Question

https://gyazo.com/a674d1e6c9c80f4ba4f5da80719869e7
Okay, so I finished Part A and B, but I'm confused on part C because of the gamma function and all. I'm thinking that I should use the bisection method for this one? Any help is appreciated.

Answer 1

@Shadow

Answer 2

@Vocaloid

Answer 3

By trying to find \(x>0\) such that \(\ln (\Gamma(x)) = 2\), you are looking for solutions to \( \Gamma(x) = e^2, ~ x > 0\)


From the definition:

\(\Gamma (x)= \int_0^{\infty } dt ~~~~ t^{x - 1} ~e^{ -t } \) [<-- that's a Laplace Transform \(\mathbb F(s)\), BTW, with \(s = 1, ~~ f(t) = t^{x-1}\)]

Then:

\( \Gamma' (x )= \int_0^{\infty } dt ~~~~ t ~ e^{ - t } ~ \ln t \qquad \triangle\)


So an approximation method using a derivative like Newton in your spreadsheet would require another spreadsheet function for \( \triangle\), I think.

And yet you have a function for the log of the Gamma function in your software so you can just try and then optimise any old guess you like. That's after having plotted the function in your spreadhseet software?

So not sure I actually get the point of any of this.

I mean, even plotting that function to find the "imitialization data" is just using software. Who carries a picture of the Ln(Gamma) function around in their head?!

So, plan of action:

1. Reckon you are right - Bisection. Using numbers the software gives you.
2. Make a table of x values in your spreadhseet software from 0 upward
3. Play with that until you see values bouncing round 2
4. Refine that table into 2 tables, focussing on each hit
5. etc etc

it looks like this:

Answer 4

Or you can just do this

https://www.wolframalpha.com/input/?i=ln+(gamma(x)+)+%3D+2