Originally posted by FlyingDutchman
Do negative numbers have square roots? I don't think so, but my maths teacher says they do in a complicated way...

You mean something like sqrt(-1)?
Yes they do, but not in real numbers.
In complex numbers the sqrt(-1) = i.
i is defined that i^2 = -1.

Originally posted by FabianFnas
You mean something like sqrt(-1)?
Yes they do, but not in real numbers.
In complex numbers the sqrt(-1) = i.
i is defined that i^2 = -1.

Thanks for the attempt, but that's way too advanced for me.

Originally posted by FlyingDutchman
Thanks for the attempt, but that's way too advanced for me.

"i" is a notation to write Sqrt ( -1 )
It's simply the "imaginary number" that multiplied by itself gives the result -1

Originally posted by serigado
"i" is a notation to write Sqrt ( -1 )
It's simply the "imaginary number" that multiplied by itself gives the result -1

OK. But although that makes sense, you can't ever numerically define i. So how does it help you?

Originally posted by FlyingDutchman
OK. But although that makes sense, you can't ever numerically define i. So how does it help you?

The number itself does not exist (in real numbers).
But the concept of Sqrt (-1) is helpful is hundreds of applications in mathematics, and really simplifies things a lot!
Here's an interesting property (when I've been told this when I was in school I didn't believe or understand it).

Exp ( i * Pi ) = - 1

The exponential of Sqrt (-1) * Pi is equal to -1 !! Now that's incredible 🙂 The exponential of an imaginary number is related to the Sin and Cosin functions, but there's a lot more!

Originally posted by serigado
The exponential of an imaginary number is related to the Sin and Cosin functions, but there's a lot more!

Adding to this a bit, your trigonometric functions can be written in quite neat formulas using 'e's and 'i's. For instance,

cos(x) = [e^(i*x)+e^(-i*x)]/2

EDIT: Also, the function described in the above post is indeed incredible. I've made my girlfriend memorise it.

Originally posted by Swlabr
Adding to this a bit, your trigonometric functions can be written in quite neat formulas using 'e's and 'i's. For instance,

cos(x) = [e^(i*x)+e^(-i*x)]/2

EDIT: Also, the function described in the above post is indeed incredible. I've made my girlfriend memorise it.

Also, the function described in the above post is indeed incredible. I've made my girlfriend memorise it.
She'll dump you in 2 weeks. Give her some chocolates and flowers, maybe you can save your relation yet.

Originally posted by Swlabr
Adding to this a bit, your trigonometric functions can be written in quite neat formulas using 'e's and 'i's. For instance,

cos(x) = [e^(i*x)+e^(-i*x)]/2

EDIT: Also, the function described in the above post is indeed incredible. I've made my girlfriend memorise it.

is that e as in 2.71828 yadayada or a different symbol?

Originally posted by joe shmo
is that e as in 2.71828 yadayada or a different symbol?

that same one
e^1 = 2.7182818...
e^(i*Pi) = -1

Originally posted by serigado
that same one
e^1 = 2.7182818...
e^(i*Pi) = -1

Then that is certainly is uncanny to say the least!!😵

Originally posted by serigado
that same one
e^1 = 2.7182818...
e^(i*Pi) = -1

can someone walk me through the simplification of this? how is it solved without a calculator? How does the imaginary factor dissappear? how do we arrive at -1 longhand?😕

but before anyone waist's great time, know that i will not understand with anything involving Trig or higher....

Originally posted by joe shmo
can someone walk me through the simplification of this? how is it solved without a calculator? How does the imaginary factor dissappear? how do we arrive at -1 longhand?😕

but before anyone waist's great time, know that i will not understand with anything involving Trig or higher....

Exp[i*x] = Cos[x] + i * Sin [x]
when x=Pi,
Cos[Pi] = -1 and i*Sin[Pi] = 0

To verify that formula you can compare the Taylor Series of Exp[x] with Cos[x] and Sin[x].
Exp[x] = 1 + x + x^2 / 2 + ... + x^n / n!
so,
Exp [i*x] = 1 + i*x - x^2/2 - i*x^3/3! + ...
The odd terms will correspond to Sin[x] expansion, and the even terms to the Cos[x] expansion.
(Maybe I got a little too advanced for you now (?) )

Originally posted by serigado
Exp[i*x] = Cos[x] + i * Sin [x]
when x=Pi,
Cos[Pi] = -1 and i*Sin[Pi] = 0

To verify that formula you can compare the Taylor Series of Exp[x] with Cos[x] and Sin[x].
Exp[x] = 1 + x + x^2 / 2 + ... + x^n / n!
so,
Exp [i*x] = 1 + i*x - x^2/2 - i*x^3/3! + ...
The odd terms will correspond to Sin[x] expansion, and the even terms to the Cos[x] expansion.
(Maybe I got a little too advanced for you now (?) )

its looks awesome, but thats about all i can say....hahah

perhaps if i can get through another 5 or 6 semesters it will all make perfect sence.......perhaps lol

Originally posted by joe shmo
its looks awesome, but thats about all i can say....hahah

perhaps if i can get through another 5 or 6 semesters it will all make perfect sence.......perhaps lol

I learned this last year in high school, you only need to know how to differentiate.
Every science course teaches this 1st semester. First time you see this, seems strange, but it's very logical when you get the concept.

Originally posted by serigado
I learned this last year in high school, you only need to know how to differentiate.
Every science course teaches this 1st semester. First time you see this, seems strange, but it's very logical when you get the concept.

I'm slow of the mark....never had geometry or any class that would familiarize me with trig functions. In High School I took Algebra 1 and thats it as far as math. I suppose you had Algebra 1 nocked off by the seventh grade? As a result I'm starting College level mathematics at the bottom of the heap...It also took me five years after high school to wise up and decide to give college a try. Im taking Pre Calc now, i can analyze some of the solution, but some things in there remain unstudied to this point...