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Geometric series

Here’s a technique that is well worth mastering, because geometric series come up over and over again in anything to do with multipliers, investments and also in intermediate macro. Typically you are faced with a problem which has some sort of series in it like this: S=1+x+ x^2 +x^3 +x^4 +.....x^n You have to find the sum of the series.

There’s an easy technique which sorts this out. You just multiply both sides by 1-x. Here’s why it works:
S(1-x)=[1+x+ x^2 +x^3 +x^4 +.....x^n]-[x+ x^2 +x^3 +x^4 +.....x^{n+1}]

Look at what happens now to the right hand side of the equation, pretty much everything cancels out, and you are just left with S(1-x)=1-x^{n+1} so S=\frac{1-x^{n+1}}{(1-x)}. You can multiply top and bottom of the right hand side by -1 to express it in a form which might be a bit more convenient to use for calculations: S=\frac{x^{n+1}-1}{(x-1)}

If x>1 then the series will ‘explode’ (just keep getting bigger), if -1<x<1 then the series will converge around a value.

Here's an example. If you have a fixed rate mortgage for £100000, with a constant interest rate of 5% compounded annually, over a 25 year term, what would be the monthly repayment (if it is the same every month?)

You can do this by working out what you will owe at the end of each year and seeing if you can spot the pattern.

At the end of the first year, you owe 100000(1.05)-12x. Your original debt of 100000 has increased in value by 5% but you have made 12 monthly payments of value x.

At the end of the second year, you owe (100000(1.05)-12x)(1.05)-12x. This is basically
your debt at the end of the first year multiplied by 1.05 minus the 12 monthly payments you made in the second year. This can be rewritten as 100000(1.05^2)-12x(1.05)-12x

At the end of the third year, you owe 100000(1.05^3)-12x(1.05^2)-12x(1.05)-12x

At the end of the fourth year, you owe 100000(1.05^4)-12x(1.05^3)-12x(1.05^2)-12x(1.05)-12x

Now you can see the pattern developing.
At the end of the 25th year, you owe 100000(1.05^{25})-12x(1.05^{24})-12x(1.05^{23})-......-12x. Of course as this loan is a 25 year term, at the end of the 25th year you owe nothing, so we have an expression in terms of x: 0 = 100000(1.05^{25})-12x(1.05^{24})-12x(1.05^{23})-......-12x

We can rearrange this to say 100000(1.05^{25})=12x(1.05^{24})+12x(1.05^{23})+......+12x and take a factor of 12x out of the right hand side to get 100000(1.05^{25})=12x(1.05^{24}+1.05^{23}+......+1).

When I see a series like this I like to write the terms in the bracket in increasing order, starting with 1, so I would re-write that equation as 100000(1.05^{25})=12x(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24}). Call this equation (1).

Now it is starting to look a bit more manageable. We can use that trick I explained above, to sort out the series in the bracket. Define the series as S, so that S=(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24}).

Now you just multiply both sides by (1-1.05) to get S(1-1.05)=(1+1.05+1.05^2 + 1.05^3 + .....1.05^{24})-(1.05+1.05^2 + 1.05^3 + .....1.05^{25}).

This cancels down beautifully to getS(1-1.05)=(1-1.05^{25}) and hence S=\frac{(1-1.05^{25})}{(1-1.05)}

Now we can sub that into (1) above, 100000(1.05^{25})=12x \frac{(1-1.05^{25})}{(1-1.05)}

A bit of rearranging gets it in terms of x, \frac {100000(1.05^{25})(1-1.05)}{12(1-1.05^{25})}=x so x works out as £591.27

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Categories: Geometric series, Maths
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