a non-zero continuous function with compact support and its derivatives are zero at zero
$begingroup$
I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?
real-analysis derivatives continuity
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
$endgroup$
add a comment |
$begingroup$
I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?
real-analysis derivatives continuity
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
$endgroup$
add a comment |
$begingroup$
I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?
real-analysis derivatives continuity
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
$endgroup$
I am looking for a continuous function with support compact, which its derivates are zero at zero, it means $f'(0)=0$,$f''(0)=0$..., I have seen this question All derivatives zero at a point $implies$ constant function?. So, I know that one answers is a flat function but the difference is that I want a function such that $f(0)neq0$
Maybe, that function is piecewise function or something different. I've tried to build it but I couldn't, any help, please?
real-analysis derivatives continuity
real-analysis derivatives continuity
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
asked Jan 14 at 23:20
David LlerenaDavid Llerena
254
254
add a comment |
add a comment |
2 Answers
2
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oldest
votes
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Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.
Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.
Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.
answered Jan 14 at 23:42
MindlackMindlack
4,910211
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add a comment |
$begingroup$
A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
$$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
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1
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your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.
Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.
Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.
answered Jan 14 at 23:42
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.
Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.
Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.
answered Jan 14 at 23:42
MindlackMindlack
4,910211
$endgroup$
add a comment |
$begingroup$
Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.
Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.
Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.
answered Jan 14 at 23:42
MindlackMindlack
4,910211
$endgroup$
Let $f_0(x)=0$ if $x leq 0$, $f_0(x)=e^{-1/x}$ if $x >0$. It is pretty standard to note that $f_0$ is smooth.
Consider now $f_1(x)=frac{f_0(1-x)}{f_0(1-x)+f_0(x)}$, $f$ is smooth, is $1$ when $x leq 0$, and $0$ when $x geq 1$.
Finally take $f(x)=f_1(x-1)f_1(-1-x)$. $f$ is smooth, $f(x)$ is zero as long as $x-1 geq 1$ or $-1-x geq 1$ (ie as soon as $|x| geq 2$) and is one whenever $|x| leq 1$, so $f^{(k)}(0) = 0$ if $k >0$, and $f(0)=1$.
answered Jan 14 at 23:42
MindlackMindlack
4,910211
answered Jan 14 at 23:42
MindlackMindlack
4,910211
answered Jan 14 at 23:42
MindlackMindlack
4,910211
answered Jan 14 at 23:42
MindlackMindlack
4,910211
4,910211
add a comment |
add a comment |
$begingroup$
A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
$$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
$endgroup$
1
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
add a comment |
$begingroup$
A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
$$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
$endgroup$
1
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
add a comment |
$begingroup$
A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
$$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
$endgroup$
A piecewise function could certainly work. For instance consider the function $f: mathbb R longrightarrow mathbb R$ defined by
$$f(x):=begin{cases}1&text{ if }x in [-1,1] , \2-x&text{ if } x in (1,2], \ 2+x &text{ if } x in [-2,-1), \ 0 &text{ if } x in (-infty,-2) cup (2, infty).end{cases}$$
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
answered Jan 14 at 23:37
Matt A PeltoMatt A Pelto
2,707621
2,707621
1
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
add a comment |
1
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
1
1
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
your function is not differentiable everywhere.
$endgroup$
– ersh
Jan 15 at 0:15
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
$begingroup$
I am aware but I didn't consider this an issue because the question asker didn't ask for a function that is differentiable everywhere (just continuous). If this seems like an issue, then see Mindlack's more fancy example (I am lazy).
$endgroup$
– Matt A Pelto
Jan 15 at 0:31
add a comment |
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