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# The Proof is Trivial! watch

1. (Original post by Indeterminate)
Problem 372***

Prove the following:

If

and if f is holomorophic and

then

What is D?
2. (Original post by Hodor)
What is D?
I assume it's any domain in the complex numbers on which f can de differentiable.
3. (Original post by james22)
I assume it's any domain in the complex numbers on which f can de differentiable.
Then I don't think it's true.
4. (Original post by Hodor)
Then I don't think it's true.
..Really? If you have a real function which was differentiable everywhere, and =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. is not really a special line for differentiability; complex derivatives must exist and be equal in all possible paths of approach).
5. (Original post by FireGarden)
..Really? If you have a real function which was differentiable everywhere, and =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. 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...
6. (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.
7. Most probably is the open 2-disk (and, if so, the proof is immediate).
Most probably 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.
9. (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 be a group, which acts on the set .

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

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 ? (i.e. colourings which cannot be obtained from another by rotations or reflections)
10. (Original post by FireGarden)
Anyway, some more 'different' maths. Group theory!

Problem 373***

Let be a group, which acts on the set .

Suppose have the same orbit. Prove that their stabilisers are conjugates (i.e.
Finally, some Algebra.

Solution 373

Simple.

From the definition of an orbit, if lie in the same orbit, then, there exists a that acts on such that, .

Now, suppose,

.

Therefore, we have, .

However, repeating the steps above using , we have,

.

And so, by symmetry, we can assert that, .
11. (Original post by Hodor)
X
(Original post by james22)
X
X

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
12. (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 ? (i.e. colourings which cannot be obtained from another by rotations or reflections)
Solution 376

Let 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 in .
Hence, we shall be done if we manage to find the number of orbits of in the set of colorings.
The following result is easily proved (counting in two ways, Lang's Algebra, ect.)

Proposition:

Let be a finite group acting on a finite set , and is the number of elements such that . Then the number of orbits of in is .

Since the representative does not matter, we have and hence the number of orbits of in is .
Fix . Notice that there is a bijection , call this . Also, the number of such that is the order of .

We count the number of solutions to the equation in two ways.

One the one hand, there are elements such that . On the other hand, each is counted times by the sum since there are 's in such that . Thus ; call this .
Hence, by and , we obtain the result.

We can interpret the elements of as products of cycles.

Second Proposition (too general):

Let be a product of cycles (counting cycles of length ). Then the number of colorings fixed by is , where 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 colors are available, there are essentially distinct colorings, since the elements are , , , , , , , .
Spoiler:
Show
It is left as an exercise to calculate the answer with .

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 is an matrix over a local ring, with coefficients in the maximal ideal , then is invertible.

Problem 378***

If is a surjective endomorphism of a Noetherian ring , then is an isomorphism.
13. (Original post by Lord of the Flies)
Solution 10

Hence
Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is , but how does that become ?
14. (Original post by Zishi)
Sorry, probably my maths skills have gotten rusty - I can't understand the part after "hence...". I is , but how does that become ?
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
15. (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! Many thanks.
16. Problem 379 **

How many subsets of are there such that the sum of the elements exceeds ?

Is this possible to generalise?
17. (Original post by henpen)
Problem 379 **

How many subsets of are there such that the sum of the elements exceeds ?

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.
18. Problem 380*

Would this be considered as trivial?
19. (Original post by henpen)
Problem 379 **

How many subsets of are there such that the sum of the elements exceeds ?

Is this possible to generalise?
It can be generalized. We have to consider two cases - .

(Original post by Arieisit)
Problem 380*

Would this be considered as trivial?
What do you think? Using and , we are done.
It can be generalized. We have to consider two cases - .

What do you think? Using and , we are done.
I only just noticed the problem and thought "oh, I can finally solve a problem on here". Too late though
Spoiler:
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