If you are talking about the prime number thing, I am not sure about the exact one but here is an approximation: [tex]\pi(n)\approx \int_{2}^{n}\frac{dt}{\mbox{ln}(t)}[/tex] [tex]\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\mbox{ln}(t)}[/tex] [tex]\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{\delta t}{\mbox{ln}(t)}[/tex] As the derivative of a sum is the sum of the derivatives, [tex]\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}[/tex] [tex]\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}-\frac{\delta t}{n \; {\mbox{ln}}^{2}(n)}[/tex] [tex]\frac{d}{dn}\pi(n)\approx - \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)} [/tex] So that is the approximate rate of change of the pi function of t as t changes.
This is not quite right. The proper derivative is: [tex]\frac{d}{dn}\pi(n)\approx \frac{d}{dn}\int_{2}^{n}\frac{dt}{\ln(t)} = \frac{1}{\ln(n)}[/tex]
According to Wolfram Alpha, [tex] \frac{d}{dn}(\int_{2}^{n}\frac{dt}{\ln(t)})=\frac{-n+n\; \mbox{ln}(n)+2}{n \;{\mbox{ln}}^{2}(n)}[/tex] that is, [tex] {\mbox{ln}}^{-1}(n)+(2{n}^{-1}{\mbox{ln}}^{-2}(n))-{\mbox{ln}}^{-2}(n)[/tex] But for Li(n) alone, it gives [tex]{\mbox{ln}}^{-1}(n)[/tex] which is your solution. And my solution gives [tex]- \int_{2}^{n}\frac{dt}{n \; {\mbox{ln}}^{2}(n)}[/tex] which is equal to [tex]\frac{n-2}{n\; {\mbox{ln}}^{2}(n)}[/tex] I guess all three solutions are equal to each other and thus, correct?
I believe the problem is here: [tex]\frac{d}{dn}\pi(n)\approx \lim_{\delta t \rightarrow 0}\sum_{t=2}^{n}\frac{d}{dn}\frac{\delta t}{\mbox{ln}(t)}[/tex] The derivative operator [tex]\frac{d}{dn}[/tex] 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.
As MathematicalPhysicist already said, the first mistake is when you moved d/dn to the other side of the summation symbol. This is not allowed, because the summation is dependent on n. You made another mistake when you differentiated the expression dependent on t with respect to n. Since the expression is not dependent on n, the result is zero. How did you get WolframAlpha to say that? I do not get that. The three solutions are not equal to each other, so they cannot all be correct.
Thanks. I forgot about that. Oops. I think I know what happened. It must have again considered d as constant rather than an infinitesimal.There seems to be a simpler solution using the second fundamental theorem of calculus and that would yield [tex]\frac{1}{\mbox{ln}(n)}[/tex] which is the answer given by you.
the derivative of the prime counting function is just the sum [tex] \delta (x-p) [/tex] taken over all primes 'p'
I am confused; pi is a step function and is therefore only differentiable at the "interior" of a step at which point it is 0.
yes but since the step function is discontinous delta function appear whenever the function has discontinuties, in the case of Pi function the discontinuities are located at the prime numbers