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The Proof is Trivial!

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Reply 880
Avatar for und
und
OP
16
Original post by DJMayes
I find it telling how the OP hasn't submitted a solution since about question 50, and we're now past question 120. This thread stopped being interesting (Or even accessible) for the vast majority of people quite a long time ago and now it seems very much to just be a battle of wits between Mladenov and Lord of the Flies.

(If this comes across as jealous, then that's probably because it is - I'd love to be able to even attempt the questions that are being thrown around lately! :lol: Unfortunately, the questions that are being thrown out are aimed at a very small audience indeed and just looking at the list in the OP it's very easy to see. It reminds me very strongly of what the STEP thread was like before Christmas.)

It's partly due to the fact that I really haven't had the time lately, but indeed most of the questions seem rather inaccessible. Let them have their fun, but I think we should have a mix.
Original post by natninja
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


Well I can't think of a way getting another general "general solution" especially involving definite integrals which are often used for particular solutions. Further more given no one knows what format you want the answer in, my answer is perfectly valid. It is indeed a general solution whether it's the one you wanted or not.
Original post by und
It's partly due to the fact that I really haven't had the time lately, but indeed most of the questions seem rather inaccessible. Let them have their fun, but I think we should have a mix.


I second this, there is nothing wrong with the difficult questions per se. After all it's what this thread is about, but it would be nice to have some which the sane are able to attempt.
Original post by bananarama2
Well I can't think of a way getting another general "general solution" especially involving definite integrals which are often used for particular solutions. Further more given no one knows what format you want the answer in, my answer is perfectly valid. It is indeed a general solution whether it's the one you wanted or not.


ok I'll believe you, ideally I wanted it in the form of a complementary function plus the particular integral
Here's a question everybody should be able to have a go at:

Problem 131 (*)

Find the sum of squares of all real roots of the polynomial: Find \ the \ sum \ of \ squares \ of \ all \ real \ roots \ of \ the \ polynomial:

f(x)=x5−7x3+2x2−30x+6 f(x) = x^5 - 7x^3 + 2x^2 -30x + 6

If you find it easy, please wait for somebody else to post an answer!
Alright, people have been asking for */** problems, so here are a few:

Problem 132*

∫0∞ln⁡x(x2+1)2 dx\displaystyle \int_0^{\infty} \frac{\ln x}{(x^2+1)^2}\, dx

Problem 133*

∫0∞ln⁡xx2+α2 dx    (α>0)\displaystyle \int_0^{\infty} \frac{\ln x}{x^2+\alpha^2}\, dx\;\;(\alpha >0)

Problem 134*

∫0∞(11+x2)(xy−xz(1+xy)(1+xz)) dx    (y,z∈R)\displaystyle \int_{0}^{\infty} \left(\frac {1}{1+x^{2}}\right)\left( \frac {x^{y}-x^{z}}{(1+x^{y})(1+x^{z})}\right)\, dx\;\;(y,z\in\mathbb{R})

Problem 135**

Find all nn such that:

1!+2!+3!+⋯+n!1!+2!+3!+\cdots +n! is a perfect square.

Find all mm such that:

11+12+13+⋯+1m\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+ \cdots+\dfrac{1}{m} is an integer.
(edited 10 years ago)
Original post by Mladenov
Something more on this would be highly appreciated.


If you are interested in reading the original paper, then you can find it here. It is a known result which can be found in some introductory books. More on the irreducible but locally-reducible polynomials can be found here. This is a nice read too.

As a side note, CDT is usually graduate knowledge (taught in a Part III course at Cam).
(edited 10 years ago)
Reply 887
Avatar for Mladenov
Mladenov
1
Solution 135

For the first part, looking (mod5)\pmod 5, we observe that for n>4n >4, 1!+2!+...+n!≡1!+2!+3!+4!≡3(mod5)1!+2!+...+n! \equiv 1!+2!+3!+4! \equiv 3 \pmod 5. Therefore n≤4n \le 4, and the solutions are n=1n=1 and n=3n=3.

The second part is a consequence of Bertrand's postulate:
There is a prime number pp, which lies strictly between mm and 2m2m, where m>1m>1.

Otherwise, we can multiply by the second largest power of 22 which is less than mm. Hence we would have k=12+nl\displaystyle k = \frac{1}{2}+\frac{n}{l}, where l≢0(mod2)l \not\equiv 0 \pmod 2, thus kk is not an integer.
I forgot to say that m=1m=1 is a solution.


Original post by jack.hadamard
If you are interested in reading the original paper, then you can find it here. It is a known result which can be found in some introductory books. More on the irreducible but locally-reducible polynomials can be found here. This is a nice read too.

As a side note, CDT is usually graduate knowledge (taught in a Part III course at Cam).


Thanks, these are useful articles.
I asked for some references, since I am interested in reading algebra and number theory.
(edited 10 years ago)
Solution 132

This took an eon to compute but I was hell-bent on getting my name on the OP again :colone:

∫0∞ln⁡x(x2+1)2dx=I\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{(x^{2} + 1)^{2}} dx = I

Let x=u−1⇒∫0∞ln⁡x(x2+1)2dx=−∫0∞u2ln⁡u(u2+1)2dux = u^{-1} \Rightarrow \displaystyle\int_{0}^{\infty} \dfrac{\ln x}{(x^{2} + 1)^{2}} dx = - \displaystyle\int_{0}^{\infty} \dfrac{u^{2} \ln u}{(u^{2} + 1)^{2}} du

Consider ∫0∞ln⁡xx2+1dx\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + 1} dx

Letting x=tan⁡θ⇒∫0∞ln⁡xx2+1dx=∫0π/2ln⁡(tan⁡θ)dθx = \tan \theta \Rightarrow \displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + 1} dx = \displaystyle\int_{0}^{\pi /2} \ln (\tan \theta) d \theta

=∫0π/2ln⁡(sin⁡θ)dθ−∫0π/2ln⁡(cos⁡θ)dθ= \displaystyle\int_{0}^{\pi /2} \ln (\sin \theta) d \theta - \displaystyle\int_{0}^{\pi /2} \ln (\cos \theta) d \theta

=0= 0

By parts, we obtain:

∫0∞ln⁡xx2+1dx=[x(ln⁡x−1)x2+1]0∞⏟0−∫0∞2x2(ln⁡x)(x2+1)2dx⏟2I+∫0∞2x2(x2+1)2dx\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + 1} dx = \underbrace{\left[\dfrac{x( \ln x - 1)}{x^{2} + 1} \right]^{\infty}_{0}}_{0} \underbrace{-\displaystyle\int_{0}^{\infty} \dfrac{2x^{2} (\ln x)}{(x^{2} + 1)^{2}} dx}_{2 I} + \displaystyle\int_{0}^{\infty} \dfrac{2x^{2}}{(x^{2} + 1)^{2}} dx

∫0∞ln⁡xx2+1dx=2I+∫0∞2x2(x2+1)2dx=0⇒I=−∫0∞x2(x2+1)2dx=−π4\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + 1} dx = 2I + \displaystyle\int_{0}^{\infty} \dfrac{2x^{2}}{(x^{2} + 1)^{2}} dx = 0 \Rightarrow I = -\displaystyle\int_{0}^{\infty} \dfrac{x^{2}}{(x^{2} + 1)^{2}} dx = -\dfrac{\pi}{4}
(edited 10 years ago)
Original post by Mladenov

I asked for some references, since I am interested in reading algebra and number theory.


There is a good chunk of algebra and number theory before you actually get to algebraic number theory.
What books do you use now? It may be best if you complement them with some expository articles.
:ninja: No one saw that rookie mistake :tongue:
Solution 133

∫0∞ln⁡xx2+α2dx\displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + \alpha^{2}} dx

Let x=αtan⁡θ⇒dx=(x2+α2)dθα,∞→π2,0→0x = \alpha \tan \theta \Rightarrow dx = \dfrac{(x^{2} + \alpha^{2}) d \theta}{\alpha}, \infty \to \dfrac{\pi}{2}, 0 \to 0

⇒∫0∞ln⁡xx2+α2dx=1α∫0π/2ln⁡(αtan⁡θ)dθ\Rightarrow \displaystyle\int_{0}^{\infty} \dfrac{\ln x}{x^{2} + \alpha^{2}} dx = \dfrac{1}{\alpha} \displaystyle\int_{0}^{\pi /2} \ln (\alpha \tan \theta) d \theta

=1α(ln⁡(α)∫0π/2dθ+∫0π/2ln⁡(tan⁡θ)dθ⏟0 (see solution 132))= \dfrac{1}{\alpha} \left( \ln (\alpha) \displaystyle\int_{0}^{\pi /2} d \theta + \underbrace{\displaystyle\int_{0}^{\pi /2} \ln (\tan \theta) d \theta}_{0 \ (\text{see solution 132})} \right)

=π2⋅ln⁡αα= \dfrac{\pi}{2} \cdot \dfrac{\ln \alpha}{\alpha}
(edited 10 years ago)
Original post by bananarama2
...

You quoted me sir? :tongue:
Original post by Felix Felicis
You quoted me sir? :tongue:


:ninja: Errm. I thought I'd spotted something, but I'd just made a mistake myself.
Original post by bananarama2
:ninja: Errm. I thought I'd spotted something, but I'd just made a mistake myself.

Ah, ok :biggrin:
Reply 895
Avatar for Mladenov
Mladenov
1
Let us not belittle complex analysis.:biggrin:

Solution 133

Let us consider f(z)=log⁡2zz2+α2\displaystyle f(z)= \frac{\log^{2}z}{z^{2}+\alpha^{2}}. This has simple poles at ±iα\pm i\alpha. Hence we have ∫γR,ϵlog⁡2zz2+α2dz=∫ϵRlog⁡2zz2+α2dz+∫CR+∫Cϵ−∫ϵR(log⁡z+2iπ)2z2+α2dz\displaystyle \int_{\gamma_{R,\epsilon}} \frac{\log^{2}z}{z^{2}+\alpha^{2}} dz= \int_{\epsilon}^{R} \frac{\log^{2}z}{z^{2}+\alpha^{2}}dz+\int_{C_{R}}+\int_{C_{ \epsilon}}-\int_{\epsilon}^{R} \frac{( \log z+2i\pi)^{2}}{z^{2}+ \alpha^{2}}dz.
We note that Res(f(z),iπ)=(log⁡α+iπ2)22iα\displaystyle Res(f(z),i\pi)= \frac{( \log \alpha+ i\frac{\pi}{2})^{2}}{2i\alpha}, and Res(f(z),−iπ)=(log⁡α+i3π2)2−2iα\displaystyle Res(f(z),-i\pi) = \frac{( \log \alpha + i\frac{3\pi}{2})^{2}}{-2i\alpha}.
Using triangle inequality, we obtain ∫CR→0\displaystyle \int_{C_{R}} \to 0 as R→∞R \to \infty, and ∫Cϵ→0\displaystyle \int_{C_{ \epsilon}} \to 0 as ϵ→0 \epsilon \to 0.
Then, separating the real and imaginary parts, we get −4iπ∫0∞log⁡xx2+α2dx=−2iππlog⁡αα\displaystyle -4i\pi \int_{0}^{\infty} \frac{ \log x}{x^{2}+\alpha^{2}}dx= -2i\pi \frac{\pi \log \alpha}{\alpha}, or equivalently ∫0∞log⁡xx2+α2dx=πlog⁡α2α\displaystyle \int_{0}^{\infty} \frac{ \log x}{x^{2}+\alpha^{2}}dx= \frac{\pi \log \alpha}{2\alpha}.


Original post by jack.hadamard
There is a good chunk of algebra and number theory before you actually get to algebraic number theory.
What books do you use now? It may be best if you complement them with some expository articles.


For number theory, I use Niven's An Introduction to the Theory of Numbers, Vinogradov's Elements of Number Theory, Borevich and Shafarevic's Number Theory, and I am to buy Lang's Algebraic Number Theory. Would you suggest anything else for algebraic number theory?

Algebra - Lang's Undergraduate Algebra and Lang's Algebra, Waerden's Algebra, and Kostrikin's An Introduction to Algebra. I am planning to buy Weibel's An Introduction to Homological Algebra, since it is essential when it comes to algebraic number theory, and more specifically - class field theory.

Would you advise some books, I should be grateful.
Problem 132
Note that:
∫0∞log⁡xx2+α2dx=πlog⁡α2α[br]⇒ddα[∫0∞log⁡xx2+α2dx]=ddα[πlog⁡α2α][br]⇒∫0∞log⁡x(−2α)(x2+α2)2dx=π2(1−lnαα2)\displaystyle \int_{0}^{\infty} \frac{ \log x}{x^{2}+\alpha^{2}}dx= \frac{\pi \log \alpha}{2\alpha}[br]\Rightarrow \dfrac{d}{d \alpha}[\displaystyle \int_{0}^{\infty} \frac{ \log x}{x^{2}+\alpha^{2}}dx]=\dfrac{d}{d \alpha}[ \frac{\pi \log \alpha}{2\alpha}][br]\Rightarrow \displaystyle \int_{0}^{\infty} \frac{ \log x(-2\alpha)}{(x^{2}+\alpha^{2})^2}dx=\dfrac{\pi}{2}(\dfrac{1-ln\alpha}{\alpha^2})
Letting alpha be 1:
∫0∞log⁡x(x2+1)2dx=−π4\displaystyle \int_{0}^{\infty} \frac{ \log x}{(x^{2}+1)^2}dx=-\dfrac{\pi}{4}
Original post by ben-smith
Note that: [...]


Precisely why I put the questions in the opposite order :tongue:
Original post by Lord of the Flies
Precisely why I put the questions in the opposite order :tongue:


I'm counting it :tongue:

Altenative to 134
x--->1/x gives
I=∫0∞(11+x2)(xy−xz(1+xy)(1+xz))dx→∫0∞(11+x2)(xz−xy(1+xy)(1+xz))dx[br]∴2I=0⇒I=0I=\displaystyle\int_{0}^{\infty} \left( \dfrac{1}{1+x^{2}} \right) \left(\dfrac{x^{y} - x^{z}}{(1+x^{y})(1+x^{z})} \right) dx \rightarrow \displaystyle\int_{0}^{\infty} \left( \dfrac{1}{1+x^{2}} \right) \left(\dfrac{x^{z} - x^{y}}{(1+x^{y})(1+x^{z})} \right) dx[br]\therefore 2I=0 \Rightarrow I=0
Reply 899
Avatar for shamika
shamika
16
Original post by Mladenov

Thanks, these are useful articles.
I asked for some references, since I am interested in reading algebra and number theory.


When you say algebraic number theory, do you mean things like number fields, calculating class numbers, cyclotomic fields and that kind of thing? My experience is that the hard part is learning the prerequisite algebra, the algebraic number theory part is very straightforward.

To that end, you need a good introduction to rings. I reckon the best way to get that (and algebraic number theory) is via online lecture notes. Search for some and see what comes up

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