OCR Mei Numerical Methods, 16th May 2012

Reply 60

Gould1610

Original post by Gould1610

I couldn't make my mind up and ended up going with -0.40.

The more I think about it though, the more i think i should've just left it at -0.4

Reply 61

User944645

2

I'm glad other people think similarly...the incosistency annoys me greatly, especially as in another question we had to use the ratio of differences again to talk about what they signified about the trapezium and midpoint rules...fingers crossed that the examiners made a mistake and that the grade boundaries will be lowered accordingly...

Reply 62

Gould1610

Really need this too. Like, every past paper I could do with a fair amount of time to spare and get really really good marks,
Such a shame that the real one didn't go so well.
I need 90% overall between D1, NM and FP1, too, i got 95 in D1 in january, so if i get 100 in FP1, i actually need at least 75 in NM to meet my offer. I'd hope I haven't dropped enough marks for that to be the case but everyone here knows what OCR MEI are like.

Reply 63

User944645

2

Original post by Gould1610

Really need this too. Like, every past paper I could do with a fair amount of time to spare and get really really good marks,
Such a shame that the real one didn't go so well.
I need 90% overall between D1, NM and FP1, too, i got 95 in D1 in january, so if i get 100 in FP1, i actually need at least 75 in NM to meet my offer. I'd hope I haven't dropped enough marks for that to be the case but everyone here knows what OCR MEI are like.

I seem to be in a similar position to you...this seemed so much worse than any of the past papers...luckily I have a backup plan though - for my maths AS I got 93% in c1, 100% in C2 and 94% in S1, so provided I got 50% or more in NM and don't bomb FP1 and D1 I can swap S1 for NM and get an A overall for both maths and further maths...I just hope never to have a paper as horrible as that ever again...

Reply 64

just george

Original post by Gould1610

Really need this too. Like, every past paper I could do with a fair amount of time to spare and get really really good marks,
Such a shame that the real one didn't go so well.
I need 90% overall between D1, NM and FP1, too, i got 95 in D1 in january, so if i get 100 in FP1, i actually need at least 75 in NM to meet my offer. I'd hope I haven't dropped enough marks for that to be the case but everyone here knows what OCR MEI are like.

Why do you need 90% overall? i thought for an A* you need 90% overall in A2, and 80% overall across AS and A2? :L

Reply 65

Revent

10

Whoever asked me for a scan, just tell me any question in particular you want me to type up. My internet is REALLY slow so a scan would take forever unfortunately :frown:

Reply 66

User944645

2

It wasn't me, but could you please put on the question about numerical differentiation/rates of convergence just so I can check if it was impossible to do... :smile:

much appreciated :smile:

Reply 67

Gould1610

Original post by just george

Why do you need 90% overall? i thought for an A* you need 90% overall in A2, and 80% overall across AS and A2? :L

My original Cambridge offer mentioned 'A in Further Mathematics' amongst other things, and i was like, erm, yo, I'm not even doing A2 further maths,
And they were like, oh yeah, our mistake, and gave me a revised offer which specifically says 'At least 90% in Advanced Subsidiary Further Mathematics'

Reply 68

just george

Original post by Gould1610

My original Cambridge offer mentioned 'A in Further Mathematics' amongst other things, and i was like, erm, yo, I'm not even doing A2 further maths,
And they were like, oh yeah, our mistake, and gave me a revised offer which specifically says 'At least 90% in Advanced Subsidiary Further Mathematics'

oh fair enough :smile:

fun times, grats on your offer anyway :smile:

Reply 69

Reminisce

11

I guess they'll have to have to accept both first order method and first order convergence then?

Reply 70

Revent

10

Original post by DavidH20

It wasn't me, but could you please put on the question about numerical differentiation/rates of convergence just so I can check if it was impossible to do... :smile:

much appreciated :smile:

The function g(x) has the values shown in the table.
x= 5, g(x)= 0.820 86
x=5.1, g(x)= 0.780 82
x=5.2, g(x)=-.742 73
x=5.4, g(x)=0.672 05

i) Find three estimates of g'(5) using the forward difference method with h=0.4, 0.2 and 0.1

ii) Use these estimates to show that the forward difference method has first order convergance
iii)Give the value of g'(5) to the accuracy justified explaining your reasoning.

For part ii, I believe what you had to mention was that the difference between g'(5) for h=.2 and 0.4 was double the difference between the differences of g'(0.1) and g'(0.2)

(edited 11 years ago)

Reply 71

just george

Original post by Revent

The function g(x) has the values shown in the table.
x= 5, g(x)= 0.820 86
x=5.1, g(x)= 0.780 82
x=5.2, g(x)=-.742 73
x=5.4, g(x)=0.672 05

i) Find three estimates of g'(5) using the forward difference method with h=0.4, 0.2 and 0.1

ii) Use these estimates to show that the forward difference method has first order convergance
iii)Give the value of g'(5) to the accuracy justified explaining your reasoning.

For part ii, I believe what you had to mention was that the difference between g'(5) for h=.2 and 0.4 was double the difference between the differences of g'(0.1) and g'(0.2)

I thought for part ii you just had to write out g'(x) to the number of decimal places that you could round the two most accurate approximations (h=0.2 and h=0.1) to, or close to that, depending on how much the error is decreasing by each time, like you may end up with a value that only h=0.1 rounds to.

its hard to explain, does that make any sense? :L

Reply 72

Reminisce

11

Original post by Revent

...

Does Q5 the error question with the spreadsheet say in cell A4 (basically, the one with the very small value) say MINUS 5.5*10^-17?

Reply 73

Revent

10

Original post by Reminisce

Does Q5 the error question with the spreadsheet say in cell A4 (basically, the one with the very small value) say MINUS 5.5*10^-17?

Yes

Original post by just george

I thought for part ii you just had to write out g'(x) to the number of decimal places that you could round the two most accurate approximations (h=0.2 and h=0.1) to, or close to that, depending on how much the error is decreasing by each time, like you may end up with a value that only h=0.1 rounds to.

its hard to explain, does that make any sense? :L

No idea, I learnt the whole module in a day so my knowledge isn't very sound :colondollar:

Reply 74

just george

Original post by Revent

Yes

No idea, I learnt the whole module in a day so my knowledge isn't very sound :colondollar:

hehe, il give an example: (numbers iv made up, they dont relate to any question :smile:

)

when h=0.4, g'(5) = 0.524876
when h=0.2, g'(5) = 0.520224
when h=0.1, g'(5) = 0.517898

so you could say g'(5) is 0.52 to 2 d.p. and justify it to this many d.p. as both h=0.1 and h=0.2 round to it, and they are the two most accurate approximations.

if the difference between h=0.2 and h=0.1 was really small though, the difference between h=0.1 and h=0.05 would be even smaller (depending on the order of convergence), and so if that difference became negligible, then you could justify an answer which only h=0.1 rounds to

so in the example, you could say g'(5)=0.518 perhaps.. (not sure you could really justify it in this case because when h=0.05 it looks like it would still change quite a lot, so 0.52 would be a safer bet)

if i put some workings to the last bit:
ratio of differences:

\frac{0.517898-0.520224}{0.520224-0.524876} = 0.5

(so error is halving every time h is halved)

therefore when h=0.05, g'(5) will likely be approximately 0.516735

if that rounded to 0.518 (which obviously it doesnt) but if it did, then you could justify your answer to 0.518 to 3d.p. by stating that the most accurate approximation (h=0.1) is closer to the actual value than h=0.2

does that make any more sense? or is that just random crap? :tongue:

EDIT:
Il give an example where you could justify an answer to which only h=0.1's approximation rounds to:

when h=0.4, g'(5) = 0.524836
when h=0.2, g'(5) = 0.520224
when h=0.1, g'(5) = 0.519071

ratio of differences =

\frac{0.519071-0.520224}{0.520224-0.524836} = 0.25

so error quarters every time h is halved

so next difference will be approximately 0.25 times the previous difference. so next value is likely to still round to 0.519 to 3d.p. so you could justify the answer of g'(5) = 0.519 to 3d.p. despite the value when h=0.2 not rounding to 0.519 :smile:

(edited 11 years ago)

OCR Mei Numerical Methods, 16th May 2012

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