Derivative of the Prime Counting Function
0
Can it be sufficient to say that $$pi'(x)=sum_{p=mathrm{primes}}delta(x-p)$$
since there is a spike in $pi(x)$ at every prime and does not grow at any other value of $x$?
calculus prime-numbers dirac-delta
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
add a comment |
0
Can it be sufficient to say that $$pi'(x)=sum_{p=mathrm{primes}}delta(x-p)$$
since there is a spike in $pi(x)$ at every prime and does not grow at any other value of $x$?
calculus prime-numbers dirac-delta
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
2
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39
add a comment |
0
0
0
Can it be sufficient to say that $$pi'(x)=sum_{p=mathrm{primes}}delta(x-p)$$
since there is a spike in $pi(x)$ at every prime and does not grow at any other value of $x$?
calculus prime-numbers dirac-delta
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
Can it be sufficient to say that $$pi'(x)=sum_{p=mathrm{primes}}delta(x-p)$$
since there is a spike in $pi(x)$ at every prime and does not grow at any other value of $x$?
calculus prime-numbers dirac-delta
calculus prime-numbers dirac-delta
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
edited Nov 29 at 12:33
Davide Giraudo
125k16150259
125k16150259
asked Nov 29 '17 at 3:42
aleden
1,725411
asked Nov 29 '17 at 3:42
aleden
1,725411
asked Nov 29 '17 at 3:42
aleden
1,725411
1,725411
2
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39
add a comment |
2
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39
2
2
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39
add a comment |
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if ${p_n}_{ninmathbb{N}}$ is the sequence of primes, $pi$ is constant in each interval $(p_n, p_{n+1})$. Therefore, in these intervals, $pi' equiv 0$. In the rest of the real line, that is, on each prime $p_n$, $pi$ is not differentiable since it is not even continuous:
$$
lim_{x to p_{n+1}^-} pi(x) = n neq n+1 = lim_{x to p_{n+1}^+} pi(x)
$$
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
add a comment |
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if ${p_n}_{ninmathbb{N}}$ is the sequence of primes, $pi$ is constant in each interval $(p_n, p_{n+1})$. Therefore, in these intervals, $pi' equiv 0$. In the rest of the real line, that is, on each prime $p_n$, $pi$ is not differentiable since it is not even continuous:
$$
lim_{x to p_{n+1}^-} pi(x) = n neq n+1 = lim_{x to p_{n+1}^+} pi(x)
$$
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
add a comment |
0
if ${p_n}_{ninmathbb{N}}$ is the sequence of primes, $pi$ is constant in each interval $(p_n, p_{n+1})$. Therefore, in these intervals, $pi' equiv 0$. In the rest of the real line, that is, on each prime $p_n$, $pi$ is not differentiable since it is not even continuous:
$$
lim_{x to p_{n+1}^-} pi(x) = n neq n+1 = lim_{x to p_{n+1}^+} pi(x)
$$
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
add a comment |
0
0
0
if ${p_n}_{ninmathbb{N}}$ is the sequence of primes, $pi$ is constant in each interval $(p_n, p_{n+1})$. Therefore, in these intervals, $pi' equiv 0$. In the rest of the real line, that is, on each prime $p_n$, $pi$ is not differentiable since it is not even continuous:
$$
lim_{x to p_{n+1}^-} pi(x) = n neq n+1 = lim_{x to p_{n+1}^+} pi(x)
$$
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
if ${p_n}_{ninmathbb{N}}$ is the sequence of primes, $pi$ is constant in each interval $(p_n, p_{n+1})$. Therefore, in these intervals, $pi' equiv 0$. In the rest of the real line, that is, on each prime $p_n$, $pi$ is not differentiable since it is not even continuous:
$$
lim_{x to p_{n+1}^-} pi(x) = n neq n+1 = lim_{x to p_{n+1}^+} pi(x)
$$
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
answered Nov 29 '17 at 3:50
Guido A.
7,2351730
7,2351730
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
add a comment |
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
Ive seen the Dirac Delta function used to assign derivatives to discontinuous functions though, such as the Heaviside step function. I was wondering if it could be applied in this situation to.
– aleden
Nov 29 '17 at 3:55
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
I think I might have misinterpreted your background. I'm not knowledgeable in the Dirac delta nor measure theory, so I can't answer that. In the simplest sense possible, which my answer is referring to, $pi' equiv 0$ almost everywhere. If you were to be able to generalize this to $mathbb{R}$, you should be able to extend $pi$ to a continuous function. Again, I'm not to experienced on the subject, but $pi$ is not extendable to a continuous function, so your notion of differentiabilty should be at the very least not the standard one.
– Guido A.
Nov 29 '17 at 4:06
add a comment |
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2
Only if you were a physicist, in which case I wonder why you're interested.
– Chaitanya Tappu
Nov 29 '17 at 3:50
Just curious, i was gonna see if I could derive an integral representation of $pi(x)$.
– aleden
Nov 29 '17 at 3:52
But if there were any reasonable integral representation (using the Lebesgue measure on $mathbb R$), $pi (x)$ would have to be absolutely continuous, which it is not. If you don't know measure theory, suffice it to say that indefinite integrals are continuous, but $pi (x)$ is not.
– Chaitanya Tappu
Nov 29 '17 at 3:56
I see, but could it suffice to say that the distributional derivative of $pi(x)$ is equivalent to the sum? Sort of how the Heaviside step function's distributional derivative is the delta function.
– aleden
Nov 29 '17 at 3:58
Sure the distributional derivative of $f(x) = sum_{n=1}^x a_n$ is $sum_{n=1}^infty a_n delta(x-n)$. Note $zeta(s)$ is the Laplace transform of $sum_{n=1}^infty a_n delta(color{red}{u-ln n})$ and the Mellin transform of $sum_{n=1}^infty color{red}{frac{1}{n}}, delta(x-1/n)$.
– reuns
Nov 29 '17 at 10:39