Prove that $sqrt{(p-1)^2 p^2 + 4} = (p - 1)p + epsilon$ for $p > 1$ and $0 < epsilon < 1$.
$begingroup$
Here $p$ is prime but that does not affect the calculation. Assume that $p > 1$ is any such integer. Numerically at $p = 2$ we have $epsilon = 0.8...$ which is the maximum value of $epsilon$.
functions inequality elementary-functions
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
Here $p$ is prime but that does not affect the calculation. Assume that $p > 1$ is any such integer. Numerically at $p = 2$ we have $epsilon = 0.8...$ which is the maximum value of $epsilon$.
functions inequality elementary-functions
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
add a comment |
$begingroup$
Here $p$ is prime but that does not affect the calculation. Assume that $p > 1$ is any such integer. Numerically at $p = 2$ we have $epsilon = 0.8...$ which is the maximum value of $epsilon$.
functions inequality elementary-functions
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
$endgroup$
Here $p$ is prime but that does not affect the calculation. Assume that $p > 1$ is any such integer. Numerically at $p = 2$ we have $epsilon = 0.8...$ which is the maximum value of $epsilon$.
functions inequality elementary-functions
functions inequality elementary-functions
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
edited Dec 15 '18 at 22:12
Ovi
12.4k1040113
12.4k1040113
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
asked Dec 15 '18 at 21:45
Lorenz H MenkeLorenz H Menke
10411
10411
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You want to show that
$$sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$
Because both sides are guaranteed to be positive, we can square both sides with no problem
$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$
Subtracting
$$4 < 2p(p-1)+1$$
Rearranging
$$0 < 2p^2-2p-3$$
The quadratic on the RHS has zeroes at $approx -0.8$, and $approx1.8$, and it opens up. Therefore for $p ge 2$, the inequality holds. And this inequality is equivalent to your original one.
EDIT:
Oh and also you can show that $epsilon$ has to be positive because $p(p-1) = sqrt{p^2(p-1)^2} < sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
$endgroup$
add a comment |
$begingroup$
You have the inequality $$sqrt{1+x}le 1+frac x2$$
This can be shown easily since $1+xge 0implies 1+frac x2>0$ so we can compare their squares and this give $1+xle 1+x+frac{x^2}4$
Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<sqrt{y^2+a}le y+frac a{2y}$$
So the question is to find if $varepsilon=dfrac{a}{2y}<1iff y>dfrac a2$
For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$
and for $p=2$ just verify manually that $sqrt{8}-2approx 0.82<1$.
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You want to show that
$$sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$
Because both sides are guaranteed to be positive, we can square both sides with no problem
$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$
Subtracting
$$4 < 2p(p-1)+1$$
Rearranging
$$0 < 2p^2-2p-3$$
The quadratic on the RHS has zeroes at $approx -0.8$, and $approx1.8$, and it opens up. Therefore for $p ge 2$, the inequality holds. And this inequality is equivalent to your original one.
EDIT:
Oh and also you can show that $epsilon$ has to be positive because $p(p-1) = sqrt{p^2(p-1)^2} < sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
$endgroup$
add a comment |
$begingroup$
You want to show that
$$sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$
Because both sides are guaranteed to be positive, we can square both sides with no problem
$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$
Subtracting
$$4 < 2p(p-1)+1$$
Rearranging
$$0 < 2p^2-2p-3$$
The quadratic on the RHS has zeroes at $approx -0.8$, and $approx1.8$, and it opens up. Therefore for $p ge 2$, the inequality holds. And this inequality is equivalent to your original one.
EDIT:
Oh and also you can show that $epsilon$ has to be positive because $p(p-1) = sqrt{p^2(p-1)^2} < sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
$endgroup$
add a comment |
$begingroup$
You want to show that
$$sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$
Because both sides are guaranteed to be positive, we can square both sides with no problem
$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$
Subtracting
$$4 < 2p(p-1)+1$$
Rearranging
$$0 < 2p^2-2p-3$$
The quadratic on the RHS has zeroes at $approx -0.8$, and $approx1.8$, and it opens up. Therefore for $p ge 2$, the inequality holds. And this inequality is equivalent to your original one.
EDIT:
Oh and also you can show that $epsilon$ has to be positive because $p(p-1) = sqrt{p^2(p-1)^2} < sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
$endgroup$
You want to show that
$$sqrt{(p-1)^2 p^2 + 4} < (p - 1)p + 1$$
Because both sides are guaranteed to be positive, we can square both sides with no problem
$$(p-1)^2 p^2 + 4<(p-1)^2p^2 + 2p(p-1) +1$$
Subtracting
$$4 < 2p(p-1)+1$$
Rearranging
$$0 < 2p^2-2p-3$$
The quadratic on the RHS has zeroes at $approx -0.8$, and $approx1.8$, and it opens up. Therefore for $p ge 2$, the inequality holds. And this inequality is equivalent to your original one.
EDIT:
Oh and also you can show that $epsilon$ has to be positive because $p(p-1) = sqrt{p^2(p-1)^2} < sqrt{p^2(p-1)^2+4}$, so you need to add something positive to the LHS here to make them equal.
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
answered Dec 15 '18 at 21:58
OviOvi
12.4k1040113
12.4k1040113
add a comment |
add a comment |
$begingroup$
You have the inequality $$sqrt{1+x}le 1+frac x2$$
This can be shown easily since $1+xge 0implies 1+frac x2>0$ so we can compare their squares and this give $1+xle 1+x+frac{x^2}4$
Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<sqrt{y^2+a}le y+frac a{2y}$$
So the question is to find if $varepsilon=dfrac{a}{2y}<1iff y>dfrac a2$
For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$
and for $p=2$ just verify manually that $sqrt{8}-2approx 0.82<1$.
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
You have the inequality $$sqrt{1+x}le 1+frac x2$$
This can be shown easily since $1+xge 0implies 1+frac x2>0$ so we can compare their squares and this give $1+xle 1+x+frac{x^2}4$
Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<sqrt{y^2+a}le y+frac a{2y}$$
So the question is to find if $varepsilon=dfrac{a}{2y}<1iff y>dfrac a2$
For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$
and for $p=2$ just verify manually that $sqrt{8}-2approx 0.82<1$.
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
$endgroup$
add a comment |
$begingroup$
You have the inequality $$sqrt{1+x}le 1+frac x2$$
This can be shown easily since $1+xge 0implies 1+frac x2>0$ so we can compare their squares and this give $1+xle 1+x+frac{x^2}4$
Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<sqrt{y^2+a}le y+frac a{2y}$$
So the question is to find if $varepsilon=dfrac{a}{2y}<1iff y>dfrac a2$
For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$
and for $p=2$ just verify manually that $sqrt{8}-2approx 0.82<1$.
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
$endgroup$
You have the inequality $$sqrt{1+x}le 1+frac x2$$
This can be shown easily since $1+xge 0implies 1+frac x2>0$ so we can compare their squares and this give $1+xle 1+x+frac{x^2}4$
Similarly (for $a>0$) factorizing $y$ inside the square root gives $$y<sqrt{y^2+a}le y+frac a{2y}$$
So the question is to find if $varepsilon=dfrac{a}{2y}<1iff y>dfrac a2$
For your problem $y=p(p-1)$ and $a=4$ so we need $p(p-1)>2$ which is indeed true for $p>2$
and for $p=2$ just verify manually that $sqrt{8}-2approx 0.82<1$.
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
edited Dec 15 '18 at 23:44
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
answered Dec 15 '18 at 23:38
zwimzwim
12.5k831
12.5k831
add a comment |
add a comment |
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