Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?Example of continuous but not absolutely continuous strictly increasing functionCan you build metric space theory without the real numbers?Proof by induction: prove that if $x_0>3$ then the following sequence is strictly increasing…A continuous function $f:Bbb Rto Bbb R$ is injective if and only if it is strictly increasing or strictly decreasingIs there a “jagged” real-valued function that is “smooth” in cardinalities greater than the reals?examples of first strictly concave then convex function?Inverse of any strictly monotonic increasing function defined over a fixed domain and range.Continuity of $argmax$ of a strictly concave functionstrictly increasing function from reals to reals which is never an algebraic numberAt which value (over $mathbbR^+$) is the gamma function strictly increasing?
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Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?
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Is there an injective, monotonically increasing, strictly concave function from the reals, to the reals?
Example of continuous but not absolutely continuous strictly increasing functionCan you build metric space theory without the real numbers?Proof by induction: prove that if $x_0>3$ then the following sequence is strictly increasing…A continuous function $f:Bbb Rto Bbb R$ is injective if and only if it is strictly increasing or strictly decreasingIs there a “jagged” real-valued function that is “smooth” in cardinalities greater than the reals?examples of first strictly concave then convex function?Inverse of any strictly monotonic increasing function defined over a fixed domain and range.Continuity of $argmax$ of a strictly concave functionstrictly increasing function from reals to reals which is never an algebraic numberAt which value (over $mathbbR^+$) is the gamma function strictly increasing?
$begingroup$
I can't come up with a single one.
The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.
Any ideas?
real-analysis functions recreational-mathematics real-numbers
asked 6 hours ago
cammilcammil
1314
$endgroup$
|
show 1 more comment
$begingroup$
I can't come up with a single one.
The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.
Any ideas?
real-analysis functions recreational-mathematics real-numbers
asked 6 hours ago
cammilcammil
1314
$endgroup$
6
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
1
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
1
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago
|
show 1 more comment
$begingroup$
I can't come up with a single one.
The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.
Any ideas?
real-analysis functions recreational-mathematics real-numbers
asked 6 hours ago
cammilcammil
1314
$endgroup$
I can't come up with a single one.
The range should be the whole of the reals. The best I have is $log(x)$ but that's only on the positive real line. And there's $f(x) = x$, but this is not strictly concave. And $-e^-x$ only maps to half of the real line.
Any ideas?
real-analysis functions recreational-mathematics real-numbers
real-analysis functions recreational-mathematics real-numbers
asked 6 hours ago
cammilcammil
1314
asked 6 hours ago
cammilcammil
1314
edited 5 hours ago
cammil
asked 6 hours ago
cammilcammil
1314
asked 6 hours ago
cammilcammil
1314
asked 6 hours ago
cammilcammil
1314
1314
6
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
1
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
1
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago
|
show 1 more comment
6
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
1
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
1
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago
6
6
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
1
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
1
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
1
1
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
1
1
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
$$
f(x) = x-e^-x
$$
is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.
Even better,
$$
f(x) = 2x -sqrt1+3x^2
$$
has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
$endgroup$
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 5 hours ago
paw88789paw88789
29.4k12349
$endgroup$
add a comment |
$begingroup$
$F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.
answered 4 hours ago
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
$$
f(x) = x-e^-x
$$
is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.
Even better,
$$
f(x) = 2x -sqrt1+3x^2
$$
has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
$endgroup$
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
add a comment |
$begingroup$
$$
f(x) = x-e^-x
$$
is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.
Even better,
$$
f(x) = 2x -sqrt1+3x^2
$$
has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
$endgroup$
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
add a comment |
$begingroup$
$$
f(x) = x-e^-x
$$
is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.
Even better,
$$
f(x) = 2x -sqrt1+3x^2
$$
has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
$endgroup$
$$
f(x) = x-e^-x
$$
is such a function. Since $f''(x) = -e^-x$ is always negative, it is strictly concave, and it's not hard to show it hits every real.
Even better,
$$
f(x) = 2x -sqrt1+3x^2
$$
has $f''(x) = -3(1+3x^2)^-3/2 < 0$ everywhere and the explicit inverse $f^-1(x) = 2x+sqrt1+3x^2$, clearly defined for all $x$.
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
edited 5 hours ago
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
answered 5 hours ago
eyeballfrogeyeballfrog
6,719630
6,719630
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
add a comment |
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
$begingroup$
+1 (All hail the Hypnotoad!) Dare I ask how you found the second example? I had to work a bit even to check the inverse formula. I assume I'm missing something really neat.
$endgroup$
– Calum Gilhooley
4 hours ago
1
1
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@CalumGilhooley The idea of "linear function + concave function" was fairly straightforward. I figured an algebraic function would have a closed-form inverse (unlike the transcendental $x-e^-x$), then fiddled with the parameters until both the function and its inverse came out looking nice.
$endgroup$
– eyeballfrog
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
@eyeballfrog Upvoted. It would be great if you can add a graph or two. Human loves graph!
$endgroup$
– Apass.Jack
1 hour ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
$begingroup$
I could’ve never come up with this.
$endgroup$
– Randall
1 min ago
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 5 hours ago
paw88789paw88789
29.4k12349
$endgroup$
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 5 hours ago
paw88789paw88789
29.4k12349
$endgroup$
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 5 hours ago
paw88789paw88789
29.4k12349
$endgroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 5 hours ago
paw88789paw88789
29.4k12349
answered 5 hours ago
paw88789paw88789
29.4k12349
answered 5 hours ago
paw88789paw88789
29.4k12349
answered 5 hours ago
paw88789paw88789
29.4k12349
29.4k12349
add a comment |
add a comment |
$begingroup$
$F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.
answered 4 hours ago
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
$begingroup$
$F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.
answered 4 hours ago
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
$begingroup$
$F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.
answered 4 hours ago
zhw.zhw.
74.5k43175
$endgroup$
$F(x) = pi x+ int_0^x arctan (-t),dt$ is an example. Many more examples like this one can be constructed.
answered 4 hours ago
zhw.zhw.
74.5k43175
answered 4 hours ago
zhw.zhw.
74.5k43175
answered 4 hours ago
zhw.zhw.
74.5k43175
answered 4 hours ago
zhw.zhw.
74.5k43175
74.5k43175
add a comment |
add a comment |
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6
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
5 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
5 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
5 hours ago
1
$begingroup$
If you start with the lower right branch of the hyperbola $xy=-1$ and transform the coordinates to slope the $x$ axis upward to the right and the $y$ axis rightward toward the top, you will have another choice.
$endgroup$
– Ross Millikan
3 hours ago
1
$begingroup$
A better title is "is there a bijective convex function from the reals to reals?" (I prefer convex since "convex" is simpler and more popular than "concave")
$endgroup$
– Apass.Jack
58 mins ago