my professor refuses to give me the references for what we're studying, this video made me explain what i missed in the 4 lectures i was sick and couldnt attend, thanks Dave!
You scared me on question four of the comprehension test, you had me thinking I was supposed to tetrate XD! Thanks for these videos Professor Dave, these actually really help and give some intuition on these topics.
Thanks a lot. Just one point, from what I've been studying, I've learned that the AST only tells us whether an alternating series is convergent or not, and it doesn't tell us anything about its divergence.
I honestly can't believe I've purchased a $150 textbook for contents that could be boiled down in 2 days worth of lecture videos. I'm not saying textbook is useless. But textbook has its limit as it cannot show the reader animations of how some things are done. Textbook is really only needed for practice problems.
thats why its called conditionally convergent, because its convergency hinges on the fact that the terms ascilate from positive to negative so the sum does in fact get smaller and smaller
Because any finite number of terms don't affect convergence, all that is required is that the terms are *eventually* decreasing, i.e. b_(n+1) = k for some k. This is something that happens a lot of the time, and it was glossed over in the video. You can check the terms are eventually decreasing by showing the derivative of x^2/(x^3 + 4) is eventually negative.
10:51 but if we just apply limit n tends to infinity to (1+1/n)^n we get = (1 + 1/infinity)^infinity = (1+0)^infinity = 1 so test must be inconclusive rather than divergent right? Please correct me if I am wrong.
This is why it's terrible when teaching limits to students just learning, to do things like "plugging in infinity", as around the 5:00 mark. This kind of sloppy calculation is okay if you're experienced with limits and indeterminate forms, but it often leads students to wrong conclusions as you've done here. To answer your question, 1^(infinity) is an indeterminate form, and we cannot conclude any value for the limit. (1 + 1/n)^n does converge to e.
Thx a lot fot that video! About problems at the end of the video: Σ[(-2)^n/3^n] is it relevant just to rewrite it as Σ[(-2/3)^n] and say that: as r=-2/3 and its less then 1, then the serires is converges?
@@Nemoguzapomnit Ok, that makes sense. Greek speakers dislike the fact that English speakers shift the vowel when referring to π, so that it sounds like pie. But we have to do that, otherwise both p and π would sound identical, like the word "pea". I found out that they say their own π to sound like like pea, and the Latin p like "pay", as it is most commonly named in languages that use Latin letters. I asked you, because I thought a similar issue might persist in your language, except with three different versions of the letter with very similar names.
so as i understand, the alternative series and all the series associated with the calculus, we only need to find if they are convergent or divergent. but why do we have to do so? what is the relevance of convergence and divergence? maybe it could sound silly but this is my first interaction with the calculus so please pardon me for that.
Kinda funny how you explain concepts better in 12 minutes than multiple hours of lecture from my professor.
Actually crazy
@@banaanpropaan9344there are proofs on here that are explained in 10 minutes and better than 2 hour explanations
just to let u know ..u explain much better than my uni professor ....u r so good .........i understood all of it in 12 mins.......
Simply the BEST proffesor,you make difficult things simple
or maybe others make simple things difficult
I’m writing math and physics exams this semester and I can say it’s all self-taught thanks to you😂much love
my professor refuses to give me the references for what we're studying, this video made me explain what i missed in the 4 lectures i was sick and couldnt attend, thanks Dave!
You're the best in the business, thank you
You scared me on question four of the comprehension test, you had me thinking I was supposed to tetrate XD! Thanks for these videos Professor Dave, these actually really help and give some intuition on these topics.
You're actually a savior. Holy moly.
Thanks a lot. Just one point, from what I've been studying, I've learned that the AST only tells us whether an alternating series is convergent or not, and it doesn't tell us anything about its divergence.
I Am from India🇮🇳 I totally understand the topic
Hope you're doing well Professor ☺️
Thank you sir for your dedication! 🙏
I love you professor Dave!
Im now studying mathematics lol, i never thought i would do this.
Same, lol! I literally have no need or reason to but it's something that I feel compelled to do.
@@demonnogo You study math just because you feel like so? Woah this stuff you are studying is hard. Knowing that people study math for fun is scary.
@@ian.ambrose Im studying for fun too, im 16 lol 💀
I honestly can't believe I've purchased a $150 textbook for contents that could be boiled down in 2 days worth of lecture videos. I'm not saying textbook is useless. But textbook has its limit as it cannot show the reader animations of how some things are done.
Textbook is really only needed for practice problems.
Very interesting - always enjoy your videos - thank you
Great lectures, really enjoyed them. thanks
At the example at 4:38 the number 1/5 = 0.2 and 4/12 = 0.3 which is larger. The terms don't get smaller. How can this series converge?
thats why its called conditionally convergent, because its convergency hinges on the fact that the terms ascilate from positive to negative so the sum does in fact get smaller and smaller
Because any finite number of terms don't affect convergence, all that is required is that the terms are *eventually* decreasing, i.e. b_(n+1) = k for some k. This is something that happens a lot of the time, and it was glossed over in the video. You can check the terms are eventually decreasing by showing the derivative of x^2/(x^3 + 4) is eventually negative.
Thank you
Thank you Professor.
Thank you professor
the minute 4:40 it is not 16/70 it is 16/68
Please add all the details about limits
that's in another video earlier in the calculus playlist
@@ProfessorDaveExplains sir please but what is 2+2 and how to prove it sir please sir
10:51 but if we just apply
limit n tends to infinity to (1+1/n)^n we get
= (1 + 1/infinity)^infinity
= (1+0)^infinity
= 1
so test must be inconclusive rather than divergent right?
Please correct me if I am wrong.
One word:
E
This is why it's terrible when teaching limits to students just learning, to do things like "plugging in infinity", as around the 5:00 mark. This kind of sloppy calculation is okay if you're experienced with limits and indeterminate forms, but it often leads students to wrong conclusions as you've done here. To answer your question, 1^(infinity) is an indeterminate form, and we cannot conclude any value for the limit. (1 + 1/n)^n does converge to e.
Same answer I get
I think it's a typo at 2:25 it's "
Thx a lot fot that video! About problems at the end of the video: Σ[(-2)^n/3^n] is it relevant just to rewrite it as Σ[(-2/3)^n] and say that: as r=-2/3 and its less then 1, then the serires is converges?
r=-2/3 and |r|=2/3
How do you pronounce the difference among "p", "п", and "π", when discussing mathematics in your language?
@@carultch we don't use russian letters in math. But anyway "п" sounds like "pæ". And about "p" and "π" - in case with "p" we would say "p variable"
@@Nemoguzapomnit Ok, that makes sense.
Greek speakers dislike the fact that English speakers shift the vowel when referring to π, so that it sounds like pie. But we have to do that, otherwise both p and π would sound identical, like the word "pea". I found out that they say their own π to sound like like pea, and the Latin p like "pay", as it is most commonly named in languages that use Latin letters.
I asked you, because I thought a similar issue might persist in your language, except with three different versions of the letter with very similar names.
@@carultch it could be but we mostly never use "p" variables
so as i understand, the alternative series and all the series associated with the calculus, we only need to find if they are convergent or divergent. but why do we have to do so? what is the relevance of convergence and divergence? maybe it could sound silly but this is my first interaction with the calculus so please pardon me for that.
good question. I looked it up and the main answer i found was basically "to help you with even higher level math." whelp.
7:53 guys, why is the series on the right divergent? I heard about the p-series
Ya it should be harmonic series
💚
how does the sum of (-1)^(n-1)/n converge if the sum of 1/n diverges?
why does it in the ratio test approach e
What is the use of series?
the simplest example is drug dosage calcs
best