How Do You Solve for 'z' in a Laplace Transform Equation Using HP50g?

nenadb
Messages
1
Reaction score
0

Homework Statement


Hi, I am writing a little program (I need it for my course from Theory of Reliability) and as I am neither experienced mathematician nor programmer I faced with this problem.

What would this formula

original?v=mpbl-1&px=-1.jpg


Look like if I would sole it for 'z' not for 'Fz' ?

Homework Equations



Right now equation in my HP50g (I managed to do that) for solving for Fz looks like:
\<<
"Enter z" "" INPUT STR\->
\-> z

'Fz=1./\v/(2.*\pi)*\.S(0.,z,EXP(-(1./2.)*z^2.),z)'

"F(z)"
\->TAG
\>>

Thank you in advance,

Regards,
Nenad
 
Physics news on Phys.org
That does not seem like a Laplace transform to me.

Anyway, to solve for -\frac{1}{2}z^{2}, you could multiply both sides by \sqrt{2 \pi}, apply the fundamental theorem of calculus, and then apply the natural logarithm to both sides.
 
Your notation is bad and your question is not well stated. I assume you are defining the function F(z) = int[exp(-t^2 /2) dt: t= 0..z]/sqrt(2pi), and you want to solve the equation F(z) = p; that is, you want to find which value of z. Is that your question? There is no exact, finite expression for F(z) or it's inverse. You need to use numerical methods and/or approximations.

RGV
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top