# Harmonics

$$\sin(x)$$ and $$\cos(x)$$ are harmonic functions.Their series can be given as,

\begin{aligned} \sin x &=& x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!} + \cdots = \sum_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!} \\ \cos x &=& 1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \dfrac{x^6}{6!} + \cdots = \sum_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!} \end{aligned}

Here, we can see that both the above series have maximum magnitude -: $1$. The speciality of both the functions is that there is $+$ and $-$ signs alternatively. Thats why for any big or small values of $x$, both the above series converges to $1$ in magnitude.

Now, if we look at the expansion of $e^{(-x)}$,

$e^{(-x)} = 1-x+\dfrac{x^2}{2!}-\dfrac{x^3}{3!}+\dfrac{x^4}{4!}-\dfrac{x^5}{5!}+...$

Here also the same situation occurs. But why it is not a function like $sine$ and $cos$? Why its value is not repeated like $sine$ and $cos$? Note by Akash Shukla
4 years, 4 months ago

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The problem is that $\sin x$ and $\cos x$ do not converge, they are just bounded between $[-1,1]$.
Proving this (and periodicity) is not direct from the series expansions, irrespective of the $+$ and $-$ signs.
$e^{-x}$ is still a function, however.

- 4 years, 4 months ago

Thank you, Sir. I think the power changes the property. As in $sin$ and $cos$ the powers are odd and even, so even for any higher value of $x$, due to + and -, it will be bounded to $-1$ to $1$.

- 4 years, 4 months ago

Of course it does change the property, but how can you say that it's bounded just by looking at the signs?

- 4 years, 4 months ago

If + and - are not alternate or if + or - sign are repeated after two terms then I think it will not repeat its value.

- 4 years, 4 months ago