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# How can invading aliens access the Internet to find out all about us?

Aliens are camping on Mars and gathering their forces ready for an invasion of Earth. They wish to find out as much as they can about us before commencing hostilities.

Question

How can they access the Internet and hence get a valid email address and search Google etc.?

Bear in mind they have no Earth money to buy broadband, or presumably a way to access it. Also there is a delay (varying according to relative orbital positions) for signals to travel between Earth and Mars. They are not familiar in the details of internet protocols or security measures although they have gleaned some rough ideas from live broadcasts that they exist.

Notes

1. There is no FTL travel or communication available within or near the solar system (for reasons of physics I won't explain here). Please just assume they can't use FTL.

2. One important piece of knowledge for them is the completest sequence they can get of the human genome and of human viruses. This is so they can carry out virus warfare.

EDIT What follows is not essential reading. It's partly to say thanks.

By implication (e.g. my mentioned of delays due to distance of Mars from Earth) I believe that my question can be seen as relating only to communication directly to and from Mars. That was certainly my original intention although not explicitly stated. @Renan has given an excellent answer to this. Others however have presumed that my aliens must have the capability to send spacecraft into near Earth orbit. Although I hadn't envisaged that, it certainly gives extra information as to what the aliens might be forced to do. Therefore I am also very interested in those answers. Particularly the "we come in peace" subterfuge mentioned at the end of the answer by @Matthew

I wish I could accept several of these as the best answer. SE doesn't allow it. So I will eventually pick one but many thanks to all who have taken the trouble to answer - some with considerable knowledge and clarity.

• The Interplanetary Internet (which is a very real thing being developed) isn't up and running to Mars yet. But once it is, they could just use that. – Michael Hampton Sep 12 '20 at 3:45
• Wearing my network geek hat, hoping that this will help readers evaluate answers. First, "getting an email address" is spurious, you don't need it to /learn/ from the Internet. Second, there are protocols other than TCP so reference to TCP-specific behaviour is spurious. Third, one can learn a lot simply by monitoring traffic (precedent: the row over Google StreetView). So if, for example, the aliens could monitor the uplink to a satellite which was retransmitting Internet traffic to a remote area, they'd see both emails being sent and quoted context... from which they could learn a great deal – Mark Morgan Lloyd Sep 12 '20 at 8:36
• The Interplanetary Internet mentioned by @MichaelHampton was my first thought, but just because it is "an internet" doesn't mean that it is "the Internet " (capitalised) with full access to all websites and the capability of monitoring traffic. As a parallel, my understanding is that the US DoD runs a secure network using internet-style protocols which reaches all significant operating locations, but that does not mean that there are any points of contact between them. – Mark Morgan Lloyd Sep 12 '20 at 8:39
• There is so much about this question that demands a suspension of disbelief. There are now so many rovers and satelites on and around Mars from so many countries, that it is monitored 24/7 by amateurs around the world, just because they can, and becasue there is something to listen to. Mars can not get a squeak out without someone on Earth noticing. Unless the invaders are using an 'electromagnetic spectrum' we do not know about (Quantum entanglement? ESP? Mind reading?) for communication, someone on Earth would have 'heard' them. Sorry, but the invaders arrived about 20 years too late. – Justin Thyme the Second Sep 12 '20 at 14:51
• @Justin Thyme the Second The fact that humans can observe aliens amassing on Mars does not prevent the aliens for prepping their invasion, nor spying on Earth. Say the humans aren't sure what aliens are doing, or lack any means of stopping the aliens on Mars. – Vincent Sep 14 '20 at 4:18

# They can't

A radio signal will take between 5 to 20 minutes to go directly from Earth to Mars and back. That makes for a round trip time of between 10 to 40 minutes.

So even supposing that:

• Your martian hackers have gone through the DNS part (i.e.: the venusians provided them the IP 172.217.13.206, which is one of the valid addresses for Google as I type this);
• The aliens write a proper HTTP request (ignore the cookie part);

...This is what will happen from the martian point of view:

### t = 0 ms

Martian computer sends a TCP packet. Most details are irrelevant to the question, but it will say "SYN" (for synchronize).

### t = 620,000 ± 4,000 ms

The computer in Mars receives the SYN + ACK packet (synchronize + acknowledge).

### t = 620,001 ± 4,000 ms

The computer in Mars sends an ACK (acknowledge) packet, which serves to formalize that a connection has been opened.

### t = 683,000 ± 6,000 ms

The computer in Mars receives a second SYN + ACK packet.

### t = 745,000 ± 6,000 ms

The computer in Mars receives a third SYN + ACK packet. Earth's computers go radio silent.

And from Earth's point of view:

### t = 0 ms

Some server on Earth receives the SYN packet from Mars.

### t = 1 ms

Server sends a SYN + ACK packet to Mars.

### t = 60,001 ms

Server sends a second SYN + ACK packet.

### t = 120,001 ms

Server sends a third SYN + ACK packet.

### t = 180,001 ms

Server gives up on the connection because it has sent three ACK packets and got no response in appropriate time (a SYN + ACK packet is also an ACK packet).

### t = 603,001 ± 2,000 ms

Server receives an ACK packet from Mars and completely ignores it since it is neither for an open connection nor a request to open one. By design, the server won't even bother to respond.

You may be thinking now that after a few tries the aliens would learn to send SYN and ACK packets in the right time to fool the server on Earth. You'd need a Laplace demon for that - The SYN + ACK packet from the server on Earth comes with a randomly generated number - let's call it X. The ACK packet from Mars must include X + 1 in one of its fields.

And sure, that X number is not truly random, but only pseudorandom. You could probably find out the next number the server would pick - if you had the only machine ever reaching that server. Other people using Google will add noise here so you won't be able to guess the next X.

Even if you do guess that number, TCP is a protocol which expects a response for every packet that is sent. Preempting the confirmation of a packet that has not been sent yet is just noise - you really need to receive every packet and the ones you got. Failure to do that means the other side (client or server) is expected to send all the same packets again, but more spread apart over time (in plain English: must resend more slowly) and with more expected confirmations. The more you try to outsmart TCP, the slower your transmission will be and the more likely you are to have timeouts.

The aliens will have a better chance when they are within Earth's sphere of influence, where their latency will be much lower (I think around 4 seconds for a TCP round-trip). However, I see that the question has been edited to rule that out.

I see that a new edit removed that constraint. Check Enthus3d's answer for an idea on how to overcome the latency problem.

• @Renan I find the title of your answer misleading, in the final section you say that they can achieve lower latency near Earth's sphere of influence, so the title caption of 'they can't' is a bit misleading. – Enthus3d Sep 11 '20 at 14:35
• @Demigan SYN comes with a random number. It is actually pseudo random, but you won't always get served by the same server even for a Google Search and numbers are served for everyone, so either you have a Laplace demon or you're stuck at this point. – The Square-Cube Law Sep 11 '20 at 14:37
• Of course they can. Having the technology to invade Earth, they certainly have the technology to place a satellite in orbit that buffers incoming/outgoing requests to completely avoid this problem. – JBH Sep 11 '20 at 17:01
• You are forgetting about timeouts. TCP/IP packets with that level of latency will timeout and cause the connection to fail. They don't just have a slow internet connection, they have no connection what so ever. – Nosajimiki Sep 11 '20 at 22:00
• "they can use UDP instead of TCP" - no, they can't, at least not for HTTP[S]. (Well, not until HTTP3 anyway, but many, many sites still use HTTP1.) Most protocols don't let you choose whether to use TCP or UDP. – Matthew Sep 12 '20 at 2:43

## From Mars: One 4K Netflix Movie per Unconditional Surrender

The real answer is provided in the next section, with the use of probes, but here is the physical reason why it is impossible to access the internet directly from Mars: signal strength loss.

Regardless of how the Aliens receive weak signals from Earth, the Aliens will still need to send signals to Earth to query from Google. They will need to use a high-gain directional antenna, which narrows the beam width to heighten the signal strength and reduce interference, to send their signals towards Earth.

If the signal is sent to an omni-directional source of a Wifi enabled device (roughly 2dbi) (since no sane person would point their directional wifi extender perfectly accurately into space and towards Mars instead of their router), and the aliens use a Nasa-grade high gain 48dbi antenna, with a signal frequency of 2.8GHz, like wifi, then the signal attenuation will be 200db (which is 1/1e20 of the signal strength). This high loss in signal strength is because of the distance, and we are forced to send signals to an omnidirectional receiver rather than a directional one (a difference of 20-50db).

The most reliable signal we can realistically send will have a signal loss sitting at roughly 37dBw (67dbBm). To achieve 37dBw signal strength, the aliens will need a signal of roughly 171 dBw due to 208dBi signal loss, which will mean a signal with 1.26e+17 watts worth of power.

This means, to send a single TCP packet of 65535 bytes or 524,200 bits at 2.8GHz, to try to initiate a search on Google from a device, will need 1.87e-4 seconds worth of a signal, or 23.6e12 joules. The energy equivalent of that is roughly one nuclear bomb.

Due to a mistake in signs, originally we had come to a conclusion that two google searches would be enough for a planet-wide extinction event. This is no longer possible with the current calculations, but, hypothetically what does this mean, in terms of planet-wide extinction events? The asteroid impact that wiped out the dinosaurs sits at roughly 4.2e23 joules, so transmission of 3/4ths of the 1,200 petabytes on the internet from Google, Amazon, Facebook, and Microsoft would be on the same scale as the next extinction of the entire human race.

Humanity's satellites would quickly notice if a few signals, each on the scale of a nuclear explosion, were sent from Mars, and large craters from each google search aftershock were to continuously blast open on Mars. If the aliens were capable of harnessing this level of power, all it would take is one 4K Netflix movie on Earth, and a few countries would be gone, the Earthlings cowed into surrender.

As such, due to the heavy loss of signal strength over the long distance between Earth and Mars, it is highly unlikely the aliens would attempt to send WiFi signals from Mars. (And even then, as in Renan's answer, they would be stopped by the latency).

## Analysis and Simulation

The more efficient way would be to probe an internet satellite, from providers like xplornet. The aliens could send a probe near one of the internet satellites, and analyze or transmit signals to it.

If they are not familiar with satellites, they can gain knowledge about them by dissecting one of the many dead satellites in the graveyard orbit, or, with more risk, hijack/study a nearby Mars Rover or Orbiter, which have a ton of communication equipment.

The probe will analyze existing signals going into and out of the internet satellite, meaning there will be little risk of detection. Any encryption on the internet signals can be broken, or they can target strictly un-encrypted signals. Once they are familiar with the signals, they can simulate their own signals to the satellite, with an appropriate 'fake latency' to simulate it as coming from Earth.

The probe they send to hack the satellite will also have a two-way communication method that will send and receive signals to and from the aliens on Mars. Since the distance is far, the probe can use directional antennas to reduce signal loss, with a similar setup to how NASA handles transmission between Mars orbiters to Earth. They could also make use of alien transmission technologies, since there is no format requirement on this signal, allowing to pick techniques that are less noticeable or have a better transmission format.

## Probe's Maximum Distance from Satellite

In terms of how far the beacon should be placed, if the aliens want to completely avoid their probe being caught, the United Space Surveillance Network tracks satellites in orbit around the Earth. A crucial part of this is the GEODSS system

the GEODSS system can track objects as small as a basketball more than 20,000 miles (30,000 km) in space or a chair at 35,000 miles (56,000 km), and is a vital part of USSPACECOM's Space Surveillance Network. Distant Molniya orbiting satellites are often detected in elliptical orbits that surpass the Moon and back (245,000 miles out).

This is to say, should a beacon be placed, and surveillance is to be completely avoided, the beacon should be located beyond 245,000 miles out from the Earth, to avoid detection by the GEODSS system. The speed of light in a vacuum is 186,282 miles per second, meaning, at this placement, the latency will be around 1.315 seconds. This is a fairly high latency, potentially causing file download/upload speeds to slow down to 0.5x-10x compared to 500-10 ping, due to handshakes being needed to confirm packets have been received (dependent on the size of the packet), but this will not affect browsing the internet.

## Dealing with Latency and Packet Loss

The delay from latency won’t be a significant issue, as accessing the internet for information isn’t reliant on latency (think Netflix, Youtube, Wikipedia). To avoid timing out for requests, they can easily program the beacon to simulate handshakes and requests locally at the satellite, effectively appearing to have close to 0 ping (or, if they are avoiding the GEODSS, ~1300 ping). They can then stream out the data to mars at their leisure. Packet loss from the distance will be reliant on what technology they use for the beacon, but also shouldn’t be an issue, as they can resend requests.

They will, however, do terribly in World or Warcraft due to actual latency and packet loss from Mars, but that's what single player games are for. Perhaps they may come in peace after being unable to resist the temptation of playing such games with low ping.

Most likely, due to high traffic volume, google would be the first website they will analyze, and they will be able to quickly simulate internet requests to and from google.

With google, they will be able to easily obtain information like how to create an email account, gain access to Wikipedia, and watch online lecture videos from Youtube. This will allow them to quickly gain a general understanding of the human race and carry out their next plans.

## Dealing with Potential Issues

Q: How would they gain the know how to work on satellites?
A: By analyzing and dissecting the dead satellites in the graveyard orbit

Q: Can probes they send completely avoid detection?
A: This depends primarily on how far ahead the aliens are in dealing with electromagnetic waves. Most likely they will have no issues easily dealing with any of our primitive surveillance systems, but even without such measures, by primarily staying out of our surveillance systems' maximum effective ranges (300,000+ miles), there is very little probability of detection. In fact, the European Space Agency alone has mapped 61-89% of Mars, and many other Space Administrations/Agencies are also constantly probing Mars for various other things. Without some kind of anti-surveillance tech, it is more likely that the alien spacecraft/base on Mars will eventually be caught than the relatively small beacon.

Q: Would the humans notice their satellite is being monitored by the aliens?
A: By first analyzing existing dead satellites, then analyzing only signals going in and out of an internet satellite, detection chance is very low.

Q: Would encryption prevent the aliens from accessing the internet at all?
A: No, only some 50% of the world’s total internet traffic is encrypted, with unencrypted HTTP traffic remaining high (20-40%) in the APAC region.

Q: Would encryption deter the aliens?
A: This is probably the biggest issue, all of this depends on the alien's level of computing technology. It should still be possible to break some of the less heavily encrypted transmissions, but breaking the more complicated encryption methods would require the use of a quantum computer or equivalent in processing power.

Q: How would they deal with the minutes high latency between Earth and Mars?
A: Simulate return responses at the satellite itself, effectively achieving near-0 ping from the perspective of the internet services.

Q: How would they deal with the massive signal loss from Earth to Mars?
A: By using two directional antennas, the signal loss from one point to the other is heavily reduced. They can use a setup like NASA's space orbiters, with one low-gain and one high-gain antenna on the probe.

Q: What if the aliens are not able to send out any probes?
A: Then barring the use of any space magic, the aliens would have no reasonable way to send and receive signals due to signal loss. If they had such space magic, actually accessing the internet from Mars and dealing with TCP would also be extremely difficult; @Renan provides an excellent answer.

• Comments are not for extended discussion; this conversation has been moved to chat. – L.Dutch Sep 15 '20 at 12:07

They need to get local.

Current (actively used) internet protocols are simply incapable of dealing with the light-lag delay between Earth and Mars. Even if they could somehow build a powerful enough wireless transceiver to connect to WiFi or other internet-carrying signals, ping times of >40 minutes simply wouldn't work.

If the aliens want to interact with the internet, they could:

1. Send a probe/satellite/scout to Earth. This probe could either hijack a telecommunications satellite or land somewhere with free WiFi (like a Starbucks or a city with free WiFi)
2. Browse the internet, create Emails, look at cat pictures, etc. While many "big" services today want 2FA, you don't actually need a physical address, phone, or credit card to access almost all of the internet.

If the aliens don't want to get local and are content with "read only" access, they could:

1. Use an insanely powerful radio-telescope to pick up individual WiFi and cellular signals on Earth
2. Figure out how to decode those signals and watch what people do on the internet
3. For bonus points, they could hack WPA2 or https protocols so they can see all the data that people send
• Actually... I'm not sure about the physical plausibility of sorting out WiFi at that range (or even from orbit), but if they could, that would work really well. – Matthew Sep 11 '20 at 13:56
• " land somewhere with free WiFi", Yes but how can they use this WiFi without knowing the protocols etc. ? They could find out if they already had the internet but they don't. I don't think un-encoded live broadcasts contain this sort of detailed information, even if they were lucky enough to pick one up. – chasly - supports Monica Sep 11 '20 at 14:25
• @chasly-reinstateMonica Aliens that can figure out how to build spaceships and travel interstellar distances shouldn't have any difficulty cracking our (primitive) language and (primitive) digital security systems, even if they're starting from zero. Most likely their computational technology is so far advanced (digital AGI) that they can just press a "hack" button on their metaphorical dashboards to break anything but the strongest encryption schemes we have (which aren't used in WiFi) without any major problems – Dragongeek Sep 11 '20 at 14:29
• @Dragongeek, that's almost exactly what happened in Troy Rising. Earth's "security" was so pathetic by their standards that their standard implants just reached into any system without them even noticing they were bypassing all of our "security". – Matthew Sep 11 '20 at 18:06
• @Dragongeek if they want to receive Wifi from mars, thats 100db below the noise flour for a 1K anntena. Doing anything with that is so far into spacemagic that if you can do that then the only answer to the question is just they use spacemagic. – user1937198 Sep 12 '20 at 0:22

### They can't.

Well... not directly. You've already noted the main issue they're going to have, which is latency. Gamers think a latency of 100 ms is terrible. Well, your aliens are going to have latency measured in minutes. At that point you're talking about connections timing out due to inactivity. (Our infrastructure is based on the premise that not receiving a reply in a timely manner means something went wrong and the connection should be dropped. You literally will not be able to establish a TCP connection to anything at all, and AFAIK there isn't much you can do online with only UDP at your disposal... and even that will probably have fits with your round-trip times.)

The only way this is going to work at all is to have a really smart (read: autonomous, at least to the point of being able to operate when only receiving instructions every few hours) probe much, much closer to Earth. The moon might be close enough. Earth orbit (or even on the surface) would be much, much better.

### Another snag

This raises another issue you haven't mentioned, however, which is remaining undetected. If they're trying to interact with the internet, even by proxy, that means sending signals toward Earth. Unless they have some totally unknown form of communication technology, someone is bound to notice. We literally have people looking for that sort of thing, and despite what Hollywood would have you believe, it's really, really hard to hide this sort of thing. Their best bet is going to be very tightly focused lasers. Unfortunately, to achieve the necessary focus may require a near-Earth relay, and that's going to be hard to hide.

### What can they do?

The key is going to be passive observation. Just monitoring our satellite broadcasts is a starting point. If they can tap into satellite internet, that will give them a pile of data to sift through. Assuming they can figure out our communications protocols (e.g. IP, TCP, HTTP) without help (which they'd need to do anyway), they should be able to get a decent pile of data just from listening to what other people are doing. However, there's another problem, which is that more and more communication is happening over secure channels (SSL, VPNs). For instance, if they want to read this answer, they'll need to somehow crack the SSL private key for SE. Better hope their computer technology is way, way ahead of ours. At least detection becomes much less of an issue; it is infinitely harder to detect a transmission that is pointed away from all of our detectors. However, slipping a satellite into orbit without being noticed is also far from trivial; we monitor the sky for this sort of thing, not because we're looking for aliens specifically, but because there is a lot of stuff up there and it's a navigational hazard.

### A totally different option

The best place to hide something is in plain sight. (Or, as one of the military minds from Honorverse likes to say, "surprise is when your enemy misinterprets something he's seen all along".) Forget stealth, what you want is misdirection. Have a small group of aliens show up very noticeably. Make it clear that you aren't even attempting stealth. Park in lunar orbit and start spamming the airwaves with communication attempts until you figure out our language. Then announce that you're a tramp freighter looking for trade opportunities in a galactic backwater. I can almost guarantee that at some point someone will help you set up an internet connection (there are certainly enough crackpots with more fantasies than common sense), maybe even invite you to come visit. Make it clear that you aren't a threat. Dangling hints at advanced technology you'd be happy to share will help.

Then, once you think you've learned enough, invite over the friends you "forgot to mention".

• Regarding the secure channels, even just getting the certificate private keys wouldn't necessarily be the solution as more and more systems are rolling out Perfect Forward Secrecy, which involves an additional session key, so even if you have the key you can't decrypt the session without active interferance or breaking the session key. But if the aliens have access to 4068 bit quantum computers all of this is moot as shors algorithm can just churn through all the keys. – user1937198 Sep 12 '20 at 0:30
• Even if they had quantum computers powerful enough, the first would probably be to understand what is supposed to be sent there (i.e. understand the protocols). Note however that there is still a lot of satellite plaintext communication – Ángel Sep 12 '20 at 23:14
• The tramp freighter idea is excellent. They could even give away some non-strategic advanced technology to 'prove' goodwill. This would be great as a storyline. One problem: How would they be able to sustain the trading subterfuge? What would they want from us? Also, how would they explain beaming all the information to Mars - wouldn't we be able to detect this (especially if the communication is two-way)? All the big processing power and the bigwigs are on Mars and they will want to determine what is asked and when. – chasly - supports Monica Sep 13 '20 at 9:19
• @chasly-reinstateMonica - I too like this idea. As for what they would want from us - they can just make something up. Or maybe go for "novelty". A backwater planet like us doesn't have much in the way of useful resources, but we do have interesting finished products - traditional rugs, pottery, whatever. Maybe even some mundane things which would seem exotic back home. As for communication - I think you could place relays around the solar system and use tight beam transmissions in such a configuration that they are undetectable from earth. – Vilx- Sep 13 '20 at 11:38
• For example, one transmitter/receiver could be on the far side of the moon. Another behind the sun, etc. The transmission times would get lengthy, but that's ok. You only need to send status reports once a day anyway. – Vilx- Sep 13 '20 at 11:39

How can Aliens on Mars hack into our internet and download all of wikipedia? Easy.

1. Launch a cubesat towards Earth.
2. Cubesat acts as a relay while engineers back on Mars use it to hack one of the communications satellites. Like our Mars rover, the cube sat is capable of executing a buffer of commands (to deal with the time delay).
3. Once satelite is hacked, the hacked code on the satellite requests wikipedia over HTTP. This is relayed to the cubesat.
4. Then the cubesat transmits the data back to Mars for Analysis. Earth / Mars uplink during line of sight is about 2megabit for NASA. Wikipedia is 9.5gb. Will take 3-4 days to transfer.

After this has been processed, they know a lot about us. They will have read https://en.wikipedia.org/wiki/Comparison_of_webmail_providers, and will just pick one with low standards.

Watch Colony (on Netflix) for an example of fiction where Aliens accomplished this.

• How do the aliens know how to hack a satellite? What will the ground station do when one of their satellite request large amounts of data? – Daniel Sep 11 '20 at 13:56
• They'll have LOS issues too cause they're bound by the same laws of physics, so will probably have a similar solution, plus its the thing sending lots of EM traffic down to the planet. The ground station will forward the request on. It's not sentient. – Ash Sep 11 '20 at 13:59
• And if they pick an internet relay satellite, (the obvious choice) it wouldn't even be that abnormal. You could easily hide 2mb/s of traffic among the 100Gb/s of uplink the HughesNet Jupiter Satellite has: gigaom.com/2012/10/01/… – user1937198 Sep 12 '20 at 0:44

The main problem will be the communication. Most internet communication messages will timeout before they reach the other side, so establishing a TCP (One of the protocols making up the internet) handshake will be quite impossible. Connection quality consists of two parts: Latency (time between sending and receiving) and throughput (amount of data sent). Having all the throughput in the world means nothing if your latency is so high that it never reaches you. Perfect latency also means nothing if your throughput is essentially zero. They have to send a "proxy" probe towards earth to actually interface with the internet and guarantee low latency between the proxy probe and a ground based internet hub.

Once they have an interface close enough to earth, they need to find an open access point. Most ground based access points will not reach high enough, forget hijacking poorly configured WLAN routers. It's possible that LTE and such will reach high enough, but they might also triangulate your position and - though it's probably never checked - it risks detection if your position reads as "in orbit".

This leaves you with two options:

1. You can put something on the ground to actually access the internet and send it up to the orbit.
2. You can try to hijack one of the internet satellites.

Option 1 is easier, because open WLAN exists pretty much everywhere. As long as it's inconspicuous, you can just drop something down, access the WLAN and you're golden. You can download all of the internet, probably including the genome of ebola, easy.

Option 2 is difficult, because there are a lot of satellites around and they are usually strongly encrypted. You might have to spend a lot of time on decrypting traffic and once you've done so, the throughput is awful. Risk of detection is much lower though.

If these invaders do, indeed, have FTL, then they have a 'time machine' into the development of our Internet. They can go, say, 50 light years' distant from Earth, and pick up our first signals from the development of our Internet. Further away, they can go right back to our first radio transmissions, and Morse Code. They only need to go through 100 years of our EM history, really, to have a complete picture.

Once they have access to our EM historical record, they can build up the evolution of our protocols. Each successive iteration of our protocols built on the last, going all the way back to Morse Code, and the Teletype machines. The ACK/SYNC/CQ codes, in fact, and ASCII, go all the way back to the teletype, completely unencrypted. Handshaking protocols go all the way back to the first dot-dash Morse Code used on the Titanic.

Going back only 40 years or so, they have pre-email, pre-WWW bulletin board system transmissions. Remember AOL floppies to connect to their servers?

It would not be difficult for a civilization that can do FTL, to predict and reverse engineer our current modern protocols, based on the evolutionary record of our radio transmissions over the last 100 years.

• They would certainly be able to listen to Morse Code as that is commonly broadcast. How would they 'listen' to Teletype machines? (especially at 50 light years) – chasly - supports Monica Sep 12 '20 at 15:34
• It is known as RTTY. See junknet.net/radio/rtty, for instance. Used by the US navy, to communicate hard copy orders to naval ship captains. – Justin Thyme the Second Sep 12 '20 at 15:59
• Even our relatively crude systems can pick out a particular cell phone transmission from a crowded major city, from our spy sattelites. – Justin Thyme the Second Sep 12 '20 at 16:22

As other answers have discussed, they can't make a two-way connection to the internet from Mars, because the latency is too long for TCP.

However, they can still receive radio signals from Earth, and a lot of internet data is transmitted via satellite. Picking up stray satellite signals from Mars would be a challenge, but since these guys have the technology to travel between the stars, setting up an appropriately sized space-based antenna is probably not outside the bounds of possibility. Making it hard to detect from Earth might be a bit tricky, but in an appropriate orbit it would just look like an asteroid, and you could always use a swarm of small antennas instead of one big one.

The main challenge would be in decoding the data, both in the sense of breaking encryption and in the sense of understanding all the protocols, to work out where each packet is going and what they all mean. There's also a major challenge in sifting through such a huge quantity of data to figure out which bits are actually useful. But if their computing power is advanced enough compared to ours (which it realistically would be for an interstellar civilisation) then this all ought to be feasible as well. Then they wouldn't be able to run their own Google searches, but if someone on Earth were to run a search they're interested in, they could simply intercept the result.

• Even the signals from cell phones and home WiFi routers can be picked up by antennas that are sensitive enough, even from Mars. If the invaders can hide an entire invasion force, they can hide antennas. – Justin Thyme the Second Sep 12 '20 at 14:17

They already did, and they are connected - Information superiority is the primary leverage for any interaction with other civilizations, terrestrial or extraterrestrial, for us or for them.

Travelling is not required - Leaning on Occam's razor principle stating that, due to the high likeliness of the current knowledge of the Drake equation in combination with the exponential likeliness of the technological singularity and quantum physical research and quantum entanglement implications, our extra-terrestrial friends are everywhere and the vast majority of the highly developed civilizations do not even have to travel here at all to interact with us. The only civilizations that need to travel by FTL are still developing.

Monitor, plan and act first - An ancient highly developed civilization would with self-repelled replicating AI monitoring technology or quantum sensor grids have an almost universal civilization sensor network that would ensure complete information monitoring. Early recognizance would ensure they can align developing civilizations microclimate so that the environment would be immunologically friendly and by seeding core cultural memes slowly acculturating the indigenous population and aligning intelligence evolution so that an “invasion” would not even be required.

Connection to simple networks is simple - If one assumes they have full "physical" access to our environment, and based on that even on our own "primitive" military intelligence organizations continuously show ability to connect and break all "unbreakable" encryption schemas, connecting to our networks is simple. Our network technology and security very likely is like reading old snail mail and would more resemble reading open post card-based communication.

They are already here, and they are chatting - So, given all statements above, we can safely assume to say they are already here, they are all connected, they know everything about us, we interact with them already and most of us have no clue. “Invasion” already happened, and we did not notice as they are a friendly bunch of folks, albeit with different looks, just chatting along in our chatrooms.

• What you say could easily be true. However, it begs the question be asked, 'Who is, or are, 'they'?' Just because SOMEextremely advanced civilization is already here, does not preclude a more primitive civilization from ALSO being here. A civilization like, say, these invading aliens based on Mars. 'Old Man's War' type scenario, for instance. There is always a bigger, more powerful, 'jerk', until there isn't. – Justin Thyme the Second Sep 14 '20 at 12:44

They can't

The aliens have NOT read the RFCs that define how the bottom 5 layers of the OSI (Open System Interconnection https://en.wikipedia.org/wiki/OSI_model) model works, thus will have to deduce that the radio signal they receive represents a packetized communications protocol with headers containing metadata and data, which may be binary, or be numeric representations of ASCII, UTF-8, ... characters. Data may be encrypted in whole or in part.

See https://ietf.org/standards/rfcs/

Also see https://youtu.be/hbFq0I7YrYQ Carl Sagan's take on communicating with aliens. Now, consider explaining "Path Maximum Transfer Unit (PMTU), or the phaseover from IPv4 to IPv6 using this method.

• A civilization sufficiently advanced enough to travel light years to get to our solar system would likely have a computer system, and maybe AI, to be able to figure that out in minutes or even seconds. They likely only came here because of the radio signals they received anyway. They likely already solved the problems involved to see that it's more than just noise, so figuring out protocols like this probably wouldn't be very difficult for them. They might not be able to get 100% on web languages, but enough to get more than idea of what to expect. – computercarguy Sep 14 '20 at 21:01
• There is a reasonwhy it is called Open Systems Interface. It is anextremely rudimentary system, designed for the 'average setup'. It is about as straightforward and basic as one can get, almost intuitive for anyone with any sort of background in digital communications. It goes way back to the teletype erra. – Justin Thyme the Second Sep 15 '20 at 0:16
• I "go back to the Teletype era". The development of computing is filled with arbitrary decisions that could have gone the other way without affecting the outcome (say, the order of fields in a packet) that it's random. Given the middle of a bitstream (on the radio) consisting of multiple packet streams, how would they guess byte size? – waltinator Apr 18 at 18:52

## They've been here for decades:

A forward scouting group really did land back in the 50's-60's and escaped Area 51. They hid out listening to TV and radio chatter over the years, then started intercepting satellite communications when it started being used for TV.

Because they got tired of their rations and hunting, while avoiding being hunted in turn, the group got remote jobs to earn money through the mail. Some was stuffing envelopes, other was processing mail-in orders, some was punch card programming for universities and corporations, and some was data decryption for the US Military. It paid just enough to be able to mail-order some food, buy housing through a mediator, and get other amenities.

## They have internet access:

Then the internet became available over satellite. The scouts learned the protocols in the same way as any human did back then, by mail-ordering books and equipment, including some computers. Because the connection wasn't secure, they were able to hack the satellite ISP to give themselves free internet until they could afford to purchase it to avoid having to cover up their "suspicious activity".

When the US Patent system came Online, the alien group downloaded plans for all kinds of things and went into manufacturing with their replication systems. This brought in a bunch of money, so they expanded their operations over the years.

Now they have an underground server farm that replicates a significant part of the internet, including the dark net. And yes, the aliens are part of the dark web with being able to supply all kinds of less than legal or otherwise hard to find stuff. They also do black- and white- hat hacking, do remote CAD and video rendering for corporations & Hollywood, website hosting, crypto-currency processing, and other internet related money making enterprises.

They also make money by posting "fake" pics and video of themselves, furthering conspiracy theories to make their presence sound that much less likely, while also profiting off the merchandise and other materials.

## The aliens own satellites:

They also routinely hack satellites to send a signal back to Mars of all they've been doing as part of regular reports. They have also utilized programs like New Atlas, building their own mini-satellites for more consistent communication to the Mars base. This has allowed them to send a huge amount of data back to their superiors for processing over the years.

To keep the Federal Government at bay, they have shell and front companies handling all the money. Some of these companies are run by humans and don't even know they are fronts. They also have accountants they only communicate through via mail, phone, or internet tech to avoid exposing their alien identities.

The aliens were originally based out of Roswell, but too many people got glimpses and called in the authorities. Fortunately the aliens left before any Feds showed up, so the reports all got called "conspiracy theories". They now have a base of operations in both Reno and Las Vegas. To keep from being found out, they move between them regularly, but only in costume and when costumes are reasonably expected, like on Halloween night or when a comic con is in town. There's also a safe house in LA, but only if the main 2 locations get compromised.

• Writing the entire story for the OP, are we? – Justin Thyme the Second Sep 15 '20 at 0:42
• +1 for imagination but +0 for sticking to the terms of the question!!! – chasly - supports Monica Sep 15 '20 at 8:48
• @chasly-reinstateMonica, you'd be surprised how often people don't "stick to the terms of the question". I was taking a liberty and challenging the frame of the question, which is valid on this Stack. Also, my original aliens did stick to the original question. They didn't have money, they didn't have internet, and they had to figure out how to get around the time lag. I just didn't assume they arrived "today". – computercarguy Sep 15 '20 at 15:41
• @JustinThymetheSecond, ok, so I got a little carried away. Maybe I should use this as a seed for my own book. It looks like the OP isn't too interested in it. – computercarguy Sep 15 '20 at 15:42

The aliens hack a satellite and download whole websites. The satellite does the work of collecting the pieces and radioing them back to the mothership where they are stored on a local cached server.

It's no different to someone with a poor internet connection downloading a tv series on Netflix to a local media player and then watching it.

The internet is already full of cache servers which operate this way.

Alternatively the aliens just clone the data passing through which means they don't have to hack anything, they just watch what everyone else is doing. They could do this from Mars just by listening to the radio waves or from their own relay satellites, isolating each signal and decrypting it. Already a lot of internet traffic goes through satellites, not to mention, telephone communication, military communication, financial data etc.

All the aliens need is sensitive enough radio antennas, excellent processing power and massive amounts of storage. They'll get all the latest shows from Netflix, kids doing homework from Wikipedia right through to encrypted government communications.

• If this is easy why don't lots of people currently do this instead of buying broadband? – chasly - supports Monica Sep 14 '20 at 0:09
• Maybe because most people don't want to base their internet usage on what their neighbour is doing? If you really want to, you can tap into your neighbor's wireless and watch what they are doing. – Thorne Sep 14 '20 at 0:55
• Then again, exactly what would the alien invaders discover about us by watching gadzillion bits of video from porn sites? – Justin Thyme the Second Sep 14 '20 at 3:26
• What does Google and Facebook discover about us from watching a gazillion bits of data? – Thorne Sep 14 '20 at 3:37
• Where the best place is to advertise. and the price to charge for that advertising. It's all about traffic. – Justin Thyme the Second Sep 14 '20 at 12:46

I am not sure I quite understand the focus of the question. It seems to me the answer is elementary, my dear Watson. And pretty much the same level of technology also, my dear Sherlock. Well, maybe not that long ago, but a century and a half ago. Sherlock and Watson were, after all, apparently around in 1887, right at the beginning of the required technology.

My eight year old grandchild can access the Internet, use email, search google, yet knows absolutely nothing about Internet protocols, TCP/IP, encryption, or, really, even what the Internet is. All of this is done by the interface, and the device. Cell phones, tablets, watches, cameras, cars, even light bulbs can access the Internet. For the end user, everything about the Internet is nothing but a black box.

So the question is, how do they get access to any one of a number of devices that already connect to the Internet, remotely, without worry about the time lag, using very simple technology to do so? That is, they do not need to know anything about the Internet, if they can gain access to a device that does all of this for them, remotely, from Mars, using simple radio wave technology, and better still, a code as simple as Morse Code, and Morse Code-type handshaking.

Very often, we tend to over-think the problem, and try to figure it out from a high level high tech perspective. Sometimes, the answer is in very old technology. Remember ham radio? The short wave band, which used to be very popular with hackers? The dot-dash of Morse code? The basis for the originator of ASCII code and binary transmission? The one that was the go-to option for technoadventurers before computers and the cell phone? The system that just needs some basic equioment, made in the basement with a soldering iron, and huge antennas in the back yard? Things that could be buit from odds and sods available in 1912? If a culture can make it to Mars from some remote place in the galaxy, they can figute out how to make a basic ham amateur radio rig.

Well, it so happens that these rigs can be and are interfaced to the Internet. That's right, ham radio was not killed by the Internet, it was just made easier. Ham has transitioned to the Internet, interfaced with ham radio transceivers through publicly accessible Internet gateways. See this, for instance.

A simple communications rig, on Mars, connects through short wave to any of a number of existing Earth gateway base stations, run by HAM radio amateurs holed up in their basements, connected to the Internet, all protocols taken care of by the ham equipment.

Since ham radio broadcasts are easy to pick up, and there is constant chatter over ham radio frequencies, and these amateur radio gateways are far from a classified secret, it would be no problem for an advanced civilization to learn how to tap into the existing widespread amateur radio/Internet gateway community just by listening to non-encrypted public access short wave radio chatter. All they need is an antenna and a basic radio rig.

So this colony on Mars uses short wave radio to contact any one of these earth-based gateways, and they are into the Internet, no knowledge of protocols necessary. At least, none that they can not pick up just by listening to unencrypted ham radio traffic. They use plain old ham amateur broadcasts to send their communication to a gateway, the gateway hardware and software takes care of all of the translation/interface/handshaking/protocols to shft the communications on to the Internet, the Internet responds back to the gateway, which beams the response back to Mars using normal, rudimentary short wave.

• The only risk with Ham, is that radio amateurs are probably more likely to notice that signal is comming from Mars. Whether that is a problem for the story is different matter. – user1937198 Sep 12 '20 at 10:54
• @user1937198 It's pretty darn certain that amateur operators would detect ANY type of foreign radio transmission from Mars. Unless the Mars invaders somehow created a tight-beam leak-proof link to some Earth base station, their attempts to tap into our internet would be 100% noticed. There are now so many rovers on Mars, and circling it, that amateurs around the world are monitoring Mars 24/7 in order to pick up the signals transmitted back by these exploratory research expeditions. You can't get a squeak out of Mars without someone on Earth picking it up. – Justin Thyme the Second Sep 12 '20 at 14:12
• It does depend on bands. Amateurs are focused on the bands that we know are transmitting. A tight beam spread spectrum signal on the right obscure band could probably hide for a significant amount of time. – user1937198 Sep 12 '20 at 14:50
• @user1937198 Amateur radio enthusiasts are now into all frequencies, including ultra-high so that they can bounce signals off meteor trails. See for instance park.org/Guests/Hamradio/gb/index.html The idea is to push the limits, to do what no one else has done, to be the first. – Justin Thyme the Second Sep 12 '20 at 15:04
• Even then if they use something like the TV bands, and I doubt anyone will notice a mars signal over terestial signals if the power is low enough once it reaches earth. – user1937198 Sep 12 '20 at 15:12


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# Another potential barrier question

1. Nov 17, 2006

### holden

Since these seem to be popular around here these days, I thought I'd another to the mix :)

"A particle of mass m and energy E > 0 approaches a potential drop -V0 from the left. What is the probability that it will be reflected back if E = V0/3? The potential energy is V(x) = 0 for x<0 and V(x) = -V0 for x>=0."

So what we're looking for is R, the reflectivity constant. My book tells me a formula for 1/T for this case (E > 0, V < E), but that's for a rectangular well and includes the width of the well in the formula.. and this one is infinite.

What I've tried to do so far is using $$\frac{d^2\psi}{dx^2} + \frac{2mE}{\hbar^2}\psi = 0$$ for the Schrodinger's equation to the left of the well. I'm not really sure why I'm using this other than it's in my notes (cause V(x) is zero?). So from this I'm getting $$\psi(x) = Ae^{\frac{i\sqrt{2mE}x}{\hbar}} + Be^{\frac{-i\sqrt{2mE}x}{\hbar}}$$.

I know R = abs(B/A)^2.. but I'm not sure what to do next. I think apply the boundary conditions..? I'm really lost on how to do that.. I think I'm missing something fundamental here.

Thanks in advance for the help!

2. Nov 18, 2006

### nrqed

To the right of the drop, the solution will be
$$C e^{+i k_2 x}$$
where $$k_2 = { {\sqrt {2m(E+ V_0)}} \over \hbar}$$
Now impose that the wavefunction is continuous at x=0. This gives obviously $A+B =C$. Now impose that the derivative is continuous too. That will give a second relation relation the three coefficients and the energy. Using those two equations, you should be able to write B in terms of A and then you can get R.

Hope this helps

Patrick

3. Nov 18, 2006

### holden

It does! Thanks!! One question, just for my own understanding: why is there only a C term to the right, instead of a C and a D term? This is where I get confused on these.

4. Nov 18, 2006

### physics girl phd

I think it will help if you think about the conditions at +/-infinity.

Also -- physically:
To the left of the barrier you have a forward propagating part (exp(+number)) and a reflected part (exp(-number))... once you hit the barrier, you have just forward propagating part that tunneled into the material (this is kinda like optics).

5. Nov 18, 2006

### nrqed

A wave of the form $e^{i kx}$ is a travelling wave moving to the right (assuming k is a real positive constant). In that problem, the inicident wave comes from the left. There is a reflected wave and a transmitted wave to the right of the origin. So to the right of the origin, there can be no wave moving to the left (this could only happen if there was a second incident wave coming from x=+infinity...but if the only incident wave is coming from x=-infinity, there can be no wave travelling to the left in the x>0 region)

Glad I could help a little

6. Nov 18, 2006

### holden

That makes perfect sense. Thanks to all for the help.

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# The naive Bayes classifier

## In words...

First of all, the naive Bayes classifier should not be confused with the (optimal) Bayes classifier as it is in fact far from optimal.

The naive Bayes classifier is one of the most basic generative classifiers. It relies on a very crude approximation/assumption of conditional independence of the features. This means that the classifier assumes that the value taken by a feature does not depend on the other features.

This assumption is obviously wrong in most applications. For instance, in spam filtering, this amounts to considering that the presence of the word "earn" does not depend on the presence of the word "money". But despite this very crude approximation, the naive Bayes classifier remains rather effective.

## In pictures...

### Demo application

This demo shows how the naive Bayes classifier can be used to filter spams in mailboxes.

## In maths...

For a random variable $X$ taking values in $\X\subset\R^d$, the conditional independence assumption at the core of the naive Bayes classifier results in the following factorization of the conditional probability of $X$ given $Y$: $$P(X=\g x\ |\ Y=y) = \prod_{j=1}^d P(X_j = x_j\ |\ Y = y) ,$$ where $X_j$ is the $j$th component of $X$.

This assumption drastically eases the estimation of the generative model as shown in the example below.

### Naive Bayes classifier with real inputs

For a continuous random variable $X$ of density $p_X(\g x)$, the factorization above reads $$p_X(\g x) = \prod_{j=1}^d p_{X_j}(x_j\ |\ Y = y)$$ where $p_{X_j}(x_j\ |\ Y = y)$ is the density of the $j$th input variable (or feature) $X_j$.

Assume now that we would like to fit a Gaussian distribution over $X$ in each class $y\in\Y$. Fitting a multi-dimensional Gaussian distribution in $\R^d$ implies to estimate the values of the $d$ means and of the covariance matrix $\g \Sigma \in\R^{d\times d}$. But with the factorization above, this can be reduced to the estimation of $d$ one-domensional densities $$p_{X_j}(x_j\ |\ Y = y) = \exp\left( \frac{-(x_j - \mu_{yj})^2}{2\sigma_{yj}^2}\right),\quad j=1,\dots,d,$$ with $d$ means $$\mu_{yj} = \frac{1}{N_y} \sum_{y_i = y} x_{ij}$$ and $d$ variances $$\sigma_{yj}^2 = \frac{1}{N_y-1} \sum_{y_i = y} (x_{ij} - \mu_{yj})^2$$ for each class $y$ having $N_y$ examples in the training set $\{(\g x_i, y_i)\}_{i=1}^N$.


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# What would the orbit pattern be if the sun and planets orbited the earth?

For the Sun and other planets to orbit the Earth, Earth would need to be at the barycentre. However, there is no way it could stay at the barycentre because the Sun's gravity would immediately accelerate it away at over $270 m {s}^{- 2}$.


text:
QUESTION

SCI 100 Week 2 DQ 1

This document of SCI 100 Week 2 Discussion Questions 1 Identifying Moderate individual stressors shows the solutions to the following problems:

How do our habits and choices relate to our physical health and our ability to manage stress? Chapter 2 lists some general steps you can take to help decrease stress.

- Shore up a support system

- Improve communication skills

- Develop healthy exercise, eating, and sleeping habits

- Learn to identify and moderate individual stressors

Pick one of these categories and find an academic-based source that describes a technique/intervention that reduced

• @
• 5 orders completed
Tutor has posted answer for $5.19. See answer's preview$5.19


text:
# Improper Fraction|Definition & Meaning

## Definition

An improper fraction is called an improper fraction in which the numerator exceeds or at least equals the denominator. The equivalent decimal value of the improper fraction is 1 or greater.

## What Is an Improper Fraction?

A fraction is inappropriate or improper when the fraction’s numerator is either greater or equal to the fraction’s denominator. For instance, the fractions 5/2 and 8/5 are examples of incorrect fractions. The numerator and the denominator are the components that make up every fraction. In mathematics, two primary types of fractions are distinguished by the values of the numerator and the denominator. These are referred to respectively as proper fractions and improper fractions.

Figure 1: Representation of improper fraction.

### History

Around 1600 BC, the Egyptians were the earliest known users of fractions, making the concept quite a bit older than that. There are many distinct kinds of fractions. A fraction is considered correct when the numerator is less than the denominator, as in the case of 1/2. Contrary to this, T, if the denominator is lower than the numerator or has the same size as the numerator, then the fraction is inappropriate. One such example is the score of 16/15.

One sort of fraction is known as an improper fraction, and it is characterized by having a numerator that is either equal to or greater than that of the denominator. Its value is either one or an integer more significant than one. Incorrect fractions are typically represented in mixed numerical form and simplified in some way since it is simpler to understand mixed fractions than improper fractions.

### How To Simplify Improper Fractions?

We are now aware of what is meant by the term “improper fraction.” Let’s talk about how to simplify such fractions here

#### Step 1

First, determine whether or not the fraction given is correct.

#### Step 2

The second step is to interpret the denominator and determine the total number of parts that it divides the numerator into.

#### Step 3

Determine which factors are shared by both the numerator and the denominator.

#### Step 4

Remove the terms that are identical from both the numerator and the denominator.

## Improper Fractions and Mixed Fractions

It is a fraction in which the fraction’s numerator is either larger than or equivalent to the value of the fraction’s denominator. For instance, 9/4 and 4/3 are both examples of incorrect fractions. In terms of numbers, they are consistently either equal to or bigger than 1. On the other hand, a fraction is said to be mixed when expressed as a composite of natural numerals and a proper fraction. This type of fraction is known as a mixed fraction. It is a more straightforward representation of an incorrect fraction. From a mathematical perspective, a mixed fraction will always be more significant than one. Moreover, if the rules are followed, any mixed fraction can be written as inappropriate.

Figure 2: Representation of mixed numbers.

Reading and comparing mixed fractions to improper ones in everyday situations is typically more straightforward. This is because mixed fractions include both whole numbers and fractions. Following some fundamental methods, which you will become familiar with in the coming sections of this page, we can easily convert any improper fraction to a mixed number or any mixed fraction to an improper fraction.

## Converting Improper Fractions to Mixed Numbers

Always remember that the denominator of the improper fraction’s mixed fraction form will be the same as the denominator of the underlying fraction. Because mixed numbers are believed to be a simplified form of improper fractions, it is essential to learn to convert between the two types of numbers. The following is a list of the procedures that need to be taken to transform an improper fraction into a mixed number:

### Step 1

Firstly, divide the numerator by the denominator.

### Step 2

In the second step, you will determine the values for the quotient and the remainder.

### Step 3

To write a fraction in the form of a mixed number, arrange the values of the quotient, the remainder, and the divisor.

Figure 3: Converting improper fraction to mixed number.

## How To Solve Improper Fractions?

The process of simplifying the value of an improper fraction after it has been solved by executing arithmetic operations on it is called “solving” improper fractions. The fundamental arithmetic operators in mathematics are addition, subtraction, multiplication, and division. These are the four arithmetic operations. The only difference between solving an improper fraction and calculating any other proper fraction is that, in this case, we need to simplify the result and represent it in mixed numbers. Otherwise, solving an improper fraction is similar to solving any other proper fraction.

## Subtracting Improper Fractions

In a manner analogous to how we added the improper fractions, we can subtract them.

• The first thing to do is determine whether the denominators are the same.

• The second step is to make the denominators more reasonable.

• The final step is to remove the fractions that have been given and, if necessary, simplify them.

## Examples of Improper Fractions

### Example 1

Small fish are fed three times a day. The amount of fish that are fed is measured in terms of the total weight of the fish fed.

During the morning feeding, 3/2 of tons of fish are fed. During the afternoon, the number (or weight) of the fish fed will be 4/3 of a ton more than the ones in the morning feeding. If the total number of fish fed on a day is 27/5 of a ton, how much is fed during the feeding at night?

### Solution

Let’s assume that the weight of fish fed at night = x tons

In the morning, the weight of fish fed = 3/2 tons

In afternoon the weight = 4/3 + 3/2 = 17/6. Solving for x:

3/2 + 17/6 + x = 27/5

13/3 + x = 27/5

x = 27/5 – 13/3

x = (81-65)/ 15

x = 16/15

So, during the night feeding, 16/15 of tons of fish is fed.

### Example 2

Jame’s father is 7 1/2 feet tall. Jame is 5 feet tall. How many times taller is Jame’s father than him?

### Solution

Jame’s father height = 7 1/2 feet

Jame’s height = 5 feet

= (15/2)/ 5/1

= 15/2 * 1/5

= 15/10

= 3/2

So, Jame’s father is 3/2 times taller than him.

All images/graphs are created using GeoGebra.

5/5 - (5 votes)


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# Find polar coordinates

• Dec 10th 2009, 07:24 PM
mmattson07
Find polar coordinates
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?
• Dec 10th 2009, 07:41 PM
Prove It
Quote:

Originally Posted by mmattson07
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?

Your calculation of $r$ is correct.

To find $\theta$, first note that the point $(4, -4)$ is in the fourth quadrant. So you would need to find the focus angle, i.e. for the point $(4, 4)$, and then subtract it from $2\pi$.
So

$\theta = 2\pi -\arctan{\frac{4}{4}}$

$= 2\pi -\arctan{1}$

$= 2\pi -\frac{\pi}{4}$

$= \frac{7\pi}{4}$.
• Dec 10th 2009, 07:58 PM
mmattson07
I see. So how would that differ for r<0?
• Dec 10th 2009, 08:53 PM
Prove It
Can you ever possibly have a negative radius?
• Dec 10th 2009, 09:12 PM
mmattson07
Perhaps not. I just added pi to account for -r.
• Dec 11th 2009, 07:01 AM
Prove It
$r$ represents the distance between a point and the origin.

A distance, in other words, a length, can never be negative.

I don't see why you would add $\pi$.
• Dec 11th 2009, 07:37 AM
HallsofIvy
Quote:

Originally Posted by mmattson07
Hey I am not sure how to find the second coordinate of these polar coordinates.

a) Find polar coordinates (r, θ) of the point (4, -4), where r > 0 and 0 ≤ θ ≤ 2π.

So $r= \sqrt{(4)^2+(4)^2}$
$= \sqrt{32}$
$=4\sqrt{2}$

I tried doing
θ= $tan^{-1}(\frac{y}{x})$
θ= $tan^{-1}(\frac{-4}{4})$
θ= $tan^{-1}(-1)$
θ= $\frac{-\pi}{4}$

But this is incorrect?

That's incorrect only because you specifically said " $0\le \theta\le 2\pi$" and $-\pi/4$ is less than 0. Since a complete circle is $2\pi$, adding $2\pi$, $-\pi/4+ 2\pi= -\pi/4+ 8\pi/4= 7\pi/4$ gives you exactly the same point and is now in the correct interval: $7\pi/4$ is larger than 0 and less than $2\pi$.

Quote:

For r < 0 and 0 ≤ θ ≤ 2π.

$r=-4\sqrt{2}$

How do I find theta?


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## Miss Maple

A recent product description at Inhabitat. Read about: Miss Maple – the pendant lamp by Elisa Strozyk and the tile-tale of cozy Sherlock Home.

### 2 Responses to “Miss Maple”

1. Silvia Says:

It’s so funny how you try – as a what is apparently called a mathphysicist – to dip into that posh New Yorkian “I see only A”- design community. They may see people like you as a disport.

Silvia wrote:

It’s so funny how you try – as a what is apparently called a mathphysicist – to dip into that posh New Yorkian “I see only A”- design community. They may see people like you as a disport.

I consider Jill Fehrenbacher to be the Pamela Landy of the New Yorkian “I see only A”- design community.

comments in german, french and russian will be translated into english.
you can use LaTeX in your math comments, by using the $shortcode: [latex] E = m c^2$


text:
## Elementary Linear Algebra 7th Edition

Ans: $x=-\frac{1}{3}, y=-\frac{2}{3}$
Step 1: Simplify second equation $\frac{1}{5}x+\frac{2}{5}y=-\frac{1}{3}$ $3x+6y=-5$ $6y=-5-3x$ Step 2: Put $6y$ to one side on first equation $3y=9x+1$ $6y=18x+2$ Step 3: The other side of $6y$ are equal, so make an equation and solve $x$ $18x+2=-5-3x$ $21x=-7$ $x=-\frac{1}{3}$ Step 4: Substitute $x$ to one of the equations $9\times(-\frac{1}{3})-3y=-1$ $-3y=2$ $y=-\frac{2}{3}$


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My Math Forum Not exactly calculus but it is a sort of minimization problem
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 February 5th, 2019, 10:38 AM #1 Senior Member     Joined: Sep 2015 From: USA Posts: 2,430 Thanks: 1315 Not exactly calculus but it is a sort of minimization problem $\text{given }a,b,c >0 \text{ show that}$ $\dfrac{a^3}{b^2(5a+2b)}+\dfrac{b^3}{c^2(5b+2c)}+ \dfrac{c^3}{a^2(5c+2a)} \geq \dfrac 3 7$ The equality occurs when $a=b=c$ but how to show this is a minima? Thanks from topsquark and idontknow
 February 5th, 2019, 02:17 PM #2 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 By simple math , it is seen that the fractions have the same minimal value . One of them is $\displaystyle F(a,b)=\frac{a^3 }{b^2 (5a+2b)}$ . (by calculus find the values ) To make it easier substitute for $\displaystyle x=\frac{a}{b}\; \;$ then $\displaystyle F(a,b)=\frac{x^3 }{5x+2 } \; \; , x>0$ But I think this is not a method , maybe someone else can assist. Last edited by idontknow; February 5th, 2019 at 02:39 PM.
 February 5th, 2019, 05:12 PM #3 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 140 Google something called the Purkiss Principle. Thanks from idontknow
 February 7th, 2019, 10:33 AM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 Then how to prove the equality ?
February 7th, 2019, 11:47 AM   #5
Senior Member

Joined: Sep 2015
From: USA

Posts: 2,430
Thanks: 1315

Quote:
 Originally Posted by idontknow Then how to prove the equality ?
equality is trivial, set a=b=c

 February 7th, 2019, 04:49 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,633 Thanks: 2080 If $x = a/b$ and $y = b/c$, the expression is a function of $x$ and $y$ for which the minimum (for positive $x$ and $y$) can be found to occur when $x = y = 1$ with the help of calculus (partial differentiation). This should work, but might be a bit tedious to do.
 February 7th, 2019, 04:54 PM #7 Senior Member   Joined: Sep 2016 From: USA Posts: 609 Thanks: 378 Math Focus: Dynamical systems, analytic function theory, numerics One way to rigorously show that it is a minimizer is to check whether the Hessian matrix is positive definite. While this is guaranteed to succeed, its not very enlightening and is almost certainly not the "best" solution. Problems with this sort of symmetry are almost always some clever trick using Cauchy-Schwartz or AM-GM inequality. The change of variables proposed by Idontknow appled to all 3 fractions is probably a step in the right direction.
 February 8th, 2019, 03:39 AM #8 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 No need to know the values of a,b,c or x,y,z . $\displaystyle F=\frac{x^3}{5x+2} + \frac{y^3 }{5y+2 } +\frac{z^3}{5z+2 }$ . And 3/7 can be the minimal , if not then it is lower then minimal , so it means finding the minimal value proves the equality . Since a constraint is given , it takes time to use calculus . So the minimal occurs at the same value for the three fractions , x=y=z . Now there are many ways to get $\displaystyle a=b=c$ from $\displaystyle a/b=b/c=c/a \;$ for a,b,c > 0 . Last edited by idontknow; February 8th, 2019 at 03:45 AM.
 February 8th, 2019, 04:28 AM #9 Global Moderator   Joined: Dec 2006 Posts: 20,633 Thanks: 2080 If you don't obtain any specific values or use calculus, how would you know whether you've found a minimum, as distinct from a maximum? Thanks from idontknow
 February 8th, 2019, 07:14 AM #10 Senior Member   Joined: Dec 2015 From: somewhere Posts: 513 Thanks: 80 The minimum occurs in a point $\displaystyle t=x=y=z$ . a=bt ; b=ct ; c=at ; and by the sum $\displaystyle a+b+c =t(a+b+c) \; \;$ or $\displaystyle t=1$ .

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# Using Stroke's Theorem

1. Jun 1, 2014

### Sai-

1. The problem statement, all variables and given/known data
Use Stokes' Theorem to evaluate $\int\int curl \vec{F}\bullet d\vec{S}$ where $\vec{F}(x,y,z) = <e^{z^{2}},4z-y,8xsin(y)>$ and S is the portion of the paraboloid
$z = 4-x^{2}-y^{2}$ above the xy plane.

2. Relevant equations
Stokes Thm:$\int\int curl \vec{F}\bullet d\vec{S} = \int \vec{F}\bullet d\vec{r}$

$\vec{F}(x,y,z) = <e^{z^{2}},4z-y,8xsin(y)>$

S: $z = 4-x^{2}-y^{2}$ above the z = 0.

3. The attempt at a solution
C: $\vec{r}(t) = <2cos(t), 2sin(t), 0>$ where $0\leq t\leq2\pi$
$\vec{r}'(t) = <-2sin(t), 2cos(t), 0>$

$\vec{F}(\vec{r}(t)) = <e^{0}, (4(0) - 2sin(t), 8(2(cos(t))sin(2cos(t))>$
$\vec{F}(\vec{r}(t)) = <1, -2sin(t), 16cos(t)sin(2cos(t))>$

$\int <1, -2sin(t), 16cos(t)sin(2cos(t))> \bullet <-2sin(t), 2cos(t), 0> dt$ from 0 to 2pi
$=\int -2sin(t)-2sin(t)2cos(t) dt$ from 0 to 2pi
$=0$

Last edited: Jun 1, 2014
2. Jun 1, 2014

### LCKurtz

Some latex pointers. If you will put a backslash in front of any function in tex it prints in a much nicer font. Also you can use \cdot instead of \bullet and \nabla\times for curl. I have done that in my quote so you can see the difference. Also you can put the limits on the integral as I illustrate in one line above.

A problem like this should give some orientation for the surface, although with an answer of zero it doesn't matter much. Your work looks correct unless I have overlooked something.

3. Jun 1, 2014

### Sai-

Thank you, I will remember those hints and tips for next time!

And thanks for looking at my work, I think it is correct too; I just can't afford to miss any points on any problem, its dead week and I need all the points I can get.


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# Is it necessary that $f: G_1 \to G_2$ is an isomorphism, for $f:H\to H$ an automorphism with $G_1, G_2\le H$ of the same cardinality?

This might be a very trivial question, and I have explained what I think about it below. Let's say I have an automorphism $$f : H \to H$$. Now, let's say I take two subgroups $$G_1$$ and $$G_2$$ of $$H$$. Is it necessary that $$f: G_1 \to G_2$$ is also an isomorphism?

I think yes, cause all the elements of $$G_1$$ and $$G_2$$ are still the elements of H, and if $$f$$ is an automorphism, $$f: G_1 \to G_2$$ should necessarily be an isomorphism too.

Edit: Originally, I forgot to say that the two subgroups $$G_1$$ and $$G_2$$ have same cardinality.

If you're assuming that $$\text{Im}(f|_{G_1}) \subseteq G_2$$ (otherwise the question doesn't really make any sense), then the answer is yes if $$G_1$$ is finite and not necessarily otherwise.

If $$G_1$$ is finite, then since $$f$$ is monomorphism, $$f : G_1 \to G_2$$ is also a monomorphism and a monomorphism between two finite groups of the same cardinality is an isomorphism.

If $$G_1$$ can be infinite, we can take $$H = \bigoplus_{n \in \mathbb{N}} \mathbb{Z}, \ G_1 = \{0\} \ \oplus \ \bigoplus_{n \geq 1}\mathbb{Z}, \ G_2=H$$ and $$f = 1_H$$. Then $$f : G_1 \to G_2$$ is not surjective but $$G_1$$ and $$G_2$$ have the same cardinality.

• It's worth noting that this is consistent with my answer. – Shaun Apr 22 at 21:02
• I just found a note online similar to what I wanted to understand. In their example, the map is $f(h) = mhm^{-1}$ for some $m \in H$ and their $G_2$ was $mG_1m^{-1}$. So, I think the assumption that $\text{Im}(f|_{G_1}) \subseteq G_2$ is satisfied in that case. – Ufomammut Apr 22 at 21:03

No.

Let $$G_1=\Bbb Z_6$$, $$G_2=D_{6}$$, and $$H$$ be some suitably large $$S_n$$, $$f=1_H$$.

NB: Here $$D_6$$ is the dihedral group of six elements.

• Oh no, I forgot to say in my original question that the two subgroups have same cardinality. – Ufomammut Apr 22 at 20:44
• I've edited the answer, @Ufomammut. – Shaun Apr 22 at 20:47
• What is your automorphism here? – Mark Apr 22 at 20:47
• Because $D_6$ (with $6$ elements) is isomorphic to $S_3$ and, for sufficiently large $n$, there is a cyclic subgroup generated by an element of order six and $S_3$ is isomorphic to a subgroup, @Ufomammut. – Shaun Apr 22 at 20:52
• They are isomorphic to subgroups of $S_n$ for some large $n$, this follows from Cayley's theorem. Though here elements of $\mathbb{Z_6}$ are just not mapped to elements of $D_6$. I think there is a problem in the statement of the question. If you look at $f:G_1\to f(G_1)$ then it is an isomorphism. If you take any two random subgroups then not sure we can even talk about $f:G_1\to G_2$. – Mark Apr 22 at 20:52


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Find fourth order differential equation

Homework Statement

Suppose that a fourth order differential equation has a solution y=-9e^(3x)xcos(x).

a. Find such differential equation, assuming that it is homogeneous and has constant coefficients.

b. Find the general solution to this differential equation. x is the independent variable.

The Attempt at a Solution

I tried to find the derivatives of y (y',y'',y''',y'''').
Then, I placed unknown constants (c1, c2, c3, c4, c5) on each order and made a matrix with each row having coefficients from values containing e^(3x)cosx, e^(3x)xcosx, e^(3x)sinx, and e^(3x)xsinx. I made the columns all have the same order (such as all values from y'' on one column). I also made c1=1 since otherwise the matrix becomes 4*5 and that cannot be solved for the unknown constants. However, this did not work out.

I would sincerely appreciate your help. I mean, I can do those functions that have y=e^rx and y=cos(x), but combination of these makes my head ache.

Related Calculus and Beyond Homework Help News on Phys.org
SteamKing
Staff Emeritus
Homework Helper
Have you tried writing cos (x) in terms of complex exponentials?

LCKurtz
Homework Helper
Gold Member
Think about what the solution whose DE had this characteristic equation$$(r-a)^2+b^2=0$$would be.

HallsofIvy
Homework Helper

Homework Statement

Suppose that a fourth order differential equation has a solution y=-9e^(3x)xcos(x).

a. Find such differential equation, assuming that it is homogeneous and has constant coefficients.

b. Find the general solution to this differential equation. x is the independent variable.

The Attempt at a Solution

I tried to find the derivatives of y (y',y'',y''',y'''').
Then, I placed unknown constants (c1, c2, c3, c4, c5) on each order and made a matrix with each row having coefficients from values containing e^(3x)cosx, e^(3x)xcosx, e^(3x)sinx, and e^(3x)xsinx.
Great! Or course, in order to have $e^{3x}cos(x)$ and $e^{3x}sin(x)$ as solutions, the characteristic equation must have roots 3+ i and 3- i and so must have (x- 3- i) and (x- 3+ i) as factors. In order to have each of those times x as solutions, those must be double roots so the characteristic equation must be of the form $(x- 3- i)^2(x- 3+ i)^2= [(x- 3- i)(x- 3+ i)]^2=$$[(x- 3)^2+ 1]^2= (x^2- 6x+ 10)^2= x^4- 12x^3+ 56x^2- 12x+ 100$.

You can get the differential equation itself from that.

I made the columns all have the same order (such as all values from y'' on one column). I also made c1=1 since otherwise the matrix becomes 4*5 and that cannot be solved for the unknown constants. However, this did not work out.

I would sincerely appreciate your help. I mean, I can do those functions that have y=e^rx and y=cos(x), but combination of these makes my head ache.
"columns"? "matrix"? Too advanced for me!

Thank you all for your replies. All of your replies really helped. The first two replies led me to Euler's formula, while the third one by HallsofIvy led the rest of the problem from the point of getting the values of r using Euler's formula. Note: the -12x should be -120x for the answer, just a point out at a typo. :)


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## Heat Equation

Let $\sigma$ denote a solution to the heat equation on [0,1] under the Dirichlet boundary conditions of $f$(0)= $a$ and $f$(1)= $b$ and IVP of $\sigma$( $x$,0)= $f$( $x$), and let $u$ denote the corresponding steady state solution. Show that the lim as t approaches infinity of $\sigma$( $x$, $t$)= $u$( $x$).


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# All Questions

157 views

### Cryptanalysis of an certain algorithm

Is an attacker of an cryptographic algorithm allowed to change order of sub-keys but not changing their number in attack? Would it be called an attack on broken algorithm of cipher?


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# Seeing uniform continuity from continuity

Let us introduce the definitions with which I am working.

Definition 1 (Continuity) Let $$f:A\to\mathbb{R}$$. Then $$f$$ is continuous on $$A$$ if for all $$y\in A$$ we have that for all $$\epsilon>0$$ there exists $$\delta>0$$ such that for all $$x\in A$$ with $$|x-y|<\delta$$, we have $$|f(x)-f(y)|<\epsilon$$.

Definition 2 (Uniform Continuity) Let $$f:A\to\mathbb{R}$$. Then $$f$$ is uniformly continuous on $$A$$ if for all $$\epsilon>0$$ there exists $$\delta>0$$ such that for all $$x,y\in A$$ with $$|x-y|<\delta$$ we have $$|f(x)-f(y)|<\epsilon$$.

I notice that these definitions are very similar. However I must clarify I know that continuity does not imply uniform continuity.

My question is this: suppose I have in front of me some proof that some $$f$$ is continuous. Uniform continuity cannot allow the choice of $$\delta$$ to be dependent on the points $$x,y$$. I wondered whether if I could see that in the proof that $$f$$ is continuous, that the choice of $$\delta$$ has no dependence on $$x,y$$, that in fact I have uniform continuity as well.

In other words, if I have $$\delta$$ dependent only on $$\epsilon$$ in some continuity proof, do I automatically have that $$f$$ is also uniformly continuous, or does a separate proof need to be written?

• I'd say, as a rule, that any assertion that is not the original statement of the theorem you are proving should need a separate proof. – Gae. S. Nov 16 '19 at 23:47
• If you’re absolutely positive (and only then) that your $\delta$ depends only on $\epsilon$ and not in $y$ (including some possibly non-explicit ways for convoluted $A$ or $f$), you can just say (and it is rigorous): “we see in the above proof [of continuity] that the choice of $\delta$ depended only on the $\epsilon$, thus the previous argument actually gives uniform continuity.” I would suggest, for homework or similar, that you write a distinct proof at least once, though. – Mindlack Nov 16 '19 at 23:49
• @Gae.S. I understand what you mean but my question was, more precisely, if $\delta$ depends only on $\epsilon$, and $f$ is continuous, is $f$ automatically uniformly continuous? – Benjamin Nov 16 '19 at 23:49
• @Mindlack This is what I thought. In almost any circumstance I imagine I would prove these things distinctly anyway, however I wanted to know essentially, whether when you have $\delta$ not dependent on $y$, whether you are allowed to move the position of the "$\forall y$" in the definition of continuity. – Benjamin Nov 16 '19 at 23:53

If in your proof the value of $$\delta$$ doesn't depend on your choice of $$x$$, but uniquely on $$\varepsilon$$, then you could have fixed $$x$$ after fixing $$\varepsilon$$ and determining $$\delta$$, therefore proving absolute continuity of the function, hence you can basically use the same proof, except for fixing $$x$$ and $$y$$ after determining $$\delta$$.
No, you don't need another proof: uniform continuity is just continuity independent from the point, as you stated above. Written in first order logic symbols (exists, for all etc.) the difference is in fact the position of quantifier $$\forall y$$ and nothing more.
• So when $\delta$ does not need to depend on $y$, we are able to move it after the $\delta$? – Benjamin Nov 16 '19 at 23:51
• Continuity: $\forall\varepsilon>0\forall y\in A\exists\delta\dots$ UC: $\forall\varepsilon>0\exists\delta\forall y\in A\dots$ and therefore you move it – Lorenzo Cecchi Nov 16 '19 at 23:54


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# Building-Blocks

The building block logic is primarily accessed interactively via the Building-Block pane. This logic allows one to easily create models interactively. With each mouse click and drag in the pane, building-block commands are emitted and recorded for future reference, allowing these interactive operations to be redone/undone (see the program undo command for further details). In this section the commands and FISH functions allowing access to the building-block data are presented.


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# Solving a first-order differential equation using Laplace Transform

everyone. I am trying to solve the following differential equation using the Laplace Transform:

$$L\frac{di}{dt} + Ri = E$$

I have managed to reduce it to the following equation:

$$I(s) = \frac{E}{sL(s+R/L)}$$

But I'm having a problem with it from there. Can somebody help me solve this please? Any help would be much appreciated.

$$i(t)=Ae^{-\frac{R}{L}t}+\frac{E}{R}$$
being $A$ an arbitrary constant. Now, take a look at your Laplace transform
$$I(s)=\frac{E}{sL\left(s+\frac{R}{L}\right)}$$
and you will recognize two poles for $s=0$ and $s=-\frac{R}{L}$. The first pole will give rise to the constant term while the other one is just the exponential contribution. Indeed, here you can find a table of well-known transformed functions. This is essential in circuit analysis. Your case is the "exponential approach".


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# Linear Algebra: intersection of subspaces

1. Mar 7, 2012

### TheTangent

1. The problem statement, all variables and given/known data
I'm working on a problem that involves looking at the dimension of the intersection of two subspaces of a vector space.

2. Relevant equations
$M \subset V$
$N \subset V$
dim(M $\cap$ N)
$[\vec{v}]_{B_M}$ is the coordinate representation of a vector v with respect to the basis for M

3. The attempt at a solution
I reformulated M $\cap$ N in a bunch of different ways that would be too long to copy down here, but I finally came to this (which may or may not be useful to me in my larger problem but I'm wondering if it is valid itself):

$\vec{v}$ is itself, so it must have the same dimension in both M and N, and since the bases are ordered, for each $\vec{b}_{Mi}$ in $B_M$ for which the corresponding scalar is not zero in the linear combination of elements of $B_M$ equal to $\vec{v}$, and each $\vec{b}_{Nj}$ in $B_N$ for which the corresponding scalar is not zero in the linear combination of elements of $B_N$ equal to $\vec{v}$, if i=j then $\vec{b}_{Mi}$ and $\vec{b}_{Nj}$ are dependent

and $[\vec{v}]_{B_M}$ has zeros in the same places as $[\vec{v}]_{B_N}$

but there is a major problem here with the fact that we may have dimM ≠ dimN

2. Mar 7, 2012

### sunjin09

first imaging a basis for the intersection {b1,b2,...,bp}, then expand it to a basis of M, {b1,b2,...,bp,m1,m2,...,mq},
also expand it to a basis for N, {b1,...,bp,n1,n2,...,nr}, dimension of the intersection is p, dim(M)=p+q, dim(N)=p+r.

3. Mar 8, 2012

beautiful