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# Firmware downgrade for LG Smart TVs

I must admit, I’m baffled to see LG pull a stunt like this after having faced public backlash over their spying on their customers’ TV viewing habits. After backpedaling on the issue last year, they’ve now done a full 180 and reserve the right to record your entire viewing habits and share this data with everyone and their mom, even in countries with most likely less strict data protection laws. And here’s the kicker: if you don’t agree to these terms, many of the SmartTV features you paid for will be lost!

That’s apparently par for the course cause LG also removed the ability to switch audio tracks on local DLNA playback some firmwares ago (alledgedly due to copyright reasons). And when I set my LG TV to a non-existant gateway to stop it from sending my data to LG (my old Fritzbox does not blacklist), I could not connect to my local NAS anymore despite it being in the same network segment.

Also bad is the fact that, probably due to being embarrassed on a regular basis by all the bugs and removed features of their firmwares, LG does not offer changelogs. So when you’ve updated your TV’s firmware on the slim hope of seeing some improvements/fixes, you’ll notice the new terms you’ll be forced to agree to only after the fact! And since LG offers no way to downgrade the firmware, you’re suddenly stuck with a device with less features and more spying on you than you bargained for.

Fortunately, some Russian hackers felt the need to adress the issue and reverse engineered the firmware update process and made a small business out of giving the LG customers what they want. Others took note and devised a do-it-yourself approach to downgrade your TV (thx, nenif).

How to downgrade your LG TV’s firmware

NOTE: I’M NOT RESPONSIBLE FOR ANYTHING THAT MIGHT GO WRONG WITH THE DOWNGRADE – YOU’RE DOING THIS AT YOUR OWN RISK!

I can’t stress that enough! Apparently some people have had weird experiences with their downgrade attempts. If you have other models/versions than those mentioned here as being applicable, don’t try this or you might brick your device!

• Download LgDTVUpDater_new.zip and unzip it to your drive c:\ so that you end up with a dir called c:\LgDTVUpDater (another destination will likely result in error messages later on).
• Check out the folder “Antwort” inside the htdocs folder. That’s a translation of “response”, meaning the server’s responses people already received and listed for your convenience. Browsing the files inside you can see links to lots of older firmwares.
• Download the firmware you want to downgrade to and unzip it to C:\LgDTVUpDater\htdocs\
• Open C:\LgDTVUpDater\dns\AcrylicHosts.txt in an editor like Notepad++ and change the IP adress of snu.lge.com in the last line to your PC’s address (in my case 10.10.10.40), then save the file. As suggested in the comments, downgrading from more recent firmwares might require to add additional servers: rdvs.alljoyn.org, fi.lgtvsdp.com, us.lgtvsdp.com – one per line – with the same IP address.
• Start C:\LgDTVUpDater\startDNS.bat and then C:\LgDTVUpDater\startHTTP.bat with a double-click resulting in 2 black windows (showing that the Acrylic DNS server and the Apache server are running) and probably your firewall complaining. In that case, allow both.
• On your TV select Settings > Network Connection > Start Connection > Cancel > Manual Setup > Wired (firmware updates are huge and any disconnects over wireless might lead to bricking your TV). Keep IP MODE on Auto Input, set DNS MODE to Manual Input and enter your PC’s IP address
• Select Settings > Support > Software Update > Check Update Version but do not update! Close when the check has been performed. This check generates a file with some needed info on your TV’s model.
• Open C:\LgDTVUpDater\htdocs\in_dump.txt and copy your TV’s model name from inside the MODEL_NM-tag. For my LN5758 it’s
HE_DTV_NC4M_AFAAABAA
• Check C:\LgDTVUpDater\htdocs\ if a file of that name already exists (these are text files with a simple xml structure inside, they have no extension like .txt or .xml). If not, copy one of the other files over to that name
• Edit the new file, in my case HE_DTV_NC4H_AFAAABAA. It will look quite jumbled cause there are no line breaks. If you add them manually, it looks something like this:

• There are only a few entries to change:
IMAGE_URL: the IP of your PC
IMAGE_NAME: name of the firmware file you unzipped
CDN_URLIP: IP of your PC and name of the firmware

You can also change
UPDATE_MAJOR_VER: major version of the firmware
UPDATE_MINOR_VER: minor version of the firmware

but this should only be cosmetic cause the revision number is also stored inside the epk. The FORCE_FLAG set to Y should ensure any current version being overwritten with the one you’ve downloaded, thus enabling the downgrade in the first place.
Save the file.

• Now close the two command prompt windows and double-click again on C:\LgDTVUpDater\startDNS.bat and then C:\LgDTVUpDater\startHTTP.bat
• On your TV select Settings > Support > Software Update > Check Update Version: a new firmware is found! Update.
• When the update is through (which should be pretty fast since it’s done from your PC), the TV will reboot.
• Upon restart, you’ll get a message window with another progress bar (in my case a yellow frame with the foreign language text flipped). The firmware you just downloaded to your TV is now being installed. Once this is through, the downgrade is complete.
• For good measure you should reset the TV to factory defaults. Please note, that you will have to scan and sort channels after that. Even if you don’t reset, you’ll have to agree to this firmware’s terms of use before you can use Smart TV functionality. As mentioned above, these should be less restrictive than the ones you had to agree to before. If not, downgrade to an even older version ;)

Feel free to add model numbers of TVs you successfully downgraded or links to firmwares in the comment section.

# LG 42LN5758 – Why LG? Why indeed!

True, it’s been a while since my last entry. Regrettably that doesn’t mean I haven’t seen crappy devices over the last few years. However, of all the devices I’ve tested recently, the LG annoyed me enough to write about it, so here it goes. I figured, Amazon’s Cyber Monday might be the right time to get a bargain on a new TV for the inlaws and the feature list of the LG 42LN5758 looked appealing. Little did I know at the time of LG’s sneaky way of logging info on their users’ viewing habits as uncovered by DoctorBeet. And sure enough, my model on firmware 03.20.38 does send the filenames of all media files on an attached USB drive to an LG server:

Of course, I could block the URLs in my router (there’s not even an option in the menu to turn this off) or wait for a firmware update dealing with this once LG figured out, the public backlash wasn’t worth the extra pennies they might gain from advertisers (apparently it’s time now, please check the update at the bottom of this page). However, I’ve read a lot of complaints about firmware updates removing features with no chance of getting them back as well as LG making it next to impossible to downgrade to a previous version.

That’s obviously par for the course cause LG makes it needlessly complicated to do the initial channel scan: you have to select ALL instead of ASTRA and then NONE which is quit unintuitive if you’ve seen the menu. Hilariously enough, they seem to think their method of sorting channels on the TV with the remote involves only a few simple steps “making your life so much easier”, I kid you not. To transfer the channel list to USB to do the sorting on your PC, you have to take this clumsy approach:

– use a regular LG remote (fortunately there’s no oh-so-improved ‘Magic Remote’ or Logitech Harmony required)
– press and hold SETTINGS until the channel info appears at the top of the screen
– release immediately and as fast as you can enter ‘1105’ followed by OK
– in the upcoming Installation Menu select USB Cloning > Send to USB
– sort the channels in Chansort on your PC and repeat the process

Why does it have to be this complicated? Why can’t the channel list be copied from a regular menu? Beats me! Apparently all TV manufacturers consider this data to be so much more than a bunch of parameters, easily being stored and edited in a simple CSV file like this

Pr#;Channel Name;Favorites;Lock;Skip;Hide;Encrypted;Satellite;Ch/Tp;Freq;ONID;TSID;SymRate;SID;VPID;APID 40;CNN Int.;0;;;;;ASTRA 19.2E;;11778;1;1068;27500;28522;165;100
they make you jump through extra hoops to get it.

The same goes for USB recording: it’s embarassing to see manufacturers – in an act of preemptive obedience towards rightsholders – encrypt all lawful (!) recordings so they cannot be watched elsewhere. But what’s the point of using exotic filesystems which renders a recording USB HDD useless for anything else? The Samsung UE40D8090 uses XFS as a filesystem but writes to drives already partitioned with it. Thus I can split my drive into NTFS/XFS partitions and play back media files from the NTFS partition and record to the XFS partition, all on the same drive. Not so with the LG, even an already paritioned drive (with JFS in this case) will be formatted. The LG apparently creates an extra 50 MB ext3 partition with a 500 KB binary ‘LGDB.vol’ file on it and if even if you keep this partition/file when resizing the JFS partition and adding a new NTFS partition, the LG deems the drive unusable and wants to format it again which prevents you from using it for anything else.

Also annoying is the fact that you cannot switch inputs fast cause pressing INPUT on your remote usually opens the fancy input list where repeated presses of INPUT do cycle through the input sources which still need to be confirmed by pressing OK. Switching from USB playback to TV like this, however, does not work, you need to press BACK (which apparently manufacturers just cannot agree on where to place on a remote) several times, depending on how deep down you were in the folder structure of your USB device, to get back to the input list where only then you can select another source. Worse, to eject the USB device you first need to switch to another source to be able to call up the Quick Menu (Q.MENU on the remote) where you need to press RIGHT 11 (!) times to reach the ‘Eject USB device’ button (how’s that for quick?) cause the generally commendable first-to-last/last-to-first vertical navigation in menus inconsistently does not apply to right/left navigation.

The main issue for the inlaws however is that one of them is hard of hearing and therefore needs to listen to the TV with a wireless TV headphone while the spouse listens via the TV’s speakers (a not too uncommon arrangement, IMO). Their current CRT TV has a cinch line-out so the speakers do stay on with the volume-adjustable headphones attached. Unfortunately the LG only has a cinch line-in and the headphone jack mutes the speakers once a headphone is plugged in. And of course you cannot turn this off and therefore cannot have audio over headphones and speakers simultaneously.

So why is there an option in the audio menu to switch between speakers and headphones, you ask? My guess is that some of the geniuses at LG eventually figured out that the placement of the headphone jack (which looks like a dummy connector, by the way) on the rear side (!) of the TV would make it inaccessible with the TV being mounted to the wall. Guess how many customers did not think of this and had to unscrew the TV again. That was probably the same genius that placed the volume/channel/settings buttons on the rear side of the device too.

While waiting for LG customer support to comment on this, I found various entries on popular consumer electronics boards suggesting this feature has been requested by customers like forever. And sure enough, LG support told me the only way to get audio from speakers and headphone simultaneously would be to use a SCART to RCA adapter which I’d already tried in the meantime resulting in speakers and headphone being slightly out of sync and therefore unusable but especially as there is no sound coming out of the headphones when recordings/media files are played back from an attached USB device! Facing these facts, the support acted in true LG fashion by shifting the blame on me, suggesting they’d expect customers to conduct proper research into a device’s features prior purchase.

Of course I didn’t bother to ask how the phrase ‘AUDIO menu items are disabled when connecting a headphone’ from the manual PDF could possibly be construed to mean that simultaneous audio is impossible, nor did I bother to test the validity of the phrase ‘Optical Digital Audio Out is not available when connecting a headphone’ by buying a Toslink to Cinch converter. I simply decided to switch to another brand.

Why LG? Why indeed!

Update, December 14th
Apparently LG has finally caved and released update 04.20.29 (currently only available via the TV’s support menu) that stops sending unencrypted user data and USB filenames to LG (HBBTV data apparently is still being sent to the networks). You can verify this by downloading this dump file and browsing it in Wireshark. Right-click > save as and once downloaded rename to .zip (I can’t be bothered to purchase WordPress’ Space Upgrade just for a new filetype). While logging this file I briefly watched ARD (with HBBTV) and ZDF before switching to USB and playing back the file “LGsux.avi”.

While it’s nice to see that LG finally got it, it took them two intermediate updates (04.00.49 and 04.04.05) the latter of which forced you to agree to a new privacy policy which only was possible AFTER the upgrade! Failing to accept the new terms of the Collection and Use of Data (PDF in German), would result in not being able to adjust basic settings, like choice of satellite. And to top it all off, LG apparently changed the TTL file format in this one which means, channels cannot be sorted on the PC anymore until Chansort is updated to handle this!

In case you were wondering: NO, of course LG did NOT enable simultaneous audio over TV speakers and headphones in any of these updates. Hey, it’s only been requested for 2 years or so…


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# Fight Finance

#### CoursesTagsRandomAllRecentScores

 Scores keithphw $5,911.61 cuiting$1,089.70 Visitor $1,088.61 Skywalke...$1,070.00 Carolll $893.33 Visitor$854.70 Visitor $840.00 trungbin$803.09 Jade $785.80 alison$771.70 Visitor $760.00 zy$679.70 ninalee $669.70 Visitor$650.00 Emma Lu $650.00 Visitor$637.00 Nisrita $620.33 Visitor$603.33 Visitor $600.00 Visitor$594.05

The security market line (SML) shows the relationship between beta and expected return.

Investment projects that plot above the SML would have:

You want to buy an apartment worth $400,000. You have saved a deposit of$80,000. The bank has agreed to lend you the $320,000 as a fully amortising mortgage loan with a term of 30 years. The interest rate is 6% pa and is not expected to change. What will be your monthly payments? A student won$1m in a lottery. Currently the money is in a bank account which pays interest at 6% pa, given as an APR compounding per month.

She plans to spend $20,000 at the beginning of every month from now on (so the first withdrawal will be at t=0). After each withdrawal, she will check how much money is left in the account. When there is less than$500,000 left, she will donate that remaining amount to charity.

In how many months will she make her last withdrawal and donate the remainder to charity?

You're the boss of an investment bank's equities research team. Your five analysts are each trying to find the expected total return over the next year of shares in a mining company. The mining firm:

• Is regarded as a mature company since it's quite stable in size and was floated around 30 years ago. It is not a high-growth company;
• Share price is very sensitive to changes in the price of the market portfolio, economic growth, the exchange rate and commodities prices. Due to this, its standard deviation of total returns is much higher than that of the market index;
• Experienced tough times in the last 10 years due to unexpected falls in commodity prices.
• Shares are traded in an active liquid market.
Your team of analysts present their findings, and everyone has different views. While there's no definitive true answer, who's calculation of the expected total return is the most plausible?

Assume that:

• The analysts' source data is correct and true, but their inferences might be wrong;
• All returns and yields are given as effective annual nominal rates.

An old company's Firm Free Cash Flow (FFCF, same as CFFA) is forecast in the graph below.

To value the firm's assets, the terminal value needs to be calculated using the perpetuity with growth formula:

$$V_{\text{terminal, }t-1} = \dfrac{FFCF_{\text{terminal, }t}}{r-g}$$

Which point corresponds to the best time to calculate the terminal value?

In the Merton model of corporate debt, buying a levered company's debt is equivalent to buying risk free government bonds and:

Find the cash flow from assets (CFFA) of the following project.

 One Year Mining Project Data Project life 1 year Initial investment in building mine and equipment $9m Depreciation of mine and equipment over the year$8m Kilograms of gold mined at end of year 1,000 Sale price per kilogram $0.05m Variable cost per kilogram$0.03m Before-tax cost of closing mine at end of year $4m Tax rate 30% Note 1: Due to the project, the firm also anticipates finding some rare diamonds which will give before-tax revenues of$1m at the end of the year.

Note 2: The land that will be mined actually has thermal springs and a family of koalas that could be sold to an eco-tourist resort for an after-tax amount of $3m right now. However, if the mine goes ahead then this natural beauty will be destroyed. Note 3: The mining equipment will have a book value of$1m at the end of the year for tax purposes. However, the equipment is expected to fetch $2.5m when it is sold. Find the project's CFFA at time zero and one. Answers are given in millions of dollars ($m), with the first cash flow at time zero, and the second at time one.

A stock just paid a dividend of $1. Future annual dividends are expected to grow by 2% pa. The next dividend of$1.02 (=1*(1+0.02)^1) will be in one year, and the year after that the dividend will be \$1.0404 (=1*(1+0.02)^2), and so on forever.

Its required total return is 10% pa. The total required return and growth rate of dividends are given as effective annual rates.

Calculate the current stock price.

Alice, Bob, Chris and Delta are traders in the futures market. The following trades occur over a single day in a newly-opened equity index future that matures in one year which the exchange just made available.

1. Alice buys a future from Bob.

2. Chris buys a future from Delta.

3. Delta buys a future from Alice.

These were the only trades made in this equity index future. What was the trading volume and what is the open interest?

Which of the following quantities from the Black-Scholes-Merton option pricing formula gives the Delta of a European put option?

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# A man saves $4 more each year than the preceding year. If  post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message Director Joined: 24 Oct 2005 Posts: 659 Location: London Followers: 1 Kudos [?]: 13 [0], given: 0 A man saves$ 4 more each year than the preceding year. If [#permalink]

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A man saves $4 more each year than the preceding year. If he saves Rs 20 in the first year, after how many years will his savings amount to$1064?
a) 10
b) 12
c) 18
d) 19
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28 Nov 2005, 04:38
A man saves $4 more each year than the preceding year. If he saves Rs 20 in the first year, after how many years will his savings amount to$1064?

A(n) = A(n-1) + 4

A(1) = 20
A(2) = 20 + 4
A(3) = 20 + 4 + 4
A(4) = 20 + 4 + 4 + 4
.
.
.
A(N) = 20 + 4(N - 1)

A(1) + A(2) + A(3) + A(4) + ... + A(N)
= (20 * N) + 4(0 + 1 + 2 + 3 + ... + N-1)
= (20 * N) + 4( (N-1)N/2 )
= (20 * N) + 2(N^2) - 2N
= 2(N^2) + 18N
= 1064

N^2 + 9N = 532
N^2 + 9N - 532 = 0
(N-19)(N+28)=0

N=19 or N=-28

A year must be positive.

Thus, he needs to save for 19 years.

(E) it is.
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## Recommended Posts

Hey everyone I was wondering if you could help me out with a little issue. I am designing a 2D space adventure game and I decided that I could really add to the overall effect by adding light sources to which ships and interstellar objects would be reflective. Currently I draw the ships image, then I draw the bumpMapped image (the image with light and shadows cast on it; no color) over top of the ships actual image. I set the transparency of the bumpMapped image to a level were it doest fully cover the ships image. What I would like to do instead, is to draw a single rendered image with the combined rgb values of the bumpMapped image and the ships actual image. So for each pixel the, red, green, blue and alpha values of the bumpMapped and ship Image would be summed and averaged, giving the concatenation of the two. The code I have written for this is,
a = (((testImg.getRGB(x, y) >>> 24) & 0xFF) + ((environmentMap[nX][nY] >>> 24) & 0xFF));
r = (((testImg.getRGB(x, y) >>> 16) & 0xFF) + ((environmentMap[nX][nY] >>> 16) & 0xFF));
g = (((testImg.getRGB(x, y) >>> 8) & 0xFF) + ((environmentMap[nX][nY] >>> 8) & 0xFF));
b = (((testImg.getRGB(x, y) >>> 0) & 0xFF) + ((environmentMap[nX][nY] >>> 0) & 0xFF));

bumpImg.setRGB(x, y, (new Color(r, g, b, a)).getRGB());

this code is quite sluggish, and so really my question is. What is the most efficient way of adding two RGB values together??

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Quote:
 Original post by leif_dragonthis code is quite sluggish, and so really my question is. What is the most efficient way of adding two RGB values together??
First off, have you profiled your application to make sure that this code is indeed the bottleneck?

If you are sure that is the case, there doesn't look to be much wrong with the way you are doing things, but there are a couple of things you could do to help the performance.

Try invoking testImg.getRGB(x, y) and environmentMap[nX][nY] at the start of the function, and cache the result in a local variable, rather than calling each 4 times. If getRGB is not inlined for some reason, the function call overhead could be significant.

Try removing the whole getRGB/setRGB setup, and interpreting each image imply as a sequence of bytes (4 bytes per pixel, one each for r, g, b and a), rather than integers, to avoid all the shift and mask operations.

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Bit-shifting is extremely quick so I doubt very much that is the problem. I would cache the ARGB values as suggested by swiftcoder instead of calling the getRGB method multiple times.

Getting and setting channels in a ARGB value is easy enough:

argb = getRGB( x, y );a = argb >>> 24 & 255;r = argb >> 16 & 255;g = argb >> 8 & 255;b = argb & 255;argb = ( a << 24 ) | ( r << 16 ) | ( g << 8 ) | b;setRGB( x, y, argb );

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Quote:
 Original post by leif_dragonWhat is the most efficient way of adding two RGB values together??

Doing lots of the operation in parallel. I.e. by using graphics hardware. Or, if you have a many-core CPU, by doing the parallelism yourself.

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Quote:
 Original post by SiRobertsonBit-shifting is extremely quick so I doubt very much that is the problem.
Since all he is doing is adding individual components, the bit-shifting in unnecessary, and not doing something is generally faster than doing something [smile]

To the OP: alpha-blending is not actually accomplished by adding all the components together, instead you need use the right equation. However, as RDragon1 has mentioned, it will be much faster to let your graphics hardware do these calculations.

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Thanks for the awesome replies guys. Towards the responses about using graphics hardware/parallel processors.. I'm not sure at how I would go about doing that. I'm programming my game using java by the way.

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Would I have to multi-thread to have it render in parallel??

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# Associated occupation numbers of frozen natural orbitals (FNOs) from MP2 method

The paper Frozen Virtual Natural Orbitals for Coupled-Cluster Linear-Response Theory gives the following (simplified) description of frozen natural orbitals (FNOs).

In the MP2 method, the unrelaxed one-electron density matrix can be written in terms of spin orbitals

$$\gamma_{pq}=\langle\Psi^{1}|\{a_{p}^{\dagger}a_{q}\}|\Psi^{1}\rangle$$

where $$|\Psi^{1}\rangle$$ is the first-order correction to the Hartree-Fock wavefunction. In the MP2 based NO method, the virtual-virtual block is constructed:

$$\gamma_{ab}=\frac{1}{2}\sum t_{ij}^{ac}t_{ij}^{bc}$$

and then diagonalised:

$$\gamma V=nV$$

The eigenvectors V are the virtual NOs, and the eigenvalues n are the associated occupation numbers.

If the eigenvalues are found from a virtual-virtual (unoccupied-unoccupied) block, how can there be an associated occupation number?

• Not sure about the right answer, maybe it is related to mixing the term "occupied" for Hartree Fock – user1420303 May 27 '20 at 22:49
• @user1420303 I thinking along the lines that occupation numbers are associated with single, double, excitations into the virtual orbitals. Though I would need clarification on this. – Wychh May 28 '20 at 8:03
• I mean that when natural orbitals are constructed, their occupation number is not integer. Many of them are very close to zero or one. But they are not exactly zero or one (specially in multireference systems). Then you can classify nat orbs as occupied when they are essentially 1 or unoccupied when they are close to 0 (and eventually you can find orbitals with intermediate occupation numbers). – user1420303 May 28 '20 at 13:24


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… Data aequatione quotcunque fluentes quantitates involvente fluxiones invenire et vice versa …

## Category: Require high school maths

### A better estimate of Kempner’s series

The Kempner series recently regained some notoriety due to a Saturday Morning Breakfast Cereal comic (the last panel). The observation first appeared in a 1914 American Mathematical Monthly article, in which it was shown that the series consisting of the usual harmonic series

$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n}$

but with all the terms, whose decimal expansion includes the digit ‘9’, removed, in fact converges to some number below 80. The original proof is given in the Wikipedia article linked above, so I will not repeat it. But to make it easier to see the idea: let us first think about the case where the number is expressed in base 2. In base 2, all the positive integers has the leading binary bit being 1 (since it cannot be zero). Therefore there are no binary positive numbers without the bit ‘1’ in its expansion. So the corresponding series converges trivially to zero. How about the case of the bit ‘0’? The only binary numbers without any ‘0’ bits are

$1, 3 = (11)_2, 7 = (111)_2, 15 = (1111)_2, \ldots, 2^n - 1$.

So the corresponding series actually becomes

$\displaystyle \sum_{n = 1}^\infty \frac{1}{2^n - 1} \leq \sum_{n = 1}^\infty \frac{1}{2^{n-1}} = 2$

So somewhere from the heavily divergent harmonic series, we pick up a rapidly converging geometric series. So what’s at work here? Among all the n-bit binary numbers, exactly 1 has all bits not being 0. So the density of these kinds of numbers decays rather quickly: in base 2, there are $2^{n-1}$ numbers that are exactly n-bit long. So if a number $x$ has a binary representation that is exactly n bits long (which means that $2^{n} \leq x < 2^{n+1}$), the chances that it is one of the special type of numbers is $\frac{1}{2^{n-1}} \approx \frac{2}{x}$. This probability we can treat then as a density: replacing the discrete sum $\sum \frac{1}{n}$ by the integral $\int \frac{1}{x}\mathrm{d}x$ (calculus students may recognize this as the germ of the “integral test”) and replacing the $\mathrm{d}x$ by the density $\frac{2}{x} \mathrm{d}x$, we get the estimate

$\displaystyle \text{Binary Kempner series} \approx \int_1^\infty \frac{2}{x^2} = 2$.

Doing the same thing with the original Kempner series gives that the chances a n-digit number does not contain the digit nine to be

$\displaystyle \left(\frac89\right)\left(\frac9{10}\right)^{n-1} \approx \left( \frac{9}{10}\right)^{n}$

The length of the decimal expansion of a natural number $x$ is basically $1 + \log x$. So the density we are interested in becomes

$\displaystyle \left( \frac{9}{10}\right)^{1+\log x} ~\mathrm{d}x$

From this we can do an integral estimate

$\displaystyle \text{Kempner series} \approx 0.9 \times \int_1^\infty \left( \frac{9}{100}\right)^{\log x} ~\mathrm{d}x$

The integral can be computed using that

$\displaystyle a^{\log b} = b^{\log a}$

to get

$\displaystyle 0.9 \times \int_1^{\infty} \left( \frac{9}{100}\right)^{\log x} ~\mathrm{d}X = 0.9\times \int_1^\infty x^{\log 9 - 2} ~\mathrm{d}x = \frac{0.9}{1 - \log 9} \approx 19.66$

Notice that this estimate is much closer to the currently known value of roughly 22.92 than to the original upper bound of 80 computed by Kempner.

Kempner’s estimate is a heavy overestimate because he performed a summation replacing every n-digit long number that does not contain the digit 9 by $10^{n-1}$; this number can be many times (up to 9) times smaller than the original number. Our estimate is low because among the n-digit long numbers, the numbers that do not contain the digit 9 are not evenly distributed: they tend to crowd in the front rather than in the back (in fact, we do not allow them to crowd in the back because none of the numbers that start with the digit 9 is admissible). So if in the original question we had asked for numbers that do not contain the digit 1, then our computation will give an overestimate instead since these numbers tend to crowd to the back.

### … and scattering a quantum particle

In the previous post we shot a classical particle at a potential barrier. In this post we shoot a quantum particle.

Whereas the behaviour of the classical particle is governed by Newton’s laws (where the external force providing the acceleration is given as minus the gradient of the potential), we allow our quantum particle to be governed by the Klein-Gordon equations.

• Mathematically, the Klein-Gordon equation is a partial differential equation, whereas Newton’s laws form ordinary differential equations. A typical physical interpretation is that the state space of quantum particles are infinite dimensional, whereas the phase space of physics has finite dimensions.
• Note that physically the Klein-Gordon equation was designed to model a relativistic particle, while in the previous post we used the non-relativistic Newton’s laws. In some ways it would’ve been better to model the quantum particle using Schroedinger’s equation. I plead here however that (a) qualitatively there is not a big difference in terms of the simulated outcomes and (b) it is more convenient for me to use the Klein-Gordon model as I already have a finite-difference solver for hyperbolic PDEs coded in Python on my computer.

To model a particle, we set the initial data to be a moving wave packet, such that at the initial time the solution is strongly localized and satisfies $\partial_t u + \partial_x u = 0$. Absent the mass and potential energy terms in the Klein-Gordon equation (so under the evolution of the free wave equation), this wave packet will stay coherent and just translate to the right as time goes along. The addition of the mass term causes some small dispersion, but the mass is chosen small so that this is not a large effect. The main change to the evolution is the potential barrier, which you can see illustrated in the simulation.

The video shows 8 runs of the simulation with different initial data. Whereas in the classical picture the initial kinetic energy is captured by the initial speed at which the particle is moving, in the quantum/wave picture the kinetic energy is related to the central frequency of your wave packet. So each of the 8 runs have increasing frequency offset that represents increasing kinetic energy. The simulation has two plots, the top shows the square of the solution itself, which gives a good indication of where physically the wave packet is located. The bottom shows a normalized kinetic energy density (I have to include a normalization since the kinetic energy of the first and last particles differ roughly 10 fold).

One notices that in the first two runs, the kinetic energy is sufficiently small that the particle mostly bounces back to the left after hitting the potential.

For the third and fourth runs (frequency shift $\sqrt{2}$ and $\sqrt{3}$ respectively) we see that while a significant portion of the particle bounces back, a noticeable portion “tunnels through” the barrier: this caused by a combination of the quantum tunneling phenomenon and the wave packet form of the initial data.

The phenomenon of quantum tunneling manifests in that all non-zero energy waves will penetrate a finite potential barrier a little bit. But the amount of penetration decays to zero as the energy of the wave goes to zero: this is known as the semiclassical regime. In the semiclassical limit it is known that quantum mechanics converge toward classical mechanics, and so in the low-energy limit we expect our particle to behave like a classical particle and bounce off. So we see that naturally increasing the energy (frequency) of our wave packet we expect more of the tunneling to happen.

Further, observe that by shaping our data into a wave packet it necessarily contains some high frequency components (due to Heisenberg uncertainty principle); high frequency, and hence high energy components do not “see” the potential barrier. Even in the classical picture high energy particles would fly over the potential barrier. So for wave packets there will always be some (perhaps not noticeable due to the resolution of our computation) leakage of energy through the potential barrier. The quantum effect on these high energy waves is that they back-scatter. Whereas the classical high energy particles just fly directly over the barrier, a high energy quantum particle will leave some parts of itself behind the barrier always. We see this in the sixth and seventh runs of the simulation, where the particle mostly passes through the barrier, but a noticeable amount bounces off in the opposite direction.

In between during the fifth run, where the frequency shift is 2, we see that the barrier basically split the particle in two and send one half flying to the right and the other half flying to the left. Classically this is the turning point between particles that go over the bump and particles that bounces back, and would be the case (hard to show numerically!) where a classical particle comes in from afar with just enough energy that it comes to a half at the top of the potential barrier!

And further increasing the energy after the seventh run, we see in the final run a situation where only a negligible amount of the particle scatters backward with almost all of it passing through the barrier unchanged. One interesting thing to note however is that just like the case of the classical particle, the wave packet appears to “slow down” a tiny bit as it goes over the potential barrier.

### Shooting a classical particle…

Here’s a small animation of what happens when you try to shoot a classical particle when there’s a potential barrier. For small initial kinetic energies, the particle bounces back. For large initial kinetic energies, the particle goes over the hump, first decelerating and then accelerating in the process.

(It may be best to watch this full screen with HD if the network supports it.)

(The NumPy code is pretty simple to write for this; and it runs relatively fast. The one for my next post is a bit more complicated and takes rather much longer to run. Stay tuned!)

### An optimization problem: theme

Let’s start simple:

Question 1: What is the linear function $\ell(x)$ that minimizes the integral $\int_{-1}^1 |x^2 + x - \ell(x)| ~\mathrm{d}x$? In other words, what is the best linear approximation of $x^2 + x$ in the $L^1([-1,1])$ sense?

This is something that can in principle be solved by a high schooler with some calculus training. Here’s how one solution may go:

Solution: All linear functions take the form $\ell(x) = ax + b$. The integrand is equal to $x^2 + x - \ell(x)$ when $x^2 + x \geq \ell(x)$, and $\ell(x) - x - x^2$ otherwise. So we need to find the points of intersection. This requires solve $x^2 + x - ax - b = 0$, which we can solve by the quadratic formula. In the case where $x^2 + x - a x - b$ is signed, we see that changing $b$ we can make the integrand strictly smaller, and hence we cannot attain a minimizer. So we know that at the minimizer there must exist at least one root.

Consider the case where there is only one root in the interval (counted with multiplicity), call the root $x_0$. We have that the integral to minimize is equal to
$\displaystyle \left| \int_{-1}^{x_0} x^2 + x - ax - b~\mathrm{d}x \right| + \left| \int_{x_0}^1 x^2 + x - a x - b ~\mathrm{d}x \right|$
each part of which can be computed explicitly to obtain
$\displaystyle \left| \frac13 x_0^3 + \frac13 + \frac{1-a}{2} x_0^2 - \frac{1-a}{2} - b x_0 - b \right| + \left| \frac13 - \frac13 x_0^3 + \frac{1-a}{2} - \frac{1-a}{2} x_0^2 - b + b x_0\right|$
Since we know that the two terms comes from integrands with different signs, we can combine to get
$\displaystyle \left| \frac23 x_0^3 + (1-a) x_0^2 - (1-a) - 2b x_0 \right|$
as the integrand. Now, we cannot just take the partial derivatives of the above expression with respect to $a,b$ and set that to zero and see what we get: the root $x_0$ depends also on the parameters. So what we would do then is to plug in $x_0$ using the expression derived from the quadratic formula, $x_0 = \frac{1}{2} \left( a - 1 \pm \sqrt{ (1-a)^2 + 4b}\right)$, and then take the partial derivatives. Before that, though, we can simplify a little bit: since $x_0^3 + (1-a) x_0^2 - b x_0 = 0$ from the constraint, the quantity to minimize is now
$\displaystyle \left| - \frac13 x_0^3 - (1-a) - b x_0 \right|$
A long computation taking the $\partial_b$ now shows that necessarily $x_0 = 0$, which implies that $b = 0$. But for the range of $a$ where there is only one root in the interval, the quantity does not achieve a minimum. (The formal minimizer happens at $a = 1$ but we see for this case the integrand of the original cost function is signed.

So we are down to the case where there are two roots in the interval. Now we call the roots $x_+$ and $x_-$, and split the integral into
$\displaystyle \left| \int_{-1}^{x_-} x^2 + x - ax - b~\mathrm{d}x - \int_{x_-}^{x_+} x^2 + x - ax - b~\mathrm{d}x + \int_{x_+}^1 x^2 + x - ax - b~\mathrm{d}x \right|$
and proceed as before. (The remainder of the proof is omitted; the reader is encouraged to carry out this computation out by hand to see how tedious it is.) Read the rest of this entry »

### Find the errors!

Using methods discussed in class this term, find the mean value over $[-\pi,\pi]$ of the function $f(x) = \sin(2x) \cdot \exp [1 - \cos (2x)]$.

The conceptual parts of the question are (based on the syllabus of the course)

1. Connecting “mean value of a continuous function over an interval” with “integration”, an application of calculus to probability theory and statistics.
2. Evaluating an integral by substitutions/change of variables.
3. Familiarity with the trigonometric functions $\sin, \cos$ and their properties (periodicity, derivative relations, etc).

I was told to grade with an emphasis on the above, so I prepared a grading rubric such that the above three key ideas gave most of the points. Here’s an otherwise reasonable answer that unfortunately does not use the methods discussed in class and so would receive (close to) zero credit (luckily no students turned in an answer like this):

The function $f(x)$ satisfies $f(x) = - f(-x)$, i.e. it is odd. So the average $(f(x) + f(-x))/2 = 0$. Since for every $x\in [-\pi,\pi]$, we also have $-x \in [-\pi,\pi]$, the mean value of $f(x)$ over that interval must be zero.

Here are some responses that can get quite a good number of points* (at least more than the above answer) based on the grading rubric (I guess it means I wasn’t imaginative enough in coming up with possible student errors). (I took the liberty of combining some of the most awful bits from different answers; the vast majority of the students’ answers are not nearly that horrible**, though only one student remembered that when changing variables one also needs to change the limits of integration.) Since most students who made any reasonable attempt on the question successfully wrote down the integral

$\displaystyle \mu = \frac{1}{\pi - (-\pi)} \int_{-\pi}^\pi f(y)~\mathrm{d}y$

(which is not to say no unreasonable attempts were made: just ask the poor bloke who decided that the Mean Value Theorem must play a role in this question), I will start from there. All mistakes below are intentional on my part. What amazed me most is how many students were able to get to the correct mean value… Read the rest of this entry »

### Decay of waves I: Introduction

In the next week or so, I will compose a series of posts on the heuristics for the decay of the solutions of the wave equation on curved (and flat) backgrounds. (I have my fingers crossed that this does not end up aborted like my series of posts on compactness.) In this first post I will give some physical intuition of why waves decay. In the next post I will write about the case of linear and nonlinear waves on flat space-time, which will be used to motivate the construction, in post number three, of an example space-time which gives an upper bound on the best decay that can be generally expected for linear waves on non-flat backgrounds. This last argument, due to Mihalis Dafermos, shows that why the heuristics known as Price’s Law is as good as one can reasonably hope for in the linear case. (In the nonlinear case, things immediately get much much worse as we will see already in the next post.)

This first post will not be too heavily mathematical, indeed, the only realy foray into mathematics will be in the appendix; the next ones, however, requires some basic familiarity with partial differential equations and pseudo-Riemannian geometry. Read the rest of this entry »

### Snell’s law and geometry

Today’s post is somewhat inspired by this question on math.stackexchange. To begin with, recall Snell’s Law from geometric optics. It gives a rule for describing the propagation of light between two media of different indices of refraction: namely that across an interface where on one side the index of refraction is $n_1$ and on the other side $n_2$, the angles of incidence and refraction $\theta_1$ and $\theta_2$, as measured from the normal to the interface, satisfies the relation

$\displaystyle \frac{\sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1}$.

Here I pose two problems:

1. Given an observer situated above the interface, and a fish below the interface, find the “apparent position” of the fish according to the observer.
2. Given an observer situated above the interface, and the “apparent position” of a fish below the interface, find the actual position of the fish.

Without loss of generality we can assume that the interface is the $x$-axis, and the observer is situated at the point $(0,1)$ in the $x-y$ plane. Read the rest of this entry »

### Repeating decimals

Here’s something a bit random for Friday. (Presumably this actually is quite well-known; but having never taken a course in number theory…)

Question: Given a fraction m/n in lowest terms, and expand it in “decimal notation” in a number base b. What are the conditions on m, n, and b that guarantees that the expansion eventually consists of a single repeating “digit”? (Note, we can clearly assume 0 < m < n.)

To make the question more clear, let’s look at some examples in base b = 10. Obviously, if n = 1, then the fraction is in fact integral, and its expansion is $m.\dot{0}$ has a repeated digit 0. If n = 2, the decimal expansion is either $0.\dot0$ or $0.5\dot{0}$. Similarly n = 4, 5, 8, 10 all have terminating decimals, so repeats the digit 0 eventually. n = 3,6,9 on the other hand, will lead to repeating digits other than 0, whereas n = 7 will lead to a repetition of a string of the 6 digits …142857142857…

Here’s the solution. Suppose m/n has an expansion of the prescribed form. Recall that a “decimal expansion”

$0.a_1a_2a_3a_4\ldots$ in base $b$

is in fact a short hand for

$\displaystyle 0 + \frac{a_1}{b} + \frac{a_2}{b^2} + \frac{a_3}{b^3} \ldots$.

So the criterion specified in the question is equivalent to the condition that

There exists some integer K, a number $c < b^K$, and a digit d such that
$\displaystyle \frac{m}{n} = \frac{c}{b^K} + \frac{d}{b^K} \sum_{j = 1}^{\infty} \frac{1}{b^j}$
which corresponds to the decimal expansion
$\displaystyle 0.\underbrace{XXXXXXXXXXXX}_{\mbox{the digits given by }c} ddddddddddd\ldots$

The infinite sum on the far right can be solved: $\sum_1^\infty b^{-j} = (b-1)^{-1}$. Multiplying the expression through by $b^K$ we have

$\displaystyle \frac{m b^K}{n} = c + \frac{d}{b-1}$

Now, by redefining $c \to c\cdot b^k + d \sum_1^{k-1} b^j$, we can replace $K\to K+k$. So we can set $K$ arbitrarily large. Which means that by doing so, after setting the left hand side to lowest terms, we can “remove” from n any prime factors that also divides $b$. To be more precise: suppose $p$ is a prime such that $p^\ell | n$ and $p^{\ell+1}$ does not divide $n$. (So that p goes into n exactly $\ell$ times.) Now suppose $p|b$ also, then the fraction $b^\ell / n$, when written in lowest terms, has a denominator that cannot be divided by $p$. Repeating this for all common prime factors of n and b we can get rid of all common prime factors from the denominator. Let us denote by $n_0$ the number $n$ with all the prime divisors of $b$ removed.

Our equation then implies that there exists some integer $m'$ that is coprime with $n_0$ such that

$\displaystyle \frac{m'}{n_0} = \frac{d}{b-1}$

which means that $n_0$ must divide $(b-1)$. That is

Answer: Let $n_0$ be the number $n$ with all prime divisors of $b$ removed. Then $n_0$ must divide $(b-1)$.

For base b = 10, (b-1) = 9. This means that any n for which the decimal expansion is eventually repeating with period 1 must have the form

$n = 2^s\cdot 5^t\cdot 3^r$ where $s,t$ are non-negative integers, and $r$ is one of 0, 1, or 2.

### What is a function anyway?

I tried to teach differential geometry to a biophysicist friend yesterday (at his request; he doesn’t really need to know it, but he wanted to know how certain formulae commonly used in their literature came about). Rather surprisingly I hit an early snag. Hence the title of this entry.

Part of the problem was, as usual, my own folly. Since he is more interested in vector fields and tensor fields, I thought I can take a short cut and introduce notions more with a sheafy flavour. (At the end of the day, tangent and cotangent spaces are defined (rather circularly) as dual of each other, and each with a partial, hand-wavy description.) I certainly didn’t expect having to spend a large amount of time explaining the concept of the function.
Read the rest of this entry »

### Arrow’s Impossibility Theorem

Partially prompted by Terry’s buzz, I decided to take a look at Arrow’s Impossibility Theorem. The name I have heard before, since I participated in CollegeBowl as an undergraduate, and questions about Arrow’s theorem are perennial favourites. The theorem’s most famous interpretation is in voting theory:

Some definitions

1. Given a set of electors $E$ and a finite set of candidates $C$, a preference $\pi$ assigns to each elector $e \in E$ an ordering of the set $C$. In particular, we can write $\pi_e(c_1) > \pi_e(c_2)$ for the statement “the elector $e$ prefers candidate $c_1$ to candidate $c_2$“. The set of all possible preferences is denoted $\Pi$.
2. A voting system $v$ assigns to each preference $\pi\in\Pi$ an ordering of the set $C$.
3. Given a preference $\pi$ and two candidates $c_1,c_2$, a bloc biased toward $c_1$ is defined as the subset $b(\pi,c_1,c_2) := \{ e\in E | \pi_e(c_1) > \pi_e(c_2) \}$
4. The voting system is said to be
1. unanimous if, whenever all electors prefer candidate $c_1$ to $c_2$, the voting system will return as such. In other words, “$\pi_e(c_1) > \pi_e(c_2) \forall e\in E \implies v(\pi,c_1) > v(\pi,c_2)$“.
2. independent if the voting results comparing candidates $c_1$ and $c_2$ only depend on the individual preferences between them. In particular, whether $v(\pi,c_1) > v(\pi,c_2)$ only depends on $b(\pi,c_1,c_2)$. An independent system is said to be monotonic if, in addition, a strictly larger biased bloc will give the same voting result: if $v(\pi,c_1) > v(\pi,c_2)$ and $b(\pi,c_1,c_2) \subset b(\pi',c_1,c_2)$, then $v(\pi',c_1) > v(\pi',c_2)$ necessarily.
3. dictator-free if there isn’t one elector $e_0\in E$ whose vote always coincides with the end-result. In other words, we define a dictator to be an elector $e_0$ such that $v(\pi,c_1) > v(\pi,c_2) \iff \pi_{e_0}(c_1) > \pi_{e_0}(c_2)$ for any $\pi\in \Pi, c_1,c_2\in C$.
5. A voting system is said to be fair if it is unanimous, independent and monotonic, and has no dictators.

And the theorem states

Arrow’s Impossibility Theorem
In an election consisting of a finite set of electors $E$ with at least three candidates $C$, there can be no fair voting system.

As we shall see, finiteness of the set of electors and the lower-bound on the number of candidates are crucial. In the case where there are only two candidates, the simple-majority test is a fair voting system. (Finiteness is more subtle.) It is also easy to see that if we allow dictators, i.e. force the voting results to align with the preference of a particular predetermined individual, then unanimity, independence, and monotonicity are all trivially satisfied.

What’s wrong with the simple majority test in more than three candidates? The problem is that it is not, by definition, a proper voting system: it can create loops! Imagine we have three electors $e_1, e_2, e_3$ and three candidates $c_1,c_2,c_3$. The simple majority test says that $v(\pi,c_1) > v(\pi,c_2)$ if and only if two or more of the electors prefer $c_1$ to $c_2$. But this causes a problem in the following scenario:

$e_1: c_1 > c_2 > c_3$
$e_2: c_2 > c_3 > c_1$
$e_3: c_3 > c_1 > c_2$

then the voting result will have $v(c_1) > v(c_2)$, $v(c_2) > v(c_3)$, and $v(c_3) > v(c_1)$, a circular situation which implies that the “result” is not an ordering of the candidates! (An ordering of the set requires the comparisons to be transitive.) So the simple-majority test is, in fact, not a valid voting system.

From this first example, we see already that, in the situation of more than 2 candidates, designing a voting system is a non-trivial problem. Making them fair, as we shall see, will be impossible. Read the rest of this entry »


text:
# Division as Inverse Multiplication

Division is the opposite of multiplication. You can use it when you want to split a group into several smaller but equally large groups.

A division consists of a dividend, a colon and a divisor. When you divide, you actually answer the question “What number do I multiply by the divisor to get the dividend?”

Rule

### ImportantWordsandExpressions

 $\text{DIVIDEND}÷\text{DIVISOR}$

Let’s have a look at how division can simplify our lives. Imagine that you have $12$ bananas, and you want to split them between yourself and $3$ neighbors. How do you divide $12$ bananas by $4$? How many will each one of you get? If you don’t know how to divide, you can solve the problem by making four separate piles and put a banana in each pile, one by one, until there are no bananas left.

This can take a lot of time, and you have to write a lot. Luckily, there is another way to do it: Because you know how to multiply, you can use multiplication to distribute the bananas with the help of division! Multiplication and division are opposites in the same way plus and minus are; what one of them does, the other one can reverse.

Is it true that twelve bananas divided by four people will give you three bananas each when we know that

 $3\cdot 4=12?$

The answer is yes! You need $4$ threes to get to $12$. Because of this, dividing $12$ bananas between you and your $3$ neighbors will give you $3$ bananas each.

Let’s take a look at some examples using the times tables:

Example 1

From the times tables you know that $5\cdot 3=15$. What is $15÷3$?

You can see that $15$ is the dividend and $3$ is the divisor. Therefore, the question is: What number do you have to multiply by $3$ to get $15$? The answer is $5$, and you write it like this:

 $15÷3=5$

Example 2

From the times tables you know that $9\cdot 8=72$. What is $72÷9$?

You can see that $72$ is the dividend and $9$ is the divisor. Therefore, the question is: What number do you have to multiply by $9$ to get $72$? The answer is $8$, and you write it like this:

 $72÷9=8$

Example 3

From the times tables you know that $6\cdot 7=42$. What is $42÷6$?

You can see that $42$ is the dividend and $6$ is the divisor. Therefore, the question is: What number do you have to multiply by $6$ to get $42$? The answer is $7$, and you write it like this:

 $42÷6=7$

Math Vault

Want to solve exercises about division? Try Math Vault!


text:
# Question #07b6f

Jul 15, 2015

The temperature of the water will actually decrease.

#### Explanation:

Dissolving potassium nitrate, $K N {O}_{3}$, in water is actually an endothermic process.

An endothermic process is characterized by the fact that heat is being absorbed by the reaction.

In your case, this happens because the energy needed to dissociate the potassium cations and the nitrate anions from each other in the solid exceeds the energy released when these ions are hydrated.

The difference in energy will be obtained from the surroundings, in this case from the water. This will result in an overall temperature drop, i.e. the solution will start to cool.

Potassium nitrate's solubility graph looks like this

At ${40}^{\circ} \text{C}$, you can dissolve approximately 65 g of $K N {O}_{3}$ in 100 g of water.

So, depending on how much potassium nitrate you actually dissolve, and what the mass of the water is, you can expect the temperature of the water to drop by anywhere from a few to several degrees Celsius.

Check out this answer for a more detailed explanation on why dissolving potassium nitrate in water is an endothermic process:

http://socratic.org/questions/why-is-dissolving-potassium-nitrate-in-water-an-endothermic-process


text:
3 years ago

# High operating temperature in V-based superconducting quantum interference proximity transistors

Francesco Giazotto, Giampiero Marchegiani, Nadia Ligato, Elia Strambini, Pauli Virtanen
Here we report the fabrication and characterization of fully superconducting quantum interference proximity transistors (SQUIPTs) based on the implementation of vanadium (V) in the superconducting loop. At low temperature, the devices show high flux-to-voltage (up to 0.52 mV/Φ0) and flux-to-current (above 12 nA/Φ0) transfer functions, with the best estimated flux sensitivity ~ 2.6 μΦ0/(Hz)1/2 reached under fixed voltage bias, where Φ0 is the flux quantum. The interferometers operate up to Tbath$≃$ 2 K, with an improvement of 70% of the maximal operating temperature with respect to early SQUIPTs design. The main features of the V-based SQUIPT are described within a simplified theoretical model. Our results open the way to the realization of SQUIPTs that take advantage of the use of higher-gap superconductors for ultra-sensitive nanoscale applications that operate at temperatures well above 1 K.

Publisher URL: https://www.nature.com/articles/s41598-017-09036-0

DOI: 10.1038/s41598-017-09036-0

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## Degrees of Freedom (df)

Degrees of Freedom denotes the number of samples that a statistician has the freedom to choose. Degrees of Freedom is based on a concept that one could not have exercised his /her freedom to select all the samples.
The concept can be explained by an analogy:
X+Y = 10 (1)
In the above equation you have freedom to choose a value for X or Y but not both because when you choose one, the other is fixed. If you choose 8 for X, then Y has to be 2. So the degree of freedom here is 1.
X+Y+Z = 15 (2)
In the formula (2), one can choose values for two variables but not all. You have freedom to choose 8 for X and 2 for Y, If so, then Z is fixed, So the degree of freedom is 2. df is calculated by subtracting 1 from the size of each group. The methods of df calculation may vary with the test used.
“Degrees of freedom” refers to the number of scores that are free to vary.  It is required when one is working on estimates of population values (sample statistics).  It is often abbreviated as df.  It is the number of observations in the data collection that are free to vary after the sample statistics have been calculated.  For example, in calculating a sample standard deviation, we must subtract the sample mean from each of the n data observations in order to get the deviations from the mean.  But once we have completed the next to the last subtraction, the final deviation is automatically determined, since the deviations prior to squaring must sum to zero.  Therefore, the last deviation is not free to vary; only n-1 are free to vary.  As a rule of thumb, every estimate of a population value in a formula equals 1 degree of freedom.
The practical impact of using degrees of freedom is found when one considers small samples.  Consider the t distribution, for example.  With small samples it gets flatter, or another way of looking at it is that more area underneath the curve (region of rejection) is farther away from the mean.  This makes using degrees of freedom a more conservative and accurate, since small samples tend to underestimate the population value.
Practical Example:
Consider a situation in which the scores that make up the distribution are unknown, and I tell you to guess at the first score in the distribution of 5 numbers.  Your guesses will be wild because that score could be any number.  The same is true for the second score, third score, and the fourth score.  All of these scores have complete freedom to vary.   But, if I tell you that the first four scores in the distribution are 3,4,5, and 6; and I tell you what the mean of the distribution (5, in this case), then the last score in this distribution has to be 7.  In other words, if the mean is known, the missing score is determined by your knowledge of the other four.  Therefore n-1 scores are free to vary.  The (-1) is often called a “restriction.”  Note that by making the numerator smaller, the standard deviation becomes larger. A larger standard deviation means that the sampling distribution is flatter; and flatter means more values are farther from the mean.  It’s tougher to find significance.
In conclusion, when the statistical formula is concerned with description, degrees of freedom is n.  When the formula is concerned with inference, some restrictions apply. The idea is to adjust for a small sample size’s tendency to underestimate the population parameter.  As n gets larger, this becomes less of a problem because the distribution becomes less flat and more normal, but we still use the sample formula to calculate the statistic.


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#### General formula for nth element of the sequence 0, 1, 0, 1, …

##### Replies

###### Alice Davidson

User

( 5 months ago )

The sequence is $$f=0,1,0,1,…f=0,1,0,1,…$$

I want to find a general formula for the $$nn$$th element. The sequence starts at $$n=0n=0$$ (the $$00$$ here is not the first element $$00$$ but rather denotes the $$00$$th position).

One easy and obvious solution is: $$nn$$th $$f=nmod2f=nmod2$$. This works because even positions have $$00$$ and odd positions have $$11$$.

However, this question is part of a homework and modulus has not been discussed (or part of the syllabus or even a prerequisite). And so I am hesitant to use it.

Is there another way to solve this problem using only basic arithmetic operations (one that a beginning high schooler knows of)?

Ra
04-Sep-2019

2 Replies

Ra
05-Sep-2019

2 Replies

11-Sep-2019

0 Replies


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# HiI know that for a short solenoid (L<R) the magnetic field at the

1. Feb 3, 2012

### Niles

Hi

I know that for a short solenoid (L<R) the magnetic field at the axis is (standard EM)
$$B(z) = \frac{1}{2}\mu_0 \frac{N}{L}I(\frac{z+\frac{L}{2}}{\sqrt{(z+L/2)^2+R^2}} - \frac{z-\frac{L}{2}}{\sqrt{(z-L/2)^2+R^2}})$$
where R is the radius of the solenoid, N the number of turns along the axis and L the length. In this system, each vertical plane consists of a single turn, but say I am looking at a solenoid, where each vertical plane consists of e.g. 2 turns. First I thought about using the above equation twice, but that is wrong since it is not 2 independent solenoids.

Is it correct to regard the system simply as a collection of coils with 2 turns each? I'm not quite sure how this would work out, since this way I can't take into account the widths of each individual coil. If my description is confusing, please let me know.

Best,
Niles.

Last edited: Feb 3, 2012
2. Feb 3, 2012

### clem

Re: Coils

Your formula is for a solenoid of any length, not just L<R.
I don't understand what you mean be 'two turns'.
Why would it make a difference?

3. Feb 3, 2012

### Niles

Re: Coils

Hi, thanks for replying. I really appreciate it. What I mean by two turns is the following:

OOOOOO
OOOOOO
------------ (axis of solenoid)
OOOOOO
OOOOOO

There I have a solenoid with 12 turns in total, i.e. a 6-turn solenoid on top of another (larger radii) 6-turn solenoid. How would one calculate the B-field for that? In principle I could just add the B-field for each one of the 12 turns, right? The reason why I am asking is because I am looking for the most general way to calculate this, because I would also have to look at e.g.

OOOO
OOOOOO
OOOOOOOO
---------------------- (axis)
OOOOOOOO
OOOOOO
OOOO


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## Precalculus (6th Edition) Blitzer

The partial fraction is, $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$
$\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{A}{x}+\frac{B}{\left( x-1 \right)}+\frac{C}{{{\left( x-1 \right)}^{2}}}$ Now, multiply both sides by \begin{align} & {{x}^{2}}+2x+7=A{{\left( x-1 \right)}^{2}}+Bx\left( x-1 \right)+Cx \\ & =A\left( {{x}^{2}}+1-2x \right)+B{{x}^{2}}-Bx+Cx \\ & =A{{x}^{2}}+A-2Ax+B{{x}^{2}}-Bx+Cx \\ & =\left( A+B \right){{x}^{2}}+\left( -2A-B+C \right)x+A \end{align} Then, compare the coefficient of ${{x}^{2}},\ x$ and constant term: $A+B=1$ …… (1) $-2A-B+C=2$ …… (2) $A=7$ …… (3) And put the value of A into equation (1): \begin{align} & 7+B=1 \\ & B=-6 \end{align} Also, put the value of A and B into equation (2): \begin{align} & -2\left( 7 \right)+6+C=2 \\ & C=2+8 \\ & =10 \end{align} Therefore, $\frac{{{x}^{2}}+2x+7}{x{{\left( x-1 \right)}^{2}}}=\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$ Thus, the partial fraction of the provided expression is $\frac{7}{x}-\frac{6}{\left( x-1 \right)}+\frac{10}{{{\left( x-1 \right)}^{2}}}$.


text:
Articles

Thursday, 22 July 2021

# The TeX typesetting system

Today I made an important discovery. I learnt about TeX - a typesetting system which was designed and written by Donal Knuth and first released in 1978. It is so old, yet so powerful. It has been noted as one of the most sophisticated digital typographical systems.

As a child I was already intrigued by the complex layout of the Hebrew Talmud (Vilna edition). How did they manage to do such sophisticated typesetting without any computers? Would it be easier with todays software?

Microsoft Word didn't get me anywhere near the desired layout. Years later I would discover Adobe InDesign, and with this software I did manage to recreate the complex layout, but it was a very tedious task, requiring manual work on every single page. I later got in to scripting, and wrote some InDesign scripts to vastly speed up the process (This was how I got into programming in the first place 😀).

Now, I finally discovered TeX. And as a programmer, this seems like the best way to typeset books, as it is code based. Here is the most basic Hello world TeX document:

\documentclass{article}

\begin{document}

Hello World!

\end{document}

This will output a page with "Hello world" in the top left. Commands start with a backslash. \documentclass{...} defines the formatting for the document. In this case we want the formatting called "article". \begin{document} tells LaTeX that from here starts the content [kinda like <body> in HTML], and that everything up until here was the 'preamble' [like <head> in HTML]. Then comes a line of actual content to write on the page. Then the self explanatory \end{document}.

### Output to PDF

To generate a PDF, run the following command:

latex hello.tex

This will generate a hello.dvi file [DVI = Device Independant]. Now a PDF can be generated with this command:

dvipdf hello.dvi hello.pdf

### Hebrew Typesetting

Everything seemed amazing, until I tried to typeset Hebrew. Looks like Tex doesn't support RTL & Hebrew out of the box. Googling for solutions didn't bring up any usable results, at least for me. So if you have any knowledge how to get Tex to behave nicely with Hebrew typesetting, please reach out!


text:
# An example of a not locally compact space in $\mathbb R^2$

Are the two subspaces $X$ and $\operatorname{cl}(X)$ of Euclidean space $\mathbb R^2$ locally compact?

$$X = \{(x,\sin 1/x) \mid 0 < x \le 4\}\cup\{(x,\sin 1/x) \mid -4 \le x \lt 0\} \cup \{(0,0)\}$$

Clearly since $\operatorname{cl}(X)$ is closed and bounded so $\operatorname{cl}(X)$ is compact and thus locally compact.

However, I'm having trouble proving that $X$ is not locally compact. My conjecture is that I need to show there is no open set $U$ and compact set $C$ such that $(0,0) \in U \subset C$. But I can't come up with a rigorous proof for this part. How can I construct a written proof of this?

• If $X$ were locally compact, then for every sufficiently small $r > 0$, the set $\overline{B_r(0)}\cap X$ would be compact, in particular, a closed subset of $\mathbb{R}^2$. – Daniel Fischer Nov 28 '14 at 22:38
• Note: \bigcup is used for things like $\displaystyle\bigcup_{n=1}^\infty A_n$, and \cup for things like $A\cup B$ and $A_1\cup\cdots\cup A_n$. Also note that "a\sin b" yields $a\sin b$ and "a sin b" yields $a sin b$ without proper spacing and with "$\sin$" improperly italicized. And in $\{(x,\sin1/x)\mid 0<x\le 4\}$, the $\{\text{curly braces}\}$ are INSIDE the math environment. I did these and a variety of other edits to the question. ${}\qquad{}$ – Michael Hardy Nov 28 '14 at 22:44
• @DanielFischer I see the existence of such closed compact balls if X were locally compact but how does the fact that such closed subsets exist lead to a contradiction? – nomadicmathematician Nov 28 '14 at 22:46
• The point is that such sets are not closed in $\mathbb{R}^2$. – Daniel Fischer Nov 28 '14 at 22:51

Hint. In a compact set in the plane every sequence has a convergent subsequence. For convenience, instead of an open ball, consider a rectangular neighborhood of $(0,0)$, $U=[-\epsilon,\epsilon]^2$ where $\epsilon>0$. Take $0<\delta<\epsilon$ and try to come up with a sequence $x_n\to 0$ such that $\delta=\sin1/{x_n}$ for all $n$. What does the sequence of points in the plane $(x_n,\sin1/{x_n})$ converge to? Is the limit of this sequence going to belong to $X$?

Every open nbhd of the origin contains a set of the form

$$U_\epsilon=X\cap\Big((-\epsilon,\epsilon)\times(-\epsilon,\epsilon)\Big)$$

for some $\epsilon>0$. Show that the closure of this set in $X$ is

$$\operatorname{cl}_XU_\epsilon=X\cap\Big([-\epsilon,\epsilon]\times[-\epsilon,\epsilon]\Big)\;.$$

Then let $a=\dfrac{\epsilon}2$ and find a sequence in

$$\big(\Bbb R\times\{a\}\big)\cap\operatorname{cl}_XU_\epsilon$$

that has no convergent subsequence in $X$ (because in $\Bbb R^2$ it converges to a point not in $X$).


text:
# Schottky diode

Main Article
Talk
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Citable Version  [?]

This editable Main Article is under development and not meant to be cited; by editing it you can help to improve it towards a future approved, citable version. These unapproved articles are subject to a disclaimer.

## Contents

The Schottky diode is a two-terminal device consisting of conductive gate (for example, a metal) on top of a semiconductor body. It is named after Walter H Schottky, who proposed a detailed theory for point contact rectification in 1938.[1] The modern theory of Schottky diode operation was contributed by Hans Bethe in 1942.[2] A generic name for this structure is the metal-semiconductor diode or M/S diode.[3]

(PD) Image: John R. Brews
Planar Schottky diode with n+-guard rings and tapered oxide. The Ohmic contact to the substrate is on the left.

For low voltage applications, below 200V, silicon is used, but for higher voltages (up to 3000 V or more) silicon carbide is used to extend the breakdown voltage. These voltages are achievable only when edge breakdown is avoided, which requires special attention to edge termination designs.[4] The figure shows three strategies toward increasing the edge breakdown voltage: an extension of the metal diode contact over a tapered oxide and also an n+-guard ring and a floating guard ring. These strategies are sometimes used together, but also are used separately.

Contact to the body is made through an ohmic contact to the p-substrate consisting of a metal-to-p+ region on the surface of the diode. An ohmic contact is one with a linear current-voltage characteristic like that of a resistor.

## Applications

(PD) Image: John R. Brews
Comparison of Schottky and pn-diode current voltage curves.

The Schottky diode is used in a large variety of applications, ranging from practical devices for switching, rectification and photo-detection, to test structures for fabrication monitoring and for studies of semiconductor defects and processes.

## Operation

Three different bias cases are examined: zero bias, forward bias, and reverse bias. A simplified one dimensional analysis along a line vertically through the center of the Schottky contact is used throughout. It is imagined that the p+-ohmic contact is vertically below the Schottky contact.

As seen in the comparison with the pn-diode shown in the figure, the Schottky diode tends to turn on at lower forward voltages but rises more slowly because of the nonideality factor noted later. The Schottky diode also tends to have a higher reverse leakage current. The mechanism for current flow is different in the two devices: the pn-diode current is diffusion driven by carrier concentration differences, while the Schottky diode is driven by thermionic emission, that is, the surmounting of a potential barrier by energetic carriers. This mechanism is described further below.

(PD) Image: John R. Brews
Schottky barrier formation on p-type semiconductor. Top: isolated charge-neutral bulk materials. Center: dipole layer formed as materials interact; Fermi level difference is reduced. Bottom: final configuration following charge transfer to reach equilibrium with no applied voltage. The neutral (shaded) region in the semiconductor moves away from the interface, leaving a charged depletion region (unshaded) near the interface. Energies are in eV.

#### Zero bias

Ordinarily, the Fermi levels of different materials differ. When they are brought into electrical contact, enabling electron transfer between the materials, the work done in removing an electron from one material and placing it in the other is equal to the energy difference in the Fermi levels. Consequently, energy is released upon contact by electron transference from the material with the higher Fermi level to the material with the lower Fermi level. This charge transfer continues until the electrical charge difference means the energy gain from transfer is countered by the electrical work required against the charge difference. At this point the two Fermi levels are brought into coincidence at a common level EF and no further charge transfer occurs. This flat Fermi level situation corresponds to thermal equilibrium, and no net current flows once equilibrium is reached.

The final potential barrier between the metal Fermi level and the majority carrier band edge in the semiconductor is called the Schottky barrier, labeled φB in the figure. It is a form of contact potential. The concomitant potential drop across the semiconductor depletion layer is called the built-in potential, labeled φbi in the figure.

The formation of a Schottky barrier is shown schematically in the figure, beginning with the isolated materials (top), an intermediate dipole formation stage (center), and the completed configuration with formation of a depletion layer (bottom). The dipole layer still is present in the bottom figure, but is omitted for clarity. The figure is not intended to indicate exactly how the processes occur with increasing proximity of the materials, but merely the processes involved. A more detailed discussion of the figure follows.

The top panel shows a charge-neutral, partly filled metal energy band and a charge-neutral semiconductor valence and conduction band, chemically and electrically isolated from each other. The Fermi level in the p-type semiconductor EFp (assumed to be in units of eV) is near its valence band edge, as set by its acceptor impurity doping. (See Fermi level). The Fermi level in the metal EFm marks the half-occupancy energy level in the partly filled band of the metal.

The center panel shows schematically the formation of an immobile interfacial dipole layer that alters the Fermi level difference, and adjusts how much mobile charge transfer occurs to form the depletion layer. This dipole formation is not only a result of quantum mechanics at a discontinuity between solids, but chemical processes involving the atomic layers in what is sometimes called the interface specific region. The interfacial dipole layer reflects the fact that properties of the bulk materials by themselves do not determine the charge transfer that occurs. Rather, the chemistry of the interface, defects at the interface, and the mechanical adaptation to new atomic positions near the interface that are differently spaced than those found in bulk material, all conspire to alter the amount of charge transfer from that expected from bulk properties.[5] This interface dipole contribution is specific to the interface materials, and is different for (say) a material interface cleaved in vacuum and that for the same material mated to some other material. This interface dipole also is different for differently oriented or differently structured crystal planes. But whatever the final amount of charge transfer, charge transfer occurs and the two adjoined materials each acquire a charge.

The bottom panel shows the final configuration following mobile charge transfer between the materials. The transfer of electrons from the metal results in a very thin layer of positive charge in the metal very near the interface. Because the electron density in the metal is extremely large, the band bending in the metal is negligible in the diagram. On the other hand, the density of holes in the semiconductor is not nearly as large as the electron density in the metal, so the charging of the semiconductor is spread deeply into the material (the field effect). The semiconductor gains electrons, as indicated by the bending of the valence band edge away from the Fermi level, which increases the valence band occupancy by electrons. Differently stated, the vacancies (holes) in the valence band are reduced in number, and the charge balance in the band-bending region is lost. In this depletion layer, the holes or majority carriers are depleted (the unshaded surface region in the figure), exposing the immobile negative acceptor dopant ions to make this region charge-negative. This dopant charge results in a potential according to Poisson's equation. The potential decreases with distance toward the bulk semiconductor, and at some distance (the depletion width) the bulk properties of the semiconductor are regained and the semiconductor bulk is charge neutral.

At zero bias there is no net current, because the current from the semiconductor is balanced by that from the metal. According to the model of thermionic emission, the current flowing over an energy barrier is proportional to the carrier density next to the barrier. One can imagine the carriers in rapid thermal motion randomly directed. The carriers approaching the barrier with energies lower than the barrier height bounce off the barrier and do not cross it. Some fraction of those that are more energetic escape backward scattering to succeed in surmounting the barrier, and are lost to the far side. On the metal side, hole injection from the metal Fermi level into the semiconductor valence band is small as it requires surmounting a large energy barrier. On the semiconductor side, hole injection into the metal requires surmounting the built-in barrier height, φbi. Although the barrier on the semiconductor side is smaller than the Schottky barrier φB on the metal side, so is the carrier density. The two currents balance at zero bias, so no net current flows. These balancing current flows are much larger than in the pn-diode because thermionic emission results in larger currents than the diffusion-recombination mechanism of the pn-diode.[6]

Evidently the interface specific region between the metal and the semiconductor determines the Schottky barrier height φB. While this barrier height is very influential in determining the current, other interfacial properties also help to decide just how much current will be injected at a given barrier height, sometimes lumped into a factor called the "generalized Richardson constant", named after the Nobel Prize winning author of the original theory of thermionic emission, Owen Willans Richardson. However, in real devices thermionic emission is complicated by the effects of diffusion, barrier shape, field emission, homogeneity of the interface, and so forth.[7] Calculation of the barrier height based upon a fundamental understanding either of current balance or of charge transfer should lead to the same result, but at present such detailed calculations remain topics for research.

(PD) Image: John R. Brews
Under forward bias VF the potential drop across the depletion layer is reduced, the depletion region is narrower, and the Fermi levels are split.

#### Forward bias

Under forward bias, a net current of holes now flows from the semiconductor to the metal (or, equivalently, of electrons from the metal to the semiconductor). At zero bias there is no net current, because the current from the semiconductor is balanced by that from the metal. But under forward bias the current from the metal side is unaffected (and extremely small), while that on the semiconductor side is increased by the forward bias, as discussed next.

If a forward bias voltage is applied VF, the Fermi level of the bulk metal EFm (in eV) is raised in energy above the bulk Fermi level in the semiconductor EFp, which lowers the potential drop across the depletion layer to a value φbi−VF and increases the hole current from the semiconductor to the metal. In contrast to the lowered barrier on the semiconductor side, the alignment of the metal Fermi level relative to the semiconductor band edges at the interface is not changed by the bias: that is, φB is fixed by the processes involved in adjusting the Fermi levels to achieve equilibrium at zero applied bias. That fixed alignment means the barrier for hole injection from the metal into the semiconductor is not changed by an applied bias, and the injected hole current from the metal remains the same as at zero bias.

According to the model of thermionic emission, the electrical current injected from the semiconductor is proportional to the hole density at the interface. Using a simple Boltzmann approximation to the Fermi function, the hole density on the semiconductor side is increased under forward bias by a factor:

$p(V_F) = p(0)e^{qV_F/k_BT} \ ,$

where p(0) is the zero forward bias value of hole density on the semiconductor side. The hole current from the semiconductor to the metal therefore is increased by this factor. The current, being by convention a flow of positive charge, is therefore positive from the semiconductor to the metal, and the Schottky barrier current in forward bias becomes:

$I(V_F) = I(0)\left( e^{qV_F/k_BT}-1 \right) \ ,$

where I(0) is related to the zero-bias current exchange between the two materials.

Due to some complications of real Schottky barriers, for example, inhomogeneity (patchwork) of the interface or intervening contamination layers, the current is usually represented as:[8]

$I(V_F) = I(0)\left( e^{qV_F/nk_BT}-1 \right) \ ,$

where the factor n is called the ideality factor, and is usually in the range 1 - 2.

(PD) Image: John R. Brews
Schottky diode under reverse bias VR. The potential drop across the depletion layer is increased, and the depletion layer is widened by the reverse bias.
(PD) Image: John R. Brews
Critical electric field for breakdown versus bandgap energy in several materials.

#### Reverse bias

If a reverse bias VR is applied, the Fermi level of the metal is lowered below that of the semiconductor, and the barrier height for holes is increased on the semiconductor side by the amount of the reverse bias. The hole density is therefore decreased compared to equilibrium (the electron occupancy increases) and the current from the semiconductor falls below the equilibrium value. That means the diode current changes direction compared to the forward biased case.

If the diode were ideal, the current in reverse bias would be simply −I(0), the hole current backward injected by the metal into the semiconductor at zero bias, because the the hole current from the semiconductor side is completely suppressed by the very low hole density at the interface in reverse bias, making the injection of holes into the metal nearly zero. However, in practice, in reverse bias a leakage current is drawn due to generation of electrons and holes by defects in the depletion layer. If a large enough reverse voltage is applied, breakdown occurs, due to runaway of the generation mechanisms, or more commonly, due to edge breakdown in the high field region near the contact perimeter. A brief outline of the breakdown phenomena follows.[9]

Under reverse bias, a large electric field is present in the depletion layer, and carriers in the depletion layer are accelerated by this field. The carriers gain energy from the field, but they lose energy by creating vibrations in the semiconductor lattice, called phonons, thereby heating up the material. However, as the field increases with application of a larger reverse bias voltage, an increased fraction of the high energy carriers fail to lose energy by phonon creation, attaining energies in excess of the material bandgap. These energetic carriers, instead of losing energy to phonons, lose energy by creating electron-hole pairs, a process called impact ionization. At a crucial breakdown field, these generated carrier pairs themselves generate pairs and the process enters runaway.

A wider bandgap material like SiC has a higher intrinsic breakdown voltage than a lower bandgap material like Si because the energy to make an electron-hole pair scales with the bandgap. An electron created by impact ionization is in the conduction band, and the associated hole is in the valence band, and these two are separated in energy by at least the bandgap energy. This bandgap energy is provided by the electric field inside the depletion layer, which must accelerate the impacting carrier beyond this energy before generation can occur. The figure shows a log-log plot of field upon the bandgap energy fitted to an empirical power law with exponent 2.4.[10]

Unlike the MOSFET or the MOS capacitor, the expansion of the depletion layer width in response to an applied reverse bias is not arrested at large biases by formation of an inversion layer, a thin surface layer at the interface made up of minority carriers (electrons in a p-type semiconductor). An inversion layer cannot form, because the carriers are transmitted to the metal, and cannot be stored at the interface.

#### Transient response

A Schottky diode in forward bias conducts a current. This current can be cut off by switching to a reverse bias. The response time of the diode is simply the time it takes for the holes to move back from the edge of the narrow depletion region of forward bias to the edge of the wider depletion region of reverse bias. This switching is very rapid, the switching time given by the RC time constant of the diode, where R is the resistance of the bulk p-type material and its ohmic contact, and C is the diode capacitance. To understand the nature of this capacitance, note that the majority carriers in the semiconductor (holes in this example) are separated from the metal contact by the insulating depletion layer, where there are only immobile dopant ions, and no mobile carriers. Thus, the diode represents an electric capacitor with two plates, namely the metal and the neutral bulk material, separated by a dielectric, namely the insulating depletion region.

Compared to the MOS capacitor, or the pn-diode or the bipolar transistor, this switching time is very fast because it is not limited by the response time of minority carriers, which have to be removed from those devices by slow recombination processes.

## Barrier height

(PD) Image: John R. Brews
Schottky barrier height vs. metal electronegativity for some selected metals on n-type silicon. The straight line is a theoretical result based upon one theory of interface dipole formation, the theory of metal induced gap states.

From the very earliest work of Schottky, Mott and Davydov it was expected that the Schottky barrier height would depend upon the electronegativity of the metal contact, as that measures the tendency of the metal to grab electrons. The more electronegative the metal, the more it will resist transfer of electrons to the semiconductor (other factors being equal). For a p-type semiconductor, transfer of electrons is between the metal and the valence band of the semiconductor, and a greater electronegativity of the metal means fewer electrons for the semiconductor, or a lower barrier against hole injection from the metal to the valence band. For an n-type semiconductor, a greater electronegativity of the metal means fewer electrons for the conduction band of the semiconductor, and a higher barrier against electron transference out of the metal into the conduction band.

The figure shows the Schottky barrier against injection of electrons into the conduction band of n-type Si as a function of electronegativity of the metal. A correlation with electronegativity is observed, but the dependence is not as strong as originally thought because of the influence of the details of the interface charge distribution.[11] The theoretical straight line is:

$\varphi_B = 0.76 +0.101 ( X_m-X_{Si}) \ ,$

where all units are in eV, Xm is the Miedema electronegativity of the conducting contact, and XSi is the Miedema electronegativity of silicon. [12] The slope of 0.101eV/Miedema electronegativity unit can be compared with the original Schottky-Mott rule, which focused upon properties of the metal and semiconductor viewed in isolation, leading to a barrier given by:

$\varphi_B = \Phi_M-\chi_S \ ,$

where ΦM is the metal workfunction (energy to remove an electron from the metal) and χS is the semiconductor electron affinity (ionization energy of semiconductor). This equation leads to a slope of about 1eV/Miedema unit, much larger than the slope in the figure.

The figure also shows multiple barrier heights for the same metal are observed, corresponding to different interface structures.

(PD) Image: John R. Brews
Theoretical dependence of Schottky barrier heights for diodes using p-SiC vs. electronegativity of the metal according to Mönch[13]

Different semiconductors will lead to different slopes for such a plot. An example for diodes on p-type SiC is shown in the figure.[13] The slope of the theoretical straight line in this figure is −0.24eV/Miedema electronegativity unit. A negative slope is expected for p-type semiconductors (ignoring other factors) because the more electronegative the metal the fewer electrons it will transfer to the valence band of the semiconductor, or differently stated, the lower the barrier height for hole injection into the semiconductor.

The theory behind the low slopes of the straight lines in the figures is based upon dividing electronic states at the metal semiconductor interface into four groups: (i) electron states in the metal, called the metal-induced gap states (MIGS), which have an amplitude that decays into the semiconductor (ii) electron states in the semiconductor that decay into the metal, (iii) localized states that decay from the interface both into the metal and into the semiconductor, and finally (iv) those states that are bulk-like in both materials and decay in neither one. The theory plotted in the figure, the MIGS theory, suggests that the barrier height is determined primarily by the first group of states.[11] The original identification of these states, also known as interface-induced gap states (IFIGS), was made by Heine at Bell Labs.[14] A simplified version of this model predicts an interface charge dipole that results from the negative charge in the MIGS and the positive image charge reflected in the metal itself. This interface dipole is found to be influenced by the metal electronegativity (as is the entire chemistry of the interface specific region), but has a weaker dependence upon electronegativity than proposed by the simple Schottky-Mott rule.

The role of MIGS in explaining the interface dipole in the interface specific region is still under discussion, and alternatives are proposed like the bond polarization model with its origins in molecular studies. An overview and assessment of these theories is provided by Tung, who emphasizes that the MIGS are an interface property and depend upon both the metal and the semiconductor.[15]

Whatever the underlying origin of the interface dipole, it reduces the initial mismatch between the semiconductor and the metal Fermi levels, reducing the amount of mobile charge transfer that creates the depletion layer, thereby reducing the built-in potential drop in the semiconductor and reducing the depth of the depletion region.

## Notes

1. W. Schottky (1939). "Zur Halbleitertheorie der Sperrschicht- und Spitzengleichrichter". Zeitschrift für Physik A Hadrons and Nuclei vol. 113: pp. 813 ff. DOI:10.1007/BF01340116. Research Blogging. W. Schottky (1941). "Vereinfachte und erweiterte Theorie der Randschicht-gleichrichter". Zeitschrift für Physik A Hadrons and Nuclei vol. 118: pp. 539 ff. DOI:10.1007/BF01329843. Research Blogging. See also E. Spenke (1949). "Zur Randschichttheorie der Trockengleichrichter". Z. Physik vol. 126: pp. 67 ff. DOI:10.1007/BF01331042. Research Blogging.
2. HA Bethe (November, 1942). "MIT Radiation Lab. Report" vol. 43-12.
3. The term "Schottky diode" may be taken erroneously to refer to diffusion as the mechanism of operation as first proposed by Mott, Schottky and Davydov. However, the mechanism in most devices is a combination of thermionic emission and diffusion, as later proposed by Bethe, the so-called thermionic-diffusion theory. See Chih-Tang Sah (1991). “§560: Metal/semiconductor diode”, Fundamentals of solid-state electronics. World Scientific, p. 474. ISBN 9810206372.  and SM Sze and KK Ng (2007). “Chapter 3 Metal-semiconductor contacts”, Physics of semiconductor devices, 3rd ed. Wiley, pp. 134 ff. ISBN 0471143235.
4. B. Jayant Baliga (2005). “§3.2 Schottky diode edge terminations”, Silicon carbide power devices. World Scientific, pp. 44 ff. ISBN 9812566058.
5. For further discussion, see for example, Howard R. Huff, H. Iwai, H. Richter (2006). “Electrical properties: Partial charge transfer in interface bonds”, Silicon Materials Science and Technology X, Issue 2. The Electrochemical Society, p. 195. ISBN 156677439X. Raymond T Tung (2001). "Recent advances in Schottky barrier concepts". Materials Science and Engineering: R: Reports vol. 35 (1-3): pp. 1-138. DOI:10.1016/S0927-796X(01)00037-7. Research Blogging. David Cahen, Gary Hodes (2002-06-05). "Molecules and electronic materials". Advanced materials progress report vol. 14 (No. 11): pp. 789-798. DOI: 10.1002/1521-4095(20020605)14:11<789::AID-ADMA789>3.0.CO;2-H
6. See the discussion in the general reference by CT Sah, p. 489.
7. Raymond T Tung (November 2001). "Recent advances in Schottky barrier concepts". Materials Science and Engineering: R: Reports vol. 35 (Nos. 1-3): pp. 1-138. DOI:10.1016/S0927-796X(01)00037-7. Research Blogging.
8. The ideality factor for Schottky diodes is discussed extensively by Winfried Mönch (2010). “§3.5 Barrier heights of real Schottky contacts”, Electronic Properties of Semiconductor Interfaces. Springer, pp. 47 ff. ISBN 3642057780.
9. A discussion is found, for example, in Yuan Taur, Tak H Ning (2005). “§2.4 High-field effects”, Fundamentals of modern VLSI devices. Cambridge University Press, pp. 90 ff. ISBN 9780521559591.
10. Sadao Adachi (2005). “Figure 9.11”, Properties of group-IV, III-V and II-VI semiconductors. Wiley. ISBN 0470090324.
11. Winfried Mönch (2001). “Figure 19.18”, Semiconductor Surfaces and Interfaces, 3rd ed. Springer, p. 412. ISBN 3540679022.
12. The Miedema electronegativity unit is derived from the work functions of metal alloys and intermetallic compounds, and is thought to be more appropriate here than the electronegativity scale of Pauling based upon covalent bonds in small molecules. See A.R. Miedema (July,1973). "The electronegativity parameter for transition metals: Heat of formation and charge transfer in alloys". Journal of the less common metals vol. 32 (No. 1): pp 117-136. DOI:10.1016/0022-5088(73)90078-7. Research Blogging. The Miedema scale is related to Pauling's scale approximately as XMied = 1.93XPaul + 0.87, according to W. Mönch (2006). “Equation (2.16)”, Susanne Siebentritt, Uwe Rau, eds: Wide-gap chalcopyrites. Birkhäuser, p. 24. ISBN 3540244972.  A tabulation of Miedema electronegativites is found in AR Miedema (1992). "Energy effects and charge transfer in metal physics: modelling in real space". Physica B: Condensed Matter vol. 182: p. 14. DOI:10.1016/0921-4526(92)90565-A. Research Blogging. Some particular values are: XSi = 4.7, XCs = 1.95, XPt = 5.65.
13. Figure adapted from Winfried Mönch (2004). “Figure 9”, W. J. Choyke, Hiroyuki Matsunami, Gerhard Pensl, eds: Silicon carbide: recent major advances. Springer, p. 139. ISBN 3540404589.
14. V. Heine (1965). "Theory of surface states". Phys. Rev. vol. 138: pp. A1689-A1696. DOI:10.1103/PhysRev.138.A1689. Research Blogging.
15. Raymond T Tung (2001). "Previously cited paper; §4: Models on bonds and dipoles at MS interfaces". Materials Science and Engineering: R: Reports vol. 35 (1-3): pp. 1-138. DOI:10.1016/S0927-796X(01)00037-7. Research Blogging.


text:
# How do you solve by substitution x - 4y = -1  and 3x + 5y = 31?

Jul 13, 2015

$x = 7$
$y = 2$

#### Explanation:

We are given:

$x - 4 y = - 1$

and

$3 x + 5 y = 31$

We can do this a number of ways, but let's start with $x - 4 y = - 1$

Let's multiply $x - 4 y = - 1$ by $- 3$ to get its $x$ to look like the $x$ in $3 x + 5 y = 31$ but with an opposite sign:

$x - 4 y = - 1$

$\left(- 3\right) \left(x - 4 y\right) = \left(- 3\right) \left(- 1\right)$

$- 3 x + 12 y = 3$

Now, let's add $- 3 x + 12 y = 3$ to $3 x + 5 y = 31$ to get rid of those $x$'s:

$- 3 x + 12 y = 3$
$3 x + 5 y = 31$

$17 y = 34$

Now, divide both sides by $17$ to get:

$y = 2$

Now, just plug in the value for $y$ into $3 x + 5 y = 31$

$3 x + 5 y = 31$
$3 x + 5 \left(2\right) = 31$
$3 x + 10 = 31$

subtract both sides by $10$:

$3 x + 10 = 31$
$3 x = 21$

divide both sides by $3$:

$3 x = 21$
$x = 7$

So, $x = 7$ and $y = 2$

To check if our answers are correct, you can take either of the two given equations and plug in the values of $x$ and $y$.

Let's try this with $x - 4 y = - 1$

$x - 4 y = - 1$

$\left(7\right) - 4 \left(2\right) = - 1$

$7 - 8 = - 1$

$- 1 = - 1$