# Thread: Geometric Series or Complex Polynomial

1. ## Geometric Series or Complex Polynomial

Before I post the problem I should check that this is being posted in the right forum. Not being a mathematician myself I am not even certain of the area of maths I should be posting the question. Some say the solution would be solved with a complex polynomial and others say it is a variation on a geometric series). Apologies if this is in the wrong forum and would you kindly point me to the right one.

In non mathematical terms here is the problem. I have clients who ask me to calculate a return on their investment from existing pension portfolios. The inputs I usually receive are the dates of lump sums invested and amounts and the dates of regular monthly sums invested, their respective dates and amounts. Then I look up the current value and want to try to figure out what the return or ever something resembling a best estimate of a return would be.

example case
a) Lump sum invested in December Jan 2011 of £55,186 plus
b) Regular monthly contributions made since Jan 2011 amounting to £64,220. Monthly contribution amounts have varied between £1394 per month and £1878 per month.
c) Total value of contributions is a+b (£120,036) however b has been invested over a period of time.
d) Value today is £149,000

I would be happy taking average value of monthly contributions as a guide if that would make the problem easier to solve.

Any help would be appreciated.

2. ## Re: Geometric Series or Complex Polynomial

To find the "rate of return", I assume you want the average rate of return (as market fluctuations continuously change the instantaneous rate of return). Assume it is a constant interest rate. Assume it is compounded monthly. Now, this is a problem we can solve. Set it up like you are calculating compound interest. Let's say the client invests for $n$ months. Let's say $p_k$ is the amount the client invests in month $k$. So, $p_0$ is the principle investment: £55,186. The total value (after compounding interest) would be:

$\sum_{k=0}^n p_k\dfrac{\left(1+\dfrac{i}{12}\right)^{n-k}-1}{\tfrac{i}{12}} = £149,000$

So,
$\sum_{k=0}^n p_k\left(1+\dfrac{i}{12}\right)^{n-k} = \dfrac{149,000i}{12}+120,036$

As some have told you, this would involve solving a complicated polynomial (the variable I am using is $i$, which is "annual interest", or annual rate of return). However, there are techniques for solving complicated polynomials that take relatively very little time (especially when you know the approximate range for zeros of the polynomial).

Alternately, if you use the average monthly contribution, then you can simplify the problem slightly to:

$p_0\left(1+\tfrac{i}{12}\right)^n+p_{avg}\sum_{k=1 }^{n}\left(1+\tfrac{i}{12}\right)^{n-k} = \dfrac{149,000i}{12}+120,036$
The LHS simplifies to this:

$p_0\left(1+\tfrac{i}{12}\right)^n+p_{avg}\dfrac{ \left( 1+\tfrac{i}{12} \right)^n - 1 }{ \tfrac{i}{12} } = \dfrac{149,000i}{12}+120,036$

This is a simpler polynomial, but will still require a computer algebra system to solve.

3. ## Re: Geometric Series or Complex Polynomial

I think I need a bit of time to understand what is going on here. thank you for your reply. I was advised to post the question on business maths and am also looking at whether this can be done with excel using IRR formulas.

If I had to choose the method it would be excel however I have no way of checking the answers once excel comes up with an answer. Here is what I posted in the business maths section

Date Payments
Mar-14 £1,879
Feb-14 £1,879
Jan-14 £1,879
Dec-13 £1,879
Nov-13 £1,879
Oct-13 £1,879
Sep-13 £1,879
Aug-13 £1,879
Jul-13 £1,879
Jun-13 £1,746
May-13 £1,746
Apr-13 £1,746
Mar-13 £1,746
Feb-13 £1,746
Jan-13 £1,746
Dec-12 £1,514
Nov-12 £1,514
Oct-12 £1,514
Sep-12 £1,514
Aug-12 £1,514
Jul-12 £1,514
Jun-12 £1,514
May-12 £1,514
Apr-12 £1,514
Mar-12 £1,514
Feb-12 £1,514
Jan-12 £1,514
Dec-11 £1,401
Nov-11 £1,401
Oct-11 £1,401
Sep-11 £1,401
Aug-11 £1,401
Jul-11 £1,401
Jun-11 £1,401
May-11 £1,401
Apr-11 £1,401
Mar-11 £1,394
Feb-11 £1,394
Jan-11 £1,394

I will ask a friend to help me out with the calculations.