Intuitive reasoning that a function can't exist
$begingroup$
Sorry for the title; I had no idea how to make it concise yet informative.
I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.
Here is the statement:
There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.
I appreciate any help!
real-analysis functions
edited 1 hour ago
SvanN
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
$endgroup$
add a comment |
$begingroup$
Sorry for the title; I had no idea how to make it concise yet informative.
I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.
Here is the statement:
There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.
I appreciate any help!
real-analysis functions
edited 1 hour ago
SvanN
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
$endgroup$
1
$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
8
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago
add a comment |
$begingroup$
Sorry for the title; I had no idea how to make it concise yet informative.
I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.
Here is the statement:
There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.
I appreciate any help!
real-analysis functions
edited 1 hour ago
SvanN
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
$endgroup$
Sorry for the title; I had no idea how to make it concise yet informative.
I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.
Here is the statement:
There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.
I appreciate any help!
real-analysis functions
real-analysis functions
edited 1 hour ago
SvanN
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
edited 1 hour ago
SvanN
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
edited 1 hour ago
SvanN
1,9721422
edited 1 hour ago
SvanN
1,9721422
edited 1 hour ago
SvanN
1,9721422
1,9721422
asked 1 hour ago
Math-funMath-fun
7,0411425
asked 1 hour ago
Math-funMath-fun
7,0411425
asked 1 hour ago
Math-funMath-fun
7,0411425
7,0411425
1
$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
8
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago
add a comment |
1
$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
8
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago
1
1
$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
8
8
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I do not think it is true. Take for example
- $f(x) = 1 - x + frac12x^2 - e^{-x}$
- $f'(x) = - 1 + x + e^{-x}$
- $f''(x) = 1 - e^{-x}$
- $f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
answered 1 hour ago
HenryHenry
98.9k476164
$endgroup$
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I do not think it is true. Take for example
- $f(x) = 1 - x + frac12x^2 - e^{-x}$
- $f'(x) = - 1 + x + e^{-x}$
- $f''(x) = 1 - e^{-x}$
- $f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
answered 1 hour ago
HenryHenry
98.9k476164
$endgroup$
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
add a comment |
$begingroup$
I do not think it is true. Take for example
- $f(x) = 1 - x + frac12x^2 - e^{-x}$
- $f'(x) = - 1 + x + e^{-x}$
- $f''(x) = 1 - e^{-x}$
- $f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
answered 1 hour ago
HenryHenry
98.9k476164
$endgroup$
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
add a comment |
$begingroup$
I do not think it is true. Take for example
- $f(x) = 1 - x + frac12x^2 - e^{-x}$
- $f'(x) = - 1 + x + e^{-x}$
- $f''(x) = 1 - e^{-x}$
- $f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
answered 1 hour ago
HenryHenry
98.9k476164
$endgroup$
I do not think it is true. Take for example
- $f(x) = 1 - x + frac12x^2 - e^{-x}$
- $f'(x) = - 1 + x + e^{-x}$
- $f''(x) = 1 - e^{-x}$
- $f'''(x) = e^{-x}$
Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases
answered 1 hour ago
HenryHenry
98.9k476164
edited 48 mins ago
answered 1 hour ago
HenryHenry
98.9k476164
answered 1 hour ago
HenryHenry
98.9k476164
answered 1 hour ago
HenryHenry
98.9k476164
98.9k476164
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
add a comment |
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
1
1
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
37 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
28 mins ago
add a comment |
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$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
1 hour ago
$begingroup$
If you write down a formal proof, one can see which steps translate to intuitive reasoning and which are technicalities.
$endgroup$
– Torsten Schoeneberg
1 hour ago
8
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
1 hour ago