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Looking for some help understanding the following concept. I know when you are trying to determine the stability of fixed points for the system,

$$x'=\sin x$$

You proceed with the following steps,

$$x'=0$$ $$\sin x=0$$ $$x^{*}=k\pi$$

deriving $x'$ to determine stability would give,

$$\cos x$$

plugging in the fixed point,

$$\cos (k\pi)$$

Therefore, when $k\pi$ is even it is equal to $1$, so it would be unstable. On the other hand when $k\pi$ is odd it is equal to $-1$ so it would be stable.

However what if you were given $$x'=\cos x$$.

The derivative would be $-\sin x$ and the fixed points would still be $k\pi$ but when $k\pi$ is odd and when it is even, is it stable or unstable?

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    @Arnaldo thanks for the edit would you be able to help me with my problem? – 2017-02-15
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    I don't know about stability but taking to consideration what you wrote, in the second case the fixed point is $x^*=\frac{\pi}{2}+k\pi$ (that happens when $x'=\cos x=0$). Am I right? – 2017-02-15
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    yes that seems to be correct – 2017-02-15
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    So, $-\sin (\frac{\pi}{2}+k\pi)$ is $-1$ (the system is stable) when $k$ is even and $1$ (the system is unstable) when $k$ is odd. – 2017-02-15
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    okay so if that was just $sin$ instead of $-sin$ would the system then be unstable when k is even and stable when it is odd? – 2017-02-15
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    Taking to consideration what you said about stable and unstable, I think so! – 2017-02-15
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    awesome thanks for your help – 2017-02-15

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