# Oxford PAT 2014, Question 2

## 7 thoughts on “Oxford PAT 2014, Question 2”

1. Hi Amy, thanks for your question.

So I assume you are using r to refer to the common ratio of a geometric series (i.e. the series is a geometric series of form a + ar + ar^2 + ar^3 … , where a = 1 and r = e^-x)? This is a perfectly good way of doing it. The step between the statements modulus e^-x <1 and e^-2x is <1 is valid because e^-2x is (e^-x)^2, and for all values n, if mod(n) < 1 then n^2 < 1.

1. Amy says:

Thanks for replying. Yes, i’m referring to the common ratio. So is it valid to to say modulus e^-×<1 or do you have to square it to ensure the value is positive or something?

1. Of course the key point here is that the (modulus of the) common ratio must be less than 1 or each successive term of the series will get bigger.

I don’t think the examiners are asking you to worry about negative common ratios, because I think they are assuming that x is just a real number (if they had wanted you to think about cases where x is a complex number then they would have said so explicitly). If x is a real number (i.e. not a complex number) then e^-x is always greater than zero. So in that case modulus e^-x is the same thing as e^-x, and there is nothing for you to worry about.

2. Amy says:

Hi there. I’ve used a slightly different method to do this but come out with the correct answer. This method involves stating that modulus r < 1, therefore modulus e^-x 0. However, having looked at other solutions on the internet they determine that modulus e^-x <1 and then jump to e^-2x is <1 achieving exactly the same answer. Why is this? Thanks

3. Harry says:

what is the e-ix thing? and why do times e^-x by s?

1. Hi Harry, thanks for your question.

The sum in the question is the sum of all values of e^-ix for x = 0 to infinity. (NB maybe I should have used n instead of x here to avoid the risk of anyone thinking that i is meant to represent the square root of -1, but I don’t think it really matters).

The first part of the answer says that if x is negative then (-ix) is positive so the terms increase in size as i increases, and so the series doesn’t converge to a finite sum.

The second part of the answer uses a standard technique for finding the sums of geometric series like this. Multiply the whole sum by the value of the term when i = 1, and this just shifts all the terms up by one place. So the term when i=1 multiplied by the sum equals the sum minus the term when i=0. This gives you an equation that you can solve to find the sum.

1. Harry says:

thanks very much