How to find all solutions to the famous equation x^x=y^y (with Lambert W function)

[@blackpenredpen]

Check out the graph of y=e^W(x*ln(x)) which is NOT exactly the same as just y=x www.wolframalpha.com/input/?i=graph+of+y%3De%5Eproductlog(x*ln(x)) So, why is this? Well stay tuned.

[@lmaooopsie8894]

for x >= 1/e though, it is

[@fangamenst7965]

Why xln(x) = ln(x)*e^ln(x) ??

[@ISoldßinLadensViagraOnEbayఔ]

Can you do x^y * y^x =xy^xy

[@robloxcutter]

7:00 technically y=x because W(xlnx)=lnx, e^lnx = x

[@plasmacrab_7473]

That is really cool, I especially like the parametric equation, as we have a way of deciphering the results without requiring a calculator.

[@casey206969]

I know plus it gives cool identities like for t=3. (√3/3)^(√3/3) = (√27/27)^(√27/27).

[@rcuster93]

@blackpenredpen I’m not sure I understand how W(x*ln(x)) isn’t just x, since x=e^ln(x). And if that was the case, that would just give y=x. Did I miss something?

[@BigRedProductions11]

Robbie Custer same, commenting so I can see any replies

[@casey206969]

xe^x is not injective so W is a multivalued function

[@skulleeman]

Edit: oh I understand lol. People have good answers below. If it was W(x*e^x) then it would equal x, but it's W(x*ln(x)) instead so it doesn't. I'm not sure what you mean by x=e^ln(x) since that is just a complicated way of writing x=x.

[@sylowlover]

@@skulleeman I think they mean that due to symmetry, taking the W of both sides will just give the same result of ln(x)=ln(y). But a commenter above made a good point about branches and functions being multivalued.

[@icespirit]

W[ln(x)*x]=W[ln(x)*e^ln(x)] (the fish in this case is lnx, so) =ln(x)

[@HaydarKarhan]

Well well...

[@BigDBrian]

I prefer method 2. method 1 uses the assumption that y=tx which seems to come out of nowhere and just happens to work out. method 2 uses none of that, and ends up with a nice formula without introducing any new variables.

[@MichaelGrantPhD]

mrBorkD But the Lambert W function cannot be computed analytically. At least I can use the first method with a simple calculator, to compute a wide variety of pairs (x,y). No can do with the W function.

[@wesleydeng71]

@@BigDBrian This is a commonly used trick. You can also let y=x^t and will work out as well.

[@mathlegendno12]

I don’t understand much of the natural log, so I prefer the first one

[@louf7178]

ha ha

[@seroujghazarian6343]

Speaking of the lambert W function, I actually tried to find the exact value of W(-1) by knowing that its opposite, -W(-1) , is a solution to x=ln(x). So, I worked around it, and I got to these 2 equations: Re(-W(-1))=Im(-W(-1))×tan(Im(-W(-1))) and Re(-W(-1))=-W(-cos(Im(-W(-1))) The latter coming from the fact that x=lnxx=e^x. And it felt like I was going nowhere with this.

[@Hacker118]

what is the derivative and intergral of W (x)? i am just only curious on that

[@anegativecoconut4940]

For the record the minimum of the second method function is 1/e . How unexpected. ;-)

[@asup759]

love those coincidences with e, did you know the maximum of x^(1/x) occurs at x=e

[@mikail5682]

5:01 Yes. A fish.

[@maxreenoch1661]

I knew the lambert W function was coming.

[@drewmichael3986]

But x ln x is just equal to (ln x)*e^ln x! So W(x ln x) = ln x then you make x = y! So I prefer the non-trivial first solution. (By the way, the example you gave for the second solution makes x = y)

[@blackpenredpen]

Actually not quiet. e^W(x*ln(x)) = x when x>=1/e That's why I showed when you put 0.3 in for x you would get something that's not 0.3

[@Theraot]

Drew, careful with those factorials :P

[@sinamolavi713]

@@blackpenredpen whats the difference between x and y that lets us simplify W(y ln y) but not W(x ln x)? and if we had the problem x^x=a*y^y as a more general form can we still solve it?

[@NuptialFailures]

I know little about this function, but would there be a way to do a proof for your last video on x tetrated infinitely many times=i using the Lambert W Function?

[@leandroteles7857]

0:34 One force is equal to one force. -Redpen, Blackpen.

[@RussellSubedi]

You say trivial solutions are boring. Yet, I find trivial solutions to be the most interesting.

[@Thankmel8r]

Since the W function is multi-valued, the second approach captures all the solutions, both trivial (x=y) and non-trivial (x!=y). The Wolfram plot is deceiving, as it contains only one branch of the graph. However, the W function is not computable in closed form. On the other hand, the first approach provides all the non-trivial solutions, only, but it is computable in closed form.

[@damianmatma708]

04:20 I think we should also write the Domain for "t": t ∈ (0;1) ∪ (1; +∞) (t>0 because both "x" and "y" have the same sign (x>0 and y>0) and t≠1 since x≠y)

[@lightyagami6647]

Yes ,i thought the same thing XD

[@rotten-Z]

(1/4)^(1/4)=((1/4)^(1/2))^(1/2)=(1/2)^(1/2) Keep it simply

[@jzanimates2352]

Hey there blackpedredpen. I really enjoyed the videos you did a while ago on numbers in different bases. Could you try and do things like intergrals and derivatives in different number bases, or maybe trig functions in different bases. I thought it would be a really cool challenged for you to try

[@eddiecurrent7721]

Wouldn't the right hand side of the final equation just reduce to x, so that you're left with y=x?

[@PeaceTheBall]

W function has multiple outputs in a specific range so that's probably why we leave it with e^W(xlnx)

[@sinamolavi713]

@@PeaceTheBall doest that apply to y as well? thats very confusing

[@karammohamed5569]

Do a video on 2^x=3x

[@shoaibakterhimel5451]

really great solutions, both of them. i couldn't solve it even after trying for 2 days straight

[@darcash1738]

On the lambert one, why not also make the x in a nice form, e^lnx lnx? With this we get lny = lnx y = x Is this the same? Is there some reason this is different? When should we not try to make it in the easiest form, and how might we be losing solutions by doing this

[@adios4]

1st method was relatively simpler to input values or who will grab a calculator to calculate W(69) thanks BlackpenRedpen do a Qna pls and tell your name

[@adamcionoob3912]

Really nice. Good job.

[@pleappleappleap]

How do you calculate the value of the W function for an arbitrary input?

[@angelmendez-rivera351]

I would have preferred to write y = x*Ln(x)/W[x*Ln(x)], since division is generally simpler than exponentiation.

[@Graknorke]

Can you explain where that comes from?

[@engineeringforfuture]

The more better is the way you connect productlog with other function, so both of them solve the same problem.

[@SteveMathematician-th3co]

This is so coooool!

[@afafsalem739]

Interesting lecture

[@samharper5881]

Gorgeous mathematics.

[@i_am_anxious02]

1) again, you love aquatic animals 2) that weird face with one of the fish was there in the rectangle when you did x^y=y^x Lol

[@animesubber7136]

Plot y^y=x^x in desmos, you'll get a sword. One is y=x, and the other two must be the stuff provided above

[@JakeMarley-k6g]

That is what i thought when i first saw it. Talk about a longsword

[@physicsphysics1956]

That equation just gives you the y=x graph.

[@L1N3R1D3R]

I haven't tried, but could you use the existence of these two answers to find the elementary form of the Lambert W function for a different specific type of input, or would it just revert to W(xe^x)?

[@angelmendez-rivera351]

L1N3R1D3R You cannot obtain an elementary form of the W(x) function. When we say it cannot be done, we do not say it because it has not been done, but rather, we say it because there is essentially a theorem which proves that this function is impossible to be express with a finite combination of elementary functions and arithmetic and algebraic operations. We could ten billion years searching for a way, and it is impossible, just as impossible as finding an ending to the digit expansion of π.

[@unnwas]

0^0 = 0 or = 1? I almost replied "Obviously 1 and 0"...

[@casey206969]

For x, y > 0.

[@stephenphelps920]

0^0 is undefined

[@unnwas]

+Stephen Phelps oh... makes sense

[@unnwas]

+casey206969 I thought that before watching the video

[@alexscott8125]

@@stephenphelps920 0^0 is weird mainly due to how it is often used as being equal to 1 in mathematical problems such as differentiating or integrating power series. Having 0^0=1 tends to somehow give the correct outcome to certain problems it seems.

[@MOHNAKHAN]

Great sir...👍

[@brezhart5088]

If you draw x^x, graph, you can see infinity variety of possible way to solve this.

[@JamilKhan-hk1wl]

Yes, you can have t to be any real value

[@seroujghazarian6343]

Wait, isn't W(x×lnx)=lnx?

[@blackpenredpen]

only when x>=1/e

[@Graknorke]

So the Lambert W Function solution only works when y is above 1/e as well?

[@seroujghazarian6343]

@@blackpenredpen so W(-1/e)=-1?

[@seroujghazarian6343]

@@blackpenredpen also, isn't it the same condition for y?

[@blackpenredpen]

@@seroujghazarian6343 I will make some follow up videos later on.

[@connoruzzo4004]

Very cool! I study physics and am only getting a minor in maths, I am wondering what college courses usually introduce the w function, out of curiosity?

[@willnewman9783]

In my experience, it is not standardly introduced in any course. It is not a very important function.

[@stevethecatcouch6532]

According to Wolfram Mathworld "Banwell and Jayakumar (2000) showed that a W-function describes the relation between voltage, current, and resistance in a diode, and Packel and Yuen (2004) applied the W-function to a ballistic projectile in the presence of air resistance. Other applications have been discovered in statistical mechanics, quantum chemistry, combinatorics, enzyme kinetics, the physiology of vision, the engineering of thin films, hydrology, and the analysis of algorithms (Hayes 2005)."

[@xardasnecromancer590]

I prefer the second method. First one isn't complete. It gives only such solutions that are in the form of y = tx and what if there are different ones like y = sin(tx)? The second method is strict.

[@wassollderscheiss33]

And how do you calculate W of any value?!

[@JefiKnight]

Does either method capture all the answers?

[@antimatter2376]

that nike product placement in the thumbnail.

[@zat5176]

X to the teeth power lol

[@sarmitachowdhury3977]

Love your videos!! Can you please make a video on convergent series and ratio tests?

[@blackpenredpen]

sarmita chowdhury I have some on those already. You can search it

[@sarmitachowdhury3977]

@@blackpenredpen Thanks a lot

[@Ed-wn8hp]

Raising both sides of the equation to the 4th power will lead to 1/4 = 1/4.

[@SakiJ93]

Mmmh... why is W(x*ln(x)) = ln(x) only when x >= 1/e? is it the same for W(x*e^x) = x?

[@blackpenredpen]

Because x*ln(x) is not 1 to 1. The graph of x*ln(x) has the lowest point at x=1/e and then increase when x>1/e

[@Oberatous-Udurabas]

What would the solutions be to x^y=y^x ?

[@aaaaaa-jg2fg]

Waiting for some ways to graph it

[@blackpenredpen]

Graph the parametric equations : )

[@davidhilbert8414]

i^x=x ; i=√-1 How to find the value of x?

[@want-diversecontent3887]

Privata Kanalo i^x = x e^ln(i^x) = x e^(x*ln(i)) = x e^(x*i*(pi/2)) = x e^i((x*pi)/2) = x e^i(theta) = cis(theta) cis((x*pi)/2) = x cos((x*pi)/2) + i(sin((x*pi)/2)) = x I just can't. Nope. Nope. Somebody with actual math knowledge, please continue this for me.

[@kutuboxbayzan5967]

i^x=x i=x^(1/x) Ln (i)=ln (x)/x let x=e^u -i*pi/2=-u*e^(-u) W (-i*pi/2)=u=ln (x) X=-i*pi/(2w (-i*pi/2))

[@Patapom3]

Amazing!

[@fanamatakecick97]

Wait, if W(y*ln(y)) = ln(y), then shouldn’t W(x*ln(x)) = ln (x)?

[@alwysrite]

like the first method because you don't even need to know about Logs or Lambert to solve it.

[@saharhaimyaccov4977]

If t is -2?

[@arunlalwani7565]

Why cant we take x=y???

[@stormswindy3013]

literally anything else < FISH :D

[@GourangaPL]

Method on the right is easier (at least for me)

[@maverickreynolds]

Why the fish?

[@ToniBrasil100]

Make Sense 😂🙏

[@crazyphil7782]

Watch out, your solution is incomplete. The trivial cases are, although trivial, needed to make it complete and correct. Also, there’s no need for the product log. Just imply, by removing (0,0) and (1,1) (both are solution) from base set, that x/y = ln(y - x). Then you complete by searching all substitutions which work (like y = tx). The final step, to make it rigorous, is to prove that such constructed solution set is complete.

[@numeroVLAD]

Show us plots

[@titan1235813]

I want cheese. Anybody has some cheese? Blue cheese, please. Thank ya'll

[@Dialectic42]

t’th power... teeth power?

[@fugmopoly]

...why does he use a fish

[@migtrewornan8085]

It just seems to me that answers in terms of W(?) are just re-statements of the problem not solutions per se.

[@blackpenredpen]

Mig Trewornan Interesting enough, if you really put in some "good" values like 0.3, then you can get some cool answers out of it. I will make some follow up videos.

[@thebloxxer22]

If x could equal y, then x=1 and y=1. *Solved.*

[@GuyMichaely]

That's a big caveat you got there fam

[@ffggddss]

At the end of Part 1, your result, x = tᵗʹ⁽¹⁻⁻ᵗ⁾, y = t¹ʹ⁽¹⁻⁻ᵗ⁾, also means that tx = y = xᵗ. So you've also generated a solution for xy = yˣ. In Part 2, from where you have x ln(x) = y ln(y) take the W() of both sides in different ways. If you had treated both sides the same way, you'd have got the useless ln(x) = ln(y) so maybe you could've pointed out that you were being a bit clever there. The above fact also suggests that getting non-trivial aolutions relies on using the part of the W() "function" that is double-valued. Otherwise, you couldn't get two different answers on opposite sides. Or something like that... Fred

[@continnum_radhe-radhe]

🔥🔥🔥

[@user-mk4jq7uq9c]

X=Y

[@林黑黑-b3c]

粗框眼镜好看~

[@ИгорьКупринюк]

Why are you switch your avatar so often?

[@mathadventuress]

X=y problem solved

[@Lost_City007]

Yeah 👍

[@KaviAmanTenguriyaShaurya]

U r really awesome. Wow

[@achikaznag963]

Wait a sec that gives us x=y ( did i miss something ?)

[@Krakabraka]

1^1 = 0^0

[@anuragray2323]

GG NICE !

[@jiaming5269]

Wow

[@CharIie83]

first one is better

[@gouravmadhwal5548]

Plz integrate ln(ln(ln(lnx)))

[@blackpenredpen]

Gourav Madhwal yay!!!!