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# Harmonics

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

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

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
7 months, 3 weeks 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. · 7 months, 2 weeks 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 $$1$$. · 7 months, 2 weeks ago

Of course it does change the property, but how can you say that it's bounded just by looking at the signs? · 7 months, 2 weeks ago