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OCR MEI FP2 Thread - AM 27th June 2016

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Original post by Bunderwump
Can you remember what the show that in part (i) was?


I edited my post a few mins after, sorry. Pretty sure thats right, though it was pretty strange and could be wrong.
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leaf3
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How many more marks do you think I'll drop for not condensing my lns and so getting rid of the pi's in the first question? had to rush to correct a mistake at the end so forgot...
Where do you need to times by -1 in Q2? Don't you just times by j twice?

2sin(0.5x){sin(0.5x)-jcos(0.5x)} (* j)

2sin(0.5x){jsin(0.5x)+cos(0.5x)}

2jsin^2(0.5x) + 2sin(0.5x)cos(0.5x)

2sin(0.5x)cos(0.5x) = sin(x)

2sin^2(0.5x) = 1 - cos(x)

j(1-cos(x)) + sin(x)

j -jcos(x) + sin(x) (* j)

-1 + cos(x) + jsin(x)

= -1 + z

Can someone correct this please?
Original post by gregsonog
Where do you need to times by -1 in Q2? Don't you just times by j twice?

2sin(0.5x){sin(0.5x)-jcos(0.5x)} (* j)

2sin(0.5x){jsin(0.5x)+cos(0.5x)}

2jsin^2(0.5x) + 2sin(0.5x)cos(0.5x)

2sin(0.5x)cos(0.5x) = sin(x)

2sin^2(0.5x) = 1 - cos(x)

j(1-cos(x)) + sin(x)

j -jcos(x) + sin(x) (* j)

-1 + cos(x) + jsin(x)

= -1 + z

Can someone correct this please?


Multiplying by j twice is the same as multiplying by -1 so at the end do you not need to divide by -1 to get 1-z?
Original post by gregsonog
Where do you need to times by -1 in Q2? Don't you just times by j twice?

2sin(0.5x){sin(0.5x)-jcos(0.5x)} (* j)

2sin(0.5x){jsin(0.5x)+cos(0.5x)}

2jsin^2(0.5x) + 2sin(0.5x)cos(0.5x)

2sin(0.5x)cos(0.5x) = sin(x)

2sin^2(0.5x) = 1 - cos(x)

j(1-cos(x)) + sin(x)

j -jcos(x) + sin(x) (* j)

-1 + cos(x) + jsin(x)

= -1 + z

Can someone correct this please?


You have multiplied by j^2, so divide by j^2 at the end. Which will result in the correct 1-z.
ah sheeit thanks
hey guys for the complex series proof, i had it right but then tried to be picky and changed the (-2j)^n to 2^n for even when it should have been 2^n*(-1)^(n/2) . Due to cancelling my final answer was fine but my values of C and S were different - how many will i lose
(sorry somehow deleted for some reason )
How many marks will I lose altogether then for this and the second part? I showed that the binomial was (-1+z)^n and then proved the rest of the result and it all cancelled down
Original post by decombatwombat
Matrix multiplication is commutative or however you say it (that might not even be the right word I just remember it from FP1) so A x B x C = (A x B) x C = A x (B x C)


the word for that is associative :tongue: commutative means A x B = B x A, which of course doesn't hold for matrices
Original post by StrangeBanana
the word for that is associative :tongue: commutative means A x B = B x A, which of course doesn't hold for matrices


Ah yeah, I knew it was one of those ;D
Original post by ComputerMaths97
snip


I think we did about the same, except for the most part, instead of making several small errors I drained all my UMS into the pothole that was the binomial expansion of Q2. Seriously, why am I the only one who didn't know how to do that? :biggrin: I got as far as expressing it in terms of Z and then *does not compute*. Ah well.

Praise be to the Gods of strong normal Maths and sneaky beaky swapping modules around. :biggrin:
Original post by decombatwombat
C = 1 - nC1 cosx + nC2 cos2x - ... + nCn cos nx

S = 0 - nC1 sinx + nC2 sin2x - ... + nCn sin nx

Then show that C/S = cot(1/2 x)

Then you had to get 2sin(x/2)(sin(x/2) - jcos(x/2)) into 1-z


Why do you have to do

'Then you had to get 2sin(x/2)(sin(x/2) - jcos(x/2)) into 1-z' ?? How many marks would I lose if I forgot to do this?
Question 2 solution in full. In the exam I also neglected the (-1)^(n/2) when n is an even number and just left the entire thing as positive. Hopefully that will only lose 1 mark.

Part (i):
part i.jpg

Part (ii) (in two pictures):
Attachment not found

Attachment not found
part ii 1.jpg244.3KB
part ii 2.jpg239.2KB
Original post by Sophieoo1
Why do you have to do

'Then you had to get 2sin(x/2)(sin(x/2) - jcos(x/2)) into 1-z' ?? How many marks would I lose if I forgot to do this?


This was the first part, it didn't explicitly ask for this in the second part, but it wanted you to use that. Sorry should have made that clear.
LOL -i neglected the (-1)^(n/2) as well !! (my comment from above)
Original post by Bunderwump
Question 2 solution in full. In the exam I also neglected the (-1)^(n/2) when n is an even number and just left the entire thing as positive. Hopefully that will only lose 1 mark.

Part (i):
part i.jpg

Part (ii) (in two pictures):
Attachment not found

Attachment not found
Original post by chemphys
LOL -i neglected the (-1)^(n/2) as well !! (my comment from above)


Yeah I only realised because I saw your comment first :P
Original post by Bunderwump
Yeah I only realised because I saw your comment first :P


oh i see :smile: - i dont know why i did it- it was completely unnecessry as i had the right answer interms of j anyway- really hope its juts 1 mark
Original post by Bunderwump
Question 2 solution in full. In the exam I also neglected the (-1)^(n/2) when n is an even number and just left the entire thing as positive. Hopefully that will only lose 1 mark.

Part (i):
part i.jpg

Part (ii) (in two pictures):
Attachment not found

Attachment not found


the third attachment isnt working for me :/
(edited 7 years ago)
Original post by Sophieoo1
the third attachment isnt working for me :/


I'll put them here too:

Part (i): https://drive.google.com/open?id=0B6vNgTMtYIF9Vkt2am16aUJHaXM

Part (ii) 1: https://drive.google.com/open?id=0B6vNgTMtYIF9ZldqdGdScTg3RHc

Part (ii) 2: https://drive.google.com/open?id=0B6vNgTMtYIF9dHlPM24xdXRpR0k
Original post by Bunderwump

Thanks Ive got it - how many marks do you reckon I'll drop for not including the -1^n ??

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