=0 everywhere, wouldn't you believe that was constant?

You'd know this was constant, since the tangent to the curve of f is always parallel to the x axis - which only happens when f is constant. The above proposition says the same is true for complex functions; and by the definition of the complex derivative, it should appeal to intuition that it is 'obviously true', as any straight line in the complex plane behaves like the real line in terms of differentiation (i.e.

\mathbb{R}

is not really a special line for differentiability; complex derivatives must exist and be equal in all possible paths of approach).

Reply 2364

Hodor

Original post by FireGarden

..Really? If you have a real function

f:\mathbb{R} \mapsto \mathbb{R}

which was differentiable everywhere, and

f'

\mathbb{R}

is not really a special line for differentiability; complex derivatives must exist and be equal in all possible paths of approach).

In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...

Reply 2365

james22

Original post by Hodor

In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...

True, I think D should also be connected.

Reply 2366

Mladenov

Most probably

D

is the open 2-disk (and, if so, the proof is immediate).

Reply 2367

james22

Original post by Mladenov

Most probably

D

is the open 2-disk (and, if so, the proof is immediate).

What if it is something more unusual? I guess the proof for convex domains is just the same as in the real case, but what about concave domains? Thinking about it geometrically I think it is true but the proof is not as obvious.

Reply 2368

FireGarden

Original post by Hodor

In the real case, let A = (-1,0) u (1,2). Then I can think of a non-constant function f: A -> R with f' = 0...

Oh, sure! .. please excuse the evident jadedness from always being told D is a domain!

Anyway, some more 'different' maths. Group theory!

Problem 373***

Let

G

be a group, which acts on the set

X

.

Suppose

x,y \in X

have the same orbit. Prove that their stabilisers are conjugates (i.e.

\exists g\in G : G_y = gG_xg^{-1})

Problem 374***

How many ways can you colour the vertices of a square using at most three colours, which are distinct under the action of

D_8

? (i.e. colourings which cannot be obtained from another by rotations or reflections)

Reply 2369

MW24595

Original post by FireGarden

Anyway, some more 'different' maths. Group theory!

Problem 373***

Let

G

be a group, which acts on the set

X

.

Suppose

x,y \in X

have the same orbit. Prove that their stabilisers are conjugates (i.e.

\exists g\in G : G_y = gG_xg^{-1})

Finally, some Algebra.

Solution 373

Simple.

From the definition of an orbit, if

x, y

lie in the same orbit, then, there exists a

g \in G

that acts on

x

such that,

gx=y \Rightarrow x= g^{-1}y

.

Now, suppose,

a \in G_x, \Rightarrow ax= x [br][br]\Rightarrow gax = gx = y \Rightarrow ga(g^{-1}y)= y [br][br]\Rightarrow gag^{-1}y=y \Rightarrow gag^{-1} \in G_y

.

Therefore, we have,

g G_x g^{-1} \subset G_y

.

However, repeating the steps above using

g^{-1}y=x

, we have,

b \in G_y \Rightarrow g^{-1}bg \in G_x [br][br]\Rightarrow g^{-1}G_yg \subset G_x[br][br]\Rightarrow G_y \subset g G_x g^{-1}

.

And so, by symmetry, we can assert that,

G_y = g G_x g^{-1}

(edited 10 years ago)

Reply 2370

Indeterminate

Original post by Hodor

Original post by james22

Original post by Mladenov

Oops, as I had posted it from memory (came across it a long time ago), I made some mistakes in stating it. I'll consult the source and re-post it :smile:

Reply 2371

Mladenov

Original post by FireGarden

Problem 374***

How many ways can you colour the vertices of a square using at most three colours, which are distinct under the action of

D_8

? (i.e. colourings which cannot be obtained from another by rotations or reflections)

Solution 376

Let

S

be the set of colorings of the vertices.
Firstly, observe that two colorings are essentially distinct (i.e. distinct under the action of the dihedral 8 group) if they belong to different orbits of

D_{8}

S

.
Hence, we shall be done if we manage to find the number of orbits of

D_{8}

in the set of colorings.
The following result is easily proved (counting in two ways, Lang's Algebra, ect.)

Proposition:

Let

G

be a finite group acting on a finite set

X

, and

\alpha(g)

is the number of elements

x \in X

such that

\pi_{g}(x) = x

. Then the number of orbits of

G

X

\displaystyle \frac{1}{card (G)}\sum_{g \in G} \alpha(g)

.

Since the representative does not matter, we have

\displaystyle \sum_{t \in Gs} \frac{1}{card(Gt)} =1

and hence the number of orbits of

G

X

\displaystyle \sum_{x \in X} \frac{1}{card(Gx)}

.
Fix

x \in X

. Notice that there is a bijection

G/G_{x} \sim Gx

, call this

(1)

. Also, the number of

g \in G

such that

\pi_{g}(x)=x

is the order of

G_{x}

.

We count the number of solutions

(x,g)

to the equation

\pi_{g}(x)=x

in two ways.

One the one hand, there are

\alpha(g)

elements

x \in X

such that

\pi_{g}(x)=x

. On the other hand, each

g \in G

is counted

\alpha(g)

times by the sum

\displaystyle \sum_{x \in X} card(G_{x})

since there are

\alpha(g)

x

's in

X

such that

\pi_{g}(x)=x

. Thus

\displaystyle \sum_{x \in X} card(G_{x}) = \sum_{g \in G} \alpha(g)

; call this

(2)

.
Hence, by

(1)

and

(2)

, we obtain the result.

We can interpret the elements of

D_{8}

as products of cycles.

Second Proposition (too general):

Let

\pi

be a product of

m

cycles (counting cycles of length

1

). Then the number of colorings fixed by

\pi

n^{m}

, where

n

is the number of available colors.

Although, we do not need it, I write it in case someone is interested.

For our problem, it is sufficient to note that the vertices in a given cycle have the same color.
Thus, if

n

colors are available, there are

\frac{1}{8}(n^{4}+n+n^{2}+n^{2}+n+n^{3}+n^{3}+n^{2})

essentially distinct colorings, since the elements are

e

[1,2,3,4]

[1,2][3,4]

[1,3][2,4]

[1,4,3,2]

[1][3][2,4]

[1,3][2][4]

[1,4][2,3]

Spoiler

As a side note, we have a problem with the numeration of the problems. It starts with Indeterminate's problem 372.

Basic commutative algebra.

Problem 377***

Prove that if

M

is an

n \times n

matrix over a local ring, with coefficients in the maximal ideal

J

, then

J+M

is invertible.

Problem 378***

If

f : A \to A

is a surjective endomorphism of a Noetherian ring

A

, then

f

is an isomorphism.

(edited 10 years ago)

Reply 2372

SalBhatti

Original post by Lord of the Flies

Solution 10

\displaystyle x\to \frac{\pi}{2}-x\Rightarrow \int_0^{\frac{\pi}{2}} \frac{\sin^{17} x\,dx}{\sin^{17} x+\cos^{17} x}= \int_0^{\frac{\pi}{2}} \frac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

Hence

\displaystyle I=\frac{1}{2}\int_0^{\frac{\pi}{2}} dx=\frac{\pi}{4}

Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is

\dfrac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

, but how does that become

\frac{1}{2}\int dx

(edited 10 years ago)

Reply 2373

james22

Original post by Zishi

Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is

\dfrac{\cos^{17} x\,dx}{\sin^{17} x+\cos^{17} x}

, but how does that become

\frac{1}{2}\int dx

You have 2 experssions for I, one has sine terms on the top and the other cas cos terms. By adding them you get (sin^17+cos^17)/(sin^17+cos^17)=1 so 2I=1

Reply 2374

SalBhatti

Original post by james22

You have 2 experssions for I, one has sine terms on the top and the other cas cos terms. By adding them you get (sin^17+cos^17)/(sin^17+cos^17)=1 so 2I=1

Ahh, great! :grin:

Many thanks.

Reply 2375

henpen

Problem 379 **

How many subsets of

\{1,2,...,50\}

are there such that the sum of the elements exceeds

637

?

Is this possible to generalise?

Reply 2376

The nameless one

Original post by henpen

Problem 379 **

How many subsets of

\{1,2,...,50\}

are there such that the sum of the elements exceeds

637

?

Is this possible to generalise?

As for each set, there is a reverse set. As 637 is slightly less than half of the sum of numbers from 1 to 50(637.5), for each pair of sets, one of them is going to satisfy the condition. So it is half of all sets, namely 2^49, 562949953421312.