The discussion revolves around solving the integral ∫e^(z^2) * fn(z) dz, where fn(z) is defined as the n-th derivative of e^(-z^2). Participants explore the form of fn(z) and the implications of substituting its derivative into the integral.
Participants generally agree on the mathematical properties discussed, such as the derivative of e^(-z^2) and the simplification of the integral. However, there is some confusion expressed regarding the integration process, indicating that the discussion remains partially unresolved.
There are unresolved aspects regarding the integration of the expression after substitution, as participants express confusion about the next steps.
T.Engineer said:I need to work on solving ∫e^(z^2)* fn(z) dz.
where fn(z) = d^n/dz^n * e^(-z^2).
Thanks a lot!
cliowa said:Try working out (inductively) the form of fn(z).
T.Engineer said:you mean to find the derivative for the
fn(z) = d^n/dz^n * e^(-z^2).
and then substitue it?
cliowa said:Indeed, yes. Try f1, then f2 and so on to see what's going on.
T.Engineer said:the derivative for e^-(Z^2) is
-2 Z e^-(z^2)
is not that true?
cliowa said:Yes it is. Do you notice anything special about the integral?
T.Engineer said:Now if I will substitute the derivative of fn(z)
Which is:
-2 z * e^(-z^2)
So, the integral equation will be like this:
-2 z * e^(z^2) * e^(-z^2) dz
here I get confused and I couldn’t find the integral?
cliowa said:What is [itex]e^{a}\cdot e^{b}[/itex] equal to?
T.Engineer said:you mean it will be somthing like this
e^[(z^2)*(-z^2)]
or it maybe e^(a+b)
cliowa said:There's no guessing involved there! It's [itex]e^a*e^b=e^{a+b}[/itex].
T.Engineer said:So, the integral it will be like this:
-2 Z dz
is not that right?
cliowa said:That's correct.