# Power Series

1. Oct 27, 2013

### zzmanzz

1. The problem statement, all variables and given/known data

Given:

$$sin(x) = \frac{e^{ix}-e^{-ix}}{2}$$

Show that $$sin(x)$$ can be written as:

$$sin(x) = \sum_{n=0}^n \frac{x^{(2n+1)}}{(2n+1)!}$$

2. Relevant equations

$$e^x = \sum_{n=0}^n \frac{x^{n}}{(n)!}$$

3. The attempt at a solution

I'm unsure how to treat the imaginary number in order to get a power expansion for $$e^{ix}$$

I know that it's not the same as $$e^{ax}$$ where a is just a constant. Any help on the complex part would be greatly appreciated!

Last edited: Oct 27, 2013
2. Oct 27, 2013

### Ray Vickson

How is it different from $e^{ax}$ with $a = i$? More to the point: who told you it is different?

3. Oct 27, 2013

### zzmanzz

I'm just going by the fundamental law for imaginary multiplication. i.e.

$$i = \sqrt{-1}$$
$$i^2 = -1$$
$$i^3 = -i$$
$$i^4 = 1$$

Therefore, if a = i, the expansion has to follow the above pattern?

4. Oct 27, 2013

### zzmanzz

Wait, I guess I never got the formula for

$$e^{ax}$$ in the first place now that I think about it. Good catch. Sorry about that!

Is $$e^{ax} = [\sum_{n=0}^n \frac{(ax)^{n}}{(n)!}]$$ correct?

Last edited: Oct 27, 2013
5. Oct 27, 2013

### zzmanzz

I think it's

$$e^{ix} = \sum_{n=0}^n \frac{(ix)^{n}}{(n)!}$$

therefore,

$$\frac{e^{ix}}{2} = \frac{1}{2}\sum_{n=0}^n \frac{(ix)^{n}}{(n)!}$$

$$\frac{e^{-ix}}{2} = \frac{1}{2}\sum_{n=0}^n \frac{(-ix)^{n}}{(n)!}$$

Then I show that:

$$\frac{1}{2}\sum_{n=0}^n \frac{(ix)^{n}}{(n)!} - \frac{1}{2}\sum_{n=0}^n \frac{(-ix)^{n}}{(n)!} = \sum_{n=0}^n \frac{x^{(2n+1)}}{(2n+1)!}$$

Right?

6. Oct 27, 2013

### Dick

You would do that except you've got some bad information in the problem statement. sin(x)=(e^(ix)-e^(-ix))/(2i) not over 2. And there should be an alternating sign in the series. Or are you really trying to expand the hyperbolic sine, sinh(x)?

7. Oct 28, 2013

### zzmanzz

Sorry, it's late and I'm missing the details.

He writes that

$$sin(x) = \frac{e^{ix} - e^{-ix}}{2}$$

can be represented in power expanded form as:

$$sin(x) = \sum_{n=0}^n (-1)^n \frac{x^{(2n+1)}}{(2n+1)!}$$

so you were sight, I was missing the alternating part in the final formula. I looked it over twice and there is no 2i, just 2? Maybe he made a mistake?

8. Oct 28, 2013

### Dick

Yes, he made a mistake. e^(ix)=cos(x)+i*sin(x) and e^(-ix)=exp(i*(-x))=cos(-x)+i*sin(-x)=cos(x)-i*sin(x). So e^(ix)-e^(-ix)=(2i)*sin(x).

9. Oct 28, 2013

### zzmanzz

According to his notes: $$cosh(x) = \frac{e^{x} - e ^{-x}}{2}$$

He didn't give us sinh(x), or cos(x). Just assigned the sin(x) as I stated above with the alternating addition. It has to be 2i to cancel out in order to get the final formula.

I've been trying to cancel/combine the expansion terms for 45 mins now and couldn't get his formula.

I will bring it to his attention. Thank you!!

Last edited: Oct 28, 2013
10. Oct 28, 2013

### Dick

The cosh formula is wrong too. cosh(x)=(e^x+e^(-x))/2. Either whoever is writing the notes or transcibing the notes has some attention problems.

11. Oct 28, 2013

### zzmanzz

Wow, I'm looking at his pdf and he has cosh(x) written once in your format, and then once again with the -. This is a 3 hour lecture and he was talking for 2 hrs and then just rushed through the last part in 45 mins.

One more thing, is his formula for cos(x) = (e^(ix) + e(-ix))/2 right? Or should this also have 2i in the denominator? I will bring all of this to his attention tomorrow.

12. Oct 28, 2013

### Dick

Ah, at last a correct formula. Yes, that one works.

13. Oct 28, 2013

### zzmanzz

Thank you,

So, just to summarize:

$$sin(x) = \frac{e^{ix}-e^{-ix}}{2i}$$

$$sinh(x) = \frac{e^{x}-e^{-x}}{2}$$

$$cos(x) = \frac{e^{ix}+e^{-ix}}{2}$$

$$cosh(x) = \frac{e^{x}+e^{-x}}{2}$$

Thanks!

14. Oct 28, 2013

### Dick

Sure, and $$sinh(x) = \frac{e^{x}-e^{-x}}{2}$$ if you want to complete the list. The sin and cos formulas follow from deMoivre's expression $e^{ix}=cos(x)+isin(x)$, you don't have memorize them all you can derive them.