Try this one next: kzread.info/dash/bejne/mq6o065xYZOalJM.html

@aronbucca6777

2 жыл бұрын

Have you ever made a lesson-like video where you prove the rules of derivation and integration? I think it would be interesting

@masternobody1896

2 жыл бұрын

Lol 😆 that 1 guy want hard problem

@NeymarJR-du9rb

2 жыл бұрын

Here’s an impossible question of trigonometry: Find the value of x: (1/sin2x) + (square root 3/cos3x) = -1/square root of 3 Looks fine but none of our school maths teachers could solve so do give it a go!

@motherisape

2 жыл бұрын

Where is sandwich guy

@That_One_Guy...

2 жыл бұрын

I wonder if there's a way to solve polynomial-trig or exponential-trig function

Stop giving hard questions. You're making me feel like I'm bad at math.

@jacobjay4915

2 жыл бұрын

😂😂😂

@panashemahadzva4278

2 жыл бұрын

😂😂😂😂

@lumiere7623

2 жыл бұрын

😂😂😂

@japinabear6594

2 жыл бұрын

😂😂😂

@harrymetu2746

2 жыл бұрын

😂😂😂😂

Wow randomly scrolling through KZread and seeing this guy used to be my math teacher a year ago glad to see you are successful on KZread you are a great professor.

@siddharthagotur7449

2 жыл бұрын

Bruh he's been successful on youtube for years

@pierreaoun9600

2 жыл бұрын

@@thebigbradwolf me when nothing ever happens🙄

@joshuagonzalez3880

2 жыл бұрын

You went to pierce huh

@RonnieRjbj

2 жыл бұрын

@@joshuagonzalez3880 still go yeah

@JayTemple

2 жыл бұрын

I assumed that he was great in class, but it's cool to hear it confirmed!

I absolutely love math when I'm not the one doing it.

@mr.nobody1738

Жыл бұрын

🤣🤣🤣

@UnknownPerson-nl7te

Жыл бұрын

Bruh

@chichobar1705

Жыл бұрын

B r u h

@riyansiam6669

Жыл бұрын

Same 😂

@dhruvrathi3746

8 ай бұрын

B R U H

I just started the integration chapter in calc 1 today, can't wait to understand most of the stuff on this channel!

@ieatgarbage8771

2 жыл бұрын

Oh this particular video is almost entirely algebra

@W2wxftcxxtcrw

2 жыл бұрын

ill pray for you

@Maagiicc

2 жыл бұрын

Integration isn't too bad, good luck nevertheless

@mightycahjo4906

2 жыл бұрын

@Del Squared - دل تربيع Masya Allah, brother👍

@62mohadshaikh68

2 жыл бұрын

@Del Squared - دل تربيع yes i am also learning it... 😀😀 Thanks brother ❤️❤️

As an Engineering student and having to deal with these problem solving, I am glad that I am already done with calculus with the help of this guy's math marathon. Thanks man.

@Vase0I0

2 жыл бұрын

Don't think this is your typical engineering problem dude... 😂

@whannabi

2 жыл бұрын

@@Vase0I0 yeah, they'll be dealing with more excel sheets than actual problem solving

@piemaster831

2 жыл бұрын

@@whannabi hey man excel sheets are great for problem solving

@MrKostaspapagalos

Жыл бұрын

​@@whannabi How many excel sheets do you think it took to design and develop the platform and device you are using to watch your favourite math videos and comment your "educated" opinion ?

@sea-dyedblossom238

Жыл бұрын

@@whannabi bu-but... excel sheets are beautiful 🥲 **hugs my excel sheets tightly**

This video gives us a very important lesson that many ML practioners overlook- Context is important in Math. When it comes to fields like Machine Learning, people sometimes blindly apply techniques without evaulating context. The part about 1 not being a valid solution encapsulates that perfectly.

@wborden

Жыл бұрын

That lesson applies so well to life

@keescanalfp5143

Жыл бұрын

well the problem is perhaps actually even simpler as simple can be. at 1.54 bprp says: ..and maybe we can multiply with ln(x) on both sides .. (!), without excluding explicitly this factor being zero. right here he himself introduces an extra root, namely the case ln(x) = 0. well may we think that leaving a denominator the denominator if including a function of the unknown, that's algebra for beginners.. ?

I like functions like the Lambert W function. When there's an equation you can't solve, just invent a new function :D

@mathsman5219

2 жыл бұрын

Same With logarithmic Function. And actually same with the solution for x+2=4

@itsphoenixingtime

2 жыл бұрын

i mean, there is a point between "Inventing a new function will help us with other fields" and "hiding an approximation behind a new function that is created for the sake of solving an equation" right?

@chimaeria6887

2 жыл бұрын

I hate the Lambert function. It's just BS trying to find a 'solution' for something not possible otherwise.

@That_One_Guy...

2 жыл бұрын

@@chimaeria6887 so you're saying logarithm and inverse trigonometric function is BS too, what a fool.

@goombacraft

2 жыл бұрын

@@chimaeria6887 What, you mean like logarithms?

I was just playing around with number guesses yesterday in a sleep deprived state, and happened to find that pi^(1/3) is a very close approximation to the value you get for x. Not sure what this means, but it's cool!

@ericthegreat7805

Жыл бұрын

This equals (1/sqrt(pi))^(-2/3). IIRC this has to do with the Gamma function

@blackbomber72

Жыл бұрын

There is probably an approximation formula lurking around

@ericthegreat7805

Жыл бұрын

@@blackbomber72 oh maybe sterlings formula!

@opinionshurt2905

Жыл бұрын

I like pie.

@redisforever6952

Жыл бұрын

@@opinionshurt2905 same

Thank you for making these videos. I watch daily just to enjoy the beauty of integrals. Well explained and engaging!

I'm starting to understand things from your videos, which (for me) is incredible because before i didn't understand at all. Keep up the good work sir 👍

Great question! Loved your introduction of Lambert W function here..

The different branches of Lambert W correspond to the complex branches of the logarithm. An easier way to identify the solution, and the fact that it's unique, is graphically: we are looking for where the function f(x) = x^2 - 1 - 3 ln x takes the value 0 in the range x>0. This function tends to infinity as x-> 0+ and as x-> infinity, and by elementary calculus has a unique local minimum at x=(3/2)^(1/2), where f is negative. Hence (by the intermediate value theorem) it must have a zero on either side of the minimum. On the left, there is the easy solution x=1, which can be ruled out because (as noted in the video) it doesn't solve the original problem, which requires ln x to be non-zero. So the required solution must be bigger than square root of 1.5, and also less than 2, since f(2)>0. To get a better approximation, set x=1+y and expand as a Taylor series about y=0: f = - y ÷ 5/2 y^2 - y^3 + O(y^4). (This is a convergent series for |y|

Nothing more ominous then 36 boxes of expo markers sitting in the background.

@lornacy

2 ай бұрын

😂

Been learning cal one on my own watching this dudes vids is always an inspiration to keep on learning. Good luck with your maths chaps !!!

@blackpenredpen

2 жыл бұрын

Thanks. You should also check out the channel “just calculus” for calc 1 tutorials. That guy is okay. 😆

@xirsixussien7303

2 жыл бұрын

Checkout Proffesor Leonard. I am Electrical Engineering major and I have learnt calculus 1 from him. I am currently learning calc 2

What a fun little problem. It was nice to see the non-principle branch get utilized for a change!

Woow this was sick!! Missed these crazy videos a bit on this channel between all the Calc 1 uploads recently!! :) great vid!!

@blackpenredpen

2 жыл бұрын

Thanks! I am teaching calc 1 this semester and the last time I taught it was 2 years ago. So my mind is full of calc 1 these months.

I was so close to solving it! I knew I had to use the product log function and I knew there would be some tricky branch stuff and I got it so close to the required form but I couldn't work out to raise both sides to the -2/3 will certainly keep it in mind for next time

Fascinating video. Frankly, I did a naive geometrical interpretation and got 1.5. I pretty much went back to the definition of what an integral was as the "area under a curve" and realized if I just made the "curve" a straight line and "integrated" it's just the area of a box with a base of 2. Using X = (1.5)^(1/t) obviously just cancels out the t's and makes it a constant being integrated - thereby giving me 1.5t from 0 to 2 which gives 3. Apparently, that's a fairly close approximation, but I don't know if I just go lucky.

If you plug y = (x^2 -1)/ln(x) and y = 3 into a graphing calculator and then use the intersect function you get 1.464. This video is still super cool because he got an exact answer using some math I didn't know existed, so keep up the great work!

@DungeonNumber5

Жыл бұрын

And you, sir, were using the "good enough" approach, finding the practical solution fast. Math nerds may hate it but engineering of any kind (except maybe nuclear and space tech) is about the "good enough" values. Enjoyers of PI = 22/7, all aboard!

@zayedelahee2166

Жыл бұрын

@@DungeonNumber5 pi=3

@donmoore7785

Жыл бұрын

So we just need to carry around graphing calculators? As a math teacher, I know that isn't true.

@Ninja20704

Жыл бұрын

@@donmoore7785You can do it with a normal calculator as well. Just use like Newton’s method or something. Most scientific calculator’s have an iteration function so that once you set up the equation u can just spam the equal sign until the decimal places dont change.

@spirat14

Жыл бұрын

​@@donmoore7785 As a math teacher, you should know that desmos is a thing that exists. So yes, not only should we, we already do.

This was my introduction to the Lambert W function. Thanks! 👍 😊 I took a summer course in Special Functions years ago but the whole course was on Laplace transforms! 😠

I always forget about the fish, and even if I'd remembered, I'd have given up on it if the principal branch hadn't worked. But I rapidly came up with a reasonable answer using Newton's method, and I was satisfied. It is always fun to see the fish, though.

I feel good after seeing that you got x as + - 1 and I instantly thought of picking another branch of lambdaW function after using numerical methods to find the answer

Thank you. I did need in an exam the Lambert function, concerning the final exam of hydraulic centrals in engineering. No one reached me this function before. Hard to believe.

7:38 It was at this exact point I went "Oh no, we are going into the multi-value complex logarithm stuff, aren't we?" 😆

I love it when answers to seemingly simple problems use special functions. Can you do more problems that can be solved using special functions?

@aca4262

Жыл бұрын

🤓👈🤣🤣🤣

@Nino-eo8ey

Жыл бұрын

@@aca4262 What's wrong? You're in the comments of a math video. No idea if you were overwhelmed by the topics or something else.

@aca4262

Жыл бұрын

@@Nino-eo8ey what's wrong with what's wrong of my comment?

The answer checks out intuitively, as integral of 1^t from 0 to 2 would give 2, and integral of 2^t from 0 to 2 would give (4 - 1) / ln 2 =~ 4.32. So the answer feels like it should be just under 1.5.

@remnantotaku81

2 жыл бұрын

Interesting, I never thought of looking at it like that.

@zidannadiframadhan419

2 жыл бұрын

yeahh of course

@soupisfornoobs4081

5 ай бұрын

Nice intuitive approximation for 3/ln2 XD

I had a 99%-sure hunch Mr. Lambert was hiding somewhere behind one of those curtains in this problem! Thanks, this was fun! Fred

I really enjoyed watching that. Thank you.

At x2-1= 3lnx , you can also use graph method(plotting the equation graph and checking for any intersection of the two equation) also to find the solution.

@princetavishspan7849

Жыл бұрын

Good luck doing that with a pen

Some of my math teachers were obsessed with giving extremely long and tedious problems, so that I had to work on each problem for like 30 minutes, and some math teachers were much more forgiving and gave problems that could be solved in like 5 minutes if I knew what I was doing (like concise u-substitutions in Calculus 2, for example).

@Speed001

2 жыл бұрын

I disliked having to do u-sub after u-sub. Like 3 times. Though all of the technical math did teach me to stay weary of online calculators as there were rare occasions where they would get the wrong answer I think.

@somebodyelse9130

Жыл бұрын

@@Speed001 Doing Calc I and II, i never encountered a problem where the online calculators gave me an incorrect answer; I only encountered problems which they couldn't solve. Those were interesting, but frustrating, because those problems couldn't be solved by bluntly applying the antiderivative techniques as explained in a textbook, but required some slight deviation that the calculator didn't account for. This was really rare, though. I only found one or two problems it actually couldn't solve.

@Peter_1986

10 ай бұрын

One thing that I like to do is to try to use MatLab for as many math problems as possible; this is a great way to practise that program, while at the same time studying the math courses themselves.

I love how e is irrational, but the first 9 digits after the decimal misleads you to think e is rational and 1828 repeats forever.

@TNTErick

10 ай бұрын

that is exactly the reason why i say 2.7182818284590452 when introducing the thing to my friend

@yurenchu

8 ай бұрын

​@@TNTErick You're missing a decimal 7 there, my friend.

@TNTErick

8 ай бұрын

@@yurenchuThank you, my friend. I did miss a 7 there

this is amazing. To answer the title as well, I definitely could not solve it before watching :D

I didn't get as far as the ^-2/3 step. As such, my application of W didn't help. I tried taking a quadratic in terms of x, and that rearranged back to the same equation so I did a quadratic in terms of 1. That made it worse.

by multiplying with ln(x) we actually introduced the wrong answer (1) into the equation. Thats why I gotten into the habbit of always excluding Zeros, I could potentially multiply with while doing stuff to equations, at the Time I'm introducing them.

@danielyuan9862

Жыл бұрын

You should always have a separate case for them, too. Because they could still be solutions, but would need to be treated differently to solve.

thanks! please, explain more in detail the branches of the Lambert function, I got lost there

Whenever the math has exponentials as key part, integers become just as messy as e is by the integers standards. It's like oil and water. It makes sense that the answer is this crazy.

I don't know what is more impressive, the equation itself or the way you switch your marker.

I saw the thumbnail and tried it on paper for myself. I haven't studied the Lambert W function before, so I didn't make it that far. Instead, I actually did use Newton's method but my pocket calculator was imprecise and when plugging the result (something like 1.3ish) in to the integral's answer, I got 2.something rather than 3. That's when I went "wait, what if the question mark is a function and not some constant?". But then it's easy to put something like x=at^(1/t) and get some value for a that solves the original question. Would be interesting to try to find the families of all functions for which this can work I suppose. Great question, though, I had fun!

Cool example on "multiple" inversion of non-monotonic functions. I see there are eterogeneous comments about the option between the use of " standard" formal w function and the "direct" evaluation by means of numerical methods. Anyhow, if you study real calculus through a theoretical approach, you will see that the inverse of a continuos function is still continuos and therefore closed intervals are mapped to closed intervals (intermediate value theorem); as a matter of fact, the inverse function g(y) of a continuos function f(x) has to be always evaluated by solving equations of the type f(x)=y , i.e. F(x,y)=f(x)-y=0, with respect to the x symbol; the Dini's theorem provides all the theretical background (for instance, it describes also the derivability properties). As a matter of fact, it is due to a merely conventional tradition the fact we call the inverse function ln(y) of exp(x) "elementary", and the inverse function w(y) of x exp(x) "non elementary"... in both cases the theory states their continuity and derivability properties that can be exploited for efficient and reliable computation. For instance, for any "smooth" function (like x exp(x) ) we can simply express its inverse as a Taylor power series expansion by means of the Lagrange theorem. Obviously we can alternatively employ iterative methods (e g. Newton, dicotomic algorithms, Caccioppoli-Banach contractions...) whose correctness is founded on the continuity and completeness properties of the real topology.

@anshumanagrawal346

2 жыл бұрын

The Lambert W "function" is not a real function

@anshumanagrawal346

2 жыл бұрын

It's just make belief

@iWrInstincts

2 жыл бұрын

@Adrian Martinez Dorsett I was thinking the same thing... I am not sure what his point is.

@peterdecupis8296

2 жыл бұрын

@@anshumanagrawal346In your opinion shall the function f(x)= x*exp(x) map a real interval of x to non-real sets🤔? Perhaps in this context, we are not interested in the complex polidromic inverse function of z*exp(z)...

@anshumanagrawal346

2 жыл бұрын

@@peterdecupis8296 I don't understand what you're saying

Funny I actually was able to solve this as I encountered the Lambert W function as a young lass while I was personally interested in the equation nᵏ = kⁿ, n ≠ k. Was pretty surprising to me that there's such deep maths behind a seemingly innocent equation.

@astinnugs

Жыл бұрын

thats why the fishy has a evil face

What my approach is... integrate then.. on right hand side lnx and other will populate on left hand side . Make graphs of both and find interesting points.

This was a really fun problem to watch you solve!

It was new to me (interesting). Thank you so much *bP🖋️rP🖍️* ❤️

Aha,😂 I was confident until the first half , until you told thn answer was wrong. Needless to say I am not yet familiar with W Lambert functions. Yet, a good stretch. Thank you

I prepared for my maths exam from your 100 integrals in one shot video thanks alot 🤩

I just used the Newton method. Although, the xo approximation of 1.1 converges at 1, you get the other root of y=x^2-3lnx-1 by starting with xo=1.5. It converges to 1.464251632 with just a couple of iterations

@adrianyaguar7666

2 жыл бұрын

exactly. Newton's method is better in that it also gives an approximate result, but in a much simpler way. Anyway, the W function is very interesting :)

@TehMansYT

2 жыл бұрын

You’re so good wow

@tylerl_08

2 жыл бұрын

No idea what the fuck you just said

@pi_xi

2 жыл бұрын

W(x) is also calculated numerically, for example with the Newton method.

@eugeneimbangyorteza

Жыл бұрын

I did the exact same thing

Nice one! I love the run-through of the wrong branch of W. :)

Lol I was watching your (infinity-infinity)^infinity video when this came out

Awesome as always!!

@blackpenredpen

2 жыл бұрын

Thanks

Hi (sorry for my bad English), I accidentally saw your video and got interested in the task. As I understand it, the value to be found does not necessarily have to be a constant. Therefore, I derived the function by iteration. This function: (3t/2)^(1/t). When searching, I started from the property of degrees. We need to get rid of from t. And then choose an expression that, when integrated, will give 3. I spent about 50 minutes. ((3t/2)^(1/t))^t => 3t/2 => (after integration) ((3t^2)/4) from 0 to 2 => 3*4/4 - 3*0/4 => 3

I've not idea about maths. Literally. I only did 1 bach (spain) and i forgot everything. But i watched a lot of your videos. Dunno why. Dunno if you're doing it well or you're failing. I'm just enjoying this. Its like maths are incredible when you don't do exams about maths

Just a non-maths student wondering: Is it legit to apply W(0) on LHS to get -2/3 x^2, while applying W(-1) on RHS?

@asparkdeity8717

2 жыл бұрын

No, u can’t apply different branches of W to both sides of an equation. Like above clearly W(0) [-2/3 e^-2/3] isn’t equal to W(-1) [-2/3 e^-2/3] so u should treat W(0) as a different function to W(n) for any other branch n

@curtiswfranks

2 жыл бұрын

I think that the LHS with the other branch will resolve to the same thing because it is a symbolic expression, but I am not entirely sure. Really, this should be repeated for every branch. Then, each yielded x should be plugged into the integral and verified. But any nonreal x can be immediately discarded due to the restriction of the domain assumed at the beginning (and the fact that we are not using complex analysis techniques in the actual integration). So, I suspect that there are only a finite handful of real-valued solutions x to the W equation, all of which I further suspect were shown here. We eliminated the bad ones, so only the good ones remain, if so.

@curtiswfranks

2 жыл бұрын

Yeah, -1/e ≤ x ⇐ (W(x) ∈ ℝ, & x ∈ ℝ). Moreover, x < 0 ⇒ y ∈ {W₋₁(x), W₀(x)}, where: y e^y = x.

@EebstertheGreat

2 жыл бұрын

The W function can be understood as a multifunction, just like arcsin. There are infinitely many values of z that solve the equation sin z = 1/2 (for instance), but only one is between -π/2 and π/2, so we pick that one as the principle branch and say arcsin 1/2 = π/6. But if all we know is that x is a real number and that sin x = 1/2, it is invalid to conclude that x = π/6, because it could be π/6 plus any multiple of 2π. Suppose you were given the problem "solve sin x = 1/2, 1 Well actually even before that, we introduce two spurious solutions. We know that x² - 1 = 3 log x is well-defined and true if and only if the original integral equation holds. But after that point, he exponentiates both sides of the equation, neglecting the cases where log x is undefined. That's how we get the invalid solutions. At this point, we should note that the original equation clearly can't hold for any x ≤ 1 and from that point on assume x > 1. Doing so makes the next few steps valid. We know that -2/3 x² e^(-2/3 x²) = -2/3 e^(-2/3) if and only if the original integral equation holds (because x > 1). In the next step, he applies the W function to both sides of the equation. That step is valid, but only in one direction. It is true that if a = b, then W(a) = W(b) (using the same branch on both sides), but the reverse is not necessarily true, because the W function is not injective. Note that if you plug in x = 1 or x = -1 to the equation at the start of this paragraph, it does hold, so these spurious solutions came in earlier like I said. The problem here is not that new wrong solutions are introduced but that old right solutions are discarded because we are only looking at one branch.

@EebstertheGreat

2 жыл бұрын

@@curtiswfranks You can just look at the graph and see that there are exactly two real solutions when -1/e < α < 0, which is what he was trying to explain at the end. So you only need to look at the 0 and -1 branches.

Just curious, which particular math class is Lambert W function taught? This is the first time I've ever heard of this function, and I've been in college for 3 years working with math classes.

@shrujanbeesetty9722

2 жыл бұрын

first time i came across it was actually in physics for quantum (schrödingers) but it was just additional information that wouldnt be tested so i never really cared about it

@MKWiiLuke4TW

2 жыл бұрын

it’s not really in the curriculum I don’t think, maybe it comes up in some graduate level numerical diff eq courses but I haven’t seen it

@gasun1274

2 жыл бұрын

even throughout a math degree program this function isn't used much. ive only seen it widely used in this channel.

@RH-ro3sg

Жыл бұрын

Not even all mathematicians encounter it. At least, I don't recall encountering it during my formal master or phd education, but I have to say I was specializing in a very different field (related to combinatorial optimization). I only came across it when I was just fooling around with some equations (nothing to do with my work whatsoever) in my free time in a computer algebra system and it gave me a solution in terms of the Lambert function, which was when I decided to check it out.

@soupisfornoobs4081

5 ай бұрын

We were told about it in 11th grade maths, just as the inverse of xe^x. Dunno why it's so rarely taught

The right mixture of understanding and wondering to me!!

Taking pre calculus rn and just went over inequalities. This is gonna be a rough journey

Hard question: solve for pi without "approximating" it and/or assuming it can't be done. Think about using an annulus composed of 4a units squared and reversing the square per quadrant.

@andrewhone3346

7 ай бұрын

There are lots of ways to approximate pi. Archimides came up with some of the first (dividing up a regular polygon with 2^n sides into triangles and take increasingly large n), and this was taken quite far by ancient Chinese mathematicians. Then Ramanujan (early 20th century) came up with a much better approximation method based on modular forms, and later this was developed further into what I believe is the best algorithm due to the Chudnovsky brothers, who calculated many millions of digits on a supercomputer they built in their New York apartment. There has maybe been an additional improvement in the method (Borwein? Bailey?) but I haven't followed it closely. In any case, this amount of precision is irrelevant for any practical purposes - it is more used as a benchmark problem to test the performance of new computers.

@ongoldenpi

7 ай бұрын

@@andrewhone3346 You didn't read what was asked. Solve for pi WITHOUT approximating it and/or assuming this can not be done. It is possible to use the Pythagorean theorem to do it & express the answer as a ratio of integers/non-. Archimedes' n-gon method produces the wrong answer & 3.14159... is not the circumference of a circle whose diameter is 1.

The architect used the same equation to eliminate Neo but not 3

Fantastic integral but I very much enjoyed your presentation. 👍

omg this video literally taught me W function and its features

Comment: Stop giving 1+2 BPRP: Okay, so find √π!

As someone who is yet to be actually taught calculus, let me just be happy with chemistry and x = _/(3/2)

Can you do more detailed videos about the lambert W function branches?

I love it. You should submit a version of this in next year's Putnam.

@imabstrong3726

2 жыл бұрын

imo this is not at the putnam level. It as a routine calculation if you know about the lambert W function, which forces there to be a solution. I think putnam tries to be much more elegant than this

This is very easy man , you are making me feel like I am super good at math 😂😂😂

@blackpenredpen

Жыл бұрын

😂 good work!

Could you try solving f'(t)=2f(2t+1)-2f(2t-1), f(0)=1? (Non constant solutions)

I tried to visualize it in my head and come up with a quick guess before watching and got sqrt(2) which is actually suprisingly close.

The desired function f=e*(3/(e^2-1))^1/t The solution is obtained without using the Lambert W-function. To find the primitive of f^t, you need to answer the question, which function, when differentiating, will give a power function: (f^t)'=f^t*(ln(f)+t/f*f') It remains to solve the differential equation ln(f)+t/f*f'=1

@leif1075

2 жыл бұрын

Yea but isn't that cheating since if you don't know about this function there's no way to really derive it?

@Andrew-xz6sg

2 жыл бұрын

@@leif1075 This is just one of the options that came to my mind right away. I found a whole class of functions suitable for this integral. For example: f=[(3*n/2^n)*t^(n-1)]^1/t; n>0 f=[3/tan(2) * cos(t)^-2]^1/t ... etc.

Did you create this question or did you find it somewhere. It's a good one.

@blackpenredpen

2 жыл бұрын

I came up with it.

8:10 "negative one is just outrageous" xD

Watching him pause after he finished writing down the equation felt reassuring that people don't just immediately solve stuff like this.

If you get a problem like this on a standardized test, you can approximate the answer very closely in 20 seconds with the regular power rule: Derivative of the function: t(x)^(t-1) (integral from 0 to 2) = 2x^1 = 3. X = 2/3 or 1.5, which is very close to the correct answer of 1.46. If that's your closest option by a wide ballpark on a multiple choice test, you'll know what to choose. Some tests obviously will make the problem much more difficult to approximate in your head like that, but cut corners where you can if you want to finish in time.

These is a easier solution. (ln2^(1/t)*e^(ln2))^t This simplifies To ln2*e^(ln2*t) Which when you integrate from 0 to 2 You get ln2/ln2 [4-1] =3

Very helpful! Thank you!

Keep it up sir ! Now your questions level are good ....

The integral is equal to (?^2 - 1)/ln(?), so the equation can be rewritten as (?^2 - 1)/ln(?) = 3. This is equivalent to ?^2 - 1 = 3·ln(?) = 3/2·2·ln(?) = 3/2·ln(?^2). Let x = ?^2. Hence x - 1 = 3/2·ln(x), which is equivalent to 2/3·x - 2/3 = ln(x), which is equivalent to e^(-2/3)·exp(2/3·x) = x, which is equivalent to exp(-2/3) = x·exp(-2/3·x), which is equivalent to -2/3·exp(-2/3) = -2/3·x·exp(-2/3·x), which is equivalent to -2/3·x = W[-1, -2/3·exp(-2/3)] or -2/3·x = W[0, -2/3·exp(-2/3)] = -2/3, which means x = -3/2·W[-1, -2/3·exp(-2/3)], or x = 1.

Sir , please can you explain how to find range of the function f(x)=sqrt of (x-1)/(x²-1). Please I am very much frustrated 🥺

@gustavosenna7338

2 жыл бұрын

You have to factor out (x-1), since x²-1 = (x+1)(x-1)

@kuldeeepmeena3804

2 жыл бұрын

@@gustavosenna7338 Brother ,That's all I know !

@kuldeeepmeena3804

2 жыл бұрын

@@nextgaming4666 sorry , but this is wrong , correct range is,{0

@kuldeeepmeena3804

2 жыл бұрын

@@nextgaming4666 my answer is also same ! But it is wrong

@Convergant

2 жыл бұрын

\sqrt{\frac{x-1}{x^2-1}}=\sqrt{\frac{x-1}{(x+1)(x-1)}}=\sqrt{\frac{1}{x+1}}(x eq 1) f(x) has no real value for x-1(+), f(x)->infinity. The function monotonically decreases to 0 as x->infinity, and excludes 1/sqrt(2), so: range: (0, infinity)\(1/sqrt(2))

i procrastinate doing my maths homework with your channel!

love from india. Your videos makes the maths feel like magic, they are very informative and interesting.

Can you please try to find the integral of I={[x(π+49)]¹⁵/⁷}/π²(x^π+7)?

@thecuriouskid4481

2 жыл бұрын

Hehe, a famous one

@eresoup7229

2 жыл бұрын

Yikes

@iyannazarian866

2 жыл бұрын

@@thecuriouskid4481 can you elaborate on this a bit more ?

@dane4kka

2 жыл бұрын

@@iyannazarian866 because you can't really integrate a t^k/(t+1) dt, when k is arbitrary. But if you notice that 15/7 is quite close to pi-1, you can find an easy solution to this integral. Besides, even wolfram alpha proofs that the brute force result using hypergeometric functions is oddly accurate to our quirky substitution.

1.46425…. But to be honest, solved using Desmos.

I was going round in circles rearranging for ages thinking, this is just a number, I must be able to solve it haha >.< I was too lazy to do a graphical method although I did think about that. Nice! Now I'm curious about lambert w functions! Thanks, subbed!

Such a cool solution, to a seemingly easy problem at first glance

We should allow X to be a function of t :D Let x(t) = 2*(ln(2))^(1/t) (x(t))^t = ln(2)*2^t => the integration is just 2^t evaluated from 0 to 2 = 4 - 1 = 3 :D

Damn maths is hard but there is always ways to make it easy.

Try 0->1 integral of x^(xe^x) dx Bernoulli integral

Name a more iconic duo than blackpenredpen and lambert function.

How do calculators approximate values when using the Lambert W function? Is there some sort of infinite series or something of that sort?

@chaosredefined3834

2 жыл бұрын

Newton-Raphson method, probably.

@pf32900

2 жыл бұрын

Wikipedia has a page on the Lambert W function: en.wikipedia.org/wiki/Lambert_W_function

First thoughts: (x^2-1)/lnx=3 x^2-1=3lnx (x-1)(x+1)=3lnx Finding the zeros.of LHS and RHS we get that x=1 Maybe there're more roots, these are the first things that came to my mind

@ffggddss

2 жыл бұрын

But x = 1 doesn't work, because substituting it back into the original integral gives 2 = 3. Fred

Its very easy. Just set the RHS to integral of (3/2)t. This resolves X to e^(ln(t)/t). Or simply Tth root of t. Simple. Ignoring 3/2.

Rewrite your equation as x = sqrt(3 * ln(x) + 1). Iterate on a spreadsheet and x settles around 1.464252 for x with an initial value > 1.

I am stuck at, e(?)³ = e^(?)²

@shanthil6799

2 жыл бұрын

1??

@unhatchedegg9013

2 жыл бұрын

@@shanthil6799 0 also

@peteschupp4545

2 жыл бұрын

Just take ln and you get x^3 = x^2. You get 0=x^2 (x-1) so either x^2 or x-1 has to be 0 so you get x=0^x=1

I like hard questions !!

So I approached the problem differently and got a close answer, was wondering if the approximation was the cause of my error or not. If you follow the definition of an integral which is the surface under the function, wouldn't x = 1,5 be good since you are integrating with the born 0 to 2. Simple math but 1,5*2 = 3 and you could imagine it as two rectangle under the curve with a surface of 1,5.

Could you please make a seperate video for the 'Fish' function .. Glad if you do

Don't touch my math

Hey! Thank you for video! Where did you buy the picture on the wall? I really like it :D

@blackpenredpen

2 жыл бұрын

I designed it. You can check out my merch store in the description if you are interested

@user-be9ew1cx9e

2 жыл бұрын

@@blackpenredpen Thank you! You're a master of design :3

Such kinda equations involving exponential or log can also be solved by using Newton-Ralphson method. The same solution I have got using Newton-Ralphson method, by solving the equation x^2 - 1 -3lnx = 0 => x = 1.46.

Damn that’s crazy. Its one part about math ir really like, to explore questions and solutions like this and explore the meaning behind things like that with the professor. Not the useless, repetitive same questions for homework that just take time to solve but could by done by any idiot. Not the annoying thing of being forced to write so much and solve everything by hand. Its about exploring new interesting things instead of being bored to death with.. yeah solve this shit over and over until you can do it in your sleep. We‘re starting to learn some new things at uni thankfully but the bs is that we still have to do the same old shit in the so called practice groups we‘re actually graded on.

For the case : The integral of x^t wrt t. It implies that the integral of such form yielding the variable part-> x^t/Ln(x) is defined or valid for all non-zero (x≠0) positive integral (or generally positive reals) values of x except 1 or if Ln|x| is to replace Ln(x) so that the function is continuous for all negative integers( generally negative reals) as well, it must still be the case such that the domain of the function is restricted as 1