Solving for sin(x) with Complex Numbers

[zzmanzz]

Homework Statement

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)!}

Homework Equations

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

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!

 

[Ray Vickson]

Homework Statement

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)!}

Homework Equations

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

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!

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

 

[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?

 

[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?

 

[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?

 

[Dick]

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?

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)?

 

[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?

 

[Dick]

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?

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).

 

[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!

 

[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.

 

[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.

 

[Dick]

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.

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

 

[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!

 

[Dick]

Thank you,

So, just to summarize:

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

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

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

Thanks!

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.