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Complex numbers
(cos(x)+isin(x))^n=cos(nx)+isin(nx)
Now using maclaurine series
e^ix=1+x+x^2/2!+x^3/3!+....+x^r/r!+...
this can be used to that
e^ix=cos(x)+isin(x)
which means
e^inx=cos(nx)+isin(nx)
this is then used to show that when n=1 x=pi
e^(ipi)=-1
e^(ipi)+1=0
This result has been describe by many as the most beautiful equation in Maths.
"Sniff, It's so... Beatutiful!!!"
:sobs:
It is pretty wonderful.
Oooh... try drawing f(x) = (-1)^x.
It always disturbed me that I couldn't really draw that one properly. Now I see that the plane of my graph paper was simply intersecting a line in a conceptual three dimensional space. Makes sense.
How do you draw it?Quote:
Originally Posted by Xei
I think he means graphing it on a line.
Yeah, but I don't know how you draw it on a graph. What if you get f(0.5)=(-1)^0.5 How do you draw this?
I is sorry, I doesn't know math too well.
You need to know Euler's relationship (that the OP was talking about) to be able to do it.
What you just asked is a special case;
-1 ^ 0.5 = sqrt-1 = i.
Complex numbers (look 'em up if you don't know what you are, they're very simple really) can be written as a + bi. Alternatively you can write this as |z|*(cosx + isinx), where |z| is the length of the complex number and x is the angle it makes with the horizontal (this should be clear by looking at an Argand diagram). It turns out that (cosx + isinx) can be shown to be equal to e^ix.
If you set x = pi, you get e^ipi = cospi + isinpi = -1.
Therefore you can write f(x) = -1^x = (e^ipi)^x = e^ipix = cospix + isinpix.
Basically this traces out a circle radius 1 in the Argand plane as x increases, so if you plotted this plane against x you would get a helix.
Thought it was pretty.
Oh so it's like a 3D graph? This is what I didn't get. It never occurred to me to put the complex plane in there too. If I'm understanding this now anyway. Thanks for your explanation.Quote:
Originally Posted by Xei
Yeah; it's neat if you consider how this relates to trying to plot the function in 2D. The only answers you can get normally are integer powers of -1, where even powers are 1 and odd powers are -1. If you visualise the 3D answer, you can see that these few real answers are simply where the graph intersects the plane of the paper.
y= (-1/e)^x is a nice extension. Like a hurricane.
Which program are you using? (Unless you're one of those savants who visualize it all.) It might be smart to get one :P
e^(i*pi) = -1 is my favourite formula! Although it took me a while to get my head round it at first.
Xei: I've never thought about that kind of graph before. I can't really visualise what it would look like...