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    Problem 301**

    Prove, given  a,c > 0;\ \ ac-bd\not=0

     \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)
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    (Original post by FireGarden)
    Problem 301**

    Prove, given  a,c > 0;\ \ ac-bd\not=0

     \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)
    Is there a nicer way than the rather horrible:
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    tan both sides, and addition formula
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    (Original post by Smaug123)
    Is there a nicer way than the rather horrible:
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    tan both sides, and addition formula
    Yes. The formula arises naturally, if you consider the right objects/behaviour. I wrote the question with the nice answer in mind
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    (Original post by Smaug123)
    Is there a nicer way than the rather horrible:
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    tan both sides, and addition formula
    That's not that horrible though, it's only 3-4 lines of working.
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    (Original post by james22)
    That's not that horrible though, it's only 3-4 lines of working.
    It's not pretty, though - it's really mechanical
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    (Original post by FireGarden)
    Problem 301**

    Prove, given  a,c > 0;\ \ ac-bd\not=0

     \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)
    Solution 301

     \tan {( \arctan {\frac{b}{a}} + \arctan { \frac{d}{c}})} = \displaystyle \frac{ \frac{b}{a} + \frac {d}{c} } {1- \frac {bd}{ac} } = \frac{ ad -bc}{ac- bd}

    Here's a nice one from the Putnam.

    Problem 302**

    Let  F_0 (x) = ln x. For  n \ge  0 and  x > 0, let  F_{n+1}(x) = \displaystyle \int_0^x  F_n (t)\ dt .

    Evaluate:

     \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .
    .
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    (Original post by MW24595)
    Solution 301

     \tan {( \arctan {\frac{b}{a}} + \arctan { \frac{d}{c}})} = \displaystyle \frac{ \frac{b}{a} + \frac {d}{c} } {1- \frac {bd}{ac} } = \frac{ ad -bc}{ac- bd}

    Here's a nice one from the Putnam.

    Problem 302**

    Let  F_0 (x) = ln x. For  n \ge  0 and  x > 0, let  F_{n+1}(x) = \displaystyle \int_0^x  F_n (t)\ dt .

    Evaluate:

     \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .
    .
    Solution 302:
    Consider:
    F_1(x)=x\ln(x)-x

F_2(x)=\dfrac{x^2}{2}\ln(x)-\dfrac{3x^2}{4}
    Suppose then that: F_n(x) is of the form:
    x^n \left(\dfrac{\ln(x)}{n!} - C_n \right)
    For some constant C_n
    As it is true for F_1(x) and F_2(x) Suppose that it is also true for F_k(x)
    \Rightarrow F_{k+1}(x)=\displaystyle\int^x_0 \dfrac{t^k}{k!}\ln(t)-C_kt^k \ dt
    Let:
    u=\ln(t) \Rightarrow u'=\dfrac{1}{t}
    v'=\dfrac{t^k}{k!} \Rightarrow v=\dfrac{t^{k+1}}{(k+1)!}
    \Rightarrow F_{k+1}=\left[\dfrac{t^{k+1}\ln(t)}{(k+1)!} \right]^x_0 -\displaystyle\int^x_0 \dfrac{t^{k+1}}{(k+1)!}\times \dfrac{1}{t} \ dt -\left[C_k\dfrac{t^{k+1}}{k+1} \right]^x_0
    Observing that:
    \displaystyle\lim_{t \to 0} t^{k+1}\ln(t)=0
    \Rightarrow F_{k+1}(x)= \dfrac{x^{k+1}\ln(x)}{(k+1)!}-\dfrac{x^{k+1}}{(k+1)(k+1)!}-C_k \dfrac{x^{k+1}}{k+1}

\Rightarrow F_{k+1}(x)=x^{k+1} \left(\dfrac{\ln(x)}{(k+1)!}-C_{k+1} \right)
    Where:
    C_{k+1}=\dfrac{C_k}{k+1}+\dfrac{  1}{(k+1)(k+1)!}
    This is of the same format as was proposed:
    \therefore F_{n}(x)=x^n \left(\dfrac{\ln(x)}{n!} - C_n \right) \forall n \in \mathbb{N}
    Now using the fact that:
    C_{k+1}=\dfrac{C_k}{k+1}+\dfrac{  1}{(k+1)(k+1)!}

\Rightarrow C_{k+1}=\dfrac{1}{(k+1)!} \left(k!C_k+\dfrac{1}{k+1} \right)

\therefore C_n=\dfrac{1}{n!} \left((n-1)!C_{n-1}+\dfrac{1}{n} \right)
    Note that:
    (n-1)!C_{n-1}= \left((n-2)!C_{n-2}+\dfrac{1}{n-1} \right) and that C_0=0
    \therefore C_n=\dfrac{1}{n!}\left(\dfrac{1}  {n}+\dfrac{1}{n-1}+. . . +\dfrac{1}{3}+\dfrac{1}{2}+1 \right)=\dfrac{1}{n!}\displaystyle\sum_{  r=1}^n \dfrac{1}{r}
    So:
    F_n(x)=x^n \left(\dfrac{\ln(x)}{n!} -\dfrac{1}{n!}\displaystyle\sum_{  r=1}^n \dfrac{1}{r} \right)
    Subbing in:
    F_n(1)=1^n \left(\dfrac{\ln(1)}{n!} -\dfrac{1}{n!}\displaystyle\sum_{  r=1}^n \dfrac{1}{r}=-\dfrac{1}{n!}\displaystyle\sum_{  r=1}^n \dfrac{1}{r} \right)

\Rightarrow n!F_n(1)=-\displaystyle\sum_{r=1}^n \dfrac{1}{r}
    So:
    \displaystyle\lim_{ n \to \infty} \left(\dfrac{ n! F_n (1)} {\ln(n)}\right)=\displaystyle\lim_{ n \to \infty} \left( \dfrac{-1}{\ln(n)}\displaystyle\sum_{r=1  }^n \dfrac{1}{r} \right)
    Note that:
    \gamma = \displaystyle\lim_{n \to \infty} \left( \displaystyle\sum_{r=1}^n \dfrac{1}{r}-\ln(n) \right)

\therefore \gamma + \displaystyle\lim_{n \to \infty}\ln(n)=\displaystyle\lim_  {n \to \infty} \left( \displaystyle\sum_{r=1}^n \dfrac{1}{r}\right)

\Rightarrow \displaystyle\lim_{ n \to \infty} \left( \dfrac{ n! F_n (1)} {\ln(n)} \right)=\displaystyle\lim_{ n \to \infty} \left(\dfrac{-\gamma-\ln(n)}{\ln(n)} \right)=-1
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    (Original post by MW24595)
    Problem 302**

    Let  F_0 (x) = ln x. For  n \ge  0 and  x > 0, let  F_{n+1}(x) = \displaystyle \int_0^x  F_n (t)\ dt .

    Evaluate:

     \lim_{ n \rightarrow \infty} \displaystyle \frac{ n! F_n (1) } { \ln {n} } .
    If only they had Putnam in the UK..

    Solution 302 (2)

    Calculating the first few terms we conjecture that \displaystyle F_n(x)=\frac{1}{n} \left(xF_{n-1}(x)-\frac{x^n}{n!} \right).

    This is true for n=1 as \displaystyle F_1 (x)=xF_0 (x)-x. Integrating both sides, using integration by parts, formalizing by noting that \displaystyle \lim_{n \to 0} n \ln(n) = 0 and factorising F_{n+1}(x) on the left hand side shows that, under this hypothesis, we have that:

    \displaystyle \begin{aligned} F_{n+1} (x) = \frac{1}{n} \left( \int_0^x t F_{n-1}(t) \ dt - \frac{x^{n+1}}{(n+1)!} \right) = \frac{1}{n} \left(x \int_0^x F_{n-1} (t) \ dt - \int_0^x \int_0^y F_{n-1} (t) \ dt \ dy - \frac{x^{n+1}}{(n+1)!} \right) \end{aligned}
    \displaystyle \begin{aligned} = \frac{1}{n} \left( x F_n (x) - F_{n+1} (x) - \frac{x^{n+1}}{(n+1)!} \right) =  \frac{1}{n+1} \left( x F_n (x) - \frac{x^{n+1}}{(n+1)!} \right) \end{aligned}
    which completes the induction.

    This then gives \displaystyle \begin{aligned} n F_n (1) = F_{n-1} (x) - \frac{1}{n!}\ \Rightarrow n! F_n(1) = (n-1)! F_{n-1} (1) - \frac{1}{n} = \cdots =F_0 (1) -\left(1+\frac{1}{2}+ \cdots + \frac{1}{n} \right)= -H_n \end{aligned}

    where H_n denotes the nth harmonic number.

    Using the asymptotic relation \displaystyle \lim_{n \to \infty} H_n = \gamma + \ln (n), which follows directly from the definition of the Euler Mascheroni constant, we get:

    \displaystyle \lim_{n \to \infty} \frac{n! F_n (1)}{\ln(n)} = - \lim_{n \to \infty} \frac{\gamma + \ln(n)}{\ln(n)} = -1 \ \square as the logarithmic function is obviously divergent under this limit.

    Edit: Dammit, too slow!
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    (Original post by MW24595)
    I just came across this. There's an off reference in the section titled "Integration Problems." Read the last paragraph.

    http://en.wikipedia.org/wiki/Gamma_function
    Oh cheers, must've missed that! (That page is one of my 'most-visited' :lol:)
    (Original post by MW24595)
    Solution 300
    Nice solution broseph! Here is your prize!
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    (Original post by Jkn)
    Oh cheers, must've missed that! (That page is one of my 'most-visited' :lol:)

    Nice solution broseph! Here is your prize!
    Roflol. I'm surprised you did.
    Interesting that the reference was almost bang-on though.

    Lol, thanks. Ha

    Found anything else of interest lately?
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    (Original post by FireGarden)
    Problem 301**

    Prove, given  a,c > 0;\ \ ac-bd\not=0

     \arctan\left(\dfrac{ad+bc}{ac-bd}\right) = \arctan\left(\dfrac{b}{a}\right) + \arctan\left(\dfrac{d}{c}\right)
    I just realized.

    Alternate Solution 301

    Let  z_1 = a+ ib, and  z_2 = c + id .

    Then,  z_1 = a+ bi = \displaystyle e^{i \arctan{ \frac{b}{a}}} , and,  z_2 = \displaystyle e^{i \arctan {\frac{d}{c}}} .

    



\Rightarrow z_1 z_2 = (a+ib)(c+id) = (ac- bd) + i(ad + bc) = e^ { i \arctan{\frac{ad+bc}{ac-bd}}}



.

    But,  z_1 z_2 = \displaystyle e^{i \arctan{ \frac{b}{a}} + i \arctan{ \frac{d}{c}}} = e^ { i \arctan{\frac{ad+bc}{ac-bd}}} .

    Taking logarithms, we have:

    \displaystyle \arctan{\frac{ad+bc}{ac-bd}} = \arctan{ \frac{b}{a}} + \arctan{ \frac{d}{c}} .
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    Just posted this in one of the regular threads.. thought you guys might like it

    Problem 303**

    Give a justification for the claim  1^{\infty} > 2
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    (Original post by MW24595)
    I just realized.

    Alternate Solution 301

    Let  z_1 = a+ ib, and  z_2 = c + id .

    Then,  z_1 = a+ bi = \displaystyle e^{i \arctan{ \frac{b}{a}}} , and,  z_2 = \displaystyle e^{i \arctan {\frac{d}{c}}} .

    



\Rightarrow z_1 z_2 = (a+ib)(c+id) = (ac- bd) + i(ad + bc) = e^ { i \arctan{\frac{ad+bc}{ac-bd}}}



.

    But,  z_1 z_2 = \displaystyle e^{i \arctan{ \frac{b}{a}} + i \arctan{ \frac{d}{c}}} = e^ { i \arctan{\frac{ad+bc}{ac-bd}}} .

    Taking logarithms, we have:

    \displaystyle \arctan{\frac{ad+bc}{ac-bd}} = \arctan{ \frac{b}{a}} + \arctan{ \frac{d}{c}} .
    This is the solution I had in mind. Well done sir!
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    (Original post by FireGarden)
    Just posted this in one of the regular threads.. thought you guys might like it

    Problem 303**

    Give a justification for the claim  1^{\infty} > 2
    This question doesn't really make sense, do you mean we need to find a function, f(x), such that as x->infinity f(x)->1 but f(x)^x->c>2?
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    (Original post by james22)
    This question doesn't really make sense, do you mean we need to find a function, f(x), such that as x->infinity f(x)->1 but f(x)^x->c>2?
    You could do it that way if you like. I thought the question would make sense, since I'm asking about an indeterminate form, and thus it can take many values - the question is to find some way to approach the IF so that its value is larger than two, as you correctly interpreted. Though your approach is not the only one!
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    (Original post by FireGarden)
    Just posted this in one of the regular threads.. thought you guys might like it

    Problem 303**

    Give a justification for the claim  1^{\infty} > 2
    Solution 303:
    Note that:
    \displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x=1^{\infty}
    Now let:
    y=\displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x

\Rightarrow \ln(y)=\displaystyle\lim_{x \to \infty} x\ln \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)=\displaystyle\lim_{x \to \infty}\dfrac{\ln \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)}{\frac{1}{x}}

\Rightarrow \ln(y)=\displaystyle\lim_{t \to 0} \dfrac{\ln \left(\tan \left(\dfrac{\pi}{4} +t \right) \right)}{t}
    Applying L'Hopital's rule:
    \ln(y)=\displaystyle\lim_{t \to 0} \left(\dfrac{\csc \left(t+\frac{\pi}{4} \right) \times \sec \left(t+\frac{\pi}{4} \right)}{1}\right)
    \Rightarrow \ln(y)=\displaystyle\lim_{t \to 0} \left(\dfrac{2\sec(2t)}{1}\right  )=2

\Rightarrow y=e^2
    Now doing some bad maths:
    \displaystyle\lim_{x \to \infty} \left(\tan \left(\dfrac{\pi}{4} +\dfrac{1}{x} \right) \right)^x=1^{\infty}=e^2>2
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    ...Well that sure does it, but what a meal!

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    Mine:

     1^{\infty} = \displaystyle\lim_{x\to\infty} \left( 1+\dfrac{1}{x}\right) ^{x} = e

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    (Original post by FireGarden)
    ...Well that sure does it, but what a meal!

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    Show


    Mine:

     1^{\infty} = \displaystyle\lim_{x\to\infty} \left( 1+\dfrac{1}{x}\right) ^{x} = e

    Yeah, I came across the limit problem above earlier, so pretty much just copied and pasted it
    I considered;
    \left(\dfrac{x}{x+1} \right)^x but it didn't work so I gave up. :facepalm:
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    (Original post by Jkn)
    Problem 161**/***

    Let k be an integer greater than 1. Suppose a_0>0 and \displaystyle a_{n+1}=a_{n}+\frac{1}{\sqrt[k]{a_{n}}} for n>0

    Evaluate \displaystyle\lim_{n \to {\infty}} \frac{a^{k+1}_n}{n^k}
    I hadn't actually tried this problem when I posted it (as it was a 'problem 6' from the Putnam exam and I was hunting around for things that were going to challenge people), but I've given it a go and just managed to solve it! Really enjoyable so I hope people take a crack at it!
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    (Original post by FireGarden)
    Just posted this in one of the regular threads.. thought you guys might like it

    Problem 303**

    Give a justification for the claim  1^{\infty} > 2
    (Original post by joostan)
    Solution 303
    (Original post by FireGarden)
    ...Well that sure does it, but what a meal!

    Spoiler:
    Show


    Mine:

     1^{\infty} = \displaystyle\lim_{x\to\infty} \left( 1+\dfrac{1}{x}\right) ^{x} = e

    Spoiler:
    Show
    Not to be a ****, but surely you can both see that all of this is utterly wrong?

    If you are intending it as a 'joke problem' of the sort where the object is to spot the error (e.g. prove that 1=2), then please make sure that this is clear in the statement of the problem.
 
 
 
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