This is genuinely a very powerful test of someone's calculus skills. I guess it might be a bit overwhelming to use it during an actual test, but it can definitely be used by students as a self-check.
now the society wants you to make even *harder* question including some deadly *definite triple integrals* along with *laplace transforms* and *partial derivatives* just casually floating around in the question as a one million subscriber special Edit: alrighty here u are, at a million subs Congrats for that👍 . . . now gimme my question *pweeease*
Ive always wondered, whats the difference between regular derivatives and partial derivatives. I’ve seen it quite a number of times when he does differential equation and Feyman’s technique, but no idea what it really means. Thank you in advance.
@@Ninja20704 With sufficient degrees of freedom, like say some function plotted on x, y and z it may sometimes be practical to keep one variable constant such that we can "slice" the plane and examine a regular 2d plane for derivations or such
@@ES-qe1nh Agreed. In addition to that, a partial derivative implicitly depends on what other variables you have, since they are to be kept constant. For instance, suppose f(x,y) = x+y Then the partial derivative (I write D because of my keyboard, but I mean the partial-d) D/Dx equals Df/Dx = 1 If I reparametrize, or transform my coordinate system to new variables x and z, where x remains the same but z = x-y, then f(x,z) = 2x-z, so now suddenly Df/Dx = 2 even though we changed nothing essentially about either f or x! Alternatively, suppose that y itself is a function of x, say y(x) = x², then f(x) = x+x², then we can compute the normal full derivative as df/dx = 1+2x I guess the thing to note is that all of the derivatives of f with regard to x are different. So they are actually different beasts, not just different notations.
@@Ninja20704 they appear more frequently in physics than maths but to simplify the definition, it’s basically the derivative except it gives more pain than normal derivative
I cant express how much I love this channel. I am currently studying Soil science and agricultural chemistry and surprisingly enough the math needed for it is extremely advanced.I unfortunately lost some time and almost dropped out but right now I am determined to graduate.I started learning math by myself from the fundamentals to calculus and now I'm trying to study complex analysis by myself,and this channel just keeps me motivated.Thank you Mr. bprp!!!
Even dough I didn’t understand most of the video, I find your channel really interesting and I love to watch your videos. I can clearly see your passion to maths and your happiness during all videos. Keep going man! I really admire people like you!
With the gaussian integral you can take advantage of the fact that the integrand is an even function and the integral is bounded symmetrically, so you can change the lower bound to zero and double the result of that. That shows right away that our "half-gaussian" integral in the 'u' world is sqrt(π)/2, no worries about convergence.
I have been dealing with gamma function so much lately that as soon as I saw the limit I instantly realized "ok the limit is just x going to sqrt(pi)/2"
May i request you to make more of these all-in-one questions? I find it very amusing to solve and it was incredibly satisfying when i got the question right. This may just be the right tool for me to do brain exercises during leisure times. I love your work very much. I hope you gain an even greater reach on TH-cam and make more people understand calculus - or even give birth to a whole new generation of masters. God bless you
I didn't recognize the power series so I took it to be the integral of x^[2(2n+1)] and turned into a geometric series, then integrated to get the log version hahaha
As i can see, d²/dx² of the whole thing inside is just equal to (x^2(2n+1))/(2n+1). For the limit we know that x approaches √π/2, don't ask why cuz it's too easy. And also the limit is not undefined when x=√π/2 so we just put x=√π/2 and the thing left is the sum series of 1/4*π^(2n+1)/(2n+1) Well, i think from here u guys can solve this on your own
Very interesting problem, with a lot of concepts rolled into one! The only gripe I have, though, is that this seems to rely very heavily on the student remembering the solutions to past problems. Recognizing the Gaussian integral is pretty reasonable, but would the student be screwed if they didn’t have the Taylor series for the inverse hyperbolic tangent memorized and be able to recognize it…?
I may not know what you are doing right now and may get frustrated while trying to understand it, but I'm telling you, I WILL be back in a month and I WILL get it. Cya in 1 month, or 4 weeks, or 30 days, or 1800 hours, or 108000 minutes, or 6480000 seconds. I'll be back.
Can show the solution for the integral from 0 to 1 of ((x^2)-1)/(ln(x))? Somehow the answer equals ln(3), but any online source gives the answer in terms of the Exponential Integral, and uses numerical approximation to get a value that visually looks equal to ln(3), but it doesn’t show how to plug in the bounds to get that answer. I get that you can substitute u = ln(x) and dx = (e^u)du and so the expression becomes integral from -infinity to 0 of (e^3u - e^u)/u du. This seems to not be directly solvable in terms of real valued closed form/elementary functions. The question was on our advanced calculus quiz, and somehow the correct answer (multiple choice) was ln(3).
Your videos have saved my grade during my Differential Equations course! I was wondering if you could do a video on how to solve Boundary Value Problems for 4th Order DEs related to deflection of a beam? Thanks again!
When I saw this thumbnail I immediately remembered my Analysis 1 and Analysis 2 classes, ahhh so nostalgic 😌, it only missed some double integrals here
Could you perhaps try to solve lim x -> infinity of x/(tan((pi/2)-pi/x)) in one of your upcoming videos, I think the result will be surprising to you but I wouldn't know how to solve this using classical calculus techniques
Here's how I did it: The denominator in the limit is tan(pi/2-pi/x)=cot(pi/x)=1/tan(pi/x), by trigonometric identities. Hence, the function inside the limit is x/(1/tan(pi/x))=xtan(pi/x). Our limit is now lim(x->inf)(xtan(pi/x)). Now, we will introduce a substitution. Let t=pi/x, meaning x=pi/t. As x->inf, t->0+. Our limit becomes lim(t->0+)((pi/t)tan(t)). Taking the pi out of the limit since it's a constant gives pi*lim(t->0+)(tan(t)/t). We can rewrite tan(t)/t as (sin(t)/cos(t))/t=(sin(t)/t)*(1/cos(t)), using trigonometric identities. Our limit becomes pi*lim(t->0+)((sin(t)/t)*(1/(cos(t)))=pi*lim(t->0+)(sin(t)/t)*lim(t->0+)(1/cos(t)). We can do this because both of the resulting limits exist. The first limit, lim(t->0+)(sin(t)/t), is famously equal to 1 and blackpenredpen definitely made a video on it already. The second limit can be evaluated using direct substitution: 1/cos(0)=1. Our limit is hence equal to pi*1*1=pi. QED
To do this, we will need the Lambert W function, which blackpenredpen just seems to be obsessed with for some reason (watch his older videos). This function is defined as the function W(x) for which W(xe^x)=W(x)e^W(x)=x. In other words, W(x) is the inverse function of xe^x. Let's start with y=x^(1/x) and swap x and y to get x=y^(1/y). We want to solve for y. Taking the reciprocal of both sides gives 1/x=1/(y^(1/y)). But since 1=1^(1/y), we have 1/x=(1^(1/y))/(y^(1/y)). Using exponent properties, we get 1/x=(1/y)^(1/y). Taking the natural log on both sides, we get ln(1/x)=ln((1/y)^(1/y)). Using log properties, we get ln(1/x)=(1/y)ln(1/y). Since 1/y=e^ln(1/y), we get ln(1/x)=ln(1/y)e^ln(1/y). We are now in a good place to use the Lambert W function since the RHS is of the form ke^k. Doing that gives W(ln(1/x))=W(ln(1/y)e^ln(1/y))=ln(1/y), by the definition of the Lambert W function. We can now solve for y easily: ln(1/y)=W(ln(1/x)) 1/y=e^W(ln(1/x)) y=1/e^W(ln(1/x))=e^(-W(-ln(x)) (using log and exponent properties) Hence, the inverse function to f(x)=x^(1/x) is e^(-W(-ln(x)). QED (As an exercise, try to check that this is the correct answer using the definition of the Lambert W function.)
For the last part I didn't know the power series of inverse hyperbolic functions so I took the derivative again to get 2x/(1-x^4) and then integrated to get (1/2)(log(x^2+1)-log(1-x^2)). Probably could have also gotten this just remembering the power series for log(1+x).
hey could you or anyone in the comments show why when you find the area between the two curves y=x⅔ + √(1-x²) and y=x⅔ - √(1-x²) [the two curves which gives a heart shape when you graph them together] the area is = to pi??
The two curves go from -1 to 1, so just using area between two curves, you get integral(-1, 1) (x^2/3+sqrt(1-x^2)-(x^2/3-sqrt(1-x^2)) dx the x^2/3 cancels, and you get integral(-1, 1) (2sqrt(1-x^2)) dx and you can notice that this is the area of two semicircles with radius 1, so the area would be pi*1^2 = pi.
Here's a question for you: Imagine two curves, 1/x and another like cos(x). What constant would you have to add to the cosine curve to make it be tangent to 1/x. So, given the function cos(x) + a, determine a such that the function becomes tangent to 1/x (obviously using graphs to decide it is cheating, they should be used only to decide what's reasonable)
The derivation is pretty simple. Let's call f(x) the proposed power series of arctanh(x). Compute f'(x). This is a geometric series of x^2. So, f'(x) = 1/(1-x^2) for -1
Ah yes, a new episode of blackpenredpenbluepenpurplepengreenpen, my favourite!
😂
New episode of bprpbpppgp
Find indefinite integral of xsinx/(1+(cosx)^2)
Please help
Ahh yes, I remember it going down like this in the anime.
This is genuinely a very powerful test of someone's calculus skills.
I guess it might be a bit overwhelming to use it during an actual test, but it can definitely be used by students as a self-check.
With this all in one calculus becoming a thing, you are showing again why you are among the top mathematicians in the platform, if not the best
now the society wants you to make even *harder* question including some deadly *definite triple integrals* along with *laplace transforms* and *partial derivatives* just casually floating around in the question as a one million subscriber special
Edit: alrighty here u are, at a million subs
Congrats for that👍
.
.
.
now gimme my question
*pweeease*
Ive always wondered, whats the difference between regular derivatives and partial derivatives. I’ve seen it quite a number of times when he does differential equation and Feyman’s technique, but no idea what it really means. Thank you in advance.
@@Ninja20704 With sufficient degrees of freedom, like say some function plotted on x, y and z it may sometimes be practical to keep one variable constant such that we can "slice" the plane and examine a regular 2d plane for derivations or such
@@ES-qe1nh Agreed.
In addition to that, a partial derivative implicitly depends on what other variables you have, since they are to be kept constant.
For instance, suppose
f(x,y) = x+y
Then the partial derivative (I write D because of my keyboard, but I mean the partial-d) D/Dx equals
Df/Dx = 1
If I reparametrize, or transform my coordinate system to new variables x and z, where x remains the same but z = x-y, then f(x,z) = 2x-z, so now suddenly
Df/Dx = 2
even though we changed nothing essentially about either f or x!
Alternatively, suppose that y itself is a function of x, say y(x) = x², then f(x) = x+x², then we can compute the normal full derivative as
df/dx = 1+2x
I guess the thing to note is that all of the derivatives of f with regard to x are different. So they are actually different beasts, not just different notations.
society
youtubebu.com/watch?v=udZddgY5Cea
n the question as a one million subscriber special
@@Ninja20704 they appear more frequently in physics than maths
but to simplify the definition, it’s basically the derivative except it gives more pain than normal derivative
I cant express how much I love this channel. I am currently studying Soil science and agricultural chemistry and surprisingly enough the math needed for it is extremely advanced.I unfortunately lost some time and almost dropped out but right now I am determined to graduate.I started learning math by myself from the fundamentals to calculus and now I'm trying to study complex analysis by myself,and this channel just keeps me motivated.Thank you Mr. bprp!!!
As someone who’s education didn’t go past 3rd grade. Thank you for your videos, I’m doing my best to learn all the things I missed out on.
Try learning logarithms, matrices and algebraic whole square solutions
@@humzakhan3962 bro just passed 3rd grade and is studying high school maths,, i want dedication like him
GANBAREEEEEE
@@extreme4180 he didnt just pass 3rd grade, his education didnt go past 3rd grade, read the comment
@@doomsdaycookie7034they were joking lol
Even dough I didn’t understand most of the video, I find your channel really interesting and I love to watch your videos. I can clearly see your passion to maths and your happiness during all videos. Keep going man! I really admire people like you!
That little backtrack at 7:46 was funny XD I thought I accidentally rewound the video cuz I spaced out for a single second the first time
With the gaussian integral you can take advantage of the fact that the integrand is an even function and the integral is bounded symmetrically, so you can change the lower bound to zero and double the result of that. That shows right away that our "half-gaussian" integral in the 'u' world is sqrt(π)/2, no worries about convergence.
I have been dealing with gamma function so much lately that as soon as I saw the limit I instantly realized "ok the limit is just x going to sqrt(pi)/2"
@@MessedUpSystem lmao I just revised Laplace transform and this came by, immediately noticed it's L{sqrt(t)}(1) which is sqrt(pi)/2
Congrats on the 1M subs!! Well deserved!!
Thanks!
For the first bit, I just noticed that it’s the same as the evaluating (1/2)! Via the gamma function, which is sqrt(pi)/2
Seeing this I've really understood the meaning of:
Don't eat the whole cake in one turn, a slice by slice is good 🍰
a thing, y
youtubebu.com/watch?v=qTt8Efn8KoU
whole cake in one turn, a slice by slice is goo
You should make a playlist about generating function, specail functions and Sturm Liouville Systems
Congrulations for 1 million subscribers !!! Keep it up !
Congratulations on 999k subs!
Thank you!
Aww welcome
May i request you to make more of these all-in-one questions? I find it very amusing to solve and it was incredibly satisfying when i got the question right. This may just be the right tool for me to do brain exercises during leisure times. I love your work very much. I hope you gain an even greater reach on TH-cam and make more people understand calculus - or even give birth to a whole new generation of masters. God bless you
Can you find the radius of a circle which touches Latus rectum , axis and circumference of the parabola Y²=4aX
Calculus exam in a couple days, just what i needed!
Congratulations for attaining 1M subs. Keep moving 👍
Gamma Functions of the integral whuch x is aproaching wuld make things easier for those who have done advanced calculus. It's just sqrt(pi)/2...
Congratulations for 1M sir .
Edit-Love from India❤️
(turn over the paper)
heart attack
Thank you for promoting interest in mathematics!
Congratulations on reaching 1 Million Subscribers!
ONLY 1K LEFT UNTIL 1MIL
gotta love those ones!!
I was able to solve everything but the tanh^-1 (x^2) bc my last course never covered that 😮
Congratulations on 1 million!
Congratulations for 1 million subscribers :)
I didn't recognize the power series so I took it to be the integral of x^[2(2n+1)] and turned into a geometric series, then integrated to get the log version hahaha
Honestly, your contante is Awesome.
There's a way to find exactly (1+(2+(3+(4+..)^1/4)^1/3)^1/2)^1/1 [The sum of n n-roots of n plus the next root]
1 MILLION SUBS!!!! CONGRATS
CONGRATS FOR 1 MILLION BRO
As i can see, d²/dx² of the whole thing inside is just equal to (x^2(2n+1))/(2n+1).
For the limit we know that x approaches √π/2, don't ask why cuz it's too easy. And also the limit is not undefined when x=√π/2 so we just put x=√π/2 and the thing left is the sum series of 1/4*π^(2n+1)/(2n+1)
Well, i think from here u guys can solve this on your own
I'm in my final 4 weeks of calc 2. I have never once been taught hyperbolic trig functions so that last part flew right over my head
i havent taken any calculus classes so i dont understand anything at all, but i still like to watch
congrats for 1M subs , im here since 327 k subs
Aww thank you!!
Given is the function f(x) = -x² + 5. Find the tangent, that crosses the point P(3|10), of that function.
Y-10=-2x(x-3)
Lots of formulas kudos to all your successes and videos !!
Man congrats a Million!!
You can technically multiply the 2u with the u and use Feynmann integration.
Hey !!!!!!!
Namaste🙏
I'm challenging you
Solve the integral without using any property....
I = 2/π ∫ dx/( 1+e^sinx)(2+cos2x)
Limits from -π/4 to π/4
Zero
Very interesting problem, with a lot of concepts rolled into one! The only gripe I have, though, is that this seems to rely very heavily on the student remembering the solutions to past problems. Recognizing the Gaussian integral is pretty reasonable, but would the student be screwed if they didn’t have the Taylor series for the inverse hyperbolic tangent memorized and be able to recognize it…?
When he smiles and scoughs it is so cute
Seriously I love you guys 😊
I may not know what you are doing right now and may get frustrated while trying to understand it, but I'm telling you, I WILL be back in a month and I WILL get it. Cya in 1 month, or 4 weeks, or 30 days, or 1800 hours, or 108000 minutes, or 6480000 seconds. I'll be back.
I may be 10 months late but......
Did u understand it ?
Wow, thanks!!!
That video is very, very amazing
Congrats 1 million subscriptions！
this was a great problem. loved it
2:29 I think that should be a minus sign. It will give you the same answer at the end tho cause you square it
That integral can be solved by using the gamma function as well....ig it's easier to solve by using it..... great one btw!
this just makes me realize how much math I forgot over the last 25 years
So this problem was actually one of the easier ones we had to solve back in the day. Thank you for such a great step by step explanation!
broo what did u do back in the days damn💀💀
@@mayankchaudhary8921 probably something with differentials equations 🤣
SinQ/CosQ a 10^6
Awesome video! Thank you!
Pretty good question for exploding my head but amazing result 👏
I love this stuff!
Youmare very good at calculus,pls teach basic of calculus
Can you find the integral of sin(e^(-x^2)) from negative infinity to positive infinity??🤔🤔
Can show the solution for the integral from 0 to 1 of ((x^2)-1)/(ln(x))? Somehow the answer equals ln(3), but any online source gives the answer in terms of the Exponential Integral, and uses numerical approximation to get a value that visually looks equal to ln(3), but it doesn’t show how to plug in the bounds to get that answer. I get that you can substitute u = ln(x) and dx = (e^u)du and so the expression becomes integral from -infinity to 0 of (e^3u - e^u)/u du. This seems to not be directly solvable in terms of real valued closed form/elementary functions. The question was on our advanced calculus quiz, and somehow the correct answer (multiple choice) was ln(3).
im from biology majors and yet i love watching these videos 💀💀
Sir please solve this question √9-4√5
Plz pick up more this type problem mix all math concepts
It's just 1/2Y(1/2) the result of integral. Gamma functions
Your videos have saved my grade during my Differential Equations course! I was wondering if you could do a video on how to solve Boundary Value Problems for 4th Order DEs related to deflection of a beam? Thanks again!
Hey blackpenredpen! You left an outtake in toward the end. That sigh really got me ):
Congratulations 🎉 100 subscribes
If you expect your viewers to just know the Gaussian Integral, you should expect them to know the Gamma function.
My chemistry 2 professor used to do this.
Bro you are genius.
I am a students of Mathematics from Bangladesh 🇧🇩
I would love to see your reaction of one of your students finishing this problem (the exam) in ten minutes
When I saw this thumbnail I immediately remembered my Analysis 1 and Analysis 2 classes, ahhh so nostalgic 😌, it only missed some double integrals here
Could you perhaps try to solve lim x -> infinity of x/(tan((pi/2)-pi/x)) in one of your upcoming videos, I think the result will be surprising to you but I wouldn't know how to solve this using classical calculus techniques
Here's how I did it:
The denominator in the limit is tan(pi/2-pi/x)=cot(pi/x)=1/tan(pi/x), by trigonometric identities. Hence, the function inside the limit is x/(1/tan(pi/x))=xtan(pi/x). Our limit is now lim(x->inf)(xtan(pi/x)).
Now, we will introduce a substitution. Let t=pi/x, meaning x=pi/t. As x->inf, t->0+. Our limit becomes lim(t->0+)((pi/t)tan(t)).
Taking the pi out of the limit since it's a constant gives pi*lim(t->0+)(tan(t)/t).
We can rewrite tan(t)/t as (sin(t)/cos(t))/t=(sin(t)/t)*(1/cos(t)), using trigonometric identities. Our limit becomes pi*lim(t->0+)((sin(t)/t)*(1/(cos(t)))=pi*lim(t->0+)(sin(t)/t)*lim(t->0+)(1/cos(t)). We can do this because both of the resulting limits exist.
The first limit, lim(t->0+)(sin(t)/t), is famously equal to 1 and blackpenredpen definitely made a video on it already.
The second limit can be evaluated using direct substitution: 1/cos(0)=1.
Our limit is hence equal to pi*1*1=pi. QED
Now we are 1M family 🎉
Couldn't we just use gamma function for the limit integral?
The markers!
i like watching these even though i have no idea what is even happening
I'm curious as to what the inverse function to f(x) = x^(1/x) is. I can't solve it at all
To do this, we will need the Lambert W function, which blackpenredpen just seems to be obsessed with for some reason (watch his older videos). This function is defined as the function W(x) for which W(xe^x)=W(x)e^W(x)=x. In other words, W(x) is the inverse function of xe^x.
Let's start with y=x^(1/x) and swap x and y to get x=y^(1/y). We want to solve for y.
Taking the reciprocal of both sides gives 1/x=1/(y^(1/y)). But since 1=1^(1/y), we have 1/x=(1^(1/y))/(y^(1/y)). Using exponent properties, we get 1/x=(1/y)^(1/y).
Taking the natural log on both sides, we get ln(1/x)=ln((1/y)^(1/y)). Using log properties, we get ln(1/x)=(1/y)ln(1/y).
Since 1/y=e^ln(1/y), we get ln(1/x)=ln(1/y)e^ln(1/y). We are now in a good place to use the Lambert W function since the RHS is of the form ke^k.
Doing that gives W(ln(1/x))=W(ln(1/y)e^ln(1/y))=ln(1/y), by the definition of the Lambert W function.
We can now solve for y easily:
ln(1/y)=W(ln(1/x))
1/y=e^W(ln(1/x))
y=1/e^W(ln(1/x))=e^(-W(-ln(x)) (using log and exponent properties)
Hence, the inverse function to f(x)=x^(1/x) is e^(-W(-ln(x)). QED
(As an exercise, try to check that this is the correct answer using the definition of the Lambert W function.)
@@youngmathematician9154 its the best function of all time
@@youngmathematician9154 Thanks a lot!!!
As long as i watched this video, my brain more hot than my phone
For the last part I didn't know the power series of inverse hyperbolic functions so I took the derivative again to get 2x/(1-x^4) and then integrated to get (1/2)(log(x^2+1)-log(1-x^2)). Probably could have also gotten this just remembering the power series for log(1+x).
hey could you or anyone in the comments show why when you find the area between the two curves y=x⅔ + √(1-x²) and y=x⅔ - √(1-x²) [the two curves which gives a heart shape when you graph them together] the area is = to pi??
The two curves go from -1 to 1, so just using area between two curves, you get
integral(-1, 1) (x^2/3+sqrt(1-x^2)-(x^2/3-sqrt(1-x^2)) dx
the x^2/3 cancels, and you get
integral(-1, 1) (2sqrt(1-x^2)) dx
and you can notice that this is the area of two semicircles with radius 1, so the area would be pi*1^2 = pi.
Lesgooooo 1 million!!
I think that you can also express the summation as 2tanh¯¹(x)
Appreciating the stock of markers in that shelf 😂😂😂
Here's a question for you:
Imagine two curves, 1/x and another like cos(x). What constant would you have to add to the cosine curve to make it be tangent to 1/x.
So, given the function cos(x) + a, determine a such that the function becomes tangent to 1/x (obviously using graphs to decide it is cheating, they should be used only to decide what's reasonable)
As a kid that doesnt understand calculus entirely. My honest reaction was:
WHAT THE FUC-
This challenge is for you!🔥
Solve:
a³ + b² = 1 ;
a² + b³ = -1
Note: The solutions of these equations are real integers.
Fermat's Last Theorem. No thanks.
Trivial solution: (a, b) = (0, -1)
isnt there some stuff from ramanujan to solve this? think i have seen something similar
a = 0
b = -1
Not enough greens functions
If this is your test, I'm glad you weren't my calculus professor in college.
Would be nice if you showed us how to derive the power series for tanh^-1 (x). Still, amazing video.
The derivation is pretty simple. Let's call f(x) the proposed power series of arctanh(x).
Compute f'(x). This is a geometric series of x^2.
So, f'(x) = 1/(1-x^2) for -1
You can also use the root of unity filter on the log power series.
We need only 1K left to 1M subscribers!
snwer=one
do this integral x+sin(x)^10+tanh(x)
why we are using that identity?.... what is that identity about?... is there any source to know something deep about that identity?
My head... It's about to explode!!!!
That was so funny. I really enjoy it 🥺❤️🔥
I still struggle with integration by parts.
When you said calculus is over
Bprp: What do you said maaan
24 November is celebrated as the day of teachers for the Republic of Turkey, happy teachers' day to you too
@Realblackpenredpen. you're not even as real as your real numbers dude