exp(x) monotonic and convergent to 0 for negative xConvergent seriesIf a series is conditionally convergent, then the series of positive and negative terms are both divergentConvergent series and of positive integers and partial sums.Getting close to any real with a semi-convergent seriesConvergent series with general term $a_nb_n$partial sum of convergent seriesIf $sum a_n$ is a convergent series of real numbers then the following series are convergent.Convergent Series and RearrangementsIs monotonic and convergent series bounded by the limit?Does this non-monotonic sequence converge?
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exp(x) monotonic and convergent to 0 for negative x
Convergent seriesIf a series is conditionally convergent, then the series of positive and negative terms are both divergentConvergent series and of positive integers and partial sums.Getting close to any real with a semi-convergent seriesConvergent series with general term $a_nb_n$partial sum of convergent seriesIf $sum a_n$ is a convergent series of real numbers then the following series are convergent.Convergent Series and RearrangementsIs monotonic and convergent series bounded by the limit?Does this non-monotonic sequence converge?
$begingroup$
How is it possible that exp(x) for negative x is always close to 0 and even monotonic? I mean, look on the series representation of it: $sum_k=0^inftyfracx^kk!$
it makes sense that for positive x it behaves like that, but for huge negative numbers it's like "10 trillions - 43535 gazillions + 4435435 fartillions -...". You know what I mean,
how is that not only convergent to 0 as $xto -infty$, but also monotonic ?
real-analysis sequences-and-series convergence
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How is it possible that exp(x) for negative x is always close to 0 and even monotonic? I mean, look on the series representation of it: $sum_k=0^inftyfracx^kk!$
it makes sense that for positive x it behaves like that, but for huge negative numbers it's like "10 trillions - 43535 gazillions + 4435435 fartillions -...". You know what I mean,
how is that not only convergent to 0 as $xto -infty$, but also monotonic ?
real-analysis sequences-and-series convergence
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
2
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
How is it possible that exp(x) for negative x is always close to 0 and even monotonic? I mean, look on the series representation of it: $sum_k=0^inftyfracx^kk!$
it makes sense that for positive x it behaves like that, but for huge negative numbers it's like "10 trillions - 43535 gazillions + 4435435 fartillions -...". You know what I mean,
how is that not only convergent to 0 as $xto -infty$, but also monotonic ?
real-analysis sequences-and-series convergence
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How is it possible that exp(x) for negative x is always close to 0 and even monotonic? I mean, look on the series representation of it: $sum_k=0^inftyfracx^kk!$
it makes sense that for positive x it behaves like that, but for huge negative numbers it's like "10 trillions - 43535 gazillions + 4435435 fartillions -...". You know what I mean,
how is that not only convergent to 0 as $xto -infty$, but also monotonic ?
real-analysis sequences-and-series convergence
real-analysis sequences-and-series convergence
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 17 hours ago
Piotr
asked 17 hours ago
PiotrPiotr
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
PiotrPiotr
111
asked 17 hours ago
PiotrPiotr
111
111
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Piotr is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
2
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
2
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago
$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
2
2
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
As $exp(x)exp(-x)=1$, $exp(-x)$ must be monotone (decreasing) since $exp(x)$ is monotone (increasing). It's also clear that $lim_x rightarrow inftyleft[exp(-x)right]=lim_x rightarrowinftyleft[frac1exp(x)right]=0$.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
Empirical observations:
Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-infty$.
In the limit, you get a positive function with a root at $-infty$. Positiveness of the function explains monotonicity, as it is its own derivative.
There are two mysteries:
how the minima of the even polynomials remain positive,
why the minima/roots escape to infinity instead of converging to a finite value.
Additional remark:
For a fixed $x$, the general term $dfracx^nn!$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
As $exp(x)exp(-x)=1$, $exp(-x)$ must be monotone (decreasing) since $exp(x)$ is monotone (increasing). It's also clear that $lim_x rightarrow inftyleft[exp(-x)right]=lim_x rightarrowinftyleft[frac1exp(x)right]=0$.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
As $exp(x)exp(-x)=1$, $exp(-x)$ must be monotone (decreasing) since $exp(x)$ is monotone (increasing). It's also clear that $lim_x rightarrow inftyleft[exp(-x)right]=lim_x rightarrowinftyleft[frac1exp(x)right]=0$.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
As $exp(x)exp(-x)=1$, $exp(-x)$ must be monotone (decreasing) since $exp(x)$ is monotone (increasing). It's also clear that $lim_x rightarrow inftyleft[exp(-x)right]=lim_x rightarrowinftyleft[frac1exp(x)right]=0$.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
As $exp(x)exp(-x)=1$, $exp(-x)$ must be monotone (decreasing) since $exp(x)$ is monotone (increasing). It's also clear that $lim_x rightarrow inftyleft[exp(-x)right]=lim_x rightarrowinftyleft[frac1exp(x)right]=0$.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 16 hours ago
Quantum ChillQuantum Chill
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 16 hours ago
Quantum ChillQuantum Chill
134
answered 16 hours ago
Quantum ChillQuantum Chill
134
134
New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Quantum Chill is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
4
4
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
$begingroup$
From the series representation, this is far from clear.
$endgroup$
– Yves Daoust
16 hours ago
add a comment |
$begingroup$
Empirical observations:
Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-infty$.
In the limit, you get a positive function with a root at $-infty$. Positiveness of the function explains monotonicity, as it is its own derivative.
There are two mysteries:
how the minima of the even polynomials remain positive,
why the minima/roots escape to infinity instead of converging to a finite value.
Additional remark:
For a fixed $x$, the general term $dfracx^nn!$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
add a comment |
$begingroup$
Empirical observations:
Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-infty$.
In the limit, you get a positive function with a root at $-infty$. Positiveness of the function explains monotonicity, as it is its own derivative.
There are two mysteries:
how the minima of the even polynomials remain positive,
why the minima/roots escape to infinity instead of converging to a finite value.
Additional remark:
For a fixed $x$, the general term $dfracx^nn!$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
$endgroup$
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
add a comment |
$begingroup$
Empirical observations:
Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-infty$.
In the limit, you get a positive function with a root at $-infty$. Positiveness of the function explains monotonicity, as it is its own derivative.
There are two mysteries:
how the minima of the even polynomials remain positive,
why the minima/roots escape to infinity instead of converging to a finite value.
Additional remark:
For a fixed $x$, the general term $dfracx^nn!$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
$endgroup$
Empirical observations:
Looking at the successive partial sums, you observe that for the even degrees, the polynomial has a single minimum and remains positive. Hence the next partial sum (also the antiderivative), which has an odd degree, is monotonic and has a single root. That root must be to the left of the minimum, and for higher degrees these keep moving towards $-infty$.
In the limit, you get a positive function with a root at $-infty$. Positiveness of the function explains monotonicity, as it is its own derivative.
There are two mysteries:
how the minima of the even polynomials remain positive,
why the minima/roots escape to infinity instead of converging to a finite value.
Additional remark:
For a fixed $x$, the general term $dfracx^nn!$ increases as long as $n<x$, then decreases. This somehow explains why the minima escape, possibly at unit speed.
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
edited 16 hours ago
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
answered 16 hours ago
Yves DaoustYves Daoust
129k676227
129k676227
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
add a comment |
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
$begingroup$
This is the answer I was looking for, thank you! Maybe I should investigate this a little more, but it seems like good enough for now.
$endgroup$
– Piotr
14 hours ago
add a comment |
Piotr is a new contributor. Be nice, and check out our Code of Conduct.
Piotr is a new contributor. Be nice, and check out our Code of Conduct.
Piotr is a new contributor. Be nice, and check out our Code of Conduct.
Piotr is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
It's monotonic simply because it takes only positive values and is its own derivative.
$endgroup$
– Bernard
17 hours ago
$begingroup$
yes, I completely understand that, just the terms themselves for negative x seem... ridiculous
$endgroup$
– Piotr
17 hours ago
$begingroup$
Everything you wish to know follows from $exp(x)cdotexp(-x)=1$.
$endgroup$
– Wojowu
17 hours ago
2
$begingroup$
Unfair downvotes for a double question that makes much sense.
$endgroup$
– Yves Daoust
16 hours ago