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    (Original post by Lord of the Flies)
    I had already solved it but was eager to post a solution to 98, after that blunder. Also, it doesn't matter what order the solutions come in - this thread is not a competition, it's for fun!
    Haha I swear you never make mistakes!

    I know, I know :lol: Tbf fair I only really typed it to practice my latex
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    (Original post by Jkn)
    France vs. Bulgaria
    I would genuinely pay to see a maths-off between these two :lol: Stick L'art in there as well :lol:
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    Solution 105

    I shall not consider convergence; if somebody is interested, I can justify my approach.

    Let us see..

    From Cauchy, we have
    \begin{aligned} \displaystyle \left(\sum_{i=1}^{\infty} (-1)^{i}\frac{x^{2i+1}}{(2i)!!} \right)\left(\sum_{i=0}^{\infty} \frac{x^{2i}}{((2i)!!)^{2}} \right) &= xe^{-\frac{x^{2}}{2}}\left(\sum_{i=0}  ^{\infty} \frac{x^{2i}}{4^{i}(i!)^{2}} \right) \end{aligned}.
    Hence,
    \begin{aligned} \displaystyle \int_{0}^{\infty} \left(\sum_{i=1}^{\infty} (-1)^{i}\frac{x^{2i+1}}{(2i)!!} \right)\left(\sum_{i=0}^{\infty} \frac{x^{2i}}{((2i)!!)^{2}} \right)dx &= \sum_{i=0}^{\infty} \frac{1}{2^{2i}(i!)^{2}} \int_{0}^{\infty} x^{2i+1}e^{-\frac{x^{2}}{2}}dx \\&= \sum_{i=0}^{\infty} \frac{1}{2^{2i}(i!)^{2}} \int_{0}^{\infty} 2^{i}x^{i}e^{-x}dx \\&= \sum_{i=0}^{\infty} \frac{1}{2^{2i}(i!)^{2}} 2^{i}\Gamma(i+1) \\&= \sum_{i=0}^{\infty} \frac{1}{2^{i}(i!)} \\&= e^{\frac{1}{2}} \end{aligned}.

    Edit: I was to post a solution to problem 108, but just saw the spoiler.
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    (Original post by Felix Felicis)
    I would genuinely pay to see a maths-off between these two :lol: Stick L'art in there as well :lol:
    Me too :lol: "Student room international Olympiad". As long as I get to set the questions I'm happy

    (Original post by Mladenov)
    Edit: I was to post a solution to problem 108, but just saw the spoiler.
    Hahahaa, did you figure out why? :rolleyes:
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    (Original post by metaltron)
    I got 180 points I think for a Number Theory Question, it was Q8 but it was similar to STEP I Q1 2000.

    For Combinatorics I got 180 points too, but also got three wrong answers for one so that ruined the moment for me really.

    I think you can only get 180 pts for a question in Level 3.
    Oh yeah that one was quite funny, I took seconds! That means theres a 6/8 overlap between levels (thats level 4's 3rd hardest one atm!) :eek:

    Awesome man. Yeah level 4 only goes up to 230
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    (Original post by Mladenov)
    Splendid solution.
    You are not using continuity, so it is not necessary.

    Another approach:
    Spoiler:
    Show
    Clearly, f \equiv 1 is a solution. Suppose that there is at least one z such that f(z) \not= 1. We have f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y) for all x,y\in \mathbb{R^{+}}. Since f is bounded below and satisfies f(x+y)=f(x)+f(y), it follows that f(x)=cx, c\in \mathbb{R^{+}}. Note that c^{2}xy=cxy implies c=1. Hence all the solutions are f \equiv 1 and f(x)=x.
    We have f(xy)=f(x)f(y) and f(x+y)=f(x)+f(y) for all x,y\in \mathbb{R^{+}}.

    How did you get these?
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    Since there is some love for function based questions..

    Problem 109*/**

     f(x) = sin(x) - \displaystyle\int_0^x (x-t)f(t) dt

    Find f exactly
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    (Original post by FireGarden)
    Since there is some love for function based questions..

    Problem 109*/**

     f(x) = sin(x) - \displaystyle\int_0^x (x-t)f(t) dt

    Find f exactly
    Not sure if this is right...

    Assume x is not a function of t.

    Letting x=t gives f(t)=sint.

    So \displaystyle f(x) = sinx + \int_0^x (t-x)sint dt
    \displaystyle = sinx +[[-(t-x)cost] + sint]_0^x = 2sinx-x
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    (Original post by Jkn)
    Not sure if this is right...

    Solution 109

    Assume x is not a function of t.

    Letting x=t gives f(t)=sint.

    So \displaystyle f(x) = sinx + \int_0^x (t-x)sint dt
    \displaystyle = sinx +[[-(t-x)cosx] + sinx]_0^x = 2sinx-x
    t is simply a dummy variable for the integral, it doesn't really need to be considered all that much.

    Edit: Realising that it wasn't previously clear at all, it's not correct.
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    (Original post by FireGarden)
    t is simply a dummy variable for the integral, it doesn't really need to be considered all that much.
    Is it right then?

    Well then how can you define a dummy variable? And why is it you you can assume that x can be treated as a constant in the integral?
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    (Original post by Jkn)
    Is it right then?

    Well then how can you define a dummy variable? And why is it you you can assume that x can be treated as a constant in the integral?
    I edited my previous post.

    x is certainly a constant in the integral; giving the function f a value of x defines the limits and the x inside the integral as a constant.
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    (Original post by FireGarden)
    Since there is some love for function based questions..

    Problem 109*/**

     f(x) = sin(x) - \displaystyle\int_0^x (x-t)f(t) dt

    Find f exactly
    Partial solution (will finish if I get time later).

    Spoiler:
    Show
    Differentiate both sides to get f'(x)=cos(x)+xf(0)=cos(x)+cx

    so f(x)=sin(x)+cx^2/2+d for constants c and d. By plugging this into the original equation we may get conditions that c,d must satisfy.
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    (Original post by james22)
    Partial solution (will finish if I get time later).

    Spoiler:
    Show
    Differentiate both sides to get f'(x)=cos(x)+xf(0)=cos(x)+cx

    so f(x)=sin(x)+cx^2/2+d for constants c and d. By plugging this into the original equation we may get conditions that c,d must satisfy.
    You have carried out your method incorrectly. (Though it was the method I was expecting)
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    (Original post by FireGarden)
    I edited my previous post.

    x is certainly a constant in the integral; giving the function f a value of x defines the limits and the x inside the integral as a constant.
    Well then if it is a constant then why is it denoted most the most universally recognisable symbol for a variable?

    And if it is a constant then its derivative is not f'(x) so how is it that you would expect this?
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    Solution 109

    \displaystyle f(x)=\sin x-\int_0^x (x-t) f(t)\,dt=\sin x-\int_0^x\int_0^t f(w)\,dw\,dt

    Hence y+y''=-\sin x\Rightarrow y=\mathcal{C}\cos x+\mathcal{C}'\sin x+\frac{1}{2}x\cos x

    \displaystyle\cdots\Rightarrow \mathcal{C}'=\frac{1}{2},\; \mathcal{C}=0 \Rightarrow f(x)=\frac{1}{2}\big(\sin x+x\cos x\big)
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    (Original post by Jkn)
    Well then if it is a constant then why is it denoted most the most universally recognisable symbol for a variable?

    And if it is a constant then its derivative is not f'(x) so how is it that you would expect this?
    Do you believe  f(x) = \displaystyle\int_0^x x sin(t) dt to be an ill-defined function?

    Take some particular value,  x= \pi , and you get  f(\pi) = \displaystyle\int_0^{\pi} \pi sin(t) dt = 2\pi

    My function works similarly, and is just as well-defined.
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    (Original post by Lord of the Flies)
    Solution 109

    \displaystyle f(x)=\sin x-\int_0^x (x-t) f(t)\,dt=\sin x+\int_0^x\int_0^t f(w)\,dw\,dt

    Hence y+y''=-\sin x\Rightarrow y=\mathcal{C}\cos x+\mathcal{C}'\sin x+\frac{1}{2}x\cos x

    \displaystyle\int_0^0 f(t)\,dt= f(0)=0\Rightarrow \mathcal{C}=\frac{1}{2},\mathcal  {C}'=0 \Rightarrow f(x)=\frac{1}{2}\big(\sin x+x\cos x\big)

    Sign mistake! Wait!
    Correct. I was about to question your sinh(x), but saw your edit as I quoted you!
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    (Original post by FireGarden)
    Correct. I was about to question your sinh(x), but saw your edit as I quoted you!
    Yes I started with the wrong equation :lol:

    (Original post by Felix Felicis)
    ...
    That sig. :sexface:
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    (Original post by FireGarden)
    Do you believe  f(x) = \displaystyle\int_0^x x sin(t) dt to be an ill-defined function?

    Take some particular value,  x= \pi , and you get  f(\pi) = \displaystyle\int_0^{\pi} \pi sin(t) dt = 2\pi

    My function works similarly, and is just as well-defined.
    Hmm I suppose I'll just ponder it some more (never really since this type of function before). Thanks
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    Problem 110*

    Prove that \displaystyle (\sum_{k=1}^{\infty}{a_k}{b_k})^  2 \leq \sum_{k=1}^{\infty}{a_k}^2 \sum_{k=1}^{\infty}{a_k}^2 for all real numbers \displaystyle a_k, b_k with using Cauchy-Schwartz.

    Problem 111* (another problem I've adapted from something else. Oh how I love generalisations...)

    Let a,b and \lambda be positive integers such that  a\lambda=b^3. Find the minimum value of a+b for the different cases that arise according to the value of \lambda given that \lambda has 4 proper factors (i.e. factors not including itself and 1.)
 
 
 
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