\(\sin(x)\) 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
1 year, 11 months ago
Easy Math Editor
*italics*or_italics_**bold**or__bold__paragraph 1
paragraph 2
[example link](https://brilliant.org)> This is a quote# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"2 \times 32^{34}a_{i-1}\frac{2}{3}\sqrt{2}\sum_{i=1}^3\sin \theta\boxed{123}Comments
Sort by:
Top NewestThe 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.
Log in to reply
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 \(1\).
Log in to reply
Of course it does change the property, but how can you say that it's bounded just by looking at the signs?
Log in to reply
If + and - are not alternate or if + or - sign are repeated after two terms then I think it will not repeat its value.
Log in to reply