- #1
T.Engineer
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I need to work on solving ∫e^(z^2)* fn(z) dz.
where fn(z) = d^n/dz^n * e^(-z^2).
Thanks alot!
where fn(z) = d^n/dz^n * e^(-z^2).
Thanks alot!
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 alot!
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.
The integral of e^(z^2) is a special function called the error function, denoted as erf(z).
The general form of the integral ∫e^(z^2)* fn(z) dz is ∫e^(z^2)* f(z) dz, where f(z) is a function of z.
The integral ∫e^(z^2)* fn(z) dz can be solved using various methods, such as substitution, integration by parts, or applying specific integration rules.
The function e^(z^2) is crucial in this integral because it is the integrand and cannot be separated from the integral. It also has special properties that make it challenging to integrate.
There are many computer software programs, such as MATLAB and Wolfram Alpha, that have built-in functions for solving integrals. You can also use online integral calculators to solve the integral step by step.