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Investigating expanding squares and geometric series
By Monica Anderson (P1748) on Thursday, December 23, 1999 - 06:55 pm:

Hi! I hope someone can help me out there! I'm doing this mini project on expanding squares. What happens is that each day a square grows. E.g Day 1 there would be a 8cm sided square. so the area of the square would be 64cm squared.
Day 2 there would be still the 8cm sided square but with four additional squares attached which have 4cm sides. So the total area for day 2 would be 128cm squared.
Day 3 would be same as day 2 but with 4 extra little squares added on with sides of 2cm. Total area = 144cm squared.
Day 4 would be same as day three but with 4 more squares with sides of 1cm. Total area = 148cm squared.
Do you get what's happening? The square expands each day and is connected to the first day. I've put that in a really complicated way, what i mean is that say the first square has 4cm length then it will gain 4 additional squares which will be 2cm length.
Anyway i am investigating the area covered by the squares on different days.

i have drawn couple of graphs, but am looking to find the nth term.

64, 128, 144, 148
Can you see a term for these numbers? i can see a pattern but not a term!

So could you sort help me out, please! And could someone give me tips on producing a successful investigation.

Thank you very much!

Luv,
Monica

By Alastair Fletcher (Anf23) on Thursday, December 23, 1999 - 11:32 pm:

Hi Monica,

I think I see what you mean, and I believe that you have a problem involving geometric series. Do you know what these are? For now, I'll assume you do.

Here's what's happening -
Day 1, area = 64
Day 2, area = 64 + 4*16
Day 3, area = 64 + 4*16 + 4*4
Day 4, area = 64 + 4*16 + 4*4 + 4*1
Day 5, area = 64 + 4*16 + 4*4 + 4*1 + 4*1/4
and so on.

Now, a geometric series is a sequence a of numbers an generated by two numbers, let's call them A and R. The first term in the sequence is A, so a0 = A. Then each successive term is got by multiplying the previous term by R.
So a1 = A×R
a2 = A×R2
and so on.

Let SN be the sum of the first N terms of the sequence.
SN = A + A×R + A×R2 + ... + A×RN-1 ......(1)
= SN-1 i=0A×Ri

I hope you understand everything I've done so far, and the notation isn't confusing.

Now, R×SN = A×R + A×R2 + ... + A×RN ........(2)

So, SN - R×SN = A - A×RN by subtracting equations (1) and (2) and since all the other terms vanish.

So, SN = A × (1 - RN)/(1 - R)

Let's compare this with what you've got.
Ignore the original 64 for the moment, and you'll see that the rest of the terms form a geometric series. The first term is 4×16 = 64, and R, the common ratio is 1/4.

So on Day 2 (still ignoring the original 64), you have area 4×16, then on Day 3 you add area 4×16×1/4, on Day 4 you add 4×16×1/4×1/4 and so on.

Putting numbers in my formula above, and remembering the original 64, the area on day N is
64 + 64 × (1 - (1/4)N)/(1 - 1/4).

Now, as N®¥, (1/4)N®0. That is, as N gets bigger, (1/4)N gets smaller and smaller.

If S¥ denotes the limit of SN as N gets bigger and bigger, we see that
S¥ = 64 + 64×1/(1 - 1/4)
=149 1/3
And that is what the area of the all the squares tends to as time passes!

For your investigation, what would happen if you changed ... The area of the original square? The area of squares that get added on? The common ratio, R? How about trying different shapes, or 3D shapes? How can we be sure we always get a finite area as time passes (eg. how about if we have an original square of size A, then on day 2 we add squares of size A/2, on day 3 we add squares of size A/3, on day 4 we add squares of size A/4...)?

There are plenty of directions to go in!

If you have any other questions, or you don't understand something that I've done, then write back.

Merry Christmas!
Alastair

By Monica Anderson (P1748) on Friday, December 24, 1999 - 06:25 pm:

thank you alastair for getting back to me so quick. I'm going to overlook your comments later and hope to progress with my project. I'm not really into the mathematics mood being it christmas. Well thank you again!
Have a very merry christmas and a happy new year!
luv,
Monica

By Monica Anderson (P1748) on Sunday, December 26, 1999 - 02:14 pm:

Dear alastair,

Could you possibly explain what a geometric series is as i haven't come across it before?

Thank you,

Monica Anderson

By Alastair Fletcher (Anf23) on Monday, December 27, 1999 - 01:24 am:

Hi Monica,

I'll try and explain what a geometric series is, so here goes...

First of all, a sequence of numbers is a collection (finite or infinite) of numbers in order (ie. the position of the number in a written out list is important)
eg. (i) 1,2,3,4,5,6
(ii) 2,4,8,16,32,...
(iii) 3,-54,0.432,987,-1,...
There may or may not be any order in where the numbers appear in the sequence (as with the third example)

Now, a series is the sum of all the numbers in a sequence. So with the examples above, the corresponding series' are,
(i) 1+2+3+4+5+6
(ii) 2+4+8+...
(iii) 3-54+0.432+...

The series may sum to something finite, as in the first example, or not have a finite sum.

In fact, you have to be very careful when dealing with infinite sums, as this minor detour will show -
let S = 1+2+4+8+16+... (similar to example (ii)),
then 2S = 2+4+8+16+... = S-1.
So 2S = S-1, and S=-1, but surely 1+2+4+... doesn't equal -1?
(Don't worry about this - it's just a little warning about the dangers of infinity!)
OK, detour over!

A geometric series is a special sort of series which can be completely written down given only two things. First some examples.

(a) 1 + 2 + 4 + 8 + 16 +...
(b) 1/2 + 1/4 + 1/8 + 1/16+...
(c) 3 + 4 + 16/3 + 64/9 + 256/27 + ...

You'll see that a geometric series always has an infinite number of terms. Now the nitty-gritty... we can write down a geometric series knowing just the first term of the series and one other number called the common ratio.

In example (a), the first term is 1 and the common ratio is 2. This means that any term is two times the preceding term. So, the second term is 2 times the first term ie. 2×1=2. The third term is 2 times the second term ie. 2×2=4 and so on. The 312th term will be 2 times the 311th term (which I'll leave you to work out if you really want to!)

In example (b), the first term this time is 1/2. Can you see what the common ratio is? Something times a half gives a quarter, so that something must be a half. Checking further down the sequence confirms that the common ratio is indeed 1/2.

Can you work out what the first term and common ratio are for the third example?

Generalizing into algebra (which isn't everyone's cup of tea!), we can denote the first term by A and the common ratio by R, so the series is
A + A×R + A×R2 + A×R3 + ... = S¥ i=0A×Ri
Ignore this last bit if you want to, but it provides a general framework for geometric series.

Right, I hope that helps and if there's anything you don't understand or if you have any further questions about geometric series or anything else, then write back!

Hope you had a good Christmas,
Alastair

By Monica Anderson (P1748) on Wednesday, December 29, 1999 - 06:13 pm:

Dear Alastair,

Thank you very much for taking time to help me. I am extremely grateful and thank you dearly for it

I just have a few more queries. First of all if you go back to the first time you wrote to me regarding the geometric sequence fitting in with my investigation.

Let SN be the sum of the first N terms of the sequence.
SN = A + A×R + A×R2 + ... + A×RN-1 ......(1)
= SN-1 i=0A×Ri
I don't seem to understand what you have done here and how you got that answer.


Now, R×SN = A×R + A×R2 + ... + A×RN ........(2)
So, SN - R×SN = A - A×RN by subtracting equations (1) and (2)
and since all the other terms vanish.
I am a little confused on this part too.

If S denotes the limit of SN as N gets bigger and bigger, we see that
S = 64 + 64×1/(1 - 1/4)
=149 1/3
And that is what the area of the all the squares tends to as time passes!
I understand how you achieved this answer but don't get what you are explaining; "that is what the area of all the squares tends to do as time passes."

I am sorry to be a burden to you!

Thank you again for your help!

Have a fantastic time celebrating the New Year!

Monica

By Alastair Fletcher (Anf23) on Thursday, December 30, 1999 - 12:04 am:

Hi again Monica,

---Let's start with your first query, and I'll use as an example the geometric series with first term 1 and common ratio 1/2, so it's
1 + 1/2 + 1/4 + 1/8 +...

The sequence of terms of this geometric series is 1, 1/2, 1/4, 1/8,... (Check from last time if you need to on sequences and series) So the first term is 1, the second term is 1/2, the third term is 1/4 and so on.

In general, the n'th term is 1×1/2×1/2×...(n-1 times)...×1/2 = 1×(1/2)n-1
Note that the n'th term has the common ratio raised to the (n-1)'th power, because the first term of the sequence has the common ratio raised to the 0'th power, and anything to the power zero equals 1.

If we shorten the geometric series to just the first n terms, then we have 1 + 1/2 + ... + (1/2)n-1 and this is what I called the sum to n terms, or Sn.

I used the S notation before, which basically is a shorthand way of writing out a sum - if you haven't come across it before, then I won't use it this time, but I can explain it if you would like me to.

---Onto your second question, the sum to n terms as I gave above is
Sn = 1 + 1/2 + 1/4 + ... + (1/2)n-1 and call this equation(a). If you multiply both sides of the equation by 1/2, then you should get
1/2 × Sn = 1/2 ×(1 + 1/2 + 1/4 + ... + (1/2)n-1)
= 1/2 + 1/4 + 1/8 + ...+ (1/2)n
and call this equation(b).

If you now subtract the left hand side of equation(b) from the left hand side of equation(a), you should get Sn - 1/2 × Sn and this is just 1/2 × Sn.

If you subtract the right hand side of equation(b) from equation(a), you should get
(1 + 1/2 + 1/4 + ... + (1/2)n-1) - (1/2 + 1/4 + 1/8 + ... + (1/2)n)
= 1 + 1/2 - 1/2 + 1/4 - 1/4 +... + (1/2)n-1 - (1/2)n-1 - (1/2)n by pairing up equal terms from the two brackets, and so you end up with 1 - (1/2)n because everything else cancels.

So, putting the left hand side and right hand sides together,
1/2 × Sn = 1 - (1/2)n

I hope with this example you can see that for a general geometric series with first term A and common ratio R, if you work through as above and in the first posting,
Sn = A×(1 - Rn)/(1 - R)

---For the last part, imagine you plotted the squares on a piece of graph paper and you put the new squares on every day according to the rules. Now also imagine that you happen to be immortal and plot the squares forever (failing this, you start a cult and the members have to continue your work of adding squares far into the future...).

Anyway, you will find that however long you add squares for, you will not need more than one piece of graph paper (OK, it may have to be quite large) to fit in all the squares you may ever draw. In fact, if you somehow get to an infinite time in the future, you will find that your squares will have filled in an area of 149 1/3.

The point is that because your squares follow the pattern of a geometric series with common ratio less than 1, you will always be able to fit all the squares on one (possibly quite large) piece of graph paper.

If the common ratio was greater than or equal to 1, then you will not be able to fit all the squares on one piece of graph paper (however large it is) if you do it for long enough. If you did do this, then at some point you would have filled the universe with graph paper with squares on, and you would still be adding more!

Hope this helps - if anything doesn't make sense, then I'll try and clarify.

Have a great New Year/Century/Millennium!

Alastair

PS. D'oh! Just spotted that I used n instead of N, as before, for n terms etc. Don't think it makes too much difference though...