You are not logged in.

- Topics: Active | Unanswered

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

*Sinh(iθ)= iSin(θ)**Cosh(iθ)=Cos(θ)*

**X'(y-Xβ)=0**

Offline

**mitsy****Member**- Registered: 2006-04-11
- Posts: 4

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?

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

Good point.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

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

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

"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

Hmmm.. Is the second line supposed to say cosh(iθ) = icos(θ) ???

A logarithm is just a misspelled algorithm.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Sinh(z)+Cosh(z)=Exp(z)

**X'(y-Xβ)=0**

Offline

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

George,Y wrote:

So,

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

*Last edited by ben (2006-07-20 07:31:47)*

Offline

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

If we assume that:

Then:

As required. Is this what you mean by "why"?

"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

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

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 (2006-07-21 10:59:52)*

Offline

**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,905

And...

Don't forget the great equation:

IPBLE: Increasing Performance By Lowering Expectations.

Offline

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

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

A logarithm is just a misspelled algorithm.

Offline

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

Whence the even famouser

The 5 fundamental transcendentals, all in bed together (but no action!!)

Offline

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

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:

"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

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

Ricky wrote:

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

No problem, math is fun, right?

Consider the equation:

But why should I? Where did it come from, other than the taylor series I showed?

And so it must be that:

Why? It could be, by your reasoning, that

Anyway, let's not fall out in public. Others seem to have lost interest anyway.

Offline

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

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

The Beginning Of All Things To End.

The End Of All Things To Come.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

I'm astonished to see for far you guys haven't proved what I proposed!

Okay, this is how i got the two formulae.

Euler's formula( proven in #10)

substituding z with -z will get

simlified form:

Hence

sinh(iz)=[(1)-(2)]/2=isinz

cosh(iz)=[(1)+(2)]/2=cosz

**X'(y-Xβ)=0**

Offline

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

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 (2006-07-25 05:51:53)*

Offline

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

I just checked on this

George,Y wrote:

Sinh(iθ)= iSin(θ)Cosh(iθ)=Cos(θ)

for real x, cosh x = cos ix, and not as you wrote (how do you get thetas and stuff in here, without using LaTex?)

Offline

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

The top of every page has this bar:

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

Copy and paste.

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

OK, my proof is true if all the properties of circular and hypobolic functions are still true when the domains are complex numbers instead of real numbers.

By the way, i(x+iy)=-y+xi

I just post this to explain why sometimes Mathematica will give a result involving i when you assaign only an integration task, and that the coefficient of i is always very small. The reason, I guess, is that it did some circular to hypobolic shifting. And you can just ignore the imaginary part.

I admit the proof was insufficient.

*Last edited by George,Y (2006-07-25 18:10:15)*

**X'(y-Xβ)=0**

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

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"

**X'(y-Xβ)=0**

Offline

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

George,Y wrote:

I was not assuming the result.

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

You're right.

George, I apologize. Sorry

In case you're interested, the reason why you need to exercise extreme caution when replacing any f(x) with f(z) is that, in the complex plane, you have to check for continuity on a *circle* centred on some f(z_0) rather than on the real line segment centred on some f(x_0).

Complex analysis throws up some truly amazing theorems, all based on the requirement that functions possess what might be called isotropic derivatives

Offline

**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

It doesn't matter.

Thank you for your advice. I'll study complex analysis some day.

**X'(y-Xβ)=0**

Offline

**Patrick****Real Member**- Registered: 2006-02-24
- Posts: 1,005

I've never read it myself, but maybe you could read Complex Analysis if you want. Don't know if it's well written, just had it on my harddisc(this is the original source, I have downloaded it and kept the source url in a file ).

This is a textbook for an introductory course in complex analysis. It has been used for our undergraduate complex analysis course here at Georgia Tech and at a few other places that I know of.

*Last edited by Patrick (2006-07-26 19:33:42)*

Support MathsIsFun.com by clicking on the banners.

What music do I listen to? Clicky click

Offline