The Student Room Group

Riemann sums

Hi,
I’m attempting part a of this question but I am really not sure what I am doing, can someone please tell me if I am on the right track? The answer is clearly one but I’m not good with doing it this way and it doesn’t look my answer will end up being 1

0B893AA6-68A5-4954-97F6-FE71468ECE40.jpg.jpeg
21BF8C07-AB60-45ED-9235-2E93E0C8C701.jpg.jpeg

Scroll to see replies

Reply 1
446103C3-83F1-451F-B2CD-83B7C387CF98.jpg.jpeg
Better quality image of working of it helps
Reply 2
Probably been a bit neater to note its symmetric about pi/2 and using a centered form of the sum / formula like

But not worked it through.
Reply 3
Original post by mqb2766
Probably been a bit neater to note its symmetric about pi/2 and using a centered form of the sum / formula like

But not worked it through.


Is there anything wrong in the working? Since I end up with root 3/4, and what is the formula that you mentioned? it doesn’t seem to show up in your reply
Reply 4
Original post by grhas98
Is there anything wrong in the working? Since I end up with root 3/4, and what is the formula that you mentioned? it doesn’t seem to show up in your reply


Your working is stilll a bit blurred. The ans is 1 as you say, so you must be dropping a simple factor somewhere. The formula is in the youtube (about 12 min). Its just a way to centreing the series which makes it easier to map the num and denom to trig terms.
Reply 5
Original post by mqb2766
Your working is stilll a bit blurred. The ans is 1 as you say, so you must be dropping a simple factor somewhere. The formula is in the youtube (about 12 min). Its just a way to centreing the series which makes it easier to map the num and denom to trig terms.


Alright I guess there must be a mistake in there somewhere, may I ask is the way I evaluate the fraction near the end for N->infinity alright? For the top part I say it converges to 1 - e^pii/3 and for the bottom I try to use Taylor series ( I can’t just say 1 - e^0 )

Also how on earth do you do part b? Is there something I can look up because I’ve not actually seen a question like that before
Reply 6
Original post by grhas98
Alright I guess there must be a mistake in there somewhere, may I ask is the way I evaluate the fraction near the end for N->infinity alright? For the top part I say it converges to 1 - e^pii/3 and for the bottom I try to use Taylor series ( I can’t just say 1 - e^0 )

Also how on earth do you do part b? Is there something I can look up because I’ve not actually seen a question like that before

Can you upload a better pic?
Reply 7
Original post by mqb2766
Can you upload a better pic?


Of the question or the working?
Reply 8
Original post by grhas98
Of the question or the working?


The pictures are quite clear it is just TSR puts them at a lower quality for some reason

But for part b what is the strategy?
Reply 9
Original post by grhas98
Of the question or the working?

Working but the question aint great either. For b) you should have something like
https://math.stackexchange.com/questions/139183/differentiating-definite-integral
Reply 10
Original post by grhas98
The pictures are quite clear it is just TSR puts them at a lower quality for some reason

But for part b what is the strategy?


Can you put them on another site and link it or ...? Dont think you can attach images to dms here.
Reply 11
Original post by mqb2766
Working but the question aint great either. For b) you should have something like
https://math.stackexchange.com/questions/139183/differentiating-definite-integral


But I’m this question x is also inside the function within the integral, how might I deal with this?
Reply 12
Original post by grhas98
But I’m this question x is also inside the function within the integral, how might I deal with this?

Do one thing at a time, can you get a better pic of your workinug uploaded somewhere?
Reply 13
Original post by mqb2766
Do one thing at a time, can you get a better pic of your workinug uploaded somewhere?


Well I’m not sure where I would upload it too? I’ll try re writing the part of (a) I am interested in and hope it comes out not blurry
Reply 14
Original post by mqb2766
Do one thing at a time, can you get a better pic of your workinug uploaded somewhere?


4210FD56-5431-4EA7-8D67-66AE4CEAF395.jpg.jpeg
Reply 15
Looking back at your working, I cant help thinking your making it difficult for yourself with notation. Using
Nh = pi/3
and working it through with h and subbing at the end would simplfiy things. In particular, a few simple lines got me to something like
h [ e^(ipi/3) - e^(i2pi/3)] / [1 - e^ih]
and a bit of simple reasoning like that video seems to finish it off and take the imaginary part. There may be the odd typo in the previous, but its simpler than youre making it with your notation.

Id guess your error is coming when you pull e^ipi/3 out the Im(). Id guess your argument is that you take the real part so the next line is effectively real*im, but youd also have an im*real part which seems to be missed out.
(edited 10 months ago)
Reply 16
Original post by mqb2766
Looking back at your working, I cant help thinking your making it difficult for yourself with notation. Using
Nh = pi/3
and working it through with h and subbing at the end would simplfiy things. In particular, a few simple lines got me to something like
h [ e^(ipi/3) - e^(i2pi/3)] / [1 - e^ih]
and a bit of simple reasoning like that video seems to finish it off and take the imaginary part. There may be the odd typo in that.

Id guess youre error is coming when you pull e^ipi/3 out the Im(). Id guess your argument is that you take the real part so real*im, but youd also have an im*real part which seems to be missed out.


I’m not entirely sure what you mean, are we summing over h?

And twa that makes sense, it’s probably where the error is coming from
Reply 17
Original post by grhas98
I’m not entirely sure what you mean, are we summing over h?

And twa that makes sense, it’s probably where the error is coming from

h is the usual width so the series is
h * [e^(ipi/3) + e^(i(pi/3+h)) + e^(i(pi/3+2h)) +... + e^(i(pi/3+(N-1)h))
= h e^(ipi/3) [ 1 + e^(ih) + e^(i2h)) +... + e^(i(N-1)h) ]
= ...
so sum over an index n going from 0 to N-1, but obviously N->inf => h->0, which is the usual argument to make at the end.

Having a simple formula for the sum makes it easier to translate back to trig terms and reason about the re/im parts.
(edited 10 months ago)
Reply 18
Original post by mqb2766
h is the usual width so the series is
h * [e^(ipi/3) e^(i(pi/3 h)) e^(i(pi/3 2h)) ... e^(i(pi/3 (N-1)h))
so sum over an index n going from 0 to N-1, but obviously N->inf => h->0, which is the usual argument to make at the end.


Ah okay, and you also suggest centering it around pi/2 what exactly would that look like?
(edited 10 months ago)
Reply 19
Original post by grhas98
Ah okay, and you also suggest cantering it around pi/2 what exactly would that look like?


For me its more for thinking about using centered / trig terms on the numerator and denomator at the end. You can always centre stuff at the end by pulling out the "mid" term, so r - 1 on the denominator is
r^(1/2) [ r^(1/2) - r^(-1/2)]
and obviously the [] is a ~sin() term here
(edited 10 months ago)

Quick Reply

Latest