Proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$
I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$
Obviously
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$
so if I can show
$$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
inequality
asked Nov 29 at 23:14
WillG
46038
add a comment |
I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$
Obviously
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$
so if I can show
$$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
inequality
asked Nov 29 at 23:14
WillG
46038
add a comment |
I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$
Obviously
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$
so if I can show
$$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
inequality
asked Nov 29 at 23:14
WillG
46038
I'm looking for a simple proof that $frac{1+x^2}{n^2} geq 1-e^{-x^2/n^2}$ for all $x,n in mathbb{R}$.
My first attempt was to express the exponential as a Taylor series:
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2}-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... , .$$
Obviously
$$frac{1+x^2}{n^2} geq frac{x^2}{n^2},$$
so if I can show
$$-frac{1}{2!}frac{x^4}{n^4}+frac{1}{3!}frac{x^6}{n^6}- , ... <0,$$
then I'm done. But I'm stuck here, and also wondering if there's an even simpler way.
inequality
inequality
asked Nov 29 at 23:14
WillG
46038
asked Nov 29 at 23:14
WillG
46038
asked Nov 29 at 23:14
WillG
46038
asked Nov 29 at 23:14
WillG
46038
asked Nov 29 at 23:14
WillG
46038
46038
add a comment |
add a comment |
3 Answers
3
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oldest
votes
It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$
$$Leftrightarrow$$
$$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$
answered Nov 29 at 23:29
Rebellos
14.3k31245
add a comment |
For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
add a comment |
Set $y=x^2/n^2$. Then you want to show that
$$
frac{1}{n^2}+yge 1-e^{-y}
$$
Note that $yge0$. A standard process is to consider
$$
f(y)=frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
$$
f(y)>0
$$
for $yge0$.
answered Nov 29 at 23:41
egreg
177k1484200
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
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active
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votes
It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$
$$Leftrightarrow$$
$$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$
answered Nov 29 at 23:29
Rebellos
14.3k31245
add a comment |
It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$
$$Leftrightarrow$$
$$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$
answered Nov 29 at 23:29
Rebellos
14.3k31245
add a comment |
It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$
$$Leftrightarrow$$
$$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$
answered Nov 29 at 23:29
Rebellos
14.3k31245
It is $e^t geq t + 1$ for all $t in mathbb R$ (easily proven by elementary calculus), thus :
$$e^{-x^2/n^2} geq 1 - x^2/n^2$$
But $1/n^2 >0$ for all $n in mathbb R$, thus if you add it to the LHS it will still be greater or equal to the RHS :
$$e^{-x^2/n^2} + 1/n^2 geq 1 - x^2/n^2 Leftrightarrow x^2/n^2 + 1/n^2 geq 1-e^{-x^2/n^2} $$
$$Leftrightarrow$$
$$boxed{frac{1+x^2}{n^2} geq 1 - e^{-frac{x^2}{n^2}}}$$
answered Nov 29 at 23:29
Rebellos
14.3k31245
edited Nov 29 at 23:34
answered Nov 29 at 23:29
Rebellos
14.3k31245
answered Nov 29 at 23:29
Rebellos
14.3k31245
answered Nov 29 at 23:29
Rebellos
14.3k31245
14.3k31245
add a comment |
add a comment |
For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
add a comment |
For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
add a comment |
For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
For any $x geq 0$ we have $1-e^{-x} leq x$ $,,$ (1). Hence $1-e^{-x^{2}/n^{2}} leq x^{2}/n^{2}$ which gives the inequality you want. To prove (1) consider $1-e^{-x} -x$. Its derivative is $e^{-x}-1$ which is negative. Since the function vanishes at $0$ and is decreasing it must be $leq 0$ on $[0,infty)$.
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
answered Nov 29 at 23:29
Kavi Rama Murthy
49.3k31854
49.3k31854
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
add a comment |
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
+1 by me since we answered at the same time and the answer is as precise !
– Rebellos
Nov 29 at 23:37
add a comment |
Set $y=x^2/n^2$. Then you want to show that
$$
frac{1}{n^2}+yge 1-e^{-y}
$$
Note that $yge0$. A standard process is to consider
$$
f(y)=frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
$$
f(y)>0
$$
for $yge0$.
answered Nov 29 at 23:41
egreg
177k1484200
add a comment |
Set $y=x^2/n^2$. Then you want to show that
$$
frac{1}{n^2}+yge 1-e^{-y}
$$
Note that $yge0$. A standard process is to consider
$$
f(y)=frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
$$
f(y)>0
$$
for $yge0$.
answered Nov 29 at 23:41
egreg
177k1484200
add a comment |
Set $y=x^2/n^2$. Then you want to show that
$$
frac{1}{n^2}+yge 1-e^{-y}
$$
Note that $yge0$. A standard process is to consider
$$
f(y)=frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
$$
f(y)>0
$$
for $yge0$.
answered Nov 29 at 23:41
egreg
177k1484200
Set $y=x^2/n^2$. Then you want to show that
$$
frac{1}{n^2}+yge 1-e^{-y}
$$
Note that $yge0$. A standard process is to consider
$$
f(y)=frac{1}{n^2}+y-1+e^{-y}
$$
and note that $f(0)=1/n^2>0$. Also
$$
f'(y)=1-e^{-y}=frac{e^y-1}{e^y}>0
$$
for $y>0$. Therefore the function $f$ is strictly increasing over $[0,infty)$ and so
$$
f(y)>0
$$
for $yge0$.
answered Nov 29 at 23:41
egreg
177k1484200
answered Nov 29 at 23:41
egreg
177k1484200
answered Nov 29 at 23:41
egreg
177k1484200
answered Nov 29 at 23:41
egreg
177k1484200
177k1484200
add a comment |
add a comment |
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