integral of x^x vs integral of x^ln(x) (aren't they both impossible?)

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In this video, we will integrate x^ln(x) with a special function, erfi(x), see here for more details • how WolframAlpha defin...
and I will also give my proposal to the integral of x^x
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[@latt.qcd9221]

When you can't solve an integral, just call it a new function and name it after yourself.

[@FootLettuce]

No you can't just define a function yourself and call it the solution to the integral... BPRP: haha, x^x integral go X2(x)

[@cosmicvoidtree]

I think someone had actually proposed BPRP(x) as the integral of x^x. I don’t remember who, but I’ll see if I can find the video

[@kono152]

Bprpi(x)

[@deananderson7714]

@@FootLettucebro it was a joke

[@justsaadunoyeah1234]

((sqrt(pi))/2)bprp(x) = \int x^x dx ​@@kono152

[@sirajb6363]

blackpenredpen? blackboardwhitechalk?🤔🤔

[@blackpenredpen]

Siraj B Lol

[@erikkonstas]

White is the blackboard's black...

[@erikkonstas]

@@abp4739 Check whom he's replying to again...

[@blackpenredpen]

"between you and I" or "between you and me"?

[@MaksymCzech]

Between us :)

[@matthewg.6262]

blackpenredpen it’s “between you and me” bc without the “you” part, you wouldn’t say that something is “between I.” you would say “between me.”

[@zinoovic5947]

@@MaksymCzech lol thats what I was gonna say

[@edgarddjahoui3381]

You can try a U-sub to solve this question.

[@jannesl9128]

Wouldn't it be cool if it's "between substitution-letter and sqrt(-1)"? (:

[@pinchus2714]

I think Ti(x) for tetration integral. What's the integral of Ti(x) or X2(x)

[@ahusky4498]

Whoa there buddy, relax. The government doesn't want us to know that

[@blackpenredpen]

I like this very much!!! Maybe Ti_2(x) so that we can extend this to more

[@angelmendez-rivera351]

Unfortunately, Ti(x) already stands for the inverse tangent integral in most standard notation and software, so this is just a failure.

[@pinchus2714]

@@angelmendez-rivera351 It's Ti_2(x)

[@jamez6398]

Or you can have tetrX_n(x) for the number of times you tetrate x by x in the integral.

[@jzanimates2352]

Can we see more of this X function in future videos please? It interests me!

[@blackpenredpen]

JZ Animates I sketched its graph too. Will upload the video in a day or two.

[@jzanimates2352]

blackpenredpen awesome!

[@cpotisch]

@@blackpenredpen You never did.

[@MegaTitan64]

Oh my gosh, that's such a good idea! I'm so glad that you thought of the X function. :)

[@blackpenredpen]

MegaTitan64 thanks!!! Someone suggested Ti_2(x) for tetration. I think it’s pretty cool too

[@shaochen5821]

Riemann has entered the chat Riemann has left the chat shamefully

[@sergioh5515]

I am definitely happy.. I am enjoying the use of non elementary functions!

[@blackpenredpen]

Sergio H glad to hear!!!

[@angelmendez-rivera351]

Whether they are only non-elementary or not is in the eye of the beholder. When you work with these functions often enough, they become elementary to you. It's much like the ln and the sin and cos. They used to not be elementary.

[@emmanuelpaygar3736]

Can differentiate and still obtain x^lnx ??

[@emmanuelpaygar3736]

@@blackpenredpen can you differentiate and still obtain x^lnx ?

[@ozzymandius666]

Excellent. You are now encouraging folk to explore parts of calculus that don't exist in any texts, who knows what interesting things might be discovered?

[@roros2512]

7:34 I think that C is not necessary because the definition of X2(x) is a definite integral, so the constant is already in the 0 limit thanks for the videos, they are great

[@martindf6831]

If you assume the constant of the 0 limit you are just writing only one function because X2(0) is a simple number. You need that C because C can be every number not just one. I'm not sure if I've explained myself well. Regards from Spain

[@angelmendez-rivera351]

roros2512 It is necessary, because while X2(x) is a definite integral, the antiderivative of x^x is nor.

[@angelmendez-rivera351]

Martín dF Ditto.

[@GreenMeansGOF]

I thought that this was going to be a proposal video for your girlfriend.😂

[@blackpenredpen]

GreenMeansGO lolllll

[@androidlg7311]

He is already married to math

[@latt.qcd9221]

It is a proposal video for his girlfriend. His girlfriend is the integral of x^x.

[@78anurag]

@@latt.qcd9221 Ouch 🤕 Press F to pay respects to Bprp

[@VibingMath]

Man u have a new name now which is "X-man" for defining the X function! Can't wait to see your Integral of x^x in coming videos! Your idea should be documented and written as a math paper!

[@yaleng4597]

AHHHHHHHH! THE BLACKBOARD!!! Haven't seen it for a long time.

[@blackpenredpen]

Yea!!! I recorded in my office yesterday

[@alexdemoura9972]

whitechalkredchalk

[@jayapandey2541]

#Supreme

[@AlBoulley]

"I don't know… yeah." Love that outro.

[@SmileyHuN]

So, let's just summon Dr. Peyam for a little help!

[@CengTolga]

Now let's define 𝑋𝑛(x) for rational numbers. Then generalize it to real numbers, then to complex numbers.

[@niccologeraci9175]

Why hasn't anyone come up with a special function for the integral of x^x yet?

[@novidsonmychanneljustcomme5753]

Maybe because there's no practical use of it in mathematics/physics/etc. so far, so it doesn't need a special name...? But this is only an assumption, correct me if I'm wrong.

[@12semitones57]

niccolo geraci I don’t think that defining a special function that is the integral of x^x would have any practical use other than amusement.

[@angelmendez-rivera351]

novidsonmychannel justcommenting You are correct. Contrary to most non-elementary antiderivatives, such as the polylogarithms, the trigonometric integral functions, the elliptic integral, the logarithmic integral, the exponential integral, the Dawson integrals, the cumulative distributive functions, and the Fresnel integrals, the antiderivative of x^x has no use in practicality. Even the well-known sophomore's dream identity is relatively useless in applications.

[@joshuahillerup4290]

@@novidsonmychanneljustcomme5753 this is math, practicality is very much incidental here.

[@angelmendez-rivera351]

Joshua Hillerup Not necessarily. Unless you are studying some very obscure abstract mathematics, all of mathematics originates from the necessity to solve some problem for an application, whether that problem is for a mathematical application or a pragmatic one.

[@MercuriusCh]

blackpenredpen , I suppose the name of this function could be Ladi_2(x) , u know, Ladder two Integration function , cause we have a ladder of two x (like x^x). Of course, the number of x is index of this function and can be changed according to an amount of x)

[@angelmendez-rivera351]

mercurius channel The correct terminology for this is tetration. Ladder already has a potentially different usage in mathematics.

[@MercuriusCh]

Angel Mendez-Rivera , ok, thanks for your attention)

[@Craznar]

For X^X - I tweeted you an idea, why not use limit (t->x) of integral (x^t)?

[@blackpenredpen]

Christopher Burke I saw but I was not sure. I don’t think we can do n approaches x.

[@skylardeslypere9909]

I don't think you could do that since x is a variable and not a fixed number

[@Craznar]

@@skylardeslypere9909 both versions have solutions on Wolfram Alpha ... limit (t->x) integral (x^t) and limit(t->x) integral (t^x).

[@PlutoTheSecond]

What about lim (t->0) of integral x^(x-t)?

[@angelmendez-rivera351]

Christopher Burke Yes, but they don't mean the same thing. Also, this is an integral, so by introducing limits, you are introducing problems due to the lack of uniform convergence. Best to leave it with the original definition.

[@fanyfan7466]

Some interesting thoughts I guess. X_1(x) would just be x^2/2 right? B/c integral of t dt from 0 to x. X_0(x) I’m assuming would just be x, as I’m guessing the zero would be the integral of 1. There is a way to extend tetration to -1, which is 0. So X_-1(x)=0. There isn’t a good way for other negative integers, since the next one, -2, blows up to -infinity. For positive, rational values, we can do something similar to roots for exponents. Like, ^(.5)x (idk how else to write that) would be the inverse of ^(2)x aka x^x. So X_.5(x) would be the integral of e^W(ln(t)) from 0 to x. You could do that for any rational number. For irrational numbers, just do closer and closer rational approximations, like how we do with exponents. So, X_e(x) for instance could be defined. So, now we could potentially throw any positive number in for n in X_n(x) and it would have some meaning. What do you guys think?

[@ethannguyen2754]

With special functions, it becomes incredibly easy to define indefinite integrals for any function. It kind of feels like cheating.

[@ahusky4498]

Instead of X, use T for tetration. Keep the idea for the 2 and 3 subscripts.

[@skylardeslypere9909]

Ti(x) for tetration integral And then Ti_2(x) and Ti_3(x) etc ***not my idea***

[@angelmendez-rivera351]

Skylar Deslypere As I said elsewhere, Ti(x) is already notation reserved for the arctangent integral function, so cannot be used for x^x.

[@skylardeslypere9909]

@@angelmendez-rivera351 oh really? What would be the definition for the arctangent integral function? I've never heard of that one befroe

[@skylardeslypere9909]

@@angelmendez-rivera351 I can't find it anywhere

[@angelmendez-rivera351]

Skylar Deslypere The definition would the integral of arctan(t)/t from 0 to x. It gets used a lot to derive polylogarithm identities. It also serves the role that Si(x) serves, but with inverse trigonometric functions.

[@pedronaga1]

If I'm not mistaken the integral of x^x will result in an infinite series involving the incomplete gamma function. I think I still have the solution of it in my computer, if I find it I will post the result later.

[@TheDeaththereaper]

Can’t you approximate x^x using a Taylor series instead

[@sergioh5515]

How would you go about that?

[@sergioh5515]

Actually I guess it's possible if we rewrite the bases (maybe it'll work) but idk for sure..also the point was to derive a more exact or more pure (or a more analytical solution)...like an explicit function, not a series.

[@user-dz1px2mc7o]

You can approximate using a double sommation series : www.quora.com/Does-the-integral-of-x-x-exist/answer/Siddhant-Grover-12

[@angelmendez-rivera351]

メ乇しム尺٥ ㄈ That is not really an answer for our purposes, though.

[@angelmendez-rivera351]

neon underwood There is no Taylor series you can define for x^x unless you use complex numbers, but in that case, why would you bother to use a Taylor series and not a Laurent series? Also, a series is not really an answer.

[@General12th]

I love it! I will shout your proposal across all the lands!

[@TheSenator007]

I would call X_n(x) the Power Tower Integration Function (PoTowIn Function) of degree n.

[@203_sayantansen5]

You are an inspiration

[@benjaminbrady2385]

Tetration integral! Ti_k(x) = integral of kth tetration of t dt from 0 to x

[@blackpenredpen]

Benjamin Brady yea that’s a good idea!! Someone else mentioned that too. And someone else mentioned that ti(x) is for tangent integral... so I am not sure...

[@benjaminbrady2385]

@@blackpenredpen oh well... Anyway, the name isn't too important. It's still a cool function!

[@pranay7264]

Gotta love the enthusiasm he has.

[@yaleng4597]

I think Ti is better for tetration(or tower if you like XD) integral

[@angelmendez-rivera351]

Yale NG That notation is already reserved for the inverse tangent integral.

[@semi8883]

This'll take me a while...I'll try to see how it goes on paper, and then I'll come back to watch the video. Of course I'll try to solve x^x too! Shouldn't be worse than dealing with x^(-x).

[@angelmendez-rivera351]

Since you invented a new function, time to study it! First, I want to give credit to the user Void for the notation I am going to use to represent your function. This notation is Oi(4, 2, x). 4 refers to the fact that tetration is the fourth hyper-operator. 2 refers to the height of the hyper-operator, in this case of tetration. x refers to the upper bound of integration. In other words, Oi(n, m, x) is defined as the integral from t = 0 to t = x of H(n, m, t), where n is the hyper-operator number, m is the height of the hyper-operator, and t is the base of the hyper-operator. Now, for some special cases. Oi(4, 0, x) = x, since t^^0 = 1 for all t, and Oi(4, 1, x) = x^2/2, since t^^1 = t for all t. Little known is the awesome fact that t^^(-1) = log(t), so Oi(4, -1, x) = x·log(x) - x. Oi'(4, n + 1, x) = x^Oi'(4, n, x) trivially follows for all integer n > -2, which is every integer in the domain. Oi(4, 2, 1) has the special value calculated in the sophomore's dream identity. Question: what is the limit of Oi(4, 2, x)/x^x as x -> ♾? By L'Hôpital's theorem, the limit is the same as the limit of x^x/[x^x·(1 + ln(x))] = 1/[1 + ln(x)], and this goes to 0 as x -> ♾, implying that Oi(4, 2, x) ♾ of x!/[x^x·e^(-x)·sqrt(2πx)] is 1, so the limit in question is the same as the limit of [Oi(4, 2, x)/x^x]/[e^(-x)·sqrt(2πx)]. We can use L'Hôpital's theorem here. The derivative of the numerator is [x^(2x) - x^x·Oi(4, 2, x)·(1 + ln(x))]/x^(2x) = 1 - x^(-x)·Oi(4, 2, x)·(1 + ln(x)). We can use L'Hôpital's theorem to know what Oi(4, 2, x)·(1 + ln(x))/x^x approaches in the limit. The limit of [x^x·(1 + ln(x)) + Oi(4, 2, x)/x]/[x^x·(1 + ln(x))] = 1 + Oi(4, 2, x)/[x^(x + 1)·(1 + ln(x))] as x -> ♾ is obviously 1, so the limit of 1 - Oi(4, 2, x)·(1 + ln(x))/x^x is 0. The derivative in our numerator is -e^(-x)·sqrt(2πx) + e^(-x)·sqrt(2π/x), which still approaches 0. We can use L'Hôpital's theorem again on our derived limit to get -{[x^x·(1 + ln(x)) + Oi(4, 2, x)/x]·x^x - x^x·Oi(4, 2, x)·(1 + ln(x))^2}/x^(2x) = -{(1 + ln(x)) + [Oi(4, 2, x)/x - Oi(4, 2, x)(1 + ln(x))^2]/x^x, and now clearly the numerator goes to infinity while the denominator will still go to 0, since it is approximately exponential relative to x^x, which means the limit goes to infinity, meaning x!

[@deejayaech4519]

You might be able to actually. x^(x+1)dx = x^2*x^(x-1) = x^2*(e^xln(x))/x, we might be able to integrate (e^(xln(x)))/x in terms of Ti2 and Ei where Ti2 is the antiderivative of x^x

[@angelmendez-rivera351]

@@deejayaech4519 I have looked into it, and that seems extremely unlikely. There are no results available anywhere that suggest that this is possible, and actually trying integration by parts does not get you there within any reasonable amount of steps.

[@deejayaech4519]

@@angelmendez-rivera351 integration by parts gets you an infinite series if you apply it nievly. You have to do some rearanging after xti2(x) (e^xln(x)/x looks similar to e^x/x which does have an integral as a special function, Ei(x). But if it isnt possible with only elementry functions, common non elementry ones, and ti2(x), that rases the question of what functions are needed to integrate it.

[@venkatbabu186]

One is divergence and other convergence. To read a particular frequency you use convergence sequence and to transmit you use a divergent. Something uses both the phenomenon using motion for convergence and divergence. Vectors. Some use power sequence to do that. Which is like integral and differential.

[@angelmendez-rivera351]

Venkatesh Babu Ponnuchamy WTF?

[@wabc2336]

u²+u completing the square is one method Another method: u² + u = u(u+1) = (u+0)(u+1) = (u+1/2 - 1/2)(u+1/2 + 1/2) = (u+1/2)² - (1/2)² = (u+1/2)² - 1/4 Not very practical and only obvious in this instance, but it's nice how the same answer can be reached in another way

[@montano0222]

Me: *Doing nothing and watching math videos for fun* ThatTutorGuy: *In every video* Allow me to introduce myself

[@cyrus1774]

I love the video! I think with x^ln(x) you can just do a u sub with u = ln(x) right from the beginning, you don't have to rewrite it.

[@blackpenredpen]

Ahhhhh you are right!!!!! I am just so used that step lol

[@thedoublehelix5661]

That's what I did :)

[@neutrino5695]

Can you make a video about Airy function, please?

[@karolakkolo123]

Yes!

[@bouzianehamza4468]

Blackpenredpen , Dude you taught us what is li (x), Ei (x), Errf (x) but but ..how to calculate Ei (2) or Errfi (6) ? this has an answer?

[@karolakkolo123]

Yes, but to those values I suppose are transcendental and not representible with elementary functions, so the only way to calculate them is through infinite series representations for the integrals

[@volodymyrgandzhuk361]

So now we can integrate any linear function raised to itself. Given int (mx+q)^(mx+q) dx, with m!=0 (which is to be read as "m differs from 0", not "m factorial equals zero"). Now, we can multiply and divide our function by 1/m, getting int 1/m*m*(mx+q)^(mx+q) dx. And we can also take 1/m out of the integral: 1/m int m*(mx+q)^(mx+q) dx. Now we can make the u substitution, namely: u=mx+q, du=m dx. Then we get: 1/m int u^u du=1/m X_2(u)+c. And back to the x world: 1/m X_2(mx+q)+c.

[@alexdemoura9972]

Hyper-4 Tetration may have the following notation: ⁿa = a^^n = H₄(a,n) So why not use the Hyperoperation integral notation? Hi₄(x,n) = ∫₀ˣ H₄(t,n) dt Also since 0⁰ and Ln(0) are indeterminate forms, may we suggest start the integral from 1 instead of zero? Hi₄(x,n) = Xₙ(x) = ∫₁ˣ H₄(t,n) dt

[@matteoferraretto1012]

I really appreciated the blackboard!

[@user-fu1hc8ij3u]

Can you please proof the connection between hyperbolic function and exponential function

[@angelmendez-rivera351]

زهره الامل He already did

[@user-fu1hc8ij3u]

Please give me the title

[@DjVortex-w]

Can I use this same approach in all math tests? I just define a new function that's defined as the answer to the question at hand, and give it as the answer.

[@blackpenredpen]

WarpRulez lolll I don’t think it will work on exams.

[@DragonLoad49]

Such a clever trick to get started!

[@nickcampbell3812]

When he said "Square root of pi over four" and wrote sqrt(pi)/2 I was a bit confused. Dis he mean only like sqrt(pi/4) and just simplify the root 4 to 2?

[@JoseWui]

You know it’s serious when he breaks out the chalk and blackboard

[@GreenGoblinCoryintheHouse]

I love your videos. Keep up the good work .

[@sanjaybalaji9206]

Sir nice hair cut😄..........sir can u do some questions on newton's lebnitz theorem and limit as a summation?.....pls......

[@tom_szcz_org]

Maybe a good name for the mystery special function can be (n)pti(x) - power tower integral; n is an integer in superscript, just like shown @ 5:37 indicating the size of said power tower. PS. Love your vids and interesting math problems you present on your channel 😁

[@bloodyadaku]

While there is no elementary function that is a solution to the integral of x^x dx, it is still possible to find an exact value for any definite integral of x^x dx using power series.

[@angelmendez-rivera351]

bloodyadaku Well... only if you call non-closed-form summations "exact." Most mathematicians don't. Series typically don't count as an answer. If they did, every function would be non-elementary.

[@bloodyadaku]

@@angelmendez-rivera351 That's fair. I don't actually know the math behind it; I haven't gotten that far in my mathematics career yet. So I guess I used the wrong term.

[@dayzimlich]

white chalk red chalk yayyy!

[@soupe2000]

Let's call X2 the blackpenredpen 2 function.

[@oscartroncoso2585]

HAVE A GOOD DAY BPRP 🔥

[@blackpenredpen]

Thanks, you too Oscar!

[@7quantumphysics]

You should check out Knuth's "up arrow" notation :)

[@blackpenredpen]

7quantumphysics yea I know about that. That’s a really interesting one.

[@Mystery_Biscuits]

What about Tet(x)? Short for tetration

[@angelmendez-rivera351]

Mystery Biscuits I approve.

[@mhersaribekyan1487]

Blue chalk Red chalk White chalk

[@danji9485]

Now integrate the function for the slope of my life going downhill.

[@chrisesca9927]

Just did it, its infinite 😂

[@bhavyajain01]

Can u please solve x^(-1/2) exp(-ax) integral in limit -infinity to infinity Or in limit 0 to infinity I'm so much struggling with this...

[@kostas71374]

Is there a course or a bundle of courses propably with the same instructors like this guy that you will learn from absolute scartch level to advanced and be able to solve almost any integrals series ect . I am studying calculus 1 (mathematics 1) which are about functions,limits,series,integrals and calc 2 which is differential equations. I am studying from notes and books and sometimes watch videos for a topic.

[@crazyduck2379]

Dumb question (i warn that i may make enormous mistakes since i haven't properly studied calculus yet, i just know a thing or two about integrals):when we have e^lnx² wouldn't it bd like e^lnx times e^lnx, and if i dodn't miss someting particular e^lnx is always equal to x, therefore shouldn't e^lnx^2=x^2? (Once again i apologize profusely to those who studied the subject whose eyes have probably started bleeding from the thousand different errors i most likely made, i just had to ask)

[@DiegoMathemagician]

Chalkboard??

[@blackpenredpen]

Yea, I couldn’t find a classroom so I recorded this in my office.

[@arikahn3907]

What about a Tylor series expansion x^x solution?

[@NitronNeutron]

How satisfying seing you at a blackboard even if your channel name does not make sense now.

[@mariadelrosariogonzales7390]

Amigo!!!, vos tarde o temprano vas a integrar hasta las integrales que no se podian integrar!!!, jajaja!!!, genio!!!

[@etemkaandelibas3649]

I want to learn formulas and their proofs. Where can I find books about them ? (I have just learnt integral calculus)

[@santiagoargayo6409]

I want to know too. Have you found something?

[@princeiitb1415]

Cengage

[@ritujithmanoj2133]

You should use Hagoromo

[@vitakyo982]

I've been looking at x^x long time ago , but never did find a primitive , good luck ...

[@LDzNik]

Can you make a video about the integral from the gamma function? In my version it's x!psi(x) - k(x) + C, where k(x) = integral from x to 1 of the function x!psi(x) - Kappa function

[@joshuahillerup4290]

I like it. And my first thought was wondering if/how fractional integrals would work with this

[@helloitsme7553]

So the integral of x^x is not findables in terms of other integral based defined functions interesting

[@angelmendez-rivera351]

HelloItsMe Correct

[@lffb8803]

I feel like i'm wrong, but if you do the integral from 0 to x of x^x, won't that cause a problem because 0^0 = undefined?

[@user-nt8kt3tm1i]

It's not a problem. One point isn't important for a value of an integral, if there is no infinite limit. Limit of x^x for x->0 is 1.

[@tiziocaio101]

I have a challenge for you: sqrt(-x^2-4x)=1-x^2

[@tobiashartmann5421]

x≈-0.23677 x≈-1.7266 x≈0.98170 - 1.2175 i x≈0.98170 + 1.2175 i

[@angelmendez-rivera351]

Tizio Caio Simply square the equation to get -x^2 - 4x = x^4 - 2x^2 + 1, and rearrange to get x^4 - x^2 + 4x + 1 = 0. This is a depressed quartic, which one can solve with a standard algorithm. Rearrange to get x^4 - x^2 = -(4x + 1). Add 1/4 to the equation to get x^4 - x^2 + 1/4 = -(4x + 3/4). Notice that x^4 - x^2 + 1/4 = (x^2 - 1/2)^2. Now, add to the equation the expression 2y(x^2 - 1/2) + y^2 to obtain (x^2 - 1/2) + 2y(x^2 - 1/2) + y^2 = (x^2 - 1/2 + y)^2 = 2yx^2 - y + y^2 - 4x - 3/4 = 2yx^2 - 4x + (y^2 - y - 3/4). Find y and z such that 2yx^2 - 4x + (y^2 - y - 3/4) = [sqrt(2y)·x - z]^2 = 2yx^2 - 2·sqrt(2y)·zx + z^2 is a perfect trinomial square. In order to obtain a perfect trinomial square, y and z must satisfy 2·sqrt(2y)·z = 4 and z^2 = y^2 - y - 3/4. Thus 8y·z^2 = 16, hence y·z^2 = 2, implying y^3 - y^2 - 3y/4 = 2. This polynomial equation is equivalent to 4y^3 - 4y^2 - 3y - 8 = 0. Then z = sqrt(y^2 - y - 3/4). To solve this cubic equation in y, let y = w - (-4)/[(3)(4)] = w + 1/3. The result is 4(w + 1/3)^3 - 4(w + 1/3)^2 - 3(w + 1/3) - 8 = 4(w^3 + w^2 + w/3 + 1/27) - 4(w^2 + 2w/3 + 1/9) - 3w - 1 - 8 = 4w^3 - 13w/3 - 8(1 + 1/27) = 4w^3 - 13w/3 - 8*28/27 = 0. This implies that w^3 - 13w/12 - 56/27 = 0. Now, we use Cardano's method, by letting w = v - u, with vu = -13/36 and v^3 - u^3 = 56/27. v = -13/(36u), so v^3 - u^3 = (-13/36)^3/u^3 - u^3 = 56/27, implying u^6 + 56u^3/27 + (13/36)^3 = 0. Finally an equation we can solve trivially! Using the quadratic formula, u^3 = -56/54 (+/-) sqrt[(56/54)^2 - (13/36)^3]. Assuming +, v^3 = -56/54 + sqrt[(56/54)^2 - (13/36)^3] + 56/27 = 56/54 + sqr[(56/54)^2 - (13/36)^3]. This means w = cbrt{56/54 + sqrt[(56/54)^2 - (13/36)^3]} - cbrt{-56/54 + sqrt[(56/54)^2 - (13/36)^3]} = cbrt{56/54 + sqrt[(56/54)^2 - (13/36)^3]} + cbrt{56/54 - sqrt[(56/54)^2 - (13/36)^3]}. This means y = 1/3 + cbrt{56/54 + sqrt[(56/54)^2 - (13/36)^3]} + cbrt{56/54 - sqrt[(56/54)^2 - (13/36)^3]}. The next step is to simplify y before proceeding. 56/54 = 28/27, and (28/27)^2 = 784/729, while (13/36)^3 = 2197/46656. 46656/729 = 64, since 36^3 = 9^3*4^3 = 3^6*64 = 27^2*64, so 784/729 = 50176/46656. This means that (28/27)^2 - (13/36)^3 = 50176/46656 - 2197/46656 = 47979/46656 = 9*5331/6^6 = 3^2*5331/216^2 = (3/216)^2*5331 = 5331/72^2, hence sqrt[(28/27)^2 - (13/36)^3] = sqrt(5 331)/72. Thus, y = 1/3 + cbrt[28/27 + sqrt(5 331)/72] + cbrt[28/27 - sqrt(5331)/72]. From the cube roots, you can factor 1/27 to get y = 1/3·{1 + cbrt[28 + 27·sqrt(5 331)/72] + cbrt[28 - 27·sqtt(5 331)/72]}. You can rearrange the radicand to have a common denominator of 72, which has a factor of 8, which you can factor out of the cube roots, resulting in y = (1/6)[2 + cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}]. This is the most y can be simplified. Now, we can calculate 2y = (1/3)[2 + cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}], y - 1/2 = (1/6)[-1 + cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}], and y^2 = (1/36)[4 + 2·cbrt{[2 016 + 27·sqrt(5 331)]/9} + 2·cbrt{[2 016 - 27·sqrt(5 331)]/9} + cbrt{[4 064 256 + 729*5 331 + 4 032*27*cbrt(5 331)]/81} + cbrt{[4 064 256 + 729*5 331 - 4 032*27*cbrt(5 331)]/81} + 2·cbrt{[4 064 256 - 729*5 331]/81}] = (1/36)[30 + 2·cbrt{[2 016 + 27·sqrt(5 331)]/9} + 2·cbrt{[2 016 - 27·sqrt(5 331)]/9} + cbrt{98 155 + 1 344·sqrt(5 331)} + cbrt{98 155 - 1 344·sqrt(5 331)}]. Therefore, y^2 - y - 3/4 = -1/4 - (1/9)[cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}] + (1/36)[cbrt{98 155 + 1 344·sqrt(5 331)} + cbrt{98 155 - 1 344·sqrt(5 331)}]. This is perfect, because s = sqrt(2y) = sqrt{(1/3)[2 + cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}]}, and z = sqrt(y^2 - y - 3/4) = sqrt{-1/4 - (1/9)[cbrt{[2 016 + 27·sqrt(5 331)]/9} + cbrt{[2 016 - 27·sqrt(5 331)]/9}] + (1/36)[cbrt{98 155 + 1 344·sqrt(5 331)} + cbrt{98 155 - 1 344·sqrt(5 331)}]}. t = y - 1/2 was already calculated. With all this, we can now go and solve (x^2 + t)^2 = (s·x - z)^2. This equation implies x^2 + t = s·x - z and x^2 + t = z - s·x. Rearranging both, we get x^2 - s·x + t + z = 0 and x^2 + s·x + t - z. Using the quadratic formula to finally solve, we get x = [-s (+/-) sqrt{s^2 - 4(t - z)}]/2 and x = [s (+/-) sqrt{s^2 - 4(t + z)}]/2. Both formulas combined give x = [(+/-)1 s (+/-)2 sqrt{s^2 - 4t (-/+)1 4z}]/2. I already found the values of s, t, and z, so all which remains is substituting into this formula. You can bet your life I will not complete this last step. You can just call s, t, and z a group of special constants, and here are the solutions in terms of those constants.

[@tiziocaio101]

Angel Mendez-Rivera thank you so much!

[@mrmimeisfunny]

I like it, but I would've liked some example values. Like I know X₁(x)=xx/2 and X₂(0)=0 but what is X2(1)? I know you can compute it but I don't really know how. Also I propose using ж₂ and ж₃ instead because the Cyrillic alphabet is so underutilized in math it's criminal.

[@sinosodialajay797]

Hlo sir I am stuck in a problem which is graph of y= (-1)^x Please help me because i am confused I know this will not have more points on x,y axis but will also have curve on iota axis

[@arnabacharya349]

You know e^(i*pi) = -1, Hence, y = (e^(i*pi))^x = e^(i*pi*x) which is the unit circle in the complex plane

[@sinosodialajay797]

@@arnabacharya349 o h wow thanks Very good

[@cabbageman]

Your capital X function is a lot like "up-arrow notation", what you wrote as X2(x) = x↑↑2 = x^x. Heres a link for further reading: en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

[@thedoublehelix5661]

Technically nothing wrong with defining a new special function. My one problem to defining away x^^2 is whether or not it would be useful to consider such a function. Does X2(x) give us any insight to the underlying integral or help us solve problems in a meaningful way? I feel that X2(x) is just an unnecessary wrapper to the integral.

[@angelmendez-rivera351]

The Double Helix It's not really useful. Tetration isn't use in any applications other than googology.

[@Dionisi0]

4:08 when 4=2=e=Pi

[@ppereztorres]

name idea: Tower Function. With T sub2(x) for x^x

[@blackpenredpen]

Paulo Pérez Unfortunately That notation is taken by the Taylor polynomial with deg 2

[@angelmendez-rivera351]

It's also taken already. Ti(x) and T2(x) both refer to the arctangent integral.

[@sharanya2045]

I was having a bad day and I felt so bored. Not anymoreeee

[@sahilbaori9052]

:/

[@TheDannyAwesome]

How do you know all the X_n are independent?

[@engr.rimarc.liguan1795]

Master. What does it mean by erfi(t)?

[@plaustrarius]

Would the bounds of big x have to start at some other number? 0^0 shouldn't be definable. I'll have to try it out, great video!! Also big ups for the hyper-operation notation!!

[@volodymyrgandzhuk361]

x^x is defined in (0, +infty). The parenthesis mean that this bound is excluded. So, it's not defined in 0. If a function is differentiable in a point, then it's continuous in that point, but the opposite is not always true. So there's no reason for excluding 0 from X_2(x)

[@erikcastro6840]

Gracias chino, buen video

[@spudhead169]

How about the integral of x^-x (or 1/x^x)? WA says this converges on the definite 0-infinity interval so the indefinite integral must be possible right?

[@ismailnurmaghribi1702]

Is there any proof for showing it's constant (or +c) was equal to 0...?

[@volodymyrgandzhuk361]

c can be any real number

[@beyondscience004]

Thats interesting.i'll try that new idea

[@alondayan7740]

Great idea

[@unitywithpatience2559]

No problem I will solve that integral in future for you😇

[@meme_engineering4521]

White chalk, red chalk

[@joseantoniosanchesgarcia9220]

i have a challenge for you, can you solve an integral with the left hand and a differential equation with the right hand at the same time?

[@elix1350]

Can't we take ln both sides? so it becomes ln x multiplied by x? and go through it using integration by parts?

[@matrixstuff3512]

Why not try looking at the power series to see if anything can be done in that way?

[@matrixstuff3512]

You could also look ad the derivatives of your function and figure out what differential equation it solves and maybe express it in terms if some sort of bessel function

[@angelmendez-rivera351]

Kyle W It doesn't solve any differential equation that you can relate to a Bessel function. And the only way you can do this with power series is by using nested summations.

[@matrixstuff3512]

@@angelmendez-rivera351 I realized that (both) once I sat down and looked at it. But I still think looking at it as a solution to a differential equation might be useful

[@IronAceSUB]

I'm working on taking the derivative of the gradient function of f(x,y)=x^y over the direction vector . I think this allows me to substitute y's for x's since we're on the y=x line. I just want the function of the intersection of z=x^y and y=x... maybe I just need to use a system of equations instead of all this calculus 😅

[@noether9447]

Why does the integral of x^x at 0 is considered to be 0? It could have been anything??.

[@justabunga1]

Even though at x=0 is not defined at that point but has a limit that goes to 1, it is not an improper integral. Both limits of integration have the same endpoint. That's why it is always equal to 0.

[@mathswithpana]

Hello there. How can I use Wolfman. I have installed Wolfman Mathematica Installed 11.3

[@meiz1795]

4:08 "Square root of pi over *4*"

[@alexnguyen8394]

hey blackpenredpen do you have an email that I could send questions or concepts to?

[@elliottmanley5182]

I suggest HI(x) for Hyperoperation Integral.

[@azizlokhandwala2345]

BlackredPen my proposal is to check whether lnx /(x^x) is integrable from 1 to inf or not

[@volodymyrgandzhuk361]

Let's call it Chow's function!