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P: 3,243
 Quote by dimension10 If you are talking about the prime number thing, I am not sure about the exact one but here is an approximation: $$\pi(n)\approx \int_{2}^{n}\frac{dt}{\mbox{ln}(t)}$$ $$\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\mbox{ln}(t)}$$ $$\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{\delta t}{\mbox{ln}(t)}$$ As the derivative of a sum is the sum of the derivatives, $$\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}$$ $$\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}-\frac{\delta t}{n \; {\mbox{ln}}^{2}(n)}$$ $$\frac{d}{dn}\pi(n)\approx - \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)}$$ So that is the approximate rate of change of the pi function of t as t changes.
I believe the problem is here:
$$\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}$$

The derivative operator $$\frac{d}{dn}$$ should operate on the sum.
I am not even sure what it means to have delta t inside a sum where t is a dumby variable, and then taking a limit of it as it approaches zero, the notation here is quite problematic.