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@lapicethelilsusboy491

Жыл бұрын

I'll ask again: Are you okay?

@leonardobarrera2816

Жыл бұрын

Thanks for the video

@jamespat7975

Жыл бұрын

How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?

@Abdul-ot3lu

Жыл бұрын

Is brilliant worh

@hustler3of4culture3

Жыл бұрын

Brilliant

Log triangles are naturally very hard to manipulate, on account of their large size and weight.

@fasebingterfe6354

Жыл бұрын

I agree

@U014B

Жыл бұрын

I never wood've thought about that.

@theabyss5647

Жыл бұрын

I think it's not because they're heavy but because of their temperature. We're taking about an ln triangle and liquid nitrogen is not mechanically problematic.

@davidbrisbane7206

Жыл бұрын

Log triangles are good for the environment. They trap a lot of CO2 in them.

@chitlitlah

Жыл бұрын

"naturally" But natural logs are a bit easier to work with than other logs.

Fun fact: If you try the same thing with sine you will get x=pi/6 and with cosine x=pi/4

@joaomatos6598

Жыл бұрын

How?

@oom_boudewijns6920

Жыл бұрын

Probs u use trig. Identities

@AlchemistOfNirnroot

Жыл бұрын

For a sin(x), sin(2x) and sin(3x) triangle and then you got the cos(x) solution as a result of the trig identity?

@astha_yadav

Жыл бұрын

If u differentiate at the right triangle square law thing, removing the powers, then raise from e as powers removing ln , then solve the quad eqn u get 3/2 which is a correct soln Edit: actually there is some thing wrong with this method though i haven't figured out what I accidentally checked for lnx + ln2x = ln3x rather than the square form, so the soln is wrong Not deleting incase someone wishes to help out

@oom_boudewijns6920

Жыл бұрын

@@astha_yadav who asked🤣he asks about the trig version

You should change the side length to ln(3x), ln(4x) and ln(5x) to make people to remind of the famous 3-4-5 right angled triangle. :)

@artsmith1347

Жыл бұрын

WolframAlpha gives two solutions for log^2(3 x) + log^2(4 x) = log^2(5 x) x≈0.25848 x≈0.67166

@thexoxob9448

8 ай бұрын

The 0.25 solution doesn't work because 3 times that is less than 1.. which means length is negative, which is impossible

@MasterofNoobs69

7 ай бұрын

@@thexoxob9448it is possible with complex numbers, and you are then squaring it to make it real. The i-1-0 triangle is an example of absurd triangles you can create like this. The math works out, even if the geometry doesn’t.

@zzciobzz2963

5 ай бұрын

​@@thexoxob9448less than 1 isn't negative. it's between 0 and 1

The "Fading In" Intro is so much better!

@blackpenredpen

Жыл бұрын

Thanks!!!

5:12 the nice little detail about completing the square here is if you do NOT simplify ln(2) - ln(3) to ln(2/3), but add [ln(2) - ln(3)]^2 to [ln(3)]^2 - [ln(2)]^2 on the RHS, the [ln(2)]^2 will CANCEL 10:00 an approximation is x = 3.8549

@Jack_Callcott_AU

Жыл бұрын

@CaradhrasAiguo49 I agree, that is the number I got! 👍

@Rex-xj4dj

Жыл бұрын

I did that but still got about 2.45

@abhishankpaul

6 ай бұрын

I agree with your result

I finally managed to get to the solution of the problem all by my self I feel so proud, it is all thanks to your videos

@davidbrisbane7206

Жыл бұрын

I only feel relieved when I solve maths problems.

great video! I have been watching you since 2018 and your content is constantly getting better! good job mr. bprp.

@dimitrispapadakis2122

Жыл бұрын

Είμαστε συνονόματοι και έχουμε την ίδια εικόνα προφίλ :)

@junaidhasrat11

Жыл бұрын

@@dimitrispapadakis2122 don't tell me this is your alt account

@blackpenredpen

Жыл бұрын

Thank you!

@dimitrisg63

Жыл бұрын

@@junaidhasrat11 no hahaha

BPRP, you are my favorite math teacher. Thanks for another video.

@blackpenredpen

Жыл бұрын

Thank u!

0:00 Bro came from imaginary world to real world

@Kulshummaam

11 күн бұрын

😂😂😂

I started working this out myself until I reached the quadratic in LnX at which point I realized there was a much easier way to find the solution. All I had to do was watch the video and bprp would work it out for me :)

Wow, wonderful topic and excellent presentation!

this is a beautiful problem, your presentation is so impeccable I have watched it several times🙋🏻‍♂️

It's simply fascinating how the quadratic formula pops up like this. More than once a non-scientist/engineer/mathematician has said to me, "They made us memorise the quadratic formula in school. Why? Where will that ever be relevant?" And the answer is... well, everywhere! If you could pick only one formula to memorise, I think this would be a strong choice!

@Someniatko

Жыл бұрын

It's even better to understand how to derive this formula! It's pretty easy!

@cristianrdz7667

11 ай бұрын

@@Someniatko Yeah, is easy

Your videos are amazing!!

I've started watching your vids since a month and the way u explain is so cool. i could understand understand calculus at the age of 14 thanks to you! #yay

@blackpenredpen

Жыл бұрын

Glad to hear 😃

Thanks for this vid. I solved something similar inspired by this problem: instead the sides of the triangle were cosh x, cosh(2x) and cosh(3x). Took a lot of algebraic manipulation but the final answer was pretty cool. Maybe another video?

@otiswebb5783

Жыл бұрын

There are 2 real solutions for x

I saw your great older video on x^x. Would you consider making a video on plotting x^x (in 3 dimensions) for Real input and complex output? I tried to sketch the full 3d curve, with the x axis being Real and running perpendicular to the complex plane which is used for the output of x^x. So the x axis is the Real input; the y axis is the Real output and the z axis is the imaginary output. So the y and z axes form the complex plane output of the Real x input. So you have a simple exponential-looking 2d curve for positive x, it crosses the real y axis (or has a limit at x=0) at y=1, but the curve then becomes a complex shrinking spiralling "vase" shape for negative x. It's the "smoothness" of the curve as it crosses the y axis and changes from Real 2d to Complex 3d that I can't visualize. Would you consider making a video on this 3d graph and discuss the 3d smoothness of the real-complex transition at x=0 ? i.e. what's the limit of the 3d angle of the complex curve at x=0. AND: on this graph is x^x at x=0, a forbidden indeterminate point or is it equal to 1 ?

At 9:09 you discard the negative sqrt. But this leads to a positive value of x: (3/2) exp( -sqrt( ln(3/2) ln(9) ) ) = .536676; (3/2) exp(+-sqrt( ln(3/2) ln(9) ) ) = 3.854877

@shadowgamer6383

11 ай бұрын

Even though it's a positive value of x, the side length of the triangle which is ln x will become negative. And we can't have triangle with negative sides

@shreyaschaturvedi8851

10 ай бұрын

​@@shadowgamer6383exactly

0:00 Just a man coming out of the blue with a Blue pen and Red pen and a sweet Log problem for us .

The approximate value of x is 3.85488

@Smosh7i

3 ай бұрын

What about x = 0.583676

@computernerd1101

3 ай бұрын

@@Smosh7i That does work algebraically, but if x < 1, then ln(x) < 0. Geometrically, it doesn't make sense for the edge of a triangle to have a negative length.

I love these videos

This x is actually a solution to log(x)^2 + log(2x)^2 = log(3x)^2 for any log base greater than one. Bases between one and zero satisfy the equasion but they don't make a right triangle as log(x) would be negative. The other solution of the quadratic formula will give the right answer for bases between one and zero.

You are brilliant👍 ☺ I love your dedication

This was a really good log rule refresher lol

I used to hate math, but this guy has somehow an interesting way of explaining things, so I somehow just got hooked lol 😂

@-guitarhero

Жыл бұрын

same, he’s the reason I’m obsessed with math too

Just a tip for quadratic equations: use simplified form of the solutions when coefficient of the linear term is even as it was the case here, i.e,: ax + 2bx + c =0 => x= [-b +|- sqrt(b^2 - ac)]/a 😉 With a=1, even simpler x= - b +|- sqrt(b^2 - c)

@anastasissfyrides2919

Жыл бұрын

Much more preferable to divide by the common factor than memorizing yet another formula

@kangalio

Жыл бұрын

i know it as x²+px+q => -p/2±sqrt((p/2)²-q)

@antonyqueen6512

Жыл бұрын

@@anastasissfyrides2919 it’s not memorising new formula, it’s simplifying the 2’s

@NoNameAtAll2

Жыл бұрын

- b/2 you forgot to divide b by 2

@antonyqueen6512

Жыл бұрын

@@NoNameAtAll2 no I didn’t. That’s the whole point. It is simplified. You don’t have the division by 2. The coefficient of at x is even: 2b, thus -2b/2a= -b/a and sqrt[(2b)^2 - 4ac]/2a= sqrt[b^2 - ac]/a With the coefficient a of x^2 being a=1 you have the simplified solution as indicated in the comment above ☝️

Please please please please please do another marathon session. Really need it. Calculus. Maybe Laplace, Fourier, Bessel etc. Please?

Note: x needs to be greater than 1.5 as sum of two sides need to be greater than third side or the difference between 2 sides needs to be less than the third side.

Amazing creativity

I love how you just pop into existence in the beginning

Has anyone noticed the WIZARDRY at the first 3 seconds of the video?!? I haven't yet watched this but the first few seconds scared the bejezsus outta me. Why did they do that? The editor must have had a chuckle.

Thanks

Thank you

Dear bprp, I have a question, and I'd appreciate it if you solve it in your next video: Find the max/min value for sinA*sinB*sinC where A, B, and C are three angles in a triangle (A+B+C=pi) Thank you

@simonwillover4175

Жыл бұрын

Picks complex A, B, C

@bebizambi392

Жыл бұрын

Possible solution?: since A, B and C are angles which form a triangle, you could take C=π-(A+B). Then, sinC= sin(A+B) due to allied angles. Resulting expression is sinA*sinB*sin(A+B). I used maxima and minima for above expression using partial derivatives and got the answer.

@davidp4427

Жыл бұрын

Help me out here. A + B + C = 180° so 180° = pi ??? Am I missing something?

@nguyenphungdunganh3941

Жыл бұрын

@@davidp4427 radians since we're adults now

This video was soo satisfying, because I always realised what he was about to do, split seconds before he actually did it

The sudden chimpmunk voice jump scared me at 8:45

Almost the same as BPRP direct analysis, notice that: log(2x) - log(x) = log(2) log(2x) + log(x) = log(2x^2) from which the product gives the difference of squares, [log(2x)]^2 - [log(x)]^2 = log(2)log(2x^2) = log(2)[log(2)+2log(x)] ...eq(1) from the triangle pythagoras, [log(2x)]^2 + [log(x)]^2 = [log(3x)]^2 = [log(x) + log(3)]^2 ...eq(2) eq(2) - eq(1) gives, 2[log(x)]^2 = [log(x) + log(3)]^2 - log(2)[log(2)+2log(x)] a quadratic in log(x), let u = log(x), u^2 - 2[log(3/2)]u - log(6)log(3/2) = 0 solve for u using quadratic formula and your done x = e^(1/2){ 2log(3/2)+- sqrt[4[log(3/2)]^2 +4log(6)log(3/2)] } etc..

In my take, I factored out the 4 from the square root resulting in the product of the square root and 2 then I factored out 2 in the numerator and cancelled it with the 2 in the denominator. When you didn't do the same I expected you would have some twist so I was afraid if I have to rewrite it again.

Well educative

I believe I messed up somewhere along my working and don't feel like restarting. But from a number theory perspective, couldn't this be solved through Euclid's formula? Often used only with integers, but it applies to the real numbers too. If a^2 + b^2 = c^2. Where: a = m^2-n^2 = lnx b = 2mn = ln2x c = m^2+n^2 = ln3x We can raise everything to the power of e. Then rearrange for x in each equation. And set 3x to be equal to the sum of each equation. (3x = e^a + e^b + e^c). I'm not sure where to go from here though, but I haven't worked through far enough to think about that section, and I'm too lazy to do it since I already mucked up once.

Great stuff here. When I saw the thumbnail my first thought was IS THIS POSSIBLE? Any other type of functions we can use for the sides of the right angled triangle? What about e^x?

@oenrn

Жыл бұрын

He did e^x in another video.

I took me a while to notice how seamlessly he was switching between red and black and now I’m extremely jealous

what a incredible video..

I got the same answer, I wanted to check it before watching this video. Checking it with algebra by putting the solution back into the original equation proved difficult, much harder than the actual problem in fact.

If you solve the generalized problem of using sides ln(nx), ln((n+1)x), and ln((n+2)x), you get: x = (n+2)/(n (n+1)) * e^sqrt(2 * ln((n+2)/n) * ln((n+2)/(n+1)))

blackpenredpen always comes up with interesting problems.

That was actually a pretty fun problem.

nice job

I thought he said, "Love triangle" 😂🤣🤣

7:50 Best part: SHWOO!

X is approximately 3,8549 and is a transcendental number!

@shivamchouhan5077

Жыл бұрын

Actually it is 3.854765

@desiaasm

Жыл бұрын

@@shivamchouhan5077 Yeah I just rounded it mate

@shivamchouhan5077

Жыл бұрын

@@desiaasm But you added comma (,) instead of dot(.) So your answer was 38549

@nuclear3011

Жыл бұрын

@@shivamchouhan5077 in some countries (like Poland, where I live, for example) people use commas to mark the decimal point and use dots in big numbers e.g. 1.000.000

@shivamchouhan5077

Жыл бұрын

@@nuclear3011 Oh thanks for telling I didn't know that one, btw this can lead to calculations errors in some cases.

Tbh in calc 2 it wasnt the calc that got me but the occaisonal algebra trick

Hi thanks for the vid have a nice day it was great

Wow!! I solved it by a different method (but your method is much simpler and shorter) and got this answer x = 3^[a(a+ sqrroot2)] Where a = sqrroot {[log₃(3/2)]} I thought my answer is wrong ,but after using the calculator, I found that my answer is correct !! 🥳🥳

You forgot to distribute the square power in the b^2 of the cuadratic formula. (2ln(2/3)^2 is 4(ln(2/3))^2, not 4ln(2/3) as you say in the video

Now I'm wondering whether there are any "log-Pythagorean triples", i.e. integers a, b, c such that (ln a)^2 + (ln b)^2 = (ln c)^2. If there are, how would one go about finding them?

@Utesfan100

Жыл бұрын

Bonus points if you use Lambert's W function

Gracias.👍🏽

I love your videos so much! I wish as a young nerd interested in math I had such wonderful resources available to me. Unfortunately I was a young nerd is 80's America, before the internet was common let alone KZread and pretty much the worst time to be young nerd interested in math. :-/

After the second step just replace lnx by u and then continue that would be easier

I knew I was wrong when the answer I got was super long, roughly 3 times longer than the one you got. I checked both of their exact values to make sure it wasn't just a different way of expressing the same value and it wasn't :(

Hypotenuse and legs are on the both side of the triangle.

Yea, nice problem🥳🤔

you should simplify the exponential of the square root.

After a very exhaustive effort, I got the following solution: x=e^{-ln(2/3) (+/-) sqrt[2ln(3/2)ln3]}

Bprp's top 10 catchphrases 1. Let's do some math for fun! 2. Oh my god! Looks pretty crazy! 3. Wouldn't it be nice... 4. Don't forget the plus C! 5. Today, we have the integral of... 6. Let's go to the complex world! 7. I don't like to be at the bottom, I like to be on the top. 8. Bring this down down! 9. Don't worry, don't worry. 10. The best friend of the black pen is the red pen.

ln(3x)^2 = ln(2x)^2 + ln(x)^2 Then take the derivative of both sides above and cancel common factors ln(3x) = ln(x) + ln(2x) ln(3x) = ln(2x^2) ln(3x/2x^2) = 0 ln(3/(2x)) = 0 then raise to the power of e 3/(2x) = 1 x = 3/2

@AndreyNsk89

Жыл бұрын

Applying derivatives to both sides of the equation does not produce an equivalent equation. For example equation x = x + 1 doesnt have solutions, but if you take derivatives it will become 1 = 1, i.e. it is correct for all x.

@e.s.r5809

Жыл бұрын

The other issue here (besides the one Andrey pointed out-- I think you meant square root, not derivative) is that you still have to take (ln3 + lnx)^2 = ln3ln3 + 2ln3lnx + lnxlnx. (p + q)^2 =/= p^2 + q^2 sqrt(p^2 + q^2) =/= p + q (p + q)^2 = p^2 + 2pq + q^2 By Pythagoras (substituting lnx = z, ln3 = a, ln2 = b for the sake of everyone's sanity) we expand our brackets and reach: z^2 + 2az + a^2 = 2z^2 + 2bz + b^2 Gathering, simplifying, and substituting back a and b: z^2 + ln(4/9)z + ln(6)ln(2/3) = 0 As you can see, it's a quadratic with two solutions and no common factors to cancel. If you took the exponential of both sides now you'd reach an impasse (or at least something very gross). Remember you have to take exp of the entire expression, and by log rules, those multiplying lns would end up as powers: alog(b) = log(b^a) lnxlnx = ln( x^(lnx) ) 1 = exp{ ln[ x^(ln(x)) × x^ln(4/9) × ... Grim! 😅

Pretty cool algebra 2 problem.

Challenge: find x such that: log(ax)^2 + log(bx)^2=log(cx)^2 for arbitrary a,b,c

@europeankid98

Жыл бұрын

x = panda

I solved it by splitting ln(2x) into ln2+lnx and ln3x=ln3+lnx Then by pythagoras theorem we get (ln2+lnx)²+(lnx)²=(ln3+lnx)² (ln2)²+2(ln2(lnx)+2(lnx)²=(ln3)²+2(ln3)(lnx)+(lnx)² (ln2)²-lnsolved it by splitting ln(2x) into ln2+lnx and ln3x=ln3+lnx Then by pythagoras theorem we get (ln2+lnx)²+(lnx)²=(ln3+lnx)² (ln2)²+2(ln2(lnx)+2(lnx)²=(ln3)²+2(ln3)(lnx)+(lnx)² (2ln2lnx)-(2ln3lnx)+(lnx)²=(ln3)²-(ln2)² ln(x)(2ln2-2ln3)+lnx²=ln(3^ln(3)÷2^ln(2)) (lnx)²+ln(4/9)lnx-ln(3^ln3÷2^ln2)=0 Sub Y=lnx Y²+ln4/9Y-ln(3^ln3/2^ln2)=0 One solution is Y≈1.3493 Since Y=lnx X=e^Y=e^1.3493 X≈3.855 Other solution Y≈-0.5384 X≈e^-0.5384≈0.584

@captainkarma7374

Жыл бұрын

Please dont tell me you typed all that 💀

@alienbsg

Жыл бұрын

@@captainkarma7374 was tedious but that's how proofs are lmao

@asparkdeity8717

Жыл бұрын

X=0.58 not a solution as then ln(X) < 0 for one of the triangle sides

@NoNameAtAll2

Жыл бұрын

@@asparkdeity8717 why is that a problem? if x

@asparkdeity8717

Жыл бұрын

@@NoNameAtAll2 this problem represents side lengths of triangles, all of which must be positive, yet lnX < 0 so isn’t a solution; u just need one extra sentence saying this solution should be disregarded

Can you explain some math famous problems like the zeta function or something like that

@alikanan7011

Жыл бұрын

I like this idea

x ≈ 3.85488

0:00 It’s true, bprp has super speed.

cant I use the power rule for logs at 2(ln2)(lnx) so that 2lnx^(ln2) => lnx^(2ln2) => lnx^(ln4)?

@manavrana225

Жыл бұрын

That 2 goes to power of not power of lnx so it wiil be (ln(x²))^(ln2)

0:00 man just phased into existence to teach me math 😭

I was really hoping for it to work for all X

Can you do 100 Linear Algebra video

Your t-shirt made me think that the golden ratio will appear in the answer...

Let x= e^u for an easier time... Great video though :D

Sir some of your videos have a unique black board which can move up and down, I have seen such type of board in Oxford Mathematics youtube channel, can you please tell me name of the type of your that board???? In my country I never saw that. I want to build one myself.. take love sir

@stephenbeck7222

Жыл бұрын

Pretty common in older university buildings with large lecture halls. I think the videos you are talking about were filmed in a classroom at Berkeley, where Dr Peyam teaches. Michael Penn might be the KZread guy to ask about advice for building a chalk board.

Why don't you set y-lnx from the beginning?

Ln9 can be 2ln3 so u simplify 2 and 1/2 and you'll get (ln3)² and you simplify the square root and ull get e^(ln3) and you simplify more and you'll finally get 9/2

How to solve this integral question ? Integral [ ( 1/x^2 * (1+x^4)^0.5) ] dx ?

What song is used during the Brilliant ad?

Guys this solution works! My love triangle problem is gone, thanks to this.

Shouldn’t the final solution be x = (3/2)+e^(sqrt(ln(3/2)*ln9)) ?

This is not too hard. Just apply the Ln formula, solve a quadratic equation, and take the exp. A year 11 should be able to do it. Takes less than a page. Great exercise and great content.

I like crazy pythagorean triple questions. I have one for you... Can you find a pythagorean triple (a, b, c) such that (1/c, 1/b, 1/a) is also a pythagorean triple?

@charlesstimler9276

Жыл бұрын

The golden ratio rules!

@whocares12372

Жыл бұрын

What is the answer plz

@endeavourer1073

Жыл бұрын

.

Neat problem - more subtle than I at first thought - which was that you would be proving an identity. The solution can be summarized as follows : put a(x) = ln( x ) ; b(x) = ln( 2*x ) ; c(x) = ln( 3*x ) ; if a(x)^2 + b(x)^2 = c(x)^2 -> x = { 3*N/2, 3/(2*N) } where N = e^y ; y = sqrt(-2*ln(2)*ln(3)+2*ln(3)^2) -> N = 2.569917715.. x = { 0.5836762755, 3.854876572 } The open question is : when are mathematicians going to admit that not only are calculators here to stay but also math software ?

that's ok!

What about exponential triangle problem, exp(x), exp(2x), exp(3x) ?

@davidhowe6905

Жыл бұрын

I tried this just now; first of all, thought it was impossible - then noticed my basic algebra error! I got x = 0.2406 (4 decimal places). Similar method; use Pythagoras then simplify to get quadratic in exp(2x) giving exp(2x) = (1 + sqrt(5))/2 (I think this is correct)

you know the class is gonna be good when the teacher literally teleports in the room

3.90(aproximately)

Nice good job blprp

you can simplify a litle further because ln(a)ln(b) = ln(a^b) so ln(3/2)ln(9) = ln((3/2)^9) and then you can find x=3/2 * e^sqrt(ln(19683/512)), looking at that now I can actually see why you didn't but I don't wanna waste the time I spent making this comment

*appears out of nowhere* *starts explaining math*

If u are bored solve this find range of a for all value of y lie in R as y = (ax²+3x-4)/(3x-4x²+9) . . . . Ans- a€(1,7)

At 7:01 isn’t wrong? Because 2/3 /6 is equals 4. It’s same as 2/3 / 1/6 so we flip second and change the operation into multiplication

@Noctarc

Жыл бұрын

0.6666.../6 = 0.1111... not 4

@D7mh76

Жыл бұрын

@@Noctarc oh yes i get it now, thank you. I wrote it as 2/3.1/6 but forget that it is 2/3//6-double lines is for longe division line-.

dang the intro is smooth

@blackpenredpen

Жыл бұрын

Thanks!

Idea for the math for fun: calculate sin(e) This will be literally 'approximation of sine'