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    (Original post by 0x2a)
    Surely if \pi^2 is rational then \pi^3 is irrational?
    Yep, that's the same answer as mine, effectively I read the question as "show that if pi is irrational, then either pi^2 is irrational or pi^3 is irrational".
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    (Original post by Smaug123)
    Yep, that's the same answer as mine, effectively I read the question as "show that if pi is irrational, then either pi^2 is irrational or pi^3 is irrational".
    Ah right, one or the other. My bad. It's similar to showing that either e + \pi or e\pi (or some other combination of some transcendental numbers) need to be irrational.
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    (Original post by PAPADAPADOPOLOUS)
    The question is 'it is well known that he number pie is irrational, prove by contradiction that at least one of pie squared or pie cubed is irrational.' 5 marks
    I'm not surprised.
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    Can anyone help me with Quantative Rsearch methods question?
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    (Original post by 0x2a)
    Problem 478*

    (i) Prove that if
     |x - x_0| < min\left(\dfrac{\epsilon}{2(|y_0  | + 1)} , 1\right) and |y - y_0| < \dfrac{\epsilon}{2(|x_0| + 1)} .

    then |xy - x_0y_0| < \epsilon .

    *Note that: min(a,b) = a if  a < b .

    (ii) Prove that if y_0 \neq 0 and

    |y - y_0| < min\left(\dfrac{|y_0|}{2},\dfrac  {\epsilon|y_0|^2}{2}\right) .

    then  y \neq 0 and

     \left|\dfrac{1}{y} - \dfrac{1}{y_0}\right| < \epsilon .

    (iii) Replace the question marks with expressions involving  \epsilon, x_0, y_0 so that the conclusion will be true:

    If y_0 \neq 0 and

     |y - y_0| < \  ? and |x - x_0| < \  ?

    then  y \neq 0 and

    \left|\dfrac{x}{y} - \dfrac{x_0}{y_0}\right| < \epsilon.
    I'll have a crack at part i), maybe the others tomorrow, its too late.

    Spoiler:
    Show


     |x - x_0| < min\left(\dfrac{\epsilon}{2(|y_0  | + 1)} , 1\right) \Rightarrow |x - x_0| < 1, \, \, |x - x_0| < \dfrac{\epsilon}{2(|y_o| + 1)}

    |y - y_0| < \dfrac{\epsilon}{2(|x_o| + 1)}

     |x - x_0| \geq |x| - |x_0| \Rightarrow |x| - |x_0| < 1 \Rightarrow |x| < |x_0| + 1

     \dfrac{\epsilon}{2(|x_0 + 1|)} > 0 \Rightarrow |y - y_0||x| < \dfrac{\epsilon (|x_0| + 1)}{2(|x_0| + 1)} \Rightarrow |xy - xy_0| < \dfrac{\epsilon}{2}

     |x - x_0||y_0| < \dfrac{\epsilon |y_0|}{2(|y_0| + 1)}

     \dfrac{|y_0|}{|y_0| + 1} < 1 \Rightarrow |xy_0 - y_0x_0| < \dfrac{\epsilon}{2}

     |xy_0 - x_0y_0| + |xy - xy_0| \geq |xy_0 - x_0y_0 + xy - xy_0|

     |xy_0 - x_0y_0| + |xy - xy_0| \geq |xy - x_0y_0|

     |xy_0 - x_0y_0| + |xy - xy_0| < 2 \cdot \dfrac{\epsilon}{2} \Rightarrow |xy-x_0y_0| < \epsilon

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    This thread seems empty, so I thought I'd post a couple of problems.

    Problem 482**

     \displaystyle \int^{1}_{0} \dfrac{\arctan x}{x+1} \ dx

    Problem 483***

     \displaystyle \int^{\pi}_{-\pi} \dfrac{e^{\sin x + \cos x}\cos (\sin x )}{e^{\sin x} + e^{x}} \ dx
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    (Original post by newblood)
    Problem 480 ** (not much probability theory in A-Level)

    Let  X_1, X_2,...,X_n be i.i.d r.v's with uniform distribution on  [1,2] .

    Determine a unique value,  \Delta , such that \lim_{n\to \infty} Pr(a <(X_1X_2...X_n)^{1/n} < b) = 1 iff  a< \Delta <b
    Is it
    Spoiler:
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    4/e
    ?

    I'll type up my "solution" tomorrow if I'm right, I've only done a little probability though so the chances are I'm far off.
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    (Original post by Flauta)
    Is it
    Spoiler:
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    4/e
    ?

    I'll type up my "solution" tomorrow if I'm right, I've only done a little probability though so the chances are I'm far off.
    Just found this, subscribing
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    (Original post by ThatPerson)
    This thread seems empty, so I thought I'd post a couple of problems.

    Problem 482*

     \displaystyle \int^{1}_{0} \dfrac{\arctan x}{x+1} \ dx
    I've been labouring over 482 for a while now, any hints? Is the problem typed correctly? I've tried substitution, IBP, pretty much everything that's in my not-so-sophisticated toolbox. :confused:
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    (Original post by Zacken)
    I've been labouring over 482 for a while now, any hints? Is the problem typed correctly? I've tried substitution, IBP, pretty much everything that's in my not-so-sophisticated toolbox. :confused:
    Does your toolbox contain trig substitutions?

    Specifically the substitution
    Spoiler:
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     x = \tan \theta ?

    Also in hindsight, the difficulty rating should've been ** as there is one property that isn't usually taught, though it's appeared in STEP a few times.
    Spoiler:
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     \displaystyle \int^{b}_{a} f(x) \ dx = \int^{b}_{a} f(a+b-x) \ dx


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    (Original post by ThatPerson)
    Does your toolbox contain trig substitutions?

    Specifically the substitution
    Spoiler:
    Show
     x = \tan \theta ?

    Also in hindsight, the difficulty rating should've been ** as there is one property that isn't usually taught, though it's appeared in STEP a few times.
    Spoiler:
    Show
     \displaystyle \int^{b}_{a} f(x) \ dx = \int^{b}_{a} f(a+b-x) \ dx


    I did try, that, couldn't get far with it though. I'll have another go.
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    For 483 I have reduced it to \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

    but I'm not sure how to solve this one.
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    (Original post by james22)
    For 483 I have reduced it to \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

    but I'm not sure how to solve this one.
    Yeah same I gave up at that point, apparently the answer is  \pi or something which is quite nice but I think you need to use results from the exponential integral to get the answer
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    (Original post by jjpneed1)
    Yeah same I gave up at that point, apparently the answer is  \pi or something which is quite nice but I think you need to use results from the exponential integral to get the answer
    It looks like complex analysis would work, but I'm not sure how yet.
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    (Original post by james22)
    For 483 I have reduced it to \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

    but I'm not sure how to solve this one.

    (Original post by jjpneed1)
    Yeah same I gave up at that point, apparently the answer is  \pi or something which is quite nice but I think you need to use results from the exponential integral to get the answer
    I can't be bothered to dig out some of my notes, so this may not be rigorous:

    \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)} = \displaystyle\int^\pi_0 \left[\dfrac{1}{2} \exp(\frac{1}{2}(e^{-ix}-e^{ix})) + \dfrac{1}{2} \exp(\frac{1}{2}(e^{ix}-e^{-ix}))\right]e^{\cos(x)}

     = \displaystyle\int^\pi_0 \left[\dfrac{1}{2} \exp(\frac{1}{2}(e^{-ix}-e^{ix})) + \dfrac{1}{2} \exp(\frac{1}{2}(e^{ix}-e^{-ix}))\right] e^{\frac{1}{2}(e^{-ix}+e^{ix})}

    After multiplying through and cancelling you end up with

    = \displaystyle\int^\pi_0 \dfrac{1}{2} \left( e^{e^{-ix}} + e^{e^{ix}}\right)

    =  \dfrac{1}{2} \displaystyle\int^\pi_0 \left( \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}}{n!} + \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{inx}}{n!} \right)

     = \dfrac{1}{2} \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}+e^{inx}}{n!}

     = \displaystyle\int^\pi_0  \displaystyle\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{n!}

     = \displaystyle\sum_{n=0}^{\infty} \displaystyle\int^\pi_0 \dfrac{\cos(nx)}{n!} (by MCT)

     =  \displaystyle\sum_{n=0}^{\infty} \dfrac{\sin(\pi n)}{n!n}

     = \displaystyle\lim_{n \rightarrow 0} \dfrac{\sin(\pi n)}{n!n}

     = \pi
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    (Original post by Noble.)
    I can't be bothered to dig out some of my notes, so this may not be rigorous:

    \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)} = \displaystyle\int^\pi_0 \left[\dfrac{1}{2} \exp(\frac{1}{2}(e^{-ix}-e^{ix})) + \dfrac{1}{2} \exp(\frac{1}{2}(e^{ix}-e^{-ix}))\right]e^{\cos(x)}

     = \displaystyle\int^\pi_0 \left[\dfrac{1}{2} \exp(\frac{1}{2}(e^{-ix}-e^{ix})) + \dfrac{1}{2} \exp(\frac{1}{2}(e^{ix}-e^{-ix}))\right] e^{\frac{1}{2}(e^{-ix}+e^{ix})}

    After multiplying through and cancelling you end up with

    = \displaystyle\int^\pi_0 \dfrac{1}{2} \left( e^{e^{-ix}} + e^{e^{ix}}\right)

    =  \dfrac{1}{2} \displaystyle\int^\pi_0 \left( \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}}{n!} + \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{inx}}{n!} \right)

     = \dfrac{1}{2} \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}+e^{inx}}{n!}

     = \displaystyle\int^\pi_0  \displaystyle\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{n!}

     = \displaystyle\sum_{n=0}^{\infty} \displaystyle\int^\pi_0 \dfrac{\cos(nx)}{n!} (by MCT)

     =  \displaystyle\sum_{n=0}^{\infty} \dfrac{\sin(\pi n)}{n!n}

     = \displaystyle\lim_{n \rightarrow 0} \dfrac{\sin(\pi n)}{n!n}

     = \pi
    I'm not sure about that jump in the last 3 lines. How do you get the sum being equal to the limit?
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    (Original post by james22)
    I'm not sure about that jump in the last 3 lines. How do you get the sum being equal to the limit?
    Well \sin(n \pi) = 0 \ \ \forall n \in \mathbb{N} while the denominator is non-zero for all non-zero n, so the only non-zero term is when n = 0
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    Solution 482:

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     I = \displaystyle\int_0^1 \dfrac{\arctan(x)}{x+1} \ dx

    Firstly, we don't like the arctan here, so we get rid of it - use the substitution  x = \tan(\theta) :

     \Rightarrow I = \displaystyle\int_0^{\frac{\pi}{  4}} \dfrac{\theta \sec^2(\theta)}{1+\tan(\theta)} \ d\theta

    We don't particularly like sec and tan either, so get rid of those too:

     I = \displaystyle\int_0^{\frac{\pi}{  4}}\dfrac{\theta}{\cos^2(\theta)  +\sin(\theta)\cos(\theta)} \ d\theta

    We can clean up the bottom a bit using double-angle formulae, and then use the substitution  u=2 \theta to obtain the following:

     2I =\displaystyle\int_0^{\frac{\pi}  {2}} \dfrac{u}{1+\cos(u)+\sin(u)} \ du

    Use the substitution  y = \frac{\pi}{2} - u to then obtain:

     4I = \frac{\pi}{2}\displaystyle\int_0  ^{\frac{\pi}{2}} \dfrac{1}{1+\cos(u)+\sin(u)} \ du

    This is the key step insofar as that it gets rid of the ugly bit - now we only have one type of function to deal with. I think there are a few ways to complete this now, such as Weierstrass substitution. However, for this solution, an easy way is to just repeat the substitutions we did in reverse to obtain:

     4I = \frac{\pi}{2} \displaystyle\int_0^1 \dfrac{1}{x+1} \ dx \Rightarrow I = \frac{\pi}{8}\log(2)

    Moral: Symmetry (in this question, sine and cosine in the interval  [0, \frac{\pi}{2}] ) is a powerful tool; use it if you can.

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    (Original post by james22)
    For 483 I have reduced it to \displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

    but I'm not sure how to solve this one.
    Spoiler:
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    One approach is to consider  \displaystyle \Re\left(\displaystyle \int^{\pi}_{0} e^{e^{ix}} \ dx \right) .

    I'm sure Complex Analysis would also work, but I haven't studied it yet.
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    (Original post by ThatPerson)
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    One approach is to consider  \displaystyle \Re\left(\displaystyle \int^{\pi}_{0} e^{e^{ix}} \ dx \right) .

    I'm sure Complex Analysis would also work, but I haven't studied it yet.
    How can you evaluate that integral though?
 
 
 
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