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The Proof is Trivial!

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Original post by Sketch
Is this the (***) solution? If not, I don't have any knowledge of double integrals at A Level! :lol:


Pf! I'm trying to stay away from this thread because I need to work and you quote me back here! :tongue: Yes it is. A similar method, again with double integrals, is to make all the limits independent, so that the switch is more intuitive:

∫0∞arctan⁡ex−arctan⁡xx dx=∫0∞∫1e11+t2x2 dt dx=∫1e∫0∞11+t2x2 dx dt=∫1eπ2t dt=π2\displaystyle\begin{aligned}\int_0^{\infty} \frac{\arctan ex-\arctan x}{x}\,dx &= \int_0^{\infty}\int_1^{e}\frac{ 1}{1+t^2x^2}\,dt \,dx \\ &=\int_1^{e}\int_{0}^{\infty} \frac{1}{1+t^2x^2}\,dx \, dt=\int_1^{e}\frac{\pi}{2t}\,dt=\frac{\pi}{2}\end{aligned}

Original post by Jkn
For ****s sake :lol: classic physicist :lol:


Actually, it's a very standard switch for double integrals, and what he has written is perfectly rigorous. And I'm sorry for picking on you again, but it looks like you are commenting something you don't know much about - the substitution is not r2=x2+y2r^2=x^2+y^2 at all.
Original post by Mladenov
As you want another proof for problem 128, try to prove the following result:

Problem 130*

n2n+1((2n)!!)2((2n−1)!!)2(2n+1)<(∫0∞e−x2dx)2<n2n−1(2n−3)!!)2(2n−1)((2n−2)!!)2π2\begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}}{((2n-1)!!)^{2}(2n+1)} < \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} < \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\frac{\pi}{2} \end{aligned}.
Then note that e−x2e^{-x^{2}} is even.


Well it's so obvious when you put it like that(!)
Reply 862
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Jkn
11
Original post by Lord of the Flies

Actually, it's a very standard switch for double integrals, and what he has written is perfectly rigorous. And I'm sorry for picking on you again, but it looks like you are commenting something you don't know much about - the substitution is not r2=x2+y2r^2=x^2+y^2 at all.

****s sake hahaha :lol:

Exactly, I have no idea what a double integral even is (asides from it meaning integrating something twice). But it would be nice for such a step to be justified in terms of regular integration :tongue: Looking back on my post, I suppose what I'm really doing is requesting a proof/derivation of this "double integration" technique :tongue:

Oh. What is it then?
Reply 863
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Jkn
11
Original post by Mladenov
As you want another proof for problem 128, try to prove the following result:

Problem 130*

n2n+1((2n)!!)2((2n−1)!!)2(2n+1)<(∫0∞e−x2dx)2<n2n−1(2n−3)!!)2(2n−1)((2n−2)!!)2π2\begin{aligned} \displaystyle \frac{n}{2n+1}\frac{((2n)!!)^{2}}{((2n-1)!!)^{2}(2n+1)} < \left(\int_{0}^{\infty} e^{-x^{2}}dx \right)^{2} < \frac{n}{2n-1}\frac{(2n-3)!!)^{2}(2n-1)}{((2n-2)!!)^{2}}\frac{\pi}{2} \end{aligned}.
Then note that e−x2e^{-x^{2}} is even.

Hmm, looks suspiciously similar to the series I derived (it included double factorial notation). I'll give it a go :lol: (It doesn't require double integrals does it?)
(edited 10 years ago)
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Mladenov
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Original post by Jkn
Hmm, looks suspiciously similar to the series I derived (it included double factorial notation). I'll give it a go :lol: (It doesn't require double integrals does it?)


It does not.
Original post by Jkn
****s sake hahaha :lol:

Exactly, I have no idea what a double integral even is (asides from it meaning integrating something twice). But it would be nice for such a step to be justified in terms of regular integration :tongue: Looking back on my post, I suppose what I'm really doing is requesting a proof/derivation of this "double integration" technique :tongue:

Oh. What is it then?


Spoiler

(edited 10 years ago)
Reply 866
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Jkn
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Original post by Mladenov
It does not.

Hmm.. seems a bit of a beast. I've simplified the LHS in terms of the maclaurin's arctan expansion so I can pretty much prove it. But were you suggesting I start with the middle? (Or trap it between the multiples of π\pi? :tongue:)
Original post by Lord of the Flies

Spoiler


Hmm, sounds interesting. I will add it to my post-STEP reading list :smile:

Well yes it has always been obvious that they cover it but the equivalence isn't! Anyway... I need to stop asking people to rigorously prove things they regard as assumed knowledge (I have a habit of doing that :lol:)
Reply 867
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Mladenov
1
Original post by Jkn
...


Shall, I recommend you commence with e−x2e^{-x^{2}}.
I am not sure which was more of an adventure - doing the problem, or typing the solution.

Solution 130

First we prove an intermediate result:

yn=∫0π/2cos⁡nt dt⇒yn=(n−1)∫0π/2sin⁡2tcos⁡n−2t dt=(n−1)yn−2−(n−1)yn⇒yn=n−1nyn−2⇒{yn=(2k−1)!!(2k)!! y0    (n=2k)yn=(2k)!!(2k+1)!! y1    (n=2k+1)(y0=π2     y1=1)\displaystyle \begin{aligned}y_n=\int_0^{\pi/2}\cos^nt\,dt\Rightarrow y_n&=(n-1)\int_0^{\pi/2}\sin^2 t\cos^{n-2}t\,dt\\[7pt]&=(n-1)y_{n-2}-(n-1)y_{n}\\[7pt]&\qquad\Rightarrow y_n=\frac{n-1}{n}y_{n-2}\\[7pt]&\qquad\Rightarrow\left\{ \begin{array}{l l} y_n=\dfrac{(2k-1)!!}{(2k)!!}\,y_0\;\;(n=2k)\\[10pt] y_n=\dfrac{(2k)!!}{(2k+1)!!}\,y_1\;\;(n=2k+1)\end{array}\right. \\[7pt]&\qquad\qquad\big(y_0=\dfrac{\pi}{2}\,\;\;y_1=1\big)\end{aligned}

∙\bullet Lower bound:

x≠0:  ex>x+1⇒e−nx2>(1−x2)nx\neq 0:\; e^x>x+1\Rightarrow e^{-nx^2}>(1-x^2)^n. Integrate and set x=sin⁡t:x=\sin t:

Unparseable latex formula:

\displaystyle\begin{aligned} \int_0^{1}e^{-nx^2}dx>\int_0^1 (1-x^2)^n\,dx\Rightarrow \int_0^{1}e^{-nx^2}dx>\int_0^{\frac{\pi}{2}} \cos^{2n+1} t\,dt=\frac{(2n)!!}{(2n+1)!!}



Observe that ∫01e−nx2dx<∫0∞e−nx2 dx=1n∫0∞e−x2 dx\displaystyle \int_0^1 e^{-nx^2}dx<\int_0^{\infty} e^{-nx^2}\,dx=\frac{1}{\sqrt{n}}\int_0^{\infty} e^{-x^2}\,dx and hence:

n((2n)!!(2n+1)!!)2<(∫0∞e−x2 dx)2\displaystyle n\left(\frac{(2n)!!}{(2n+1)!!} \right)^2<\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2

∙\bullet Upper bound:

Similarly, ex>x+1⇒e−nx2<(x2+1)−ne^{x}>x+1\Rightarrow e^{-nx^2}<(x^2+1)^{-n}. This time set x=tan⁡tx=\tan t and integrate directly over the interval of interest:

Unparseable latex formula:

\displaystyle \begin{aligned} \int_0^{\infty}e^{-nx^2}dx<\int_0^{\infty} (x^2+1)^{-n}\,dx\Rightarrow \int_0^{\infty}e^{-nx^2}dx<\int_0^{\frac{\pi}{2}} \cos^{2n-2} t\,dt=\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2}



As above, this yields:

(∫0∞e−x2 dx)2<n((2n−3)!!(2n−2)!!π2)2\displaystyle \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2< n\left(\frac{(2n-3)!!}{(2n-2)!!}\frac{\pi}{2} \right)^2



Now it is well-known that 21⋅23⋅43⋅45⋯=lim⁡n→∞12n+1((2n)!!(2n−1)!!)2=π2\displaystyle \dfrac{2}{1}\cdot \dfrac{2}{3}\cdot \dfrac{4}{3}\cdot \dfrac{4}{5}\cdots = \lim_{n\to\infty}\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2=\frac{\pi}{2} hence:

Unparseable latex formula:

\displaystyle\begin{aligned} \frac{n}{2n+1}\left[\frac{1}{2n+1} \left(\frac{(2n)!!}{(2n-1)!!} \right)^2\right]<\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2<\frac{\pi^2}{4}\cdot\frac{n}{2n-1}\left[\frac{1}{2n-1} \left(\frac{(2n-2)!!}{(2n-3)!!} \right)^2\right]^{-1}



Unparseable latex formula:

\displaystyle\begin{aligned} \frac{1}{2}\left(\frac{\pi}{2} \right)\leq\left(\int_0^{\infty} e^{-x^2}\,dx\right)^2\leq\frac{\pi^2}{4}\cdot\frac{1}{2} \left( \frac{2}{\pi}\right)\Rightarrow \left(\int_0^{\infty} e^{-x^2}\,dx\right)^2=\frac{\pi}{4}



Thus ∫−∞∞e−x2 dx=π\displaystyle\int_{-\infty}^{\infty} e^{-x^2}\,dx=\sqrt{\pi}

Mladenov I think there is a typo in your question, the π2\frac{\pi}{2} should be included in the square as well.
(edited 10 years ago)
Reply 869
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Mladenov
1
Original post by Lord of the Flies
Solution 130


Well done. :biggrin:


Original post by Lord of the Flies
Mladenov I think there is a typo in your question, the π2\displaystyle \frac{\pi}{2} should be included in the square as well.


Yes, it is a typo, I apologise.
Original post by bananarama2
Solution 129

ddx1k∫0xf(x′)sinh⁡(k(x−x′))dx′ \frac{d}{dx} \frac{1}{k} \int_0^{x} f(x')\sinh (k(x-x')) dx'

=1kf(x)sinh⁡(k(x−x))−f(0)1ksinh⁡(kx)×0+1k∫0xkf(x′)cosh⁡(k(x−x′))dx′= \frac{1}{k} f(x) \sinh (k(x-x)) - f(0) \frac{1}{k}\sinh(kx) \times 0 + \frac{1}{k} \int_0^x k f(x') \cosh (k(x-x')) dx'

=∫0xf(x′)cosh⁡(k(x−x′))dx′= \int_0^x f(x') \cosh (k(x-x')) dx'

ddx∫0xf(x′)cosh⁡(k(x−x′))dx′ \frac{d}{dx} \int_0^x f(x') \cosh (k(x-x')) dx'

Unparseable latex formula:

= f(x)\cosh (k(x-x)) - 0 + \int_0^x kf(x')\shin (k(x-x')) dx'



=y1′′ = y_1''

SO

y1′′−k2y y_1'' - k^2y

Unparseable latex formula:

= f(x)\cosh (k(x-x)) + \int_0^x kf(x')\shin (k(x-x')) dx' - \int_0^x kf(x')\shin (k(x-x')) dx'



=f(x) = f(x)


hmmmm not sure about your multiplication by zero in the second line though you appear to get the correct answer so I'll give you that.

For the more general case

PROBLEM <some number>**

Find the most general solution to the linear second order differential equation:

y''+ay'+by=f(x)

(there is absolutely no need to apply the wronskian btw so don't if you can help it)

Spoiler

Seeing as you appear from an earlier post to have done multiple integrals, there is a far simpler way to approach this problem :smile:
(edited 10 years ago)
Original post by und
Problem 1*

Prove that if the function f\displaystyle f is continuous on [a,b]\displaystyle [a,b] and ∫abf(x)g(x)dx=0\displaystyle \int^b_af(x)g(x) dx = 0 for any continuous function g\displaystyle g, then f=0\displaystyle f=0 on [a,b]\displaystyle [a,b].


You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.
(edited 10 years ago)
Reply 873
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Mladenov
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Original post by k9markiii
You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.


I do not know what you are trying to say, but not all integrable functions are continuous. For example, you can take any bounded function such that the set of its discontinuities has measure zero.
Original post by Mladenov

Edit: For the sake of completeness, I would like to add that we can use Chebotarev's density theorem to show that there are many primes qq which satisfy the condition. In other words, we look for those primes qq for which the degree of the splitting field of PP over FqF_{q} is pp.
What do you think about the case when deg⁡(P)=pα\deg (P) = p^{\alpha}, where α>1\alpha > 1.


In fact, infinitely many. If you play with Chebotarev's theorem for a while, then you will discover that such an irreducible P∈Z[X]P \in \mathbb{Z}[X], which has a root in Z/qZ\mathbb{Z}/q\mathbb{Z} for all but finitely many primes qq, must have degree 11. An even more striking fact is that for every P∈Z[X]P \in \mathbb{Z}[X], with deg⁡(P)≥1\deg(P) \geq 1, there is a certain number mm, determined by the coefficients of the irreducible factors of PP, such that PP has a root in Z/nZ\mathbb{Z}/n\mathbb{Z} for all n∈Nn \in \mathbb{N} if, and only if, it has a root in Z/mZ\mathbb{Z}/m\mathbb{Z} (with one more condition).

You don't need the prime degree assumption. A monic P∈Z[X]P \in \mathbb{Z}[X], with deg⁡(P)>1\deg(P) > 1, whose reduction in Z/pZ[X]\mathbb{Z}/p\mathbb{Z}[X] has a root for all primes pp must be reducible over Z\mathbb{Z}. This can be proved by playing with the Galois group and only a weaker form of Frobenius' density theorem.

A good example to keep in mind, though, is x4+1x^4 + 1. This is irreducible over Z\mathbb{Z}, but reducible in Z/pZ\mathbb{Z}/p\mathbb{Z} for all primes pp. The condition for having a root is stronger: the congruence x4≡−1 (p)x^4 \equiv -1\ (p) has no solutions for primes p≡3 (4)p \equiv 3\ (4).
Original post by Lord of the Flies
...and the lecturer is French!....


Makes me think of this :colone:

Original post by natninja
hmmmm not sure about your multiplication by zero in the second line though you appear to get the correct answer so I'll give you that.

For the more general case

PROBLEM <some number>**

Find the most general solution to the linear second order differential equation:

y''+ay'+by=f(x)

(there is absolutely no need to apply the wronskian btw so don't if you can help it)


Can I check the answer first? I get

y=eβx∫(eαx−βx∫f(x)e−αxdx)dx y=e^{\beta x} \int \left(e^{\alpha x - \beta x } \int f(x) e^{-\alpha x} dx \right)dx

α=−a+a2−4b2β=−a−a2−4b2 \alpha = \frac{-a+\sqrt{a^2-4b}}{2} \qquad \beta= \frac{-a-\sqrt{a^2-4b}}{2}
Original post by bananarama2
Can I check the answer first? I get

y=eβx∫(eαx−βx∫f(x)e−αxdx)dx y=e^{\beta x} \int \left(e^{\alpha x - \beta x } \int f(x) e^{-\alpha x} dx \right)dx

α=−a+a2−4b2β=−a−a2−4b2 \alpha = \frac{-a+\sqrt{a^2-4b}}{2} \qquad \beta= \frac{-a-\sqrt{a^2-4b}}{2}


the answer looks a fair bit nicer than that... but your alpha and beta are correct, your answer should be a sum of roots and you shouldn't have integrals with respect to x but with respect to a dummy variable, they should also be definite integrals
(edited 10 years ago)
Original post by natninja
Seeing as you appear from an earlier post to have done multiple integrals, there is a far simpler way to approach this problem :smile:


That is a solution to problem 130 :wink:

(would you mind spoilering that quote, so that it takes up less space on the page?)

Original post by k9markiii
You can't use integration over a non-continuous function anyway though. Although you could do it with the chain rule to show you can integrate part of it.


Yes you can.

Original post by ukdragon37
Makes me think of this :colone:


:lol:

I like this one too.
(edited 10 years ago)
Reply 879
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Mladenov
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Original post by jack.hadamard
An even more striking fact is that for every P∈Z[X]P \in \mathbb{Z}[X], with deg⁡(P)≥1\deg(P) \geq 1, there is a certain number mm, determined by the coefficients of the irreducible factors of PP, such that PP has a root in Z/nZ\mathbb{Z}/n\mathbb{Z} for all n∈Nn \in \mathbb{N} if, and only if, it has a root in Z/mZ\mathbb{Z}/m\mathbb{Z} (with one more condition).


Something more on this would be highly appreciated.

Original post by jack.hadamard

You don't need the prime degree assumption. A monic P∈Z[X]P \in \mathbb{Z}[X], with deg⁡(P)>1\deg(P) > 1, whose reduction in Z/pZ[X]\mathbb{Z}/p\mathbb{Z}[X] has a root for all primes pp must be reducible over Z\mathbb{Z}. This can be proved by playing with the Galois group and only a weaker form of Frobenius' density theorem.


Some difficulties arise here; but it is indeed splendid result.:tongue:

Original post by jack.hadamard
In fact, infinitely many. If you play with Chebotarev's theorem for a while, then you will discover that such an irreducible P∈Z[X]P \in \mathbb{Z}[X], which has a root in Z/qZ\mathbb{Z}/q\mathbb{Z} for all but finitely many primes qq, must have degree 11.


Here is what I think.

Spoiler

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