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OCR FP3 June 2009

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Reply 60
Avatar for gyyy2807
gyyy2807
14
Also, for the intergration part of qu7, did anybody get

1/3 X ln (root2) where a = 1/3 and b = root 2
Reply 61
Avatar for charlie123
charlie123
samski1
can anyone give a solution for 2(ii) i disproved it by introducing a new variable that was theta-pi for when theta was over pi. but this gave a negative coefficient on the whole thing.

here's 4:

Spoiler



whats 2 ii ;s question i think i got it?
Reply 62
Avatar for around
around
13
Well, the constant term of your linear factor has to divide 1, so that narrows down your choices slightly.

plus (t-1) was obviously not a solution (co-efficents didn't sum to zero) so you were left with a pretty easy choice.
Reply 63
Avatar for Xei
Xei
6
Also, for the intergration part of qu7, did anybody get

1/3 X ln (root2) where a = 1/3 and b = root 2

Pretty sure that's right. :smile:
gyyy2807
Also, for the intergration part of qu7, did anybody get

1/3 X ln (root2) where a = 1/3 and b = root 2


Yes, though written as 16ln⁡2\dfrac{1}{6} \ln 2.
Reply 65
Avatar for milisa
milisa
yay phew lol i was goin mad then haha
Reply 66
Avatar for gyyy2807
gyyy2807
14
Glutamic Acid
Yes, though written as 16ln⁡2\dfrac{1}{6} \ln 2.

oh crap. do you reckon it'll be ok if i just left it as 1/3 ln (root2)?
Reply 67
Avatar for samski1
samski1
charlie123
whats 2 ii ;s question i think i got it?


prove closure of the group which had reiθre^{i\theta}

i said, if you multiply 2 together and theta is greater than pi then you have:

rei(ϕ+π)re^{i(\phi+\pi)}
where phi = theta - pi

then you can use e^ipi = -1 :

=−rei(ϕ)=-re^{i(\phi)}

wooohoooo disproof!
Reply 68
Avatar for Xei
Xei
6
The question said prove it. Disproving it isn't proving it.

Basically I just had pqe^(a+b+n2pi)i which is obviously of the same form.
gyyy2807
oh crap. do you reckon it'll be ok if i just left it as 1/3 ln (root2)?


Should be fine; they didn't specify that b has to be an integral or rational.
Reply 70
Avatar for samski1
samski1
i know it said prove it :frown: was a sarcastic woohooo :P

how did you get the n2pi?
Reply 71
Avatar for Xei
Xei
6
Oh okay lol :tongue:

You have to subtract 2pi to get it back into the given range. That's fine because you're essentially dividing by e^2pi which is 1.
Reply 72
Avatar for abacathoo
abacathoo
mr m here?
To my chagrin, I couldn't sneak the paper out at the end of the exam to upload here.
Reply 74
Avatar for samski1
samski1
ahhhhh i was trying to subtract pi to get back in the range. didnt think of 2pi lol... i had the right jist of it though...

for part (iii) did u have e^(2gamma - 2pi) ?

any solutions on 1 floating about. i was glad that the modulus was 1 for sure but still not sure if my answers were right. fractions like 25/18 popping to mind...
samski1
ahhhhh i was trying to subtract pi to get back in the range. didnt think of 2pi lol... i had the right jist of it though...

for part (iii) did u have e^(2gamma - 2pi) ?

any solutions on 1 floating about. i was glad that the modulus was 1 for sure but still not sure if my answers were right. fractions like 25/18 popping to mind...


Yeah, I had ei(2γ−2π)e^{i(2 \gamma - 2 \pi)}.
Reply 76
Avatar for Xei
Xei
6
samski1
ahhhhh i was trying to subtract pi to get back in the range. didnt think of 2pi lol... i had the right jist of it though...

for part (iii) did u have e^(2gamma - 2pi) ?

any solutions on 1 floating about. i was glad that the modulus was 1 for sure but still not sure if my answers were right. fractions like 25/18 popping to mind...

I think I recognise both of those things.
Reply 77
Avatar for samski1
samski1
just scanning and uploading. is it legal for me to post them here?
Reply 78
Avatar for NesQuiK.
NesQuiK.
OP
13
Glutamic Acid
Yeah, I had ei(2γ−2π)e^{i(2 \gamma - 2 \pi)}.

********. Had -2pi, forgot to times it by i
ARGH!
samski1
just scanning and uploading. is it legal for me to post them here?


Morally legal, yes.

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