offset in the quantum */
item = (long)*f_pos / itemsize;
rest = (long)*f_pos % itemsize;
s_pos = rest / quantum; q_pos = rest % quantum;
/* follow the list up to the right position (defined elsewhere) */
dptr = scull_follow(dev, item);
if (dptr == NULL || !dptr->data || ! dptr->data[s_pos])
goto out; /* don't fill holes */
/* read only up to the end of this quantum */
if (count > quantum - q_pos)
count = quantum - q_pos;
if (raw_copy_to_user(buf, dptr->data[s_pos] + q_pos, count)) {
retval = -EFAULT;
goto out;
}
*f_pos += count;
retval = count;
out:
mutex_unlock(&dev->mutex);
return retval;
}
ssize_t scull_write(struct file *filp, const char __user *buf, size_t count,
loff_t *f_pos)
{
struct scull_dev *dev = filp->private_data;
struct scull_qset *dptr;
int quantum = dev->quantum, qset = dev->qset;
int itemsize = quantum * qset;
int item, s_pos, q_pos, rest;
ssize_t retval = -ENOMEM; /* Value used in "goto out" statements. */
if (mutex_lock_interruptible(&dev->mutex))
return -ERESTARTSYS;
/* Find the list item, qset index, and offset in the quantum. */
item = (long)*f_pos / itemsize;
rest = (long)*f_pos % itemsize;
s_pos = rest / quantum;
q_pos = rest % quantum;
/* Follow the list up to the right position. */
dptr = scull_follow(dev, item);
if (dptr == NULL)
goto out;
if (!dptr->data) {
dptr->data = kmalloc(qset * sizeof(char *), GFP_KERNEL);
if (!dptr->data)
goto out;
memset(dptr->data, 0, qset * sizeof(char *));
}
if (!dptr->data[s_pos]) {
dptr->data[s_pos] = kmalloc(quantum, GFP_KERNEL);
if (!dptr->data[s_pos])
goto out;
}
/* Write only up to the end of this quantum. */
if (count > quantum - q_pos)
count = quantum - q_pos;
if (raw_copy_from_user(dptr->data[s_pos]+q_pos, buf, count)) {
retval = -EFAULT;
goto out;
}
*f_pos += count;
retval = count;
/* Update the size. */
if (dev->size < *f_pos)
dev->size = *f_pos;
out:
mutex_unlock(&dev->mutex);
return retval;
}
/* Beginning of the scull device implementation. */
/*
* Empty out the scull device; must be called with the device
* mutex held.
*/
int scull_trim(struct scull_dev *dev)
{
struct scull_qset *next, *dptr;
int qset = dev->qset; /* "dev" is not-null */
int i;
for (dptr = dev->data; dptr; dptr = next) { /* all the list items */
if (dptr->data) {
for (i = 0; i < qset; i++)
kfree(dptr->data[i]);
kfree(dptr->data);
dptr->data = NULL;
}
next = dptr->next;
kfree(dptr);
}
dev->size = 0;
dev->quantum = scull_quantum;
dev->qset = scull_qset;
dev->data = NULL;
return 0;
}
int scull_release(struct inode *inode, struct file *filp)
{
printk(KERN_DEBUG "process %i (%s) success release minor(%u) file\n", current->pid, current->comm, iminor(inode));
return 0;
}
/*
* Open and close
*/
int scull_open(struct inode *inode, struct file *filp)
{
struct scull_dev *dev; /* device information */
dev = container_of(inode->i_cdev, struct scull_dev, cdev);
filp->private_data = dev; /* for other methods */
/* If the device was opened write-only, trim it to a length of 0. */
if ( (filp->f_flags & O_ACCMODE) == O_WRONLY) {
if (mutex_lock_interruptible(&dev->mutex))
return -ERESTARTSYS;
scull_trim(dev); /* Ignore errors. */
mutex_unlock(&dev->mutex);
}
printk(KERN_DEBUG "process %i (%s) success open minor(%u) file\n", current->pid, current->comm, iminor(inode));
return 0;
}
/*
* The "extended" operations -- only seek.
*/
loff_t scull_llseek(struct file *filp, loff_t off, int whence)
{
struct scull_dev *dev = filp->private_data;
loff_t newpos;
switch(whence) {
case 0: /* SEEK_SET */
newpos = off;
break;
case 1: /* SEEK_CUR */
newpos = filp->f_pos + off;
break;
case 2: /* SEEK_END */
newpos = dev->size + off;
break;
default: /* can't happen */
return -EINVAL;
}
if (newpos < 0)
return -EINVAL;
filp->f_pos = newpos;
return newpos;
}
struct file_operations scull_fops = {
.owner = THIS_MODULE,
.llseek = scull_llseek,
.write = scull_write,
// .unlocked_ioctl = scull_ioctl,
.open = scull_open,
.release = scull_release,
};
/*
* Set up the char_dev structure for this device.
*/
static void scull_setup_cdev(struct scull_dev *dev, int index)
{
int err, devno = MKDEV(scull_major, scull_minor + index);
cdev_init(&dev->cdev, &scull_fops);
dev->cdev.owner = THIS_MODULE;
dev->cdev.ops = &scull_fops;
err = cdev_add (&dev->cdev, devno, 1);
/* Fail gracefully if need be. */
if (err)
printk(KERN_NOTICE "Error %d adding scull%d", err, index);
else
printk(KERN_INFO "scull: %d add success\n", index);
}
void scull_cleanup_module(void)
{
int i;
dev_t devno = MKDEV(scull_major, scull_minor);
/* Get rid of our char dev entries. */
if (scull_devices) {
for (i = 0; i < scull_nr_devs; i++) {
scull_trim(scull_devices + i);
cdev_del(&scull_devices[i].cdev);
}
kfree(scull_devices);
}
/* cleanup_module is never called if registering failed. */
unregister_chrdev_region(devno, scull_nr_devs);
printk(KERN_INFO "scull: cleanup success\n");
}
int scull_init_module(void)
{
int result, i;
dev_t dev = 0;
/*
* Get a range of minor numbers to work with, asking for a dynamic major
* unless directed otherwise at load time.
*/
if (scull_major) {
dev = MKDEV(scull_major, scull_minor);
result = register_chrdev_region(dev, scull_nr_devs, "scull");
} else {
result = alloc_chrdev_region(&dev, scull_minor, scull_nr_devs, "scull");
scull_major = MAJOR(dev);
}
if (result < 0) {
printk(KERN_WARNING "scull: can't get major %d\n", scull_major);
return result;
} else {
printk(KERN_INFO "scull: get major %d success\n", scull_major);
}
/*
* Allocate the devices. This must be dynamic as the device number can
* be specified at load time.
*/
scull_devices = kmalloc(scull_nr_devs * sizeof(struct scull_dev), GFP_KERNEL);
if (!scull_devices) {
result = -ENOMEM;
goto fail;
}
memset(scull_devices, 0, scull_nr_devs * sizeof(struct scull_dev));
/* Initialize each device. */
for (i = 0; i < scull_nr_devs; i++) {
scull_devices[i].quantum = scull_quantum;
scull_devices[i].qset = scull_qset;
mutex_init(&scull_devices[i].mutex);
scull_setup_cdev(&scull_devices[i], i);
}
return 0; /* succeed */
fail:
scull_cleanup_module();
return result;
}
module_init(scull_init_module);
module_exit(scull_cleanup_module);
#### Knowledge Point 1 -- Classification of Drivers
Drivers are divided into three categories: character devices, block devices and network interface. The above code is an example of character devices, and the other two will be discussed later.
As shown above, brw-rw-- -- the permission bar, block devices starts with "b" and the character devices starting with "c".
#### Knowledge Point 2 -- The Major and Minor Numbers
The major number is used to distinguish the driver. In general, the same major number indicates that it is controlled by the same driver.
Multiple devices can be created in one drive, distinguished by minor numbers. The major and minor numbers determine a driver device together (as shown above).
brw-rw---- 1 root disk 8, 0 Dec 17 13:02 sda
brw-rw---- 1 root disk 8, 1 Dec 17 13:02 sda1
The major number of equipment sda and sda1 is 8, and one minor number is 0 and the other minor number is 1.
#### Knowledge Point 3 -- How the Driver Provides the API
In my mind, the interface provided by the driver is /dev/xxx, and under Linux, "everything is about file", so the operation of the driver device is actually the operation of the file and the driver is used to define/open/read/write...what /dev/xxx will happen. The API of driver you can think is all about file operations.
What file operations are there? They are all defined in the file_operations structure of the kernel <linux/fs.h>[5] header file.
In the code I illustrated above:
struct file_operations scull_fops = {
.owner = THIS_MODULE,
.llseek = scull_llseek,
.write = scull_write,
.open = scull_open,
.release = scull_release,
};
I declare a structure and assign it. Except for the owner, the values of other members are function pointers.
Then I used cdev_add to register the file operation structure with each driver in the scull_setup_cdev function.
For example, if I perform "open" operation on the driver device, I will execute the scull_open function, which is equivalent to "hooking" the open function in the system call.
#### Knowledge Point 4 -- Generate the Corresponding Device under /dev
Compile the above code, get scull.ko, then sign it, and finally load it into the kernel via insmod.
Check if it is loaded successfully:
Although the driver has been loaded successfully, it does not create a device file in the /dev directory. We need to manually use mknod for device linking:
### Summary
In this example, there is no operation on the actual physical device, just simply use kmalloc to apply for a block of memory in the kernel space.
No more details about the code, which can be found by looking up the header files or Google.
Here I would like to share my way of learning the development of drivers: read books to understand the basic concept firstly, and then look up for the details when you need to use them.
For example, I don't need to know what API the driver can provide, and all I need to know is that the API provided by the drivers is all about file operations. As for the file operations, currently I only need to open, close, read and write. I will look up for more file operations when necessary.
### About Knownsec & 404 Team
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Knownsec's specialties include network attack and defense integrated technologies and product R&D under new situations. It provides visualization solutions that meet the world-class security technology standards and enhances the security monitoring, alarm and defense abilities of customer networks with its industry-leading capabilities in cloud computing and big data processing. The company's technical strength is strongly recognized by the State Ministry of Public Security, the Central Government Procurement Center, the Ministry of Industry and Information Technology (MIIT), China National Vulnerability Database of Information Security (CNNVD), the Central Bank, the Hong Kong Jockey Club, Microsoft, Zhejiang Satellite TV and other well-known clients.
404 Team, the core security team of Knownsec, is dedicated to the research of security vulnerability and offensive and defensive technology in the fields of Web, IoT, industrial control, blockchain, etc. 404 team has submitted vulnerability research to many well-known vendors such as Microsoft, Apple, Adobe, Tencent, Alibaba, Baidu, etc. And has received a high reputation in the industry.
The most well-known sharing of Knownsec 404 Team includes: KCon Hacking Conference, Seebug Vulnerability Database and ZoomEye Cyberspace Search Engine.
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# Unique factorization
Main Article
Discussion
Related Articles [?]
Bibliography [?]
Citable Version [?]
This editable Main Article is under development and subject to a disclaimer.
In mathematics, the unique factorization theorem, also known as the fundamental theorem of arithmetic states that every positive whole number can be expressed as a product of prime numbers in essentially only one way. For instance, ${\displaystyle 12=2\times 2\times 3}$, and 2 and 3 are prime numbers. In addition, if you manage to factor 12 into primes again, the only possibility is that there are three factors, two of them being 2 and one being 3 as above.
Unique factorization allows us to consider the prime numbers as "atoms" from which all other whole numbers can be assembled through multiplication. Unique factorization is the foundation for most of the structure of whole numbers as described by elementary number theory. For instance, the assumption that many electronic financial transactions are secure is based on the belief that the product of two very large prime numbers is difficult to factor. If several other prime factorizations were possible, perhaps some having small prime factors, finding a factorization might be much easier and such security methods would be ineffective.
## Applications
Unique factorization imparts useful structure to the whole numbers. For instance, it guarantees that any two positive whole numbers have a greatest common divisor (gcd) and a least common multiple(lcm). In fact, knowing the prime factorization of two numbers makes it much easier to find their gcd and lcm than not knowing. Also, unique factorization is implicit in the statement of many results, such as the Chinese remainder theorem.
## History
A result about unique factorization appeared in book IX of Euclid's Elements. However, it did not apply to all whole numbers, but only to those that can be written as the product of distinct primes.
The unique factorization property of the integers was implicitly understood by many intervening mathematicians, but most mathematics historians ascribe the first precise statement and proof this property to Carl Friedrich Gauss about 2000 years after Euclid. A precise statement and proof may be found on the "advanced" subpage.
Several factors contribute to the huge amount of time between the results of Euclid and Gauss. First, many mathematicians probably considered this to be an axiom of the integers that did not require proof — it was intuitively obvious to them. Second, a precise proof of the property from more fundamental principles is surprisingly difficult, so the mathematical abilities of many mathematicians before Gauss would not have been enough to produce a proof. Third, in his advanced work, Gauss encountered systems of "numbers" like the integers for which unique factorization did not hold. The failure of unique factorization to hold in other contexts showed that unique factorization cannot be taken for granted, and is a property that should be proved from more fundamental principles.
This failure of unique factorization in other contexts is a rather advanced topic, but a discussion of it is provided at the page on unique factorization domains.
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# Math Help - Another log question
1. ## Another log question
Alright, I'm trying this problem and I am completely lost. Are there any websites to understand logs? I went to the tutorial in the helpful math sites section and after I got so far into, I was really lost.
Anyways.. here's the question. Unlike the other one, I can't figure this out.
All of these have the subscript of 2.
log6-log15+log20.
I tried first dividing the 6 by 15, and then multiplying the answer by 20 and ended up with 8. However, the answer in the back is 3, so I know I'm doing something wrong? I just don't get these things...
2. Originally Posted by Chinnie15
Alright, I'm trying this problem and I am completely lost. Are there any websites to understand logs? I went to the tutorial in the helpful math sites section and after I got so far into, I was really lost.
Anyways.. here's the question. Unlike the other one, I can't figure this out.
All of these have the subscript of 2.
log6-log15+log20.
I tried first dividing the 6 by 15, and then multiplying the answer by 20 and ended up with 8. However, the answer in the back is 3, so I know I'm doing something wrong? I just don't get these things...
Hi!
$log_2(6) - log_2(15) + log_2(20) = log_2(\frac{6}{15}20) = log_2(8)$
So far you were right. However, you only know so far that this expression is equal to $log_2(8)$. Let's say $x=log_2(8)$, then we need to find x!
From the definition of logarithms,
$a^{log_a(n)} = n \Rightarrow 2^{log_2(8)} = 8 \Rightarrow log_2(8) = 3$, since $2^3 = 8$
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# Tag Info
2
If a body of mass m hanged on a string is moving, let uniformly, on a circle fixed relatively to the ground, then an observer G on the ground uses the 2nd Newton Law : $$\mathbf{F}=m\cdot \mathbf{a} \tag{01}$$ and finds the relation between the force $\mathbf{F}$ and the acceleration $\mathbf{a}$. For observer G there exists a "real" force, the tension ...
1
Note that you have to swing the pail with a certain minimum speed for the water to stay in. That minimum speed is such that when the pail is at the top of the arc, the rope accelerates the pail downward faster than gravity accelerates the water downward. Otherwise, the water falls out.
0
If water particles move in a circular path its because of some net force towards the centre. This net force is usually called "centripetal force". Don't ever put centrifugal force into the description. It is not a force, but just a name of the "feeling" that your body (or in this case water particles) want to move out of the circular motion but can't.
0
The loss of energy (efficiency<1) manifests itself in two ways Friction. When doing free body diagrams add frictional forces at the sliding contacts and find which coefficient of friction $\mu$ gives you the efficiency values you expect. Structural damping (hysteresis). This is more difficult to put in the equations of motion, because they assume ...
0
First of all, the transmission of force between any two pairs of mating gears happens due to surface contact between the two and hence through friction. Next, any loss is a dissipation in energy. Though I am not that well-versed in vibration analysis, I know that an periodic dissipation of energy is accounted in the force-equation using a damping ...
0
Newtons 2nd law only holds in an inertial frame. If you use the center of mass of the rolling object as your frame of reference then the frame will be accelerating and F=MA won't hold. However if you fix the frame of reference to a relatively stationary point, say the surface of the earth, it will. Even in this case the frame is not truly inertial, but ...
1
So the definition of angular momentum should be this: "Angular momentum is the product of the angular velocity of the body or system and its moment of inertia with respect to the rotation axis, and that is directed along the rotation axis". That's not a useful definition at all, because (i) it does not specify what this "moment of inertia" thing is, and ...
0
I think the error occurs where you state : Now because we found that both disks have the same amount of kinetic energy (and the same mass), that means that they have the same translational speed. (In fact, my professor also did a demo of this in class and we observed that they did have the same speed). The two discs have the same total KE but ...
2
The angular velocity is related to the tangential speed $v_t$ by: $$\omega = \frac{v_t}{r}$$ However the velocity $v$ being used here is not the tangential speed. It is the linear velocity $v_l$ of the centre of mass of the disk relative to the floor. The question has chosen a value for the angular velocity of the disk that appears similar to the ...
1
$\omega=\frac{v}{r}$ is used for rotating without slipping. Your disc is rotating and slipping. So, it is possible that $\omega=\frac{v}{3r}$.
3
I am not sure what you actually want to ask. So I would recommend that you put more effort into your question. Assuming that this is the setup that you have, a scissor in gray and the cross section of the rope in orange: Say the toque applied to the joint of the scissors is $\tau$. What is this force $F$, then? The distance of the rope contact point to ...
0
Rolling of a circular body, on a flat surface, and without sliding results in $v_\text{tan}=v_\text{cm}$ where $v_\text{tan}=ωr$ is the tangential speed of any point on the rim of the body in the center-of-mass frame of reference. This is understood by studying the motion in the c.m. frame: there, the flat surface has velocity $v_\text{cm}$ (backwards). The ...
0
Are you confused about how you get into this? \begin{align} v & = \omega\,r \\ ({\rm m/s}) & = ({\rm rad/s})\,({\rm m}) = ({\rm m\,rad/s}) \end{align} Radians are not units with dimensions. They can be seen as $({\rm rad}) = ({\rm m/m})$ like arc length to radius. This makes to above right hand side equivalent to the left hand side.
-1
This is all true because the ratio of a circle's diameter to circumference is constant (pi). The unit of radian is actually chosen so that l = r.theta. So if you go theta radians around a circle you travel r.theta. For the angular to linear velocities, think of a disc rotating at an angular speed omaga. then the further out from the centre of the disc you ...
0
@Prayas Agrawal No I'm not missing that - I'm just explaining the original error. Of course the total work = the KE. In fact since the floor is frictionless, the disc remains stationary and just spins since the force has no line of action through the CoM. In the case of friction the force of the weight of the disc provides a torque on the ground via the ...
1
"As we can see from the picture, both disks have the same force being applied to them and they also go the same distance d→" This is the erroneous assumption - the 2 discs do not go the same distance. Some of the distance that the rope is pulled will rotate disc 2 as it unravels. As a result the liner distance is less and the balance of work goes into ...
1
While you are correct saying disk 2 has rotational kinetic energy, you are missing that no matter the situation, since ground is frictionless, the work done be external force(F, in this case), is same in both cases. Thus by work energy theorem, $$Work=change in KE$$.Thus since work in case 1 equals work in case 2 thus both disk have same kinetic energies.
0
When connected to the centre of mass of disc 1 the force causes an acceleration of the centre of mass and the work done by the force is $Fd$ where $d$ is the displacement of the centre of mass and the force $F$. The translational kinetic energy of the disc increases by an amount $Fd$. When the force acts on the rim of disc 2 the centre of mass of the disc ...
0
When youare talking about angular momentum about any axis(inluding one that passes through center of mass), you carry out the "formula":$\vec{L}=\vec{L_{com}}+I_{com}\vec{\omega}$ where direction of $\vec{L_{com}}$ and $I_{com}\vec{\omega}$ is to be kept in mind, during vector summation.While we calculate $\vec{L}$, about COM, $\vec{L_{com}}$ becomes zero ...
0
When you talk about a rigid body the distance between any two points is fixed. That implies angular velocity w is same for all points. Now ony moment of inertia changes: I=Icm+md^2 [d is distance between cm and the point] This is the parallel axis theorem. Get I and angular momentum=Iw
0
The angular momentum will be different: however, you will be able to calculate it with the parallel axis theorem. Measure the distance from your new axis to the center of mass and call it $d$. Your new rotational inertia $I$ can be calculated from the rotational inertia around the center of mass $I_{\mathrm{cm}}$ using the formula: I = I_{\mathrm{cm}} + ...
1
The proof is very simple and comes right form the definition. let $R(t)$ be a rotation matrix as a function of time. $R$ is an orthogonal matrix so its inverse is equal to its transpose: $I = R(t)R^T(t).$ ($I$ is the identity matrix) The time derivative of the above equation is $0 = \frac{d[R(t)R^T(t)]}{dt} = ... 0 The 1.5 m radius you'll use for the moment of inertia. The 0.1 m radius of the axle will relate the linear speed to the angular speed. 1 But why must$d\hat{u}$be orthogonal to$\vec{\Omega}$too (i.e. be tangential to a circle orthogonal to$\vec{\Omega})$? To get such a precession there must be a clockwise torque in the plane of the screen acting on the system which means that the torque vector must be pointing into the screen. That torque produces a change in the angular momentum ... 1 I think sections 4.1.2 and 4.1.3 of this lecture on dynamics explains it by looking at each component separately. Since$\frac{{\rm d}}{{\rm d}t} \sin \theta = \dot{\theta} \cos\theta$and$\frac{{\rm d}}{{\rm d}t} \cos \theta = -\dot{\theta} \sin\theta$the components of${\rm d}u$are perpendicular to$u$. Our first step is to choose cartesian axes, ... 1 Assume that your external force$F$was applied at some distance$r$from the centre of mass of the spool. To satisfy the rolling condition the linear acceleration of the centre of mass$a$must equal the radius$R$of the spool times its angular acceleration about the centre of mass$\alpha$.$a = R \alpha\$ Just suppose that there is no friction and ...
Top 50 recent answers are included
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# A simple broblem on combinatorics
How many destinct permutations of the letters of the word BRILLIANT can be made?
×
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Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5. We Provide Step by Step Answer of Exercise, MCQs, Numericals Practice Problem Questions of Exercises, Subjective Upthrust and Archimedes Class-9, Visit official Website CISCE for detail information about ICSE Board Class-9 Physics.
## Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5
-: Select Exercise :-
Practice Problems I
Practice Problems II
Objective Questions
Subjective Questions
### Practice Problems I
Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5
Practice Problem 1:
Question 1.
A solid of density 2700 kgm’ and of volume 0.0015 m3 ¡s completely immersed in alcohol of density 800 kgm’.Calculate :
1. Weight of solid in SI system.
2. Upthrust on solid in SI system.
3. Apparent weight of solid in alcohol.
4. Will the apparent weight of solid be less or more, if it is immersed completely in brine solution? Give a reason, [g = 10 ms-2]
Density of solid = ρ= 2700 kgm3
Volume of solid = V = 0.0015 m3
Density of alcohol = ρ’ = 800 kgm3
1. Mass of solid = m = V x p
m = 0.0015 x 2700 = 4.05 kg
Weight of solid = mg = 4.05 x 10 = 40.5 N
2. Volume of alcohol displaced = Volume of solid
V = 0.0015 m3
Mass of alcohol displaced = m’ = V x p’
m’ = 0.0015 x 800 = m’ = 1.2 kg
Upthrust = Weight of alcohol displaced
= m’g= 1.2 x 10= 12N
3. Apparent weight of solid in alcohol
= Actual weight of solid – Upthrust
= 40.5 -12 = 28.5 N
4. When a solid is immersed completely in brine solution, then upthrust acts on it in upward direction, as a result, its apparent weight of solid will be less than actual weight of solid will be less than actual weight of solid.
Question 2.
A stone of density 3000 kgm3 is lying submerged in water of density 1000 kgm3. If the mass of stone in air is 150 kg, calculate the force required to lift the stone. [g = 10 ms2]
Density of stone =ρ = 3000 kgm3
Density of water = ρ’ = 1000 kgm3
Mass of stone = m = 150 kg
Acceleration due to gravity = g = 10 ms-2
Actual weight of stone = mg = 150 x 10 = 1500 N
Volume of water displaced = Volume of stone
V = 0.05 m3
Mass of water displaced = m’ = V x p’
m’ = 0.05 x 1000 = 50 kg
Upthrust = m’g = 50 x 10 = 500 N
Force required to lift the stone
= Actual weight of stone – upthrust
= 1500 -500 = 1000 N
Question 3.
A solid of area of cross-section 0.004 m2 and length 0.60 m is completely immersed in water of density 1000 kgm3. Calculate :
1. Wt of solid in SI system
2. Upthrust acting on the solid in SI system.
3. Apparent weight of solid in water.
4. Apparent weight of solid in brine solution of density 1050 kgm3.
[Take g = 10 N/kg; Density of solid = 7200 kgm3]
Area of cross-section of solid = A = 0.004 m2
Length of the solid = l = 0.60 m
Density of water = p’ = 1000 kgm3
Acceleration due to gravity = g = 10 ms-2
Density of solid = p = 7200 kgm3
(1) Volume of solid = V = A x l
V = 0.004 x 0.60 = 0.0024 m3
Mass of solid = m = V x p
m = 0.0024 x 7200 = 17.28 kg
Weight of the solid = mg = 17.28 x 10 = 172.8 N
(2) Volume of water displaced = Volume of solid
= V = 0.0024 m3
Mass of water displaced = m’ = V x p’
m’ = 0.0024 x 1000 = 2.4 kg
Upthrust = Weight of water displaced
= m’g = 2.4 x 10 = 24 N
(3) Apparent weight of solid=Actual weight of solid – upthrust
= 172.8-24= 148.8 N
(4) Density of brine solution =ρb= 1050 kgm3
Volume of brine solution displaced = Volume of solid = V
V = 0.0024 m3
Mass of brine solution displaced
= mb = V x ρb = 0.0024 x 1050
mb = 2.52 kg
Upthrust acting on solid in brine solution = Weight of brine solution displaced -mbg
= 2.52 x 10 = 25.2 N
Apparent weight of solid in brine solution
= Actual weight – Upthurst
= 172.8-25.2= 147.6 N
Practice Problem 2: (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
Question 1.
A solid of density 7600 kgm3 is found to weigh 0.950 kgf in air. If 4/5 volume of solid is completely immersed in a solution of density 900 kgm3, find the apparent weight of solid in liquid.
Weight of solid in air = 0.950 kgf
∴ Mass of solid in air = m = 0.950 kg
Density of solid =ρ = 7600 kgm3
Apparent weight of solid in liquid
= Actual weight – Upthrust
= 0.950-0.09 = 0.860 kgf
Question 2.
A glass cylinder of length 12 x 10-2 m and area of crosssection 5 x 10-4 m2 has a density of 2500 kgm-3. It is immersed in a liquid of density 1500 kgm-3, such that 3/8. of its length is above liquid. Find the apparent weight of glass cylinder in newtons.
Length of glass cylinder = l = 12 x 10-2 m
Area of cross-section = A = 5 x 10-4 m2
Volume of glass cylinder = V = A x l
V= 5 x lo-4 x 12 x 10-2
V= 0.00006 m3
Acceleration due to gravity = g = 9.8 m/s2
Density of glass cylinder = ρ = 2500 kgm3
Mass of glass cylinder = m = V x ρ
m = 0.00006 x 2500 m
= 0.15 kg
Weight of glass cylinder = mg = 0.15 x 10=1.5 N
Mass of liquid displaced by glass cylinder = V’ x ρ’
m’= 0.0000375 x 1500
m’ = 0.05625 kg
Upthrust = Weight of liquid displaced by the glass cylinder
= m’g = 0.05625 x 10 = 0.5625 N
Apparent weight of glass cylinder in liquid = Actual weight of glass cylinder – Upthrust
= 1.5 – 0.5625 = 0.9375 N
Practice Problems 3:
Question 1.
A solid weighs 0.08 kgf in air and 0.065 kgf in water.Find
(1) R.D. of solid
(2) Density of solid in SI system. [Density of water = 1000 kgm3]
Weight of solid in air = 0.08 kgf
Weight of solid in water = 0.065 kgf
Density of water = 1000 kgm3
(1) Relative density (RD) of solid
= Weight of solid in air
wt. of solid in air-wt. of solid in water
Question 2.
A solid of R.D. = 2.5 is found to weigh 0.120 kgf in water. Find the wt. of solid in air.
Relative density of solid = R.D. = 2.5
Weight of solid in water = W’ = 0.120 kgf
Weight of solid in air = W = ?
Question 3.
A solid of R.D. 4.2 is found to weigh 0.200 kgf in air. Find its apparent weight in water.
Relative density of solid = R.D. = 4.2
Weight of solid in air = W = 0.200 kgf
Practice Problems 4:
Question 1.
A sinker is found to weigh 56.7 gf in water. When the sinker is tied to a cork of weight 6 gf, the combination is found to weigh 40.5 gf in water. Calculate R.D. of cork.
Weight of sinker in water = 56.7 gf
Weight of cork = 6 gf
Wt. of sinker in water + Wt. of cork in air
= 56.7 + 6 = 62.7 gf …(1)
Wt. of cork in water + Wt. of sinker in water = 40.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of cork in air – Wt. of cork in water = 62.7 – 40.5 = 22.2 gf
Question 2. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A solid lighter than water is found to weigh 7.5 gf in air. When tied to a sinker the combination is found to weigh .If the sinker alone weighs 72.5 gf in water, find R.D. of solid.
Weight of solid in air = 7.5 gf
Weight of sinker in water = 72.5 gf
Wt. of sinker in water + Wt. of solid in air
= 72.5 + 7.5 = 80.0 gf …(1)
Wt. of solid in water + Wt. of sinker in water = 62.5 gf …(2)
Subtract eq. (1) from eq. (2)
Wt. of solid in air – Wt. of solid in water = 80- 62.5 = 17.5 gf.
Practice Problems 5:
Question 1.
An aluminium cube of side 5 cm and RD. 2.7 is suspended by a thread in alcohol of relative density 0.80. Find the tension in thread.
Side of an aluminium cube = l = 5 cm
Volume of aluminium cube = V = Z3 = (5)3 =125 cm3
Relative density of aluminium = R.D. = 2.7
Relative density of alcohol = R.D. = 0.80
Density of water = 1 g cm-3
Density of aluminium =ρ = 2.7 g cm-3
Mass of aluminium = V x ρ
m = 125 x 2.7 = 337.5 g
Wt. of aluminium cube acting downwards = 337.5 gf
Volume of alcohol displaced = Volume of cube = V = 125 cm3
Upthrust due to alcohol = V x ρalcohol x g
Question 2.
A cube of lead of side 8 cm and R.D. 10.6 is suspended from the hook of a spring balance. Find the reading of spring balance. The cube is now completely immersed in sugar solution of R.D. 1.4. Calculate the new reading of spring balance.
Length of side of cube = l = 8 cm
Volume of cube =β= (8)3 = 512 cm3
V = 512 cm3
Relative density of lead cube = R.D. = 10.6 Relative density of sugar solution = R.D. 1.4
Density of water = 1 g cm-3
Practice Problems II
Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5
Practice Problems 1:
Question 1.
A hollow cylinder of copper of length 25 cm and area of cross-section 15 cm2, floats in water with 3/5 of its length inside water. Calculate :
(1) apparent density of hollow copper cylinder.
(2) wt. of cylinder.
(3) extra force required to completely submerge it in water.
(1)
Length of hollow cylinder of copper = hcu= l= 25 m
Length of hollow cylinder of copper inside water
Question 2.
A cork cut in the form of a cylinder floats in alcohol of density 0.8 gcm-3, such that 3/4 of its length is outside alcohol. If the total length of cylinder is 35 cm and area of cross-section 25 cm2, calculate :
(1) density of cork
(2) wt. of cork
(3) extra force required to submerge it in alcohol
Practice Problems 2:
Question 1.
A cylinder made of copper and aluminium floats in mercury of density 13.6 gem-3, such that 0.26th part of it is below mercury. Find the density of solid.
Density of mercury = ρHg =13.6 g.cm-3
Density of solid cylinder = ρsolid = ?
0.26th part of the cylinder is below mercury
Let Vsolid = Volume of solid cylinder
Volume of mercury displaced by immersed part of the solid cylinder
Question 2.
An iceberg floats in sea water of density 1.17 g cm 3, such that 2/9 of its volume is above sea water. Find the density of iceberg.
Density of sea water =ρw = 1.17 g cm-3
Density of solid ice berg =ρi = ?
Practice Problems 3: (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
Question 1.
A wooden block floats in alcohol with 3/8 of its length above alcohol. If it is made to float in water, what fraction of its length is above water? Density of alcohol is 0.80 g cm-3.
Let length of wooden block = x
Question 2.
A hollow metal cylinder of length 10 cm floats in alcohol of density 0.80 g cm-3, with 1 cm of its length above it. What length of cylinder will be above copper sulphate solution of density 1.25 g cm-3?
Length of hollow metal cylinder = x= 10 cm
1 cm length of cylinder is above the alcohol
∴Length of the cylinder below alcohol = (10-1) = 9cm
Density of alcohol =ρalcohol = 0.80 g cm-3
Density of copper sulphate solution = ρCuS04 = 1.25 g cm-3
When block floats in alcohol By the law of floatation :
Length of metal block below copper sulphate solution = 5.76 cm
So, length of metal block above copper sulphate solution
= 10 – 5.76 = 4.24 cm
Practice Problems 4:
Question 1.
What fraction of an iceberg of density 910 kgm-3 will be above the surface of sea water of density 1170 kgm-3 ?
Ans.
Let volume of iceberg = Vi = x
Volume of iceberg inside sea water = Vw
Density of iceberg = ρi = 910 kmg-3
Density of sea waer = ρw = 1170 kgm-3
By law of floatation:
Weight of icebeig = Weight of sea water displaced by the iceberg
Question 2.
What fraction of metal of density 3400 kgm-3 will be above the surface of mercury of
density 13600 kgm-3, while floating in mercury?
Density of metal =ρm = 3400 kgm-3
Density of mercury = pHg = 13600 kgm-3
Let volume of metal = x
and volume of metal inside mercury = y
By law of floatation:
Weight of mercury displaced by metal = wt. of metal
Practice Problems 5:
Question 1.
A balloon of volume 1000 m3 is filled with a mixture of hydrogen and helium of density 0.32 kgm-3. If the fabric of balloon weighs 40 kgf and the density of cold air is 1.32 kgm-3, find the tension in the tope, which is holding the balloon to ground.
Volume of balloon = V = 1000 m3
Density of mixture of hydrogen and helium =ρ = 0.32 kgm3
Density weight of empty balloon = 40 kgf
Density of cold air = p’ = 1.32 kgm3
Volume of balloon=Volume of mixture of hydrogen and helium gas
= Volume of cold air displaced by balloon
= V= 1000 m3
Weight of mixture of hydrogen and helium gas in balloon = Vρg
= 1000 x 0.32 x g = 320 kgf
Down thrust = Weight of empty balloon + Weight of mixture of hydrogen and helium gas
= 40+ 320 = 360 kgf
Upthrust = Weight of cold air displaced by balloon = Vρ’g
= 1000 x 1.32 x g= 1320kgf
Tension in the rope = Upthrust – downthrust
= 1320-360 = 960 kgf
Question 2.
A balloon of volume 800 cm3 is filled with hydrogen gas of density 9 x 10-5 gem-3. If the empty balloon weighs 0.3 gf and density of air is 1.3 x 10-3 gem-3, calculate the lifting power of balloon.
Volume of balloon = V = 800 cm3
Density of hydrogen gas = pH = 9 x 10-5 g cm-3
Weight of empty balloon = 0.3 gf
Density of air = ρa = 1.3 x 10-3 g cm-3
Weight of hydrogen gas in balloon = VρHg
= 800 x 9 x 10-5 x g
= 72 x 10-3gf
= 0.072 gf
Volume of balloon = Volume of hydrogen gas in the balloon = Volume of air displaced by balloon.
Downthrust = Wt. of empty balloon + wt. of hydrogen gas in balloon
= 0.3 + 0.072 = 0.372 gf
Upthrust = Wt. of air displaced by balloon
= Vρag
= 800 x 1.3 x 1o-3 x g
= 8 x 1.3 x 1o-1 x g
= 10.4 x 10-1 x g = 1.04 gf
Lifting power of balloon = Upthrust – Down thrust
=1.04-0.372 = 0.668 gf
Question 3.
A balloon of volume 120 m3 is filled with hot air, of density 38 kg-3. If the fabric of balloon weighs 12 kg, such that an additional equipment of wt. x is attached to it, calculate the magnitude of Density of cold air is 1.30 kgm-3.
Volume of balloon = V = 120 m3
Density of hot air = ρhot air = 0.38 kgm-3
Mass of empty balloon = 12 kg
Weight of the empty balloon = 12 kgf
Weight of the additional equipment attached with the balloon
=x kgf
Density of cold air = ρcoldair = 1.30 Kgm3
Volume of balloon=Volume of hot air inside the balloon=Volume
of cold air displaced by balloon = V = 120 mWeight of hot air = Vphotair g
= 120 x 0.38 x g = 45.6kgf
Weight of empty balloon + Weight of hot air inside the balloon +
Weight of equipment = Downthrust
12 + 45.6 + x = Downthrust
Downthrust = 57.6 + x
Upthrust = Weight of cold air diplaced by balloon
=Vρcoldair g
= 120 x 1.30 xg= 156kgf
By law of floatation :
Downthrust = Upthrust
57.6+x= 156
x= 156-57.6
x =98.4 kgf
Practice Problems 6: (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
Question 1.
A test tube weighing 17 gf, floats in alcohol to the level P. When the test tube is made to float in water to the level P, 3 gf of the lead shots are added in it. find the R.D. of alcohol.
When tube floats in alcohol :
Weight of test tube = 17 gf
By law of floatation :
Weight of alcohol displaced by test tube = Weight of test tube = 17 gf
When tube floats in water :
When test tube is made to float in water to the same level, as in alcohol then 3g lead stones are added in it.
.’. Weight of test tube = 17 gf + 3 gf = 20 gf
Weight of water displaced by test tube = 20 gf
Volume of alcohol displaced = Volume of water displaced
Question 2.
A test tube loaded with lead shots, weighs 150 gf and floats upto the mark X in water. The test tube is then made to float in alcohol. It is found that 27 gf of lead shots have to be removed, so as to float it to level X. Find R.D. of alcohol.
When tube floats in water :
Weight of test tube = 150 gf
By law of floatation:
Weight of water displaced = Weight of test tube = 150 gf
When test tube floats in alcohol :
When test tube is made to float in alcohol, then 27 gf of lead shots have to removed, so that it can float upto the same level as in water.
∴Weight of test tube in alcohol = 150 – 27 = 123 gf
By law of floatation:
Weight of alcohol displaced by test tube = Weight of test tube in alcohol = 123 gf
As volume of alcohol displaced = Volume of water displaced
∴ R.D. of alcohol
(A) Objective Questions
Multiple Choice Questions.
### Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5
Select the correct option.
1.The force experienced by a body when partially or fully immersed in water is called :
(a) aparent weight
(b) upthrust
(c) down thrust
(d) none of these
2. When a body is floating in a liquid :
(a) The weight of the body is less than the upthrust due to immersed part of the body
(b) The weight of body is more than the upthrust due to the immersed part of the body
(c) The weight of body is equal to the upthrust due to the immersed part of the body
(d) none of the above
3. With the increase in the density of the fluid, the upthrust experienced by a body immersed in it :
(a) decreases
(b) increases
(c) remains same
(d) none of these
4. The apparent weight of a body in a fluid is :
(a) equal to weight of fluid displaced
(b) volume of fluid displaced
(c) difference between its weight in air and weight of fluid displaced
(d) none of the above
5. The phenomenon due to which a solid experiences upward force when immersed in water is called :
(a) floatation
(b) buoyancy
(c) density
(d) none of these
6. When an object sinks in a liquid, its :
(a) buoyant force is more than the weight of object
(b) buoyant force is less than the weight of object
(c) buoyant force is equal to the weight of the object
(d) none of the above
7. The SI unit of density is :
(a) gem-3
(b) kgem-3
(c) kgm-3
(d) gm-3
8. When a body is wholly or partially immersed in a liquid, it experiences a buoyant force which is equal to :
(a) volume of liquid displaced by it
(b) weight of liquid displaced by it
(c) both (a) and (b)
(d) none of the above
9. The ratio between the mass of a substance and the mass of an equal volume of water at 4°C is called :
(a) relative density
(b) density
(c) weight
(d) pressure
10. A body has density 9.6 gcm -3. Its density in SI system is :
(a) 96 kgm-3
(b) 960 kgm-3
(c) 9600 kgm-3
(d) 96,000 kgm-3
Ans:
Explanation :
Density = 9.6 g cm-3
(B) Subjective Questions
### Goyal Brothers Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5
Question 1.
A wooden block floats in water with two third of its volume submerged.
(1) Calculate density of wood.
(2) When the same block is placed in oil, three quarter of its volume is immersed in oil. Calculate the density of oil.
Question 2.
A.metal cube of 5cm edge and relative density 9 is suspended by a thread so as to be completely immersed in a liquid of relative density 1.2. Find the tension in the thread.
Volume of metal cube = (side)3 = 53 = 125cm3
Density of cube = 9 g cm-3
Weight of cube acting downward = mg = v × d
F↓= 125 x 9 = 1125 gf
Density of liquid dL– 1.2 g cm-3
.’. Upthrust due to liquid in the upward direction.
F2↑ = v dL= 125 x 1.2 = 150.0 gf Tension in the string = Net downward force = F – F2
= 1125- 150 = 975 gf= 9.75 N
Question 3.
A weather forecasting plastic balloon of volume 15 mcontains hydrogen of density 0.09 kgm-3. The volume of equipment carried by the balloon is negligible compared to its own volume. The mass of the empty balloon is 7.15 . kg. The balloon is floating in air of density 1.3 kgm-3.
(1) Calculate the mass of hydrogen in balloon.
(2) Calculate the mass of hydrogen and the balloon.
(3) If the mass of equipment is x kg, write down the total mass of hydrogen, the balloon and the equipment,
(4) Calculate the mass of air displaced by balloon.
(5) Using the law of floatation, calculate the mass of equipment.
Volume of Hydrogen V = 15 m3
Density of hydrogen = d= 0.09 kg m-3
(1) Mass of hydrogen in balloon = Vd= 15 × 0.09
= 1.35 kg
The mass of empty balloon alone = 7.15 kg
(2) The mass of hydrogen and balloon = 1.35 + 7.15
= 8.50 kg
Mass of equipment = x kg
(3) Total mass of hydrogen + Balloon + Equipment
= (8.50 + :c) kg
Density of air = 1.3 kg m-3
(4) Mass of air displaced by balloon = v x d=15 x 1.3
= 19.5 kg
(5) According to law of floatation
Total downward wt. = UPTHRUST
8.5+x= 19.5
Mass of equipment x = 11 kg
Question 4.
(a) State the principle of floatation.
(b) The mass of a block made of certain material is 1.35 kg and its volume is 1.5 x 10-3 m3.
1. Find the density of block.
2. Will this block float or sink? Give reasons for your answer.
(a) PRINCIPLE OF FLOATATION : “When a solid is floating in a fluid, the weight of whole solid acting vertically downward at its CENTRE OF GRAVITY, is equal to the weight of fluid displaced by the IMMERSED part of solid acting upward, at its CENTRE OF BUOYANCY or at the centre of the BULK OF LIQUID displaced.”
OR
“The weight of a floating body is equal to the weight of the liquid displaced by its SUBMERGED part.”
(b) Mass of block = m = 1.35 kg
Volume of block = V = 1.5 x 10-3 m3
(1)
(2) Density of block (900 kgm-3) is less than density of water (1000 kgm-3)
.’. Block will float in water
Question 5.
(a) State Archimedes’ Principle.
(b) A block of mass 7 kg and volume 0.07 m3 floats in a liquid of density 140 kg/m3. Calculate :
1. Volume of block above the surface of liquid.
2. Density of block.
(a) ARCHIMEDES’ PRINCIPLE : “Whenever a body is immersed in a liquid (fluid), wholly or partially, it loses weight equal to the weight of liquid displaced by it.”
(b) Mass of block = m = 7 kg
Volume of block = V = 0.07 m3
Density of liquid =ρl = 140 kgm-3
Let V’ = Volume of block immersed in the liquid
By law of floatation:
Weight of block = Weight of liquid displaced by the immersed pa
Question 6. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
(a) A body whose volume is 100 cm3 weighs 1 kgf in air. Find its weight in water.
(b) Why is it easier to swim in sea water than in river water?
(a) Volume of body = V = 100 cm3 = 10-4 m3
Weight of body in air = 1 kgf
Density of water = pw = 1000 kgm3
We know volume of water displaced = Volume of body
= V = 10-4 m3
Upthrust = Weight of water displaced by body = Vpwg
= 10-4 x 1000 x g .
= 10-1 kgf =0.1 kgf
Weight of body in water=Weight of body in air – Upthrust
= 1-0.1 =0.90 kgf
(b) With smaller portion of man’s body submerged in sea water, the wt. of sea water displaced is equal to the total weight of body.
While to displace the same weight of river water, a larger portion of the body will have to be submerged in water. It is easier for man to swim in sea water.
Question 7.
Why does a ship made of iron not sink in water, while an iron nail sinks in it?
Density of iron is more than density of water, therefore weight of iron nail is more than wt. of water displaced by it and nail SINKS. While shape of iron ship is made in such a way that it displaces MORE WEIGHT OF WATER than its own weight. Secondly the ship is HOLLOW and THE EMPTY SPACE contains AIR which makes the AVERAGE DENSITY OF SHIP LESS THAN THAT OF WATER and hence ship floats on water.
Question 8.
A solid of density 5000 kgm-3 weighs 0.5 kgf in air. It is completely immersed in a liquid of density 800 kgm-3. Calculate the apparent weight of the solid in liquid.
Density of solid, ds = 5000 kg m-3
Weight of body in air = 0.5 kgf
mg = 0.5 kgf
∴ m – 0.5 kg
(1) Apparent weight of the solid in water = 0.5 – 0.08
= 0.42 kgf
(2) Apparent weight of body in liquid of density 800 kg m-3 is zero.
Density of solid is less than density of liquid i.e. upthrust is more than weight of body.
Question 9.
(a) A body dipped in a liquid experiences an upthrust. State the factors on which the upthrust depends
(b) While floating, is the weight of body greater than, equal to or less than upthrust?
Ans.
(a) Factors on which upthrust depends are :
1. Volume of body immersed in fluid.
Upthrust is maximum when body completely immersed in the fluid.
2. Density of the fluid.
Upthrust α density of fluid
Larger the density of the fluid, large will be the upthrust acting on the body.
(b) When the body floats then weight of the body is equal to the upthrust acting on the body.
Question 10.
A sinker is first weighed alone under water. It is then tied to a cork and again weighed under water. In which of the two cases weight under water is less and why?
Weight of sinker, when tied to a cork, under water is less than that when it is alone weighed under water. Because cork displaces more water than its own weight and hence large upthrust acts on the sinker.
Question 11. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A solid weighs 105 kgf in air. When completely immersed in water, it displaces 30,000 cm3 of water, calculate relative density of solid.
Weight of solid in air =105 kgf
Volume of solid = Volume of water displaced = 30000 cm3
= 30000 x 10-6 m3 = 0.03 m3
pw = Density of water = 1000 kgm-3
Wt. of water displaced by solid .
Question 12. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A test tube loaded with lead shots weighs 25 gf and floats upto the mark X in water. When the test tube is made to float in brine solution, it needs 5 gf more of lead shots to float upto level X. Find the relative density of brine solution.
When test tube floats is water :
Weight of test tube = 25 gf
By law of floatation
Weight of water displaced = Weight of test tube = 25 gf
When test tube floats in brine solution, it needs 5 gf more of lead shots to float upto same level as in water.
Weight of test tube = 25 + 5 = 30 gf
By law of floatation :
Weight of brine solution displaced = Weight of test tube = 30 gf As, volume of brine solution displaced = Volume of water displaced
Question 13.
A wooden block is weighed with iron, such that combination just floats in water at room temperature. State your observations when :
(1) water is heated above room temperature
(2) water is cooled below 4°C. Give reasons to your answers in (1) and (2).
(1) We know density of water decreases with rise in temperature and hence upthrust decreases.
(2) Density of water is maximum at 4°C. When water cooled below 4°C, then its density and hence upthrust acting on it decreases. So, wooden block weighed with iron, sinks more than earlier.
Question 14
A rubber ball floats in water with 2/7 of its volume above the surface of water. Calculate the average relative density of rubber ball.
Let volume of rubber ball = V
Question 15. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A cube of ice whose side is 4.0 cm is allowed to melt. The volume of water formed is found to be 58.24 cm3. Find the density of ice.
Side of ice cube = l = 4 cm
Volume of ice cube = V = β = (4)3 = 64 cm3
Volume of water = V = 5824 cm3
Density of ice cube = ρi= ?
Density of water = ρw = 1 gem-3
By law of floatation:
Volume of ice cube x Density of ice=Volume of water x Density of water
64 x ρi = 58.24 x 1
Question 16.
A jeweller claims to make ornaments of pure gold of relative density 19.3. A customer buys from him a bangle of weight 25.25 gf. The customer then weighs the bangle under water and finds its weight 23.075 g with the help of suitable calculations explain whether the bangle is of pure gold or not.
[R.D. of gold is 11.61 and bangle is not of pure gold] Ans. R.D of pure gold = 19.3 …given
Question 17.
(a) When a piece of ice floating in water melts, the level of water inside the glass remains same. Explain.
(b) An inflated balloon is placed inside a big glass jar which is connected to an evacuating pump. What will you observe when the evacuating pump starts working? Give a reason for your answer.
(a) A piece of ice displaces an amount of water equal to its own weight. Volume of water displaced is equal to volume of submerged part of ice cube. When ice cube melts, its volume decreases and gets occupied in that volume of water which is displaced by it. As a result, level of water inside the glass remains same when piece of ice (ice cube) melts.
(b) When evacuating pump starts working, pressure inside the glass jar reduces. As the pressure inside the balloon is more than pressure outside the balloon inside the glass jar, so balloon with burst.
Question 18.
(a) A trawler is fully loaded in sea water to maximum capacity. What will happen to this trawler, if moved to river water? Explain your answer.
(b) A body of mass 50 g is floting in water. What is the apparent weight of body in water? Explain your answer.
(a) Density of sea water is more than the density of river water.So river water offers less upthrust to the trawler as compared to sea water.So, when a trawler is fully loaded sea water to maximum capacity, is moved to river water, it will sink.
(b) Mass of body = 50 g
Apparent weight of body = Weight of body in air – Weight of water displaced by body.
But when a body floats, then weight of body in air is equal to the weight of water displaced by the body.
=> Apparent weight of the body = 0
Question 19.
A body of mass ‘m’ is floating in a liquid of density ‘p’
(1) what is the apparent weight of body?
(2) what is the loss of weight of body?
Mass of body = m
Density of liquid = ρ
(1) Apparent weight of body = Weight of body in air – Weight of liquid displaces by body.
When a body floats in the liquid, then weight of the body in a liquid is equal to weight of liquid displaced by the body.
=> Apparent weight of body = 0
(2) Loss in weight of body is equal to the weight of liquid displaced by the body.
Question 20. (Goyal Brothers Upthrust and Archimedes’ Principle Class-9)
A block of wood of volume 25 cm3 floats in water with 20 cm3 of its volume immersed in water.
Calculate :
(1) density of wood
(2) the weight of block of wood.
Volume of wooden block = V = 25 cm3
Volume of wooden block immersed in water = 20 cm3
Volume of water displaced by wooden block=volume of wooden
block immersed in water = 20 cm3
Density of water = ρwater = 1 gcm-3
Density of wooden block =ρwood = ?
By law of floatation:
Volume of wooden block x Density of wood=Volume of water displaced x Density of water
Question 21.
A solid body weighs 2.10 N. in air. Its relative density is 8.4. How much will the body weigh if placed
(1) in water,
(2) in liquid of relative density 1.2?
Weight of solid body in air = 2.10 N
R.D. of solid = 8.4
Weight of body in water = Weight of body in air – Weight of water displaced by body
= 2.10 -0.25 = 1.85 N
(2) Upthrust due to water = Weight of water displaced by body
= 0.25 N
Upthrust due to liquid = Upthrust due to water ×R.D. of liquid
= 0.25 x 1.2 = 0.30 N
Weight of body in liquid = Weight of body in air – Upthrust due to liquid
= 2.10-0.30= 1.80 N
-: End of Goyal Brothers : Upthrust and Archimedes’ Principle Class-9 ICSE Physics Ch-5:-
Thanks
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# Music Theory/Fundamentals of Common Practice Music
## Notation
This section is intended for those without any experience with Western music notation and its associated practices.
Humans with the ability to hear experience sound when delicate structures inside their ears detect waves of pressure traveling through the air (or any other medium to which they are mechanically coupled). When these waves are regular1 and oscillates at a constant speed, we recognize a tone or note. 2 Notes are the basic elements of Western music. They are the most common thing we hear in "songs," in almost every genre, period, or style of music in existence (in western society).
Before we talk about how musicians know what notes to "play," we need to determine one thing. With the exception of the human voice and electronic instruments, how many notes should a piano, for instance, play? How many notes should a violin play? These are important questions to answer because if one instrument can play, say, 15 notes, but another can only play 12, then the notes we hear are going to be different, which means musicians cannot collectively play the same notes of a given song. To find the optimal number, we need to balance "our wants versus our limits." In music, we want as many diverse notes as possible. However, instruments that we cannot make will never allow us to make any "organized sounds." Ergo, we want to follow this formula:
${\displaystyle {\text{limations of our instruments}}\leq {\text{number of notes}}\leq {\text{diversity of notes}}}$
It turns out, unsurprisingly, we found our answer: 12 different tones. However, what does "different" mean in this context?
## Pitch
What are "different notes"? To answer that, we need to look back at the definition of "notes". Looking carefully, we find that a note is simply a sound we hear after air molecules are vibrated. The air molecules are vibrated by sound waves. Ergo, we define a tone as a sound wave that oscillates some number of times per second at a constant rate. Let's call the number of oscillations per a given amount of time a frequency. What happens if we increase the frequency of a note? Well, we make it oscillate faster. However, since oscillations determine the note we hear, we, therefore, have a "different note." Eureka, a "different note" simply means a different frequency of a note. In fact, based on the following information, we can create a model:
A frequency determines the "sound" we hear from a "note." To measure a frequency, we use a unit called Hertz (${\displaystyle {\text{Hz}}}$), which is equal to one oscillation per second. Let's define a "common note" at ${\displaystyle 440{\text{ Hz}}}$.3 Compacting the peaks of the wave, we find that the frequency increases or the number of oscillations per second, ${\displaystyle {\text{Hz}}}$, increases; we hear this "highness" of the original note of ${\displaystyle 440{\text{ Hz}}}$. Separating the peaks of the wave, the number of oscillations per second, ${\displaystyle {\text{Hz}}}$, decreases; the resulting sound we hear is of a "lower" note of the original ${\displaystyle 440{\text{ Hz}}}$.
A "low" note or a "high" is the number of Hertz that is either deducted or added onto the original note. The "lowness" or "highness" of a note is the pitch. Although this is not a rigorous definition, according to philosophy, for our purposes this is good enough.
The next time you hear a series of notes, think of all the science at play. When we hear series of higher and lower notes, either singly or in groups, perhaps spaced out at different points in time and of different durations, with the variation in their frequencies, spacing, and lengths patterned in ways that correspond to established rules, we recognize the sound of music.
## Writing and Analysis
Written music is presented on one or more staves, which represent a set of notes that may be played in time across one or more instruments. Most contemporary music will consist of a pair of staves, as per this example, showing treble and bass clefs. Staves are divided into regular periods, known as measures or bars. The duration of a measure in a given piece of music is specified with a time signature, which is shown next to the clef. Contemporary music often uses 4/4 and 3/4 time signatures.
Sometimes, music written for fretted instruments such as guitars will be presented as a tablature rather than a standard musical notation on a staff. A tablature is a more direct way of showing how the instrument should be played, listing finger positions rather than notes. Keyboarded instruments such as the piano offer a simple translation from musical notes to finger positions, while stringed or valved instruments require the player to be able to translate notes into positions along strings, or specific valves that must be opened.
## Footnotes
1. Imagine a wave in the ocean but two-dimensional. In that wave, you have an amplitude – the highest displacement of a wave compared to when it is "flat" – and a trough – the lowest displacement of a wave compared to when it is "flat." Say you ask the following question: "How many ripples oscillate in that wave every second?" This question is very often asked that scientists have defined this as a term called frequency, which is again, the number of oscillations that happen in a wave every given amount of time. Ergo, when the waves are oscillating at a frequency that is non-changing, we call that wave regular.
2. The explanation of waves presented here is a little simplified. A sound wave is called a longitudinal wave, which is when the oscillation of a wave is parallel to the direction it is traveling in. All waves must originate from a vibration somewhere, with soundwaves being no exception. The reason we hear sounds at all is due to the vibrations of air molecules. If we did not have air molecules or water molecules in the air, sound cannot be heard. This is the reason why you cannot hear anything in space.
3. We picked the "common note" arbitrarily.
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Extending the basis of a T-invariant subspace
1. Sep 11, 2013
Bipolarity
Let $T: V → V$ be a linear map on a finite-dimensional vector space $V$.
Let $W$ be a T-invariant subspace of $V$.
Let $γ$ be a basis for $W$.
Then we can extend $γ$ to $γ \cup S$, a basis for $V$, where $γ \cap S = ∅$, so that $W \bigoplus span(S) = V$.
My question:
Is $span(S)$ a T-invariant subspace of $V$?
I've been trying to prove it is, but am not sure. I would like some assistance, so I know where I might steer my proof. Thanks!
EDIT: Never mind just solved it. It's false!
BiP
Last edited: Sep 11, 2013
2. Sep 11, 2013
Erland
This is not true. As a simple counterexample, let $V=\Bbb R^2$, and let ($\bf e_1,e_2$) be the standard basis in $\Bbb R^2$, and let $W=span\{\bf e_1\}$, and let $T$ be defined by its action on the basis vectors: $T(\bf e_1\rm)=\it T(\bf e_2\rm)=\bf e_1$. Then, $W$ is $T$-invariant, and we can take $S=\{\bf e_2\it\}$, and it satisfies your assumptions, but $span \,S$ is not $T$-invariant.
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# Find arrays with distinct subarray sums
Consider an array A of length n. The array contains only integers in the range 1 to s. For example take s = 6, n = 5 and A = (2, 5, 6, 3, 1). Let us define g(A) as the collection of sums of all the non-empty contiguous subarrays of A. In this case g(A) = [2,5,6,3,1,7,11,9,4,13,14,10,16,15,17]. The steps to produce g(A) are as follows:
The subarrays of A are (2), (5), (6), (3), (1), (2,5), (5,6), (6,3), (3,1), (2,5,6), (5,6,3), (6,3,1),(2,5,6,3), (5,6,3,1), (2,5,6,3,1). Their respective sums are 2,5,6,3,1,7,11,9,4,13,14,10,16,15,17.
In this case all the sums are distinct.
However, if we looked at g((1,2,3,4)) then the value 3 occurs twice as a sum and so the sums are not all distinct.
For each s from 1 upwards, your code should output the largest n so that there exists an array A of length n with distinct subarray sums.
Your code should iterate up from s = 1 giving the answer for each s in turn. I will time the entire run, killing it after one minute.
Your score is the highest s you get to in that time.
In the case of a tie, the first answer wins.
Examples
The answers for s = 1..12 are n=1, 2, 3, 3, 4, 5, 6, 6, 7, 8, 8, 9.
Testing
I will need to run your code on my ubuntu machine so please include as detailed instructions as possible for how to compile and run your code.
• s = 19 by Arnauld in C
• I am quite certain that there is no such rule. Currently there is nothing stopping anyone from just writing a simple print statement for the first 12 values that you provided and having a score of 12. It is almost certainly true that the largest value will use up most of the time, which is why I think my suggestion makes more sense. It's (of course) fine if you disagree, but I'm pretty sure you are just making your challenge much more annoying to answer for no reason. – FryAmTheEggman Nov 8 '18 at 23:21
• @FryAmTheEggman Hard-coding the output is a standard loophole. – JungHwan Min Nov 9 '18 at 1:12
• I downvoted this challenge because the nature of this challenge encourages some extent of hard-coding the solutions, and requiring the answer to "actually compute the output" is a non-observable requirement; there is no clear distinction between what is computed and what is hard-coded. – JungHwan Min Nov 9 '18 at 3:31
• @JungHwanMin Oh no! I am confused as the scoring system is basically the same as codegolf.stackexchange.com/questions/174407/… and codegolf.stackexchange.com/questions/165504/… and codegolf.stackexchange.com/questions/157049/… and... (and so on). What can I change to make you happy with my question? – Anush Nov 9 '18 at 10:45
• @JungHwanMin What's the difference between hard-coding the number 9 and hard-coding an array of length 9? It's good practice to avoid unobservable requirements whenever possible, but it's rather difficult for some kinds of challenges (fastest-code, anything related to quines, etc.). – Dennis Nov 9 '18 at 15:48
# C (gcc), s = 19
This is basically a port of my Node.js answer. It goes one step further with the default compiler options and two steps further with -O2 or -O3.
#include <stdio.h>
#include <string.h>
#include <stdint.h>
void search(uint8_t * arr, int n, uint8_t * sum, int * max, char * str, int depth) {
int i, j, k, s;
char tmp[16];
// do we have a better array?
if(depth > *max) {
for(str[0] = '\0', i = 0; i < depth; i++) {
sprintf(tmp, "%d ", arr[i]);
strcat(str, tmp);
}
*max = depth;
}
// try to append i = 1 to n to the current array
for(i = 1; i <= n; i++) {
// provided that doing so does not produce a sum that was already encountered
if(!sum[i]) {
for(sum[s = i] = 1, j = depth; j && !sum[s += arr[j - 1]]; j--) {
sum[s] = 1;
}
if(!j) {
// this is valid: set the value in arr[] and do a recursive call
arr[depth] = i;
search(arr, n, sum, max, str, depth + 1);
}
// whether the recursive call was processed or not, we have to clear the flags
// that have been set above
for(sum[s = i] = 0, k = depth - 1; k >= j; k--) {
sum[s += arr[k]] = 0;
}
}
}
}
int solve(int n, char * str) {
int max = 0; // best length so far
int hi = n * (n + 1) / 2; // highest possible sum for n
uint8_t sum[hi + 1]; // encountered sums
uint8_t arr[n]; // array
memset(sum, 0, (hi + 1) * sizeof(uint8_t));
search(arr, n, sum, &max, str, 0);
return max;
}
/////////////////////////////////////////////////////////////////////////////////////
void main() {
char str[128];
int n, res;
for(n = 1; n <= 19; n++) {
res = solve(n, str);
printf("N = %d --> %d with [ %s]\n", n, res, str);
}
}
Try it online!
• Thanks again! I wonder what speedups will be possible. – Anush Nov 9 '18 at 19:40
• @Anush Just compiling with -O3 allows to reach $N=19$ on TIO. – Arnauld Nov 9 '18 at 22:10
• Could you add the main function to your answer so people can just copy and paste? – Anush Nov 10 '18 at 19:13
• @Anush Sure. Done. – Arnauld Nov 10 '18 at 20:56
# Node.js, s = 17
Just a simple recursive search to get the ball rolling. Returns both the length and a valid array.
solve = n => {
var max = 0, // best length so far
best, // best array so far
sum = {}; // encountered sums
(search = a => {
var i, j, k, s;
// do we have a better array?
if(a.length > max) {
max = a.length;
best = [...a];
}
// try to prepend i = 1 to n to the current array
for(i = 1; i <= n; i++) {
// provided that doing so does not produce a sum that was already encountered
if((sum[j = 0, s = i] ^= 1) && a.every(n => sum[j++, s += n] ^= 1)) {
search([i, ...a]);
}
// reset the flags that have been toggled above
for(sum[s = i] ^= 1, k = 0; k < j; k++) {
sum[s += a[k]] ^= 1;
}
}
})([]);
return [ max, best ];
}
Try it online!
• Thank you for the first answer! – Anush Nov 9 '18 at 16:55
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General linear group:GL(2,Z)
View a complete list of particular groups (this is a very huge list!)[SHOW MORE]
Definition
The group $GL(2,\mathbb{Z})$ is defined as the group of invertible $2 \times 2$ matrices over the ring of integers, under matrix multiplication. Since the determinant is multiplicative and the only invertible integers are $\pm 1$, this can equivalently be defined as:
$\left \{ \begin{pmatrix}a & b \\ c & d \\\end{pmatrix} \mid a,b,c,d \in \mathbb{Z}, ad - bc = \pm 1 \right \}$.
This is a particular case of a general linear group over integers, which in turn is a particular case of a general linear group.
The subgroup of matrices of determinant one is special linear group:SL(2,Z), and it is a subgroup of index two.
Arithmetic functions
Function Value Explanation
order Infinite (countable)
exponent Infinite (countable)
derived length not defined
Frattini length not defined Has a free non-abelian subgroup, so not solvable.
Group properties
Property Satisfied Explanation Comment
abelian group No
nilpotent group No
solvable group No
perfect group No
GAP implementation
The group can be defined using GAP's GeneralLinearGroup function, as:
GL(2,Integers)
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# I need some help with some basic material
1. Oct 1, 2004
I can't seem to figure these problems if anyone can give me the concept and start me off that would be great.
Problem one: A car is traveling at a constant speed of 27 m/s on a highway. At the instant this car passes an entrance ramp, a second car enters the highway from the ramp. The second car starts from rest and has a constant acceleration. What acceleration must it maintain so that the two cars meet for the first time at the next exit which is 1.8 km away?
Problem two: A bus stop to pick up riders. A woman is running at a constant velocity of +5.0 m/s in an attempt to catch the bus When she is 11 m from the bus, it pulls away with a constant acceleration of +1.0 m/s^2. From this point, how much time does it take her to reach the bus if she keeps running with the same velocity?
*edit*
Oh sorry i posted this after i read the before you post thread Once again Im sorry for posting Homework help here
Last edited: Oct 1, 2004
2. Oct 1, 2004
### Pyrrhus
For the first problem consider the formulas
$$vt = x$$
$$\frac{1}{2}at^2 = x$$
You go the speed for the car and the distance x they will both travel, what can you do with that?
For the second problem
She will do
$$vt = x + 11$$
while the bus will do
$$\frac{1}{2}at^2 = x$$
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# Anything that ends in 0 can be divided by (there is more than one answer)
###### Question:
Anything that ends in 0 can be divided by (there is more than one answer)
### A person is earning PHP 600 per day to do a certain job Express the total salary as a function of the number n of days that the person
a person is earning PHP 600 per day to do a certain job Express the total salary as a function of the number n of days that the person works...
### Diego wrote 6/4 * 8/3 =48/4 .Explain what's Diego's mistakes was and how you know the equation is not true.
Diego wrote 6/4 * 8/3 =48/4 .Explain what's Diego's mistakes was and how you know the equation is not true....
### The area of a triangle can be found using the formula:Area=1/2=base=height Find the area of the triangle pictured below
The area of a triangle can be found using the formula: Area=1/2=base=height Find the area of the triangle pictured below where the measurements are given in meters, (M)2m5m ...
### What kind of evidence should a writer use to support a claim or counter claim?
What kind of evidence should a writer use to support a claim or counter claim?...
### How did the exploited of the vikings in europe affect the history of europe and its peoples?
How did the exploited of the vikings in europe affect the history of europe and its peoples?...
### If a formal parameter is a nonconstant reference parameter, during a function call, its corresponding actual parameter must be
If a formal parameter is a nonconstant reference parameter, during a function call, its corresponding actual parameter must be a(n) ....
### Aschool group of 3 adults and 34 students is visiting an aquarium. admission is $16 per adult and$9
Aschool group of 3 adults and 34 students is visiting an aquarium. admission is $16 per adult and$9 per student. which expression could be used to find the total cost of admission, in dollars...
### Which of the following sentences contains a verb in the future tense? a. they will take the dog to the
Which of the following sentences contains a verb in the future tense? a. they will take the dog to the park with them. b. she really should begin to pack. c. jean's daughter has begun to drive. d. by that time, he will have been gone for three days....
### What is an example of what happens when two objects with forces and mass collide with one another.
What is an example of what happens when two objects with forces and mass collide with one another....
### Pls help ha ing trouble on this one
Pls help ha ing trouble on this one $Pls help ha ing trouble on this one$...
### 5) what happened in november 1917 when the government of the russian empire under czar nicholas was
5) what happened in november 1917 when the government of the russian empire under czar nicholas was taken over by the bolsheviks? the soviet union signed a treaty with germany to support the central powers and fight against the allies for the remainder of the war. the soviet union signed an armisti...
### Can i rawr at your gf/bf. rawr XD -snules-
Can i rawr at your gf/bf. rawr XD -snules-...
### Each of the planets also spins about an axis running through it. true or false
Each of the planets also spins about an axis running through it. true or false...
### Acid halides reside at the top of the stability ladder making them difficult to synthesize. examine
Acid halides reside at the top of the stability ladder making them difficult to synthesize. examine the reaction below and draw the starting material needed to form the indicated product. be sure to include all lone pair electrons and formal charges ci: pcly 1st attempt d see periodic table see hin...
### Kirk traded his new zealand dollars for mexican pesos at an exchange rate of 1 new zealand = 8.8352 mexiacan pesos. he gave the exchanger
kirk traded his new zealand dollars for mexican pesos at an exchange rate of 1 new zealand = 8.8352 mexiacan pesos. he gave the exchanger a total of 2600 new zealand dollars. approximatly how many pesos does kirk have...
### What united the american people together in the decades following the american revolution?
What united the american people together in the decades following the american revolution? was it effective?...
### Which inference is best supported by these lines? . . . I must Once in a month recount what thou hast been, Which thou forget'st.
Which inference is best supported by these lines? . . . I must Once in a month recount what thou hast been, Which thou forget'st. This dam'd witch Sycorax, For mischiefs manifold and sorceries terrible To enter human hearing, from Argier, Thou know'st, was banish'd: for one thing she did They would ...
### There are 7 ushers and 11 technicians helping at the Harper middle school fall play .What is the ratio of technicians to all players
There are 7 ushers and 11 technicians helping at the Harper middle school fall play .What is the ratio of technicians to all players...
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# 37 - Electric Field and Potential Questions Answers
Find potential energy of disc having charge density sigma uniformly distributed on its surface,radius R
Joshi sir comment
Let potential of a disc of radius x is V. For this, see video given below
Now potential energy of this disc with a circular layer of charge
Joshi sir comment
A hollow spherical shell of radius r has a uniform charge density sigma. It is kept in a cube of edge 3r such that the center of the cube coincides with the centre of the shell. The electric flux coming out of a face of the cube is
Sir pls sove
Joshi sir comment
if any issue, inform
A solid conducting sphere of radius R is placed in a uniform electric field E as shown in figure-1.441. Due to electric field non uniform surface charges are induced on the surface of the sphere. Consider a point A on the surface of sphere at a polar angle θ from the direction of electric field as shown in figure. Find the surface density of induced charges at point A in terms of electric field and polar angle θ.
Joshi sir comment
To nullify the electric field inside, the conducting sphere creates a situation in which electric field inside due to charge on sphere would be same as that of given electric field. Now this is the case, which is same as Irodov 3.17.
See the video now
Finally
a small ball of mass 1kg and a charge 2/3 uC is placed at the center of a uniformly charged sphere of radius 1m and charge 1/3 mC. a narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. the sphere is made to rotate about its vertical diameter at a constant rate of 1/2pi revolutions per second. find the spedd w.r.t ground with which the ball slide out from the groove. neglect any magnetic force acting on ball?
Joshi sir comment
Joshi sir comment
Initially the spheres a and b are at potential be a and b respectively now sphere b is earthd by closing the switch the poential of a will become
Joshi sir comment
No meaning can be incurred by this language. Rectify it for getting answer
Joshi sir comment
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# Category Archives: Programming
Programming is my life! As I’m working as a programmer, it more or less is. And I’m also coding in my free-time.
Languages I’m proficient in include R, Java, Python and C++. But I’m also capable in writing Bash or Perl scripts, even though I don’t like the second one that much.
In this category of my blog I will generally present code to you. Be it code in form of the implementation of an algorithm or an application. The two languages Iyou will probably find most of here are R and C++. Also because they are easy to integrate into each other. But this might change over time, as programming languages and their use change.
If you want some algorithm explained, just write me. I will see, what I can do!
## Bookmark-Coloring Algorithm
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# And Writing Documentations In R
I wrote this function months ago, while writing the report for my seminar and I wanted to my make a post about this Bookmark-Coloring Algorithm ever since, but I never found the right chance. As I yesterday in the evening wrote some documentation for the Raspository, I thought it would be nice to make a post about this specific topic. Therefor I thought that using the Bookmark-Coloring Algorithm as an example would be nice.
## The Bookmark-Coloring Algorithm
This specific algorithm’s use is to generate a personal pagerank. In more detail this algorithm calculates given a starting website the chance to end up at other websites. But this idea is applicable to other fields as well. In my post about network-based integration through heat diffusion I showed you a similar method applied to a network of multi-omic data. On the same data-set you could use the Bookmark-Coloring Algorithm.
The basic idea behind the Bookmark-Coloring Algorithm is that you have some color that diffuses through a network, which is in my opinion equivalent to heat diffusing in a network. Correct me, if I’m wrong.
I implemented the algorithm following the paper about it by Pavel Berkhin. More precisely I implemented the Algorithm 2 from the paper. So let me show my implementation:
require(igraph)
require(dequer)
BCA<- function(graph, v, retentionCoefficient = 0.4, tol = 0.001){
# initialise vector of transition chances
p<-c()
p[V(graph)] <- 0
q <- queue()
pushback(q, v)
# initialise vector that indicates how much color is in one node
colorInVertex <- c()
colorInVertex[v] <- 1
# execute as long queque q has elements
while(length(q) > 0){
i <- pop(q)
w <- colorInVertex[i]
# use up the color in node
colorInVertex[i] <- NA
p[i] <- p[i] + retentionCoefficient * w
# if all color is used up continuew to next element in queque
if(w < tol){
next
}
# execute for all neighbors
for(j in neighbors(graph, i, mode = "out")){
if(!is.na(colorInVertex[j])){
colorInVertex[j] <- colorInVertex[j] +
((1 - retentionCoefficient) * w/degree(graph, i, mode = "out"))
}else{
# initialise color in neighbor
pushback(q, j)
colorInVertex[j] <- (1 - retentionCoefficient) * w/degree(graph, i, mode = "out")
}
}
}
return(p)
}
I wrote some comments, that hopefully help you to understand, what’s going on. That’s also the first part of documentation. It’s the step you’ll probably do, while writing your code and in my opinion it’s always useful.
So are we done with the documentation? No. Not, if we want to this function into a package.
## roxygen2 documentation
roxygen2 is a nice package allowing you to write in-line documentation in R. I won’t go to much into detail about it here as there are lot of online sources for it1, but I will show you a short example, how to do it!
Now let me show, how to write a documentation for the BCA function, I will go over all the specified tags.
#' Bookmark Coloring Algorithm
#'
#' @aliases BookmarkColoringAlgorithm
#'
#' @description This function calculates a teleportation vector from a given
#' starting node to other nodes in a given network.
#'
#' @export BCA
#' @import dequer
#' @import igraph
#'
#' @param graph an object of type \code{\link[igraph]{igraph}}.
#' @param v a starting vertex from the above graph. Can be either its identifier
#' or a igraph.vs object.
#' @param retentionCoefficient the restart probability for each node.
#' @param tol a tolerance treshold, indicating what the smalltest value of color
#' is, that should propagate further
#'
#' @return a preference/teleportation vector
#'
#' @references \insertRef{Berkhin2006}{Raspository}
#'
#' @examples
#' library(igraph)
#' g <- make_ring(5)
#' preferenceVector <- BCA(g, 1)
BCA <- function(graph, v, retentionCoefficient = 0.4, tol = 0.001){
OK, let’s go over some of the tags…
In the first line of course you have the title of the function. Additional to the description, you can also add a details tag, where description should give a short overview over the method, you could include theoretical background in the details.
Then needed packages are imported. roxygen2 will convert them into lines in your NAMESPACE file.
With params and return you should shortly describe their types and what they should contain.
For the following use of the references tag, you also need to import the Rdpack package and include a “REFERENCES.bib” in the inst folder with the regarding BibTeX entries. In my opinion you should always use this, when implementing some method from some source… Be it a book or a paper. Rdpack imports those references automatically from your BibTeX file into the documentation files.
Last, but not least I included a runnable example. This is important to give your user a starting point on how to use your function, but furthermore it is a test for the function. Each time your package is built, example code is run. So you will be notified, if there are any errors.
But we will go more into automated testing another time. Because there is of course more you can do. But you should always write example code, if your function is visible to the user.
After writing this documentation in your file, you have to compile it by writing:
roxygen2::roxygenise()
## Availability Of The Code
You can access a maintained version of the code in my GitHub repository Raspository under R/BCA.R.
## Salt And Pepper Noise
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# And Measuring Noise
Using the nomenclature developed in yesterday’s post I will today also implement a method for creating salt and pepper noise in images. This noise simulates dead pixels by setting them either to the lowest or highest grey value, in our case 0 or 1.
First let’s install and load the Raspository, where the methods and class developed yesterday are located in.
if (!requireNamespace("devtools", quietly = TRUE))
install.packages("devtools")
devtools::install_github("David-J-R/Raspository")
library(Raspository)
Without further ado let me show you how to implement the salt and pepper noise.
saltAndPepperNoise <- function(object, percentage = .2){
# select the indices to set to 0 or 1 at random
indices <- sample(length(object@current), length(object@current) * percentage)
# draw zeros and ones from a binomial distribution
values <- rbinom(length(indices), 1, 0.5)
object@current[indices] <- values
return(object)
}
OK, our animal guest test subject today will be an owl.
set.seed(1)
owl <- imageBWFromJpeg("owl.jpg")
plot(owl)
It looks surprised, doesn’t it? It probably knows about the salt and pepper noise we will add right away.
owlNoise <- saltAndPepperNoise(owl)
plot(owlNoise)
Uhm, this looks really annoying now. The other kind of noises hadn’t been that disrupting. Just like dead pixels on your screen, they make you feel bad.
## Introducing Two Measures of Noise
But what would be nice now as well, would be to have some measure of the noise. Let’s just use the mean squared error for it. I’ll implement a function that can compare two pictures or the current and original picture in an imageBW object depending on the input parameters.
MSE <- function(object, other = NULL){
if(is.null(other)){
errorMatrix <- object@original - object@current
}else{
errorMatrix <- other@current - object@current
}
squaredErrorMatrix <- errorMatrix ^ 2
return(mean(squaredErrorMatrix))
}
Nice! Now we finally have a measure for our noise. Of course this will be especially helpful later, when we want to evaluate the performance of reconstruction methods.
So let’s test it.
MSE(owlNoise)
## [1] 0.05545031
MSE(owlNoise, owl)
## [1] 0.05545031
As expected1 both function calls leave us with the same result.
Let’s also implement the peak-signal-to-noise ratio, which gives us back a decibel value. The lower the value the more noise you have in the picture. You can calculate it according to the following formula, which I will also “simplify” a bit.
$$PSNR(x, y) = 10 \cdot log_{10} \left( \frac{MAX^{2}}{MSE(x, y)} \right) \$$
I hope you know the logarithm rules. 😛
$$PSNR(x, y) = 10 \cdot log_{10}(MAX^{2}) – 10 \cdot log_{10}(MSE(x, y)) \$$ $$PSNR(x, y) = 20 \cdot log_{10}(MAX) – 10 \cdot log_{10}(MSE(x, y))$$
Where x and y are the two pictures to compare and MAX is the maximum value a pixel can take on.
And for MAX being one in our case we get:
$$PSNR(x, y) = – 10 \cdot log_{10}(MSE(x, y))$$
Subsequently, it’s now time to implement this measure as well:
PSNR <- function(object, other = NULL){
mse <- MSE(object, other)
return(-10 * log(mse, base = 10))
}
That was pretty straightforward. So let’s test it!
PSNR(owlNoise)
## [1] 12.56096
It’s a relatively low value, that indicates a lot of noise. In my opinion this value is more intuitive, probably because of use of the logarithm function. Indeed the MSE value alone didn’t seem that big.
But that’s it for now. There will be probably further stuff next week. But probably not that much.
Thanks for reading and have nice remaining weekend. 🙂
## Availability Of The Code
You can access a maintained version of the code in my GitHub repository Raspository under R/imageNoise.R.
But as I will expand the code, it will also probably grow there.
## Multiplicative Noise And S4 Classes
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# Let’s Add Some More Noise
Like in my yesterday’s post mentioned, there are more kinds of noise. So let’s first load today’s picture and then add some multiplicative noise!
library(jpeg)
pigeonBW <- pigeon[,,1]
plot(as.raster(pigeonBW))
Oh look it’s a pigeon on a lantern. I wonder what this little guy was thinking? But let’s not dwell on this. Rather let’s think how we can scramble up this picture!
I will first try multiplicative normal distributed noise, which is pretty straight forward, if you followed yesterday’s post.
set.seed(1)
pigeonNormalNoise <- pigeonBW * (1 + rnorm(length(pigeonBW), sd = sd(pigeonBW)))
pigeonNormalNoise[pigeonNormalNoise > 1] <- 1
pigeonNormalNoise[pigeonNormalNoise < 0] <- 0
plot(as.raster(pigeonNormalNoise))
Can you see the difference to the additive noise from yesterday? Indeed the darker areas of the pictures seem nearly untouched from the noise. That’s because the intensity of the noise is now dependent on the intensity of the pixel. And intensity is what dark pixels are lacking… So to speak.
But it’s really annoying and redundant to have to write the same stuff over and over again, isn’t it? So let’s do, what a Computer Scientist is best in. Let’s generalize some methods.
But where to start?
In my view it would be useful to have some kind of an (black and white) image class, which saves the image and operations on it. Hence the process would also be reproducible, which is always nice.
imageBW <- setClass("imageBW", slots=list(original="matrix", current="matrix", operations="list"))
imageBWFromJpeg <-function(pathToJpeg){
require(jpeg)
imageBW <- image[,,1]
return(new("imageBW", original = imageBW, current = imageBW, operations = list()))
}
plot.imageBW <- function(object){
plot(as.raster(object@current))}
As an illustration I also overwrote the default plot function. This makes our lives even easier. In the next step I’ll implement the different noise functions. There are probably more generalized ways of doing so, but for now this will suffice. I also haven’t come up with a good way of storing the done operations yet. So I’ll probably also do that another day.
cropPixels<- function(object){
object@current[object@current > 1] <- 1
object@current[object@current < 0] <- 0
return(object)
}
slot(object, "current") <- slot(object, "current") +
runif(length(slot(object, "current")), min = -1, max = 1)
return(cropPixels(object))
}
addNormalNoise <- function(object, sd = NULL){
if(is.null(sd)){
object@current <- object@current + rnorm(length(object@current), sd = sd(object@current))
}else{
object@current <- object@current + rnorm(length(object@current), sd = sd)
}
return(cropPixels(object))
}
multiplyUnifNoise <- function(object){
object@current <- object@current * (1 + runif(length(object@current), min = -1, max = 1))
return(cropPixels(object))
}
multiplyNormalNoise <- function(object, sd = NULL){
if(is.null(sd)){
object@current <- object@current * ( 1 + rnorm(length(object@current),
sd = sd(object@current)))
}else{
object@current <- object@current * ( 1 + rnorm(length(object@current),
sd = sd))
}
return(cropPixels(object))
}
For the future I would also like to have this working with infix operators. Meaning that I could do stuff like image <- image + normalNoise(...) or image <- image * (1 + normalNoise(...)) so that I have a neat little grammar for working with images. However for the moment those functions will do the job. Now let us make use of the newly implemented methods and add a lot of noise to the picture.
image <- imageBWFromJpeg("pigeon.jpg")
image <- multiplyUnifNoise(image)
plot(image)
Haha, there’s not much left of this poor pigeons. But you can still see its and the lamp’s outlines. And I’m anyway quite certain that there are algorithms out there, that could reconstruct a lot of the original image. You will see that later on. 🙂
## Availability Of The Code
You can access a maintained version of the code in my GitHub repository Raspository under R/imageOneChannel.R.
But as I will expand the code, it will also probably grow there.
## How to Add Noise To Images
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# Or Let’s Make Some Noise!!
As I wrote in one of my last posts there will be some content about image processing and computer vision. And following my second lecture of it, here we go! Fortunately R has some capabilities of handling pictures, of which I will take use.
For now the topic will just be, how to add noise to images, which can be useful to you, e.g. if you want to prove the performance of an algorithm for handling noise. So we’re simulating noise. Natural noise in images can come from different sources… The most common of it would be due to high ISO values or/and long exposure times. So in general in low light conditions.
However I’ll only work with black and white pictures for today. Maybe color comes at some point in the future. 🙂
First I’ll show you how to load pictures into R and how to print them again. It’s quite easy tbh:
library(jpeg)
class(goose)
## [1] "array"
dim(goose)
## [1] 600 897 3
As the picture is black and white we only need one dimension of it. And we can just plot it by transforming it into a raster object:
gooseBW <- goose[,,1]
plot(as.raster(gooseBW))
Cute this goose, isn’t it? 🙂 Now let’s destroy this picture with noise! Nyehehehehe1.!
To begin with I will add some uniform distributed noise. As the values in the matrix are between 0 and 1, I will add values between -1 and 1
set.seed(1)
gooseUniformNoise <- gooseBW + runif(length(gooseBW), min = -1, max = 1)
But we can’t plot this right, as the matrix now also contains values above 1 and below 0. I’ll fix this by cropping the boundaries and then plot.
gooseUniformNoise[gooseUniformNoise > 1] <- 1
gooseUniformNoise[gooseUniformNoise < 0] <- 0
plot(as.raster(gooseUniformNoise))
Interesting how you can still make out the goose in spite of the noise? Now let’s do the same with normal distributed noise. I’ll take the standard deviation of the black and white picture.
gooseNormalNoise <- gooseBW + rnorm(length(gooseBW), sd = sd(gooseBW))
gooseNormalNoise[gooseNormalNoise > 1] <- 1
gooseNormalNoise[gooseNormalNoise < 0] <- 0
plot(as.raster(gooseNormalNoise))
In the lecture we also did multiplicative and impulse noise. However I won’t do them in this post, maybe another time.
Anyway, do you already see, why I like R so much? 🙂 If you do Linear Algebra and Statistics stuff it’s just super comfy!
I’ll show you probably even more reasons why, when I demonstrate you, how to denoise picture again. But that’s it for today.
Have a nice weekend!
## Map Plots About the Global Burden of Disease
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# A practical example
Like promised in another post I will show you how to do a map plots with R. For this purpose I will use the ggmap package, which makes this a relatively easy task.
But before I begin with the actual code, let me give you a short motivation
## Why to use map plots
Motivations for using map plots can be various. For example if you’re a journalist and let’s say you want to visualize a kind of events (like say earthquakes) in a regional context, this is a very demonstrative way of visualizing your data.
Or if you want to present some kind of data about places or countries map plots are always a good option.
The first time I did a map plot was actually part of an awesome lecture I had back in Munich at the TUM. Afterwards I got the chance to use this skill right away in the next semester for my Bachelor’s thesis.
As some of you might know the area, where I applied the algorithm which I improved and implemented for my thesis, was mental disorders. During the writing process of my thesis, I found it a good starting point in my thesis and the accompanying presentation to emphasize the prevalence of mental disorders in the world. In order to do so I used a map plot.
That’s basically also what I will do now, but this time with the case of cancer.
But on a side note I’m not saying that you should do those kinds of plots for each thesis or presentation regarding a disease topic. It’s just one possible starting point for it and not necessarily the best.
So please don’t just mindlessly copy, what I’m doing here. 🙂
## Getting the data for dissease related map plots
First let’s load all the packages we will need for this little exercise.
library(XLConnect)
library(data.table)
library(ggplot2)
library(ggthemes)
library(maps)
XLConnect is a package for loading excel sheet, which we will need.
That I like to use data.table you probably already noticed. It’s just super fast and comfy for some procedures and it has some nice synergies with ggplot2.
The maps package contains as the name suggests map data, which can be used to plot. Alternatively one could also use ggmap.
And ggthemes contains a neat theme for maps, which I will use.
First let’s load our world map. This data.table contains region names and the boundaries of those regions as longitudes and latitudes. ggplot can plot those as polygons.
mapdata <- data.table(map_data("world"))
knitr::kable(mapdata[1:5])
longlatgrouporderregionsubregion
-69.8991212.4520011ArubaNA
-69.8957112.4230012ArubaNA
-69.9421912.4385313ArubaNA
-70.0041512.5004914ArubaNA
-70.0661212.5469715ArubaNA
OK, done. Now we need to download the data on 2004’s mortality from the WHO.
download.file("www.who.int/entity/healthinfo/global_burden_disease/gbddeathdalycountryestimates2004.xls", "gbd.xls")
tmp <- readWorksheetFromFile(file = "gbd.xls", sheet = "Deaths 2004")
causes <- tmp$Col1[14:143] countries <- unname(tmp[6,7:198]) deathRates <- tmp[14:143,7:198] You should probably take a look at the Excel file yourself to understand it and what I’m doing later. The file is made for humans to look at it and not directly for machines to read it. Which is why we have to do some cleaning and transforming. In my experience as a Bioinformatics students this is something you have to do almost always. Even if you have a machine readable format, there’s no perfect data-set. You will always have some missing data or have to transform your data in some way. And this isn’t necessarily a trivial step. Often you will spend a lot of time here. And that’s good. If cleaning data was trivial, then we wouldn’t need data scientist. ## Cleaning data To begin with we have to transform the death rates to numeric values… Because they’re characters (strings) right now. For this purpose we have to also replace the separating comma at the thousand position. You see? What’s done to make the data more human readable, makes it less machine readable. That’s often the case. Then we set the column names to the countries and transform the matrix together with the vector of causes to a data.table. deathRatesNum <- matrix(as.numeric(gsub(",", "", as.matrix(deathRates))), nrow = dim(deathRates)[1]) ## Warning in matrix(as.numeric(gsub(",", "", as.matrix(deathRates))), nrow = ## dim(deathRates)[1]): NAs introduced by coercion colnames(deathRatesNum) <- countries DT <- data.table(causes = causes, deathRatesNum) Now we want a clean or also called long data-set. In this new data set we will have only three columns. Two variables (causes and region), which uniquely identify the value death rate. Similar to a database we can also set those variable columns as keys, which makes it very fast searchable. DTclean <- melt(DT, id.vars = "causes", variable.name = "region", value.name = "deathRate") setkey(DTclean, causes, region) Next let us see, if we have some regions in our data.table that aren’t in our map. DTclean[!region %in% mapdata$region, unique(region)]
## [1] Antigua and Barbuda
## [2] Brunei Darussalam
## [3] Congo
## [4] Côte d'Ivoire
## [5] Democratic People's Republic of Korea
## [6] Iran (Islamic Republic of)
## [7] Lao People's Democratic Republic
## [8] Libyan Arab Jamahiriya
## [9] Micronesia (Federated States of)
## [10] Republic of Korea
## [11] Republic of Moldova
## [12] Russian Federation
## [13] Saint Kitts and Nevis
## [14] Saint Vincent and the Grenadines
## [15] Serbia and Montenegro
## [16] Syrian Arab Republic
## [17] The former Yugoslav Republic of Macedonia
## [19] Tuvalu
## [20] United Kingdom
## [21] United Republic of Tanzania
## [22] United States of America
## [23] Venezuela (Bolivarian Republic of)
## [24] Viet Nam
## 192 Levels: Afghanistan Albania Algeria Andorra ... Zimbabwe
As expected, there are 24 regions from the WHO sheet not in the mapdata. Even though there’s probably a more elegant solution, I will change them manually. It’s a work that has to be done once. For this purpose it’s probably only necessary to fill it in for the big countries. So this is bearable.
DTclean[region == "Brunei Darussalam", region := "Brunei"]
DTclean[region == "Congo", region := "Republic of Congo"]
DTclean[region == "Democratic People's Republic of Korea", region := "North Korea"]
DTclean[region == "Iran (Islamic Republic of)", region := "Iran"]
DTclean[region == "Côte d'Ivoire", region := "Ivory Coast"]
DTclean[region == "Lao People's Democratic Republic", region := "Laos"]
DTclean[region == "Libyan Arab Jamahiriya", region := "Libya"]
DTclean[region == "The former Yugoslav Republic of Macedonia", region := "Macedonia"]
DTclean[region == "Micronesia (Federated States of)", region := "Micronesia"]
DTclean[region == "Republic of Moldova", region := "Moldova"]
DTclean[region == "Republic of Korea", region := "South Korea"]
DTclean[region == "Russian Federation", region := "Russia"]
DTclean[region == "Serbia and Montenegro", region := "Serbia"]
DTclean[region == "Syrian Arab Republic", region := "Syria"]
DTclean[region == "United Republic of Tanzania", region := "Tanzania"]
DTclean[region == "United Kingdom", region := "UK"]
DTclean[region == "United States of America", region := "USA"]
DTclean[region == "Venezuela (Bolivarian Republic of)", region := "Venezuela"]
DTclean[region == "Viet Nam", region := "Vietnam"]
And yea of course the work isn’t done completely yet. We also should check if there are regions in the mapdata, that aren’t in the WHO data-set. This could be due to various reasons… One being, that a region isn’t a member of the WHO and therefore the WHO doesn’t publish data on them.
Or more likely that a country from the WHO data-set span more than one region on the map, Serbia and Montenegro being such a case.
However I’m lazy now and I won’t do this today. How about you doing it and writing me a comment? 😛 Let it be a team1 effort.
## Making the map plots
OK, before we do the actual plotting let’s first calculate for how much percentage of all deaths in each country cancer is the cause. In detail I do this by joining the data.table with itself.
On a side note: W000 is the WHO code for all death causes combined and W060 for Malignant neoplasms, which is a more formal name for cancer.
Then we need to join the data.table with the map on the region name.
DTcaused <- DTclean[causes == "W000"][DTclean[causes == "W060"], on = "region"][, .(region, percentageCaused = i.deathRate / deathRate)]
deathrateMap <- mapdata[DTcaused, on = "region", allow.cartesian=TRUE, nomatch = 0]
And finally we can do our plot. For this purpose we first plot all regions in grey and as overlay we fill the countries, that we have data on, with a color between grey and red depending on how frequent cancer as a death cause is.
g <- ggplot() + geom_polygon(data = mapdata, aes(long, lat, group = group), fill = "grey")
g <- g + geom_polygon(data = deathrateMap, aes(long, lat, group = group, fill = percentageCaused))
g <- g + scale_fill_gradient(low = "grey", high = "red", aesthetics = "fill", name = "Percentage of\ndeaths caused\nby cancer")
g + ggthemes::theme_map()
And of course there’s one thing about this plot that could be misleading. Given that regions with missing data and very low prevalence of cancer deaths will both be grey, you hopefully see the potential problem here?
It’s not necessarily wrong or bad to do so. But I hope you recognize how someone could make a plot this way to mislead his audience. That’s why I recommend when it comes to looking at plots not only to think about, what is shown, but also what isn’t shown. Since no large data-set is complete… So ask the person who presents it to you, how she/he handled missing data points.
So what does this map actually say? From my perspective I don’t think anything surprising. At the moment, this data set captured, cancer was (and probably still is) mostly a problem of industrialized countries and it doesn’t seem to be connected to geography primarily (Can you see how Israel, Japan and South Korea pop up?).
Although the difference between the USA and Canada could be something interesting.
But this map, in my opinion, shows very clearly that cancer is one of the leading causes of death in the developed world, which also is the reason, why we also spend so much money on researching it.
However the main purpose of this post was to show you, how to make such plots and not discuss the reasons of different causes of mortality.
Ultimately I hope that this post has helped you.
Of course it is important that you mention your sources (cite them if you write a paper). This is because your approach has to be reproducible and you have to give those people, who did the preliminary work, credit for it.
In R you can get the proper citations for the packages you used the following way:
citation("ggmap")
##
## To cite ggmap in publications, please use:
##
## D. Kahle and H. Wickham. ggmap: Spatial Visualization with
## ggplot2. The R Journal, 5(1), 144-161. URL
## http://journal.r-project.org/archive/2013-1/kahle-wickham.pdf
##
## A BibTeX entry for LaTeX users is
##
## @Article{,
## author = {David Kahle and Hadley Wickham},
## title = {ggmap: Spatial Visualization with ggplot2},
## journal = {The R Journal},
## year = {2013},
## volume = {5},
## number = {1},
## pages = {144--161},
## url = {https://journal.r-project.org/archive/2013-1/kahle-wickham.pdf},
## }
citation("maps")
##
## To cite package 'maps' in publications use:
##
## Original S code by Richard A. Becker, Allan R. Wilks. R version
## by Ray Brownrigg. Enhancements by Thomas P Minka and Alex
## Deckmyn. (2018). maps: Draw Geographical Maps. R package version
## 3.3.0. https://CRAN.R-project.org/package=maps
##
## A BibTeX entry for LaTeX users is
##
## @Manual{,
## title = {maps: Draw Geographical Maps},
## author = {Original S code by Richard A. Becker and Allan R. Wilks. R version by Ray Brownrigg. Enhancements by Thomas P Minka and Alex Deckmyn.},
## year = {2018},
## note = {R package version 3.3.0},
## url = {https://CRAN.R-project.org/package=maps},
## }
##
## ATTENTION: This citation information has been auto-generated from
## the package DESCRIPTION file and may need manual editing, see
## 'help("citation")'.
You get the idea. Also cite the other packages, if you use them in your publication or thesis.
The output is in bibtex format. So I hope you know what to do with it. 😛
Of course the data on the global burden of disease you have to cite as well. Thus I’ll give you the formatted citation for it:
WHO. (2004). The global burden of disease: 2004 update: causes of death. 2004 Update, 8–26.
And last, but not least, please also mention me. This however is not a necessity, but a sign of respect towards my work. By all means respect is an important thing, unfortunately not often enough given in our society.
## Network-Based Integration Through Heat Diffusion
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# Re-Implementing netICS Heat Diffusion In R
OK, I successfully procrastinated this blog entry about heat diffusion now for too long, even though the code for it has been waiting for over a month. :’D So it’s now time for it! I promised it! And it’s time for it now. And I guess it doesn’t need to be perfect, because the whole idea of this blog is to put things out now, rather than waiting for some future moment, where I can do it perfectly.
So that being said, let’s start. 🙂
In my last post I already talked about the integration of multi-omics data. This time I will show you one method called netICS from a more or less pure coding perspective. It uses Heat Diffusion for finding so-called mediator genes in multi-omics data.
You can find the original code written in Matlab here and the corresponding paper here, which I would suggest you reading, if I aroused your interest in the topic.
Thus let’s just dive straight into the matter!
Because we are cool kids we don’t download the data “manually”, but do it right away in R.
And of course we do this with the super fast and comfy fread function from the data.table package, which I would suggest you to learn, when doing R.
We load the data directly from the Github of the original netICS:
library(data.table)
col.names = c("Gene", "Sample"), header = FALSE)
header = FALSE, col.names = "Gene")
header = FALSE, col.names = c("Gene", "pval"))
header = FALSE, col.names = c("Gene", "pval"))
The network used in the method is stored as a Matlab object. But ff course there’s also a package that does this job for us! After that we have to extract the adjacency matrix from it. Because we will need this adjacency matrix later.
library(R.matlab)
adjacencyMatrix<-tmp$adj.lar.com rm(tmp) And that already was it with loading the data! ## Combing p-values In my last post I already talked about combining p-values. We need to implement it to combine the p-values from the rnaDiffExp and rppa. Maybe you will also see now, why I like data.tables so much. I’ve commented the code for you. combineDifferentialExpressions<-function(diffExp1, diffExp2){ # Combine both data.tables through an outer join mergedDiff<-merge(diffExp1, diffExp2, by ="Gene", all=TRUE) # if the first p-value is missing take the second one mergedDiff[is.na(pval.x) & !is.na(pval.y), pval := pval.y] # if the second p-value is missing take the first one mergedDiff[!is.na(pval.x) & is.na(pval.y), pval := pval.x] # if both are present combine them by Fisher's method and perform a chi-square test mergedDiff[!is.na(pval.x) & !is.na(pval.y), pval := pchisq(-2 * (log(pval.x) + log(pval.y)) , 4)] return(mergedDiff[,.(Gene, pval)]) } I really love the syntax of data.tables in R I have to say. 😀 Now let’s execute it and afterwards adjust the p-values with the FDR method and then take a look at it: diffExp <- combineDifferentialExpressions(rnaDiffExp, rppa) diffExp[, p.adjusted:=p.adjust(pval, method = "fdr")] knitr::kable(diffExp[1:5]) Genepvalp.adjusted A1BG0.00000000.0000000 A1CF0.88511070.9505269 A2BP10.22216470.2792292 A2LD10.91032690.9695794 A2M0.00000000.0000000 ## Heat Diffusion I won’t go into detail about heat diffusion and its theoretical background at this place. Let me just summarize the idea or rather goal short. We’re transforming the adjacency matrix first to a transition matrix, with which you might be familiar from Markov Chains. Basically each cell in the matrix stands for a transition between two genes (in our case). Here R can also shine with its natural Linear Algebra capabilities. normaliseAdjacencyMatrix <- function(adjacencyMatrix){ return(adjacencyMatrix %*% diag(1/colSums(adjacencyMatrix))) } You see… Isn’t that basically a beautiful one-liner? In the next step I implement the transformation of this matrix to a diffusion matrix. This matrix basically represents the connectivity and network topology. Multiplying a vector of starting probabilities with it, gives us back a stationary distribution. Which basically means after initially many transitions in the network we will end up at a certain set of vertices in the network with a certain probability. The restart probability indicates how likely it is that we will be “reported” back to our starting point instead of a transition in a step. So let’s implement it: performInsulatedHeatDiffusion <- function(adjacencyMatrix, restartProbability){ temperature <- diag(dim(adjacencyMatrix)[1]) - (1 - restartProbability) * adjacencyMatrix return(restartProbability * solve(temperature)) } We will run through this network in both directions. Basically, one time from the direction of mutated genes and the other time from the differentially expressed ones. So let’s wrap this up in another function. We set the default restart probability to 0.4. netICS <- function(adjacencyMatrix, networkGenes, mutationData, diffExpGenes, restartProbability = 0.4){ # only keep mutated and differentially expressed genes, that are present in the network mutationInNetwork <- mutationData[networkGenes, on="Gene", nomatch = 0] diffExpInNetwork <- diffExpGenes[networkGenes, on="Gene", nomatch = 0] # calculating the regular diffusion matrix connectivity <- performInsulatedHeatDiffusion(normaliseAdjacencyMatrix(adjacencyMatrix), restartProbability) # calculating the diffusion matrix in the opposite direction connectivityBackward <- performInsulatedHeatDiffusion(normaliseAdjacencyMatrix(t(adjacencyMatrix)), restartProbability) # ranking mediator genes result <- prioritize(connectivity, connectivityBackward, networkGenes, mutationData, mutationInNetwork, diffExpInNetwork) return(result) } At this point you might have already noticed that we haven’t done anything yet with the diffusion matrix. For this reason we will implement the prioritize function, which will do this for us. ## Gene Prioritization First we implement a helper function for the diffusion to avoid redundant code: diffuseSample<-function(connectivity, networkGenes, S){ # find the positions of the input genes in the matrix positions <- networkGenes$Gene %in% S$Gene #multiply them with equal probability with the diffusion matrix weights <- rep(1/sum(positions), sum(positions)) %*% connectivity[positions,] return(as.vector(weights)) } S here is the list of genes with which we multiply the diffusion matrix. The starting probability is for that matter uniformally distributed between them. Afterwards we write our actual prioritization function. The diffusion in this case is only once executed for the differentially expressed genes, but sample-wise for the mutated genes. Subsequently, we combine both probabilities sample-wise by multiplying and rank them. Those ranks are then combined by three different methods, them being median, sums and rho-scores. For the last one the RobustRankAggreg package is used: prioritize<-function(connectivity, connectivityBackward, networkGenes, mutationData, mutationInNetwork, diffExpInNetwork){ # diffuse for differential expression, if differential expression isn't sample-wise if(!"Sample" %in% colnames(diffExpInNetwork)){ Ed<-diffuseSample(connectivityBackward, networkGenes, diffExpInNetwork) } scores <- data.table() for(sample in unique(mutationData$Sample)){
# diffuse for mutated genes
Em<-diffuseSample(connectivity, networkGenes, mutationInNetwork[Sample == sample])
# diffuse for differential expression, if differential expression is sample-wise
if("Sample" %in% colnames(diffExpInNetwork)){
Ed<-diffuseSample(connectivityBackward, networkGenes, diffExpInNetwork[Sample == sample])
}
# Multiply scores
E <- Em * Ed
scores <- rbind(scores, data.table(Gene = networkGenes$Gene, Sample = sample, Score = E)) } # rank scores scores[, rank.in.sample:=rank(-Score), by=.(Sample)] # combine ranks library(RobustRankAggreg) ranks<-scores[, .(med=median(rank.in.sample), sum=sum(rank.in.sample), rho = rhoScores(rank.in.sample/max(rank.in.sample))), by=.(Gene)] return(ranks) } Phew… The work is done! Now it’s time to execute the code. If you do this at home you might notice, that it will take a while until it’s finished. That’s because the calculation of the diffusion matrix is very expensive, when it comes to processing time. Your computer has to solve a system of linear equations for it, which is not linear. But don’t ask me in which O it exactly lies. 🙂 result<-netICS(adjacencyMatrix = adjacencyMatrix, networkGenes, mutationData = mutationData, diffExpGenes = diffExp[p.adjusted < 0.05, .(Gene)]) knitr::kable(result[order(sum)][1:10]) Genemedsumrho RPS6KB196874489.50e+00 AKT377077330.51e-07 AKT194478520.50e+00 SERPINE181881454.50e+00 TP53102684433.51e-07 MAP2K284386530.51e-07 CDKN2D75886534.50e+00 CREB1108087179.51e-07 VEGFA98787678.50e+00 E2F188588245.50e+00 I will probably use this code at some point of the future again, as Network theory is one of my areas of interest and it’s pretty nice to have that lying around. This method, heat diffusion, is by the way very similar to Pageranking… The stuff that made Google big. 🙂 ## Some Stuff About The Near Future I will probably next week do a post about making map plots. As I have to do one anyway and I figured it would be nice to document it here, as it could be quite useful to some of you. And in the first week of April I will be on a Hackathon (my very first!) about Metabolomics in R. I will probably keep you posted about that then and I’m looking so forward to it. So see you hopefully soon! ## Availability Of My Code You can access a maintained version of the code of the correct version in my GitHub repository Raspository under R/NetICS.R. Please follow and like us: ## Alternating Least Squares Warning: count(): Parameter must be an array or an object that implements Countable in /homepages/10/d771928198/htdocs/clickandbuilds/RandomThoughtsScienceandProgramming/wp-content/plugins/kblog-metadata/kblog-author.php on line 332 # How To Implement Alternating Least Squares In R And How Not To Do It So it’s time for my first actual content. And like predicted in my Hello World post, it will be something implemented in the R programming language. More precisely it’s a Machine learning algorithm called Alternating Least Squares. But first, before we indulge ourselves in code, let me tell you why this algorithm is of interest for me and what it does. ## Introduction I’ve been working now for a few months as part of my research assistant job on the Bioconductor package BEclear . I won’t go into detail about the package, you only need to know that it uses something called a Latent Factor Model to impute1 missing data in data-sets. Let’s say you have a matrix D containing missing values. The rows in the matrix stand for features and the columns for samples. Then you could assume that the matrix $$D_ {ij}$$ is modeled by both features and sample specific effects in the following way: $$D_ {ij} = L_ {i}^ {T} \times R_ {j}$$ Where $$L_i$$is the feature specific matrix and $$R_j$$ the sample specific matrix. For the imputation of missing values you now try to estimate those two matrices, the latent factors, from the existing values. Methods based on this assumption are already applied in variety of fields. Like with batch effects in DNA Methylation in the case of the BEclear package or in recommender systems for e.g. Netflix like described in a paper, which helped me a lot in understanding this topic. To estimate the latent factors there are different methods. One of them, implemented in the BEclear package is a gradient descent. Another method for it is Alternating Least Squares (ALS), which I wanted to implement on my own.2 The lecture notes of Hastie et al served as my source for implementing this method. I highly recommend you reading both the paper from Koren et al and those lecture notes, if you want to know more about the theoretical background and also the applications of those methods. You just need to know some Linear Algebra, then they should be easy enough to understand in my opinion. But as a short summary… ALS tries to estimate the feature and sample matrix by alternating fixing one of them and then calculating the other one by solving the the system of equations and you do this in general until convergence. In Gradient Descent on the other hand both matrices are estimated at the same time. ## How Not To Implement Alternating Least Squares As a start I will show you my first faulty try in implementing the Alternating Least Squares. Maybe you will learn something by me sharing it with you. And you should try to guess what my mistake was. As my implementation reuses a function from BEclear34, you have to install this package first. For this purpose I guess it’s easiest if you just install it from GitHub via the following lines of code: if (!requireNamespace("devtools", quietly = TRUE)) install.packages("devtools") devtools::install_github("David-J-R/BEclear") And now let’s come to the first implementation of the Alternating Least Squares algorithm. See if you can find its problem. I tried to comment all those steps so that the code should be comprehensible. But if not, please let me know! 🙂 imputeALSfaulty<- function(data, lambda = 0.75, r = 10, iters = 20){ # We require the BEclear package, because we're using its loss function require(BEclear) D <- data D[is.na(D)] <- 0 # We initialise L and R with random values L <- matrix(rnorm(nrow(data) * r), nrow(data), r) / sqrt(r) R <- matrix(rnorm(r * ncol(data)), r, ncol(data)) / sqrt(r) currLoss <- BEclear:::loss(L,R, 1, data)$loss
print(currLoss)
for(i in 1:iters){
# L and R are determined by solving the following system of the equations
# We repeat this step iters-times
L <- t(solve(R %*% t(R) + diag(lambda,r), R %*% t(D)))
R <- solve(t(L) %*% L + diag(lambda,r), t(L) %*% D)
currLoss <- BEclear:::loss(L,R, 1, data)$loss print(currLoss) } # L and R are multiplied to get the estimated values D <- L %*% R # Missing values are replaced with estimated value for (i in seq_len(nrow(data))) for (j in seq_len(ncol(data))) { if (is.na(data[i, j])) { data[i, j] <- D[i, j] } } return(data) } Now let me show you the problem with this implementation, if you haven’t recognized it on your own by now. First we will load the example data and functions from the BEclear package to generate ourselves a sample data-set with missing values: library(BEclear) data("BEclearData") batchEffect <- calcBatchEffects( data = ex.data, samples = ex.samples, adjusted = TRUE, method = "fdr") mdifs <- batchEffect$med
pvals <- batchEffect$pval summary <-calcSummary(mdifs, pvals) cleared.data <- clearBEgenes(ex.data, ex.samples, summary) And then we run the faulty version of ALS on it. The printed output of the function is the loss of the current solution during each iteration. result <- imputeALSfaulty(cleared.data, iters = 10) ## [1] 2586.68 ## [1] 101.8086 ## [1] 95.60281 ## [1] 95.29458 ## [1] 95.21404 ## [1] 95.20139 ## [1] 95.20632 ## [1] 95.21329 ## [1] 95.21869 ## [1] 95.22233 ## [1] 95.2247 If we now take a look at the imputed values, you can see what’s wrong: boxplot(result[is.na(cleared.data)]) They’re all pretty close to zero. That’s because we set the missing values to zero. This way the solve method “tries” to generate R and L the way that the missing values are also very close to zero. Of course we don’t want that… This way we could just set the missing values right away to zero. ## How To Implement Alternating Least Squares Finally let me show you an implementation that actually does, what it should do. And again if something is unclear, don’t hesitate to ask me! imputeALScorrect<- function(data, lambda = 0.75, r = 10, iters = 80){ # We require the BEclear package, because we're using its loss function require(BEclear) # We initialise L and R with random values L <- matrix(rnorm(nrow(data) * r), nrow(data), r) / sqrt(r) R <- matrix(rnorm(r * ncol(data)), r, ncol(data)) / sqrt(r) currLoss <- BEclear:::loss(L,R, 1, data)$loss
print(currLoss)
for(iter in 1:iters){
# Now we iterate over the feature dimmension of L
for(i in 1:dim(L)[[1]]){
# We determine the revealed entries for the feature
# And subset the data and R so to only retain the revealed entries
revealedEntries <- !is.na(data[i,])
y <- as.matrix(data[i, revealedEntries])
x <- R[,revealedEntries]
# We solve the linear equation for the feature
L[i,] <- as.vector(solve(x %*% t(x) + diag(lambda, r), x %*% y))
}
# We iterate over the sample dimmension of R
for(j in 1:dim(R)[[2]]){
# We determine the revealed entries for the sample
# And subset the data and L so to only retain the revealed entries
revealedEntries <- !is.na(data[,j])
y <- as.matrix(data[revealedEntries, j])
x <- L[revealedEntries,]
# We solve the linear equation for the sample
R[,j] <- as.vector(solve(t(x) %*% x + diag(lambda, r), t(x) %*% y))
}
currLoss <- BEclear:::loss(L,R, 1, data)$loss print(currLoss) } # L and R are multiplied to get the estimated values D <- L %*% R # Missing values are replaced with estimated value for (i in seq_len(nrow(data))) for (j in seq_len(ncol(data))) { if (is.na(data[i, j])) { data[i, j] <- D[i, j] } } return(data) } A further advantage of this implementation is, that it is relatively easy to write a parallelised version of it. Maybe I will show you that in one of my next posts. After I overheard a conversation at the university that R is apparently bad for this, I feel almost challenged to do so. Now let’s take a look at the imputed values. We just take the sample data-set from before for this cause. result <- imputeALScorrect(cleared.data, iters = 10) ## [1] 2571.072 ## [1] 109.301 ## [1] 99.38027 ## [1] 97.17519 ## [1] 95.42625 ## [1] 94.00547 ## [1] 92.83838 ## [1] 91.87368 ## [1] 91.07338 ## [1] 90.40794 ## [1] 89.85372 boxplot(result[is.na(cleared.data)]) Now that looks more like real data… Doesn’t it? But to be sure let’s compare it to the predicted values by the BEclear package. For the comparison we calculated the Root Mean Squared Error: library(Metrics) result.BEclear <- imputeMissingData(cleared.data) ## INFO [2019-02-08 12:17:10] Starting the imputation of missing values. ## INFO [2019-02-08 12:17:10] This might take a while. ## INFO [2019-02-08 12:17:10] BEclear imputation is started: ## INFO [2019-02-08 12:17:10] block size: 60 x 60 ## INFO [2019-02-08 12:17:10] Impute missing data for block 1 of 4 ## INFO [2019-02-08 12:17:10] Impute missing data for block 2 of 4 ## INFO [2019-02-08 12:17:11] Impute missing data for block 3 of 4 ## INFO [2019-02-08 12:17:11] Impute missing data for block 4 of 4 rmse(result.BEclear[is.na(cleared.data)], result[is.na(cleared.data)]) ## [1] 0.03196931 Well the difference isn’t that big. But of course for assessing the accuracy of the method an elaborate evaluation would be needed. However for something I coded just for fun I’m satisfied with this first look. ## Addendum: Biases Just for fun let’s also add biases to our model, like described by Koren et al, to further improve our algorithm. The idea behind the bias is to capture the variability of the data that arises from the features or samples alone, while the two matrices L and R capture the variability that arises from the interaction of features and samples together. In other words by introducing the biases we unburden L and R a bit. We use a method, where the biases for each entry in the data-set are the sum of the overall average over all values and the average difference from this average of the corresponding column and row. And to save valuable computation time we just subtract this bias for each value from a copy of each value and use this transformed matrix for further calculations. Of course we have to add the bias later again. And here we go with the improved implementation: imputeALSBias<- function(data, lambda = 0.75, r = 5, iters = 10, use.biases=TRUE){ # We require the BEclear package, because we're using its loss function require(BEclear) # copy the data D <- data # We initialise L and R with random values L <- matrix(rnorm(nrow(data) * r), nrow(data), r) / sqrt(r) R <- matrix(rnorm(r * ncol(data)), r, ncol(data)) / sqrt(r) currLoss <- BEclear:::loss(L,R, 1, D)$loss
print(currLoss)
if(use.biases){
# we calculate the biases
biasData<-mean(data, na.rm = TRUE)
biasRows<-rowMeans(data - biasData, na.rm= TRUE)
biasCols<-colMeans(data - biasData, na.rm= TRUE)
# subtract the biases from the data
D <- D - biasData - biasRows
D <- t(t(D) - biasCols)
}
for(iter in 1:iters){
# Now we iterate over the feature dimmension of L
for(i in 1:dim(L)[[1]]){
# We determine the revealed entries for the feature
# And subset the data and R so to only retain the revealed entries
revealedEntries <- !is.na(D[i,])
y <- as.matrix(D[i, revealedEntries])
x <- R[,revealedEntries]
# We solve the linear equation for the feature
L[i,] <- as.vector(solve(x %*% t(x) + diag(lambda, r), x %*% y))
}
# We iterate over the sample dimmension of R
for(j in 1:dim(R)[[2]]){
# We determine the revealed entries for the sample
# And subset the data and L so to only retain the revealed entries
revealedEntries <- !is.na(D[,j])
y <- as.matrix(D[revealedEntries, j])
x <- L[revealedEntries,]
# We solve the linear equation for the sample
R[,j] <- as.vector(solve(t(x) %*% x + diag(lambda, r), t(x) %*% y))
}
currLoss <- BEclear:::loss(L,R, 1, D)\$loss
print(currLoss)
}
# L and R are multiplied to get the estimated values
D <- L %*% R
if(use.biases){
# we add the biases again
D <- t(t(D) + biasCols)
D <- D + biasData + biasRows
}
# Missing values are replaced with estimated value
for (i in seq_len(nrow(data)))
for (j in seq_len(ncol(data)))
{
if (is.na(data[i, j])) {
data[i, j] <- D[i, j]
}
}
return(data)
}
Testing this implementation, if you wish, is now your turn! 🙂
Maybe at some later point I will compare the performance and correctness of various different settings of this functions5. But for now that’s enough. Of course there are more sophisticated bias models we could think of. But one could even think of bias models like biases that are also determined by the Alternating Least Squares method during each iteration.
So we won’t run out of things to do any time soon.
## Conclusion
Yea, what’s the conclusion? I think it’s quite simple… Don’t be lazy, while coding!
OK, OK… I will say a bit more. I think what you can learn from the faulty example is that you should always think what your code is actually doing and take a look at the results to see, if something is fishy. Other than that I hope that you learned something and I could show you that some methods used in Machine Learning aren’t that complicated.
For now my implementation of ALS is still worse, when it comes to run time, in comparison to the Gradient Descent implemented in the BEclear package. But I also spend a lot of time optimizing the second. And maybe I will show you in a future blog how to optimize it. As this is my first blog post I would highly welcome feedback from you! 🙂
So have a nice day and until next time!
## Availability Of The Code
You can access a maintained version of the code of the correct version in my GitHub repository Raspository under R/imputeALS.R.
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# Generalist models on Paperspace: Images, text, audio and video combined
A new class of deep learning called a generalist model is capable of running on images, text, audio, video and more all at the same time. Here we explore the capabilities of 3 of these models: Perceiver IO, Data2vec, and Gato. We show how to run Perceiver IO on Paperspace.
17 hours ago • 12 min read
Bring this project to life
What if instead of training lots of models for many different tasks, we only had to train one model? Then that model does all of our tasks.
While it sounds fanciful, there are examples already available where the same model can perform many different tasks, running on images, text, audio, video, and more.
Such generalist models (aka. generalist agents or multi-modal models) are becoming increasingly able to take on a large number of different use cases. This means that in the future we may not need to manage many models, but can run end-to-end artificial intelligence on just a few, or even one.
Simplifying end-to-end AI in this way would open it up to use by many more people, because the remaining model or models could potentially be put within easy-to-use no-code interfaces, while retaining their full problem-solving ability.
So while we have not yet seen the one model to rule them all, it is worth exploring their current capabilities.
Here, we will review 3 of them:
• Perceiver IO
• Data2vec
• Gato
and, specifically, we will show how to run Perceiver IO on Paperspace.
Perceiver IO and Gato are from DeepMind, and Data2vec is from Meta AI.
The days of true artificial general intelligence (AGI) are still some way off, but in future these may be remembered as taking some of the steps towards it.
## Why are they general?
Each of the 3 models presented here differs in its details from the others, but the underlying idea that makes them general is that the data passed into them is converted into the same form regardless of input. So images, text, audio, etc., all get converted into the same underlying form that the models can then be trained on.
The models then output predictions on the data in this form that get converted back to actionable outputs in an inversion of the transformation used for the inputs.
The changeable components of the models become just the input and output for dealing with their data, while the models themselves retain the same form.
## Why are they important?
The importance of these models lies not just in their potential successors a long way down the road paving the way to an artificial general intelligence, but in their more near-term utility:
• Fewer models to manage: one model for many tasks instead of multiple models strung together in a dataflow simplifies management, such as when routing data, versioning, deploying, checking accountability, data I/O versus training, and so on.
• Simpler interfaces: having a general model with swappable input and output means it will become easier to create more general end-to-end interfaces and applications, opening their use to less technical and even no-code users.
• They can outperform specialist models: aside from being more general, there are several instances of these generalist models outperforming previous specialist models on the same tasks, so the process is not only simpler but can also produce better results.
Having a combination of abilities in one place, and being able to deal with many types of data at once, means they can also be combined in novel ways. For example, not only creating a video, but also captioning it and speaking the words, giving greater accessibility for all audiences. The possibilities multiply as abilities are added.
## The 3 models
Now lets' see a general overview of what each of the models is doing. For more in-depth descriptions, see the original blogs and papers referenced below.
### Perceiver IO
Perceiver IO (blog, HF repository, paper, Wikipedia) is based on a transformer neural network. Originally used for natural language processing (BERT, etc.), transformers were then applied to computer vision, e.g., the vision transformer ViT, and now as generalist models. They work by using attention, which means that the parts of the data found to be more important in training receive greater weight and have more compute time spent on them.
The input data, regardless of source such as images, text, audio, etc., is mapped to a latent space of lower dimension. In other words, the network itself is not modality-specific, and all data passed in proceeds in the same form. A latent space means that items in the original data that resemble each other are closer together in the new space. The model is using cross-attention, which is the use of attention to map the data to different numbers of dimensions. The use of attention to reduce the size of the data in this way is also known as an attention bottleneck. Specific features of the latent data are associated to their parts of the input, keeping what passes through the network in sequence with the original data.
Because all data passes through the network in the same form, it means that all the outputs are also the same form. The outputs therefore then have to be interpreted in terms of the task the user wants to perform, and transformed back appropriately. This is done by the user passing in a query that combines with the latent data in the network to produce outputs in the desired form. This encoding of data at the start then decoding at the end resembles what is done in the widely used class of models known as encoder-decoder neural networks.
The use of latent data in the network, which is of lower dimension than the original data, means that the network can scale to handle large inputs and outputs without becoming too slow, unlike transformers.
The diagram from the original DeepMind blog shows a useful overview of the Perceiver setup:
Perceiver IO has been shown to match the performance of other models, such as BERT for language and ResNet50 for images, plus is state-of-the-art on AudioSet audio+video data and Sintel video for optical flow. Further demonstrated applications include classifying 3D point clouds and videogaming.
### Data2vec
Announced in January 2022 by Meta AI (aka. Facebook), Data2vec (blog, repository, paper) is the first self-supervised algorithm that works for multiple different types of data.
Self-supervised algorithms are important, because they are able to train on data that does not have labels. There is a lot more unlabeled data out there than labeled data. Self-supervised models learn by observing their environment directly.
But, as they point out in the blog, the way that other self-supervised models learn depends a lot upon the type of input data. So being able to learn from all types of data in one model is a step forward from this, and opens up wider usage of self-supervised learning.
The types of data that Data2vec is shown working on are not quite as general as Perceiver IO, but they still include images, text, and speech.
In a similar idea to Perceiver IO, Data2vec achieves its generality by transforming all input data into a particular representation, again the latent space encoding of the most important parts. The model then learns on this representation in a self-supervised manner, rather than the original data.
The learning is done by the combination of a teacher network and a student network. The teacher turns whatever the input data is into a target representation, then the student predicts the representation with part of the data hidden. This is then iterated until the student network is able to predict the data. Currently the exact method of turning the data into its representation depends on whether it is images, text, or speech.
The animation from the original Data2vec blog gives a schematic illustration of the learning process:
Because all types of input data are formed into a specific representation, the model's outputs are also in the form of this representation. This then has to be translated back to the external format to see the model's predictions directly.
Like Perceiver IO, the general model here not only equals the performance of previously specialized models, but outperforms them for computer vision and speech tasks. The NLP text tasks are also done well. The data they tested it on includes ImageNet for vision, LibriSpeech for speech, and GLUE for text.
The method is not limited to the 3 types of data studied here. So long as the data can be converted into the model's representation, it can be used.
Data2vec has a public GitHub repository, which can be run on Paperspace by pointing to it from a Gradient Notebook in the usual manner.
### Gato
Like their Perceiver IO, DeepMind's Gato (blog, paper, Wikipedia) is also based upon a transformer neural network. The difference from Perceiver IO is that a single Gato model learns all at the same time how to do many different tasks.
The tasks include playing Atari video games, captioning images, text chat with a user, stacking blocks with a robot arm, navigating simulated 3D environments, following instructions, and others.
As with Perceiver IO and Data2vec, the different data types are all processed into a single more general type to pass through the network. Here, it is a sequence of tokens:
• Text is encoded into subwords
• Images are sequences of non-overlapping patches in raster order, then normalized
• Discrete values are integer sequences
• Continuous values are floating point sequences, normalized and then binned
Examples of discrete and continuous values respectively are button presses when playing a game, and proprioceptive inputs. In total, Gato was trained on 604 different tasks.
The tokenized and sequenced data is passed through an embedding function. What the function does depends on the type of input data, and the result is the vectors input into the model. This again follows the idea of all types of input being converted into the same representation within the model.
The diagram from the original blog entry shows a sketch of how things pass through, and some of the tasks that can be performed:
The model is deployed by applying the inverse of the encoding + sequencing that was performed on its inputs to the model's outputs to give its next action to perform. It is sent an initial observation of the environment, then it acts, observes the new environment, acts again, and so on.
Gato's performance relative to more specialized models depends upon which task it is performing. Because results are presented for 604 different tasks, we don't attempt to quantify them all here, but very roughly, the same trained model with the same weights gets most of the way to expert performance for most tasks. This is different to Perceiver IO and Data2vec, where sometimes those models were outperforming the state-of-the-art, but this model is doing the most different things.
There is therefore room for improvement on any single task, but this is an example of a general method where all of it parts will benefit from ever increasing computing power in the future. The model size presented was held down to be usable for real-time robotics as one of its tasks, and so at 1.2 billion parameters it is relatively small by current standards. So it is in fact striking how many things it can do at this scale, and the scaling analysis indicates further room for improvement if it were to be trained with more parameters.
Of the 3 models discussed here, Gato is the most proto-AGI-like because it is one model that can do many things all at once. And indeed, the paper devotes more text than the papers for the other two models to discussing its broader context in the evolution and safety of AI.
For now it remains much closer to existing deep learning models than to a future superintelligence. But these models will only get better as time goes on.
Gato doesn't have a public version yet, so it is not available to run on Paperspace.
## Generalist models on Paperspace
Bring this project to life
Now let's look at how to run a generalist model on Paperspace. We will focus on Perceiver IO. Click the link above or follow the instructions below to launch Perceiver IO in a Gradient Notebook.
The original JAX-based content from DeepMind has been implemented on PyTorch in an excellent GitHub repository and blog on Hugging Face by Niels Rogge. While JAX can be run on Paperspace, we use this PyTorch repository as our route to running Perceiver IO.
To run it on Paperspace, start a Gradient Notebook in the usual way. Use the PyTorch runtime, and all options on their default settings, except for:
The GitHub repository for this Workspace contains many other models besides Perceiver, so to run it, navigate to the Perceiver directory, where we will find five .ipynb Jupyter notebooks.
Each one can be opened and run using Run All, or by stepping through the cells.
We recommend restarting the notebook kernel after a run so that the GPU does not run out of memory on a subsequent run. We can also check the GPU memory usage at any time by using either the metrics tab on the left-hand GUI navigation bar, or nvidia-smi in the terminal.
💡
Note: For Perceiver_for_Multimodal_Autoencoding.ipynb, we will need to run pip install accelerate before running the notebook. Either do so from the command line in the terminal, or add a cell to the notebook containing !pip install accelerate. If we choose the cell method, restart the notebook kernel before proceeding to run the rest of the cells, so that accelerate can be imported.
The Perceiver IO functionalities shown in the 5 notebooks are as follows:
• Fine_tune_Perceiver_for_text_classification.ipynb: This shows text classification and runs on a subset of the Internet Movie Database (IMDB) dataset, performing binary classification of whether a movie review is good or bad. Perceiver IO doesn't need to tokenize text first, so the bytes are provided directly as input. It goes through from fine-tuning training to showing inference on the test set, and the prediction for the review being good or bad.
• Fine_tune_the_Perceiver_for_image_classification.ipynb: This shows image classification into 10 classes using a subsample of the CIFAR-10 dataset. The model trained on ImageNet has its final output replaced to classify on CIFAR-10. The mode evaluates to 60s or 70s percentage accuracy, but it could be trained on more data, with more augmentation than the notebook shows by default. (As mentioned earlier, when fully trained Perceiver IO can match the performance of other models that were designed specifically for images.)
• Perceiver_for_Optical_Flow.ipynb: This uses 2 frames of the Sintel video dataset, with optical flow being to interpolate between them at the pixel level. The resulting flow is then visualized using Gradio. While use of the Perceiver simplifies many tasks, this one simplifies versus a specialized setup more than most.
💡
Note: If you want to use Gradio, make sure that the launcher has "Share" set to True by changing the line iface.launch(debug=True) to iface.launch(debug=True, share=True). Otherwise, you will not be able to open the popup window, as Gradient doesn't currently support port forwarding.
• Perceiver_for_Multimodal_Autoencoding.ipynb: Multimodal autoencoding means that the model is working with more than one type of input data at the same time. In this notebook, the UCF101 dataset is loaded, using its video, audio, and class label data. Multimodal pre- and post-processing is used, and the resulting output shows the reconstructed video, audio, and predicted class labels for what the video is showing.
• Perceiver_for_masked_language_modeling_and_image_classification.ipynb: This shows masked language modeling and image classification on the same model architecture, resulting in correct text and classification outputs.
## Conclusions
We have introduced the generalist models Perceiver IO, Data2vec, and Gato, and shown Perceiver IO running on Paperspace. It performs tasks related to images, text, audio, and video. Each one is straightforward to load up and run, in the usual Paperspace manner.
These models represent a step forward in generality compared to most models. Particularly significant is that they not only generalize the earlier models, but in some cases also outperform them.
While a true artificial general intelligence is still some way in the future, in the shorter term these models may both simplify and broaden the reach and capability of AI and end-to-end data science.
The Perceiver IO Hugging Face content, like their blog, is based upon the original content from DeepMind, but implements it in PyTorch + .ipynb rather than JAX + .py.
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# Thread: Boundary pts Vs. limit pts.
1. ## Boundary pts Vs. limit pts.
Hello everyone
I would like to know if there is any difference in an "analytic" pov between a limit point and a boundary point... Or answer me: ]a,b[ ... a and b are the boundary points and the limit points??
2. Originally Posted by rebghb
I would like to know if there is any difference in an "analytic" pov between a limit point and a boundary point... Or answer me: ]a,b[ ... a and b are the boundary points and the limit points??
Consider the set $M = (0,1] \cup \left\{ 2 \right\}$.
The boundary points of $M$ are $0,~1,~\&~2$.
But $2$ is not a limit point of $M$.
However, any point in $[0,1]$ is a limit point of $M$.
3. okay i got that... what can we say about 2 then?? is it isolated?? and is set M bounded?? (ps im nwe to this so patience please)
4. A boundary point that is not a limit point is an isolated point.
If there is a $B>0$ such that $M\subseteq [-B,B]$ then the set is bounded.
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### Home > CCG > Chapter 8 > Lesson 8.1.2 > Problem8-23
8-23.
Multiple Choice: Which equation below is not a correct statement based on the information in the diagram?
1. $3x + y = 180°$
1. $2x-1°=4°-x$
1. $2x-1°=5y-10°$
1. $2x-1°+3x=180°$
Read the Math Notes box in Lessons 2.1.1 and 2.1.4 about angle pair relationships.
B because $2x-1º$ and $4º-x$ are a linear pair so $2x-1º+4°-x=180$, not $2x-1º=4º-x$.
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# Text Environments: Vertical and Horizontal
How does one obtain two text environments in TeX - one vertical, one horizontal, as in the attached image? It would be useful to be able to specify two text environments per page akin to something like this:
\vertical{This text appears on the side}
\horizontal{This text appears horizontally}
While being able to specify the width available to text appearing vertically and having the horizontal text naturally flow down the page naturally - as if were a document with shortened page width. I tried implementing this with minipage and rotation but was not able to create coherent environments. Is there a 'natural way' of doing this?
• Will the content only remain on a single page? Dec 14 '17 at 6:45
• @Werner I was only ever planning on having a single page Dec 14 '17 at 17:31
I tried implementing this with minipage and rotation but was not able to create coherent environments.
This should be enough.
\documentclass{article}
\usepackage{graphicx,lipsum}
\begin{document}
\rotatebox[origin=c]{90}{%
\begin{minipage}{.7\textheight}
\lipsum[2]
\end{minipage}}
\hfill
\begin{minipage}[c][.7\textheight][t]{.65\linewidth}
\lipsum[1]
\end{minipage}
\end{document}
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Horizontal fixing of Tikz nodes
(Please feel free to offer a better and more precise subject title.)
I have two vertical chains with different count of nodes in it. But I want to visualize that some nodes of the right belong to one of the left. They need to appear in the same row/line. In the example code below you see Y and two nodes should appear on the right side of it.
The problem here is that I can not handle the right chain like it has 4 nodes and join them. Using \matrix make it impossible to use chain.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{matrix,positioning,chains,scopes}
\begin{document}
\begin{tikzpicture}
[
myright/.style={
draw},
myleft/.style={
myright,
on chain,
fill=cyan!30}
]
% LEFT
{[start chain=L going below,
every node/.append style=myleft]
\node {X};
\node {Y};
\node {Z};
}
% RIGHT
{[start chain=R going below]
\node [myright,right=of L-1] {belong to X};
\matrix [right=of L-2]
{
\node [myright] {belong to Y};\\
\node [myright] {also belong to Y};\\
};
\node [myright,right=of L-3] {belong to Z};
}
\end{tikzpicture}
\end{document}
btw: This question is related to this one about drawing PRISMA flow chart.
• If you place two nodes in branch R and both belong to Y, they cannot both be aligned horizontally with Y because R is going below. And what about the final node? Where is C which it should align with? – cfr Jan 28 '16 at 22:23
• The TikZ manual has an example which combines a matrix of nodes with chains, by the way. – cfr Jan 28 '16 at 22:26
• The example mentioned by cfr is in page 70. – Ignasi Jan 29 '16 at 9:51
• The thing is, it isn't clear what you are trying to do. You cannot align 2 nodes horizontally with Y and align those 2 nodes vertically with the node with content belong to X. The way you describe the nodes as children of nodes on the left suggests that your branches should really be going right, with the main chain going below rather than the main chain going right with branches going below. (Or if this is really a tree, maybe a tree structure would work better.) Right now, you appear to want a spatio-temporal impossibility. The TikZ gurus are good. But not that good ;). – cfr Jan 29 '16 at 20:43
• @Ignasi Page 545 in the version I have. – cfr Jan 29 '16 at 22:31
As I understand the current version of this question, the picture you've drawn places the nodes correctly but does not allow them to be joined with the ease of chained nodes, because the right column of nodes are not chained. Also, perhaps you want some separation between the two nodes belonging to Y, though I'm not sure.
Note that the best way to do this kind of thing is very heavily dependent on the details of the diagram in question, your own familiarity with different aspects of TikZ and different libraries, and whether the diagram is a one-off or one of a sequence of similar diagrams. There's a balance to be struck here between efficiency/elegance in coding the diagram (which might involve a great deal of effort to set up) and ease of set up (if any).
Just given the current example, if I've understood the issue correctly, I would try something like this.
1. Create the left hand chain of nodes, including a branch from X going right.
2. Place the topmost right node on the branch from X.
3. Position a matrix to the right of Y, aligning with the top right node.
4. Position the bottom right node to the right of Z, aligning with the top right node.
5. Create the second, right hand chain, adding the nodes on the right to the chain and joining them as they are added.
This is a little messy in terms of naming and mix of methods but you end up with two chains of nodes which can be easily joined and referred to. Note that there is no problem with a node being in more than one chain or a matrix and a chain or....
If you do not want to separate the nodes belonging to Y, delete the row sep. If you do not want them joined, delete the relevant [join].
\documentclass[tikz,border=10pt]{standalone}
\usetikzlibrary{matrix,positioning,chains,scopes}
\begin{document}
\begin{tikzpicture}
[
myleft/.style={
draw,
fill=cyan!30}
]
% LEFT
{[start chain=L going below,
every on chain/.append style={myleft},
every node/.append style={on chain},
]
\node {X};
{[start branch=X going right,
every on chain/.append style={fill=none}
]
\node {belong to X};
}
\node {Y};
\node {Z};
}
\matrix (m) [row sep=2.5pt, matrix of nodes, every node/.append style={draw}] at (L-2 -| L/X-2)
{
belong to Y\\
also belong to Y\\
};
\node (z2) [draw] at (L-3 -| L/X-2) {belong to Z};
% RIGHT
{[start chain=R going below]
\chainin (L/X-2);
\chainin (m-1-1) [join];
\chainin (m-2-1) [join];
\chainin (z2) [join];
}
\end{tikzpicture}
\end{document}
• (+1) .. @cfr, in the first question (on which OP provide link), the desired picture was very different from those in this question. So I'm after four attempting to help OP completely confused, what he like to have. Use matrix for nodes with common "blue" nodes" is very good! – Zarko Jan 29 '16 at 23:34
• @Zarko Yes. I know. The question was and, really, still is very unclear. (See also my comments on the question.) I am still not certain this is what is wanted, but it seemed at least a little clearer. I notice that C is no longer an issue in this version. I'm pleased about that because I never could figure out where or what C was ;). – cfr Jan 29 '16 at 23:49
Now your example is far more simple as picture present on given link. Main differences is, that the right branch has only vertically spaced nodes. Considering this is probably on grid option what you looking for:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{calc,positioning,chains,scopes}
\begin{document}
\begin{tikzpicture}[
node distance = 10mm and 30mm,
on grid,
start chain = A going below,
start chain = B going below,
myright/.style = {draw, minimum height=4ex, minimum width=33mm,
on chain=A},
myleft/.style = {draw, fill=cyan!30, minimum height=4ex,
on chain=B}
]
% LEFT
\begin{scope}[every node/.style={myleft}]
\node {X}; % name=B-1
\node {Y};
\node {Z};
\end{scope}
% RIGHT
\begin{scope}[every node/.style={myright}]
\node [right=of B-1] {belong to X}; % name=A-1
\node {belong to Y};
\node {also belong to /};
\node {belong to no one};
\end{scope}
\end{tikzpicture}
\end{document}
In above MWE distances of grid is determined by node distance. Nodes are anchored to the grid with its center. This means, that if they aren't of the same width, they will not have aligned left (or right) borders.
Two chains are selected for simplest node naming (left are B-1, B-2, B-3, right are A-1, A-2, A-3 and A-4).
To this two branches it is easy add one more branch on the right (as has mentioned picture), but more effort is needed, if on the of the right branch are two nodes ...
Addendum (1): Partly considered images on question flowchart-tikz, the upgrade of the above MWE is:
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,chains,positioning,scopes}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{3mm}
\makeatletter
%---------------------------------------------------------------%
\tikzset{
suspend join/.code={\def\tikz@after@path{}},
node distance = 13mm and 44mm,
on grid = true,
start chain = A going below,
start chain = B going below,
MR/.style = {% My Right
draw, minimum height=4ex, text width=31mm,
inner sep=1mm, align=center, % left?
on chain=A},
ML/.style = {% My Left
draw=cyan!60!black, rounded corners, fill=cyan!30,
minimum width=4ex, inner sep=1mm,
node contents={\rotatebox{90}{#1}},
on chain=B},
arrow/.style = {thick,-{Triangle[]}},
}
%---------------------------------------------------------------%
\makeatother
\begin{document}
%---------------------------------------------------------------%
\begin{tikzpicture}
% LEFT BRANCH
\node [ML=X]; % name=B-1
\node [ML=Y];
\node [ML=Z];
\node [ML=WWW];
\node [ML=QQ];
% RIGHT BRANCH
\begin{scope}[every node/.style={MR,join=by arrow}]
\node [draw=none,
right=of B-1] {}; % name=A-1, auxiliary node which serves
% as placeholder for real left and right node
% (A-L) and (A-R) defined latter in
% the TOP part of this code
\node [suspend join] {belong to Y};
\node {belong to Z};
\node {belong to WWW, however text in this node has three lines};
\node {belong to QQ};
\end{scope}
% TOP ROW LEAVES (horizontally are not on grid)
\begin{scope}[node distance=2mm,
every node/.style={MR}]
\node (A-L) [ left = of A-1.center] {left top leave};
\node (A-R) [right = of A-1.center] {right top leave};
\end{scope}
% RIGHT LEAVES
\node (C-1) [MR, right = of A-3] {upper right leave};
\node (C-2) [MR, right = of A-4] {lower right leave};
% ARROWS NOT DETERMINED BY "JOIN" MACRO
\path[arrow] (A-L) edge (A-2) (A-R) edge (A-2)
(A-3) edge (C-1) (A-4)edge (C-2);
\end{tikzpicture}
%---------------------------------------------------------------%
\end{document}
which gives:
The contains (i hope so) enough comments that is its structure is clear. Mai goal in code writing has been its concise and to be clear. In comparison with the first MEW, it has novelty suspend join, by which is simplified use of join macro in right branch of flowchart.
Addendum (2): Your question is like *Variable structure System, for which seems not fulfill Filipov's criteria (and consequently without sliding to some equilibrium point ...) .
Let assume:
• that second and third node (from top down) in right branch are have common blue node in the left branch
• height of these nodes in right branch is not fixed
• blue node (in the left branch) span distance from top of upper node to bottom of lower node (in right branch, to which it belong)
This can be achieved with following:
• measuring distance, which blue node should span and separeteli design this node
• introduce fake nodes for maintain left branch on grid
For this I add two libraries: calc and fit, define fake nodes and real left nodes. The code has enough comments (I hope so), that it is easy to anderstand, what it doing.
\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{arrows.meta,calc,chains,fit,positioning,scopes}
\usepackage[active,tightpage]{preview}
\PreviewEnvironment{tikzpicture}
\setlength\PreviewBorder{3mm}
\makeatletter
%---------------------------------------------------------------%
\tikzset{
suspend join/.code={\def\tikz@after@path{}},
node distance = 13mm and 44mm,
on grid = true,
start chain = A going below,
start chain = B going below,
MR/.style = {% My Right node
draw, minimum height=4ex, text width=31mm,
inner sep=1mm, align=center, % left?
on chain=A},
ML/.style = {% My Left node
draw=cyan!60!black, rounded corners, fill=cyan!30,
minimum width=4ex, inner sep=1mm,
node contents={\rotatebox{90}{#1}},
on chain=B},
FL/.style = {%Fake Left node
node contents={\rotatebox{90}{\phantom{X}}},
on chain=B},
RL/.style = {%Fake Left node
draw=cyan!60!black, rounded corners, fill=cyan!30,
minimum width=4ex, inner xsep=0pt,
label=center:\rotatebox{90}{#1},
node contents={}},
arrow/.style = {thick,-{Triangle[]}},
}
%---------------------------------------------------------------%
\makeatother
\begin{document}
%---------------------------------------------------------------%
\begin{tikzpicture}
% LEFT BRANCH
\node [ML=X]; % name=B-1
\node [FL]; % auxiliary node which maintain branch on grid
% real nodes will take a place latter
\node [FL];
\node [ML=WWWW];
\node [ML=QQ];
% RIGHT BRANCH
\begin{scope}[every node/.style={MR,join=by arrow}]
\node [suspend join,
right=of B-1] {belong to X}; % name=A-1,
\node {belong to Y};
\node {also belong to Y};
\node {belong to WWWW, however text in this node has three lines};
\node {belong to QQ};
\end{scope}
% REAL BLUE NODE, instead of FL
\path let \p1 = (B-2 |- A-2.north),
\p2 = (B-2 |- A-3.south),
\n1 = {veclen(\y2-\y1,\x2-\x1)} in
node[RL=Y,
minimum height=\n1,
fit=(B-2) (B-3)];
% RIGHT LEAVES
\node (C-1) [MR, right = of A-3] {upper right leave};
\node (C-2) [MR, right = of A-4] {lower right leave};
% ARROWS NOT DETERMINED BY "JOIN" MACRO
\path[arrow] (A-3) edge (C-1) (A-4)edge (C-2);
\end{tikzpicture}
• Sorry, I don't made my point clear. Please see my code. There are two "child nodes" for Y. Your exampel just fitted simple one node to anoathor on the horizonatl axes. – buhtz Jan 29 '16 at 20:34
• Sorry, my crystal ball is still broken ... if I understand you now correctly, than -- according to my picture -- nodes "belong to Y" and "belong to Z" has one blue node, which span distance from top of blue node "Y" to bottom of node "Z"? If this is true, than this node can not be placed on the grid anymore, consequently, the code will become quite more complicated ... – Zarko Jan 29 '16 at 20:59
• I totaly adjusted my question with new code and a screenshot. – buhtz Jan 29 '16 at 21:10
• Not adjusted ... you changed your question! If the figure, you now show in question generated by your code, what is thenwhat is wrong with your solution? – Zarko Jan 29 '16 at 21:20
• What does the on grid do, especially in the first version? Does it do anything not done by the chains already? – cfr Jan 29 '16 at 23:50
|
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Laser-Phase Dependence for Electron Capture in Laser-Assisted Proton -- Hydrogen Collisions
Thomas Niederhausen, Uwe Thumm
( James R. MacDonald Laboratory, Kansas State University, Manhattan, KS 66506-2604, USA)
We calculate electron capture probabilities for ion--atom collisions in a strong laser field ($5\times10^{13}~\mathrm{W/cm^2}$) by numerically solving the 3-dimensional time--dependent Schr\"{o}dinger equation. For circularly polarized laser fields and an impact energy of $1.2~\mathrm{keV}$, we find a substantial modification of the electronic dynamics in the $p$--$H$ collision system as compared to field-free collisions. In particular, we observe a strong dependence on the laser phase and the impact parameter for electron capture, which can be explained using semi-classical arguments.
This work was supported by the Chemical Sciences, Geosciences and Biosciences Division,
Office of Basic Energy Sciences, Office of Science, U.S. Department of Energy.
Submitted to DAMOP, May 2006 in Knoxville, TN.
|
|
# Loops in tikz for accessing labels
I would like to create something like this drawing using Tikz:
I found a nice tikz example that creates something similar:
\documentclass[border=10pt]{standalone}%
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
\foreach \n in {0,...,4} {
\foreach \k in {0,...,\n} {
\node at (\k-\n/2,-\n) {${\n \choose \k}$};
}
}
\end{tikzpicture}
\end{document}
I thought it would be trivial to change the label at a node by accessing the label on a list, something like lst[\n], where lst contains monomials, but it turned out to be more complicated than I thought. Besides, I couldn't find a way to add the exponent programatically. Is this possible with tikz using loops, or do I have to create an ad-hoc solution?
\documentclass[border=10pt]{standalone}%
\usepackage{ifthen}
\usepackage{tikz}
\usetikzlibrary{math}
\begin{document}
\newcommand{\exponent}[2]{\ifthenelse{#2=0}{}{\ifthenelse{#2=1}{$#1$}{$#1^{#2}$}}}
\begin{tikzpicture}
\foreach \x in {0,...,7} {
\foreach \n in {0,...,\x} {
\pgfmathparse{int(subtract(\x,\n))}
\edef\xexp{\pgfmathresult}
\edef\nexp{\n}
\node[inner sep = 8pt] (X\xexp-N\nexp) at (2*\n-\x,-\x) {\ifthenelse{\xexp=0 \AND \nexp=0}{$1$}{\exponent{\xi}{\xexp}\exponent{\eta}{\nexp}}};
\node[inner sep = 16pt] (dX\xexp-N\nexp) at (2*\n-\x,-\x) {}; % d: dummy
}
}
\draw[blue, thick, dashed] (X0-N0.north) -- (X1-N0.west) -- (X1-N1.south) -- (X0-N1.east) node[inner sep = 0] (S1) {} -- cycle;
\draw[red, thick, dashed] (dX0-N0.north) -- (dX4-N0.west) -- (dX4-N1.south) -- (dX3-N1.south) --
(dX1-N3.south) -- (dX1-N4.south) -- (dX0-N4.east) node[inner sep = 0] (S4) {} -- cycle;
\node[right = 0.5cm, above = 0.5cm] (S1t) at (S1.west) {$S_1(\Omega_{\rm st}^{(q)})$};
\draw[blue] (S1t) edge[thick, out=270, in=90, ->] (S1);
\node[right = 0.5cm, above = 0.5cm] (S4t) at (S4.west) {$S_4(\Omega_{\rm st}^{(q)})$};
\draw[red] (S4t) edge[thick, out=270, in=90, ->] (S4);
\end{tikzpicture}
\end{document}
• Impressive! That’s exactly what I was looking for. Thanks! – aaragon Jan 23 '18 at 19:20
• @aaragon Happy to help – caverac Jan 23 '18 at 20:29
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