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

sinx=xx33!+x55!x77!+=n=0(1)nx2n+1(2n+1)!cosx=1x22!+x44!x66!+=n=0(1)nx2n(2n)! \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 -: 11. The speciality of both the functions is that there is ++ and - signs alternatively. Thats why for any big or small values of xx, both the above series converges to 11 in magnitude.

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

e(x)=1x+x22!x33!+x44!x55!+...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 sinesine and coscos? Why its value is not repeated like sinesine and coscos?

Note by Akash Shukla
5 years, 2 months ago

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

Ameya Daigavane - 5 years, 2 months ago

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Thank you, Sir. I think the power changes the property. As in sinsin and coscos the powers are odd and even, so even for any higher value of xx, due to + and -, it will be bounded to 1-1 to 11.

Akash Shukla - 5 years, 1 month ago

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

Ameya Daigavane - 5 years, 1 month ago

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

Akash Shukla - 5 years, 1 month ago

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