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?
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 3 hours ago
cammilcammil
1214
$endgroup$
add a 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 3 hours ago
cammilcammil
1214
$endgroup$
5
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
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
1 hour ago
add a 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 3 hours ago
cammilcammil
1214
$endgroup$
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 3 hours ago
cammilcammil
1214
asked 3 hours ago
cammilcammil
1214
edited 3 hours ago
cammil
asked 3 hours ago
cammilcammil
1214
asked 3 hours ago
cammilcammil
1214
asked 3 hours ago
cammilcammil
1214
1214
5
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
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
1 hour ago
add a comment |
5
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
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
1 hour ago
5
5
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
1
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
1
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
3 hours ago
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
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
1 hour 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
1 hour ago
add a 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 3 hours ago
eyeballfrogeyeballfrog
6,664630
$endgroup$
$begingroup$
+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
2 hours ago
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 3 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 2 hours ago
zhw.zhw.
74.5k43175
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
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 3 hours ago
eyeballfrogeyeballfrog
6,664630
$endgroup$
$begingroup$
+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
2 hours 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 3 hours ago
eyeballfrogeyeballfrog
6,664630
$endgroup$
$begingroup$
+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
2 hours 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 3 hours ago
eyeballfrogeyeballfrog
6,664630
$endgroup$
$$
f(x) = x-e^-x
$$
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 3 hours ago
eyeballfrogeyeballfrog
6,664630
edited 2 hours ago
answered 3 hours ago
eyeballfrogeyeballfrog
6,664630
answered 3 hours ago
eyeballfrogeyeballfrog
6,664630
answered 3 hours ago
eyeballfrogeyeballfrog
6,664630
6,664630
$begingroup$
+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
2 hours 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
2 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
2 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
2 hours ago
add a comment |
$begingroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 3 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 3 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 3 hours ago
paw88789paw88789
29.4k12349
$endgroup$
How about
$f(x)=left{beginarraycc ln(x+1)& &xge 0\1-e^-x& &x<0endarrayright.$
answered 3 hours ago
paw88789paw88789
29.4k12349
answered 3 hours ago
paw88789paw88789
29.4k12349
answered 3 hours ago
paw88789paw88789
29.4k12349
answered 3 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 2 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 2 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 2 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 2 hours ago
zhw.zhw.
74.5k43175
answered 2 hours ago
zhw.zhw.
74.5k43175
answered 2 hours ago
zhw.zhw.
74.5k43175
answered 2 hours ago
zhw.zhw.
74.5k43175
74.5k43175
add a comment |
add a comment |
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5
$begingroup$
$f(x) = -e^-x$?
$endgroup$
– Daniel Schepler
3 hours ago
1
$begingroup$
@DanielSchepler I was just about to write the same, +1.
$endgroup$
– Michael Hoppe
3 hours ago
$begingroup$
Sorry, I should have made clear, it should map to the whole of the reals. (What's the mathematical term for that?)
$endgroup$
– cammil
3 hours ago
1
$begingroup$
@cammil a surjection (i.e. a function whose range is equal to its codomain).
$endgroup$
– Jake
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
1 hour ago