Solving A Non-Standard Equation

[@kappasphere]

One thing that's been bugging me is that 1/2 - ln(sqrt(2)/2) can be simplified to 1/2 - ln(2^0.5) + ln(2) = 1/2 - 1/2 ln(2) + ln(2) = 1/2 (1 + ln(2))

[@usdescartes]

Better than that, even. 1/2 (1 + ln 2) = 1/2 (ln e + ln 2) = 1/2 ln 2e = ln√(2e) That's actually kinda nice-looking!

[@xyz.ijk.]

That was excellent. Thank you. I may have some catch-up work to do, but overall I understood where you were going.

[@SyberMath]

Glad it was helpful!

[@Vemu]

What's the program you're using for writing?

[@guystuart4095]

Yes, the app for drawing the graphs. Very nice graphs!

[@guystuart4095]

I see it, desmos

[@tharunsankar4926]

Desmos

[@leesweets4110]

ChatGPT is doing his scripts now.

[@SyberMath]

Notability

[@RashmiRay-c1y]

Squaring both sides and following some straightforward manipulations, we can write x = -a -1/2 W(-2e^(-2a)), where W is the Lambert W function. For a less than roughly .7, there won't be any real solutions. But for a =1, W_(-1) gives x=0 and W_(0) gives x = -.8. Thus we will expect 2 solutions.

[@yoav613]

Nice! I think it is more simple to solve this if you notice that this eq is: e^(2x)-x=a, then using simple calc we see that the function e^(2x)-x has 1 minimum point at (-0.5ln2, 0.5(1+ln2)). Therfore for tangent a=0.5(1+ln2),and if a>0.5(1+ln2) there are 2 intersection points and ifa

[@leif1075]

What about other solutions if a equals 1 x can equal zero?

[@premkumarsr4021]

Beautiful. Lovely. No words to express my happiness

[@SyberMath]

Thanks a lot 😊🤩🧡

[@popitripodi573]

Very interesting equation ❤❤❤❤

[@SyberMath]

Glad you think so!

[@bassem.al-ashour]

When squatting both sides, you get e^2x=x+a (×+a)=e^2×......divide both sides by e^2x to get (x+a)e^(-2×)=1.....multiply both sides by e^(-2a) to get (×+a)e^(-2x-2a)=e^(-2a) (×+a)e^[-2(x+a)]=e^(-2a)......multiply both sides by -2 to get -2(×+a)e^[-2(x+a)]=-2e^(-2a)......Lumbert both sides to get -2(×+a)=W(-2e^(-2a)) x+a=-[W(-2e^(-2a))]/2 x=-[W(2e^(-2a))]/2-a

[@scottleung9587]

Cool!

[@ilanbar1970]

Nice!

[@SyberMath]

Thanks!

[@ardiris2715]

ChatGPT suggested the bisection route, and then quit. (:

[@AbdulRahmanAttari]

❤❤

[@SyberMath]

Thank you! 💗

[@srividhyamoorthy761]

Ooh interesting Edit very interesting

[@SyberMath]

Glad you think so!

[@attila3028]

if you solved what is x in terms of a

[@oakpope]

And so, you didn't solve the equation.

[@jonasvuillemin9412]

U didn’t solve it, u have 2 other possibilty depending on the value of a and btw u didn’t prove that u can derivate g

[@barakathaider6333]

👍

[@sphakamisozondi]

This is a delicious problem. Where can we get these types of math problems?

[@talberger4305]

X=-0.5×W(-2e^(-2a))-a

[@rubensramos6458]

Exactly.

[@GreenMeansGOF]

Same

[@tetramur8969]

I want Lambert's W function to be more popular and widely used

[@GreenMeansGOF]

@@tetramur8969 The people have spoken, Syber😎

[@srividhyamoorthy761]

@@tetramur8969 do agree with dat

[@ParaNozek]

ln(2e)/2

[@leesweets4110]

Cant find books in English? HA, aint that the truth. You cant find videos on youtube in English, either, really. You either have a heavy accent or youre literally speaking Hindi or Japanese. The most advanced mathematics videos in English on youtube only go up to the addition of fractions.

[@allanmarder456]

The solution using the Lambert W function is x= -[W(-2/(e^(2a)) +2a]/2. . To prove this, first square the original equation and get e^(2x) -x = a. Now let z be such that z*(e^z) = -2/(e^(2a)) Then x= -[z+2a]/2 So e^2x = e^(-[z+2a]) = e^(1/((e^z)*(e^2a)) =(1/e^z) * (z*(e^z)/(-2) = -z/2. Next subtract x and get = -z/2 - (-(z+2a/2)) = -z/2 +z/2 + a. =a. So for example if e^x = sqrt(x+2). we get W(-2/(e^4)) = =-.038052 and -4.895084 ( 2 answers). and x= -(.038052+4)/2 = - 1.98097 and x = - (-4.895084+4)/2 =.44754. Checking ( to 4 decimals) e^(-1.98097)=.1379 sqrt(-1.98097+2) =.1379. e^.44754 =1.5644 sqrt(2+.44754) = 1.5644.

[@guystuart4095]

The number of parentheses in your expression for x do not match up. There are three “(“ and just two “)”.

[@allanmarder456]

@@guystuart4095 Corrected

[@guystuart4095]

I am trying to learn more about the Lambert W function. Can you give me a hint on why you chose z such that z*e^z = -2/e^2a. I am kind of stuck on this point. Just a hint? Thanks, Guy

[@guystuart4095]

I figured another way to find Lambert W function. Square both sides of original equation and then div both sides by e^2x. Define y=x+a and manipulate until you have -2y e^-2y = -2e^-2a. Take the W function of both sides and you are home free.

[@allanmarder456]

@@guystuart4095 Thanks for your reply. There are many different ways to manipulate an equation to make it work with the Lambert function. If you search online you can find many examples of types of equations that can be solved via Lambert. As you have shown sometimes it's just a question of manipulating things until you get what you want. That's what I do. Again thanks for your comment.