(Taylor's theorem) Proving that $\sin(x) = \sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$

Question

I'm starting a class on Advanced Mathematics I next semester and I found a sheet of the class'es 2012 final exams, so I'm slowly trying to solve the exercises in it or find the general layout. I will be posting a lot of questions with the exercises I find challenging, and I would like to ask for any help or methodologies that will make it easier for me to solve.

I understand the whole ordeal is categorized as "homework", but any assistance would be appreciated, as I'm completely clueless and I would like to be prepared.

The following exercise is $Ex. 10$, graded for $10\%$.

Define Taylor's Theorem and prove that $\sin(x) = \sum\limits_{n=0}^{\infty}\dfrac{(-1)^{n}x^{2n+1}}{(2n+1)!}$.

From the class'es page, I've got the definition of Taylor's theorem along with the following equation that resembles a Sum:

$$f(x)=f(x_0)+\frac{f'(x_0)}{1!}(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+\frac{f'''(x_0)}{3!}(x-x_0)^3+\cdots+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n+\frac{f^{(n+1)}(x_0)}{(n+1)!}(x-x_0)^{n+1}$$

Apart from the similarity, I do not know how to continue. Do I take the second part of the equation and try to find a closed type that leads to sin(x)?

Answer 1

Notice that

$$f(x) = sin x \ \ \ \ \ \ \ \ f(0)= 0\\ f'(x) = cos x \ \ \ \ \ \ f'(0)=1 \\ f''(x) = -sinx \ \ f''(0)=0 \\ f'''(x) = -cosx \ \ f'''(0) = -1 \\ f^{(4)}(x)=sinx \ \ \ f(0)=0 $$

The derivatives repeat in a cycle of four, from Taylor series definition we have that $$\sum_{n=0}^{\infty} \frac{f^{(n)}(a)(x-a)^{n}}{n!}$$ Take $a=0$ and the result follows.

Answer 2

HINTS/GUIDANCE

You need to know that if you differentiate $\sin x$ you get $\cos x$, and if you differentiate $\cos x$ you get $-\sin x$. So the derivatives toggle backwards and forwards between $\cos x$ and $\sin x$ but you need care with the minus signs.

Obviously take $x_0=0$. The answer clearly has powers of $x$, not powers of $(x-x_0)$ and anyway the lhs has no visible $x_0$, so $x_0$ must be $0$.

And of course you need to know that $\sin 0=0,\cos 0=1$.

Answer 3

Hint

Consider $\cos(x)+i\sin(x)=e^{ix}$. Use the Taylor expansion of $e^y$ around $y=0$; in the result, replace $y$ by $ix$ and collect together the real and imaginary parts. So, knowing $e^y$, you have both Taylor expansions ($\cos(x)$) and ($\sin(x)$).

Added later

$$e^y=\sum_{k=0}^\infty \frac{y^k}{k!}$$ $$e^{ix}=\sum_{k=0}^\infty \frac{(ix)^k}{k!}=\sum_{k=0}^\infty \frac{i^k x^k}{k!}=\sum_{k=0}^\infty \frac{i^k x^k}{k!}$$ Now spplit the summation over the odd and even values of $k$. So,$$e^{ix}=\sum_{n=0}^\infty \frac{i^{2n} x^{2n}}{2n!}+\sum_{n=0}^\infty \frac{i^{2n+1} x^{2n+1}}{(2n+1)!}=\sum_{n=0}^\infty \frac{i^{2n} x^{2n}}{2n!}+i\sum_{n=0}^\infty \frac{i^{2n} x^{2n+1}}{(2n+1)!}$$ $$e^{ix}=\sum_{n=0}^\infty \frac{(-1)^{n} x^{2n}}{2n!}+i\sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}=\cos(x)+i \sin(x)$$