Finding The Inverse of x + ln(x)

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[@charliebrownfri]

So much effort ❤ in teaching us with politeness and great wit, and people complaining about “pi’s” and “e’s” (not the numbers, unfortunately); stop this world, I wanna get out 😮

[@maxhagenauer24]

f(x) = x + ln(x)
y = x + ln(x)
f^-1(x) means swap x and y:
x = y + ln(y)
e^x = e^(y + ln(y))
e^x = e^y * e^ln(y)
e^x = y * e^y
y = W_n (e^x) where n is a natural number = f^-1(x)

[@leif1075]

Butnthsts a CHEST lambert isnonly uf you knownit or.dont

[@maxhagenauer24]

@@leif1075 What?

[@ProactiveYellow]

So the left inverse is pretty easy:
f(x)=x+lnx
f(x)=ln(e^x)+ln(x)
f(x)=ln(xe^x)
e^f(x)=xe^x
W(e^f(x))=x
g(x)=w(e^x), g(f(x))=x
Composing the other way gives
f(g(x))=w(e^x)+ln(w(e^x))
=Ln(e^g(x))+ln(g(x))
=ln(w(e^x)e^(w(e^x)))
=ln(e^x) (I think)
=x, thus g(x) should be a total inverse, if I recall my properties of the prod-log correctly.

[@ramenguy6827]

Hey can you tell me what's that w function?
I couldn't convert it to explicit form but how that w works?

[@ProactiveYellow]

@@ramenguy6827 the product-log or lambert-W function is defined as the inverse of xe^x. This means that W(xe^x)=x, and w(x)e^(w(x))=x. I don't know how to define it in other terms, unfortunately.

[@berndkru]

@@ramenguy6827 The w function cannot be expressed in terms of elementary functions. See Wikepedia "Lambert W function"

[@nevoitzhak2092]

You are correct
By definition w(xe^x)=x
Substituting w(x) for x gives w(w(x)e^w(x))=w(x)
Hence w(x)e^w(x) = x and you are correct after replacing x with e^x

[@bayareapianist]

1) x>0 must have been given in the problem, because ln(×) doesnnot exist for x

[@gdtargetvn2418]

wasn't that obvious

[@SyberMath]

1- that's the domain. I don't think it needs to be given
2- I hadn't heard of it, either, but I find it interesting

[@maxvangulik1988]

e^f(x)=xe^x
W(e^f(x))=x
f^-1(x)=W(e^x)

[@ashishraje5712]

Wonderful

[@SyberMath]

Thank you

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@freddyalvaradamaranon304]

Muy interesante video sobre funciones inversas con logaritmos, muchas gracias por compartir tan buena explicación. 😊😊❤😊😊.

[@SyberMath]

My pleasure. Thank you!

[@giuseppemalaguti435]

f(x)=x+lnx=y...x=e^(y-W(e^y))

[@edgardojaviercanu4740]

beautiful!

[@SyberMath]

Thank you!

[@Why553-k5b_1]

let f^-1(x)=g(x)
then g(x+ln(x))=x and g(x)+ln(g(x))=x
so g(x)=x-ln(g(x)
so x=f(x-ln(g(x)))=(x-ln(g(x)))+ln(x-ln(g(x)))
idk how to solve this equation

[@phill3986]

😁🔥👍😁🔥👍

[@rob876]

x + ln x = y
xe^x = e^y
W(xe^x) = W(e^y)
x = W(e^y)
finv(x) = W(e^y)

[@leif1075]

But lambert is just toi know it or dont so technically not a real solution right?

[@RoseWilson-u2t]

Miller Jennifer Young Gary Miller Anthony

[@miguelaphan58]

...justo lo que se necesita para raspar a todo el salón..incluyendo al profe !!

[@DhdhBdhx-m4z]

Sancs veru mach

[@Mario_Altare]

3:02 Spoiler alert 😀
3:03 Oops

[@SyberMath]

😄😄

[@rajeevlochan5499]

Hello bro

[@SyberMath]

What's up?

[@pleappleappleap]

Please get a pop filter for your microphone. Your "P" sounds hurts my ears on headphones.

[@user-lg6fq1yt4g]

🙄

[@SyberMath]

sorry. i don't know how that works

[@user-lg6fq1yt4g]

@@SyberMath relax we like your videos the way they are.

[@pleappleappleap]

@@SyberMath Google "microphone pop filter" It's just a bit of mesh that sits in front of your microphone and stops the puff of your breath when you say "P", "F", "T" sounds and the line. It costs very little and stops thumping sounds from coming out people's headphones. Most importantly, it doesn't change the sound of your voice.

[@swapanpal9801]

what is the meaning of W( )?

[@SyberMath]

W(xeˣ) = x
W(x) is the inverse of f(x) = xeˣ

[@almanduku9043]

Thanks for question & answer too