To Summarize: I'm working on a problem from an academic journal and would be grateful for somebody to check my working.
Hi folks, I'm currently trying to replicate a method for finding value of a parameter after implementing the Levenberg-Marquardt algorithm, by following an academic paper. I have already implemented the algorithm to find estimates for three parameters. By following the paper, I'm left with the following equation:
$\frac{a}{b}$=(($1-2e^\frac{-(c)}{d}$+$ e^\frac{-(f+c)}{d})$)$(\frac{1-e^{\frac{-g}{h}}}{1-e^\frac{-g}{d}}$)-(($\frac{i}{b}$cos$\alpha$)$(e^{\frac{-(f+c)}{d}}))$
Which I then need to solve for the variable 'd'. I have replaced the variables with letters for simplicity.
What I was hoping for, is for somebody to check my answer to see if the working makes sense. Thanks a lot for your time.
Multiplying both sides by $1-e^{\frac{-g}{d}}$ and $b$:
$\frac{a}{b}(1-e^{\frac{g}{d}})b=(1-2e^{\frac{-c}{d}} +e^{\frac{-(f+c)}{d}})(1-e^{\frac{-g}{h}})-(icos\alpha)(e^{\frac{-(f+c)}{d}})$
Then multiplying out brackets and simplifying:
$a=ae^{\frac{-g}{d}}=1-2e^{\frac{-c}{d}} + e^{\frac{-(f+c)}{d}} +e^{\frac{-g}{h}} +2e^{\frac{-g}{h}}e^{\frac{-c}{d}}-e^{\frac{-g}{h}}e^{\frac{-(f+c)}{d}} -icos\alpha . e^{\frac{-(f+c)}{d}}$
Then simplifying out exponentials:
$ln(a)-(ln(a)-\frac{g}{d})=ln(1)-(ln(2)+\frac{(-c)}{d})-\frac{(f+c)}{d} +\frac{g}{h} + ln(2) -\frac{g}{h} -\frac{c}{d} -\frac{g}{h}-\frac{(f+c)}{d}$
Then multiplying by $d$:
$dln(a)-(dln(a)-g)=dln(1)-(dln(2)-c)-f-c+\frac{gd}{h}+dln(2)-\frac{gd}{h}-c-\frac{gd}{h}-f-c$
Multiplying out brackets and taking all terms with d to one side:
$dln(a)-dln(a)-dln(1)+dln(2)-\frac{gd}{h}-dln(2)+\frac{gd}{h}+\frac{gd}{h}=-g+c-c-f-c-f-c$
Then simplifying:
$-dln(1)+\frac{gd}{h} = -g-2c-2f$
Factoring out $d$ on LHS:
$d(-ln(1)+\frac{g}{h}) = -g-2c-2f$
Then finally, solving for d:
$d=\frac{-g-2c-2f}{-ln(1)+\frac{g}{h}}$
Thanks a lot!