A question about a possible solution to A5 from Putnam 2018
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Can the following function exist: a smooth function $f:Bbb{R}toBbb{R}$ satisfies the following condition: $f^{(n)}(0)=0$ for all $n$, but there exists a sequence of points $x_nto 0$ such that $f^{(n)}(x_n)toinfty$. This is in reference to problem A5 of Putnam 2018. If I can prove that such a function cannot exist, I will have proved the Putnam problem.
real-analysis contest-math
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
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add a comment |
$begingroup$
Can the following function exist: a smooth function $f:Bbb{R}toBbb{R}$ satisfies the following condition: $f^{(n)}(0)=0$ for all $n$, but there exists a sequence of points $x_nto 0$ such that $f^{(n)}(x_n)toinfty$. This is in reference to problem A5 of Putnam 2018. If I can prove that such a function cannot exist, I will have proved the Putnam problem.
real-analysis contest-math
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
$endgroup$
add a comment |
$begingroup$
Can the following function exist: a smooth function $f:Bbb{R}toBbb{R}$ satisfies the following condition: $f^{(n)}(0)=0$ for all $n$, but there exists a sequence of points $x_nto 0$ such that $f^{(n)}(x_n)toinfty$. This is in reference to problem A5 of Putnam 2018. If I can prove that such a function cannot exist, I will have proved the Putnam problem.
real-analysis contest-math
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
$endgroup$
Can the following function exist: a smooth function $f:Bbb{R}toBbb{R}$ satisfies the following condition: $f^{(n)}(0)=0$ for all $n$, but there exists a sequence of points $x_nto 0$ such that $f^{(n)}(x_n)toinfty$. This is in reference to problem A5 of Putnam 2018. If I can prove that such a function cannot exist, I will have proved the Putnam problem.
real-analysis contest-math
real-analysis contest-math
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
edited Dec 24 '18 at 15:54
fierydemon
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
asked Dec 24 '18 at 15:35
fierydemonfierydemon
4,61522259
4,61522259
add a comment |
add a comment |
1 Answer
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Take any $C^infty$ function $f$ such that all $f^{(n)}(0) = 0$ but
$f(x) > 0$ for $x > 0$; a standard example is
$$ f(x) = cases{exp(-1/x) & if $x ge 0$cr
0 & if $x le 0$} $$
Note that for $x > 0$ and all positive integers $n$,
$$ f(x) = int_0^x frac{(x-s)^n}{n!} f^{(n+1)}(s); ds $$
as can be seen using integration by parts.
Now $|x-s|^n/n! le x^n/n! to 0$ as $n to infty$, so there must be
$x_n in (0,x)$ with $f^{(n+1)}(x_n) ge f(x) n!/x^n$.
For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n to 0+$
as $n to infty$, because $f(x) n!/x^n to infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n in (0, y_n)$ with $f^{(n+1)}(x_n) ge n f(1)$.
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
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Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
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– fierydemon
Dec 24 '18 at 16:49
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Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Take any $C^infty$ function $f$ such that all $f^{(n)}(0) = 0$ but
$f(x) > 0$ for $x > 0$; a standard example is
$$ f(x) = cases{exp(-1/x) & if $x ge 0$cr
0 & if $x le 0$} $$
Note that for $x > 0$ and all positive integers $n$,
$$ f(x) = int_0^x frac{(x-s)^n}{n!} f^{(n+1)}(s); ds $$
as can be seen using integration by parts.
Now $|x-s|^n/n! le x^n/n! to 0$ as $n to infty$, so there must be
$x_n in (0,x)$ with $f^{(n+1)}(x_n) ge f(x) n!/x^n$.
For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n to 0+$
as $n to infty$, because $f(x) n!/x^n to infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n in (0, y_n)$ with $f^{(n+1)}(x_n) ge n f(1)$.
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
add a comment |
$begingroup$
Take any $C^infty$ function $f$ such that all $f^{(n)}(0) = 0$ but
$f(x) > 0$ for $x > 0$; a standard example is
$$ f(x) = cases{exp(-1/x) & if $x ge 0$cr
0 & if $x le 0$} $$
Note that for $x > 0$ and all positive integers $n$,
$$ f(x) = int_0^x frac{(x-s)^n}{n!} f^{(n+1)}(s); ds $$
as can be seen using integration by parts.
Now $|x-s|^n/n! le x^n/n! to 0$ as $n to infty$, so there must be
$x_n in (0,x)$ with $f^{(n+1)}(x_n) ge f(x) n!/x^n$.
For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n to 0+$
as $n to infty$, because $f(x) n!/x^n to infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n in (0, y_n)$ with $f^{(n+1)}(x_n) ge n f(1)$.
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
$endgroup$
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
add a comment |
$begingroup$
Take any $C^infty$ function $f$ such that all $f^{(n)}(0) = 0$ but
$f(x) > 0$ for $x > 0$; a standard example is
$$ f(x) = cases{exp(-1/x) & if $x ge 0$cr
0 & if $x le 0$} $$
Note that for $x > 0$ and all positive integers $n$,
$$ f(x) = int_0^x frac{(x-s)^n}{n!} f^{(n+1)}(s); ds $$
as can be seen using integration by parts.
Now $|x-s|^n/n! le x^n/n! to 0$ as $n to infty$, so there must be
$x_n in (0,x)$ with $f^{(n+1)}(x_n) ge f(x) n!/x^n$.
For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n to 0+$
as $n to infty$, because $f(x) n!/x^n to infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n in (0, y_n)$ with $f^{(n+1)}(x_n) ge n f(1)$.
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
$endgroup$
Take any $C^infty$ function $f$ such that all $f^{(n)}(0) = 0$ but
$f(x) > 0$ for $x > 0$; a standard example is
$$ f(x) = cases{exp(-1/x) & if $x ge 0$cr
0 & if $x le 0$} $$
Note that for $x > 0$ and all positive integers $n$,
$$ f(x) = int_0^x frac{(x-s)^n}{n!} f^{(n+1)}(s); ds $$
as can be seen using integration by parts.
Now $|x-s|^n/n! le x^n/n! to 0$ as $n to infty$, so there must be
$x_n in (0,x)$ with $f^{(n+1)}(x_n) ge f(x) n!/x^n$.
For each $n$, take $y_n > 0$ so that $f(y_n) n!/y_n^n = n f(1)$. We have $y_n to 0+$
as $n to infty$, because $f(x) n!/x^n to infty$ for any fixed $x$. Then taking $x = y_n$ in the previous paragraph, we get $x_n in (0, y_n)$ with $f^{(n+1)}(x_n) ge n f(1)$.
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
answered Dec 24 '18 at 16:42
Robert IsraelRobert Israel
333k23223484
333k23223484
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
add a comment |
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Just confirming, you're saying that such a smooth function can exist, and that there is no contradiction?
$endgroup$
– fierydemon
Dec 24 '18 at 16:49
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
$begingroup$
Yes, it exists.
$endgroup$
– Robert Israel
Dec 24 '18 at 18:06
add a comment |
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