Hi all! Wanna help a UA-cam education OG? Please post comments, questions and anything else on your mind in the comment section! so, don’t forget to LIKE, THUMBS UP, and SUBSCRIBE! I’d appreciate it greatly as it helps me :)
glad you like them all : ) i figured that if they suck, i would just delete them. most people seem to like them though : ) and honestly, i forget half of this stuff, so i make them for 'future me' - the one that has forgotten and re-watches old me (newer me?), so i have to be nice and soothing to my future self : )
well for one thing, He doesn't go over proofs, which are extremely abstracted. He just tells you how to do the concrete part of math: solving. The simplicity of the way he teaches plus the fact that the videos are short and convinient makes it easy to learn here. The people coming here are also in the mindset of learning too, otherwise it would be unlikely that they'd come here at all!
Dude, I was literally tweaking out about this midterm that I have to take in like 40 minutes and your videos on convergence tests totally saved my ass. Thank you, Patrick.
6:11 "The numerator is going to get big very quickly, the numerator is going to get big as well, but not as fast as the numerator." Idk why I found that so funny, probably too much studying...
You're literally my savior. I have an exam today and my professor was crap at explaining everything, so I didn't understand any of the series and sequences. I watched at least 8 of your videos and they helped a lot! Thank you so much, Patrick!!
well, it is basically knowing that the n/(n+1) is the flip of (n+1)/n; that raised to a power of n has something to do with 'e'! i am seen that limit so many times now, that i just recognize it. so to answer your question, what made me think to do that is: experience! sometimes one has to be shown what to do a few times before it sticks!
"most people seem to like them though" No No Patrick. Everybody LOVES them. You are a great man. Thousands of students around the world are more grateful than you know. Thank you for the time and effort you put into these videos. They have saved the lives of many students.
I would be so lost without you! I can honestly say I learned more over the course of watching your videos in one day than I learned the past 3 weeks in my Cal 2 class!
these videos are very much helpful i have been watching sequence and series topic for almost 8 hrs with small intervals and practicing problems in between ....i hope i will do well in exams thank you so much ....people like you should always be there ....!!
1 to any real number power is 1. the problem is if you have a function of the form [f(x)]^[g(x)] and f(x) is APPROACHING 1 (but not necessarily equal to 1). limit notation is not very good; the standard notation is to use equality, when in reality it is just 'getting close' to that number (in most cases). eventually, one figures this out and just learns to deal with it : )
series are really cool. it is one of my favorites subjects (if not my favorite subject) that one encounters in calculus. some very beautiful and useful results come from them. of course, they are also pretty tricky at times!
Patrick I love you! Your videos have helped me so much throughout calc 1 and now calc 2! Thank you for making these videos, so many of us appreciate it!!
God bless you dude! You have no idea how much these videos have helped me for my cal 2 final. the limit of my appreciation for your work approaches infinity. Great work!
@patrickJMT i like it when you go into the conceptual depths but at the same time keeping maths enjoyable, i feel really grateful to you., you just inspire me.....i like "improper integrals" in your video series very much :-)
thank you so much for being there i have been watching you videos for the whole day today about this topic..........they are very much easy to understand..thank you again!!
For the 3rd question he did, lim n->infinity (1+(1/n))^n is in fact equal to e. Set y=(1+(1/n))^n, then take the natural log of both sides. You will have ln y =n*ln(1+(1/n)) Then take the limit on both sides. lim n->infinity lny = lim n-> infinity (n * ln(1+(1/n)) Now, the goal is to rewrite it in an indeterminate form in which the L'Hopital Rule can be applied easily. lim n-> infinity [ln(1+(1/n)] / [1/n] which = 0/0. Take the derivative on the numerator and denominator and you will be left with lim n-> infinity (1/ (1+(1/n))) and that limit will equal to 1. Now we have lim n->infinity ln y = 1 lim n->infinity e^(lny)=e^1 and then replace y with y=(1+(1/n))^n after you cancel out e and the natural log. Therefore lim n->infinity (1+(1/n))^n = e lim n->infinity (1-(1/n))^n is equal to 1/e.
You did an excellent job! I only wish that I had discovered this while I was learning about this instead of after I had already finished the course. However, I never really did understand it during the course, so I was lost during the assignment and exams, so this was like me learning it for the first time. However, my calculus teacher seemed to avoid doing examples or at least examples using numbers like you used. Again, good job! I know others will benefit from it like I did.
Yeah, but stating e as the answer is more exact than saying 2.71 is the answer. For example, if for a problem you find the answer to be sqrt 2, it is much more accurate and precise to let the answer remain sqrt 2, even if you have been taught to simplify. This is because saying sqrt 2 = 1.41 is actually incorrect, even though sqrt 2 is approximately 1.41 if one is using three significant figures. So, you could indeed say the answer is approximately 1.41 (like if you are doing a physics problem where estimation>accuracy), and you would be technically correct, but it is much better to say sqrt 2. There is just no point in simplifying, it only takes more time to produce a crappier answer. So, like +poopnuggets (lol) said: to answer e is to be exact, to name some number of digits of e is the answer is less exact. It is unnecessary to simplify, much like it is unnecessary to rationalize-- just more time and work for a less correct answer. I do not know how understanding an answer can be e is equivalent to "making it," but realizing that a concept like e is more exact than its estimated number is indeed helpful in math. Because, actually, e is not a number but a constant that can be expressed precisely as the base of the natural log (ln), such that ln(e)=1 (e^1=e), and can be estimated as the infinite series of 1/n!, which does not have an exact value since infinity is not a number either, just another concept, but does give us the estimation often used (2.71).
You can also do a lot of cool stuff with e that you cannot do with anything else (like its role in probability theory), and it has unique properties (such as the derivative of e^x is e^x, and the same applies for riemann integrals). ALSO, all digits of the constant e are not known (it is irrational), though the known number of digits is constantly increasing
you can show the the limit as n->infinity of (1 + 1/n)^n = e by using l'hopitals rule. you dont use lhopital at 840 since it is not an indeterminate form; it is 1/e.
Great video! I have a test on Tuesday and your videos have been so helpful. I'd just like to let you know, though, that this video's not on your website. Thanks again!
isn't 1^(n) as the limit goes to infinity and indeterminate form, therefore applying L'H rule on 8:40? It is to my understanding that 1^infinity can either be 1, infinity, or undefined, but how do we decide on the result ? Should i always assume 1^infinity is one?? My head hurts..
Great videos, they are really helping my preps for the exams! By the way, at 5.45 you say that 0*27=27, not that it matters, the answer will still be the same.
I spent hours trying to understand my professor's instructions... yet I watched 3 of your videos, and I'm good to go. I've been barely passing Calc II, and not because I'm "dumb"... it's because I've spent an insane amount of time trying to understand HIS teaching. I should have sought other sources, and I would've done so much better in the class. I'm studying to be a teacher, and this really makes me see how there is NO "dumb" student... different students learn differently.
I thought series were going to be hard, but this whole concept so far, including doing differential eqtns with power series, and the Taylor polynomial section, has been easier than integration techniques IMO. First time in a while that I didn't actually need Patrick's videos to get it down correctly. Still watched them though, they always have something inventive to teach you.
at 5:47 why does that equal inf/27? since 0*27 = 0 then wouldn't it be inf/0(another thing I don't understand)? I guess I just don't know what limit law makes that valid.
OMG I agree with Inferfire........it is plain scary that I understand all your root test/ratio test stuff, I have my final tomorrow and before these videos I thought I was screwed because i have absolutely NO idea what my prof is babbling about....I wanna seriously thank you so much!!!
at 4:42 when you separated the denominator 3^(1+3n) into 3 * 3^n, i didn't understand that, because if you were to put it back together wouldn't it be 9^n?
Just wanted to ask about example #2 at 5:25 in the video, you said it was infinity over 27, but wouldn't it be infinity over 0? If it was infinity over zero would you do L' Hopital's Rule or how would you finish it?
Bruh. If I pass my Calculus 2 exam. I’m so sending some cash your way and making a donation. You definitely deserve it! I’ve been using your videos since pre Cal. I don’t know where I’d be if it wasn’t for your videos. Thank you so much!😭
6:08 [The second time he says "numerator," he means "denominator," but I still find this bit very amusing] "If you start plugging in large values for 'n,' the *numerator* is gonna get big very quickly; the *numerator* is gonna get big as well but not as fast as the *numerator.*
I understand that if you have a function raised to the nth power, it is a good idea to use the root test. However, what about the other tests: integral test, sequence test, comparison test, and ratio test? What should you look for in a function to determine which of the other tests you should use?
@patrickJMT Thanks for all your videos!! Its much easier for me to study math when I can pause and rewind the lecture :) But I have a question, say I have a sum of (x/n)^n what do I do with x? at the end I got: lim (x/n) do I say that it converges for x
been struggling through series so bad -_-... this video made me realize how simple it actually is... i swear lectures and textbooks complicate things....
For the last example on this video it is easier to use L'Hopitals on that limit by taking the ln of the limit instead of having to memorize that that limit equals e. You can show why it equals e.
I think the second example has a small error. The denominator has the (3^3)(3^1/n). You say that this becomes 27 as n goes to infinity. I'm pretty sure that it becomes (9)(1) not (9)(3) since 3^0 is 1 not 3. Doesn't matter because the numerator is infinity, but still a comment is warranted.
It's correct because: 1=A/A for any A so: 1=(1/n)/(1/n) He just multiplied it by 1 so it doesn't change the result. Example: 2/3 = (2/3) * [(1/2) / (1/2)]= (2*(1/2)) / (3*(1/2)) = (1/(3/2)) Ignore the example if it's confusing. He multiplied by 1. That's just it.
@js6781535 No, it looks like it will at first glance but the number that is being raised to the n'th power is approaching 1 as n goes to infinity. ie - (1+Very Small Number)^(Very Large Number). As he points out, this is the definition of the number e or approximately 2.718 ...or 2.7 1828 1828 45 90 45 (broken up into the memorization pattern to the 16th significant figure, like anyone would ever do that)... Still absolutely convergent either way, but 1/e was correct.
Hi all! Wanna help a UA-cam education OG? Please post comments, questions and anything else on your mind in the comment section! so, don’t forget to LIKE, THUMBS UP, and SUBSCRIBE! I’d appreciate it greatly as it helps me :)
Almost 13 years later.... still amazing. Thank you so much for this. It's taught timelessly.
yes
Damn man i was 3 years old when he uploaded this video, and now look at me watching this video after all these years! UA-cam is great
glad you like them all : )
i figured that if they suck, i would just delete them. most people seem to like them though : )
and honestly, i forget half of this stuff, so i make them for 'future me' - the one that has forgotten and re-watches old me (newer me?), so i have to be nice and soothing to my future self : )
Still depressed that I've learned more from an hour of your videos than a month of class.....
well for one thing, He doesn't go over proofs, which are extremely abstracted. He just tells you how to do the concrete part of math: solving. The simplicity of the way he teaches plus the fact that the videos are short and convinient makes it easy to learn here. The people coming here are also in the mindset of learning too, otherwise it would be unlikely that they'd come here at all!
@5ANASHEEROA do you mean michael jordan? was he in the middle east?
Dude, I was literally tweaking out about this midterm that I have to take in like 40 minutes and your videos on convergence tests totally saved my ass. Thank you, Patrick.
6:11 "The numerator is going to get big very quickly, the numerator is going to get big as well, but not as fast as the numerator." Idk why I found that so funny, probably too much studying...
You're literally my savior. I have an exam today and my professor was crap at explaining everything, so I didn't understand any of the series and sequences. I watched at least 8 of your videos and they helped a lot! Thank you so much, Patrick!!
"Tired of scrounging youtube for math help?"
No, I've found this channel. Why would anyone pay for online tutoring videos if they've found patrickJMT?
7 years down the line and us bloaks still get that stupid ad lol
well, it is basically knowing that the n/(n+1) is the flip of (n+1)/n; that raised to a power of n has something to do with 'e'! i am seen that limit so many times now, that i just recognize it. so to answer your question, what made me think to do that is: experience!
sometimes one has to be shown what to do a few times before it sticks!
"most people seem to like them though" No No Patrick. Everybody LOVES them. You are a great man. Thousands of students around the world are more grateful than you know. Thank you for the time and effort you put into these videos. They have saved the lives of many students.
I would be so lost without you! I can honestly say I learned more over the course of watching your videos in one day than I learned the past 3 weeks in my Cal 2 class!
these videos are very much helpful i have been watching sequence and series topic for almost 8 hrs with small intervals and practicing problems in between ....i hope i will do well in exams
thank you so much ....people like you should always be there ....!!
1 to any real number power is 1.
the problem is if you have a function of the form [f(x)]^[g(x)] and f(x) is APPROACHING 1 (but not necessarily equal to 1).
limit notation is not very good; the standard notation is to use equality, when in reality it is just 'getting close' to that number (in most cases). eventually, one figures this out and just learns to deal with it : )
series are really cool. it is one of my favorites subjects (if not my favorite subject) that one encounters in calculus. some very beautiful and useful results come from them.
of course, they are also pretty tricky at times!
just started studying this in calc II and your video was very helpful. sound quality was good and you made this topic less frightening!
@remirap no, i am long done with school.
I love the videos - they're so clear and really add to what I try to learn in class (we just go too quickly!)
Thanks for doing this!
Patrick I love you! Your videos have helped me so much throughout calc 1 and now calc 2! Thank you for making these videos, so many of us appreciate it!!
God bless you dude! You have no idea how much these videos have helped me for my cal 2 final. the limit of my appreciation for your work approaches infinity. Great work!
So good man. Still helps people out after 11 years
just so you know your the man. . .helped me through my calc II class with relative ease. Best vids on the net for sure.
@patrickJMT i like it when you go into the conceptual depths but at the same time keeping maths enjoyable, i feel really grateful to you., you just inspire me.....i like "improper integrals" in your video series very much :-)
thank you so much for being there i have been watching you videos for the whole day today about this topic..........they are very much easy to understand..thank you again!!
For the 3rd question he did,
lim n->infinity (1+(1/n))^n is in fact equal to e.
Set y=(1+(1/n))^n, then take the natural log of both sides. You will have
ln y =n*ln(1+(1/n)) Then take the limit on both sides.
lim n->infinity lny = lim n-> infinity (n * ln(1+(1/n))
Now, the goal is to rewrite it in an indeterminate form in which the L'Hopital Rule can be applied easily.
lim n-> infinity [ln(1+(1/n)] / [1/n] which = 0/0.
Take the derivative on the numerator and denominator and you will be left with lim n-> infinity (1/ (1+(1/n))) and that limit will equal to 1.
Now we have lim n->infinity ln y = 1
lim n->infinity e^(lny)=e^1 and then replace y with y=(1+(1/n))^n after you cancel out e and the natural log. Therefore lim n->infinity (1+(1/n))^n = e
lim n->infinity (1-(1/n))^n is equal to 1/e.
thank you
Geez you make it much easier to understand than my professor...great stuff!!
You did an excellent job! I only wish that I had discovered this while I was learning about this instead of after I had already finished the course.
However, I never really did understand it during the course, so I was lost during the assignment and exams, so this was like me learning it for the first time. However, my calculus teacher seemed to avoid doing examples or at least examples using numbers like you used.
Again, good job! I know others will benefit from it like I did.
you know you made it when you say "The answer is the number e"
e is a number though
Yeah, but stating e as the answer is more exact than saying 2.71 is the answer. For example, if for a problem you find the answer to be sqrt 2, it is much more accurate and precise to let the answer remain sqrt 2, even if you have been taught to simplify. This is because saying sqrt 2 = 1.41 is actually incorrect, even though sqrt 2 is approximately 1.41 if one is using three significant figures. So, you could indeed say the answer is approximately 1.41 (like if you are doing a physics problem where estimation>accuracy), and you would be technically correct, but it is much better to say sqrt 2. There is just no point in simplifying, it only takes more time to produce a crappier answer.
So, like +poopnuggets (lol) said: to answer e is to be exact, to name some number of digits of e is the answer is less exact. It is unnecessary to simplify, much like it is unnecessary to rationalize-- just more time and work for a less correct answer. I do not know how understanding an answer can be e is equivalent to "making it," but realizing that a concept like e is more exact than its estimated number is indeed helpful in math. Because, actually, e is not a number but a constant that can be expressed precisely as the base of the natural log (ln), such that ln(e)=1 (e^1=e), and can be estimated as the infinite series of 1/n!, which does not have an exact value since infinity is not a number either, just another concept, but does give us the estimation often used (2.71).
You can also do a lot of cool stuff with e that you cannot do with anything else (like its role in probability theory), and it has unique properties (such as the derivative of e^x is e^x, and the same applies for riemann integrals). ALSO, all digits of the constant e are not known (it is irrational), though the known number of digits is constantly increasing
(1 + 1/n)^n = e ? i dont get that part and also, why dont we apply lopital rule at 8:40
?
you can show the the limit as n->infinity of (1 + 1/n)^n = e by using l'hopitals rule. you dont use lhopital at 840 since it is not an indeterminate form; it is 1/e.
patrickJMT thank you patrick, you're the best, I did take my quiz today and it was fantastic. Thank you so much
my pleasure! glad to hear the quiz went well!
omg, thanks a ton patrick!! i have a maths exam next week, luv ur videos so much
Another few videos to once again get ready for yet another Calc 2 test. Thx Patrick you tht man.
good luck!
Great video! I have a test on Tuesday and your videos have been so helpful.
I'd just like to let you know, though, that this video's not on your website.
Thanks again!
isn't 1^(n) as the limit goes to infinity and indeterminate form, therefore applying L'H rule on 8:40? It is to my understanding that 1^infinity can either be 1, infinity, or undefined, but how do we decide on the result ? Should i always assume 1^infinity is one?? My head hurts..
zChosenUA-camr 1^infinity = e
Mikail Constant Bilyamin I don’t think so lol I think it’s just 1
glad to help it make sense for you!
Great videos, they are really helping my preps for the exams! By the way, at 5.45 you say that 0*27=27, not that it matters, the answer will still be the same.
dude ur a lifesaver. Thank you for all this. You make so much easier
6:12 "the numerator is going to get big quickly, the numerator is going to get big as well, but not as quick at the numerator"
lmao
lol
I'm looking like a dumbass laughing in the middle of the library right now...
but only the 1/n term goes to zero so that we are left with (3^0)(3^3) = 1(27) = 27
i need to start paying you for all the nice ratings and comments : )
my pleasure my friend! hope the exam turns out well
I'm confused at 7:52, why did he move (1/n) onto both the numerator and denominator after distributing it already??
JocoFitness He is doing a clever form of "one"
Rule number one: never and I mean NEVER question Patrick
i love you! thank you so much. ive been watching your videos all day for my final tomorrow. i think ill pass it because of you! :)
ops, thanks!
i did that in another video too : )
i am making sure everyone is paying attention!
I spent hours trying to understand my professor's instructions... yet I watched 3 of your videos, and I'm good to go. I've been barely passing Calc II, and not because I'm "dumb"... it's because I've spent an insane amount of time trying to understand HIS teaching. I should have sought other sources, and I would've done so much better in the class. I'm studying to be a teacher, and this really makes me see how there is NO "dumb" student... different students learn differently.
@patrickJMT no theres a place called jordan located in the middle east. You are making such a huge a positive feedback
Your last example really helped me as I wondered if we would have to use the root test twice.
I know this is an over ten year old video but Patrick I hope you're having a good day
Sean Martin lmfao
Thank you so much for all your videos . Your videos have truely helped me.
...and btw... THANK YOU. Your instruction is fantastic. I'm so grateful to have found these videos.
I thought series were going to be hard, but this whole concept so far, including doing differential eqtns with power series, and the Taylor polynomial section, has been easier than integration techniques IMO. First time in a while that I didn't actually need Patrick's videos to get it down correctly. Still watched them though, they always have something inventive to teach you.
Thank you so much!
I've been having trouble with Series! It'll certainly help my final today!!
Thank you so much for making this easy to understand! Fantastic.
wow. im at UT too.. and the whole time I wondered where patrick was at and it was right here.. same world!! Just show Austin rules!
at 5:47 why does that equal inf/27? since 0*27 = 0 then wouldn't it be inf/0(another thing I don't understand)? I guess I just don't know what limit law makes that valid.
damn dude I was just asking a question.
EDIT: ignore this. some asshat was being an asshat and his comment got deleted.
broadswordslayer lim(n→inf) x^(1/n) should be 1, x for any real number
you can use a calculator to try it.
you sound like the hippie teacher on beavis and butthead. On a serious note your videos are very helpful. Thank you for posting!
i think you need to flip your fraction (n+1)/n
YOU ROCK. I have a calc test this upcoming monday
@BlueColourPencils awww, thanks! go texas!
thanks again my friend!!
OMG I agree with Inferfire........it is plain scary that I understand all your root test/ratio test stuff, I have my final tomorrow and before these videos I thought I was screwed because i have absolutely NO idea what my prof is babbling about....I wanna seriously thank you so much!!!
@armidylano44 well, i used to bore people at a major university as well, so maybe it is all the same
Wow! incredible... especially the very last part of the root test!
@BornAtTheBar yep, got a masters
I love seeing all of the steps. Thanks!
Patrick , could you explain why , with root test in example 3 ( 1+ 1/n) ^n=e . I really forgot what is this e means.
Thank you.
at 4:42 when you separated the denominator 3^(1+3n) into 3 * 3^n, i didn't understand that, because if you were to put it back together wouldn't it be 9^n?
Just wanted to ask about example #2 at 5:25 in the video, you said it was infinity over 27, but wouldn't it be infinity over 0? If it was infinity over zero would you do L' Hopital's Rule or how would you finish it?
The power goes to zero and any number raised to the power of 0 is 1. So the denominator is 1 multiplied by 3^3.
I like how at 6:12 you referred to the numerator twice just to test if folks are paying attention. ;)
6:09 was pretty funny. :) thanks so much for the help though man. You have been a lifesaver.
Bruh. If I pass my Calculus 2 exam. I’m so sending some cash your way and making a donation. You definitely deserve it! I’ve been using your videos since pre Cal. I don’t know where I’d be if it wasn’t for your videos. Thank you so much!😭
You rock, man. Better teacher than my Calculus professor. And I go to a major university too! =O
Very helpful. Much better than my cal II teacher who assumes we know everything.
@patrickJMT
Hey, I noticed you're from Austin. I'm actually at UT right now =)
Anyway, thanks again. King of youtube math, you are.
I love how I got a video saying that videos from the 2000s math are not educational enough when this was a really good video.
Thank you for your videos! Everything is so clear.
Isnt the limit of 3^(1/n) as n-> infinity supposed to be 1???? at 5:32
+TheDiningTable if n is infinity, then 1/n will be 0. and 3^0 is 1.
Thank you helped a lot. Especially the last example!
your videos are brilliant patrick, just brilliant :)
by the way, i wanted to ask if n-th term theorem could be used on your 3rd example here.
man youu are a LIFE SAVER !!! I wish u were my TEACHER !!!
Awesome video dude, this helped me a ton thank you
God bless you Patrick
6:08 [The second time he says "numerator," he means "denominator," but I still find this bit very amusing]
"If you start plugging in large values for 'n,' the *numerator* is gonna get big very quickly; the *numerator* is gonna get big as well but not as fast as the *numerator.*
Those are fantastic questions. I would love to see a video just to have those answered. Is there a possibility of having this done?
How did you get 1/e? Confused on that part, unfortunately.
lim (1+1/n)^n=e
Thanks! I love how detailed you are. Keep it up!
woah, i never thought of that. thanks bud, this is gonna help with my exam i got tomorrow!
no problem
I understand that if you have a function raised to the nth power, it is a good idea to use the root test. However, what about the other tests: integral test, sequence test, comparison test, and ratio test? What should you look for in a function to determine which of the other tests you should use?
@Raxarax i was being a bit ' tongue in cheek ' with that comment
it really helps me,thanks Patrick
@patrickJMT
Thanks for all your videos!! Its much easier for me to study math when I can pause and rewind the lecture :)
But I have a question, say I have a sum of (x/n)^n
what do I do with x?
at the end I got:
lim (x/n)
do I say that it converges for x
been struggling through series so bad -_-... this video made me realize how simple it actually is... i swear lectures and textbooks complicate things....
For the last example on this video it is easier to use L'Hopitals on that limit by taking the ln of the limit instead of having to memorize that that limit equals e. You can show why it equals e.
I think the second example has a small error. The denominator has the (3^3)(3^1/n). You say that this becomes 27 as n goes to infinity. I'm pretty sure that it becomes (9)(1) not (9)(3) since 3^0 is 1 not 3. Doesn't matter because the numerator is infinity, but still a comment is warranted.
thank you patrick. it really helps me a lot
It's correct because:
1=A/A for any A so: 1=(1/n)/(1/n) He just multiplied it by 1 so it doesn't change the result.
Example: 2/3 = (2/3) * [(1/2) / (1/2)]= (2*(1/2)) / (3*(1/2)) = (1/(3/2))
Ignore the example if it's confusing. He multiplied by 1. That's just it.
Awesome video man. I definitely learned a lot.
well, 'e' is used in the continuous compounding formula.
and dont go hatin' on poor ole ' e ', it just wants to be useful like all the other numbers!
@js6781535 No, it looks like it will at first glance but the number that is being raised to the n'th power is approaching 1 as n goes to infinity. ie - (1+Very Small Number)^(Very Large Number). As he points out, this is the definition of the number e or approximately 2.718 ...or 2.7 1828 1828 45 90 45 (broken up into the memorization pattern to the 16th significant figure, like anyone would ever do that)... Still absolutely convergent either way, but 1/e was correct.
You just saved my life man thank you so much