Math

[klatukatt]

Ah proofs... I miss proofs...
(And Proof! Hey CC! *nudge nudge wink wink*)

My favorite was the Quadratic Formula. Ahem, if I may...

IF
a(x^2) + bx + c = 0

THEN

x=

-b (+or-) √[(b^2) - 4ac]
..............2a...............


If you followed that, I love you.


I love math. Unfortunatly, at the level I am at, the next class I would take would be specifically designed for those aspiring to be mathemeticians.

I'm not mathemetician material.

[Jonathan]

1200-year-old problem "easy"

A British mathematician claims to have solved the problem that occurs when you try to divide any number by zero. This is what he did:

0^0

= 0^(1-1)

= 0^1 · 0^(-1)

= (0/1)^1 · (0/1)^(-1)

= 0/1 · 1/0

= 0/0

= a new number called "nullity"

Very strange if you ask me. As I see it, the professor has just given a name to an undefined number. Easy indeed.

Btw, does zero to the power of zero equal one? 0º=1 or not?

[klatukatt]

I think 0^0 should equal one...

[Jonathan]

That's what my calculator says too. But why?

Anything to the power of zero equals one. But at the same time, zero to the power of anything equals zero. So how the heck does one define 0^0?

[Gwaimir Windgem]

To say that something is "too such and such a power" means it is multiplied by itself so many times. When you multiple zero by itself any number of times, you get zero. 0 x 0 = 0. 0 x 0 x 0 = 0. Et cetera. It stands to reason.

I don't know why something to the power of zero would equal one, however...it seems that, saying that you multiply it by itself zero times, you would either say it is the original itself, or it is zero; I lean more towards the former. Why is it said to equal one?

[Jonathan]

Originally Posted by Gwaimir Windgem

I don't know why something to the power of zero would equal one, however... Why is it said to equal one?

It's due to a law of exponents.

(x^a) * (x^b) can be written x^(a+b)
If b = 0 we get
(x^a) * (x^0) = x^(a+0) = x^a

This means that x^0 = 1

The question is weather it still equals 1 for x = 0

edit: Or you can just enter any number into the nullity equation a couple of posts above. x for instance. You'll get that:

x^0 = x/x = 1

[mithrand1r]

Originally Posted by Jonathan

That's what my calculator says too. But why?

Anything to the power of zero equals one. But at the same time, zero to the power of anything equals zero. So how the heck does one define 0^0?

Zero raised to the Zero power is undefined.

(0º) is undefined.

At least under conventional mathematics.

With calculus:

lim(x^x) as (x goes to 0) (from the positive side) is equal to e^0 which is equal to 1

(I think this is why my TI-89 graphing calculator gives tells me that 0^0 = 1)

My calculator says that 0^(-1) is undefined and 0^(1) = 0

yet 0^( 1 + [-1] ) = 1

[mithrand1r]

Originally Posted by Jonathan

1200-year-old problem "easy"

A British mathematician claims to have solved the problem that occurs when you try to divide any number by zero. This is what he did:

0^0

= 0^(1-1)

= 0^1 · 0^(-1)

= (0/1)^1 · (0/1)^(-1)

= 0/1 · 1/0

= 0/0

= a new number called "nullity"

Very strange if you ask me. As I see it, the professor has just given a name to an undefined number. Easy indeed.

Btw, does zero to the power of zero equal one? 0º=1 or not?

While I think I understand the concept of "nullity", are there any applications of nullity?

Nullity * Nullity = ????

3 + Nullity = ???

Nullity^Nullity = ???

What will a computer do with Nullity?

I am just curious.

(Interesting article.)

[Jonathan]

Originally Posted by mithrand1r

While I think I understand the concept of "nullity", are there any applications of nullity? ... What will a computer do with Nullity?

I haven't heard of any applications (if there even are any). As for computers, the mathematician seemed to think they'd be less eager to crash if they have a hunch about what "nullity" if they happen to divide by zero. Which is silly, computers are almost always programmed to check for division by zero anyway.

From one thing to another, I've never come across any applications of the imaginary number i. I'm sure there must be some though. Does anyone know?

[Millane]

Originally Posted by trolls' bane

Okay, I feel that we (or at any rate, I) am in dire need of a Math thread. I seem to remember there being one, but the closest I could find was a Math Problem thread, which far from encompasses the rather broad scope that discussion of math requires.

I think to start, I would like it if someone could attempt to teach me how to do a formal proof, as I have learned little in that regard from Geometry.

please tell me that you need this help for some specific schooling purpose, and not for "fun"... People that find maths "fun" creep the **** out of me
people keep telling me to take a maths subject next year so i will have something practical at the end of my degree. well **** practicality if it means doing more maths

[klatukatt]

Math is fun. *creepy smile*

How you doin?

[mithrand1r]

Originally Posted by Jonathan

I haven't heard of any applications (if there even are any). As for computers, the mathematician seemed to think they'd be less eager to crash if they have a hunch about what "nullity" if they happen to divide by zero. Which is silly, computers are almost always programmed to check for division by zero anyway.

From one thing to another, I've never come across any applications of the imaginary number i. I'm sure there must be some though. Does anyone know?

It is used in electrical engineering, IIRC. I am not sure how, but I think that is what my friend said.

Here are two qutoes from websites and what they say on the subject.

http://www.newton.dep.anl.gov/newton...th/MATH033.HTM

Use of imaginary numbers

Author: charles l paulk
What are imaginary numbers used for? What are some of their practical
applications?

Response #: 1 of 2
Author: tee
"Practical" is often in the eye of the beholder. From the perspective of
algebra, complex numbers give us a complete system for finding the roots of
polynomials. Since numerous applications are based on polynomial models in
theory, complex numbers play a part in all of these. For example, in
electrical engineering we find complex roots in circuit theory where the
polynomial is part of the model equation for simple circuits. Alternating
current relates to complex root cases in the polynomial model. In mechani-
cal engineering, the same type of model relates to vibrations with wavelike
results connected to the complex root cases.

Response #: 2 of 2
Author: asmith
And in physics, it turns out quantum mechanics uses complex numbers for just
about everything - the wave functions of particles have a complex amplitude,
including a real and "imaginary" part, and both are essential.

http://mathforum.org/library/drmath/view/53606.html

Originally Posted by Dr. Math

Applications of Imaginary Numbers

Date: 10/14/97 at 18:40:14
From: Beatka Zakrzewski
Subject: Imaginary Numbers in the work force (applications)

Dear Dr. Math,

I know that you have already received tens of questions about
imaginary numbers but I can't seem to find a straight answer to mine.
Where are imaginary numbers used today in real life, as in the work
force or other areas that use math? I have heard that they are used
in electrical technologies - if so, could you specifically tell me in
which areas?

Thank you.
Sincerely, Beatka Zakrzewski

Date: 10/14/97 at 19:03:06
From: Doctor Tom
Subject: Re: Imaginary Numbers in the work force (applications)

Hi Beatka,

I answered almost exactly the same question for a teacher a couple of
years ago and saved my reply. I'll send it to you. Unfortunately, as
you'll see from the description, they are often used with more
sophisticated math than you know yet, but if you ignore the details
below and concentrate on the problem descriptions, they may make some
sense to you.

I teach Algebra II and Trigonometry in Gilroy High School,
Gilroy, CA. The students' ages are 15-17.

Their question is: Who uses imaginary numbers in the real world?

The students learn to calculate with i and complex numbers of
the form a + bi. The book indicates that electrical engineers
use imaginary numbers, but use j instead of i. What do they
*do* with the numbers? And does anyone else actually use them?


First off, electrical engineers use "j" because "i" is almost always
used to mean "current" in their equations. Where most of us would
write "a + bi", they'd write "a + bj", but it means the same thing.

Electrical engineers often have to solve what are called "differential
equations," which are a bit hard to explain without knowing a bit
about calculus. Basically, a differential equation relates functions
to their rates of growth. The solution to a differential equation is
usually a function, not a number.

As a specific example, suppose you have a snowplow that keeps piling
up more and more snow in front of it so that the farther it goes, the
heavier the load it is pushing, and the heavier the load, the slower
it goes, and the slower it goes the slower the pile of snow in front
of it grows.

You can (with a differential equation) relate the amount of snow at a
given time t (call it A(t)) to the velocity of the plow, and the
equations can be solved to give the function A(t) at all times t.

But often, it's easier to solve differential equations in the domain
of complex numbers because the equations are a lot nicer, but you know
that the solution you care about is just the real part of the
solution. It's difficult to give an example without some calculus.

I can, however, show you a nice example that may make it clear that
working in the domain of complex numbers is easier in some cases than
working strictly in the reals. If you're studying the complex numbers
and trigonometry at the same time, in theory you can follow the next
steps, but if you can't, don't worry; just look at the messiness of
the calculation - I'm trying to show how going to the domain of
complex numbers can drastically simplify a problem.

Try to work out the following: (cos x + i*sin x)^3 . Multiply this out
and reduce it to the simplest form. I get this:

cos^3 x + 3*i*cos^2 x * sin x - 3*cos x * sin^2 x - i*sin^2 x.

(It's a nice exercise in algebra, complex numbers, and trigonometry.)

Let's just look at the real part (you can do the same sort of thing
with the imaginary part):

cos^3 x - 3*cos x * sin^2 x

= cos x * ( cos^2 x - sin^2 x) - 2*cos x * sin^2 x

Now, since cos(2x) = cos^2 x - sin^2 x, and sin(2x) = 2*sin x * cos x,
we can re-write the equation above as:

cos x * cos(2x) - sin x * sin(2x).

Now, cos(3x) = cos(x + 2x) = cos x * cos(2x) - sin x * sin(2x), so the
real part is just cos(3x). With a similar amount of ugly calculation,
we can get that the imaginary part is i*sin(3x).

So the answer is: cos(3x) + i*sin(3x).


I assume you understand exponents = x^2 = x * x, x^3 = x * x * x, and
so on. Well, it turns out that there's a special number called e
which is equal (approximately) to 2.71828182845... which satisfies the
following equation (in the complex numbers:

e^(ix) = cos(x) + i*sin(x)

So the original problem I stated was to find the cube of the number on
the left:

[(e^(ix)]^3 = e^[i(3x)] = cos(3x) + i*sin(3x),

so you're done in a single step.

Another important application of complex numbers to the real world is
in physics. In quantum mechanics, one cannot say with precision where
a particle is. You can only give an probability distribution of its
position in space. And the only way to calculate the distributions is
using complex variables. Unfortunately, this is even harder to
explain, but such calculations have to be done for almost any
calculations about nuclear reactions.

Finally, something that may not be precisely an "application," but one
that you can easily experiment with, is that a certain class of
complex numbers behave as rotation operators.

For example, draw the usual real and imaginary axes, and plot any
point on it (say 3 + 5i) Multiply this number by i, and you get
(-5 + 3i). If you plot this new point, you'll find that it is the
original point rotated about the origin by 90 degrees counter-
clockwise. This works for ANY complex number. Multiply by i, and
you'll rotate it by 90 degrees.

Now, take any complex number, and multiply it by cos(45) + i*sin(45)
(in degrees), which is about (.707 + .707*i). This rotates points
clockwise by 45 degrees. And there's nothing special about 45 degrees.
Multiply any complex number by cos(x) + i*sin(x), and you'll rotate
the number about the origin by an angle x.

In the same way, adding a fixed complex number is equivalent to a
translation, and multiplying by a real number expands or contracts the
values. By combinations of rotations, translations, and expansions/
shrinkages, you can do most 2-dimensional computer graphics operations
on objects in a plane.

Unfortunately, in 3 dimensions, it's not so easy, and the easy way
involves something called matrix multiplication.

I know none of these is easy to understand, but if you want to do them
eventually, you've got to have a solid foundation in the basic
operations on complex numbers - just as you can't do algebra until
you learn to add and subtract real numbers.

-Doctor Tom, The Math Forum
Check out our web site! http://mathforum.org/dr.math/

[Jonathan]

Originally Posted by mithrand1r

It is used in electrical engineering, IIRC. I am not sure how, but I think that is what my friend said.

Here are two quotes from websites and what they say on the subject.

http://www.newton.dep.anl.gov/newton...th/MATH033.HTM


http://mathforum.org/library/drmath/view/53606.html

Good post. Thanks for sharing!

Now that you say it, I remember that a friend told me too that complex numbers are used in electrical engineering. I had no idea about the quantum mechanics though.

Anyone but me who thinks Euler's formula below is quite remarkable?

e^(pi*i) + 1 = 0

[Gwaimir Windgem]

Originally Posted by Jonathan

It's due to a law of exponents.

(x^a) * (x^b) can be written x^(a+b)
If b = 0 we get
(x^a) * (x^0) = x^(a+0) = x^a

This means that x^0 = 1

The question is weather it still equals 1 for x = 0

edit: Or you can just enter any number into the nullity equation a couple of posts above. x for instance. You'll get that:

x^0 = x/x = 1

That's what happens when you break it down into those formulae and rules, at the expense of the signification of what you are saying; you come up with silly things like that. And the nullity problem. To speak of any number or zero divided by zero, or to the zeroth power really signifies nothing.

[mithrand1r]

Originally Posted by Jonathan

Good post. Thanks for sharing!

Now that you say it, I remember that a friend told me too that complex numbers are used in electrical engineering. I had no idea about the quantum mechanics though.

Anyone but me who thinks Euler's formula below is quite remarkable?

e^(pi*i) + 1 = 0

It is interesting since it has π and e and 1 and 0 and i and the =.

It has a little of everything.

[Jonathan]

Originally Posted by Gwaimir Windgem

That's what happens when you break it down into those formulae and rules, at the expense of the signification of what you are saying; you come up with silly things like that. And the nullity problem. To speak of any number or zero divided by zero, or to the zeroth power really signifies nothing.

True. It is by convention that any number raised to the power of zero equals one. That is (a^0 = 1). This is to make the operation consistent with algebraic rules even though the operation itself is meaningless.
Same thing about the nullity problem. But if there's any application of "nullity", it would be interesting to know.

Originally Posted by mithrand1r

It is interesting since it has π and e and 1 and 0 and i and the =.

It has a little of everything.

It also has addition, multiplication and exponentiation.
Sine and cosine are also related since e^(X*i) = Cos(X) + i Sin(X)

edit: I found a very educational site that deals with common mathematical fallacies and stuff.
Zero Saga

[Jonathan]

[JaeSon15]

So here's a cool thing mathematical fact that I'm surprised no one's brought up in this thread, and it's actually something even research mathematicians take into account today. Said imprecisely, infinity can be greater than infinity!

When you're determining the cardinality (size) of a set with a finite number of elements, you can just count the number of elements.

Here's a simple example:

Set B = {John, Mike, Tim, Dan, Steve}
Set G = {Kim, Monica, Jane, Jessica}

Set B has 5 elements, and Set G has 4 elements. Therefore, set B is larger.

But, when you're dealing with infinite sets, you can't simply count the elements. The best you can do is try to compare elements between sets by forming a correspondence. We can actually apply this to the example above. Let's form a correspondence between Set B and G: John-Kim, Mike-Monica, Tim-Jane, Dan-Jessica... But wait! Set B has a leftover! It must be larger.

This is very similar to how mathematicians compare infinite sets. So, let me get to the main result. The real numbers is an infinite set that is actually greater in cardinality than the natural numbers! The proof of this is called Cantor's Diagnolization Argument. This was actually really trippy stuff back in the day; not only did Cantor almost go insane thinking about it, he received threats from abroad.

[katya]

Infinity is always weird... I was thinking earlier that there are infinite numbers between 1 and 2, but there are more numbers between 1 and 3. Something like that.
It's like in elementary where we would say "infinity plus 1"

[trolls' bane]

I wonder if Achilles has passed the tortoise in the race yet...