I'm working on some past exam questions, and I am struggling with the second part of this question:
Define the formal power series by the formulas:
$$\sin(z) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!}z^{2n-1}, \space \cos(z) = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!}z^{2n} $$
(a) Show $\exp(iz) = \cos(z) + i\sin(z)$
(b) Use this relation to derive expressions for the formal power series $\sin(kz)$ and $\cos(kz)$ in terms of $\sin(z)$ and $\cos(z)$ , both of which may occur in each expression and $k$ is a positive integer.
I think I have done the first part correctly (although I am not sure if it is stringent enough) but I am struggling with part (b), and would greatly appreciate some help.
Here is what I have so far:
Given the formulas for $\sin(z)$ and $\cos(z)$, I simply write out a few terms for both:
$$ \cos(z) = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - ...$$ and $$ i\sin(z) = iz - \frac{iz^3}{3!} + \frac{iz^5}{5!} - ...$$ Then adding these together, I get:
$$ \cos(z) + i\sin(z) = 1 + iz - \frac{z^2}{2!} - \frac{iz^3}{3!} + \frac{z^4}{4!} + \frac{iz^5}{5!} - ...$$ which can be re-written as:
$$ \cos(z) + i\sin(z) = 1 + iz + \frac{(iz)^2}{2!} + \frac{(iz)^3}{3!} + \frac{(iz)^4}{4!} + \frac{(iz)^5}{5!} + ...= \exp(iz)$$ Is this a valid way to show (a)?
EDIT:
For part(b), I have made some progress, thanks to the help below. But I don't think my answer is complete.
From (a), I have $$\exp(iz) = \cos(z) + i\sin(z) \implies \exp(ikz) = \exp(iz)^k = \cos(kz) + i\sin(kz) = (\cos(z) + i\sin(z))^k $$
Using the binomial theorem:
$$ \cos(kz) + i\sin(kz) = (\cos(z))^k + {k\choose1}(\cos(z))^{k-1}i\sin(z) + {k\choose2}(\cos(z))^{k-2}(i\sin(z))^2 +...+{k\choose{k-1}}(\cos(z))(i\sin(z))^{k-1} + (i\sin(z))^{k}$$
Collecting the coefficients gives me:
$$\cos(kz) = (\cos(z))^k - {k\choose2}(\cos(z))^{k-2}(\sin(z))^2 + {k\choose4}(\cos(z))^{k-4}(\sin(z))^4 -...$$ and
$$ \sin(kz) = {k\choose1}(\cos(z))^{k-1}(\sin(z)) - {k\choose3}(\cos(z))^{k-3}(\sin(z))^3 + {k\choose5}(\cos(z))^{k-5}(\sin(z))^5 - ... $$
Writing these in summation notation I get:
$$\cos(kz) = \sum_{n=0} (-1)^n {k\choose{2n}}(\cos(z))^{k-(2n)}(\sin(z))^{2n} $$ and
$$\sin(kz) = \sum_{n=0} (-1)^n {k\choose{2n+1}}(\cos(z))^{k-(2n+1)}(\sin(z))^{2n+1} $$
However, I am not sure what my upper limit summation should be? From testing a few numbers for $k$, it seems to $\lfloor{\frac{k}{2}}\rfloor$ But I'm not sure if this is correct, as I have never really seen this before.
Also, since it is supposed to be an expression for a formal power series (does that mean the limit of summation should be $\infty$? But then that doesn't really make sense with my expression?
Help!
The key is to note that we have the following: $$ \cos(kz) + i\sin(kz) = e^{ikz} = (e^{iz})^k = (\cos(z) + i\sin(z))^k $$ If you then look at the real and imaginary parts of this expression, you should be able to figure something out.
As for the fact that $e^{ikz} = (e^{iz})^k$:
Let us sketch a proof more generally that $e^{a+b}= e^ae^b$. Since it would then follow, e.g., that $e^{iz}e^{iz} = e^{iz + iz}$ as desired (for $k = 2$)
In this case:
$$ e^{a+b} = \sum_{k=0}^\infty \frac{(a+b)^k}{k!} \\ = \sum_{k=0}^\infty \frac{1}{k!}\sum_{m=0}^k{k \choose m}a^{k-m}b^m \\ = \sum_{k=0}^\infty \frac{1}{k!}\sum_{m=0}^k\frac{k!}{(k-m)!m!}a^{k-m}b^m \\ = \sum_{k=0}^\infty \sum_{m=0}^k\frac{1}{(k-m)!m!}a^{k-m}b^m $$
Now, take a look at the definition of the product of two power series, say $e^a$ and $e^b$. Using that, and a little bit of induction, the claim should follow.