Post History
#5: Post edited
by
Pavel Kocourek
·
2023-01-25T08:14:28Z (almost 2 years ago)
Summarized the question.
Is the really only one solution?
- Is $f(x)=\sin(x)$ the unique function satisfying $f'(0)=1$ and $f^{(n)}(\Bbb R)\subset [-1,1]$ for all $n=0,1,\ldots$?
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is smaller than $1-k$ (i.e. $a\in[k-1,1-k]$). As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.
- > **Question.** Is there a function $f:\Bbb R \to \Bbb R$ with $f'(0)=1$ and $f^{(n)}(x)\in [-1,1]$ for all $n=0,1,\ldots$ and $x\in \Bbb R$, other than $f(x)=\sin(x)$?
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is smaller than $1-k$ (i.e. $a\in[k-1,1-k]$). As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.
#4: Post edited
by
Grove
·
2022-09-10T21:57:17Z (about 2 years ago)
Chnaged formatting
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is $<1-k$. As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is smaller than $1-k$ (i.e. $a\in[k-1,1-k]$). As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.
#3: Post edited
by
Grove
·
2022-09-10T21:55:22Z (about 2 years ago)
Described that you can add a constant to the functions solving the orgonal problem
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Edit: I just realised that the set of solutions can be expanded by adding an $a$ whose absolute value is $<1-k$. As that doesn't change the derivative(s), it also doesn't change the argument for why only one function satisfies the additional constraint. I.e. none of these solutions are solutions to the extended problem.
#2: Post edited
by
Grove
·
2022-09-06T21:08:11Z (about 2 years ago)
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $0$ and $1$, i.e. $f^{(n)}: \mathbb R\to [0,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
- Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
- Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $-1$ and $1$, i.e. $f^{(n)}: \mathbb R\to [-1,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
- It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
- If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
- My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...
#1: Initial revision
by
Grove
·
2022-09-06T15:22:23Z (about 2 years ago)
Is the really only one solution?
Going through some old notes, I stumbled upon this question, I don't remember where it take from/what inspired me to look at it. If this is an obvious copy of something posted/covered elsewhere, I hope that can be forgiven.
Consider an infinitely often differentiable function on the reals $f$, that satisfies that all derivatives of $f$ (including $f$ itself) take values between $0$ and $1$, i.e. $f^{(n)}: \mathbb R\to [0,1]$ for $n\in \mathbb N_0$ (the 0'th derivative being the function itself).
It's easy to find infinitely many functions satifying that: $f(x)=k\sin(jx+y)$ for $k,j\in [0,1]$ and $y\in [-\tfrac \pi j,\tfrac \pi j[$.
If we then add the additional requirement that $f'(0)=1$, only one of those solutions are good ($k=j=1$, $y=0$). (Calculating $f'(x)=jk\cos(jx+y)$ and then realising that every factor has to be $1$ for the product to be $1$)
My feeling is that with so many functions satisfying the requirements before adding the final requirement, there should be more than one with it. So is something wrong? With my feeling? Does a lot of other functions exist that satisfies the initial requirements and differentiate "better" (in terms of providing solutions to this)? Have I just calculated $f'$ wrongly? Or...