# zetamcfc practices LaTeX Watch

1. Well seen as I'm horrible at LaTeX I thought I could practice by writing out some random things that interested me. Will kick off with some naive set theory.

Consider a function , mapping a domain to a co-domain

Now consider a family of sets (Where means 'such that' )

Then the image

Proof: Let

If for some and for some

for some

However,

Proof: Let

If for some

'But why only the implication on line 2?'

Well, we shall consider a family of sets who's intersection is the empty set, (That is they are disjoint)

Whereas the intersections of their images is a non-empty set

Then, Thus we can only show and not equality.

A nice result I think. Probably missing a step or an assumption somewhere somewhere

Next a simple inequality on the Bell numbers where in the number of partitions of a set with elements.

Claim:

Proof: Let be a set with elements, then the power set of contains elements that is .

If we then take, the power set of the power set of , .

We see that, the power set of the power set of contains elements.

But not every element of, the power set of the power set of is a partition of

Thus

Hopefully I can try some new things each day. (Typo's will probably be rife)
2. keep going ...looking good!
3. I got a headache thanks
4. Well for today I thought I would do something a bit more fundamental , here it goes.

Theorem: If is a non-constant polynomial over , then there exists such that . (FToA)

Proof: So we shall suppose that there exists no that satisfies therefore for some the set

is non-empty, the map defined by is continuous, so is compact.

But for a subset on this is the same as being closed and bounded. It then follows that attains its minimum on and by the definition of this is also its minimum value on

Assume the minimum is obtained at . Then

, and by assumption .

Consider as runs around a small circle centred at , from this we will derive a contradiction.

Let . Expand in powers of to get

where

Here are specific complex numbers, (will leave this as a 'challenge' )

Clearly , and we are assuming this is non-zero, so . If then is constant, contrary to hypothesis. So some . Let be the smallest integer from which . In let for small . Then

Therefore

Let for . Since and we have . Setting we see that . Now

Set , so that . Then , and

But , so for suffciently small we have

contradicting the definition of . Therefore such that
5. ,we now sub ,

to get,

,removing constants we obtain,

we now consider the case that s is complex, which therefore gives us an equation for its conjugate, , that being,

,

we then add/sub the two equations together,

,

then

,we then define in polar and Cartesian form, so, ,applying these we get,

,

these reduce to, and ,

only the second of these we shall use.

That being the second one. we now set thus meaning ,

we must also use the Euler reflection formula, ,using what we know, now we get a much more simple formula,

,

now we take the limit as ,so we have ,

the right hand side having had some added, but all mathematically consistent.

Evaluating the RHS we get

as we know ,from this we have,

6. Claim: is transcendental.

Proof: Assume is not transcendental.

Then

Without the loss of generality we can assume and

Define

Where p is an arbitrary prime.

Then is a polynomial in .

Putand note that

Hence for any

Multiply by and sum over to get

from the equation supposedly satisfied by by

We claim that for each is an integer, and that this integer is divisible by unless and . To establish the claim we just apply Liebniz's rule; the only non-zero term arising when come from the factor being differentiated exactly p times. Since , all such terms are integers divisible by .

In the exceptional case , the frist non-zero term occurs when , and thenThe next non-zero terms are all multiples of .

The value of equation is thereforefor some . If , then the integer is not divisible by . So fot sufficiently large primes the value of the equation is an integer not divisible by , hence not zero.

Now estimating the integral. If thenso

which tends to zero as , thus we have a contradiction .

Therefore is transcendental which implies it is also irrational.
7. (Original post by zetamcfc)
Is that the Gamma function I see my personal favourite!
8. (Original post by zetamcfc)
Lovely.

Personally, I'd prefer over though.
9. Not done a post in a few days, here goes.

Claim: The cardinality of is the cardinality of the continuum.

Proof: We have, defines a bijection

So,

But and

Where means has the same cardinality as

Some thing to think of is, what is the cardinality of and how would one show that ?
10. Not done this in a while but always good to keep practicing

Find

First consider Where is a complex valued function. And is the semicircle traced in the anti-clockwise direction. (Probably should use R then take the limit to infinity, but hey ho)

We can see that the integrand has 3 simple poles at

Then we know by the residue theorem that:

To work out the Residues we will prove the following,

Lemma: Let , where g and h are both holomorphic in a neighborhood of c, and where . Then,

Proof:

So we can continue the solution, to find that,

Thus by the Residue theorem:

Therefore:

11. Lovely! Remember that Latex has a built in imaginary part function: (it uses Frankfurt characters, weird...) and you can always write to make it not look like the variables R, e and s bunched together.

But looks really good! (and really scary, is that complex analysis? )
12. (Original post by Zacken)
Lovely! Remember that Latex has a built in imaginary part function: (it uses Frankfurt characters, weird...) and you can always write to make it not look like the variables R, e and s bunched together.

But looks really good! (and really scary, is that complex analysis? )
Yup, but not too challenging. Ty for the ideas.
13. Claim: when n>1. (I posed the question on TPIT but I don't think anyone answered it.)

I'm interested to see if there is a Real Analysis proof for this question if anyone could give such a proof i'd be very happy

Sketch proof (Not that rigorous, but saves time ):

Consider, for n>1

So we have

shall be made up of 3 parts the line extending length along the real line, will be the sector-arc from to , the line from to the origin. From the way I have described we will obviously be going anti-clockwise around.

For this there will only be one pole inside, which we will call .

Now as our situation is clearer we have an equation collecting everything together,

First we see what is,

Now we can try to see what the integrals are.

First we must note that on as .

We know on is the expression we want, so we will work on, on

where we write because of the direction of integration.

So the integral now becomes,

Combining our results to the limit we have earlier, we get,

As r is a dummy variable we get the equation,

And we are done .

Zacken I'm sure you can find a Real Analysis proof for this.
14. What the actual f*** I don't understand any of this...
15. (Original post by zetamcfc)
Zacken I'm sure you can find a Real Analysis proof for this.
Dunno how real-analysis-y this is, but the first time you posted this, I did something akin to:

Use the substitution then

which I think you can then mangle into required form with some suitable Gama tricks; Euler's reflection formula should come in handy.
16. (Original post by Zacken)
Dunno how real-analysis-y this is, but the first time you posted this, I did something akin to:

Use the substitution then

which I think you can then mangle into required form with some suitable Gama tricks; Euler's reflection formula should come in handy.
nice substitution. (Remember to change limits for your 2nd integral )

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