i've always found that trigonometry is not as hard as your maths teacher makes it out to be...so why say it is so hard?

So hyperbolic functions work with i the same way that trig functions work with negatives.  That is pretty cool.

George,Y wrote:

So,

and

Why? Anyone know? z = x + iy, by the way

Last edited by ben (2006-07-21 05:31:47)

Well,
first I hadn't intended by question as a challange, I wanted to know
for myself. I've got it figured now.
Second, Ricky, I can't see what you did there. You assumed the result I
wanted to see shown in order to find what I wasn't looking for! Or have
I missed something?
Anyway, with the help of some textbook, I see it.
Note first that z = x + iy here. Note also that e^z is analytic
throughout the complex plane: it is an entire function. We expect there
to be a taylor series therefore.

Recall that the taylors for the circular trigs are

and with

applied to the above gives

which reduces to

We now want to let x = z = x + iy. Notice that even powers of i are all negative 1 or i, odd powers are positive 1 or i, alternating.
Remembering that

we will easily find that

gathering terms

which is merely to say that

It then follows that

All I need do now is remind you the expansion for e^-x is

and you should easily see (by comparing the appropriate taylors) that

And as we know e^z is an entire function, we can assume the same is true here.

Now that's something that really does deserve to be called cool!

Last edited by ben (2006-07-22 08:59:52)

krassi, I've seen that expression somewhere before... but I can't remember where... ;-)

Sorry ben.  In an attempt to write a quick post, I was too brief.

Consider the equation:

We can show, as my last post did, that:

But we also know that:

So we can say:

And so it must be that:

a really nice thing to see (atleast when i first saw it)

i wrote a terrain grapher in flash a while ago, when i made my complex number calculator, and i added a graphing function, and when i first tried it out, the first thing i noticed, was when graphing sinh and cosh, they were just 90 degree rotations in the complex plane

George,Y wrote:

I'm
astonished to see for far you guys haven't proved what I proposed!
Hence
sinh(iz)=[(1)-(2)]/2=isinz
cosh(iz)=[(1)+(2)]/2=cosz

Well you may be, and when I tell you why you won't like it!
First, you are assuming the result you are trying to prove (rarely a
good idea). You should really establish a relationship between the
circulars and the hyperbolics before taking your leap!

Second, even if we know that, if x is real, sinh ix = i sin x and cosh x = cos ix we need to do this.
Let z = x + iy, and replace x in the above by z. What do you get, for example for iz = i(x + iy)?
For cos iz?
For i sin z?

Hint: Use Osborne's Rule (know it?)

Last edited by ben (2006-07-26 03:51:53)

The top of every page has this bar:

[ ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ ° ]

Copy and paste.

I was not assuming the result.

sinh(x)=exp(x)/2-exp(-x)/2 by defination

My software gives "Cosh[i pi/3]=1/2"

It doesn't matter.
Thank you for your advice. I'll study complex analysis some day.