The Proof is Trivial!

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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:

f(x) = x^5 - 7x^3 + 2x^2 -30x + 6

If you find it easy, please wait for somebody else to post an answer!

x^5 - 7x^3 + 2x^2 -30x + 6=(x^2+3) (x^3-10 x+2)
a,b,c be the real roots:
(a^2+b^2+c^2)=(a+b+c)^2-2(ab+ac+bc)
Vieta tells us:
=0^2-2(-10)=20

Reply 901

metaltron

Original post by ben-smith

x^5 - 7x^3 + 2x^2 -30x + 6=(x^2+3) (x^3-10 x+2)
a,b,c be the real roots:
(a^2+b^2+c^2)=(a+b+c)^2-2(ab+ac+bc)
Vieta tells us:
=0^2-2(-10)=20

Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?

Reply 902

Mladenov

Original post by shamika

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

Number fields - that is it. Broadly speaking, my aims are understanding of the techniques of Galois cohomology, and their applications in global class field theory; understanding the axioms of class formation; and, to some extent, the Brauer group.
In addition, Riemann - Roch theory, more specifically Grothendieck groups and his theorem.

I have been told that Commutative Ring Theory by Matsumura is more than sufficient. What is your view?

Original post by metaltron

Well done, the trick was in factorising the polynomial, though you didn't show how you did it in the solution. Also, how can you guarantee that all the roots of the cubic are real?

Let

f(x) = x^{3}-10x+2

. Note that

f

is continuous in

\mathbb{R}

, and

f(-4)<0

f(0)>0

f(2)<0

, and

f(4)>0

Problem 136*

Let

f: \mathbb{R} \to \mathbb{R}

be a strictly increasing invertible function such that for all

x \in \mathbb{R}

we have

f(x)+f^{-1}(x)=e^{x}-1

. Prove that

f

has at most one fixed point.

Problem 137**

Find all continuous functions

f: \mathbb{R} \to \mathbb{R}

which satisfy

f(f(x))-2f(x)+x=0

for all

x \in \mathbb{R}

.

Problem 138***

Let

P(X)

be an irreducible polynomial over

\mathbb{Z}[X]

. Show that

P(XY)

is irreducible over

\mathbb{Z}[X,Y]

Reply 903

jack.hadamard

Original post by Mladenov

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.

You have some serious literature (which is good). I have used Niven's book too; it is very good. I cannot say much about Vinogradov's and Borevich and Shafarevich's books. My choices were Ireland and Rosen's book, for a more algebraic approach, and Neukirch's book (which you can possibly combine with Murty's book). I have copies of Lang's Algebra and his Undergraduate Algebra. I also try to use books that do not overlap (much) in content. I am not sure you will be needing Homological Algebra in the near future.

Take a look here (taken from) and here (taken from) for structural information. This can be useful too. As Shamika says, you will need good foundation in rings, modules and Galois theory. Lecture notes are available here and here (EDIT: and here).

I can't think of anything more right now.

EDIT: Most of these books cost a fortune. It may be best if you borrow them from a (university) library. For example, all of these are available in Cam's library.

(edited 10 years ago)

Reply 904

Mladenov

Original post by jack.hadamard

I am grateful to you; this is a considerable amount of resources.

Vinogradov's book treats analytic number theory (hence not quite relevant). It is not as deep as Apostol's book, yet there are lots of good exercises.
Borevich and Shafarevich's book is one of the best. It is a shame that this book is currently out of print. I have the Russian edition. :biggrin:

I consider Lang's Algebra a bit demanding; also he never explains his motivation.

By the way, Murty's book will be very beneficial.

Unfortunately, I will have to buy these books, as I do not have permission to borrow books from university libraries.

Reply 905

Lord of the Flies

This has been a terrible day for me, maths-wise. I shall leave 136 for someone else, since people have been asking for (*) q's.

Solution 137

f(x)=x+h(x)\Rightarrow h(x+h(x))=h(x)\Rightarrow h(x+nh(x))=h(x)\Rightarrow h=\text{cnst.}^{(*)}

^{(*)}

h

were non-constant there is an interval

[x_0,x_1]

for which

h(x_0)<h(x_1)

and

h\neq h(x_0),h(x_1)

for

x\in (x_0,x_1)

. Define

y_n=x_0+nh(x_0)

and

z_n=x_1+nh(x_1)

. Then we have

h\neq h(x_0),h(x_1)

for

x\in (y_n,z_n)

. But since

z_n-y_n\to\infty

there will exist

i

such that

y_{i+1}\in (y_i,z_i)\Rightarrow h(y_{i+1})\in (h(x_0),h(x_1))\Rightarrow \Leftarrow

(edited 10 years ago)

Reply 906

jack.hadamard

Original post by Mladenov

I consider Lang's Algebra a bit demanding; also he never explains his motivation.

You are welcome. This blog also provides a useful guide: the missing article by Sam Ruth can be found here (he has removed it for some reason). Lang's Algebra becomes useful only after you have learnt most of the basic and intermediate stuff and have gained enough intuition/maturity; it is often used as a reference and not a book to learn from. Are you going to university this year? If this is the case, then you could wait a couple of months (for some of the books).

Problem 139 / **

Given an integer-valued polynomial, can you always find some values such that their sum is divisible by

n \in \mathbb{N}

Reply 907

Felix Felicis

Original post by ben-smith

Problem 132

Is this Feynman integration?

Reply 908

Mladenov

Solution 139

Fix

n

. We work in

\mathbb{Z}/n\mathbb{Z}

.

Let

P(m)

be an integer-valued polynomial. Clearly, there are infinitely many

m_{i} \in \mathbb{Z}

i\in \mathbb{Z^{+}}

such that

P(m)

is constant over the set

\{ m_{i} | i \in \mathbb{Z^{+}} \}

. Hence, we can always choose

n

elements from the set

\{ m_{i} | i \in \mathbb{Z^{+}} \}

, and

\displaystyle \sum_{i=1}^{n} P(m_{i}) = 0

Original post by jack.hadamard

I had found Lang's Undergraduate Algebra quite easy to comprehend, and thus I decided to buy his Algebra.

Yep, I am finishing high school this year.

Problem 140*

Evaluate

\displaystyle \sum_{v=1}^{\infty} \frac{1}{2^{v}v^{2}}

.

Problem 141**

Let

f \in C^{\infty}

for

x>0

, and

0 \le (-1)^{n}f^{(n)}(x) \le e^{-x}

for all

x >0

and

n \in \mathbb{Z^{+}} \cup \{0\}

. Find

f

.

Remark:

f^{(n)}(x)

is the

n

th derivative of

f

Reply 909

Felix Felicis

Solution 140

Right, after the recent discussion on the STEP solutions thread about integral notation, I need to be careful here not to mix up bound and free variables.

Lemma (is this the right time to use a lemma? :colondollar:

)

\displaystyle\sum_{r=1}^{\infty} \dfrac{1}{r^{2}} = \dfrac{\pi^{2}}{6}

(the Basel problem for those who haven't seen it)

\displaystyle\sum_{v=1}^{\infty} \dfrac{1}{2^{v} v^{2}}

Consider

\displaystyle\sum_{v=1}^{\infty} x^{v-1}

for

|x| < 1 = \dfrac{1}{1-x}

\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}}{v} = \displaystyle\int_{0}^{x} \dfrac{1}{1-t} dt = - \ln |1-x| (*) \Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v-1}}{v} = -\dfrac{\ln |1-x|}{x}

\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}}{v^{2}} = - \displaystyle\int_{0}^{x} \dfrac{\ln |1-t|}{t} dt

The sum in question,

\displaystyle\sum_{v=1}^{\infty} \dfrac{1}{2^{v} v^{2}}

(call it

\mathcal{S}

) is for

x=\frac{1}{2}

, thus:

\mathcal{S} = - \displaystyle\int_{0}^{1/2} \dfrac{\ln |1-x|}{x} dx

. Applying

1-x \to x

we get:

- \displaystyle\int_{0}^{1/2} \dfrac{\ln |1-x|}{x} dx = - \displaystyle\int_{1/2}^{1} \dfrac{ \ln x}{1-x} dx

- \displaystyle\int_{1/2}^{1} \dfrac{ \ln x}{1-x} dx = - \displaystyle\sum_{r=0}^{\infty} \underbrace{\displaystyle\int_{1/2}^{1} x^{r} \ln x dx}_{\mathcal{I}}

By parts,

\mathcal{I} = \left[ \dfrac{x^{r+1} \ln x}{r+1} - \dfrac{x^{r+1}}{(r+1)^{2}} \right]_{1/2}^{1} = \left[ \dfrac{x^{r+1} \ln x (r+1) - x^{r+1}}{(r+1)^{2}} \right]_{1/2}^{1}

= \dfrac{-1}{(r+1)^{2}} + \dfrac{\ln 2}{2^{r+1} (r+1)} + \dfrac{1}{2^{r+1} (r+1)^{2}}

\Rightarrow \mathcal{S} = -\displaystyle\sum_{r=0}^{\infty} \left( \underbrace{\dfrac{-1}{(r+1)^{2}}}_{-\frac{\pi^{2}}{6} \text{by Lemma}} + \underbrace{\dfrac{\ln 2}{2^{r+1} (r+1)}}_{\mathcal{S}_{2}} + \underbrace{\dfrac{1}{2^{r+1} (r+1)^{2}}}_{\mathcal{S}} \right)

\mathcal{S}_{2} = \ln2 \cdot \displaystyle\sum_{r=0}^{\infty} \dfrac{1}{2^{r+1} (r+1)} = \ln 2 \cdot -\ln \left(1 - \frac{1}{2} \right)

(by

(*)

applying

x = \frac{1}{2}

)

= \ln^{2} 2

\mathcal{S} = -\left( -\dfrac{\pi^{2}}{6} + \ln^{2} 2 + \mathcal{S} \right) \Rightarrow \mathcal{S} = \dfrac{\pi^{2}}{12} - \dfrac{\ln^{2} 2}{2}

Reply 910

jack.hadamard

Original post by Mladenov

I had found Lang's Undergraduate Algebra quite easy to comprehend, and thus I decided to buy his Algebra.

In the foreword, he says that his Undergraduate Algebra and Linear Algebra provide more than enough background for a graduate course, but does not say that they are sufficient for reading his Algebra book. :tongue:

Nevertheless, many of the examples come from different areas of the undergraduate curriculum; to make full use of this book you need an undergraduate degree.

Let's now make the problem more concrete.

Extension to Problem 139 / **

i) Let

a_1, a_2, ..., a_n \in \mathbb{Z}

. Prove that there exists

I \subset \{1, 2, ..., n\}

such that

\displaystyle \sum_{i \in I} a_i \equiv 0\ (n)

.

ii) How much more difficult does it become (and how many integers do we need to start with) in order to guarantee that there will be a subset,

I

, of cardinality exactly

n

having the same property?

Reply 911

Mladenov

Solution 139 (2)

Part i)
We consider the numbers

0,a_{1}, a_{1}+a_{2}, \cdots, a_{1}+a_{2}+\cdots+a_{n}

. These are

n+1

. Hence at least two of them are congruent modulo

n

. Therefore, there exists

I

such that

\displaystyle \sum_{i \in I} a_{i} \equiv 0 \pmod n

.
Note that

I=\{1,2,\cdots,n\}

is also possible. For example, when all

a_{i}

are

\equiv 1 \pmod n

.
The following is also true:
Let

n

be an arbitrary positive integer. Suppose

a_{1},\cdots, a_{k}

are given integers, which give at least

n-k+1

distinct remainders

\pmod n

. Then there is

I \subseteq \{1,2,\cdots, k\}

such that

\displaystyle \sum_{i\in I} a_{i} \equiv 0 \pmod n

.

Part ii)

It is quite well-known that if

G

is an additively written abelian group,

card(G)=n

, for some positive integer

n

, then, if

a_{i} \in G

, for

i \in \{1,2, \cdots, 2n-1\}

, we can represent

e

- the unit element of

G

, as a sum of the elements of

S

, where

S \subset \{a_{1},a_{2}, \cdots, a_{2n-1}\}

, and

card(S) = n

.
Hence, if we commence with

2n-1

integers, there always will be

I \subset \{1,2, \cdots , 2n-1\}

card(I)=n

, such that

\displaystyle \sum_{i \in I} a_{i} \equiv 0 \pmod n

.
Another way to prove this is the following:
First: it is true when

n

is a prime number.
Second: If the result is true for

n=a,b

, where

a,b \in \mathbb{Z^{+}}

, then it is true for

n=ab

Reply 912

jack.hadamard

Original post by Mladenov

Part ii)

It is quite well-known that...

You have got quite a few rabbits in your hat. :tongue:

I intended to ask for a general comment. For the people who don't know the result, this is known as Erdős–Ginzburg–Ziv theorem (EGZ). The proof you referred to is a beautiful application of the Chevalley-Warning theorem.

In fact, John Olson generalised EGZ to any finite group (abelian or not).

Reply 913

Mladenov

Original post by jack.hadamard

The proof you referred to is a beautiful application of the Chevalley-Warning theorem.

Precisely.

Original post by jack.hadamard

In fact, John Olson generalised EGZ to any finite group (abelian or not).

Apropos, I knew not that the result is still true in the non-abelian case.

Here are some * questions.

Problem 142*

Evaluate

\displaystyle \sum_{v=1}^{\infty} \frac{x^{v} \sin v\alpha}{v}

, (

|x| <1

).
Hence, find

\displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v}

, (

0<x< 2\pi

). Can you evaluate

\displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v^{3}}

, (

0 \le x \le 2\pi

)?

Problem 143*

Prove that

\displaystyle \frac{1}{n}\sum_{v=1}^{n} \left(1+ \frac{1}{2v} \right)^{2v} \le \left(1+ \frac{1}{n+1} \right)^{n+1}

.

Problem 144*

There are several castles in one country and three roads lead from every castle. A knight leaves his castle. Traveling around the country he leaves every new castle via the road that is either to the right or to the left of the one by which he arrived. According to The Rule the knight never takes the same direction (right or left) twice in a row. Prove that some day he will return to his own castle.

Reply 914

MW24595

Original post by Mladenov

Problem 142*

Evaluate

\displaystyle \sum_{v=1}^{\infty} \frac{x^{v} \sin v\alpha}{v}

, (

|x| <1

).
Hence, find

\displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v}

, (

0<x< 2\pi

). Can you evaluate

\displaystyle \sum_{v=1}^{\infty} \frac{\sin vx}{v^{3}}

, (

0 \le x \le 2\pi

Solution 142 (so far)

Consider that, summing the geometric sequence, we have:

\displaystyle \sum_{v=1}^\infty x^ {v-1} e^{iva} = \frac {e^{ia}}{1- x e^{ia}}[br][br]\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} = \displaystyle\int_{0}^{x} \frac{e^{ia}}{1-ze^{ia}} dz[br][br]\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} = e^{ia} \left[ \frac {-1}{e^{ia}} ln (1- z e^{ia}) \right]_0^x[br][br]\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} = - ln (1- x e^{ia})

Now, suppose we have an imaginary number z, such that,

z= x+ iy= r e^{it}

(where t is the argument of z.}

[br][br]\Rightarrow ln (x+iy) = ln (r) + it[br][br]\Rightarrow Im \left[ ln (x+iy) \right] = t= arctan (\frac{y}{x})

Now, consider that,

[br][br]1- xe^{ia} = (1- x \cos a) - i \sin a[br][br]\Rightarrow Im \left[\displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v}e^{iva}}{v} \right] = - Im \left[ln (1- x e^{ia}) \right][br][br]\Rightarrow \displaystyle\sum_{v=1}^{\infty} \dfrac{x^{v} \sin {av} }{v} = - arctan (\frac{-x \sin a}{1- x \cos a}) = arctan (\frac{x \sin a}{1- x \cos a})[br][br]

Now, substituting 1 in place of x in this formula and x instead of a, we get,

[br][br]\displaystyle\sum_{v=1}^{\infty} \dfrac{ \sin {vx} }{v} = arctan ( \frac { \sin x}{1- \cos x })= arctan( \cot { \frac {x}{2}})

Let, t= pi/2 - x/2

[br][br]\Rightarrow arctan ( cot \frac {x}{2}) = arctan ( \tan t) = t = \frac { \pi - x}{2}[br][br]

(edited 10 years ago)

Reply 915

Mladenov

Original post by MW24595

Solution 142 (so far)

Well. There is only one point - you can't simply substitute

x=1

, as the radius of convergence is

|x| <1

. I mean, you need to justify it.

Reply 916

Lord of the Flies

Ach, doing my best to ignore this thread but I love series and integrals too much. Here are alternative solutions to 140 and 142:

Solution 140

Differentiation under the integral sign!

\displaystyle \sum_{v\geq 1}x^{v-1}=\frac{1}{1-x}\Rightarrow \sum_{v\geq 1}\frac{1}{2^vv^2}=-\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx

Unparseable latex formula:

\displaystyle \begin{aligned}\int_0^{\frac{1}{2}} \frac{\ln (1-x)}{x}\,dx\overset{x\to \frac{x}{2}}=\int_0^1 \frac{\ln (1-\frac{x}{2})}{x}\,dx \overset{ \text{IBP}}=\int_0^1 \frac{\ln x}{2-x}\,dt \overset{x\to 1-x}=\int_0^1 \frac{\ln (1-x)}{1+x}\,dx

\displaystyle \begin{aligned}f(t)=\int_0^1 \frac{\ln (1-tx)}{1+x}\,dx\Rightarrow f'(t) &= \int_0^1 \frac{x\,dx}{(tx-1)(1+x)} \\&= \int_0^1 \frac{dx}{tx-1}+\int_0^1 \frac{dx}{(t+1)(1+x)}-\int_0^1 \frac{t\,dx}{(tx-1)(t+1)}\\& =\frac{\ln (1-t)}{t}+\frac{\ln 2}{t+1} -\frac{\ln (1-t)}{1+t}\end{aligned}

Everything falls into place now:

\displaystyle f(1) =\int_0^1 \frac{\ln (1-t)}{1+t}dt=\int_0^1 \frac{\ln (1-t)}{t}\,dt+\int_0^1\frac{\ln 2}{t+1}\,dt-\int_0^1 \frac{\ln (1-t)}{1+t}\,dt

\displaystyle -f(1) =\frac{1}{2}\left(-\int_0^1 \frac{\ln (1-t)}{t}\,dt-\int_0^1\frac{\ln 2}{t+1}\right)\,dt=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}

Solution 142

We can evaluate the sum directly:

\displaystyle\begin{aligned} g_n = \sum_{v=1}^n \frac{\sin vx}{v}=\sum_{v=1}^n \int_0^x \cos vt \,dt =\frac{1}{2}\int_0^x \csc \frac{t}{2}\sin \left(\frac{t}{2}(2n+1) \right)-1\,dt \end{aligned}

The trick is now to use that

\csc \frac{t}{2}\sim \frac{2}{t}

near

0

to get rid of a bunch of stuff:

\displaystyle\begin{aligned} g_n = \underbrace{\int_0^x \frac{1}{t}\sin \left(\frac{t}{2}(2n+1) \right) dt}_{\alpha_n}+\frac{1}{2} \underbrace{\int_0^x \left(\csc \frac{t}{2}-\frac{2}{t}\right)\sin \left(\frac{t}{2}(2n+1) \right)dt}_{\beta_n}-\frac{x}{2}\end{aligned}

n\to\infty:

\alpha_n= \displaystyle\int_0^x \frac{1}{t}\sin \left(\frac{t}{2}(2n+1) \right) dt\overset{\frac{t}{2}(2n+1)\to t}=\int_0^{\frac{x}{2}(2n+1)} \frac{\sin t}{t}\,dt\to \frac{\pi}{2}

Unparseable latex formula:

\begin{aligned} \beta_n = \displaystyle \left[\frac{\cos \left(\frac{t}{2}(2n+1)\right) \left(\csc\frac{t}{2}-\frac{2}{t}\right)}{2n+1}\right]_0^{x} +\frac{1}{2n+1}\int_0^x\cos \left( \frac{t}{2}(2n+1)\right)\left( \cot \frac{t}{2} \csc\frac{t}{2}-\frac{4}{t^2}\right)\,dt\to 0

Hence

\displaystyle g_n\to \sum_{v\geq 1}\frac{\sin vx}{v}=\frac{\pi-x}{2}

Furthermore, observe that:

Unparseable latex formula:

\displaystyle\begin{aligned} \int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \sum_{v\geq 1}\frac{\sin vt_{2k}}{v}\,dt_{2k}\cdots dt_1=\frac{1}{2}\int_0^x\int_0^{t_1}\cdots \int_0^{t_{2k-1}} \pi -t_{2k}\, dt_{2k}\cdots\, dt_1

gives:

Unparseable latex formula:

\displaystyle\begin{aligned} (-1)^k\sum_{v\geq 1} \frac{\sin vx}{v^{2k+1}}+\sum_{i=1}^{k} \left( \frac{(-1)^{k+i+1}x^{2(k-i)+1}}{(2(k-i)+1)!}\sum_{v\geq 1}\frac{1}{v^{2i}} \right)= \frac{x^{2k}}{2}\left(\frac{\pi}{(2k)!}-\frac{x}{(2k+1)!}\right)