Why is $x+e^{-x}>0$ for all $x in mathbb{R}$? The 2019 Stack Overflow Developer Survey...
I looked up a future colleague on LinkedIn before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?
What is the motivation for a law requiring 2 parties to consent for recording a conversation
How long do I have to send payment?
If a poisoned arrow's piercing damage is reduced to 0, do you still get poisoned?
Why isn't airport relocation done gradually?
Landlord wants to switch my lease to a "Land contract" to "get back at the city"
Carnot-Caratheodory metric
Are there any other methods to apply to solving simultaneous equations?
Where does the "burst of radiance" from Holy Weapon originate?
How to reverse every other sublist of a list?
Spanish for "widget"
What does "sndry explns" mean in one of the Hitchhiker's guide books?
Should I write numbers in words or as numerals when there are multiple next to each other?
What does "rabbited" mean/imply in this sentence?
Springs with some finite mass
Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?
Where to refill my bottle in India?
Pristine Bit Checking
What is the use of option -o in the useradd command?
Could JWST stay at L2 "forever"?
Why is the maximum length of OpenWrt’s root password 8 characters?
The difference between dialogue marks
Output the Arecibo Message
aging parents with no investments
Why is $x+e^{-x}>0$ for all $x in mathbb{R}$?
The 2019 Stack Overflow Developer Survey Results Are InSimplest or nicest proof that $1+x le e^x$What is the limit of this function as $n$ tends to infinity? (something to the power of e)Limit question: $lim_{xto infty}f(x),;; lim_{xto -infty}f(x)$?Does $displaystylelim_{ntoinfty}frac{n}{n+1} = 1$?Calculus - Finding the rate of change of an edge of a cube.Calculus -as X approaches O or infinityProve limit exists for an almost monotonic bounded function, involves convexity and square integrabilityCan I not use L'hopital rule here?What do second order partial derivatives mean in graphs?Find the simplest counterexample against exchanging limit and summationIf $f : [a, infty) → Bbb R$ is monotonically decreasing and the integral $int_0^infty f(x) ,dx$ is convergent, then $lim_{x→infty} f(x) = 0$.
$begingroup$
Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?
I think I am missing something really simple here. Please help.
calculus limits
edited Mar 21 at 6:31
David G. Stork
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
$endgroup$
add a comment |
$begingroup$
Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?
I think I am missing something really simple here. Please help.
calculus limits
edited Mar 21 at 6:31
David G. Stork
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
$endgroup$
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
1
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
4
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
1
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50
add a comment |
$begingroup$
Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?
I think I am missing something really simple here. Please help.
calculus limits
edited Mar 21 at 6:31
David G. Stork
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
$endgroup$
Denote $f(x) = x + e^{-x}$. Note that $f(0) = 1$ and $f'(x) = 1-e^{-x}$. That means
$$lim_{xrightarrow -infty} f'(x) = -infty.$$
So if the rate of change of $f(x)$ keeps decreasing exponentially fast to negative infinity, why is $lim_{xrightarrow -infty} f(x) = +infty$?
I think I am missing something really simple here. Please help.
calculus limits
calculus limits
edited Mar 21 at 6:31
David G. Stork
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
edited Mar 21 at 6:31
David G. Stork
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
edited Mar 21 at 6:31
David G. Stork
12.2k41836
edited Mar 21 at 6:31
David G. Stork
12.2k41836
edited Mar 21 at 6:31
David G. Stork
12.2k41836
12.2k41836
asked Mar 21 at 4:42
PaichuPaichu
804616
asked Mar 21 at 4:42
PaichuPaichu
804616
asked Mar 21 at 4:42
PaichuPaichu
804616
804616
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
1
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
4
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
1
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50
add a comment |
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
1
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
4
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
1
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
1
1
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
4
4
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
1
1
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
add a comment |
$begingroup$
Take
$x < y < 0; tag 1$
then
$f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$
it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that
$s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$
if we now choose
$y < y_0, tag 4$
then
$f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$
we re-arrange this inequality:
$f(x) - f(y) > -M(y - x) = M(x - y), tag 6$
$f(x) > f(y) + M(x - y); tag 7$
we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,
$displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$
and hence
$displaystyle lim_{x to -infty}f(x) = infty tag 9$
as well.
We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$x + e^{-x} > 0, forall x in Bbb R; tag{10}$
for
$x ge 0, tag{11}$
$e^{-x} > 0 tag{12}$
as well, hence we also have
$x + e^{-x} > 0; tag{13}$
for
$x < 0, tag{14}$
we may use the power series for $e^{-x}$:
$e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$
then
$x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$
since every term on the right is positive when $x < 0$. End of Note.
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.
Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Note that
- $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$
The last inequality follows directly by Taylor:
$$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156341%2fwhy-is-xe-x0-for-all-x-in-mathbbr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
add a comment |
$begingroup$
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
add a comment |
$begingroup$
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
$endgroup$
You have got your directions mixed up.
When the derivative is negative, the function is decreasing from left to right i.e. its graph is coming down from left to right. Which is the same as , going right to left its graph is going up or it is increasing, which is the phenomena observed here.
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
answered Mar 21 at 4:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
40.3k33678
40.3k33678
add a comment |
add a comment |
$begingroup$
Take
$x < y < 0; tag 1$
then
$f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$
it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that
$s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$
if we now choose
$y < y_0, tag 4$
then
$f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$
we re-arrange this inequality:
$f(x) - f(y) > -M(y - x) = M(x - y), tag 6$
$f(x) > f(y) + M(x - y); tag 7$
we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,
$displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$
and hence
$displaystyle lim_{x to -infty}f(x) = infty tag 9$
as well.
We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$x + e^{-x} > 0, forall x in Bbb R; tag{10}$
for
$x ge 0, tag{11}$
$e^{-x} > 0 tag{12}$
as well, hence we also have
$x + e^{-x} > 0; tag{13}$
for
$x < 0, tag{14}$
we may use the power series for $e^{-x}$:
$e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$
then
$x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$
since every term on the right is positive when $x < 0$. End of Note.
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
Take
$x < y < 0; tag 1$
then
$f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$
it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that
$s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$
if we now choose
$y < y_0, tag 4$
then
$f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$
we re-arrange this inequality:
$f(x) - f(y) > -M(y - x) = M(x - y), tag 6$
$f(x) > f(y) + M(x - y); tag 7$
we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,
$displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$
and hence
$displaystyle lim_{x to -infty}f(x) = infty tag 9$
as well.
We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$x + e^{-x} > 0, forall x in Bbb R; tag{10}$
for
$x ge 0, tag{11}$
$e^{-x} > 0 tag{12}$
as well, hence we also have
$x + e^{-x} > 0; tag{13}$
for
$x < 0, tag{14}$
we may use the power series for $e^{-x}$:
$e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$
then
$x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$
since every term on the right is positive when $x < 0$. End of Note.
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
$endgroup$
add a comment |
$begingroup$
Take
$x < y < 0; tag 1$
then
$f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$
it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that
$s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$
if we now choose
$y < y_0, tag 4$
then
$f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$
we re-arrange this inequality:
$f(x) - f(y) > -M(y - x) = M(x - y), tag 6$
$f(x) > f(y) + M(x - y); tag 7$
we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,
$displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$
and hence
$displaystyle lim_{x to -infty}f(x) = infty tag 9$
as well.
We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$x + e^{-x} > 0, forall x in Bbb R; tag{10}$
for
$x ge 0, tag{11}$
$e^{-x} > 0 tag{12}$
as well, hence we also have
$x + e^{-x} > 0; tag{13}$
for
$x < 0, tag{14}$
we may use the power series for $e^{-x}$:
$e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$
then
$x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$
since every term on the right is positive when $x < 0$. End of Note.
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
$endgroup$
Take
$x < y < 0; tag 1$
then
$f(y) - f(x) = displaystyle int_x^y f'(s); ds = int_x^y (1 - e^{-s}); ds; tag 2$
it is clear that, for any $M < 0$, there exists $y_0 < 0$ such that
$s < y_0 Longrightarrow 1 - e^{-s} < M; tag 3$
if we now choose
$y < y_0, tag 4$
then
$f(y) - f(x) = displaystyle int_x^y (1 - e^{-s}); ds < int_x^y M ; ds = M(y - x); tag 5$
we re-arrange this inequality:
$f(x) - f(y) > -M(y - x) = M(x - y), tag 6$
$f(x) > f(y) + M(x - y); tag 7$
we now fix $y$ and let $x to -infty$; then since $M < 0$ and $x - y < 0$ for $x < y$,
$displaystyle lim_{x to -infty} f(y) + M(x - y) = infty, tag 8$
and hence
$displaystyle lim_{x to -infty}f(x) = infty tag 9$
as well.
We note that (9) binds despite the fact that $f'(x) < 0$ for $x < 0$; though this derivative is negative, when $x$ decreases we are "walking back up the hill," as it were; as $x$ decreases, $f(x)$ increases.
Note Added in Edit, Wednesday 20 March 2019 10:39 PM PST: We may also dispense with the title question and show
$x + e^{-x} > 0, forall x in Bbb R; tag{10}$
for
$x ge 0, tag{11}$
$e^{-x} > 0 tag{12}$
as well, hence we also have
$x + e^{-x} > 0; tag{13}$
for
$x < 0, tag{14}$
we may use the power series for $e^{-x}$:
$e^{-x} = 1 - x + dfrac{x^2}{2!} - dfrac{x^3}{3} + ldots; tag{15}$
then
$x + e^{-x} = 1 + dfrac{x^2}{2!} - dfrac{x^3}{3!} + ldots > 0, tag{16}$
since every term on the right is positive when $x < 0$. End of Note.
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
edited Mar 21 at 5:46
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
answered Mar 21 at 5:29
Robert LewisRobert Lewis
48.9k23168
48.9k23168
add a comment |
add a comment |
$begingroup$
Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.
Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.
Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
$endgroup$
add a comment |
$begingroup$
Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.
Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
$endgroup$
Note that $f''(x) = e^{-x} >0$ so $f$ is strictly convex.
Since $f'(0) = 0$, we see that $f(x) ge f(0) = 1$ for all $x$.
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
answered Mar 21 at 4:51
copper.hatcopper.hat
128k561161
128k561161
add a comment |
add a comment |
$begingroup$
Note that
- $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$
The last inequality follows directly by Taylor:
$$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
$endgroup$
add a comment |
$begingroup$
Note that
- $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$
The last inequality follows directly by Taylor:
$$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
$endgroup$
add a comment |
$begingroup$
Note that
- $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$
The last inequality follows directly by Taylor:
$$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
$endgroup$
Note that
- $forall x in mathbb{R} : x+e^{-x}>0 Leftrightarrow forall x in mathbb{R} :e^{-x}>-x Leftrightarrow forall x in mathbb{R} :color{blue}{boxed{e^{x}>x}}$
The last inequality follows directly by Taylor:
$$color{blue}{e^x} = 1+x + underbrace{frac{e^{xi}}{2}x^2}_{geq 0} color{blue}{> x}$$
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
answered Mar 21 at 7:03
trancelocationtrancelocation
13.7k1829
13.7k1829
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3156341%2fwhy-is-xe-x0-for-all-x-in-mathbbr%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Addressing the title, for all $x, e^{-x}ge1-x$ so $x+e^{-x}ge1>0$
$endgroup$
– J. W. Tanner
Mar 21 at 4:45
1
$begingroup$
To show $x+e^{-x}> 0$, you want to show $e^{-x}> -x$. In fact, we have $e^{t}ge t+1$ for all real $t$. So $e^{-x}ge -x+1> -x$ for all real $x$, as desired.
$endgroup$
– Minus One-Twelfth
Mar 21 at 4:46
4
$begingroup$
Roughly speaking - if you are visualising $x$ going from $0$ to $-infty$, then the graph is not decreasing but increasing exponentially fast, because you are going "backwards".
$endgroup$
– David
Mar 21 at 4:49
1
$begingroup$
Thank you. That's what I was missing. Going backward. :(
$endgroup$
– Paichu
Mar 21 at 4:50