Laplace transform of any derivative

[@VibingMath]

Yay! Love that cancellation of minus sign in the integration by parts process!

[@schweinmachtbree1013]

3:10 we need some assumption on y^(n-1) here since y^(n-1) could blow up faster than e^{-st} goes to zero as t -> ∞. Similarly we need assumptions on y^(n-2), y^(n-3), ..., y', y for the other limits. Looking on wikipedia, it is sufficient to put a condition on just y^(n), namely that it be "of exponential type", whatever that means lol (and of course y needs to be n times differentiable, but this is implicit since we're talking about its nth derivative)

[@MuPrimeMath]

"Of exponential type" means that the norm of the function output is bounded above by an exponential function; clearly this is a sufficient condition for the e^(-st) term to go to zero for sufficiently large s.

[@fadiadaghestani3158]

Appreciate your efforts on approximation Q. How about laplace wrt 2 independent variables x1, x2 ?

[@sandeepsahanicodes]

Thank you @Mu Prime Math

[@emiliodaza2902]

thank you so much you're incredible

[@ozzydozzy4116]

Yoooo! Awesome video!

[@whysoserious0609]

Thank you!

[@egillandersson1780]

Nice !

[@calculusandmathematicslearning]

You are so beautiful love u