Question

Recent explorer of the lambert function and a question on here sparked in interest in me to do the following tutorial: Solving a linear(t)=exponential(t) for the variable t algebraically using the lambert function.

Answer 1

\[ \text{ How to solve something of the form (for } t \text{):} \\ at+b=c(d)^t \\ \text{ using the Lambert Function } \\
\text{ Left side is a linear expression whereas } \\ \text{ the right side is a exponential expression}\\ \text{ anyways I will solve this one in general } \\
\text{ First step: factor out the left hand side } \\ \text{ we want to factor out the coefficient of } t \\ at+b=a(t+\frac{b}{a}) \\ \text{ so now we have the equation as } a(t+\frac{b}{a})=c(d)^t \\
\text{ Second step: divide both sides by } c \\ \frac{a}{c}(t+\frac{b}{a})=d^{t} \\
\text{ Third step: sub} t+\frac{b}{a} \text{ as } u \\ \text{ so if } u=t+\frac{b}{a} \text{ then } t=u-\frac{b}{a} \\ \text{ So our equation looks like } \frac{a}{c}u=d^{u-\frac{b}{a}} \\
\text{ Fourth step: multiply both sides by } d^{\frac{b}{a}} \\ \text{ this gives us } \\ \frac{a}{c}u d^{\frac{b}{a}}=d^u \\
\text{ Fifth step: multiply } u^{-1} \text{ on both sides } \\ \frac{a}{c} d^{\frac{b}{a}}=u^{-1}d^u \\
\text{ Sixth step: rewrite } d^u \text{ as } (\frac{1}{d})^{-u} \\ \text{ so we have} \\ \frac{a}{c} d^{\frac{b}{a}}=u^{-1}(\frac{1}{d})^{-u}

\text{ Seventh step: “flip” both sides } \\ \frac{c}{a} d^{-\frac{b}{a}}=u(\frac{1}{d})^u \\

\text{ Eight step: recall } e^{\ln(\frac{1}{d})}=\frac{1}{d} \\ \frac{c}{a} d^{-\frac{b}{a}}=u(e^{\ln(\frac{1}{d}})^u

\text{ Ninth step: Multiply both sides by} \ln(\frac{1}{d}) \\ \ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}}=\ln(\frac{1}{d}) u(e^{\ln(\frac{1}{d})})^u
\text{ Tenth step: Take } W( ) \text{ (this is the Lambert function } \\ \text{ of both sides } \\ \text{ Note: recall } W(be^b)=b \\ W(\ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}})=W(\ln(\frac{1}{d}) u(e^{\ln(\frac{1}{d})})^u) \\ \text{ which can be written as } W(\ln(\frac{1}{d}) \frac{c}{a} d^{-\frac{b}{a}})=\ln(\frac{1}{d})u
\text{ Eleventh step: Divide both sides by } \\ \ln(\frac{1}{d}) \text{ aka} -\ln(d) \\
\frac{-1}{\ln(d)}W(\ln(\frac{1}{d}) \frac{c}{a}d^{-\frac{b}{a}})=u \\
\text{ Twelfth step: recall the sub we made earlier } u= t+\frac{b}{a} \\
\text{ So the solution is } t=\frac{-1}{\ln(d)}W(\ln(\frac{1}{d}) \frac{c}{a}d^{-\frac{b}{a}})-\frac{b}{a}\]

Answer 2

And I should probably place some restrictions on the constant values in my problem.

Answer 3

\[a \neq 0 \\ d \in (0,1) \cup (1,\infty)\]

Answer 4

what if a is 0 though?
\[b=c(d)^t\]
this equation should be tons simpler
\[\frac{b}{c}=d^t \\ \ln(\frac{b}{c})=\ln(d^t) \\ \ln(\frac{b}{c})=t \ln(d) \\ \frac{1}{\ln(d)} \ln(\frac{b}{c})=t \]
As we see the Lambert function isn't needed for a=0.

Answer 5

What if d=1?
Well you no longer have an exponential expression on the right hand side. This equation is tons easier than the last.
\[at+b=c \\ at=c-b \\ t=\frac{c-b}{a}, a \neq 0\]

Answer 6

Anyways I might play more with Lambert and see if I can post anything else at a later time.

Answer 7

one more restriction for my first problem c is not 0

Answer 8

Thats really a nice general method to express/solve using product log function! Did you come up with that method ? xD

Answer 9

Yeah. I had to do a few playings...
Though I did kind of get an idea for the sub here http://en.wikipedia.org/wiki/Lambert_W_function .

Answer 10

@Kainui taught me this some time back... I want to review but looks its gonna take time as I don't seem to remember much

Answer 11

Yeah I kind of remember him I think mentioning something about Lambert

Answer 12

\[\large x^{x^{x^\vdots}}=?\]

Answer 13

I didn't try to get it back then. But the question asked today may be kind of interested though I don't think it was what the asker actually intended. http://openstudy.com/users/freckles#/updates/557a14dce4b07028ea5fb518

Answer 14

i'm playing with the thingy you have there
I already know the answer because I look at math world or whatever it is called
but I want to see if I can derive it

Answer 15

\[\large x^{x^{x^\vdots}}= t \implies x^t = t \]

thats in general form that you derived earlier

Answer 16

guess i can start with Fifth step straight

Answer 17

\[x^t=t \\ x=e^{\ln(x)} \\ e^{\ln(x) \cdot t}= t \\ t^{-1} e^{\ln(x) \cdot t}=1 \\ t^{-1} e^{\ln(x) \cdot t \cdot (-1)(-1)}=1 \\ t^{-1}e^{\ln(x^{-1} )\cdot t(-1)}=1 \\ (te^{\ln(x^{-1})t})^{-1}=1 \\ te^{\ln(x^{-1} )\cdot t}=1 \\ \ln(x^{-1})te^{\ln(t^{-1}) \cdot t} =\ln(x^{-1}) \\ \ln(x^{-1} )t=W(\ln(x^{-1}))\]
so
\[t=\frac{W(\ln(x^{-1}))}{\ln(x^{-1})}\]

Answer 18

but maybe I made an algebraic mistake

Answer 19

oh no actually I think that looks good

Answer 20

the string of exponents of x can be replaced by t
and that equation holds because both t's represent an infinite exponent string of x's
if you know what I mean?

Answer 21

Nice! wolf gives the exact same expression
http://www.wolframalpha.com/input/?i=solve+x%3Dt%5Ex

Answer 22

Yes its just another form of infinite sequence... it has to converge in the first place for our analysis to make any sense

Answer 23

not bad
and not much of a difference from what I was doing earlier with the other equation

Answer 24

oh wait it is the same thing lol

Answer 25

that is an exponent=linear

Answer 26

i'm dumb :p

Answer 27

I think I got threw off by the power tower

Answer 28

Haha! im looking at discrete lambert w function, its more interesting than the continuous version xD

Answer 29

I will have to look at that tomorrow
I must go tonight
thanks for the fun power tower

Answer 30

thanks again for sharing the general method.. gnite!

Answer 31

Here are two great sites I found when learning about the product log:

http://www.had2know.com/academics/lambert-w-function-calculator.html
https://luckytoilet.wordpress.com/tag/product-log/

There are some exercises on the first link, which I think are fun to try to work out without cheating.

Answer 32

\[W=xe^{-W } \\ W'=e^{-W}-W'xe^{-W} \\ W'+W'xe^{-W}=e^{-W } \\ W'(1+xe^{-W})=e^{-W} \\ W'=\frac{e^{-W}}{1+xe^{-W}} \\ W'=\frac{1}{e^{W}+x} \\ \text{ but since } W=xe^{-W} \implies e^W=\frac{x}{W} \\ \text{ then } W'=\frac{1}{\frac{x}{W}+x}=\frac{W}{x+Wx}\]
ok just wanted to do that one thing myself
:p

Answer 33

Haha also might as well find the indefinite integral too ;P

Answer 34

\[\int\limits W(x) dx \\ \text{ Let } y=W(x) \\ \text{ then } dy=W'(x) dx \\ W'(x)=\frac{W(x)}{x+W(x)x}=\frac{y}{x+yx} \\ \text{ but if } y=W(x) \text{ then } W^{-1}(y)=x \\ \text{ and } W^{-1}(y)=ye^y \\ \text{ so } x=ye^{y} \\ W'(x)=\frac{y}{ye^y+y^2e^y} \\ \text{ back \to } dy=W'(x) dx \\ \frac{dy}{W'(x) }=dx \\ \frac{ye^y+y^2e^y}{y} dy=dx \\ (e^y+ye^y) dy=dx \\ \int\limits W(x) dx=\int\limits y (e^y+ye^y)dy\]

Answer 35

\[\int\limits W(x) dx=\int\limits (ye^y+y^2e^y) dy \\ ye^y-e^y+y^2e^y-2ye^y+2e^y+C \text{ by IBP}\]

Answer 36

\[=-ye^y+e^y+y^2e^y+C\]

Answer 37

now we need to write back in terms of x

Answer 38

\[x=ye^y \\ \int\limits W(x) dx=x+e^y+yx+C\]

we have to figure how to take care of the exp(y) and the y thingy there
\[y=W(x)\]

\[\int\limits W(x) dx=x+e^{W(x)}+W(x) \cdot x+C\]

Answer 39

hmm their answer looks a little different than mine

Answer 40

oh I forgot to bring down my negative factor for that one term

Answer 41

\[\text{ So it should read } \\ \int\limits W(x) dx=-x+e^{W(x)}+W(x) \cdot x+C\]

and I could have replaced exp(W) with x/W like they did
but don't really see a reason to