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How to prove $e^t-1 > (e-1) t^2$.
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I want to prove
$$e^t-1 ge (e-1) t^2, qquad tin[0,1].$$
For $tin[0,1]$, we have $t^n le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$
calculus inequality
edited Jul 27 at 1:50
Nosrati
19.2k41544
asked Jul 26 at 6:42
Birendra Singh
1036
add a comment |Â
up vote
0
down vote
favorite
I want to prove
$$e^t-1 ge (e-1) t^2, qquad tin[0,1].$$
For $tin[0,1]$, we have $t^n le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$
calculus inequality
edited Jul 27 at 1:50
Nosrati
19.2k41544
asked Jul 26 at 6:42
Birendra Singh
1036
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove
$$e^t-1 ge (e-1) t^2, qquad tin[0,1].$$
For $tin[0,1]$, we have $t^n le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$
calculus inequality
edited Jul 27 at 1:50
Nosrati
19.2k41544
asked Jul 26 at 6:42
Birendra Singh
1036
I want to prove
$$e^t-1 ge (e-1) t^2, qquad tin[0,1].$$
For $tin[0,1]$, we have $t^n le t^m$ whenever $m>n.$ I want to use this inequality in the proof of above statement but unable to proceed for terms beyond $t^2.$
calculus inequality
edited Jul 27 at 1:50
Nosrati
19.2k41544
asked Jul 26 at 6:42
Birendra Singh
1036
edited Jul 27 at 1:50
Nosrati
19.2k41544
edited Jul 27 at 1:50
Nosrati
19.2k41544
edited Jul 27 at 1:50
Nosrati
19.2k41544
19.2k41544
asked Jul 26 at 6:42
Birendra Singh
1036
asked Jul 26 at 6:42
Birendra Singh
1036
asked Jul 26 at 6:42
Birendra Singh
1036
1036
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
1
down vote
accepted
The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=big(exp(t)-1big)-(texte-1),t^2=sum_k=1^infty,fract^kk!-sum_k=1^infty,fract^2k!=(t-t^2)-sum_k=3^infty,fract^2(1-t^k-2)k!$$
for $tin(0,1)$.
That is,
$$g(t):=fracf(t)t(1-t)=1-sum_k=3^infty,fract,left(1+t+t^2+ldots+t^k-3right)k!> 1-sum_k=3^infty,frac(k-2)k!$$
for every $tin(0,1)$.
Now,
$$sum_k=3^infty,frac(k-2)k!=sum_k=3^infty,frac1(k-1)!-2,sum_k=3^infty,frac1k!=left(texte-2right)-2left(texte-frac52right)=3-texte,.$$
That is,
$$g(t)> 1-(3-texte)=texte-2>0,.$$
The claim immediately follows. In fact, this proof gives a stronger inequality:
$$exp(t)geq 1+ (texte-1),t^2+(texte-2),t(1-t)=1+(texte-2),t+t^2text for every tgeq 0,.$$
answered Jul 26 at 7:30
Batominovski
23k22777
add a comment |Â
up vote
1
down vote
Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $tin[0,1]$,
$$f''(t)=e^t-2(e-1)leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $tin[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+tcdot 1)geq f(0)+tcdot f(1)=0.$$
answered Jul 26 at 6:51
Robert Z
83.9k954122
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
add a comment |Â
up vote
1
down vote
Proof
Denote $$f(x)=x^t,~~~~~t in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(xi)(e-1),~~~xi in (1,e).$$
Namely, $$e^t-1=txi^t-1(e-1).$$
Thus, $$xi^t-1 geq e^t-1 geq tgeq 0,$$which is desired.
answered Jul 26 at 12:14
mengdie1982
2,835216
add a comment |Â
up vote
0
down vote
Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=frac1pmsqrt1-4z2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^frac1pmsqrt1-4z2-1-(e-1)left(frac1pmsqrt1-4z2right)^2$$
seems to have a Taylor development with positive-only coefficients.
But this needs to be proven.
answered Jul 26 at 7:01
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
This is only a hint, because it’s not difficult to explain the rest.
So, the tangent on $,displaystylefrace^t-1t,$ in $,t=1,$ is $,e-2+t,$ .
We get $,displaystylefrace^t-1tgeq e-2+t geq (e-1)t,$ .
answered Jul 26 at 9:59
user90369
7,606925
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=big(exp(t)-1big)-(texte-1),t^2=sum_k=1^infty,fract^kk!-sum_k=1^infty,fract^2k!=(t-t^2)-sum_k=3^infty,fract^2(1-t^k-2)k!$$
for $tin(0,1)$.
That is,
$$g(t):=fracf(t)t(1-t)=1-sum_k=3^infty,fract,left(1+t+t^2+ldots+t^k-3right)k!> 1-sum_k=3^infty,frac(k-2)k!$$
for every $tin(0,1)$.
Now,
$$sum_k=3^infty,frac(k-2)k!=sum_k=3^infty,frac1(k-1)!-2,sum_k=3^infty,frac1k!=left(texte-2right)-2left(texte-frac52right)=3-texte,.$$
That is,
$$g(t)> 1-(3-texte)=texte-2>0,.$$
The claim immediately follows. In fact, this proof gives a stronger inequality:
$$exp(t)geq 1+ (texte-1),t^2+(texte-2),t(1-t)=1+(texte-2),t+t^2text for every tgeq 0,.$$
answered Jul 26 at 7:30
Batominovski
23k22777
add a comment |Â
up vote
1
down vote
accepted
The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=big(exp(t)-1big)-(texte-1),t^2=sum_k=1^infty,fract^kk!-sum_k=1^infty,fract^2k!=(t-t^2)-sum_k=3^infty,fract^2(1-t^k-2)k!$$
for $tin(0,1)$.
That is,
$$g(t):=fracf(t)t(1-t)=1-sum_k=3^infty,fract,left(1+t+t^2+ldots+t^k-3right)k!> 1-sum_k=3^infty,frac(k-2)k!$$
for every $tin(0,1)$.
Now,
$$sum_k=3^infty,frac(k-2)k!=sum_k=3^infty,frac1(k-1)!-2,sum_k=3^infty,frac1k!=left(texte-2right)-2left(texte-frac52right)=3-texte,.$$
That is,
$$g(t)> 1-(3-texte)=texte-2>0,.$$
The claim immediately follows. In fact, this proof gives a stronger inequality:
$$exp(t)geq 1+ (texte-1),t^2+(texte-2),t(1-t)=1+(texte-2),t+t^2text for every tgeq 0,.$$
answered Jul 26 at 7:30
Batominovski
23k22777
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=big(exp(t)-1big)-(texte-1),t^2=sum_k=1^infty,fract^kk!-sum_k=1^infty,fract^2k!=(t-t^2)-sum_k=3^infty,fract^2(1-t^k-2)k!$$
for $tin(0,1)$.
That is,
$$g(t):=fracf(t)t(1-t)=1-sum_k=3^infty,fract,left(1+t+t^2+ldots+t^k-3right)k!> 1-sum_k=3^infty,frac(k-2)k!$$
for every $tin(0,1)$.
Now,
$$sum_k=3^infty,frac(k-2)k!=sum_k=3^infty,frac1(k-1)!-2,sum_k=3^infty,frac1k!=left(texte-2right)-2left(texte-frac52right)=3-texte,.$$
That is,
$$g(t)> 1-(3-texte)=texte-2>0,.$$
The claim immediately follows. In fact, this proof gives a stronger inequality:
$$exp(t)geq 1+ (texte-1),t^2+(texte-2),t(1-t)=1+(texte-2),t+t^2text for every tgeq 0,.$$
answered Jul 26 at 7:30
Batominovski
23k22777
The inequality is trivially an equality for $t=0$ or $t=1$. Define $$f(t):=big(exp(t)-1big)-(texte-1),t^2=sum_k=1^infty,fract^kk!-sum_k=1^infty,fract^2k!=(t-t^2)-sum_k=3^infty,fract^2(1-t^k-2)k!$$
for $tin(0,1)$.
That is,
$$g(t):=fracf(t)t(1-t)=1-sum_k=3^infty,fract,left(1+t+t^2+ldots+t^k-3right)k!> 1-sum_k=3^infty,frac(k-2)k!$$
for every $tin(0,1)$.
Now,
$$sum_k=3^infty,frac(k-2)k!=sum_k=3^infty,frac1(k-1)!-2,sum_k=3^infty,frac1k!=left(texte-2right)-2left(texte-frac52right)=3-texte,.$$
That is,
$$g(t)> 1-(3-texte)=texte-2>0,.$$
The claim immediately follows. In fact, this proof gives a stronger inequality:
$$exp(t)geq 1+ (texte-1),t^2+(texte-2),t(1-t)=1+(texte-2),t+t^2text for every tgeq 0,.$$
answered Jul 26 at 7:30
Batominovski
23k22777
edited Jul 26 at 15:44
answered Jul 26 at 7:30
Batominovski
23k22777
answered Jul 26 at 7:30
Batominovski
23k22777
answered Jul 26 at 7:30
Batominovski
23k22777
23k22777
add a comment |Â
add a comment |Â
up vote
1
down vote
Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $tin[0,1]$,
$$f''(t)=e^t-2(e-1)leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $tin[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+tcdot 1)geq f(0)+tcdot f(1)=0.$$
answered Jul 26 at 6:51
Robert Z
83.9k954122
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
add a comment |Â
up vote
1
down vote
Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $tin[0,1]$,
$$f''(t)=e^t-2(e-1)leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $tin[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+tcdot 1)geq f(0)+tcdot f(1)=0.$$
answered Jul 26 at 6:51
Robert Z
83.9k954122
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $tin[0,1]$,
$$f''(t)=e^t-2(e-1)leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $tin[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+tcdot 1)geq f(0)+tcdot f(1)=0.$$
answered Jul 26 at 6:51
Robert Z
83.9k954122
Alternative (and straightforward) approach. Note that $f(t)=e^t-1 -(e-1) t^2$ is concave in $[0,1]$ because for $tin[0,1]$,
$$f''(t)=e^t-2(e-1)leq e-2e+2=-e+2<0$$
and $f(0)=f(1)=0$. Hence, for $tin[0,1]$,
$$e^t-1 -(e-1) t^2=f(t)=f(0+tcdot 1)geq f(0)+tcdot f(1)=0.$$
answered Jul 26 at 6:51
Robert Z
83.9k954122
edited Jul 26 at 7:32
answered Jul 26 at 6:51
Robert Z
83.9k954122
answered Jul 26 at 6:51
Robert Z
83.9k954122
answered Jul 26 at 6:51
Robert Z
83.9k954122
83.9k954122
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
add a comment |Â
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
This is also the approach I would select, but the OP is explicitely asking about another one.
– Did
Jul 26 at 7:04
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
@Did Yes, I did not notice this OP's request.
– Robert Z
Jul 26 at 7:07
add a comment |Â
up vote
1
down vote
Proof
Denote $$f(x)=x^t,~~~~~t in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(xi)(e-1),~~~xi in (1,e).$$
Namely, $$e^t-1=txi^t-1(e-1).$$
Thus, $$xi^t-1 geq e^t-1 geq tgeq 0,$$which is desired.
answered Jul 26 at 12:14
mengdie1982
2,835216
add a comment |Â
up vote
1
down vote
Proof
Denote $$f(x)=x^t,~~~~~t in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(xi)(e-1),~~~xi in (1,e).$$
Namely, $$e^t-1=txi^t-1(e-1).$$
Thus, $$xi^t-1 geq e^t-1 geq tgeq 0,$$which is desired.
answered Jul 26 at 12:14
mengdie1982
2,835216
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Proof
Denote $$f(x)=x^t,~~~~~t in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(xi)(e-1),~~~xi in (1,e).$$
Namely, $$e^t-1=txi^t-1(e-1).$$
Thus, $$xi^t-1 geq e^t-1 geq tgeq 0,$$which is desired.
answered Jul 26 at 12:14
mengdie1982
2,835216
Proof
Denote $$f(x)=x^t,~~~~~t in [0,1].$$
Then by Lagrange's Mean Value Theorem, we have $$f(e)-f(1)=f'(xi)(e-1),~~~xi in (1,e).$$
Namely, $$e^t-1=txi^t-1(e-1).$$
Thus, $$xi^t-1 geq e^t-1 geq tgeq 0,$$which is desired.
answered Jul 26 at 12:14
mengdie1982
2,835216
answered Jul 26 at 12:14
mengdie1982
2,835216
answered Jul 26 at 12:14
mengdie1982
2,835216
answered Jul 26 at 12:14
mengdie1982
2,835216
2,835216
add a comment |Â
add a comment |Â
up vote
0
down vote
Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=frac1pmsqrt1-4z2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^frac1pmsqrt1-4z2-1-(e-1)left(frac1pmsqrt1-4z2right)^2$$
seems to have a Taylor development with positive-only coefficients.
But this needs to be proven.
answered Jul 26 at 7:01
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=frac1pmsqrt1-4z2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^frac1pmsqrt1-4z2-1-(e-1)left(frac1pmsqrt1-4z2right)^2$$
seems to have a Taylor development with positive-only coefficients.
But this needs to be proven.
answered Jul 26 at 7:01
Yves Daoust
111k665203
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=frac1pmsqrt1-4z2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^frac1pmsqrt1-4z2-1-(e-1)left(frac1pmsqrt1-4z2right)^2$$
seems to have a Taylor development with positive-only coefficients.
But this needs to be proven.
answered Jul 26 at 7:01
Yves Daoust
111k665203
Hint:
By the transformation
$$x(1-x)=z,$$ or
$$x=frac1pmsqrt1-4z2$$ we fold the axis and merge the two roots at $z=0$.
Then the two branches of the function
$$f(z):=e^frac1pmsqrt1-4z2-1-(e-1)left(frac1pmsqrt1-4z2right)^2$$
seems to have a Taylor development with positive-only coefficients.
But this needs to be proven.
answered Jul 26 at 7:01
Yves Daoust
111k665203
edited Jul 26 at 8:30
answered Jul 26 at 7:01
Yves Daoust
111k665203
answered Jul 26 at 7:01
Yves Daoust
111k665203
answered Jul 26 at 7:01
Yves Daoust
111k665203
111k665203
add a comment |Â
add a comment |Â
up vote
0
down vote
This is only a hint, because it’s not difficult to explain the rest.
So, the tangent on $,displaystylefrace^t-1t,$ in $,t=1,$ is $,e-2+t,$ .
We get $,displaystylefrace^t-1tgeq e-2+t geq (e-1)t,$ .
answered Jul 26 at 9:59
user90369
7,606925
add a comment |Â
up vote
0
down vote
This is only a hint, because it’s not difficult to explain the rest.
So, the tangent on $,displaystylefrace^t-1t,$ in $,t=1,$ is $,e-2+t,$ .
We get $,displaystylefrace^t-1tgeq e-2+t geq (e-1)t,$ .
answered Jul 26 at 9:59
user90369
7,606925
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is only a hint, because it’s not difficult to explain the rest.
So, the tangent on $,displaystylefrace^t-1t,$ in $,t=1,$ is $,e-2+t,$ .
We get $,displaystylefrace^t-1tgeq e-2+t geq (e-1)t,$ .
answered Jul 26 at 9:59
user90369
7,606925
This is only a hint, because it’s not difficult to explain the rest.
So, the tangent on $,displaystylefrace^t-1t,$ in $,t=1,$ is $,e-2+t,$ .
We get $,displaystylefrace^t-1tgeq e-2+t geq (e-1)t,$ .
answered Jul 26 at 9:59
user90369
7,606925
answered Jul 26 at 9:59
user90369
7,606925
answered Jul 26 at 9:59
user90369
7,606925
answered Jul 26 at 9:59
user90369
7,606925
7,606925
add a comment |Â
add a comment |Â
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