## freckles one year ago 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.

1. freckles

$\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}$

2. freckles

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

3. freckles

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

4. freckles

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.

5. freckles

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. $at+b=c \\ at=c-b \\ t=\frac{c-b}{a}, a \neq 0$

6. freckles

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

7. freckles

one more restriction for my first problem c is not 0

8. rational

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

9. freckles

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 .

10. rational

@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

11. freckles

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

12. rational

$\large x^{x^{x^\vdots}}=?$

13. freckles

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

14. freckles

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

15. rational

$\large x^{x^{x^\vdots}}= t \implies x^t = t$ thats in general form that you derived earlier

16. rational

17. freckles

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

18. freckles

but maybe I made an algebraic mistake

19. freckles

oh no actually I think that looks good

20. freckles

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?

21. rational

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

22. rational

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

23. freckles

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

24. freckles

oh wait it is the same thing lol

25. freckles

that is an exponent=linear

26. freckles

i'm dumb :p

27. freckles

I think I got threw off by the power tower

28. rational

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

29. freckles

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

30. rational

thanks again for sharing the general method.. gnite!

31. Kainui

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.

32. freckles

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

33. Kainui

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

34. freckles

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

35. freckles

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

36. freckles

$=-ye^y+e^y+y^2e^y+C$

37. freckles

now we need to write back in terms of x

38. freckles

$x=ye^y \\ \int\limits W(x) dx=x+e^y+yx+C$ we have to figure how to take care of the exp(y) and the y thingy there $y=W(x)$ $\int\limits W(x) dx=x+e^{W(x)}+W(x) \cdot x+C$

39. freckles

hmm their answer looks a little different than mine

40. freckles

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

41. freckles

$\text{ So it should read } \\ \int\limits W(x) dx=-x+e^{W(x)}+W(x) \cdot x+C$ and I could have replaced exp(W) with x/W like they did but don't really see a reason to