How many 0s does 10! have? Well, it needs to divide by 10^n, or 2^n × 5^n. So, how many 5s are in it? 2. Hence there can be at most 2 0s. There is a 4, or two 2s. Therefore, there are two zeroes in 10!.
31:00 that f(1)+f(2)+f(3)+f(4) sum is actually the upper Riemann sum so it's still valid. Also if you take the average of the upper and lower Riemann sums then you get 26 which has an error of less than one from the actual integral.
I did that exponential integral recently to, not while singing Christmas music though. Andrew is right though you can use Residue Theorem if you want to, it’s more fun if you go that direction with it imo.
That zero thing was so funny! I thought it was 11 at first before I realized about prime factoring. Then I was stupid and kept track of 2s and 5s until I realized at about 25. I got while on one foot though!
24:46 + 28:52 ) I'm a second year ChemE and next semester I have Applied Thermodynamics... I can't for the professor to mention fugacity and then all these memes about it will just pop into my brain lol XD
Get Zach Star to write it then. He’s been putting out comedy gold daily on his second channel. I almost forget he’s an engineer and not just a comedian
I think you need to freshen up on your numerical analysis. f(0)+f(1)+f(2)+f(3) and f(1)+f(2)+f(3)+f(4) are equally acceptable approximations to the integral from zero to four. We can get a better approximation by avaraging the two, which results in 26.
how is question 7 a function though? it has multiple possible outputs for a single input unless the bottom circles are not drawn at all because phi (~1.62) is not in either interval [pi/2, 2pi] or [pi, 5pi/4]
18:00 that's a shame he couldn't do proof by contradiction, because it's such an "unmathy" way yet is easy to understand, and has a troll element. If 0.(9) is less than 1 you can put a number between them in 0.(9) < X < 1, then for whatever X picked you just extend left side with enough 9s to make it bigger.
@@ammyvl1 But you can write it in a way that makes it work, no? We are deriving the statement from a property of the real numbers after all. I am fairly sure it's possible to formalize it. If not, please elaborate why? I am still studying and this turns out to be fairly interesting.
@@tilmanv.5816 This statement is not derived from a property of the real numbers. It's derived from your intuition on decimal expansions. It doesn't follow from some theorem about the real numbers, but rather the fact that "well you can add more 9's making it larger", which really doesn't constitute a valid proof. Plus, it references properties about a poorly defined number in the first place. We don't have a proper definition for 0.999... here, making these statements impossible to validate. As for the matter of formalizing, how do you formalize the statement of "then for whatever X picked you just extend left side with enough 9s to make it bigger"? There's not a very good way to do this, without just restarting from scratch. I will concede, using the sequence {0.9, 0.99, 0.999, ...} is very useful, but only because that's the only proper way to define this number (the series definition basically does the same thing, except with a limit baked in, instead of a sequence). For a proper proof, first they define 0.999... as the supremum of the sequence a_n=0.(9)_n = {0.9, 0.99, 0.999, 0.9999, ...}, (0.999... is the smallest number greater than {0.9, 0.99, 0.999, etc.}) then prove that 1 is the supremum (least upper bound) of this sequence as well. Since the supremum is unique, this implies 0.999...=1 (this is paraphrasing, by the way. The real proof is a bit longer). The reason why this constitutes a valid proof is because the only property it references (aside from trivial facts about inequalities, and trivial algebra of real numbers) is the axiom of completeness (a supremum necessarily exists), which is an axiom for the real numbers. This is rather pedantic, however, and these intuitions are much more fun than the formal mathematics required to prove such a statement.
@@ammyvl1 Okay, first of all, thanks for your detailed answer! I noticed that I should have been more clear about what I meant. I didn't mean to say, that by claiming the mentioned property is true, the proof is already valid. What I meant is that this "property" can be used as some kind of basis to build your proof on. The actual proof I was thinking about worked basically exactly like the one you mentioned. But now it got me thinking. The statement that between any real numbers exists atleast 1 other real number can definitely be proven. I even think a prof of mine did just that. With the statement proven, couldn't we derive the statement from the axioms of an ordered field?
For the area under the curve of x²+1 I got 25.333 repeating using Reimann Sums. It’s been awhile since I’ve done those 😂😂 almost forgot my Sums of Powers lol
The meme with calculating the number of zeros in 100! is much harder to do by hand, I'd say. Phrased as such, it implies that, say, the number of zeros in the number 1337069420 is 2, so you have to account for the interior zeros somehow. Y'all did end up getting it correct that 10^24 is the largest power of 10 that can divide 100!, but there seems to be 6 of those so called interior zeros, making the actual tally 30
21:00 This is the exact method We are thought to convert infinitely repeating decimals into fractions and when i first encountered this one i was thinking that i made a mistake somewhere
for some reason i have calculated sqrt(i) in my head and got the exact answer without doing much math just got 90 degrees and imagined halving the angle on the complex plane, so you get magnitude of 1, but you get 45 degrees which basically makes that triangle we all know, from that point it's just easy pythagoras idk how relevant is it, but ig it's cool to know for more basic complex equations
papi you can go further than η(2). since η(s) = (1 - 2^(1-s))ζ(s), η(2) = ( 1 - 2^(1-2))ζ(2) = ½ζ(2). But due to Euler ζ(2) is π²/6, so our integral is π²/12
The 100 factorial was wrong because other 0's do exist that are not at the rightmost(the amount you guys calculated). I would like to see you make a video solving the problem.
If you think about it, the matrix determinant one at 17:00 is so obvious because it is clearly singular (the rows are just multiples of each other). Funny how that surprised Jens :D
Problem 3 did not specify that the zeros were trailing zeros. There are 24 trailing zeros in 100! but a number can have a nontrailing zero if it is not a multiple of ten.
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“Once a year, Santa comes early. That’s what she said.” Oh boy, this is going to be a good one
:DDD
Mrs. Clause* Unless you're into that
ayo?
@@bobatuzi9799 “come on”
-Google translate
oh man that had me laughing oh boy jolly
"10 is 10, 9 is basically 10, 8 is basically 10, 7.... is basically 10..."
"so yea 10! has 5 zeros"
i died lMAO
Confusing zeros with sig figs is very interesting for sure.
well yes, because it's basically impossible to do, as it's not asking the number of zeros *at the (right) end* of 100!
How many 0s does 10! have?
Well, it needs to divide by 10^n, or 2^n × 5^n. So, how many 5s are in it? 2. Hence there can be at most 2 0s. There is a 4, or two 2s. Therefore, there are two zeroes in 10!.
@@lucas29476As for 100!, there are 24 5's (double-counting multiples of 25=5^2), and 50 multiples of 2 between 1 and 100. Hence, 24 0's.
@@cxpKSipit actually had 30 zeroes(it just does).
Try using your method to calculate the number of zeroes in 51*2 = 102 to see where it goes wrong.
Scholars will be forever indebted to Andrew for his proof that 1=1
:DDD
Everyone would fall in love in maths if all teachers become like you😇
:3
@@PapaFlammy69 UwU
that or they fall in love with the teacher 🥵😍
@@harleyspeedthrust4013 both😘
:D
"boy remember Boltzmann" never fails to make me laugh 😂😂😂
“If the question was something else, I would have been right.” 😂😂😂
Andrew: Am I allowed to swear
Also Andrew: I lost... I lost my... I lost my crap
Jens: *That was so fucking hilarious*
As a chem engineer currently taking physical chem, I can indeed confirm that fugacity is defined by the chemical potential
Nice.
DO you hate it too? :^)
@@PapaFlammy69 absolutely! Should’ve been a physics major... :)
Same I’m cheme as well fugacity is terrible
Isn't chem engineering just the worst😂
Why is Andrew unironically such a good singer???
xD
Thank you papa for allowing Andrew to record 2 videos in one week
31:00 that f(1)+f(2)+f(3)+f(4) sum is actually the upper Riemann sum so it's still valid. Also if you take the average of the upper and lower Riemann sums then you get 26 which has an error of less than one from the actual integral.
I did that exponential integral recently to, not while singing Christmas music though. Andrew is right though you can use Residue Theorem if you want to, it’s more fun if you go that direction with it imo.
How? The Intergrand has infinititely many singularities on the positive imaginary axis.
that randy marsh “I didn’t hear no bell” one fucking killed me
:D
17:05 the trick was to realize rows (or cols) are linearly dependent
Ah you noticed too but earlier than me ;-)
the obligatory Andrew Dotson video is always good
That zero thing was so funny!
I thought it was 11 at first before I realized about prime factoring. Then I was stupid and kept track of 2s and 5s until I realized at about 25. I got while on one foot though!
22:37 Ahh yes. That was so smooth. 😂😂😂
I can feel the Jingle bells going through my vain.
And what the fk was Andrew doing with factorials Lmao
Wait halal newton is there
Why does Jens sound so sad while singing the Christmas songs? You can hear the pain in his voice.
cause Christmas songs are pain lol
cause Christmas is not coming home
Maguire and Andrew ❌
Smort Papa Flammy and Handsome Andrew Dotson ✔️
lesgooo ever since the first YLYM came out, i’ve been waiting for a reboot :D this is gonna be an hour of top tier STEM dankness, i love it !
7:45 “we are both going strong,” I hate to break it to you, but NNN has already passed 🤣🤣🤣🤣
Y’all are free now 😂😂😂
Bro I legit screamed "yaaas" when I saw this video in my feed... It's really fun when u guys colab from each other's basements
:D
I really love these kind of videos, really. U can learn a lot while literally laughing at memes. Never stop making them papa :3
24:46 + 28:52 ) I'm a second year ChemE and next semester I have Applied Thermodynamics... I can't for the professor to mention fugacity and then all these memes about it will just pop into my brain lol XD
These collabs are bangers
:)
Do more collab skits like the “Mathematicians in physics class” I miss them
It's seriously so damn hard to come up with those ^^'
Get Zach Star to write it then. He’s been putting out comedy gold daily on his second channel. I almost forget he’s an engineer and not just a comedian
@@cosmicvoidtree Trueeee honestly the channel zach star himself I quite enjoy
Hello *Papa Flammy Mathy*
*Papa* and *Andrew* ?! ... Great
Its all fun till the laughter starts...then it is "Schluss mit lustig!!!"
:D
I don't know jingle bells either.
Not in english, nor in french.
Not any christmass song either. I feel you.
haha :D
I think you need to freshen up on your numerical analysis. f(0)+f(1)+f(2)+f(3) and f(1)+f(2)+f(3)+f(4) are equally acceptable approximations to the integral from zero to four. We can get a better approximation by avaraging the two, which results in 26.
One is the upper and one the lower sum, yes. Maybe the upper sum would've been the better choice after all =)
Way to help w climate change! Love ~ Andrew’s Mom 💕
:)
Woaaaa hiii
@@judedavis92 Hi! Happy Holidays!
The sequel we all wanted
and needed.
Are you sure it’s “you laugh, you math”? Seems more like “you laugh, you meth” to me 🤣🤣🤣🤣🤣
Dude, your board is soo clean!
27:09 how wonderful is the mind of a physicist
24:23 omg new Christmas carol just dropped 🔥🔥🔥.
xD
This is entertainment. Thanks, guys.
Jens could you do a series on Differential geometry any time soon? Me and many more would appreciate that..
Would Simpson's rule count as not using integrals? Because that gives you an exact answer when the function is a quadratic.
sure lol
Love this video series! I was worried it wasn't going to come back
:))
2:21 ok perfect
...it was anything but perfect
how is question 7 a function though? it has multiple possible outputs for a single input
unless the bottom circles are not drawn at all because phi (~1.62) is not in either interval [pi/2, 2pi] or [pi, 5pi/4]
18:00 that's a shame he couldn't do proof by contradiction, because it's such an "unmathy" way yet is easy to understand, and has a troll element. If 0.(9) is less than 1 you can put a number between them in 0.(9) < X < 1, then for whatever X picked you just extend left side with enough 9s to make it bigger.
That is 100% how I would do it tho. It's easy to understand and easy to prove. Why would anyone make easy things hard for nor reason?
@@tilmanv.5816 because this isn't a proper proof. it just uses an appeal to intuition to work, rather than rigorously proving it
@@ammyvl1 But you can write it in a way that makes it work, no? We are deriving the statement from a property of the real numbers after all. I am fairly sure it's possible to formalize it. If not, please elaborate why? I am still studying and this turns out to be fairly interesting.
@@tilmanv.5816 This statement is not derived from a property of the real numbers. It's derived from your intuition on decimal expansions. It doesn't follow from some theorem about the real numbers, but rather the fact that "well you can add more 9's making it larger", which really doesn't constitute a valid proof. Plus, it references properties about a poorly defined number in the first place. We don't have a proper definition for 0.999... here, making these statements impossible to validate.
As for the matter of formalizing, how do you formalize the statement of "then for whatever X picked you just extend left side with enough 9s to make it bigger"? There's not a very good way to do this, without just restarting from scratch. I will concede, using the sequence {0.9, 0.99, 0.999, ...} is very useful, but only because that's the only proper way to define this number (the series definition basically does the same thing, except with a limit baked in, instead of a sequence).
For a proper proof, first they define 0.999... as the supremum of the sequence a_n=0.(9)_n = {0.9, 0.99, 0.999, 0.9999, ...}, (0.999... is the smallest number greater than {0.9, 0.99, 0.999, etc.}) then prove that 1 is the supremum (least upper bound) of this sequence as well. Since the supremum is unique, this implies 0.999...=1 (this is paraphrasing, by the way. The real proof is a bit longer).
The reason why this constitutes a valid proof is because the only property it references (aside from trivial facts about inequalities, and trivial algebra of real numbers) is the axiom of completeness (a supremum necessarily exists), which is an axiom for the real numbers.
This is rather pedantic, however, and these intuitions are much more fun than the formal mathematics required to prove such a statement.
@@ammyvl1 Okay, first of all, thanks for your detailed answer!
I noticed that I should have been more clear about what I meant. I didn't mean to say, that by claiming the mentioned property is true, the proof is already valid. What I meant is that this "property" can be used as some kind of basis to build your proof on. The actual proof I was thinking about worked basically exactly like the one you mentioned.
But now it got me thinking. The statement that between any real numbers exists atleast 1 other real number can definitely be proven. I even think a prof of mine did just that. With the statement proven, couldn't we derive the statement from the axioms of an ordered field?
Asks if he can swear, says crap, absolute gigachad
For the area under the curve of x²+1 I got 25.333 repeating using Reimann Sums. It’s been awhile since I’ve done those 😂😂 almost forgot my Sums of Powers lol
:D
The meme with calculating the number of zeros in 100! is much harder to do by hand, I'd say. Phrased as such, it implies that, say, the number of zeros in the number 1337069420 is 2, so you have to account for the interior zeros somehow. Y'all did end up getting it correct that 10^24 is the largest power of 10 that can divide 100!, but there seems to be 6 of those so called interior zeros, making the actual tally 30
Damn - Andrew can sing lol
This video is a national treasure
I love this series so much :DD
Numpy is crazy
All you need is the random library
Calculate the number of zeros in 100! ...
Me, an intellectual: Ah, yes, I see exactly two zeros.
21:00
This is the exact method We are thought to convert infinitely repeating decimals into fractions and when i first encountered this one i was thinking that i made a mistake somewhere
7:49 Meanwhile most people can't even hear past 19000hz... [insert skull emoji]
23:03 LMFAO That laugh
Just imagine his 11 year olds finding his channel
:D
@@PapaFlammy69 they will be scarred
Jesus Christ you made this one way harder than the physics one lol.
:DDD
As someone who has always struggled with math, and has to take Calc II next year, watching you prove that 1=.99999… was awesome.
23:05 Papa Flammy the absolute chad!!!!
:D
Definitely best intro of this year...
:D
Hell yea I love these videos lmao
Ahh yes, P=P(small + andrew) make perfect sense now
Question failure: shouldn't it have been how many *trailing* zeros are in 100! rather than total zeros? Otherwise, much more difficult question.
I took Discrete Math last semester.... This is giving PTSD of the countless proofs we had to do!
for some reason i have calculated sqrt(i) in my head and got the exact answer without doing much math
just got 90 degrees and imagined halving the angle on the complex plane, so you get magnitude of 1, but you get 45 degrees which basically makes that triangle we all know, from that point it's just easy pythagoras
idk how relevant is it, but ig it's cool to know for more basic complex equations
18.55 i legit had this on my gcse higher paper
bruh
0:18 Aye, I can do this one. Just a bit of geometric series and done. (Haven't watched the video yet.)
yas :)
That's what she said... before finding another infinity boi.
papi you can go further than η(2). since η(s) = (1 - 2^(1-s))ζ(s), η(2) = ( 1 - 2^(1-2))ζ(2) = ½ζ(2). But due to Euler ζ(2) is π²/6, so our integral is π²/12
yup :) But the answer was good enough for me, since we've already derived its value on the channel hehe :)
Q: 7) was a bit wrong, the interval of the second testicle should be [ π, (5/2)π ]
Ah ye, you're probably right lol ^^'
Man Andrew’s got bars
2:25 isn’t every KFC meal is a dead little chicken?
I lost at the thumbnail
:D
18:30 I just covered a way to do this in calc 2 with series!
The 100 factorial was wrong because other 0's do exist that are not at the rightmost(the amount you guys calculated). I would like to see you make a video solving the problem.
What drawing tablet do you use?
Gaomon PD1560
"Try not to enjoy ur videos" is the hardest challenge
:D
Hey! Regular phi is OG!!!
Wich notetaking software are you using on your tablet?
I want more
I love these videos
This is a special video xD
NICE 👍
i was waiting for this
If you think about it, the matrix determinant one at 17:00 is so obvious because it is clearly singular (the rows are just multiples of each other). Funny how that surprised Jens :D
True! Another way to see it is that column1 + column2 = column3 so they are not linearly independent
Love your chanel
What is the name of the drawing tablet you're using?
Gaomon PD1560 :)
what exactly is this device called that they are using to work out the problems? never seen this before
Mine is the Gaomon PD1560 :)
Amazing video
Certified hood classic
Saw a yawning dog in this video
Today was a good day
:3
At 32:00 why dont you just use f(0.5), f(1.5), f(2.5) and f(3.5) that way it will kind of compensate itself and the answer is pretty close
5:19 but Andrew, I thought you said 3 was no more?
23:50 you should have sung padoru padoru
mcquarrie is a good undergrad quantum text but chemistry focused. I agree waiting to introduce Dirac notation till appendix is a dick move.
23:04 yeah interesting
I understood that i don't jingle bells either
Where is the math-ish Christmas song "A story often told"?
in my videos?
@@PapaFlammy69 in this video, I thought you would sing while evaluating integral from question 8
ohhhhhh, nah, not this time :D
Problem 3 did not specify that the zeros were trailing zeros. There are 24 trailing zeros in 100! but a number can have a nontrailing zero if it is not a multiple of ten.
We chem eng use fugacity for our calculations.
But can you laugh in Math?
Awesome video tho
Oh no - Flamme u got something wrong :(
#31:30
ER hat die Obersumme gemacht, das ist genau so ok wie die Untersumme ...