Function

There has to be a special relationship between the two amplitudes, no?

Not just 'smaller'

Hi everyone! I'm back with a question;

I missed a few posts, so it's probable that the answer to my question is already somewhere is this maze:

Imagine a hollow body of perfect revolution (let's say, for instance, a paraboloid as a result of a revolution of a parabola of the form [math]y=x^2[/math]).

Imagine the body produces a sound wave as a consequence of it being tapped.

Let the average sound waves outside the object follow the equation

[math]y=A\cdot \sin{\left(\frac{2\pi}{T}t\pm \frac{2\pi}{\lambda}x\right)}[/math]

What will the equation of the average sound waves inside the object look like? Or what would just the amplitude be?

Hello everyone

 

In a dissertation for maths, I've mentioned once "quadratrix"... Can't really explain what it is, but I think I know what it is, and I've discovered what seems to be a pretty beautiful property of the quadratrix of a circle (commonly known as the quadratrix of Dinostratus)

 

So I thought: what if I 'completed' the quadratrix? (i.e. completing the circle and the quadratrix, resulting in a horizontal flip of the quadratrix over x=0, a vertical one over y=0, and a horizontal one of the vertically flipped one over x=0)

The result: an eye-shaped geometrical figure.

 

If you're eager to see this 'completed' quadratrix, you can ask me and I will try to upload it somewhere (.ggb, to be opened with GeoGebra).

In this file, I've also drawed what I like to call 'sub-quadratrices' (the blue lines). (There are also lines which I'd like to call 'super-quadratrices', those are the lines outside of the quadratrix)

My conclusion of doing this: if the number of equal parts (now there are 10, for more information, seek information on the quadratrix of Dinostratus) are going to infinity, you become this eye-shaped figure; the less equal parts you have, the more you will get like 2 strange flower petals above and under y=0, with the eye-shaped 'fulfilled' quadratrix around it (touching it at the extreme values ( (0,r) and (0,-r) with r the radius of the circle).

 

And then I thought: this figure I 'made' is so special, so unique per radius of the circle, that it must have some special properties. First thing I thought of, was the area.

So then I started playing with the numbers I had:

 

Let A be the area of this fulfilled quadratrix. In the file (i.e. for r = 10), A = 175.8804176

Note that this won't be the exact area of the fulfilled quadratrix, for it's only measured with 10 equal parts the quadratrix is based on...

 

The area of the circle (i.e. for r = 10) = 314.1592654

Dividing the latter area by the former results in 1.786209458 = B

Dividing [math]\pi[/math] by this number results in 1.758804176 = C

The square root of [math]\pi[/math] is 1.772453851

 

Then I thought that this square root is about the mean of B and C:

(B+C)/2 = 1.772506817

 

Not such a big difference, right?

 

And then I thought: if this is right, I should be able to find a nice formula to express the area of the fulfilled quadratrix in function of r:

 

[math]\frac{\left(\frac{r^2\pi}{A}+\frac{A\pi}{r^2\pi}\right)}{2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow \frac{r^4\pi+A^2}{2Ar^2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow 2Ar^2\sqrt{\pi}=r^4\pi+A^2[/math]

[math]\Leftrightarrow A^2-2Ar^2\sqrt{\pi}+r^4\pi=0[/math]

 

Giving a discriminant of 0 and a final result of [math]A=\sqrt{\pi}r^2[/math].

 

Nice enough, in my opinion. It's always possible that I'm not right, but, in my opinion, it's very beautiful that this discriminant is 0, so that must mean something!

 

Can someone tell me whether I'm right or not? Can't find any appropriate information on the internet..

Are there other special properties the (fulfilled) quadratrix has?

 

Thanks!

 

Function

200th post

Dang.. Press edit and a new post pops up... 200th post gone to waste..

So the A(quadratrix)/A(circle) = 1/sqrt(pi). Conclusion: the area of a circle is sqrt(pi) times as large as that of its quadratrix.

Ich bin Deutsch nicht. Ich studierte Deutsch, als ich ein Kind war und rutschige am Wochenende. (Google übersetzen ist mein Freund.)

Ahzo.. Google Translate's everyone's friend

But I speak Dutch, not German

Zo ziet een zin in het Nederlands eruit. Het lijkt misschien een beetje op het Duits, maar is toch anders.

Now let's quit the chit-chat and leave the thread for anyone wanting to comment on-topic

Ach du liebe!! Ich werde die Korrektur vornehmen.

Aaaaah warum sprichst du Deutsch im Wochenende xd

Bist du Deutsch, vielleicht?

Sehr güt! I'm honored.

gut* (sorry, it's not my habit to correct people; but since this is a "corrective-constructive" forum.. I take my chance

Lesson of the day:

Very good = heel goed (or zeer goed) in Dutch (my mother tongue )

Thanks Function. I was thinking to cover the unknown with 'little'. Also, the unfamiliarity of the name Fermat with readers has a hint of mystery that may coax them to read the dissertation. Those familiar with Fermat will likely get the references to proof, little, and mighty. The title is just your hook and once you have the reader you can expand your different points whether you explicitly relate them to the title or leave them to figure it out when they finish. "Ahhhhh...that's what he meant!" Best of luck.

I used the title "Pierre de Fermat: unknown notable of the 17th century mathematics" - dissertation about Pierre de Fermat, his accomplishments and his little theorem.

Used the acorn thingy in my conclusion

 

Let's unpack this a little by annotating as we go.

 

p is prime, [latex]a\in\mathbb{N}, p\nmid a[/latex]

Let [latex]A=\{a,2a,3a,\cdots ,(p-1)a\}[/latex]

Now, we want to show that these p-1 numbers in A are all the nonzero residues mod p. In other words, no two of these are equal mod p. So, assume that two of them are equal.

Let [latex]ra\equiv sa\pmod{p}[/latex] with [latex] 1 \leq r, s \leq p-1[/latex]

Then: ra=mp+R and sa=np+R

What is R? This is a little messy. I think I would just skip that line entirely since there's an easier way to go. Note that if [latex]ra\equiv sa\pmod{p}[/latex] then

[latex]p |(ra - sa) = (r - s) a[/latex]

Then you can apply the theorem that [latex]p \mid ab[/latex] implies that either [latex]p \mid a[/latex] or [latex]p \mid b[/latex], plus the fact that [latex]p \nmid a[/latex] to conclude that

[latex]p | (r - s)[/latex]

So [latex]r\equiv s\pmod{p}[/latex]

But r and s are between 1 and p - 1, so the only way they can be equivalent mod p is if r = s. In other words, the set A contains a complete set of nonzero residues mod p in some order. That's what they mean by:

The elements of A must thus all be different and congruent with the elements of the set [latex]B=\{1,2,3,\cdots ,(p-1)\}[/latex]. The sequence is not important.

And now tell us the rest

 

R is the remainder.

I get everything until you get to "In other words, the set A ..."

Why is this? I do see that if both r and s are between 1 and p-1, they can only be equivalent modulo p if they're equal... But what does this prove?

Only thing it shows, as I see it, is that all the elements of A are different from each other; but this was kinda logic? I mean, if this is the reason of that step, to show that the elements of A are different... Why? Isn't it just logic that a isn't equal to 2a, to 3a and so on?

EDIT: I see what you're doing, and I get it.. So you're showing that ra and sa can only be equivalent mod p, if and only if r and s are equal? But why not just say that a isn't equal to 2a? and so on?

RE-EDIT: oh no wait, I think I get it. Thanks!

The rest of the proof is rather easy:

Since the elements of A and B are, in some order, equivalent modulo p, the product of all elements of A should be equivalent modulo p with the products of all elements of B.

Work it out, and you get Fermat's little theorem

It may just be noting that, given our a and p, the first p - 1 positive multiples of a will, mod p, give us the first p - 1 natural numbers, though not necessarily in order.

 

As an example, consider a = 8 and p = 5. Then the first p - 1 = 4 multiples of a are 8, 16, 24 and 32. These are congruent (mod 5) to 3, 1, 4 and 2, respectively.

Now that, seems very plausible. Thanks, John!

It's awkward wording, but I think B just shows that the a's are Natural numbers.

In an explanation of the proof, it says: division of elements of A and division of elements of B by p, will result in equal remainders...

Off the cuff I think it just means that it doesn't matter in what order you choose test cases. For example, you could let p=7 and let a=2 first, even though 7 is not the first prime and 2 is not the first a.

what about the congruency of elements of A to those of B?

hello everyone

Let me get straight to the point:

[math]p[/math] is prime, [math]a\in\mathbb{N}[/math], [math]p\nmid a[/math];

[math]A=\{a,2a,3a,\cdots ,(p-1)a\}[/math]

Let [math]ra\equiv sa\pmod{p}[/math]

Then: [math]ra=mp+R[/math] and [math]sa=np+R[/math]

[math](r-s)a=(m-n)p[/math]

So [math]p\mid (r-s)[/math]

So [math]r\equiv s\pmod{p}[/math]

The elements of [math]A[/math] must thus all be different and congruent with the elements of the set [math]B=\{1,2,3,\cdots ,(p-1)\}[/math]. The sequence is not important.

I found these steps of the proof (which continues after this) on the internet, and I don't really get the statement I put bold...

Can someone explain this to me? Thanks!

Function

What; me worry!? So I put your new number into the engine. It gave a dozen results again. I'll just give a couple and you can look at the rest there. (Have you used that tool before? She's a peach I tell ya! )

 

pi^2/8~~1.233700

 

pi^(1/5)~~1.257274

 

3-sqrt(3)~~1.267949

 

http://www.wolframalpha.com/input/?i=1.27

 

Edit: missed that A(circle)/A(quadratrix) = 1.780183697254104 edit. Found an interesting close-fit with e & pi for it. >> e^(1+e-pi)~~1.7801349

I know Wolfram|Alpha and she's just a beauty For 1.27, however, I can't seem to find any number that's 'special' enough to mention (except for pi/4, that's a special one, now, isn't it)..

The one you found for A(circle)/A(quadr.), however, is indeed very interesting!

Sorry; copy/paste error. Should read 6+1/e. See my edit of first reply too if you missed it.

Don't sweat it. We're all human

Alright, numbers update:

A(circle)/A(quadratrix) = 1.780183697254104

p(circle)/p(quadratrix) = 1.268487635981663 (see the edit of my original post; how beautiful is it that it is the same?)

1/e~~6.367879

I doubt this It does mention sqrt(3)/e, however..

Keeping in mind that it does concern a circle, in some way, I think pi is more special here than e.

Hello everyone

I have a rather odd question for you this evening.

Yesterday I posted a topic on the quadratrix (you should read it if you'd like to know more about it) and now I'd like to know if there's something special about the number 6.37, or a number that's really close to it, because, that's the place that the quadratrix of a circle with radius 10 intersects the x-axis...

I've tried using square root of pi, pi squared, square root of 10, and much more... I've even used the constant e...

But I can't to find an expression, using methematical constants and 10, which results in about 6.37...

The closest point I find is 2*pi...

Could someone help me in my search for the 'formula'?

Thanks!

Function

EDIT: Hmmm... I happen to find on a physics-like website (http://www.electronics-tutorials.ws/accircuits/average-voltage.html) that 0.637 would be [math]\frac{2}{\pi}[/math], so the intersectional point of the quadratrix would then be [math]\left(\frac{2r}{\pi},0\right)[/math]

Can anyone confirm this?

---

Alright, I've got another one for you: the area, included by the quadratrix and the right part of the square, divided by the area, bordered by the quadratrix and the left part of the square, equals about 1.27.

Anything special about this one?

Fermat: Proof that mighty oaks from little acorns grow

Now that's a powerful title I might do something with that! It'd be, however, nice to put it in the title somehow that he's too unknown?

you know the pH of a weak acid and you know its concentration: 0.1 M.

For a weak acid, the pH can be calculated as follows (you should've already seen this formula, else you couldn't solve this (easily)):

[math]pH=\frac{1}{2}\left(pK_a-\log{c_A}\right)[/math]

with [math]c_A[/math] the concentration of the acid. You should know what [math]pK_a[/math] is in function of [math]K_a[/math], so now it should not be too hard to calculate [math]K_a[/math]

Hello

For math, we're making a dissertation about Pierre de Fermat's life, his small theorem (extended information: use of the theorem, proof, ...) and other accomplishments in the world of mathematics (short information: what are his theorems, domains, and if are they proven).

Now, I still need to put a title on it, as it were for a thesis. I can't seem to find an appropriate, nice title, though... Could someone help me?

I think Fermat's a bit too unknown for who he's been in the mathematical world, so perhaps this could be an element in the title?

(A subtitle explaining really short the title can also be used)

Thanks!

Function

Hello everyone

In a dissertation for maths, I've mentioned once "quadratrix"... Can't really explain what it is, but I think I know what it is, and I've discovered what seems to be a pretty beautiful property of the quadratrix of a circle (commonly known as the quadratrix of Dinostratus)

So I thought: what if I 'completed' the quadratrix? (i.e. completing the circle and the quadratrix, resulting in a horizontal flip of the quadratrix over x=0, a vertical one over y=0, and a horizontal one of the vertically flipped one over x=0)

The result: an eye-shaped geometrical figure.

If you're eager to see this 'completed' quadratrix, you can ask me and I will try to upload it somewhere (.ggb, to be opened with GeoGebra).

In this file, I've also drawed what I like to call 'sub-quadratrices' (the blue lines).

My conclusion of doing this: if the number of equal parts (now there are 10, for more information, seek information on the quadratrix of Dinostratus) are going to infinity, you become this eye-shaped figure; the less equal parts you have, the more you will get like 2 strange flower petals above and under y=0, with the eye-shaped 'fulfilled' quadratrix around it (touching it at the extreme values ( (0,r) and (0,-r) with r the radius of the circle).

And then I thought: this figure I 'made' is so special, so unique per radius of the circle, that it must have some special properties. First thing I thought of, was the area.

So then I started playing with the numbers I had:

Let A be the area of this fulfilled quadratrix. In the file (i.e. for r = 10), A = 175.8804176

Note that this won't be the exact area of the fulfilled quadratrix, for it's only measured with 10 equal parts the quadratrix is based on...

The area of the circle (i.e. for r = 10) = 314.1592654

Dividing the latter area by the former results in 1.786209458 = B

Dividing [math]\pi[/math] by this number results in 1.758804176 = C

The square root of [math]\pi[/math] is 1.772453851

Then I thought that this square root is about the mean of B and C:

(B+C)/2 = 1.772506817

Not such a big difference, right?

And then I thought: if this is right, I should be able to find a nice formula to express the area of the fulfilled quadratrix in function of r:

[math]\frac{\left(\frac{r^2\pi}{A}+\frac{A\pi}{r^2\pi}\right)}{2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow \frac{r^4\pi+A^2}{2Ar^2}=\sqrt{\pi}[/math]

[math]\Leftrightarrow 2Ar^2\sqrt{\pi}=r^4\pi+A^2[/math]

[math]\Leftrightarrow A^2-2Ar^2\sqrt{\pi}+r^4\pi=0[/math]

Giving a discriminant of 0 and a final result of [math]A=\sqrt{\pi}r^2[/math].

Nice enough, in my opinion. It's always possible that I'm not right, but, in my opinion, it's very beautiful that this discriminant is 0, so that must mean something!

Can someone tell me whether I'm right or not? Can't find any appropriate information on the internet..

Are there other special properties the (fulfilled) quadratrix has?

Thanks!

Function

Hello everyone

I discovered something amazing: physics in daily life; more specific: electromagnetism.

So I took the plug of my pc speakers (10V, 0.5A, 50/60Hz), put the volume on the maximum, and experienced first of all that the speakers emitted a sound when the plug was touching my hand. If I took my hand very close to the keyboard of my laptop, another sound was emitted. If I put my hand on my microscope (plugged in, 5V, 1A), another sound. My desk lamp of 11W, 230V caused the loudest sound. When I touch the lamp, but not the plug, another, very weak sound, gets louder when I close in to the plug with my other hand.

This is rather logic, to me: B is proportional to I, so it's logic that the lamp causes a louder sound.

Now this, I find strange:

The sound is louder when I touch the plug and bring my hand near the lamp, than when I directly bring the plug close to the lamp.

Can someone tell me why it makes sound when I just touch the plug? Is it because I'm sitting in some sort of magnetic field?

Can someone also explain to me the thing I find strange?

Does this actually have anything to do with magnetic field?

Thanks!

Function

Thanks for your help

My physics teacher had explained to me why it is so, and now I understand it.

Here's the explanation:

vampares

Note that I'm in my last year of high school, about to participate in the approval exam for med school. Nothing was meant to be as complicated as you make it.

Assume that the alleles A(a) and B(b) are on the same chromosome.

Let AaBb be the genotype of the first organism. Possible gametes are AB, Ab, aB and ab. Which of those are recombinant and which are not?

[math]x[/math] is just a random, even number between 0 and 100, expressing the recombinant frequency, expressed in percentage or centimorgan. I just called it [math]x[/math] because I was tired of always using the same "20 cM"

Forgive me my behaviour, which might seem arrogant, but what you say is irrelevant to my question: I'd like to know which one of the possible gametes of organism 1: AB, Ab, aB, ab; are recombinant, and also for organism 2: aB, ab, aB, ab to decide the chance of getting one of the genotypes in the table I gave in #1.

What you say is undoubtedly very interesting, but has little to do with my question, as far as I'm aware of.