QTTDTOT

What don't you get?

>>8532894
usually the equations start out not even close to the form they need to be in to solve them. I was wondering if anyone had any tricks, or tips on what to look for when you are given something crazy and expected to make something out of it.

>>8532917
learn what the standard transformations are, like [1] and [t^n]
For things like cos or sinh use the exponential forms.

You can just directly integrate and evaluate it

>>8532877
The expression for which you're trying to find the inverse transform is usually rational (i.e. a ratio of polynomials in s).

In that case, the first step is to factor the denominator. The second step is partial fraction expansion, i.e. re-write the expression as the sum of e_i/(s-r_i) terms where the r_i are the roots. If you have repeated roots, then you need to use e1/(s-r)+e2/(s-r)"2 etc (as e1/(s-r)+e2/(s-r) will only give you (s-r) in the denominator rather than (s-r)^2).

Typically, the degree of the numerator is one less than that of the denominator, so the e_i will be polynomials of degree one less than the denominator.

Once you've done that, the individual fractions will usually have fairly straightforward inverse transforms, and L^-1{F+G}=L^-1{F}+L^-1{G}.

If the polynomials have high degree, you will often have to find the roots of the denominator numerically.

Finding the e_i for high-degree polynomials is usually done using the residue method:

Consider f(s)/g(s) = e1/(s-r1)+e2/(s-r2)+e3/(s-r3)+... where the denominator g(s)=((s-r1)(s-r2)(s-r2)...) with all roots distinct. The numerator f(s) is
f(s) = e1/(s-r1)+e2/(s-r2)+e3/(s-r3)+...
= e1(s-r2)(s-r3)... + e2(s-r1)(s-r3)... + e3(s-r1)(s-r2)... + ...

Note that each term has a factor of (s-r_i) for all r_i except the one matching e_i, so f(r_i) will have all terms except one equal to zero, leaving you with f(r_i)=e_i(s-r1)(s-r2)...., so e_i=f(r_i)/(g(s)/(s-r_i)).

>>8532877
Is Schrodinger's cat the quantization of the Monty Hall Problem? Or would the uncertainty principal come into play somehow?