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    Problem 381***

    Let A be an nxn real matrix with the modulus of each entry strictly less than 1/n. Is I-A invertable in all cases? If not, when is it invertable.
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    (Original post by keromedic)
    I only just noticed the problem and thought "oh, I can finally solve a problem on here". Too late though
    Spoiler:
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    \displaystyle \int sinxcos2x=\dfrac{1}{2} \int sin 3x -sinx
    =\dfrac{1}{6}(3cosx-cos3x)+C
    I solved it another way by substituting the identity for cos2x which makes it sinx(cos^2x-sin^2x) and integrating sinxcos^2x as -cos^3x/3 which left me with -sin^3x=-sinx(1-cos^2x)=-sinx+sinxcos^2x and the integral of this is cosx-cos^3x/3 giving me an integral of cosx-2cos^3x/3 +c in total.
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    (Original post by Mladenov)
    As Nikolai Nikolov, the leader of our mathematics team, says: Why simple, when it could be complicated?!
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    Methinks that I have digested quite well his advice.
    So weird, he tutors at my college (Univ)!
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    Problem 382**

    Let a_1 \in (1,2) and a_{k+1}=a_k+\frac{k}{a_k}. Prove that there are at most two terms in the sequence that add to make an integer.
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    Problem 383**

    Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of \$P each year. At the end of n years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of \alpha \% per annum of the total sum in the fund.

    Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the n^t^h year is \$\frac{PR(R^n -1)}{R - 1} where R=(1+\frac{\alpha}{100})
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    Problem 384***

    It seems I incorrectly starred this one. Here's a better rewording:

    Prove that if a,b,\frac{a^2+b^2}{1+ab} \in \mathbb{Z} , then \frac{a^2+b^2}{1+ab}  is a perfect square.
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    (Original post by henpen)
    Problem 384**

    Prove that for some a,b there exists an n such that
    \frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.
    Do yoyu mean for all a,b?
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    (Original post by henpen)
    Problem 384**

    Prove that for some a,b there exists an n such that
    \frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.
    What are you trying to ask? if it is just to find integers a,b,n satisfying this equation then it is absolutely trivial; a=b=n=1 is an easy solution.

    If, on the other hand, you are asking the form that question is usually asked in (Which is to prove that if  \frac{a^2+b^2}{1+ab} is an integer for integers a,b then it is a perfect square) then you should rewrite it to make things explicit and add another star to the difficulty as said question was infamously difficult for the IMO at its time of conception.
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    (Original post by henpen)
    Problem 384**

    Prove that for some a,b there exists an n such that
    \frac{a^2+b^2}{1+ab}, a, b, n \in \mathbb{Z} \Rightarrow \frac{a^2+b^2}{1+ab}=n^2.
    I believe this was the hardest IMO question ever. In year 1988 I think.


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    Okay, so I'm new to this thread :hello:

    This was in a book I'm reading and I thought it was a really beautiful result, I'm sorry if anyone has asked a question like this before. This might be possible with just *, I haven't tried it.

    Problem 385***


    0<k<1

Find in terms of k \displaystyle\int^{2\pi}_0 \frac{1}{1-2k\cos \theta +k^2}\ d\theta
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    (Original post by Flauta)
    Okay, so I'm new to this thread :hello:

    This was in a book I'm reading and I thought it was a really beautiful result, I'm sorry if anyone has asked a question like this before. This might be possible with just *, I haven't tried it.
    Welcome to the thread! :cool:


    Problem 385***


    0<k<1

Find in terms of k \displaystyle\int^{2\pi}_0 \frac{1}{1-2k\cos \theta +k^2}\ d\theta
    Solution 385

    \displaystyle\begin{aligned} \int_0^{2\pi} \frac{dx}{k^2 - 2k\cos{x} + 1} \ dx & \overset{t = \tan{\frac{x}{2}}}= \int_{-\infty}^{\infty} \frac{2dt}{(k+1)^2 t^2 + (k-1)^2} \\ \\ & = \frac{2\arctan{\left(\frac{k+1}{  k-1} t\right)}}{k^2 - 1}\bigg|_{-\infty}^{\infty} \end{aligned}

    Now, note that 0 < k < 1 \implies k - 1 < 0 so we will use the fact that \arctan\left(\frac{k+1}{k-1} t\right) = -\arctan \left(\frac{k+1}{1-k} t \right) as \arctan{x} is an odd function so that the \frac{k+1}{k-1} does not affect the sign of the argument of the arctan. Thus the integral is given by:

    \dfrac{2}{1-k^2} \left(\arctan{\left(\frac{k+1}{1-k} t \right)} \bigg|_{-\infty}^{\infty}\right) = \dfrac{2\pi}{1-k^2}
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    (Original post by Felix Felicis)
    Welcome to the thread! :cool:
    x
    Thank you!

    I didn't know it could be done that way, correct answer though
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    (Original post by Arieisit)
    Problem 383**

    Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of \$P each year. At the end of n years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of \alpha \% per annum of the total sum in the fund.

    Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the n^t^h year is \$\frac{PR(R^n -1)}{R - 1} where R=(1+\frac{\alpha}{100})
    I THINK I've got it?

    Solution 383

    

S_n=\frac{PR(R^n-1)}{R-1}

S_{n+1}=R(\frac{PR(R^n-1)}{R-1}+P)

S_{n+1}=PR(\frac{R(R^n-1)}{R-1}+1)

S_{n+1}=PR(\frac{R^{n+1}-R+R-1}{R-1})

S_{n+1}=PR(\frac{R^{n+1}-1}{R-1})

n+1=k

S_{k}=\frac{PR(R^k-1)}{R-1}



True for n=1 because



S_1=\frac{PR(R-1)}{R-1}

S_1=PR

    Is that correct?
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    (Original post by Flauta)
    I THINK I've got it?

    Solution 383

    

S_n=\frac{PR(R^n-1)}{R-1}

S_{n+1}=R(\frac{PR(R^n-1)}{R-1}+P)

S_{n+1}=PR(\frac{R(R^n-1)}{R-1}+1)

S_{n+1}=PR(\frac{R^{n+1}-R+R-1}{R-1})

S_{n+1}=PR(\frac{R^{n+1}-1}{R-1})

n+1=k

S_{k}=\frac{PR(R^k-1)}{R-1}



True for n=1 because



S_1=\frac{PR(R-1)}{R-1}

S_1=PR

    Is that correct?
    You are on the right path :yy: but the proof is not complete.

    I could finish it off for you but I wouldn't rob you of the joy of doing it yourself

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    (Original post by Arieisit)
    Problem 383**

    Ms. Janis Smith takes out an endowment policy with an insurance company which involves making a fixed payment of \$P each year. At the end of n years, Janis expects to receive a payout of a sum of money which is equal to her total payments together with interest added at the rate of \alpha \% per annum of the total sum in the fund.

    Show, by mathematical induction or otherwise, that the total sum in the fund at the end of the n^t^h year is \$\frac{PR(R^n -1)}{R - 1} where R=(1+\frac{\alpha}{100})
    When I was in school I had fun spending a few days investigating this class of problems.

    Solution 383**
    Spoiler:
    Show

    We take the "otherwise" approach. Let S_n denote the size of the fund after n years. Hence S_0 = 0 and S_{k+1} = R\left(S_{k} + P\right) for all k. For a particular year n, expanding the recurrence gives S_n = RP + R^2P + \dots + R^nP, which is an n-term geometric series with initial value RP and common ratio R. By the formula for the first n terms of a geometric series, S_n = \frac{RP(R^n -1)}{R - 1}
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    (Original post by ukdragon37)
    When I was in school I had fun spending a few days investigating this class of problems.

    Solution 383**
    Spoiler:
    Show

    We take the "otherwise" approach. Let S_n denote the size of the fund after n years. Hence S_0 = 0 and S_{k+1} = R\left(S_{k} + P\right) for all k. For a particular year n, expanding the recurrence gives S_n = RP + R^2P + \dots + R^nP, which is an n-term geometric series with initial value RP and common ratio R. By the formula for the first n terms of a geometric series, S_n = \frac{RP(R^n -1)}{R - 1}
    This is the way I did it as well. It's more "fun" than the induction

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    (Original post by Arieisit)
    This is the way I did it as well. It's more "fun" than the induction

    Posted from TSR Mobile
    I disagree, induction is pretty damn exciting, like an expedition to try to get it back into the original form

    Did I miss out a step or something? Reading over it and can't see where my mistake is.

    EDIT: Oh I see it now, nevermind. I was a bit too brief. Thanks
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    Okay so I need more LaTeX practice, and I still think problems like this are cool. We had to numerically integrate this in maths today, was dreadfully inaccurate aha. There're various ways of doing this.

    Problem 386**/***

    

Find \displaystyle\int^{2\pi}_0 \frac{1}{5-3\sin x} dx

    Hints
    Spoiler:
    Show
    Use substitution t=\tan {\frac{x}{2}}
    OR
    do a bit of complex analysis
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    (Original post by Flauta)
    Okay so I need more LaTeX practice, and I still think problems like this are cool. We had to numerically integrate this in maths today, was dreadfully inaccurate aha. There're various ways of doing this.

    Problem 386**/***

    

Find \displaystyle\int^{2\pi}_0 \frac{1}{5-3\sin x} dx

    Hints
    Spoiler:
    Show
    Use substitution t=\tan {\frac{x}{2}}
    OR
    do a bit of complex analysis
    As |\sin(x)|\le 1,

    \displaystyle \frac{1}{1-\frac{3}{5}\sin(x)}= \sum_{k=0}^\infty\left(\frac{3}{  5}\sin(x)\right)^2


    \displaystyle \frac{1}{5}\sum_{k \ge 0} \left(\frac{3}{5}\right)^k\int_0  ^{2 \pi}\sin^k(x)dx=\frac{1}{5}\sum_  {k \ge 0} \left(\frac{3}{5}\right)^k I_k,
    integrate the integral by parts, finding that (!! is the double factorial)
    \displaystyle I_{2k+1}=0, I_{2k}=2 \pi \frac{(2k-1)!!}{(2k)!!},

    where we define (-1)!!=(0)!!=1 for ease (Is this standard? The binomial expansions of (1-x)^{\frac{-n}{2}} is much neater to write if (-n)!!=1.),

    \displaystyle \frac{2\pi}{5}\sum_{k \ge 0} \left(\frac{3}{5}\right)^{2k} \frac{(2k-1)!!}{(2k)!!}.

    Note that
    \displaystyle\sqrt{1-x}^{-1}=\sum_{k \ge 0}\frac{(2k-1)!!}{(2k)!!}x^k.

    Thus the integral reduces to

    \displaystyle\frac{2\pi}{5}\left  (\sqrt{1-\left\frac{3}{5}\right^2}^{-1}\right)=\frac{\pi}{2}.
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    (Original post by henpen)
    .
    I'm not going to pretend to understand anything you've written there, but that is the correct answer! Guess there are more than 2 ways of doing it
 
 
 
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