Mixing
proof of following inequality
$begingroup$
could some one prove the following inequality,I think it is based on Taylor's expansion but I don't how to start
$$1-x geq e^{-x-x^2} $$
and also for what value of x does it hold ?
another question is that I know that $1-x geq e^{-x}$, so is the following also true $1-x geq e^{-x-x^2-x^3}$ so how do we prove such inequalities?
inequality
edited Jan 17 at 8:59
Bernard
121k740116
asked Jan 17 at 8:41
viruviru
107214
$endgroup$
add a comment |
$begingroup$
could some one prove the following inequality,I think it is based on Taylor's expansion but I don't how to start
$$1-x geq e^{-x-x^2} $$
and also for what value of x does it hold ?
another question is that I know that $1-x geq e^{-x}$, so is the following also true $1-x geq e^{-x-x^2-x^3}$ so how do we prove such inequalities?
inequality
edited Jan 17 at 8:59
Bernard
121k740116
asked Jan 17 at 8:41
viruviru
107214
$endgroup$
$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49
add a comment |
$begingroup$
could some one prove the following inequality,I think it is based on Taylor's expansion but I don't how to start
$$1-x geq e^{-x-x^2} $$
and also for what value of x does it hold ?
another question is that I know that $1-x geq e^{-x}$, so is the following also true $1-x geq e^{-x-x^2-x^3}$ so how do we prove such inequalities?
inequality
edited Jan 17 at 8:59
Bernard
121k740116
asked Jan 17 at 8:41
viruviru
107214
$endgroup$
could some one prove the following inequality,I think it is based on Taylor's expansion but I don't how to start
$$1-x geq e^{-x-x^2} $$
and also for what value of x does it hold ?
another question is that I know that $1-x geq e^{-x}$, so is the following also true $1-x geq e^{-x-x^2-x^3}$ so how do we prove such inequalities?
inequality
inequality
edited Jan 17 at 8:59
Bernard
121k740116
asked Jan 17 at 8:41
viruviru
107214
edited Jan 17 at 8:59
Bernard
121k740116
asked Jan 17 at 8:41
viruviru
107214
edited Jan 17 at 8:59
Bernard
121k740116
edited Jan 17 at 8:59
Bernard
121k740116
edited Jan 17 at 8:59
Bernard
121k740116
121k740116
asked Jan 17 at 8:41
viruviru
107214
asked Jan 17 at 8:41
viruviru
107214
asked Jan 17 at 8:41
viruviru
107214
107214
$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49
add a comment |
$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49
$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49
$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$:
$$f'(x)=(-1-2x)e^{-x-x^2},quad f''(x)=bigl(-2+(-1-2x)^2bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$
so $f''(x)<0$ between the roots of $4x^2+4x-1$, $;frac{1pmsqrt 2}2$.
Therefore $;1-x>e^{-x-x^2}$ for all $xinbigl(frac{1-sqrt 2}2,0bigr)cupbigl(0,frac{1+sqrt 2}2bigr) $.
answered Jan 17 at 9:35
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
Both inequalities they are wrong. Try $x=1$.
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
add a comment |
$begingroup$
Equality leads to transcendental equations, which will require numerical solutions.
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$:
$$f'(x)=(-1-2x)e^{-x-x^2},quad f''(x)=bigl(-2+(-1-2x)^2bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$
so $f''(x)<0$ between the roots of $4x^2+4x-1$, $;frac{1pmsqrt 2}2$.
Therefore $;1-x>e^{-x-x^2}$ for all $xinbigl(frac{1-sqrt 2}2,0bigr)cupbigl(0,frac{1+sqrt 2}2bigr) $.
answered Jan 17 at 9:35
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$:
$$f'(x)=(-1-2x)e^{-x-x^2},quad f''(x)=bigl(-2+(-1-2x)^2bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$
so $f''(x)<0$ between the roots of $4x^2+4x-1$, $;frac{1pmsqrt 2}2$.
Therefore $;1-x>e^{-x-x^2}$ for all $xinbigl(frac{1-sqrt 2}2,0bigr)cupbigl(0,frac{1+sqrt 2}2bigr) $.
answered Jan 17 at 9:35
BernardBernard
121k740116
$endgroup$
add a comment |
$begingroup$
$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$:
$$f'(x)=(-1-2x)e^{-x-x^2},quad f''(x)=bigl(-2+(-1-2x)^2bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$
so $f''(x)<0$ between the roots of $4x^2+4x-1$, $;frac{1pmsqrt 2}2$.
Therefore $;1-x>e^{-x-x^2}$ for all $xinbigl(frac{1-sqrt 2}2,0bigr)cupbigl(0,frac{1+sqrt 2}2bigr) $.
answered Jan 17 at 9:35
BernardBernard
121k740116
$endgroup$
$y=1-x$ is the equation of the tangent at $(0,1)$. Now the function $f(x)=mathrm e^{-x-x^2}$ is concave in a neighbourhood of $x=0$:
$$f'(x)=(-1-2x)e^{-x-x^2},quad f''(x)=bigl(-2+(-1-2x)^2bigr)e^{-x-x^2}=(4x^2+4x-1)e^{-x-x^2},$$
so $f''(x)<0$ between the roots of $4x^2+4x-1$, $;frac{1pmsqrt 2}2$.
Therefore $;1-x>e^{-x-x^2}$ for all $xinbigl(frac{1-sqrt 2}2,0bigr)cupbigl(0,frac{1+sqrt 2}2bigr) $.
answered Jan 17 at 9:35
BernardBernard
121k740116
answered Jan 17 at 9:35
BernardBernard
121k740116
answered Jan 17 at 9:35
BernardBernard
121k740116
answered Jan 17 at 9:35
BernardBernard
121k740116
121k740116
add a comment |
add a comment |
$begingroup$
Both inequalities they are wrong. Try $x=1$.
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
add a comment |
$begingroup$
Both inequalities they are wrong. Try $x=1$.
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
add a comment |
$begingroup$
Both inequalities they are wrong. Try $x=1$.
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
$endgroup$
Both inequalities they are wrong. Try $x=1$.
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
answered Jan 17 at 8:56
Michael RozenbergMichael Rozenberg
105k1892197
105k1892197
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
add a comment |
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
no , but $1-x geq e^{-x-x^2}$ holds for x= $frac{1}{4}$
$endgroup$
– viru
Jan 17 at 9:10
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
$begingroup$
@viru We need to prove these inequalities and they are wrong.
$endgroup$
– Michael Rozenberg
Jan 17 at 10:13
add a comment |
$begingroup$
Equality leads to transcendental equations, which will require numerical solutions.
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Equality leads to transcendental equations, which will require numerical solutions.
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
Equality leads to transcendental equations, which will require numerical solutions.
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
$endgroup$
Equality leads to transcendental equations, which will require numerical solutions.
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 9:27
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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$begingroup$
"another question is that I know that $1−xgeq e^{−x}$" I don't think this is true.
$endgroup$
– denklo
Jan 17 at 8:49