Financial Mathematics : Annuity loan with a different first period? Edit:Typo in $\LaTeX$ transcription.
Edit 2: Please note that my assumption here is wrong. The correct equation is available in my separate answer.
When trying to teach myself some finance math I've stumbled to get the exact same decimals for the annuity payment as the paper I'm reading. Basically they are discussing an annuity loan with an extended first period. They only give the answer not the calculation. When revalidating their calculations I end up with some decimals wrong for the annuity compared to the result in the paper.
The first period is ('d' days + 1 full period).
The paper doesn't give the equation they are using to calculate the annuity so I derived it myself as follows.
My thinking is that this should yield the basic discounting eq below where I simply extend the first period with a proportion of the period that the days extension represents and add full periods after that. Here $PV$ =The present value of the loan, $r_p$ is then period interest, $N$=the total number of period in the loan. The year is assumed to have 365 days in the paper as well.
$$
{\it PV}=\sum_{k=1}^N {\it x}\, (1+r_p)^{-k-12d\,/\,365}
$$
which directly solves for the annuity payment x to
$$
{\it x}={\frac {{\it PV}\,{\it r_p}}{1-\left( {\it r_p}+1 \right) ^{-N}}\left( {\it r_p}+1 \right) ^{{\frac {12\,d}{365}}}}
$$
With the same numerical values as in the paper (extending the first period by 3 days)
$PV=200,000,\, rp=0.06/12,\, N=240,\,d=3$
I get 1433.56715 with the equation above but the paper states (with their equation - which they don't give) 1433.56521.
In other parts of the paper I get the exact same results as they for d=0. This gives me reason to believe my derivation above is flawed and that they use a slightly different equation.
Having giving it some thoughts I fail to spot my logical error. Unfortunately they don't discuss this further in the paper.
(The finance package (payodd()-function) in matlab gives a third result stating this should be 1433.6780 ... but don't give details about how this is calculated - useless ..)
Have anyone an idea of where I go wrong in my thinking?
 A: You are right. Computed with illimited precision and then converted to decimals, the result is
$$x=1433.5671466104108579$$ But take care : writing $rp=\frac{0.06}{12}
$ is not the same as writing $rp=\frac{6}{1200}$.
Computing exactly $\frac {dx}{dr}$ gives a value of $138666.20$. So, if instead of
$rp=\frac{0.06}{12}$ the program computes for  $rp=\frac{0.06}{12}+10^{-7}$ it makes a difference of $0.0139$ for $x$
A: As i suspected my assumption of deriving the equation from a plain discounting was wrong. The correct equation is given below. I derived this equation by algebraically evaluating some terms in a annuity table and setting up the boundary condition that the total amortisation sums up to the PV (present value), i.e. the loan. The resulting pattern made it easy to spot the underlying equation which is the one given below.
Strangely enough I haven't been able to find this equation in any of the financial literature i searched. However, it must be a well known formula. For $d=0$ it collapses to the standard discount calculation of an annuity. However, I noticed that several approaches uses iteration to numerically calculate a suitable value.
$$
x = {\frac {{\it PV}\,{\it r_p}\, \left( 1+{\it r_p} \right) ^{N-1}}{
 \left( 1+{\it r_p} \right) ^{N}-1} \left(  \left( 1+{\frac {12\,d}{365
}} \right) {\it r_p}+1 \right) }
$$
Using this equation with the values
$PV=200,000,r_p=0.06/12,N=240,d=3$
gives $x=1433.5652$ which is the numerically correct value.
The above equation will give some more insight in the impact of a 'd' days extension of the first period by by rewriting it as
$$
x={\frac {{\it PV}\,{\it r_p}}{1- \left( 1+{\it r_p} \right) ^{-N}}}\cdot {\frac {1 + \lambda\,{\it r_p}}{1+{\it r_p}}}
$$
where
$$
\lambda=\left( 1+{\frac {12\,d}{365}}\right)
$$
The above equation can be seen to collapse to
$$
x={\frac {{\it PV}\,{\it r_p}}{1- \left( 1+{\it r_p} \right) ^{-N}}}
$$
for $d=0$ which is consistent the standard annuity equation.
Now all that remains is for me to logically understand why my first approach in my OP is wrong :-)
