Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

you know, often the algebraic behavior of numbers can be used to perform some nifty tricks for things like summation. For an alternating series we can use (-1)^n to change from even to odd numbers.

The fact that an imaginary number squared equals a negative real number might be usefull in creating some functions with unusual behavior.

For instance, what if we defined a function f(x) to return a complex number a + bi where a and b are some function of x, and g(x) returns only the real part of f(x) and not the imaginary part? Generally the portion bi would be ignored except when even powers of i appear, then portions of the imaginary part are added to the real part. The function could have some very unusual behavior!

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

For an alternating series we can use (-1)^n to change from even to odd numbers.

I believe that's supposed to read negative and positive.

Let z = a + bi, where a and b are real numbers.

Re(z) = a

Im(z) = b

It may be important to note that there really aren't any such things as "even powers" of i:

So we can always reduce i into a power of 1.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

whoops! Yeah meant to right positive and negative.

yes technically even or odd powers of i can always be reduced to either i, 1, -i, or -1 but they are still even or odd powers of i, is that not true?

say f(x) = i^2x

and h(x) = i^2(x - 1)

and let j(x) and g(x) be some functions that return real values.

then in ∑ f(n) g(n) + h(n) j(n) the functions g(x) and j(x) would "take turns"

but again, I defined the imaginary functions to only return the real part.

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

yes technically even or odd powers of i can always be reduced to either i, 1, -i, or -1 but they are still even or odd powers of i, is that not true?

No, because all complex numbers must take on the form: a + bi. It's how they are defined.

For example: .5 / .7 is certainly a rational number, but it isn't in rational *form*, an integer over another integer.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

well couldn't 5 + i^4 be written as 6 + 0i?

Not sure I see your point.

A logarithm is just a misspelled algorithm.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yes, but 5 + i^4 is not in "complex form", so to say. 6 + 0i is.

It's like saying that y = 0x^2 + 3x + 5 is a quadratic equation.

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I think I see your point from the last statement. But uh... so what? All I said was a function that would return the real part of a complex number (Whether the imaginary coefficient is zero or not) so it can have a place value on the numberline.

I read that all real numbers are also complex numbers (where b is 0) but the reverse is not true.

A logarithm is just a misspelled algorithm.

Offline

**luca-deltodesco****Member**- Registered: 2006-05-05
- Posts: 1,470

mikau wrote:

I read that all real numbers are also complex numbers (where b is 0) but the reverse is not true.

well yes, i would think so.

you could think of the reals as being a special case of the complex numbers

the integers are a special case of the reals, like the rationals

the natural numbers are a special case of the integers

would that mean that like the integers are a subset of the reals?

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

you could think of the reals as being a special case of the complex numbers

the integers are a special case of the reals, like the rationals

the natural numbers are a special case of the integers

That is correct, but historically backwards.

The integers are *an extension of* the naturals. The rationals are an extension of the the integers. The reals are an extension of the rationals. The imaginary are an extension of the reals.

Hopefully I should have an article coming out on this soon.

would that mean that like the integers are a subset of the reals?

Yep.

Offline

**ben****Member**- Registered: 2006-07-12
- Posts: 106

luca-deltodesco wrote:

would that mean that like the integers are a subset of the reals?

Yes. And that's a cool thing to do, for the reals are what is called *not* algebraically closed.

This simply means that an equality like...I dunno....x² + 2 = 0 has no solution in **R**. This is found in **C** (it's 2i, by the way)

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Naturals, integers, rationals, and reals are all not algebraically closed. But the complex numbers are. That makes them special.

Offline

**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

For an alternating series we can use (-1)^n to change from even to odd numbers.

I believe that's supposed to read negative and positive.

Ah, but there's a neat trick here.

Let the group 2 = {0, 1}, and let P: X → 2 be the parity map on X. Define P: x |→ 0 if x is odd, P: x |→ 1 if x is even. Now {±1} is isomorphic to the group 2 = {0, 1} so we may replace, say 0 with -1 and 1 with 1. Thus odd/even and -/+ are the same thing, up to isomorphism, as the usual parity rules of arithmetic easily show

Let z = a + bi, where a and b are real numbers.

Yes, but Mikau's original point was about imaginary numbers, not complex numbers

So we can always reduce i into a power of 1.

I don't think I quite agree, but you may be right. But note the two following bits of trivia (just for fun):

.........

see the pattern?

Note also that the 4th power of all of these guys is +1!!. Is this true of all integer powers of i? Of course!

*Last edited by ben (2006-08-02 04:10:21)*

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Yes, but Mikau's original point was about imaginary numbers, not complex numbers

My mistake. I always use those term interchangably when they really aren't. But it still holds that we would not call i^2 imaginary, right?

I don't think I quite agree, but you may be right.

Well, assuming that we are talking about integer powers.

Offline

**ben****Member**- Registered: 2006-07-12
- Posts: 106

Ricky wrote:

But it still holds that we would not call i^2 imaginary, right?

That is a subtle point, right enough. I suspect it would be a matter of taste. Hmm. Actually maybe it's a bit silly to even talk in these terms. The imaginaries are not a field, a ring or even a group, as we agreed.

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

ben wrote:

This simply means that an equality like...I dunno....x² + 2 = 0 has no solution in R. This is found in C (it's 2i, by the way)

Not only 2i.

Try -2i

IPBLE: Increasing Performance By Lowering Expectations.

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Wouldn't it be ±√(2) i

Offline

**ben****Member**- Registered: 2006-07-12
- Posts: 106

krassi_holmz wrote:

Not only 2i.

Try -2i

Ricky wrote:

Wouldn't it be ±√(2) i

Oh dear

Offline

**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

I recall one calculus test I took in highschool:

2x = 2

x = 2

Everyone does it.

Offline

Pages: **1**