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The Proof is Trivial!

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Original post by 0x2a
Surely if π2\pi^2 is rational then π3\pi^3 is irrational?

Yep, that's the same answer as mine, effectively :smile: 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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Avatar for 0x2a
0x2a
2
Original post by Smaug123
Yep, that's the same answer as mine, effectively :smile: 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+πe + \pi or eπe\pi (or some other combination of some transcendental numbers) need to be irrational.
(edited 9 years ago)
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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Avatar for SRaina
SRaina
Can anyone help me with Quantative Rsearch methods question?
Original post by 0x2a
Problem 478*

(i) Prove that if
∣x−x0∣<min(ϵ2(∣y0∣+1),1) |x - x_0| < min\left(\dfrac{\epsilon}{2(|y_0| + 1)} , 1\right) and ∣y−y0∣<ϵ2(∣x0∣+1)|y - y_0| < \dfrac{\epsilon}{2(|x_0| + 1)} .

then ∣xy−x0y0∣<ϵ|xy - x_0y_0| < \epsilon .

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

(ii) Prove that if y0≠0y_0 \neq 0 and

∣y−y0∣<min(∣y0∣2,ϵ∣y0∣22)|y - y_0| < min\left(\dfrac{|y_0|}{2},\dfrac{\epsilon|y_0|^2}{2}\right) .

then y≠0 y \neq 0 and

∣1y−1y0∣<ϵ \left|\dfrac{1}{y} - \dfrac{1}{y_0}\right| < \epsilon .

(iii) Replace the question marks with expressions involving ϵ,x0,y0 \epsilon, x_0, y_0 so that the conclusion will be true:

If y0≠0y_0 \neq 0 and

∣y−y0∣< ? |y - y_0| < \ ? and ∣x−x0∣< ?|x - x_0| < \ ?

then y≠0 y \neq 0 and

∣xy−x0y0∣<ϵ\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

(edited 9 years ago)
This thread seems empty, so I thought I'd post a couple of problems.

Problem 482**

∫01arctan⁡xx+1 dx \displaystyle \int^{1}_{0} \dfrac{\arctan x}{x+1} \ dx

Problem 483***

∫−ππesin⁡x+cos⁡xcos⁡(sin⁡x)esin⁡x+ex dx \displaystyle \int^{\pi}_{-\pi} \dfrac{e^{\sin x + \cos x}\cos (\sin x )}{e^{\sin x} + e^{x}} \ dx
(edited 9 years ago)
Original post by newblood
Problem 480 ** (not much probability theory in A-Level)

Let X1,X2,...,Xn X_1, X_2,...,X_n be i.i.d r.v's with uniform distribution on [1,2] [1,2] .

Determine a unique value, Δ \Delta , such that lim⁡n→∞Pr(a<(X1X2...Xn)1/n<b)=1\lim_{n\to \infty} Pr(a <(X_1X_2...X_n)^{1/n} < b) = 1 iff a<Δ<b a< \Delta <b


Is it

Spoiler

?

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.
(edited 9 years ago)
Original post by Flauta
Is it

Spoiler

?

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 :smile:
Original post by ThatPerson
This thread seems empty, so I thought I'd post a couple of problems.

Problem 482*

∫01arctan⁡xx+1 dx \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:
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

(edited 9 years ago)
Original post by ThatPerson
Does your toolbox contain trig substitutions?

Specifically the substitution

Spoiler



I did try, that, couldn't get far with it though. I'll have another go. :smile:
For 483 I have reduced it to ∫0πcos⁡(sin⁡(x))ecos⁡(x)\displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

but I'm not sure how to solve this one.
Original post by james22
For 483 I have reduced it to ∫0πcos⁡(sin⁡(x))ecos⁡(x)\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 π \pi or something which is quite nice but I think you need to use results from the exponential integral to get the answer
Original post by jjpneed1
Yeah same I gave up at that point, apparently the answer is π \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.
Original post by james22
For 483 I have reduced it to ∫0πcos⁡(sin⁡(x))ecos⁡(x)\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 π \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:

∫0πcos⁡(sin⁡(x))ecos⁡(x)=∫0π[12exp⁡(12(e−ix−eix))+12exp⁡(12(eix−e−ix))]ecos⁡(x)\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)}

=∫0π[12exp⁡(12(e−ix−eix))+12exp⁡(12(eix−e−ix))]e12(e−ix+eix) = \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

=∫0π12(ee−ix+eeix)= \displaystyle\int^\pi_0 \dfrac{1}{2} \left( e^{e^{-ix}} + e^{e^{ix}}\right)

=12∫0π(∑n=0∞e−inxn!+∑n=0∞einxn!)= \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)

=12∫0π∑n=0∞e−inx+einxn! = \dfrac{1}{2} \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}+e^{inx}}{n!}

=∫0π∑n=0∞cos⁡(nx)n! = \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{n!}

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

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

=lim⁡n→0sin⁡(πn)n!n = \displaystyle\lim_{n \rightarrow 0} \dfrac{\sin(\pi n)}{n!n}

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

∫0πcos⁡(sin⁡(x))ecos⁡(x)=∫0π[12exp⁡(12(e−ix−eix))+12exp⁡(12(eix−e−ix))]ecos⁡(x)\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)}

=∫0π[12exp⁡(12(e−ix−eix))+12exp⁡(12(eix−e−ix))]e12(e−ix+eix) = \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

=∫0π12(ee−ix+eeix)= \displaystyle\int^\pi_0 \dfrac{1}{2} \left( e^{e^{-ix}} + e^{e^{ix}}\right)

=12∫0π(∑n=0∞e−inxn!+∑n=0∞einxn!)= \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)

=12∫0π∑n=0∞e−inx+einxn! = \dfrac{1}{2} \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{e^{-inx}+e^{inx}}{n!}

=∫0π∑n=0∞cos⁡(nx)n! = \displaystyle\int^\pi_0 \displaystyle\sum_{n=0}^{\infty} \dfrac{\cos(nx)}{n!}

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

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

=lim⁡n→0sin⁡(πn)n!n = \displaystyle\lim_{n \rightarrow 0} \dfrac{\sin(\pi n)}{n!n}

=π = \pi


I'm not sure about that jump in the last 3 lines. How do you get the sum being equal to the limit?
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⁡(nπ)=0  ∀n∈N\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
(edited 9 years ago)
Solution 482:

Spoiler

(edited 9 years ago)
Original post by james22
For 483 I have reduced it to ∫0πcos⁡(sin⁡(x))ecos⁡(x)\displaystyle\int^\pi_0 \cos(\sin(x)) e^{\cos(x)}

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


Spoiler

Original post by ThatPerson

Spoiler



How can you evaluate that integral though?

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