Binomial question in A Level maths exam

Myself and @Surfer Rosa were discussing a recent Edexcel A Level maths question and I thought it warranted its own thread.

In the first part of the question you find the first three terms of the expansion of

\displaystyle \frac{1}{\sqrt{4-x}}

which are

\displaystyle \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256}

The question shows you how three possible values can be substituted in to find an approximation to

\sqrt{2}

:

x=-14

:

\displaystyle \frac{1}{\sqrt{4-x}} = \frac{1}{\sqrt{18}} = \frac{\sqrt{2}}{6}

x=2

:

\displaystyle \frac{1}{\sqrt{4-x}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}

x=-\frac{1}{2}

:

\displaystyle \frac{1}{\sqrt{4-x}} = \frac{1}{\sqrt{\frac{9}{2}}} = \frac{\sqrt{2}}{3}

Then there were two 1 mark questions:

Without evaluating your expansion,

(i) state, giving a reason, which of the three values of

x

should not be used

(ii) state, giving a reason, which of the three values of

x

would lead to the most accurate approximation to

\sqrt{2}

Our discussion was about (ii). Smaller in magnitude values of

x

do provide better approximations when there are a few terms but is it obvious without proof that

x=-\frac{1}{2}

is better given the different multipliers that lead to

\sqrt{2}

? Should this be worth more marks or do you think it's something that doesn't require proof?

I wonder what people think

@DFranklin @ghostwalker @RDKGames

Reply 1

DFranklin

18

Original post by Notnek

Our discussion was about (ii). Smaller in magnitude values of

x

do provide better approximations when there are a few terms but is it obvious without proof that

x=-\frac{1}{2}

is better given the different multipliers that lead to

\sqrt{2}

? Should this be worth more marks or do you think it's something that doesn't require proof?

I wonder what people think

@DFranklin @ghostwalker @RDKGames

The first missing term in the expansion is going to be Cx^4 (for some C). As such, the improvement by dividing x by 4 is 1/4^4, or 1/;256. It's clear this is going to dominate any small constant multipliers.

But at this level I think it's enough to say the series will converge more rapidly for smaller x, though it's pretty hard to guess what the examiners are thinking these days.

Edit: it appears I've fallen victim (again) to replying without having the actual equation visible while writing the reply. x^4 should be x^3 and 256 should be 64. The rest still holds.

(edited 4 years ago)

Reply 2

ghostwalker

17

Original post by Notnek

Our discussion was about (ii). Smaller in magnitude values of

x

do provide better approximations when there are a few terms but is it obvious without proof that

x=-\frac{1}{2}

is better given the different multipliers that lead to

\sqrt{2}

? Should this be worth more marks or do you think it's something that doesn't require proof?

I wonder what people think

I would say, perhaps naively:

Given that x=-1/2 is a quarter of x=2 in magnitude, i.e. a factor of 1/4, then this more than offsets the multiply by 3 rather than by 2 (factor of 1.5 between them) to get to

\sqrt{2}

.

So, two facts required to draw the conclusion, so 2 marks, IMO.

(edited 4 years ago)

Reply 3

Surfer Rosa

10

C+Ped from the other thread:

I think there might be more to it [...] because of the (rational) multipliers of sqrt(2).

This is what I came up with, though granted it seems like a lot for 1 mark.

The expansion is a(x) + r(x) ie the three term expansion plus remaining terms.

For x = 2

sqrt(2) = 2*a(2) + 2*r(2)

For x = -1/2

sqrt(2) = 3*a(-1/2) + 3*r(-1/2)

Comparing the remainders indicates how close the approximation is

abs(3*r(-1/2)) < abs(2*r(2))

Therefore -1/2 is a closer approximation

This was met with the response: Now, can you motivate the last disequality of the two absolute values?

Right let's see, let r(x) = mx^3 + nx^4+...

and then, for instance, compare r(2) and (3/2)*r(-1/2).

It's still not completely cleared up but I'd like to think I've got my mark.

Reply 4

Notnek

OP

21

Original post by DFranklin

But at this level I think it's enough to say the series will converge more rapidly for smaller x, though it's pretty hard to guess what the examiners are thinking these days.

This was my thinking too but I wasn't sure about what level of detail they required for 1 mark. If they expect more than a simple "smaller x" argument then it seems harsh for 1 mark at A Level and I expect most students will lose the mark.

Reply 5

Surfer Rosa

10

Given it was the last part of the question and they'd already asked one type of validity question, and given the level of insight/detail sometimes required last year I think they might have been aiming at mention of the remainder term. (There's also the possibility they've not thought it through.)

I do think the multipliers have to be considered, but once you do it's difficult to know where to stop.

I can't see what a plausible and rigorous-enough 1 mark response would be.

Original post by Notnek

This was my thinking too but I wasn't sure about what level of detail they required for 1 mark. If they expect more than a simple "smaller x" argument then it seems harsh for 1 mark at A Level and I expect most students will lose the mark.

Reply 6

DFranklin

18

Original post by Notnek

This was my thinking too but I wasn't sure about what level of detail they required for 1 mark. If they expect more than a simple "smaller x" argument then it seems harsh for 1 mark at A Level and I expect most students will lose the mark.

It's not very reasonable for them to expect more, IMHO.

I know I said "well, the first missing term is Cx^3, so the error will scale like x^3", but that's not actually that obvious (i.e. it's blatantly not true near the limit of convergence). In this case, the underlying justification is either "well, looking at the Lagrange remainder I know the error scales like x^3 / (1-x)^K (for some K I can't remember), so all is well", or "the ratio of successive binomial coefficients is tending to 1 from below, so the behavior is at least as good as a GP in (x/4), and so the error in truncating isn't going to be worse than the sum of that gp starting with the x^3 term". And that's just to claim "the error isn't bigger than this". To be sure that the truncation isn't unexpectedly good (for x = 2), you'd really need to consider the x^3 and x^4 terms (i.e. you'd need to show that error of expanding to x^4 was small, and therefore missing the x^3 term gives a "real" error). And either of this is totally unreasonable to accept anyone to come up with at A-level IMHO.

[Or to put it another way, they might be looking for an argument like mine, but given it's not "really" correct, I'm not sure it's reasonable to penalize someone for not giving it].

(edited 4 years ago)

Reply 7

T!gger34

15

Isn't this something to do with the rate or speed of convergence?
The "quicker" you can get the remainder r(x) to 0, the better your approximation.
This would surely indicate that the smaller you can select x to be, the faster it will converge.

Reply 8

DFranklin

18

Original post by dextrous63

Isn't this something to do with the rate or speed of convergence?
The "quicker" you can get the remainder r(x) to 0, the better your approximation.
This would surely indicate that the smaller you can select x to be, the faster it will converge.

What you say is the intuitive answer, and I do think it's what was expected. But there are two complicating issues here:

To get your final result for sqrt(2), you have to multiply the series expansion by a constant. The constant when x = 1/2 is 3 and when x = 2 it's 2. So there is an issue that for the smaller x, you're multiplying the error in the expansion by a bigger constant. The smaller x does "win" over the bigger constant here, but how obvious is this?

The 2nd issue is that "the series converges faster for small x" doesn't necessarily mean "the series is a better approximation for small x" (over a particular bigger value of x). The Taylor series for y = cos x is y = 1 -x^2/2! + ... If we made a first order approximation, we'd just get y = 1. Note that this is a better approximation to cos x when x = 2pi than when x = 0.01 In our case, it takes a bit of thought to justify that the same thing can't happen here (i.e. to show that x = 2 doesn't give an unexpectedly good approximation relative to x = 1/2). Actual analysis of "how big the error is" is generally only covered at degree level.

Reply 9

T!gger34

15

Original post by DFranklin

Actual analysis of "how big the error is" is generally only covered at degree level.

Indeed. I might dig out my old copy of Green's book from the loft at some stage :wink:

He lectured me at Warwick back in 1983 on such matters. Incredible man. Wrote the whole course on the board with no notes whatsoever with references to other lemmas by name and number. Must have had some kind of photographic memory. :smile:

I digress though :colondollar:

Edit - Green's book is here

(edited 4 years ago)

Binomial question in A Level maths exam

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