Equation with prime numbers
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I am reading a paper which, in order to prove a result about graphs, states that if $p$ is a prime, then $$frac{1}{2}(p^2-1)(p-1) approx frac{1}{2} (p^2-1)^{3/2}.$$
In other words,
$$frac{1}{2}(p^2-1)(p-1) = bigg(frac{1}{2}-o(1)bigg)(p^2-1)^{3/2}.$$
This is stated without further explanation, but I can't see why it is true. Can anyone help?
Thank you in advance.
number-theory prime-numbers
asked Nov 16 at 13:40
Ilefen
465215
add a comment |
up vote
0
down vote
favorite
I am reading a paper which, in order to prove a result about graphs, states that if $p$ is a prime, then $$frac{1}{2}(p^2-1)(p-1) approx frac{1}{2} (p^2-1)^{3/2}.$$
In other words,
$$frac{1}{2}(p^2-1)(p-1) = bigg(frac{1}{2}-o(1)bigg)(p^2-1)^{3/2}.$$
This is stated without further explanation, but I can't see why it is true. Can anyone help?
Thank you in advance.
number-theory prime-numbers
asked Nov 16 at 13:40
Ilefen
465215
1
Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
1
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am reading a paper which, in order to prove a result about graphs, states that if $p$ is a prime, then $$frac{1}{2}(p^2-1)(p-1) approx frac{1}{2} (p^2-1)^{3/2}.$$
In other words,
$$frac{1}{2}(p^2-1)(p-1) = bigg(frac{1}{2}-o(1)bigg)(p^2-1)^{3/2}.$$
This is stated without further explanation, but I can't see why it is true. Can anyone help?
Thank you in advance.
number-theory prime-numbers
asked Nov 16 at 13:40
Ilefen
465215
I am reading a paper which, in order to prove a result about graphs, states that if $p$ is a prime, then $$frac{1}{2}(p^2-1)(p-1) approx frac{1}{2} (p^2-1)^{3/2}.$$
In other words,
$$frac{1}{2}(p^2-1)(p-1) = bigg(frac{1}{2}-o(1)bigg)(p^2-1)^{3/2}.$$
This is stated without further explanation, but I can't see why it is true. Can anyone help?
Thank you in advance.
number-theory prime-numbers
number-theory prime-numbers
asked Nov 16 at 13:40
Ilefen
465215
asked Nov 16 at 13:40
Ilefen
465215
asked Nov 16 at 13:40
Ilefen
465215
asked Nov 16 at 13:40
Ilefen
465215
asked Nov 16 at 13:40
Ilefen
465215
465215
1
Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
1
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05
add a comment |
1
Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
1
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05
1
1
Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
1
1
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
$$lim_{p rightarrow infty}frac{p-1}{sqrt{p^2-1}}=lim_{p rightarrow infty}frac{1}{frac{p}{sqrt{p^2-1}}}=lim_{p rightarrow infty}sqrt{1-frac{1}{p^2}}=1$$
Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.
answered Nov 16 at 13:45
asdf
3,669519
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
add a comment |
up vote
2
down vote
Dividing by $(p^2-1)$ yields $(p-1) approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} approx (p+1)^{1/2}$.
answered Nov 16 at 13:46
Stockfish
41726
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$$lim_{p rightarrow infty}frac{p-1}{sqrt{p^2-1}}=lim_{p rightarrow infty}frac{1}{frac{p}{sqrt{p^2-1}}}=lim_{p rightarrow infty}sqrt{1-frac{1}{p^2}}=1$$
Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.
answered Nov 16 at 13:45
asdf
3,669519
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
add a comment |
up vote
2
down vote
$$lim_{p rightarrow infty}frac{p-1}{sqrt{p^2-1}}=lim_{p rightarrow infty}frac{1}{frac{p}{sqrt{p^2-1}}}=lim_{p rightarrow infty}sqrt{1-frac{1}{p^2}}=1$$
Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.
answered Nov 16 at 13:45
asdf
3,669519
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
add a comment |
up vote
2
down vote
up vote
2
down vote
$$lim_{p rightarrow infty}frac{p-1}{sqrt{p^2-1}}=lim_{p rightarrow infty}frac{1}{frac{p}{sqrt{p^2-1}}}=lim_{p rightarrow infty}sqrt{1-frac{1}{p^2}}=1$$
Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.
answered Nov 16 at 13:45
asdf
3,669519
$$lim_{p rightarrow infty}frac{p-1}{sqrt{p^2-1}}=lim_{p rightarrow infty}frac{1}{frac{p}{sqrt{p^2-1}}}=lim_{p rightarrow infty}sqrt{1-frac{1}{p^2}}=1$$
Use this to see where the $o(1)$ comes from. Also, note that I've never used that is $p$ prime.
answered Nov 16 at 13:45
asdf
3,669519
answered Nov 16 at 13:45
asdf
3,669519
answered Nov 16 at 13:45
asdf
3,669519
answered Nov 16 at 13:45
asdf
3,669519
3,669519
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
add a comment |
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
Thank you and that's actually a very good point. The proof generally uses the fact that $p$ is a prime, but I guess the assumption is not needed in this specific step. My bad.
– Ilefen
Nov 16 at 14:02
add a comment |
up vote
2
down vote
Dividing by $(p^2-1)$ yields $(p-1) approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} approx (p+1)^{1/2}$.
answered Nov 16 at 13:46
Stockfish
41726
add a comment |
up vote
2
down vote
Dividing by $(p^2-1)$ yields $(p-1) approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} approx (p+1)^{1/2}$.
answered Nov 16 at 13:46
Stockfish
41726
add a comment |
up vote
2
down vote
up vote
2
down vote
Dividing by $(p^2-1)$ yields $(p-1) approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} approx (p+1)^{1/2}$.
answered Nov 16 at 13:46
Stockfish
41726
Dividing by $(p^2-1)$ yields $(p-1) approx ((p-1)(p+1))^{1/2}$, i.e. $ (p-1)^{1/2} approx (p+1)^{1/2}$.
answered Nov 16 at 13:46
Stockfish
41726
answered Nov 16 at 13:46
Stockfish
41726
answered Nov 16 at 13:46
Stockfish
41726
answered Nov 16 at 13:46
Stockfish
41726
41726
add a comment |
add a comment |
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Stop playing with prime numbers,you are not a kid anymore xd
– GK A
Nov 16 at 13:44
Hey, I am trying to prove the statement in order to construct a $K_{2,2}$ graph, which is a VERY grownup thing to do!!
– Ilefen
Nov 16 at 13:51
It would help to have some additional context here showing what the paper does with the asserted approximation, because both sides are also approximately equal to simply ${1over2}p^3$ (for large $p$). My guess is that the expression ${1over2}(p^2-1)^{3/2}$ gets used to simplify some other expression.
– Barry Cipra
Nov 16 at 13:54
Well, roughly speaking there is a theorem which states that, given a graph on $n$ vertices, there exists a constant $c$ such that the upper bound for a specific property of the graph is $cn^{3/2}$. What follows is an example of the theorem for $n=(p^2-1)$, in which we try to prove that $c=1/2$. I'm not sure how all this is helpful though...
– Ilefen
Nov 16 at 14:00
1
It helps explain why the paper would both to state the approximation as they did. Basically, when you're working with large numbers, the highest power dominates everything else, so all the $-1$'s don't really matter.
– Barry Cipra
Nov 16 at 14:05