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Original post by physicsmaths
Yh that is what i meant, when the powers are 1/2 and it is 2 sequences holders becomes cauchy schwarz.
I would cautious of such questions if they ask me some crazy indepth questions about Holders inequality ill be like 'sorry i only know the basic proofs mate'. Atleast I am fairly certain I won't need this in my 30 minute test 😂.


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Yeah, I don't even know the proof lol. I've never used it before. This is definitely a hard question though.
Actually, I think I've done a proof for integer p and q, but not for real ones... I guess that would require limits everywhere.
Original post by Renzhi10122
Yeah, I don't even know the proof lol. I've never used it before. This is definitely a hard question though.
Actually, I think I've done a proof for integer p and q, but not for real ones... I guess that would require limits everywhere.


Yh I thought, Cauchy is normally in R^3 so this is R^N so maybe holders or an argument similar, so I just used the previous and summed to n and used the p+q=PQ! the second was definitely easier I thought! first part although I had something similar ina step paper! here they had not given anything so would have been sad if I did not know jensens inequality.


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(edited 8 years ago)
Original post by physicsmaths
Yh I thought, Cauchy is normally in R^3 so this is R^N so maybe holders or an argument similar, so I just used the previous and summed to n and used the p+q=PQ! the second was definitely easier I thought! first part although I had something similar ina step paper! here they had not given anything so would have been sad if I did not know jensens inequality.


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Cauchy is R^3??? It's in R^n. But yeah, you use the previous result. I spent a while on that question, I didn't really think that using jensen's would have been the best method, but I guess that that is what they wanted.
Original post by Renzhi10122
Cauchy is R^3??? It's in R^n. But yeah, you use the previous result. I spent a while on that question, I didn't really think that using jensen's would have been the best method, but I guess that that is what they wanted.

Yh it is R^n lol I was thinking of R^3 since I had seen some proof of Cauchy in step II 2006 and extended it to R^n but this had nothing to do with this lol. This is what happens when you have a brain fart.
Original post by joostan
Paper 3 Question 1:

Spoiler


Paper 3 Question 2:

Spoiler


Paper 3 Question 6:

Spoiler



Shouldn't the numerator be cos and not sin after the u sub?

I also divided the top and bottom by cos to get

1/1+tan and then used t =tan theta/2

Which gave me I=1 :/

What've I done wrong?


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Original post by I am Ace
Shouldn't the numerator be cos and not sin after the u sub?

I also divided the top and bottom by cos to get

1/1+tan and then used t =tan theta/2

Which gave me I=1 :/

What've I done wrong?


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Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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(edited 8 years ago)
Original post by physicsmaths
Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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ImageUploadedByStudent Room1449618553.411040.jpg


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ImageUploadedByStudent Room1449618487.312996.jpg131.0KB
ImageUploadedByStudent Room1449618511.850222.jpg153.2KB
Original post by physicsmaths
Na what he has done is right with the numerator, the result actually holds for 1/(1+tan^n(x))
Could you post your working? It might be since x=pi/2 cosx=0, but I am not too sure as the result holds backwards.

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The bottom 2 are fails I tried to delete


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Original post by I am Ace


Lol ur gna be so pissed, double angle tan formula is wrong.


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Original post by physicsmaths
Lol ur gna be so pissed, double angle tan formula is wrong.


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Brutal. It was so pretty as well


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Original post by I am Ace
Brutal. It was so pretty as well


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Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems


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Original post by physicsmaths
Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems


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Well done on spotting it, I'd never have realised! Ben though I've been meaning to fix that plus to a minus on my formula sheet for ages! Finally bit me in the arse


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https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
What did people get for Q4 a,b?


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Original post by physicsmaths


(a) (1/A)^(1/5)
(b) 1 + (1/2A), -1 + 1/(2A), (-1/A)^(1/3).

(they may accept +-1 for the first two solutions for (b))
Original post by Zacken
(a) (1/A)^(1/5)
(b) 1 + (1/2A), -1 + 1/(2A), (-1/A)^(1/3).

(they may accept +-1 for the first two solutions for (b))


How did u get to them? I have made a mistake somewhere.....


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Original post by physicsmaths
How did u get to them? I have made a mistake somewhere.....


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You get A=1x5−x3A = \frac{1}{x^5 - x^3} after some simplification .

For small A, you know that x has to be huge (from your graph, so x^3 can be thrown away in comparison to x^5) hence x = (1/A)^(1/5).

For big A, you know x lies somewhere between -1 and 0, so x^5 is smaller than x^3 and you can throw that away, hence A = 1/(-x^3) and solve for x, then looking at the graph, you notice solutions at x = +-1 roughly, so use x = +-1 + delta x and then approximate using binomial shiz, etc...

(I just put +-1)
Original post by Zacken
You get A=1x5−x3A = \frac{1}{x^5 - x^3} after some simplification .

For small A, you know that x has to be huge (from your graph, so x^3 can be thrown away in comparison to x^5) hence x = (1/A)^(1/5).

For big A, you know x lies somewhere between -1 and 0, so x^5 is smaller than x^3 and you can throw that away, hence A = 1/(-x^3) and solve for x, then looking at the graph, you notice solutions at x = +-1 roughly, so use x = +-1 + delta x and then approximate using binomial shiz, etc...

(I just put +-1)


Well I don't like that question lol. I just said infinity and 0- lol.


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Original post by physicsmaths
Well I don't like that question lol. I just said infinity and 0- lol.


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I quite enjoyed the question, I enjoyed the paper in general (the second paper, not so much), I think I got 5 full solutions out - but I freaking hated how they put Jensens inequality in the previous question, that was so unfair! :eek:
Original post by Zacken
I quite enjoyed the question, I enjoyed the paper in general (the second paper, not so much), I think I got 5 full solutions out - but I freaking hated how they put Jensens inequality in the previous question, that was so unfair! :eek:


Well you should done more olympiad stuff :wink: haha, Yh It was a nice paper! Paper 2 was good aswell I thought, I got Q1,2,3,5 and 7(kind of) for 7 ax+B is the remainder as they didn't say that at the start which made me wonder otherwise it does not make sense.


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Anyone have the maths and physics paper 1 Q 4 curve sketching solutions?


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