e^ix = cos x + isin x and related concepts

By Andrew Hodges on Wednesday, January 09, 2002 - 06:54 pm:

How is the result e^ix = cos x + isin x
proved?

By Arun Iyer on Wednesday, January 09, 2002 - 07:32 pm:

There are many proofs....you can always prove it using the series for sine,cosine and e^x ...

but i actually like the one by calculus...

let y=cosx+isinx......(1)
differentiating w.r.t x we get,
dy/dx=-sinx+icosx.....(2)
we know that i² =-1
therefore (2) becomes,
dy/dx=i² sinx+icosx
dy/dx=i(cosx+isinx)
dy/dx=iy.....(from 1)
i.e
dy/y=idx

integrating the above equation we get,
logy=ix+c.......(3)

from 1 we can see that when x=0,y=1....
substituting in (3) we get,
c=0

therefore,
logy=ix
y=e^ix
i.e cosx+isinx=e^ix

hence proved

love arun

By Julian Pulman on Wednesday, January 09, 2002 - 08:04 pm:

it's provable by series expansions as well and comparing the expansions of e^x and sinx, cosx..I think Arun's proof which I've come across before is far more beautiful.

By Yatir Halevi on Wednesday, January 09, 2002 - 09:25 pm:

I know that this is a stupid question, but still I'll ask it:

Let's say that we would define that 360 degrees is 4 (and not

2 π

), after all radians were defined this way (with

π

) because it was more convenient this way.

So we would get that $e^{i 2}$ is equal to $- 1$ and not $e^{i π}$ .
Where did I go wrong?

Yatir

By David Loeffler on Wednesday, January 09, 2002 - 09:37 pm:

Yatir,

I remember thinking that when I first saw that proof. The answer is that if we do this the derivative of sin x isn't cos x any more. (That's the main reason why radians are used).

David

By Philip Ellison on Wednesday, January 09, 2002 - 09:50 pm:

On a slight tangent... what are grads used for? I think that there are 400 per complete revolution, but my maths teachers don't know anything about them.... are they used for certain engineering purposes or what?

By James Oldfield on Wednesday, January 09, 2002 - 10:32 pm:

I think grades are used because they are a decimal system, as opposed to degrees which is sexagesimal.

As you might know, what this means is that an angle can be written as x° y' z", where x is the integral number of degrees, y is the "minutes" (1/60 of a degree), and z is the "seconds" (1/60 of a minute). So for example 1/3 of 10° is 3° 20'.

On the other hand, 1/3 of 10grad is just 3.33grad (to 2dp). This decimal way of thinking is more modern, ie not in fitting with the imperial ratios of 12, 14 and other bizarre numbers. Of course grades aren't used very much in practice because we can just treat degrees as decimal anyway, as most people do.

Jim

By Yatir Halevi on Thursday, January 10, 2002 - 06:07 am:

Is it because, that only in radians the following identity exists:
sin ?

Yatir

By David Loeffler on Thursday, January 10, 2002 - 12:39 pm:

Yes, that's right.

David

By Philip Ellison on Thursday, January 10, 2002 - 02:13 pm:

Thanks James... do you know why angles are written using a sexagesimal system anyway? (I think that the Babylonians used base 60... might this have some relevance). This doesn't really have any importance... it'd just be interesting to know. Thanks

By Philip Ellison on Thursday, January 10, 2002 - 02:16 pm:

Why doesn't that identity hold in degrees (or am I being dense here)?

By Yatir Halevi on Thursday, January 10, 2002 - 02:56 pm:

Philip, Yes it were the Babylonians who uses base 60 (and I think maybe the Sumerians before them as well), 60 was an important number for them. And they started defining angles, they used 360 as the amount of degrees for a whole circle, then each degress is divides into 60 minutes and each minute into 60 seconds. They used that in Astronomical calculations as well and this is how it came into time as well.

The proof of this identity is based on the fact that a radian is defined as the angle made by to radii making an arc having a length of the radius.
If you whish, I'll be happy to supply the proof...

Yatir

By Philip Ellison on Thursday, January 10, 2002 - 04:53 pm:

Thanks... I'm assuming that it doesn't hold for degrees because I'm being stupid with infinite series (ie. just because they both tend to 0 doesn't mean that the fraction tends to 1, as 0/0 is undefined)

By Yatir Halevi on Thursday, January 10, 2002 - 05:35 pm:

It's ok, I'm not really good with them either...
Actually the proof of this doesn't include any knowledge in this kind of things...:
This proof is for when we reach 0 from the positive numbers but the following could be done to when it reaches from the negative numbers...

Look at the following segment of a circle:
segment

X is the angle in radians.
The radius of the circle (OA OB) is equal to 1
Both AD and BC make an angle of 90 degrees with OB.
It is easy to see that:

and because R=1 (radius):

and therefore:
sinx < x < tanx
divide by sinx> 0 and get:

and this is equal to:

since cos0=1,
lim x--> 0 (cosx) is also equal to 1.
So we get:

Thus proving our point.

Yatir

By Philip Ellison on Friday, January 11, 2002 - 07:52 pm:

Excellent proof Yatir... why use horrible infinite series when it can be proved far more elegantly using simple trig.!
However, there was one bit that I didn't understand: why does the proof show that the limit is greater than 1 AND less than 1? Surely that means that it can't exist. Unless, of course, this is valid for infinite series.
Thanks

By Brad Rodgers on Friday, January 11, 2002 - 07:55 pm:

Yatir's case only deals with

θ

being nonzero. As

θ

goes to zero, the greater than becomes greater than or equal to, and similarly, the less than becomes less than or equal to, therefore providing equality.
Brad

By Kerwin Hui on Friday, January 11, 2002 - 08:46 pm:

The reason why we use the infinite series definition is that it can be extended relatively easily (how would you define

\sin (z)

for complex

z

if you only have the opp/hyp definition?). It is far easier to start with the infinite series definition and work our way to prove that this implies the alternative definition than the other way round. Note that in Arun's proof by differential equation that we have put the infinite series 'under the carpet', when we use the property

$dy / dx = k y \Rightarrow y = C \exp (k x)$

We have implicitly assumed the properties of $\exp (x)$ , which of course, come from the infinite series definition

$\exp (x) = \sum_{r = 0}^{\infty} x^{r} / r!$

Further, we assume that the differential equation we considered has unique solution. This may not necessarily be the case. For example, consider the DE

$dy / dx = 3 y^{2} / 3$ , $y (0) = 0$

This DE has infinitely many solutions. You can check that $y = 0$ for all $x$ is a solution, and so is $y = x^{3}$ . In fact, we have

$y = (x - b)^{3}$ , $x < b$

$y = 0$ , $b \leq x \leq a$

$y = (x - a)^{3}$ , $x > a$

is a solution provided $b < 0 < a$ .

We can alternatively start with the definition

$y = \cos (x)$ is the unique solution to $y'' + y = 0$ , $y (0) = 1$ ;

$y = \sin (x)$ is the unique solution to $y'' + y = 0$ , $y (0) = 0$ .

and work our way up to prove that we have the result $e^{i x} = \cos (x) + i \sin (x)$ . (the fact that the solution is unique is, of course, provable by considering the integral of $y'^{2}$ .).

Kerwin