Question

I've got some tough math stuff for you guys if you're interested. This is centered around a function called the product log function, also known as the Lambert W function.

Answer 1

So the old logarithm looks like this\[\LARGE y=e^x \\ \LARGE \ln (y) = x\] the product logarithm looks like this: \[\LARGE y=xe^x \\ \LARGE W(y)=x\]
We can do a handful of fancy things with it, like integrate, differentiate, or use it to solve problems that we couldn't normally solve before.

For instance, what is the inverse of the function: \[\LARGE f(x)=2x+\ln(x)\] Well we know the inverse satisfies this property: \[\LARGE x=2[f^{-1}(x)]+\ln[f^{-1}(x)]\] So if we raise these both as powers of e we get: \[\LARGE e^x=e^{2[f^{-1}(x)]+\ln[f^{-1}(x)]}=e^{2[f^{-1}(x)]}e^{\ln[f^{-1}(x)]}\]Now it simplifies more to: \[\LARGE 2e^x=2f^{-1}(x)e^{2f^{-1}(x)}\] We can take the product log of both sides to get: \[\LARGE W(2e^x)=2f^{-1}(x)\] And now we have a formula for the inverse of f(x)!\[\LARGE f^{-1}(x)=\frac{W(2e^x)}{2}\]

Now here's another one, try to find the inverse of this function! \[\LARGE g(x)=x^2e^x\] Good luck. =)

Answer 2

@ganeshie8 @ikram002p @BSwan @iambatman @Astrophysics @dan815 @Somy @abb0t @yourMom @justKidding @stopReadingAlreadyPlx

Answer 3

if (x, y) is a point on inverse function, then, (y, x) will be a point on the actual function :

x = y^2e^y

we need to solve/ express y in terms of product log function

Answer 4

x = y^2*e^y

take sqrt both sides

sqrt(x) = y*e^(y/2)

sqrt(x)/2 = (y/2)*e^(y/2)

(y/2) = W(sqrt(x)/2)

Answer 5

does that work xD

Answer 6

nice