Hi all! Wanna help a TH-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.
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!
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!!
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 : )
"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.
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...
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!
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!
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!!
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!
@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 :-)
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.
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.
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!
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.
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
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!!
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.
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.
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!!!
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.
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.
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!😭
been struggling through series so bad -_-... this video made me realize how simple it actually is... i swear lectures and textbooks complicate things....
Ratio test is usually when you have factorials and exponential stuff. Yes all this stuff is raised to the n^2, but if you threw in n+1 for all the n's, simplifying it would be a nightmare. Whenever you see an entire series raised to n, it's best to use the root test because you will eliminate the n or reduce it's power like in the example this dude did.
good way to remeber the tests (R.eally R.eally A.mazing D.octor In C.anada) D: Divergence Test A: Alternating Series Test R: Ratio Test R: Root Test I: Integral Test C: Comparison Test
dude i haven't even reached at this level of math yet , and im just sitting here and watching it....idky but who knows maybe it will help me out in the future
Are you talking about the part: n/(3^1/n *3^3)? If you look at 3^1/n as n goes to infinity, it becomes 3^(1/infinity). That equals 3^0, remember this equals 1. Your denominator just becomes 1*27
@patrickJMT i mentioned the same thing on one of your other videos also, i think the reason why people think it is boring in a classroom is because computer are just fun, or some kind of drug that emits out of our computers, lol
due you mean why he raised it to the (1/n)? and if thats the question its because the equation has you find the limit of [an] to the nth root which is the same as raising the problem to (1/n). an example is the the square root of 4 which can be written as 4^(1/2). hope this helps :)
first i would like to say you have nice picture, second is so if you change n in terms of X then divide all numerator and denominator by x then you know the rules from there in top u will get x/x which is one down you divide 3/x which is zero time 3^1/x is zero so u will get 1/0 !
@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
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.
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.*
It was not a mistake...it is 3^(1/n) as n->infinity so it would be 3^0 which is 1. It doesnt matter anyway...it was just to make it clear. Thanks listening to me :)
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Hi all! Wanna help a TH-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! TH-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!
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.
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!
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
just started studying this in calc II and your video was very helpful. sound quality was good and you made this topic less frightening!
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 : )
"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.
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...
So good man. Still helps people out after 11 years
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!
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!
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!
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!!
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 so you know your the man. . .helped me through my calc II class with relative ease. Best vids on the net for sure.
my pleasure my friend! hope the exam turns out well
@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 :-)
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
Your last example really helped me as I wondered if we would have to use the root test twice.
Geez you make it much easier to understand than my professor...great stuff!!
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.
glad to help it make sense for you!
Another few videos to once again get ready for yet another Calc 2 test. Thx Patrick you tht man.
good luck!
omg, thanks a ton patrick!! i have a maths exam next week, luv ur videos so much
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!
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.
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
dude ur a lifesaver. Thank you for all this. You make so much easier
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!!
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.
@patrickJMT no theres a place called jordan located in the middle east. You are making such a huge a positive feedback
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...
i need to start paying you for all the nice ratings and comments : )
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.
YOU ROCK. I have a calc test this upcoming monday
ops, thanks!
i did that in another video too : )
i am making sure everyone is paying attention!
but only the 1/n term goes to zero so that we are left with (3^0)(3^3) = 1(27) = 27
@5ANASHEEROA do you mean michael jordan? was he in the middle east?
Wow! incredible... especially the very last part of the root test!
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!!!
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!
I've been having trouble with Series! It'll certainly help my final today!!
i think you need to flip your fraction (n+1)/n
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.
I love seeing all of the steps. Thanks!
I like how at 6:12 you referred to the numerator twice just to test if folks are paying attention. ;)
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.
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 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! :)
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!😭
Very helpful. Much better than my cal II teacher who assumes we know everything.
Thank you so much for all your videos . Your videos have truely helped me.
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!
been struggling through series so bad -_-... this video made me realize how simple it actually is... i swear lectures and textbooks complicate things....
...and btw... THANK YOU. Your instruction is fantastic. I'm so grateful to have found these videos.
@remirap no, i am long done with school.
@Raxarax i was being a bit ' tongue in cheek ' with that comment
you sound like the hippie teacher on beavis and butthead. On a serious note your videos are very helpful. Thank you for posting!
Ratio test is usually when you have factorials and exponential stuff. Yes all this stuff is raised to the n^2, but if you threw in n+1 for all the n's, simplifying it would be a nightmare. Whenever you see an entire series raised to n, it's best to use the root test because you will eliminate the n or reduce it's power like in the example this dude did.
good way to remeber the tests (R.eally R.eally A.mazing D.octor In C.anada)
D: Divergence Test
A: Alternating Series Test
R: Ratio Test
R: Root Test
I: Integral Test
C: Comparison Test
thanks again my friend!!
It's the exponent for 3^(1/n) that equals 0. As in you will get 3^0 which is equal to 1.
Thank you so much for making this easy to understand! Fantastic.
dude i haven't even reached at this level of math yet , and im just sitting here and watching it....idky but who knows maybe it will help me out in the future
Are you talking about the part: n/(3^1/n *3^3)? If you look at 3^1/n as n goes to infinity, it becomes 3^(1/infinity). That equals 3^0, remember this equals 1. Your denominator just becomes 1*27
no problem
God bless you Patrick
You rock, man. Better teacher than my Calculus professor. And I go to a major university too! =O
woah, i never thought of that. thanks bud, this is gonna help with my exam i got tomorrow!
@patrickJMT i mentioned the same thing on one of your other videos also, i think the reason why people think it is boring in a classroom is because computer are just fun, or some kind of drug that emits out of our computers, lol
it really helps me,thanks Patrick
Thank you helped a lot. Especially the last example!
Thanks, no-one does analysis on youtube as well as you do!
man youu are a LIFE SAVER !!! I wish u were my TEACHER !!!
due you mean why he raised it to the (1/n)? and if thats the question its because the equation has you find the limit of [an] to the nth root which is the same as raising the problem to (1/n). an example is the the square root of 4 which can be written as 4^(1/2). hope this helps :)
@patrickJMT No not at all. Your vids just drive me crazy to learn maths whole day long!
@BlueColourPencils awww, thanks! go texas!
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!
6:09 was pretty funny. :) thanks so much for the help though man. You have been a lifesaver.
first i would like to say you have nice picture, second is so if you change n in terms of X then divide all numerator and denominator by x then you know the rules from there in top u will get x/x which is one down you divide 3/x which is zero time 3^1/x is zero so u will get 1/0 !
@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
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.
Normally you can use both ratio and root for these questions. it best to use root test for questions like (n)^n
Thanks! I love how detailed you are. Keep it up!
You are a gentleman and scholar.
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.*
how about: THANKS PATRICK and youtube... ? : )
Awesome video dude, this helped me a ton thank you
It was not a mistake...it is 3^(1/n) as n->infinity so it would be 3^0 which is 1.
It doesnt matter anyway...it was just to make it clear.
Thanks listening to me :)
You just saved my life man thank you so much
Thank you for your videos! Everything is so clear.
thank you patrick. it really helps me a lot
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.
patrick you rock man
@BornAtTheBar yep, got a masters