Imagine this question being on your final exam. You go through all of these steps and calculations just to forget the +C at the end which single-handedly causes you to fail the exam.
This is completely off topic, but I think I came up with an interesting problem. Given the recursive functions: f(n) = af(n-1) + bg(n-1) g(n) = ag(n-1) + bf(n-1) Where a=0.5 and a+b=φ Show that both f and g are fibbonacci sequences. I'll put my solution at the end. Stumbled upon this fact while trying to solve for g for any a or b... idk how others may attempt to solve it but I wrote them as a discrete time convolution. I got this equation for g(n): g(n) = (b^2-a^2)g(n-2)+2ag(n-1) Was trying to solve for f or g to write sin(nt) in terms of sin(t) and cos(t)... but accidentally made a mistake, one of the equations were supposed to add and the other was supposed to subtract the b term.... was doing this in the first place because, I was hoping the equation for sin(nt) would help solve the integral for (sin(Nt)cot(t/2)+cos(Nt))cos(nt) where N and n are two non-negative integers. Which I plugged into a couple integral calculators... and it didn't give me a solution. When you evaluate that integral in respect to t from -π to π, and divide by 2π, its supposed to be an inverse discrete-time fourier transform that should equal 1 for all |n|
I’m a sophomore in high school and have no idea what he’s doing but I still watch a bunch of his videos because they are interesting and relaxing. Keep up the good work👍
i was in your position a couple years ago. i’m now in calc 2, watched this whole video and actually understood all of it, and i’m very proud of myself. keep studying and you’ll get there💪
I've always wanted to be a teacher, but after watching you, it made me feel so dumb that I'm no longer sure I'm capable of it. I hope I can be as intelligent as you are one day.
It is necessary to replace the variable in time to avoid radicals. ∫6dx/(x^4-6)=6^(1/4)∫ dt/(t^4-1). t=x/6^(1/4). Here the fraction under the integral sign is so simple that it can be decomposed into elementary fractions without the method of indefinite coefficients. 1/(t^4-1)=(1/2) *{1/(t^2-1)-1/(t^2+1)}=(1/2)*{(1/2)[1/(t-1)-1/(t+1)]-1/(t^2+1)}= = (1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)
A young man as cool as a Shaolin monk with a Kung Fu Master beard and has a Pokemon ball in his hand explaining math problems... This combination made my yung hollow mind literally explode.
Can you please make a video that introduces hyperbolic functions and some sample problems with them? I’ve never known anything about them and would love to learn more about them!
Just took my Calculus CLEP exam and got the score I needed; 64 ain't great, but i get my credit so it's cool. I've never been confident in Math, and quite frankly I went into the exam expecting to fail- I'm quite happy that wasn't the case. Im very grateful for all the videos you make, and passing this exam was definitely thanks to you lol. Keep it up!
The partial fraction decomposition can be done without a system of equations if you consider that 1 = 1/(2*sqrt(6))*[(x^2+sqrt(6))-(x^2-sqrt(6))]. This improved memory efficiency allowed me do the problem in my head reasonably fast. Also, BPRP was running out of room and the video was already long so he used hyperbolic trig functions. But you can decompose 1/(x^2-sqrt(6)) with the same trick or Heaviside cover up.
STUDYING HARD NOW, HE'S GOTTA RECLAIM ALL OF THE MATH THAT ESCAPED FROM HIS BRAIN INTEGRALS, DERIVATIVES AND PARTIAL FRACTIONS HE'S MASTERED THEM ALL NOW HE'S READY FOR ACTIONNNN MATH-E-MATIC! AUT-O-MATIC! FEELS LIKE MAGIC WHEN YOU DERIVE JUST TO STAY ALIVEEEEE! MATH-E-MATIC! AUT-O-MATIC! MATH-E-MATIC! AUT-O-MATIC! *music fades out as you ride off on a motorcycle*
I’m 41 and just learning calculus. I just sat there with wide eyes and all shocked at how much work it’s needed. Damn. I think my brain almost literally exploded. Lol.
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 how I can solve this Question
here is a challenge: find f(x) such that f'(x) = f(x+1) oh and i forgot something: there is a x such that f(x) = 1 (just multiply the solution by some number to get f(x) = 1 if you don't have it already). This is just to exclude f(x) = 0
You have a limit, a derivative, a series, but the complicated integral still takes up more than half the time and half the space on the board. Classic.
The final answer can be further simplified into: -[(∜6)/2]coth⁻¹[x/(∜6)] - [(∜6)/2]tan⁻¹[x/(∜6)] + C however, what if the question is: (Without the restrictions/without the previous infinite series question) ∫[6/(x⁴-6)]dx for -∜6 < x < ∜6 Since coth⁻¹[x/(∜6)] is not defined for -∜6 < x < ∜6, what expression would it be? !
bro hes getting serious with these problems. I love your videos BPRP. Keep up these amazing videos. Im a high school student in AP Calculus AB. You help me alot with studying! Thanks man!
@@Freddy-pp4nc because they are saying to. That is the structure of the question. If it is in first principles form, you need to do it as such. If it's in d/dx form, you can just derive it. With first principles at least you know where you're working to, so it isn't completely abstract algebra.
At first, this question seemed pretty daunting but when I started solving I thought this question was really simple. I solved inside equations pretty quickly but when I got to the integration I was like "bruh f*ck this shit". In my opinion, the integration isn't difficult, it's just very lengthy.
You could have added the dx in the original problem. It was probably written in white. Excellent problem. I can solve it today, thanks to what I’ve learned from you over the past few years. I may have told you before that as a physicist, my first year of calculus was an “honors” class where all we did was proofs. The only think I really loved about that class was that we used Apostle and I fell in love with proofs by induction.
I may be wrong with the following thought, but here goes: if inside a fraction, both numerator and denominator have the same sign - in our case, both are positive - its absolute value is the fraction itself. Which means that |6/(x^4)|sqrt_4(6), and not |x|>sqrt_4(6).
I don't really know what your issue is, but I hope this helps. The series is a geometric series (of the form a^r, where |a| should be less than 1 for the series to converge. If the series starts from n=1, then the value of the series is a/(1-a). If you're confused about the inequality, then it's 6/|x^4|< 1, 61). Hope this helps.
For the first step, the inside of the limit, Couldn't you have used l'hopital's rule to do it quicker? The outcome matches what you got, but I was unsure if the methodology holds up the same
Why do americans/english people always want to use l'hôpital's rule ??? It's not that efficient, and here, recognizing that you have the definition of the derivative is much quicker. In general, taylor expansions or equivalents are much more useful than l'hôpital's rule, which is quite outdated. I really don't understand why some people only learn to do limits using this rule.
Wolfram actually gives me a solution: x = -2 + exp( ( log( t /(t^2 +2) ) *( 1 - t^2 ))/(t^2+1) + log(2)*(1 - t^2) /(t^2 +1) ) where t:=tan(x/2) and a numeric value of x=2.66536 + 2*n*pi .... maybe because of this video Wolfram updated their algorithm. (it's 6th December now and vid was posted 26th Nov )
What is the minimun number of pen colours a bprp needs to write an equation such that none of adjacent terms have the same colour? Is it like the 4 colour map problem?
I think there is an error in the last expression. The red part in factor of coth-1 and tan-1 should be (root4(6))^2 instead of just root4(6). So after simplification you get (-1/2)coth-1(x/root4(6)) - (1/2)tan-1(x/root4(6)) + c
I feel he did the partial integral in a way that was more work, but it could be the only way. I thought we could just do partial and take g’ = 6, g = 6x, h= (x^4 - 6)^-1, h’ = 0.25ln(x^4 - 6) and the do hg -integral(h’g) for the answer. Why wouldn’t this work?
Hello sir. A friend of mine got an equation he couldn't respond in an exam and i'm sure the solution isn't that hard. It's cos(z^2)=-4 Thank you if you can respond this one 😁
When you reduce the fraction at 9:47, couldn't you have canceled out factors of sqrt(6) instead? Then you would have gotten -sqrt(6)/2 with the denominator already rationalized.
In Euclid's proof of infinite primes it is stated that the product of the first n primes plus one is not divisible by any of those primes. Seems intuitive right? Wouldn't it then make sense that also the product of the first n primes MINUS one would also not be divisible by - again - any of those primes? However this isn't true at all. Could you _please_ shed some light onto this problem I am facing?
hello bprp. there's a limit that's been puzzling me for a while: lim(x,y)--->(0,0) (xy^3)/(x^2+y^6) if you try many directions like polynomail functions or y=sinx or y=e^x-1 the limit = 0 but if you saw the graph for that two variable function you see that the limit does not exist at (0,0) so from what direction is the limit not equal to 0?
Hey blackpenredpen, i have a problem ive tried to solve, but no answer yet. It goes like this LOGa(x) = a^x , you need to solve for an a value so tangency occurs (both equations kiss eachother) Ive got the value to be about 1.45, but its not satisfying to estimate it lol Thanks:)
![Noob To Max With DRAGON REWORK In Blox Fruits [FULL MOVIE]](http://i.ytimg.com/vi/LnBMOinoOvA/mqdefault.jpg)
Do you like my videos edited or unedited?
unedited
Sir do you conduct zoom meetings also.. please share to me also 😞😞😢😋😋
Un
Unedited
Unedited but u should speed up erasing and writing way too long things
You know things go serious if you see bprp using 5 colors on a question
Now he’s BPRPBPRPPP
@@Claymore_II i had a stroke guessing it xD
1. that was actually funny
2. you somehow have about twice as many likes as THE PINNED COMMENT BPRP MADE
HOW
@@ananas-nevermind As you can see, many other comments are hearted by bprp. I exchanged the heart for 500 likes.
jk lol 😂
@@Claymore_II bprpbpppgp
bp¹=black pen
rp=red pen
bp²=blue pen
pp=purple pen
gp=green pen
Imagine this question being on your final exam. You go through all of these steps and calculations just to forget the +C at the end which single-handedly causes you to fail the exam.
Bruh you forgor the dx
Bro if I am forgetting only the c part in this question...... Then I am sure.... I will be able to solve other questions easily 🤣🤣🤣🤣
It's called stuying ET (Elektrotechnic) and IT (Informationtechnic)...
Then you have a garbage math teacher/professor if your only mistake of forgetting the +C at the end causes you to fail the entire exam.
@@user-yr3uj6go8i all or nothing
13:19 LMFAO I FELT THAT “why am I doing this” 💀
yeah it sounds so genuine lol
Fermat : "I have discovered a truly marvelous proof of this, which this board is too narrow to contain."
blackpenredpen : "hold my beer"
He’s always energetic and I love it.
you can see the passion through his eyes
This is completely off topic, but I think I came up with an interesting problem.
Given the recursive functions:
f(n) = af(n-1) + bg(n-1)
g(n) = ag(n-1) + bf(n-1)
Where a=0.5 and a+b=φ
Show that both f and g are fibbonacci sequences.
I'll put my solution at the end.
Stumbled upon this fact while trying to solve for g for any a or b... idk how others may attempt to solve it but I wrote them as a discrete time convolution. I got this equation for g(n):
g(n) = (b^2-a^2)g(n-2)+2ag(n-1)
Was trying to solve for f or g to write sin(nt) in terms of sin(t) and cos(t)... but accidentally made a mistake, one of the equations were supposed to add and the other was supposed to subtract the b term.... was doing this in the first place because, I was hoping the equation for sin(nt) would help solve the integral for (sin(Nt)cot(t/2)+cos(Nt))cos(nt) where N and n are two non-negative integers. Which I plugged into a couple integral calculators... and it didn't give me a solution.
When you evaluate that integral in respect to t from -π to π, and divide by 2π, its supposed to be an inverse discrete-time fourier transform that should equal 1 for all |n|
Challenge in Maths 🏆🏆🔥🔥
ruclips.net/video/VoZZetC5GfU/видео.html
All that effort for 12 likes lol
@@dizzyd7315 lol
@@dizzyd7315 Other people are illiterate.
Oh yeah it's just the inverse hyperbolic cotangent of x over the fourth root of 6, I could have guessed that
For sure🤣🤣❤️
It’s crazy the shit they make you remember in calc but then in any science they give formula sheets longer than the test
You forgot the plus c
I’m a sophomore in high school and have no idea what he’s doing but I still watch a bunch of his videos because they are interesting and relaxing. Keep up the good work👍
Same, I really enjoy these kind of videos :D
i was in your position a couple years ago. i’m now in calc 2, watched this whole video and actually understood all of it, and i’m very proud of myself. keep studying and you’ll get there💪
I did the same thing, been learning from him for a while
You know it's gonna be hard when he doesn't recommend calc teachers to use this in a test at the start of the video
lol
@blackpenredpen ngl I'd pass out if I saw that question 😳😳
@@blackpenredpen Challenge in Maths 🏆🏆🔥🔥
ruclips.net/video/VoZZetC5GfU/видео.html
"blackpenredpen"->"blackpenredpenbluepenpurplepengreenpen"
Peak character development ladies and gentlemen
then comes penbrow
Limits, derivatives and integration in one question. It should be called fundamental question of the calculus...Complete package.. Amazing
Don’t forget the sigma!
Wait. Where's the dx?
That's -100 score for you.
My first thought also.
@@xinpingdonohoe3978 same
5 minutes for the limit, differentiation and series; and 13 minutes for integration... Well...
more like 5 minutes to do partial fractions and 3 minutes to do the integral
I've always wanted to be a teacher, but after watching you, it made me feel so dumb that I'm no longer sure I'm capable of it. I hope I can be as intelligent as you are one day.
Just keep practicing man. Everything comes on point to the ones who really know what they want. Just giving you some strength ❤️!
Remember that he’s been teaching for about a decade at a massive institution
It is necessary to replace the variable in time to avoid radicals.
∫6dx/(x^4-6)=6^(1/4)∫ dt/(t^4-1). t=x/6^(1/4).
Here the fraction under the integral sign is so simple that it can be decomposed into elementary
fractions without the method of indefinite coefficients.
1/(t^4-1)=(1/2) *{1/(t^2-1)-1/(t^2+1)}=(1/2)*{(1/2)[1/(t-1)-1/(t+1)]-1/(t^2+1)}=
= (1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)
6^(1/4)* ∫ [(1/4)*1/(t-1)- (1/4)*1/(t+1)- (1/2)*1/(t^2+1)]dx=
= 6^(1/4)* {(1/4)*ln|t-1|-(1/4)*ln|t+1|-(1/2)* atan(t)} +C=
=(6^(1/4)/4)* ln|(t-1)/(t+1)|-(6^(1/4)/2)*atan (t) +C =
=(a/4)* ln|(x-a)/(x+a)|-(a/2)*atan (x/a) +C . a= 6^(1/4).
A young man as cool as a Shaolin monk with a Kung Fu Master beard and has a Pokemon ball in his hand explaining math problems... This combination made my yung hollow mind literally explode.
Idk if it’s easier but I factored out the 1/6 on the top and bottom from the beginning to lessen the burden of roots to get 1/((x^4/6)-1).
First time in history I'm seeing this guy forget the integration dx in the question itself.
Ah yes, blackpenredpenbluepengreenpenpurplepen
Can you please make a video that introduces hyperbolic functions and some sample problems with them? I’ve never known anything about them and would love to learn more about them!
An old video here ruclips.net/video/aC5cYc7XhIs/видео.html
@@blackpenredpen thanks!!!
Wait until you hear about elliptic functions.
I like these mega problems! They're great review.
Glad to hear : )
Watching your videos, I actually started loving math !
Heyy I solved this on my own before I went further into the video. I'm proud of myself 😍
Partial fractions could be used in a different way by adding and subtracting x^2 (x squared) and 6^(1/2) (square root of 6)
Just took my Calculus CLEP exam and got the score I needed; 64 ain't great, but i get my credit so it's cool. I've never been confident in Math, and quite frankly I went into the exam expecting to fail- I'm quite happy that wasn't the case. Im very grateful for all the videos you make, and passing this exam was definitely thanks to you lol. Keep it up!
Never heard of it until looking it up. Seems very silly they wouldn’t just give you the AP Exam.
The partial fraction decomposition can be done without a system of equations if you consider that 1 = 1/(2*sqrt(6))*[(x^2+sqrt(6))-(x^2-sqrt(6))]. This improved memory efficiency allowed me do the problem in my head reasonably fast. Also, BPRP was running out of room and the video was already long so he used hyperbolic trig functions. But you can decompose 1/(x^2-sqrt(6)) with the same trick or Heaviside cover up.
My thoughts exactly.
This made me realize how much I've forgotten about partial fraction decomposition 🥲 time for a study montage
I got all the way to the integral on my own and said "I do math for fun, but I draw the line at PFD."
STUDYING HARD NOW, HE'S GOTTA RECLAIM
ALL OF THE MATH THAT ESCAPED FROM HIS BRAIN
INTEGRALS, DERIVATIVES AND PARTIAL FRACTIONS
HE'S MASTERED THEM ALL NOW HE'S READY FOR ACTIONNNN
MATH-E-MATIC!
AUT-O-MATIC!
FEELS LIKE MAGIC WHEN YOU DERIVE
JUST TO STAY ALIVEEEEE!
MATH-E-MATIC!
AUT-O-MATIC!
MATH-E-MATIC!
AUT-O-MATIC!
*music fades out as you ride off on a motorcycle*
I’m 41 and just learning calculus. I just sat there with wide eyes and all shocked at how much work it’s needed. Damn. I think my brain almost literally exploded. Lol.
A technician takes X hours to visit the stores in a couple of streets in one shift, and when he finishes his tour, another technician revisits the same stores again, needing other X hours to finish the second shift, and so on. For example, if the first technician started his shift at 1:00pm, he finishes it at 7:00pm and the second technician starts his shift at 7:00pm and finishes it at 1:00am, and so on. The supervisors used to make a quick meeting with all technicians twice a day at 1:00 am and 1:00 pm, so they need to finish their tours exactly 1:00. Identify the mathematical notation for the number of shifts should be made by the technicians in order to achieve this, write the name of this mathematical value, and find it for two tours, one with X=7 and another with X=11 how I can solve this Question
Thank you I have tears of joy now
here is a challenge: find f(x) such that
f'(x) = f(x+1)
oh and i forgot something: there is a x such that f(x) = 1 (just multiply the solution by some number to get f(x) = 1 if you don't have it already). This is just to exclude f(x) = 0
These are the kinds of questions my General Chemistry I and II professor was fond of - the entire final exam in one question.
This guy is gonna change my career!
Your videos are everything
An absolute monster
It makes me so happy to see someone this exited about maths! Keep up the entertaining content😁👍🏼!
Integration of 1/sin^7x + cos^7x possible
You have a limit, a derivative, a series, but the complicated integral still takes up more than half the time and half the space on the board. Classic.
The final answer can be further simplified into:
-[(∜6)/2]coth⁻¹[x/(∜6)] - [(∜6)/2]tan⁻¹[x/(∜6)] + C
however, what if the question is: (Without the restrictions/without the previous infinite series question)
∫[6/(x⁴-6)]dx for -∜6 < x < ∜6
Since coth⁻¹[x/(∜6)] is not defined for -∜6 < x < ∜6, what expression would it be?
!
I really felt the "Why am I doing this" at 13:20 on deeper levels than is comprehensible.
More all one calculus questions thanks)
The students: "don't worry he said he would put only one ex. on the exam"
The exercise: are you sure about that 😂😂🤣🤣
Hahaha!
bro hes getting serious with these problems. I love your videos BPRP. Keep up these amazing videos. Im a high school student in AP Calculus AB. You help me alot with studying! Thanks man!
Glad to hear it, thank you : )
@@blackpenredpen yessirrrr
coefficients can be simplify to -(6)^(1/4)
I lost it at the integration part
Until then i was really happy that things were going smooth
All i know is f(x)=x² so f'(x)=2x
There's no dx at the end of the integral at the top :( xD
You're right.
It is incomplete
the very first limit: isn't the definition of derivative?
derivative of x^(-2) is -2/x^3
Yes u are right. He mentioned it in the video
But when it's in first principles form, you should do it as an actual algebra question.
@@xinpingdonohoe3978 why, unless they say you have to why not save time
@@Freddy-pp4nc because they are saying to. That is the structure of the question. If it is in first principles form, you need to do it as such. If it's in d/dx form, you can just derive it. With first principles at least you know where you're working to, so it isn't completely abstract algebra.
I just realized how much more I have to learn
At first, this question seemed pretty daunting but when I started solving I thought this question was really simple. I solved inside equations pretty quickly but when I got to the integration I was like "bruh f*ck this shit". In my opinion, the integration isn't difficult, it's just very lengthy.
Erase part of the board!!! Also at the end, you could simplify square root of 6 / (2 * 4th root of 6) to just (4th root of 6) / 2.
You could have added the dx in the original problem. It was probably written in white.
Excellent problem. I can solve it today, thanks to what I’ve learned from you over the past few years.
I may have told you before that as a physicist, my first year of calculus was an “honors” class where all we did was proofs. The only think I really loved about that class was that we used Apostle and I fell in love with proofs by induction.
13:26 “why am i doing this”. that’s how i feel when i do implicit differentiation
13:19
Love your content helps with my love of mathematics, regards
I love how you look at it so proudly when you're done, like it's your child. 😂
😆
Great content man. As always
Congratulations from a French student. 🇫🇷❤️
i never thought i would be so satisfied by an math equation :D
I am glad to hear that : )
13:19 - My reaction when it's 4 am and I'm only watching these videos
It's nearly 2am and I've been up for over an hour trying to do this. I lay restless in bed thinking about this question
Awosome
I may be wrong with the following thought, but here goes: if inside a fraction, both numerator and denominator have the same sign - in our case, both are positive - its absolute value is the fraction itself. Which means that |6/(x^4)|sqrt_4(6), and not |x|>sqrt_4(6).
I love this!
Wait, when solving the Σ, how did x⁴ leave since the fraction is (6/x⁴)'? Shouldn't it be something else?
he says it's raised to the FIRST POWER, so it's just ^1
I don't understand your issue with the series. What's wrong with him answer?
I don't really know what your issue is, but I hope this helps.
The series is a geometric series (of the form a^r, where |a| should be less than 1 for the series to converge. If the series starts from n=1, then the value of the series is a/(1-a).
If you're confused about the inequality, then it's 6/|x^4|< 1,
61).
Hope this helps.
isnt (6/x^4)' × x^4 = -24/x ?
why would it be =6 ?
it happens at 4:12?
He did nxn-1 in the previous step
just learned what's the derivative of coth^-1. Great video !
06:32 It also can be nice way to use Heavyside cover up method, I guess. :)
Blackpenredpen Collab with purplepen greenpen and bluepen
Hey, isn’t the limit just the definition of the derivative of 1/x^2?
Yep
For the first step, the inside of the limit, Couldn't you have used l'hopital's rule to do it quicker? The outcome matches what you got, but I was unsure if the methodology holds up the same
Why do americans/english people always want to use l'hôpital's rule ??? It's not that efficient, and here, recognizing that you have the definition of the derivative is much quicker. In general, taylor expansions or equivalents are much more useful than l'hôpital's rule, which is quite outdated. I really don't understand why some people only learn to do limits using this rule.
If you have 1/(0/1), would that be undefined, or would it be 0 because (a/b)/(c/d)=(a/b)×(d/c) meaning 1×(0/1)=0
"Why am I doing this?" Chen Lu contemplating why he is attempting to solve a problem involving almost every core concept of calculus LOL....
Wolfram actually gives me a solution: x = -2 + exp( ( log( t /(t^2 +2) ) *( 1 - t^2 ))/(t^2+1) + log(2)*(1 - t^2) /(t^2 +1) )
where t:=tan(x/2) and a numeric value of x=2.66536 + 2*n*pi .... maybe because of this video Wolfram updated their algorithm. (it's 6th December now and vid was posted 26th Nov )
Is this ASMR? I get the feels just by watching this
Your awsome Professor)
its amazing sir.. your videos always enlighten me.....
"Why am I doing this 😞" was the best part 😂😂😂
Somebody get this man a bigger whiteboard! xD
Sorry for the stupid question, why ∑ (6/x^4)^n is equals to (6/x^4) / [1 - (6/x^4)] and not to 1 / [1 - (6/x^4)] ?
Bc n starts at 1
You're missing a dx after the integral as soon as it starts
You are a madman!
At 10:00 it would be better 6/√6 = √6 and then B=√6/2
Couldn't you have simplified the answer since √6 / ∜6 = 6^(1/2) / 6^(1/4) = 6^(1/2 - 1/4) = 6^(1/4) = ∜6 ?
Hello which one is biger e^x and x^n when x->+∞
What is the minimun number of pen colours a bprp needs to write an equation such that none of adjacent terms have the same colour? Is it like the 4 colour map problem?
I think there is an error in the last expression. The red part in factor of coth-1 and tan-1 should be (root4(6))^2 instead of just root4(6).
So after simplification you get (-1/2)coth-1(x/root4(6)) - (1/2)tan-1(x/root4(6)) + c
I'm gonna learn calculus and come back here to see if I can understand more than 20% of this video
I feel he did the partial integral in a way that was more work, but it could be the only way. I thought we could just do partial and take g’ = 6, g = 6x, h= (x^4 - 6)^-1, h’ = 0.25ln(x^4 - 6) and the do hg -integral(h’g) for the answer. Why wouldn’t this work?
Ah yes, my favorite episode of blackpenredpenbluepengreenpenpurplepen
Which is greater x^y ! Or y^x ! Given that x and y are real numbers? Pls answer this
the "why am i doing this" is so real
Hello sir.
A friend of mine got an equation he couldn't respond in an exam and i'm sure the solution isn't that hard.
It's cos(z^2)=-4
Thank you if you can respond this one 😁
When you reduce the fraction at 9:47, couldn't you have canceled out factors of sqrt(6) instead? Then you would have gotten -sqrt(6)/2 with the denominator already rationalized.
gonna need this for when i retake calc 2… but on a side note, where did you get the pokeball cover for your mic?
In Euclid's proof of infinite primes it is stated that the product of the first n primes plus one is not divisible by any of those primes. Seems intuitive right? Wouldn't it then make sense that also the product of the first n primes MINUS one would also not be divisible by - again - any of those primes? However this isn't true at all. Could you _please_ shed some light onto this problem I am facing?
hello bprp. there's a limit that's been puzzling me for a while: lim(x,y)--->(0,0) (xy^3)/(x^2+y^6)
if you try many directions like polynomail functions or y=sinx or y=e^x-1 the limit = 0 but if you saw the graph for that two variable function you see that the limit does not exist at (0,0) so from what direction is the limit not equal to 0?
How about 10 questions in one??
at 9:12 wouldn't that be equal to -6 and not 6
Hey blackpenredpen, i have a problem ive tried to solve, but no answer yet. It goes like this
LOGa(x) = a^x , you need to solve for an a value so tangency occurs (both equations kiss eachother)
Ive got the value to be about 1.45, but its not satisfying to estimate it lol
Thanks:)