Question

Can the inverse of \(f(x)\) where: \[
f(x)=x+e^x
\]be expressed as an elementary function?

Answer 1

I feel like, since you;re asking this an Honorary Professor of Mathematics, that this is really a question or a challenge posed to people like me on here :O

Answer 2

Titles and SmartScores only hold so much value on OS. Not saying he isn't smart as hell, but you shouldn't base people's intellectual abilities on their SmartScore.

Answer 3

Regardless, I'll try my best to see though.
An inverse function, notated as \[\Large f^{-1}(x)\]
Can be found by swapping out x for y and solving for y

Answer 4

I have also seen him @Jhannybean ask/answer questions, I know he know he knows his stuff. I should've prefaced him with that instead

Answer 5

\[\Large f(x)=x+e^{x}\]

Answer 6

\[\Large \color{red}{f^{-1}(x)=f(y)}\color{blue}{=y+e^{y}}\]

Answer 7

Since "elementary function" is a subjective term, yes.

Answer 8

What is the proof?

Answer 9

Proof that "elementary function" is a subjective term or the algebraic manipulation to get to the inverse? =D

Answer 10

here's ur answer
http://www.wolframalpha.com/input/?i=inverse+of+y%3Dx%2Be%5Ex

Answer 11

http://en.wikipedia.org/wiki/Lambert_W_function

Answer 12

(no)

Answer 13

Wow, good thing I didn't continue typing lol

Answer 14

I actually don't like Wolfram Alpha's answer, there's a better way to represent that.

Answer 15

@Kainui ?!

Answer 16

I wonder how it got that answer.

Answer 17

Here's how you can solve it though for those who are interested, W(x) is the inverse of xe^x so we have: \[\Large x=W(x)e^{W(x)}\] since f(f^-1(x))=x as usual.

Now let's actually solve the problem though. \[\Large x=y+e^y \\ \Large e^x = e^{y +e^y} \\ \Large e^x = e^y e^{e^y} \\ \Large W(e^x) = e^y \\ \Large \ln(W(e^x))=y\] So there we go, all one algebra step at a time. Now let's say you want WA's answer, well look at the simple relation we began with when we plugged lambert's W into it's inverse, I'll do some algebra on that: \[\Large a= W(a)e^{W(a)} \\ \Large \ln(a) = \ln(W(a))+\ln(e^{W(a)}) \\ \Large set \ a =e^x \\ \Large x=\ln(W(e^x))+W(e^x) \\ and there we go.\]

Answer 18

This is probably one of my favorite functions because it is incredibly useful and simple. There are several fun uses for it too if anyone's interested I can answer any questions. =D

Answer 19

I believe W(x) should be considered an elementary function for the record. It opens up a lot of equations to being solved algebraically.

Answer 20

\(\color{blue}{\text{Originally Posted by}}\) @Kainui
Here's how you can solve it though for those who are interested, W(x) is the inverse of xe^x so we have: \[\Large x=W(x)e^{W(x)}\] since f(f^-1(x))=x as usual.

\(\color{blue}{\text{End of Quote}}\)

Can you start by explaining that part? I'm confused.

Answer 21

I had a strong inkling that we had to use logarithms somewhere but wasnt sure where, or how to apply them.

Answer 22

Sorry thanks for pointing it out, I don't know why I even started with that, I should have left that to the second part. At any rate, I'll explain it through an example since it's sort of hard to wrap your mind around.

Let's consider \[\Large f(x) =e^x\] We know this function is bijective, and we define that inverse to be the natural logarithm \[\Large f^{-1}(x) = \ln(x)\] They have this nice property that \[\Large f(f^{-1}(x))=f^{-1}(f(x))=x \\ \Large e^{\ln x}=\ln(e^x)=x\] Graphically this means if we draw the line y=x, they are a reflection of each other. Or if you drew it on clear paper and flipped the paper over and swapped the x and y axes, it would be the same graph.

Answer 23

Similarly, the lambert product log is just like that except slightly different,
Let's now consider \[\Large f(x) =xe^x\] We know this function is bijective, and we define that inverse to be the product logarithm (or Lambert's W function) \[\Large f^{-1}(x) = W(x)\] They have this nice property that \[\Large f(f^{-1}(x))=f^{-1}(f(x))=x \\ \Large W(x)e^{W(x)}=W(xe^x)=x\] Graphically this means if we draw the line y=x, they are a reflection of each other. Or if you drew it on clear paper and flipped the paper over and swapped the x and y axes, it would be the same graph.