What causes delta functions to appear in the derivative of the Pi function?

In summary: The second fundamental theorem of calculus states that the derivative of a function at a point is equal to the slope of the tangent to the graph at that point.
  • #1
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What is the derivative of the pi function
 
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  • #2
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.
 
  • #3
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]
 
  • #4
I like Serena said:
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]

Then where did I make a mistake?
 
  • #5
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?
 
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  • #6
dimension10 said:
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.

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.
 
  • #7
dimension10 said:
Then where did I make a mistake?

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.
dimension10 said:
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]

How did you get WolframAlpha to say that?
I do not get that.
dimension10 said:
I guess all three solutions are equal to each other and thus, correct?

The three solutions are not equal to each other, so they cannot all be correct.
 
  • #8
I like Serena said:
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.

Thanks. I forgot about that.

I like Serena said:
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.

Oops.
I like Serena said:
How did you get WolframAlpha to say that?
I do not get that.
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.
 
  • #9
the derivative of the prime counting function is just the sum

[tex] \delta (x-p) [/tex] taken over all primes 'p'
 
  • #10
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.
 
  • #11
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
 

1. What is the derivative of the pi function?

The derivative of the pi function is equal to zero, as it is a constant value and does not change with respect to any variable.

2. Can the pi function be differentiated?

No, the pi function cannot be differentiated as it is a constant value and does not have a variable to differentiate with respect to.

3. What is the significance of the derivative of the pi function?

The derivative of the pi function is significant in calculus and mathematics as it represents the rate of change of a constant value. It also helps in solving problems related to motion and optimization.

4. How does the derivative of the pi function relate to the area of a circle?

The derivative of the pi function is related to the area of a circle through the formula for finding the area of a circle: A = πr². The derivative of this formula is 2πr, which is the circumference of the circle. This shows that the derivative of the pi function is the rate of change of the area of a circle with respect to its radius.

5. Is the derivative of the pi function used in real-life applications?

Yes, the derivative of the pi function is used in real-life applications such as engineering, physics, and economics. It helps in solving problems related to optimization, motion, and rate of change.

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