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Prove the inequality $frac{e^x+e^{-x}}{2} leq e^{x^2/2}$ for all real numbers $x$. [duplicate]
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This question already has an answer here:
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
3 answers
How do I prove what's written in the title? I was able to get an incomplete proof for the case $x>2$. Here's my try: Use $e^x = sum_{j=1}^{infty} frac{x^j}{j!}$.
Now we can see that if $x$ is a real number, then: $$e^{x^2/2}-e^x/2-e^{-x}/2=sum_{j=1}^infty frac{frac{x^{2j}}{2^{j-1}}-x^j-(-x)^j}{2j!}$$ and we need to show this is positive. Now, if $j$ is odd, then the numerator is just $x^{2j}/2^{j-1}$ which is always greater than zero. But if $j$ is even, multiply the numerator by $2^{j-1}$, and check if the result is positive. The result is $x^{2j} -2^{j} cdot x^j=x^{j}(x^j-2^j)$ So if $x$ is greater then $2$, we get a positive result, but otherwise, we don't. So how do I continue for other values of $x$?
real-analysis inequality taylor-expansion
asked Jan 21 at 18:01
OmerOmer
3619
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marked as duplicate by Martin R, Michael Rozenberg inequality
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Jan 21 at 21:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
3 answers
How do I prove what's written in the title? I was able to get an incomplete proof for the case $x>2$. Here's my try: Use $e^x = sum_{j=1}^{infty} frac{x^j}{j!}$.
Now we can see that if $x$ is a real number, then: $$e^{x^2/2}-e^x/2-e^{-x}/2=sum_{j=1}^infty frac{frac{x^{2j}}{2^{j-1}}-x^j-(-x)^j}{2j!}$$ and we need to show this is positive. Now, if $j$ is odd, then the numerator is just $x^{2j}/2^{j-1}$ which is always greater than zero. But if $j$ is even, multiply the numerator by $2^{j-1}$, and check if the result is positive. The result is $x^{2j} -2^{j} cdot x^j=x^{j}(x^j-2^j)$ So if $x$ is greater then $2$, we get a positive result, but otherwise, we don't. So how do I continue for other values of $x$?
real-analysis inequality taylor-expansion
asked Jan 21 at 18:01
OmerOmer
3619
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marked as duplicate by Martin R, Michael Rozenberg inequality
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Jan 21 at 21:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21
add a comment |
$begingroup$
This question already has an answer here:
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
3 answers
How do I prove what's written in the title? I was able to get an incomplete proof for the case $x>2$. Here's my try: Use $e^x = sum_{j=1}^{infty} frac{x^j}{j!}$.
Now we can see that if $x$ is a real number, then: $$e^{x^2/2}-e^x/2-e^{-x}/2=sum_{j=1}^infty frac{frac{x^{2j}}{2^{j-1}}-x^j-(-x)^j}{2j!}$$ and we need to show this is positive. Now, if $j$ is odd, then the numerator is just $x^{2j}/2^{j-1}$ which is always greater than zero. But if $j$ is even, multiply the numerator by $2^{j-1}$, and check if the result is positive. The result is $x^{2j} -2^{j} cdot x^j=x^{j}(x^j-2^j)$ So if $x$ is greater then $2$, we get a positive result, but otherwise, we don't. So how do I continue for other values of $x$?
real-analysis inequality taylor-expansion
asked Jan 21 at 18:01
OmerOmer
3619
$endgroup$
This question already has an answer here:
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
3 answers
How do I prove what's written in the title? I was able to get an incomplete proof for the case $x>2$. Here's my try: Use $e^x = sum_{j=1}^{infty} frac{x^j}{j!}$.
Now we can see that if $x$ is a real number, then: $$e^{x^2/2}-e^x/2-e^{-x}/2=sum_{j=1}^infty frac{frac{x^{2j}}{2^{j-1}}-x^j-(-x)^j}{2j!}$$ and we need to show this is positive. Now, if $j$ is odd, then the numerator is just $x^{2j}/2^{j-1}$ which is always greater than zero. But if $j$ is even, multiply the numerator by $2^{j-1}$, and check if the result is positive. The result is $x^{2j} -2^{j} cdot x^j=x^{j}(x^j-2^j)$ So if $x$ is greater then $2$, we get a positive result, but otherwise, we don't. So how do I continue for other values of $x$?
This question already has an answer here:
Proving that $frac{e^x + e^{-x}}2 le e^{x^2/2}$
3 answers
real-analysis inequality taylor-expansion
real-analysis inequality taylor-expansion
asked Jan 21 at 18:01
OmerOmer
3619
asked Jan 21 at 18:01
OmerOmer
3619
asked Jan 21 at 18:01
OmerOmer
3619
asked Jan 21 at 18:01
OmerOmer
3619
asked Jan 21 at 18:01
OmerOmer
3619
3619
marked as duplicate by Martin R, Michael Rozenberg inequality
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Jan 21 at 21:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin R, Michael Rozenberg inequality
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Jan 21 at 21:11
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21
add a comment |
$begingroup$
Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21
$begingroup$
Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21
$begingroup$
Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21
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5 Answers
5
active
oldest
votes
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Your computation is flawed. The expansion of $cosh{x}$ is $sum_{j geq 0}{frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $sum_{j geq 0}{frac{x^{2j}}{2^j cdot j!}}$.
So you just need to prove that $j!2^j < (2j)!$ for each $j$.
answered Jan 21 at 18:04
MindlackMindlack
4,885211
$endgroup$
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
|
show 1 more comment
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Let $f(x)=frac{x^2}{2}-lnfrac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)geq0$ for $xgeq0.$
We see that $$f'(x)=x-frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}geq0.$$
Id est, $$f'(x)geq f'(0)=0,$$
$$f(x)geq f(0)=0$$ and we are done!
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
$endgroup$
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
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– Omer
Jan 21 at 18:35
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
add a comment |
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Notice:
$$frac12(e^x+e^{-x})<frac12(e^x+1)$$
and for $x>2.11$ that $e^frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
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Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^xgeq 1+x$ is a little weak here. However the inequality
$$e^xgeq 1+x+frac{x^2}{2}$$
is of help here.
It suffices to show
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}+frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
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Noting that $cosh x = frac{e^x + e^{-x}}{2}$, you may proceed as follows:
- Note that $cosh (-x) = cosh x$ and $e^{frac{(-x)^2}{2}} = e^{frac{x^2}{2}}$
$ Rightarrow$ It is enough to show the inequality for $x geq 0$.- For $x= 0$ you have equality.
$ Rightarrow$ It is enough to show that $left(e^{frac{x^2}{2}} - cosh x right)' geq 0$ for $x>0$.- $left(e^{frac{x^2}{2}}right)' = xe^{frac{x^2}{2}} = sum_{n=0}^{infty}frac{x^{2n+1}}{2^ncdot n!}$
$left(cosh x right)' = sinh x = sum_{n=0}^{infty}frac{x^{2n+1}}{(2n+1)!}$
$$frac{1}{2^ncdot n!} geq frac{1}{(2n+1)!}Leftrightarrow frac{(n+1)cdots (2n+1)}{2^n} geq 1 mbox{ True!}$$
From this the inequality follows.
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
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add a comment |
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your computation is flawed. The expansion of $cosh{x}$ is $sum_{j geq 0}{frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $sum_{j geq 0}{frac{x^{2j}}{2^j cdot j!}}$.
So you just need to prove that $j!2^j < (2j)!$ for each $j$.
answered Jan 21 at 18:04
MindlackMindlack
4,885211
$endgroup$
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
|
show 1 more comment
$begingroup$
Your computation is flawed. The expansion of $cosh{x}$ is $sum_{j geq 0}{frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $sum_{j geq 0}{frac{x^{2j}}{2^j cdot j!}}$.
So you just need to prove that $j!2^j < (2j)!$ for each $j$.
answered Jan 21 at 18:04
MindlackMindlack
4,885211
$endgroup$
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
|
show 1 more comment
$begingroup$
Your computation is flawed. The expansion of $cosh{x}$ is $sum_{j geq 0}{frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $sum_{j geq 0}{frac{x^{2j}}{2^j cdot j!}}$.
So you just need to prove that $j!2^j < (2j)!$ for each $j$.
answered Jan 21 at 18:04
MindlackMindlack
4,885211
$endgroup$
Your computation is flawed. The expansion of $cosh{x}$ is $sum_{j geq 0}{frac{x^{2j}}{(2j)!}}$, but the expansion of $e^{x^2/2}$ is $sum_{j geq 0}{frac{x^{2j}}{2^j cdot j!}}$.
So you just need to prove that $j!2^j < (2j)!$ for each $j$.
answered Jan 21 at 18:04
MindlackMindlack
4,885211
answered Jan 21 at 18:04
MindlackMindlack
4,885211
answered Jan 21 at 18:04
MindlackMindlack
4,885211
answered Jan 21 at 18:04
MindlackMindlack
4,885211
4,885211
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
|
show 1 more comment
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
I understand you expansion of $e^{x^2/2}$, but the expansion of $e^{x}+e^{-x}$ all over 2 would be $sum_{j=1}^{infty} frac{x^j+(-x)^j}{2j!}$, so why is it enough to prove the inequality you have shown (I know how to prove it, I just don't understand why this helps)? How did you get to that expansion of cosh?
$endgroup$
– Omer
Jan 21 at 18:11
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@Omer you may prefer to use the expansion of $cos$ and the fact that $cosh(x) = cos(ix)$
$endgroup$
– Calvin Khor
Jan 21 at 18:17
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@CalvinKhor I did not learn that, I have a calculus 1 exam soon so all my knowledge is calculus 1. we only learnt that for all x, $e^x=sum_{j=1}^{infty} frac{x^j}{j!}$
$endgroup$
– Omer
Jan 21 at 18:19
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
@Omer then just directly compute. For $j$ odd and $j$ even, you may see some terms disappear, and some terms have a factor of 2. Note that your $j$ is not the same $j$ as in the expansion of Mindlack
$endgroup$
– Calvin Khor
Jan 21 at 18:23
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
$begingroup$
Omer: your sum for $e^x$ starts at $j=0$.
$endgroup$
– Mindlack
Jan 21 at 18:27
|
show 1 more comment
$begingroup$
Let $f(x)=frac{x^2}{2}-lnfrac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)geq0$ for $xgeq0.$
We see that $$f'(x)=x-frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}geq0.$$
Id est, $$f'(x)geq f'(0)=0,$$
$$f(x)geq f(0)=0$$ and we are done!
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
$endgroup$
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
add a comment |
$begingroup$
Let $f(x)=frac{x^2}{2}-lnfrac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)geq0$ for $xgeq0.$
We see that $$f'(x)=x-frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}geq0.$$
Id est, $$f'(x)geq f'(0)=0,$$
$$f(x)geq f(0)=0$$ and we are done!
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
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I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
add a comment |
$begingroup$
Let $f(x)=frac{x^2}{2}-lnfrac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)geq0$ for $xgeq0.$
We see that $$f'(x)=x-frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}geq0.$$
Id est, $$f'(x)geq f'(0)=0,$$
$$f(x)geq f(0)=0$$ and we are done!
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
$endgroup$
Let $f(x)=frac{x^2}{2}-lnfrac{e^x+e^{-x}}{2}.$
Since $f$ is an even function, it's enough to prove that $f(x)geq0$ for $xgeq0.$
We see that $$f'(x)=x-frac{e^x-e^{-x}}{e^x+e^{-x}}$$ and
$$f''(x)=1-frac{(e^x+e^{-x})^2-(e^x-e^{-x})^2}{(e^x+e^{-x})^2}=frac{(e^x-e^{-x})^2}{(e^x+e^{-x})^2}geq0.$$
Id est, $$f'(x)geq f'(0)=0,$$
$$f(x)geq f(0)=0$$ and we are done!
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
answered Jan 21 at 18:27
Michael RozenbergMichael Rozenberg
107k1894199
107k1894199
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
add a comment |
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
$begingroup$
I was looking for answer using taylor series because I want to practice on it, but this is also great and even easier! Thank you.
$endgroup$
– Omer
Jan 21 at 18:35
1
1
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
$begingroup$
You are welcome, Omer.
$endgroup$
– Michael Rozenberg
Jan 21 at 18:36
add a comment |
$begingroup$
Notice:
$$frac12(e^x+e^{-x})<frac12(e^x+1)$$
and for $x>2.11$ that $e^frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
Notice:
$$frac12(e^x+e^{-x})<frac12(e^x+1)$$
and for $x>2.11$ that $e^frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
$endgroup$
add a comment |
$begingroup$
Notice:
$$frac12(e^x+e^{-x})<frac12(e^x+1)$$
and for $x>2.11$ that $e^frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
$endgroup$
Notice:
$$frac12(e^x+e^{-x})<frac12(e^x+1)$$
and for $x>2.11$ that $e^frac{x^2}{2}>e^x+1$
See if you can show it holds for $0<x<2.11$ (Note, negativity is irrelevant as both sides are even functions)
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
answered Jan 21 at 18:21
Rhys HughesRhys Hughes
6,9741530
6,9741530
add a comment |
add a comment |
$begingroup$
Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^xgeq 1+x$ is a little weak here. However the inequality
$$e^xgeq 1+x+frac{x^2}{2}$$
is of help here.
It suffices to show
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}+frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
$endgroup$
add a comment |
$begingroup$
Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^xgeq 1+x$ is a little weak here. However the inequality
$$e^xgeq 1+x+frac{x^2}{2}$$
is of help here.
It suffices to show
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}+frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
$endgroup$
add a comment |
$begingroup$
Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^xgeq 1+x$ is a little weak here. However the inequality
$$e^xgeq 1+x+frac{x^2}{2}$$
is of help here.
It suffices to show
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}+frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
$endgroup$
Since both functions are even, we can focus on where $x$ is nonnegative. The inequality $e^xgeq 1+x$ is a little weak here. However the inequality
$$e^xgeq 1+x+frac{x^2}{2}$$
is of help here.
It suffices to show
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}leq e^{x^2/2}$$
which is much more approachable with derivatives. Of course, this doesn't hold for all $x$ due to exponential growth. However, it's fairly easy to show that it holds on $[0,1.5]$.
The inequality
$$cosh(x)leq 1+frac{x^2}{2}+frac{x^4}{8}+frac{x^6}{96}$$
holds on a much larger set--- a little more than $[0,10]$. This is just as simple as the preceding approach.
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
edited Jan 21 at 19:24
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
answered Jan 21 at 19:02
Robert WolfeRobert Wolfe
5,94922763
5,94922763
add a comment |
add a comment |
$begingroup$
Noting that $cosh x = frac{e^x + e^{-x}}{2}$, you may proceed as follows:
- Note that $cosh (-x) = cosh x$ and $e^{frac{(-x)^2}{2}} = e^{frac{x^2}{2}}$
$ Rightarrow$ It is enough to show the inequality for $x geq 0$.- For $x= 0$ you have equality.
$ Rightarrow$ It is enough to show that $left(e^{frac{x^2}{2}} - cosh x right)' geq 0$ for $x>0$.- $left(e^{frac{x^2}{2}}right)' = xe^{frac{x^2}{2}} = sum_{n=0}^{infty}frac{x^{2n+1}}{2^ncdot n!}$
$left(cosh x right)' = sinh x = sum_{n=0}^{infty}frac{x^{2n+1}}{(2n+1)!}$
$$frac{1}{2^ncdot n!} geq frac{1}{(2n+1)!}Leftrightarrow frac{(n+1)cdots (2n+1)}{2^n} geq 1 mbox{ True!}$$
From this the inequality follows.
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Noting that $cosh x = frac{e^x + e^{-x}}{2}$, you may proceed as follows:
- Note that $cosh (-x) = cosh x$ and $e^{frac{(-x)^2}{2}} = e^{frac{x^2}{2}}$
$ Rightarrow$ It is enough to show the inequality for $x geq 0$.- For $x= 0$ you have equality.
$ Rightarrow$ It is enough to show that $left(e^{frac{x^2}{2}} - cosh x right)' geq 0$ for $x>0$.- $left(e^{frac{x^2}{2}}right)' = xe^{frac{x^2}{2}} = sum_{n=0}^{infty}frac{x^{2n+1}}{2^ncdot n!}$
$left(cosh x right)' = sinh x = sum_{n=0}^{infty}frac{x^{2n+1}}{(2n+1)!}$
$$frac{1}{2^ncdot n!} geq frac{1}{(2n+1)!}Leftrightarrow frac{(n+1)cdots (2n+1)}{2^n} geq 1 mbox{ True!}$$
From this the inequality follows.
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
$endgroup$
add a comment |
$begingroup$
Noting that $cosh x = frac{e^x + e^{-x}}{2}$, you may proceed as follows:
- Note that $cosh (-x) = cosh x$ and $e^{frac{(-x)^2}{2}} = e^{frac{x^2}{2}}$
$ Rightarrow$ It is enough to show the inequality for $x geq 0$.- For $x= 0$ you have equality.
$ Rightarrow$ It is enough to show that $left(e^{frac{x^2}{2}} - cosh x right)' geq 0$ for $x>0$.- $left(e^{frac{x^2}{2}}right)' = xe^{frac{x^2}{2}} = sum_{n=0}^{infty}frac{x^{2n+1}}{2^ncdot n!}$
$left(cosh x right)' = sinh x = sum_{n=0}^{infty}frac{x^{2n+1}}{(2n+1)!}$
$$frac{1}{2^ncdot n!} geq frac{1}{(2n+1)!}Leftrightarrow frac{(n+1)cdots (2n+1)}{2^n} geq 1 mbox{ True!}$$
From this the inequality follows.
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
$endgroup$
Noting that $cosh x = frac{e^x + e^{-x}}{2}$, you may proceed as follows:
- Note that $cosh (-x) = cosh x$ and $e^{frac{(-x)^2}{2}} = e^{frac{x^2}{2}}$
$ Rightarrow$ It is enough to show the inequality for $x geq 0$.- For $x= 0$ you have equality.
$ Rightarrow$ It is enough to show that $left(e^{frac{x^2}{2}} - cosh x right)' geq 0$ for $x>0$.- $left(e^{frac{x^2}{2}}right)' = xe^{frac{x^2}{2}} = sum_{n=0}^{infty}frac{x^{2n+1}}{2^ncdot n!}$
$left(cosh x right)' = sinh x = sum_{n=0}^{infty}frac{x^{2n+1}}{(2n+1)!}$
$$frac{1}{2^ncdot n!} geq frac{1}{(2n+1)!}Leftrightarrow frac{(n+1)cdots (2n+1)}{2^n} geq 1 mbox{ True!}$$
From this the inequality follows.
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
answered Jan 21 at 18:39
trancelocationtrancelocation
12.6k1826
12.6k1826
add a comment |
add a comment |
$begingroup$
Another one: math.stackexchange.com/q/882422/42969.
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– Martin R
Jan 21 at 19:21