Recent content by bsdz

Another suggestion but please double check. Since y = g(x). I.e. y = x , x < L y = x-L , x => L Then E(y) = E(g(x)) = \int {g(x) f(x)} dx = \int_{-\infty}^{L} {x f(x)} dx + \int_{L}^{\infty} {(x-L)f(x)} dx where I am saying m(x) is equivalent to E(x). I am a bit rusty so...

In that case, it would look near identical to my original calculation and option 2. Thanks.

Thanks. That confirms option 2. I am developing a software library that automates changing of variables so sadly I can't easily quit using partials. However, I would be interested in any other algorithmic way of doing this that will work with most/all PDE/ODEs.

Hi I have a question regarding a PDE and change of variable. I can follow through the algebra but I have a problem deciding what route to take after I use the chain rule at a later point. I have an expression: - \frac{\partial^2 f}{\partial y^2} and would like to make the variable...

Consider using the roots of unity. i.e. x^4 - 14 = 0 x = e^\frac{2 \pi i k}{4} \sqrt[4]{14} ; k = 0, 1, 2, 3 That's from memory. You'll need to double check it.