inequality with power function
$begingroup$
Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$
Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)
inequality
edited Nov 27 '18 at 19:47
user376343
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
$endgroup$
add a comment |
$begingroup$
Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$
Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)
inequality
edited Nov 27 '18 at 19:47
user376343
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
$endgroup$
$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16
add a comment |
$begingroup$
Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$
Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)
inequality
edited Nov 27 '18 at 19:47
user376343
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
$endgroup$
Hello. Let $0<bleq1$ and $n in mathbb{N}$ with $ngeq1$, does this inequality hold?
$(n-1)^b leq n^b-1$
Ive tried quite a few combinations of $n$ and $b$ and it seems to be true but I dont know how to prove it (if its true)
inequality
inequality
edited Nov 27 '18 at 19:47
user376343
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
edited Nov 27 '18 at 19:47
user376343
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
edited Nov 27 '18 at 19:47
user376343
3,4033826
edited Nov 27 '18 at 19:47
user376343
3,4033826
edited Nov 27 '18 at 19:47
user376343
3,4033826
3,4033826
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
asked Nov 27 '18 at 19:29
StefanWKStefanWK
727
727
$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16
add a comment |
$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16
$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
add a comment |
$begingroup$
The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
add a comment |
$begingroup$
The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
$endgroup$
The inequality is actually the other way around. You can consider the function
$$f(x)=(x-1)^b-x^b+1$$
well-defined for every $xgeq 1$. Then
$$f'(x)=b(x-1)^{b-1}-bx^{b-1}$$
and since $xgeq 1$ we have $0leq x-1leq x$ so that $f'(x)geq 0$ for all $xgeq 1$ (here we use $b-1leq 0$) Hence $f$ is increasing. Moreover
$$f(1)=0$$
so that $f(x)geq 0$ for every $xgeq 1$ that is
$$(x-1)^bgeq x^b-1$$
for all $xgeq 1$, in particular if $xinmathbb{N}$.
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
edited Nov 27 '18 at 21:48
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
answered Nov 27 '18 at 19:38
Olivier MoschettaOlivier Moschetta
2,8311411
2,8311411
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
add a comment |
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
$begingroup$
Either my above counter-example or your proof must be wrong...
$endgroup$
– Martin R
Nov 27 '18 at 19:48
1
1
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
Olivier your idea is good, but you maybe forgot that $b-1$ is negative. You might want to fix you proof starting the sign of $f'.$
$endgroup$
– user376343
Nov 27 '18 at 19:51
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
$begingroup$
True enough, I'll change it!
$endgroup$
– Olivier Moschetta
Nov 27 '18 at 21:47
add a comment |
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$begingroup$
For $b=0.5$ and $n=2$ you get $sqrt 1 le sqrt 2 - 1$, which is wrong.
$endgroup$
– Martin R
Nov 27 '18 at 19:47
$begingroup$
It is the way around, see math.stackexchange.com/questions/1990936/….
$endgroup$
– Martin R
Nov 27 '18 at 20:00
$begingroup$
alright thanks for your help
$endgroup$
– StefanWK
Nov 27 '18 at 20:16