A Series of Interest
David W. Cantrell
payment, i the interest rate, A the loan amount, and N the number of
payments. A series is presented here for the interest rate.
In the aus.mathematics newsgroup recently, someone asked about solving such
an equation for i. Ken Pledger mentioned some appropriate links, such as
Stan Brown's <http://oakroadsystems.com/math/loan.htm> (see formula (2)
there) and the sci.math FAQ entry
<http://db.uwaterloo.ca/~alopez-o/math-faq/node76.html>. The formula cannot
be solved for the interest rate in closed form in terms of elementary
functions. Apparently the most common approach to determining i is to use a
numerical technique, such as Newton's method. But an alternative technique
is to use a series:
Letting u = (PN/A - 1)/(N + 1),
i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 - (N-1) (2N+1) (11N+7) u^4/135
+ (N-1) (2N+1)^2 (13N+11) u^5/405 -+...)
which I obtained by reversion of series. Surely this series must be well
known to those who deal with such matters often; I would greatly appreciate
references to it. (Of course, more terms of the series could be given here,
but references should make doing so unnecessary.)
Examples:
1. In example 11 from Stan Brown's page, in which P = $200,000,
A = $2,800,000 and N = 19, using just the five terms of the series shown
above, we obtain i = 3.2611% as the approximate interest rate. (For
comparison: The approximation given by Stan was 3.26%. Using an accurate
numerical technique, we find that i = 3.2596...%)
2. In the example at the end of the sci.math FAQ entry, in which P = $50,
A = $10,000 and N = 260, using just the five terms of the series shown
above, we obtain i = 10.9648% as the approximate interest rate. (For
comparison: The final approximation given in the FAQ entry was 10.997%. But
using an accurate numerical technique, we find that i = 10.9624...%)
David W. Cantrell
David W. Cantrell
> A common financial formula is P = iA/(1 - (1+i)^(-N)), where P denotes
> the payment, i the interest rate, A the loan amount, and N the number of
> payments. A series is presented here for the interest rate.
>
> In the aus.mathematics newsgroup recently, someone asked about solving
> such an equation for i. Ken Pledger mentioned some appropriate links,
> such as Stan Brown's <http://oakroadsystems.com/math/loan.htm> (see
> formula (2) there) and the sci.math FAQ entry
> <http://db.uwaterloo.ca/~alopez-o/math-faq/node76.html>. The formula
> cannot be solved for the interest rate in closed form in terms of
> elementary functions. Apparently the most common approach to determining
> i is to use a numerical technique, such as Newton's method. But an
> alternative technique is to use a series:
>
> Letting u = (PN/A - 1)/(N + 1),
>
> i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 - (N-1) (2N+1) (11N+7) u^4/135
> + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...)
>
> which I obtained by reversion of series. Surely this series must be well
> known to those who deal with such matters often; I would greatly
> appreciate references to it. (Of course, more terms of the series could
> be given here, but references should make doing so unnecessary.)
Perhaps it's not well known. But in any event, I found out just today
that the series had been already published:
H. E. Stelson, "Note on finding the interest rate" _Amer. Math.
Monthly_ 60:10 (Dec. 1963) 703-705.
As I did, he obtained the series by reversion, and did not discuss
convergence or mention the form of the general term. He then expressed
the interest rate as a continued fraction and, by considering
convergents, obtained "some very excellent approximations". I may
discuss his and various other approximations in another thread soon.
David Cantrell
Valeri Astanoff
> David W. Cantrell <DWCan...@sigmaxi.org> wrote:
> > A common financial formula is P = iA/(1 - (1+i)^(-N)), where P denotes
> > the payment, i the interest rate, A the loan amount, and N the number of
> > payments. A series is presented here for the interest rate.
> >
> > In the aus.mathematics newsgroup recently, someone asked about solving
> > such an equation for i. Ken Pledger mentioned some appropriate links,
> > such as Stan Brown's <http://oakroadsystems.com/math/loan.htm> (see
> > formula (2) there) and the sci.math FAQ entry
> > <http://db.uwaterloo.ca/~alopez-o/math-faq/node76.html>. The formula
> > cannot be solved for the interest rate in closed form in terms of
> > elementary functions. Apparently the most common approach to determining
> > i is to use a numerical technique, such as Newton's method. But an
> > alternative technique is to use a series:
> >
> > Letting u = (PN/A - 1)/(N + 1),
> >
> > i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9 - (N-1) (2N+1) (11N+7) u^4/135
> > + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...)
David,
Could you please post the code that could generate your (remarkable)
formula at any order?
Thanks in advance,
Valeri
David W. Cantrell
> DWCan...@sigmaxi.org (David W. Cantrell) wrote in message
> news:<24fe76d6.04090...@posting.google.com>...
> > David W. Cantrell <DWCan...@sigmaxi.org> wrote:
> > > A common financial formula is P = iA/(1 - (1+i)^(-N)), where P
> > > denotes the payment, i the interest rate, A the loan amount, and N
> > > the number of payments. A series is presented here for the interest
> > > rate.
> > > Letting u = (PN/A - 1)/(N + 1),
> > >
> > > i = 2( u - (N-1) u^2/3 + (N-1) (2N+1) u^3/9
> > > - (N-1) (2N+1) (11N+7) u^4/135
> > > + (N-1) (2N+1)^2 (13N+11) u^5/405 -+ ...)
>
> David,
> Could you please post the code that could generate your (remarkable)
> formula at any order?
First, I'll note again that it's not really _my_ series. Yes, I obtained it
independently, on the day that I started this thread. But I have since
found out that the series -- well, to be precise, just the first four terms
of it -- were given by H. E. Stelson in a note in the Monthly in 1953.
Below my signature is code for Mathematica which can be used to generate
the series. Note that its output is not quite as neat looking as what I had
posted. Instead of being in terms of N and my u, it's in terms of N and
s = P/A. But it should then be simple to get from Mathematica's output to
the form I had posted. While I'm doing this, since some people who don't
have Mathematica or an equivalent CAS might be interested in more terms of
the series, I'll go ahead now and show the output giving ten terms. (Of
course, just change the 10 in the code to a larger integer if you want
still more terms of the series.)
Regards,
David Cantrell
-----------------------
In[1]:=
Normal[Simplify[InverseSeries[Series[i/(1-(1+i)^(-N)),{i,0,10}],s]]]
Out[1]=
(2*N*(-(1/N) + s))/(1 + N) - (2*(-1 + N)*N^2*(-(1/N) + s)^2)/(3*(1 + N)^2)
+ (2*N^3*(-1 - N + 2*N^2)*(-(1/N) + s)^3)/(9*(1 + N)^3)
- (2*N^4*(-7 - 18*N + 3*N^2 + 22*N^3)*(-(1/N) + s)^4)/(135*(1 + N)^4)
+ (2*N^5*(1 + 2*N)^2*(-11 - 2*N + 13*N^2)*(-(1/N) + s)^5)/(405*(1 + N)^5)
- (2*N^6*(-41 - 253*N - 447*N^2 - 43*N^3 + 484*N^4 + 300*N^5)*(-(1/N)
+ s)^6)/(2835*(1 + N)^6)
+ (2*N^7*(-301 - 2823*N - 7803*N^2 - 6421*N^3 + 4272*N^4 + 9252*N^5 +
3824*N^6)*(-(1/N) + s)^7)/(42525*(1 + N)^7)
- (2*N^8*(-311 - 5860*N - 23694*N^2 - 34952*N^3 - 8455*N^4 + 30876*N^5
+ 32428*N^6 + 9968*N^7)*(-(1/N) + s)^8)/(127575*(1 + N)^8)
+ (1/(1148175*(1 + N)^9))* (2*N^9*(719 - 31924*N - 196606*N^2 - 426364*N^3
- 346849*N^4 + 132824*N^5 + 463256*N^6 + 325664*N^7
+ 79280*N^8)*(-(1/N) + s)^9)
- (1/(189448875*(1 + N)^10))* (2*N^10*(512983 - 2225859*N - 27338082*N^2
- 82514430*N^3 - 109075749*N^4 - 36930111*N^5 + 77552880*N^6
+ 109981632*N^7 + 58322064*N^8 + 11714672*N^9)*(-(1/N) + s)^10)