Turn on thread page Beta
    Offline

    8
    ReputationRep:
    (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 😂.


    Posted from TSR Mobile
    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.
    Offline

    18
    ReputationRep:
    (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.


    Posted from TSR Mobile
    Offline

    8
    ReputationRep:
    (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.


    Posted from TSR Mobile
    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.
    Offline

    18
    ReputationRep:
    (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.
    Offline

    3
    ReputationRep:
    (Original post by joostan)
    Paper 3 Question 1:
    Spoiler:
    Show
    Proof by picture seems to be the best way to go, without any fancy expressions for the \zeta function.
    A quick sketch should show that:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx \leq \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s}.
    We can compute the integral:
    \displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx = -\dfrac{1}{s-1} \left[ \dfrac{1}{x^{s-1}} \right] _1^{\infty}=\dfrac{1}{s-1}
    Also from the sketch we see that:
    \displaystyle\sum_{n=2}^{\infty} \dfrac{1}{n^s} \leq \displaystyle\int_2^{\infty} \dfrac{1}{(x-1)^s} \ dx
    Hence:
    \displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n^s} \leq 1+\displaystyle\int_1^{\infty} \dfrac{1}{x^s} \ dx
    Observing that 1+\dfrac{1}{s-1}=\dfrac{s}{s-1} gives the result:
    \dfrac{1}{s-1} \leq 1+2^{-s}+3^{-s}+... \leq\dfrac{s}{s-1}

    Paper 3 Question 2:
    Spoiler:
    Show
    I=\displaystyle\int_0^1 \dfrac{dx}{x+\sqrt{1-x^2}} \ dx
    Let x=\sin(\theta)
    \Rightarrow I = \displaystyle\int_0^{\frac{\pi}{  2}} \dfrac{\sin(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta

\theta \mapsto \dfrac{\pi}{2}-\theta

\RIghtarrow I=\displaystyle\int_0^{\frac{\pi  }{2}} \dfrac{\cos(\theta)}{\sin(\theta  )+\cos(\theta)} \ d\theta 

\Rightarrow 2I=\displaystyle\int_0^{\frac{ \pi}{2}} \ d\theta  = \dfrac{\pi}{2}

\therefore I=\dfrac{\pi}{4}.

    Paper 3 Question 6:
    Spoiler:
    Show
    False.
    In decimal expansion, an n digit number N may be written as:
    N=\displaystyle\sum_{r=0}^n a_r 10^r \ , \ a_r \in \{0,1, . . . ,9 \}
    Let x=10^n-1 for some n \in \mathbb{N}.
    Then:
    x^2=(10^n-1)^2=10^{2n}-2(10^n)+1= \left( \displaystyle\sum_{r=n+1}^{2n} 9(10)^r \right) + 8(10^n)+1.
    Certainly then, we have n digits a_r \not= 0,1.
    Hence if n>1000 we have a square number with more than 1000 digits a_r \not= 0,1.
    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?


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    (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?


    Posted from TSR Mobile
    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.

    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (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.

    Posted from TSR Mobile
    Name:  ImageUploadedByStudent Room1449618553.411040.jpg
Views: 178
Size:  151.4 KB


    Posted from TSR Mobile
    Attached Images
      
    Offline

    3
    ReputationRep:
    (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.

    Posted from TSR Mobile
    The bottom 2 are fails I tried to delete


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    Lol ur gna be so pissed, double angle tan formula is wrong.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    Lol ur gna be so pissed, double angle tan formula is wrong.


    Posted from TSR Mobile
    Brutal. It was so pretty as well


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    (Original post by I am Ace)
    Brutal. It was so pretty as well


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


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    (Original post by physicsmaths)
    Lol, I have had plenty of amazing proofs like yours disproving well known mathematical theorems


    Posted from TSR Mobile
    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


    Posted from TSR Mobile
    Offline

    18
    ReputationRep:
    https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
    What did people get for Q4 a,b?


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by physicsmaths)
    https://www1.maths.leeds.ac.uk/~read/TQC1.pdf
    What did people get for Q4 a,b?


    Posted from TSR Mobile
    (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))
    Offline

    18
    ReputationRep:
    (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.....


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by physicsmaths)
    How did u get to them? I have made a mistake somewhere.....


    Posted from TSR Mobile
    You get A = \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)
    Offline

    18
    ReputationRep:
    (Original post by Zacken)
    You get A = \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.


    Posted from TSR Mobile
    Offline

    22
    ReputationRep:
    (Original post by physicsmaths)
    Well I don't like that question lol. I just said infinity and 0- lol.


    Posted from TSR Mobile
    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:
    Offline

    18
    ReputationRep:
    (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 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.


    Posted from TSR Mobile
    Offline

    3
    ReputationRep:
    Anyone have the maths and physics paper 1 Q 4 curve sketching solutions?


    Posted from TSR Mobile
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 18, 2017
Poll
Favourite type of bread

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.