Mathematics!!

[Artinum]

Wouldn't philosophy be further to the left, being applied sociology, the study of the ethics and morals that sociology/theology have developed?

 

Depends. Philosophy covers a lot of areas. Epistemology would be way over to the right, as would most thoughts on the nature of reality. Stuff about how we think and feel diverged from philosophy a while back and became psychology. Go back far enough, and pretty much all scientific disciplines were once part of philosophy.

[Sam Ashen]

I don't know about imaginary, but that translates to about 9.8 horsepower.  

 

Um.  Why did you convert that to horsepower?  Just.  Why?

 

Nope.  Not a motor I was talking about.  Not a bad guess.

[Oramac]

Um.  Why did you convert that to horsepower?  Just.  Why?

 

Ask Tesla.  (I actually really hate electric vehicles, but it's applicable here.  Though to be fair, it's usually just a conversion of torque into horsepower rather than converting watts to horsepower.  But either way, it's a fun thought experiment.)

[Sam Ashen]

Ask Tesla.  (I actually really hate electric vehicles, but it's applicable here.  Though to be fair, it's usually just a conversion of torque into horsepower rather than converting watts to horsepower.  But either way, it's a fun thought experiment.)

 

Off topic, but since you just said you hate electric vehicles, I think they will be the eventual replacement.  However.  It will never take off in this country until they become so obviously superior that NASCAR is pointless.

[Oramac]

Off topic, but since you just said you hate electric vehicles, I think they will be the eventual replacement.  However.  It will never take off in this country until they become so obviously superior that NASCAR is pointless.

 

Oh I'm sure they'll eventually replace everything.  That doesn't stop me from hating them though.  For a myriad of reasons that would just derail the hell out of this thread.  

 

I was really just using that as an example application (if somewhat archaic) for converting wattage to horsepower.  

[insanity]

Hmm.  I am looking at....

2615 Volts Line to Ground

32.31 Amps

7300 Watts

(per phase)

 

What's wrong with imaginary numbers?

My first introduction to imaginary numbers was the teacher standing in front of the class and the first thing she said was "These are imaginary numbers, they don't exist" I spent the rest of the year mindfucked hard, I like things I can grasp, whether it's concepts, or solid objects, that was a math class that traumatized me when it came to solving problems.

 

After my EE degree, I can manipulate them and what not. ... still don't like them.

 

Also... google gave me a Lathe... :-) But I doubt that's right...

[Gainsdalf the Whey]

Speaking of horse power, it's not actually based on horses relly. Watt did his measurements with ponies and ballparked that a horse was 50% more powerful than a pony. I had been taught it was donkeys in college, but wiki says ponies.

 

I also found this fun list of silly measurement units during my fact checking: https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement

[AugustaAdaByron]

The foot should be on that list too.

[insanity]

Speaking of horse power, it's not actually based on horses relly. Watt did his measurements with ponies and ballparked that a horse was 50% more powerful than a pony. I had been taught it was donkeys in college, but wiki says ponies.

 

I also found this fun list of silly measurement units during my fact checking: https://en.wikipedia.org/wiki/List_of_humorous_units_of_measurement

Beard-Second.... *nods approvingly*

[Artinum]

My first introduction to imaginary numbers was the teacher standing in front of the class and the first thing she said was "These are imaginary numbers, they don't exist" I spent the rest of the year mindfucked hard, I like things I can grasp, whether it's concepts, or solid objects, that was a math class that traumatized me when it came to solving problems.

 

After my EE degree, I can manipulate them and what not. ... still don't like them.

 

They exist alright. "Imaginary" is a misnomer. They just exist at right angles to "real" numbers. I first learned of them in terms of complex numbers, which have a real and imaginary component, and they instantly slotted into my head as "coordinates with algebra".

[insanity]

They exist alright. "Imaginary" is a misnomer. They just exist at right angles to "real" numbers. I first learned of them in terms of complex numbers, which have a real and imaginary component, and they instantly slotted into my head as "coordinates with algebra".

Now I know they exist, but at the time it was just the letter "i" defined as the Square root of -1.... and we went on with the assumption that "i" is "just there". Complex numbers came later, but by then I was done and was on my way to simply passing the class, with a D... I don't think I hit imaginary numbers again until a good 12 years later in college the second time, this time though they gave me something I could grasp in the form of electricity.

[Sam Ashen]

They exist alright. "Imaginary" is a misnomer. They just exist at right angles to "real" numbers. I first learned of them in terms of complex numbers, which have a real and imaginary component, and they instantly slotted into my head as "coordinates with algebra".

 

This sounds about right.  Except maybe it is an "imaginary" right angle instead of a "real" right angle.  The description is good.

 

 

 

Now I know they exist, but at the time it was just the letter "i" defined as the Square root of -1.... and we went on with the assumption that "i" is "just there".

 

I usually just say they exist by definition.  More often I use "j" instead of "i."  Right now, I'm not sure why.  Or I use them interchangeably and not really think about it.

 

 

 

I don't think I hit imaginary numbers again until a good 12 years later in college the second time, this time though they gave me something I could grasp in the form of electricity.

 

Pretty much this.  So over here, we use complex numbers to generically describe Impedance.

 

Speaking of complex numbers - Is there anything in Excel to support complex numbers or would I have to do it by some brute force method?

[AugustaAdaByron]

They exist alright. "Imaginary" is a misnomer. They just exist at right angles to "real" numbers. I first learned of them in terms of complex numbers, which have a real and imaginary component, and they instantly slotted into my head as "coordinates with algebra".

 

That's how they exist in my head too. As points of a 2-dimensional space. In contrast to the real numbers which exist as points of a 1-dimensional space.

[goodbye_farewell]

Speaking of complex numbers - Is there anything in Excel to support complex numbers or would I have to do it by some brute force method?

Excel has "IMAGINARY" and "COMPLEX" functions. Also IMPRODUCT, IMSUM, IMSUB, well you get the idea. Yes Excel can handle complex numbers. Though I typically use MATLAB...

[AugustaAdaByron]

If I may shift the focus away from imaginary and complex numbers for a while... that's a fun trivia that I was thinking about in the bus today. I was discussing it with a friend in summer who was asking me to explain it to him.

 

Say that you have a set of n different keys without any knowledge as to what each one of the key unlocks.

The probability to find the correct key for your keyhole on your k-th attempt is independent of k and is equal to 1/n.

[goodbye_farewell]

If I may shift the focus away from imaginary and complex numbers for a while... that's a fun trivia that I was thinking about in the bus today. I was discussing it with a friend in summer who was asking me to explain it to him.

 

Say that you have a set of n different keys without any knowledge as to what each one of the key unlocks.

The probability to find the correct key for your keyhole on your k-th attempt is independent of k and is equal to 1/n.

Why would it not be 1/(n-k)?  Are you replacing the incorrect key back into the mix??

[AugustaAdaByron]

No, the fact that you find it on the k-th attempt implies that you didn't find it on the previous k-1 attempts so you have to take into account the probability to not find it on all the previous attempts.

 

I'll write the answer in a spoiler. I hope it makes sense. I can't write mathematics without a math editor.

 

So in your first attempt you were unlucky and chose one of the n-1 bad keys. Your probability for that is (n-1)/n. Then you discard the key.

In the second attempt you were unlucky again and chose one of the n-2 bad keys. Your probability for that is now (n-2)/(n-1).

 

The probability that you'll be unlucky on the i-th attempt is the n-i / n-i+1 

So when you reach the (k-1)-th attempt the probability you were unlucky is (n-(k-1))/(n-(k-2)).

You are lucky on your last attempt and the probability for that is 1/(n-(k-1)).

 

Notice that for every i>1 the denominator of the fraction of the i-th attempt is the numerator of the fraction of the (i-1)-th attempt.

 

The events are independent so the probability to find it on the k-th attempt is the product of their probabilities which is

(n-1 / n)(n-2 / n-1)(n-3 / n-2)...(n-(k-1) / n-(k-2)) (1 / n-(k-1)).

 

And everything cancels out apart from 1 and n.

[Sam Ashen]

Say that you have a set of n different keys without any knowledge as to what each one of the key unlocks.

The probability to find the correct key for your keyhole on your k-th attempt is independent of k and is equal to 1/n.

 

The probability is 0, assuming none of the keys fit the keyhole.  

[Oramac]

Say that you have a set of n different keys without any knowledge as to what each one of the key unlocks.

The probability to find the correct key for your keyhole on your k-th attempt is independent of k and is equal to 1/n.

 

For some reason this makes me think of drop rates in games like Warcraft.  People somehow think a 1% drop rate means if they kill a boss 100 times they'll get whatever item they're looking for, not realizing that each attempt is a single 1% chance.  

[AugustaAdaByron]

The probability is 0, assuming none of the keys fit the keyhole.  

 

OK, fair point. You do know that in the set of n keys there is exactly one that unlocks the keyhole.

[Sam Ashen]

Why would it not be 1/(n-k)?  Are you replacing the incorrect key back into the mix??

 

 

No, the fact that you find it on the k-th attempt implies that you didn't find it on the previous k-1 attempts so you have to take into account the probability to not find it on all the previous attempts.

 

I'll write the answer in a spoiler. I hope it makes sense. I can't write mathematics without a math editor.

 

So in your first attempt you were unlucky and chose one of the n-1 bad keys. Your probability for that is (n-1)/n. Then you discard the key.

In the second attempt you were unlucky again and chose one of the n-2 bad keys. Your probability for that is now (n-2)/(n-1).

 

The probability that you'll be unlucky on the i-th attempt is the n-i / n-i+1 

So when you reach the (k-1)-th attempt the probability you were unlucky is (n-(k-1))/(n-(k-2)).

You are lucky on your last attempt and the probability for that is 1/(n-(k-1)).

 

Notice that for every i>1 the denominator of the fraction of the i-th attempt is the numerator of the fraction of the (i-1)-th attempt.

 

The events are independent so the probability to find it on the k-th attempt is the product of their probabilities which is

(n-1 / n)(n-2 / n-1)(n-3 / n-2)...(n-(k-1) / n-(k-2)) (1 / n-(k-1)).

 

And everything cancels out apart from 1 and n.

 

Maybe I misunderstood, but the answer in the spoiler does not seem to match.  Let's say I picked a card from a 52-card deck and I asked you to guess what it was.  (We are also assuming you are not a magician and already know the answer.)  Just for simplicity, let's say I just ask you to guess the suit.  So 4 suits.

 

Probability your first guess is correct:  1 in 4.

Probability your second guess is correct:  1 in 4.

Probability your third guess is correct:  1 in 4.

Probability your fourth guess is correct:  1 in 4.

 

The key is in the wording:

 

The probability to find the correct key for your keyhole on your k-th attempt is independent of k and is equal to 1/n.

 

She did not say "by."  She said "on."  Very clever.  

---------------------

 

For some reason, this makes me think of some fun I was having with one of my high school math teachers.  He claimed to be able to guess the number a number I can think of between 1 and 10 in 4 guesses, if I respond if the number is higher or lower.  I already knew how this worked and decided to have fun with this one, and thought of a number between 1 and 10.

 

Too long ago, so I will just assume he went fully binary and started with:

8 - Lower

4 - Lower

2 - Higher

3 - Higher

 

"Then the number you are thinking of is not between 1 and 10."

 

"Yes it is.  You did not specify an integer number."

 

EDIT: So obviously - What number was I thinking of?

 

...........which will bring up the next question...........

[AugustaAdaByron]

Was the number π? (or am I just hungry and thinking of it?)

 

I like the binary search trick.

 

You ask someone to guess an integer between 1 and x and tell them that you can find the number in log(x) guesses as long as when you guess they say higher or lower if you don't find the exact number.

 

Start with guessing x/2. If they say higher you only need to search in the interval [x/2+1,x]. If they say lower you search in the interval [1,x/2-1]. Each time you half your search space. (That's where the log comes from.)

 

I like asking people to think of a number between 1 and 128 and tell them I'll find it in 7 guesses. It seems more impressive.

[Sam Ashen]

Was the number π? (or am I just hungry and thinking of it?)

 

Very good!  And that brings up the next question.  I always use 3.1415927 for Ï€.  So I guess that means when I was in high school, we had 8-digit calculators.  What does everybody else use?

 

I like the binary search trick.

 

You ask someone to guess an integer between 1 and x and tell them that you can find the number in log(x) guesses as long as when you guess they say higher or lower if you don't find the exact number.

 

log(x)/log(2)

 

OR ln(x)/ln(2)

 

Start with guessing x/2. If they say higher you only need to search in the interval [x/2+1,x]. If they say lower you search in the interval [1,x/2-1]. Each time you half your search space. (That's where the log comes from.)

 

I like asking people to think of a number between 1 and 128 and tell them I'll find it in 7 guesses. It seems more impressive. 

 

 

Yes, that works, too.  I just decided to choose the highest power of 2 that is within the range.  Integer number.  Of course.

[caeliter]

Maybe I misunderstood, but the answer in the spoiler does not seem to match.  Let's say I picked a card from a 52-card deck and I asked you to guess what it was.  (We are also assuming you are not a magician and already know the answer.)  Just for simplicity, let's say I just ask you to guess the suit.  So 4 suits.

 

Probability your first guess is correct:  1 in 4.

Probability your second guess is correct:  1 in 4.

Probability your third guess is correct:  1 in 4.

Probability your fourth guess is correct:  1 in 4.

 

That is not correct unless you're replacing the card.

 

Probability of the first guess being correct is 1/4

Probability of second guess is either 12/51 or 13/51

Probability of third guess is 11/50 or 12/50 or 13/50

Probability of 4th guess 10/49, or 11/49 or 12/49 or 13/49

 

If the guess is either fixed or determined randomly, then the odds can be calculated:

Prob of 1st being correct: 1/4

Prob of 2nd being correct: 1/4*12/51 + 3/4*13/51 = 51/204 = 17/68 = 1/4

Prob of 3rd being correct: 12/51*1/4*11/50 + 39/51*1/2*12/50 + 12/51*3/4*13/50 + 39/51*1/2*13/50 = (132+ 936 + 168 + 1014)/10200 = 2250/10200 = 225/1020 = 45/204 = 15/68

(I'm definitely too lazy to calculate the 4th guess)

 

The change doesn't really become noticeable until several more cards are selected.

 

For some reason this makes me think of drop rates in games like Warcraft.  People somehow think a 1% drop rate means if they kill a boss 100 times they'll get whatever item they're looking for, not realizing that each attempt is a single 1% chance.  

 

Well it's true that if you've killed the boss 99 times, the odds of getting it on the 100th attempt are no better than they were on the first.

 

If you were to dedicate yourself to repeating the process 100 times (only stopping when you've hit 100 or gotten the item) then you've got about a 64% chance of getting it before you stop. 300 attempts translates into a 95% chance of getting the desired item (meaning if you can do it every day for a year you're looking at about a 97% chance of getting the item)

[caeliter]

Since I didn't want to double post...

 

Here's another math thing.

 

One is more likely to win the grand prize for the lottery twice in a row buying only a single ticket (Megamillions) than he is to rebuild a solved rubix cube blindfolded from the disassembled pieces. However, the odds of his rebuilt cube being solvable are 1/12

 

*Edit because I missed a word*