OCR MEI Numerical Methods 12th June 2015

This year was fair, what about you?


Posted from TSR Mobile

Doing core 3 first warmed me up and boosted my confidence for NM.

I am certain I got most of it right, unfortunately I didn't finish question 7, I plotted the graph but then I realized they asked me to plot it between 0 and 1 so I didn't bother drawing secant method sketches.

I only got up to the x₄, I asked for extra paper to do x₅ but I only got as far as writing the formula and writing the first two digits (0.6)


With the integration question.

I said mid point rule, trapezium rule could not be done for all the points because f(0.3) and f(0.6) were not provided


At first I was going to do simpsons rule with two parabolas, I interpolated 0, 0,1 and 0.2 and integrated it exactly.

I was going to add the area under the parabola passing through 0.2, 0.4 and 0.6 but then I realized they didn't give me f(0.6) so I used mid-point rule and multiplied 0.4 by f(0.4) (forgot what it was)

So basically I added the parabola area and midpoint area to get my best estimate for integral I, that gave me 1.18 something but I was unsure to what degree of accuracy I should give.

I then went back and use Lagrange interpolation to find a cubic passing through all of the given points. I integrated between 0 and 0.6 and found the area was 1.16

So I said my best estimate was "1.2", to one decimal place.

In the second part where I had to explain what I would do different if they'd given me f(0.6), I wrote I would use simpsons rule with two parabolas.

remember the centrail drifference?
.23467
.2348
.235
not sure exact what did you give ur final answer as and why > i put 0.2 as 3 h round to it

I meant 0.24, Yeah I put that though.

You can't use central difference because you have no x values less than 1, they tell you to use forward difference

can rememer the exact answer but i got 1/u = 0.090909 and 1/v something aroud both to 5 s.f. and then calulcated it;

for secant thing i did about 7 or more iterations? you too or?

Ok.

I only had time to do four integrations, up to $x_4$ and I only wrote "0.6" for $x_5$ because the invigilators took my paper away before I could write everything else.

Not too bad other than the lagrange question which I only realised how to do at the end so ran out of time but the rest was ok i think

remember the centrail drifference?
.23467
.2348
.235
not sure exact what did you give ur final answer as and why > i put 0.2 as 3 h round to it

Good good, the weird question was on the midpoint trapezium rule estimates. I was unsure about that and there's probably many ways that people did it that are all quite valid...


Posted from TSR Mobile

I did Simpsons for the first integral (between 0 and 0.2 ) and mid-point for the second (between 0 and 0.6) then added them together,

But I also tried to interpolate everything with a lagrange polynomial (since 4 points were provided) and integrated it between 0 and 0.6.