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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.
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.

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freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[ \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}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8And I should probably place some restrictions on the constant values in my problem.

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[a \neq 0 \\ d \in (0,1) \cup (1,\infty)\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8what 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.

freckles
 one year ago
Best ResponseYou've already chosen the best response.8What if d=1? Well you no longer have an exponential expression on the right hand side. This equation is tons easier than the last. \[at+b=c \\ at=cb \\ t=\frac{cb}{a}, a \neq 0\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8Anyways I might play more with Lambert and see if I can post anything else at a later time.

freckles
 one year ago
Best ResponseYou've already chosen the best response.8one more restriction for my first problem c is not 0

rational
 one year ago
Best ResponseYou've already chosen the best response.1Thats really a nice general method to express/solve using product log function! Did you come up with that method ? xD

freckles
 one year ago
Best ResponseYou've already chosen the best response.8Yeah. 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 .

rational
 one year ago
Best ResponseYou've already chosen the best response.1@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

freckles
 one year ago
Best ResponseYou've already chosen the best response.8Yeah I kind of remember him I think mentioning something about Lambert

rational
 one year ago
Best ResponseYou've already chosen the best response.1\[\large x^{x^{x^\vdots}}=?\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8I 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.8i'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

rational
 one year ago
Best ResponseYou've already chosen the best response.1\[\large x^{x^{x^\vdots}}= t \implies x^t = t \] thats in general form that you derived earlier

rational
 one year ago
Best ResponseYou've already chosen the best response.1guess i can start with Fifth step straight

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[x^t=t \\ x=e^{\ln(x)} \\ e^{\ln(x) \cdot t}= t \\ t^{1} e^{\ln(x) \cdot t}=1 \\ t^{1} e^{\ln(x) \cdot t \cdot (1)(1)}=1 \\ t^{1}e^{\ln(x^{1} )\cdot t(1)}=1 \\ (te^{\ln(x^{1})t})^{1}=1 \\ te^{\ln(x^{1} )\cdot t}=1 \\ \ln(x^{1})te^{\ln(t^{1}) \cdot t} =\ln(x^{1}) \\ \ln(x^{1} )t=W(\ln(x^{1}))\] so \[t=\frac{W(\ln(x^{1}))}{\ln(x^{1})}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8but maybe I made an algebraic mistake

freckles
 one year ago
Best ResponseYou've already chosen the best response.8oh no actually I think that looks good

freckles
 one year ago
Best ResponseYou've already chosen the best response.8the 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?

rational
 one year ago
Best ResponseYou've already chosen the best response.1Nice! wolf gives the exact same expression http://www.wolframalpha.com/input/?i=solve+x%3Dt%5Ex

rational
 one year ago
Best ResponseYou've already chosen the best response.1Yes its just another form of infinite sequence... it has to converge in the first place for our analysis to make any sense

freckles
 one year ago
Best ResponseYou've already chosen the best response.8not bad and not much of a difference from what I was doing earlier with the other equation

freckles
 one year ago
Best ResponseYou've already chosen the best response.8oh wait it is the same thing lol

freckles
 one year ago
Best ResponseYou've already chosen the best response.8that is an exponent=linear

freckles
 one year ago
Best ResponseYou've already chosen the best response.8I think I got threw off by the power tower

rational
 one year ago
Best ResponseYou've already chosen the best response.1Haha! im looking at discrete lambert w function, its more interesting than the continuous version xD

freckles
 one year ago
Best ResponseYou've already chosen the best response.8I will have to look at that tomorrow I must go tonight thanks for the fun power tower

rational
 one year ago
Best ResponseYou've already chosen the best response.1thanks again for sharing the general method.. gnite!

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0Here are two great sites I found when learning about the product log: http://www.had2know.com/academics/lambertwfunctioncalculator.html https://luckytoilet.wordpress.com/tag/productlog/ There are some exercises on the first link, which I think are fun to try to work out without cheating.

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[W=xe^{W } \\ W'=e^{W}W'xe^{W} \\ W'+W'xe^{W}=e^{W } \\ W'(1+xe^{W})=e^{W} \\ W'=\frac{e^{W}}{1+xe^{W}} \\ W'=\frac{1}{e^{W}+x} \\ \text{ but since } W=xe^{W} \implies e^W=\frac{x}{W} \\ \text{ then } W'=\frac{1}{\frac{x}{W}+x}=\frac{W}{x+Wx}\] ok just wanted to do that one thing myself :p

Kainui
 one year ago
Best ResponseYou've already chosen the best response.0Haha also might as well find the indefinite integral too ;P

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[\int\limits W(x) dx \\ \text{ Let } y=W(x) \\ \text{ then } dy=W'(x) dx \\ W'(x)=\frac{W(x)}{x+W(x)x}=\frac{y}{x+yx} \\ \text{ but if } y=W(x) \text{ then } W^{1}(y)=x \\ \text{ and } W^{1}(y)=ye^y \\ \text{ so } x=ye^{y} \\ W'(x)=\frac{y}{ye^y+y^2e^y} \\ \text{ back \to } dy=W'(x) dx \\ \frac{dy}{W'(x) }=dx \\ \frac{ye^y+y^2e^y}{y} dy=dx \\ (e^y+ye^y) dy=dx \\ \int\limits W(x) dx=\int\limits y (e^y+ye^y)dy\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[\int\limits W(x) dx=\int\limits (ye^y+y^2e^y) dy \\ ye^ye^y+y^2e^y2ye^y+2e^y+C \text{ by IBP}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[=ye^y+e^y+y^2e^y+C\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8now we need to write back in terms of x

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[x=ye^y \\ \int\limits W(x) dx=x+e^y+yx+C\] we have to figure how to take care of the exp(y) and the y thingy there \[y=W(x)\] \[\int\limits W(x) dx=x+e^{W(x)}+W(x) \cdot x+C\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.8hmm their answer looks a little different than mine

freckles
 one year ago
Best ResponseYou've already chosen the best response.8oh I forgot to bring down my negative factor for that one term

freckles
 one year ago
Best ResponseYou've already chosen the best response.8\[\text{ So it should read } \\ \int\limits W(x) dx=x+e^{W(x)}+W(x) \cdot x+C\] and I could have replaced exp(W) with x/W like they did but don't really see a reason to
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