$mid f^{(n)} (t) mid leq frac{M(n!)^s}{R^n}$ then $1/f$ has the same property
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A friend of mine asked me the following problem :
Let $s in [1,infty)$ and $T > 0$.
We say that a function $f in C^{infty}([0,T], mathbb{R})$ is in $mathcal{K}^s(0,T)$ iff there is $M, R > 0$ such that :
$$mid f^{(n)}(t) mid leq frac{M(n!)^s}{R^n}, forall n in mathbb{N}, t in [0,T]$$
Consider a function $g$ such that $g in mathcal{K}^s(0,T)$ and such that there is $delta > 0$ such that $g(t) geq delta, forall t in [0,T]$then prove that $frac{1}{g} in mathcal{K}^s(0,T)$
I don’t know how to approach this problem. The assumption made on $g$ are difficult to understand intuitively. We understand that it means that $g$ is extremely regular but then how can I conclude that $1/g$ has the same property.
An idea is to use binomial theorem to expand the derivative but it doesn’t really work since the expression of $(1/g)^{(n)}$ depend on some of the derivative of $1/g$.
calculus real-analysis integration sequences-and-series functions
asked Nov 22 at 20:38
Interesting problems
13310
add a comment |
up vote
1
down vote
favorite
A friend of mine asked me the following problem :
Let $s in [1,infty)$ and $T > 0$.
We say that a function $f in C^{infty}([0,T], mathbb{R})$ is in $mathcal{K}^s(0,T)$ iff there is $M, R > 0$ such that :
$$mid f^{(n)}(t) mid leq frac{M(n!)^s}{R^n}, forall n in mathbb{N}, t in [0,T]$$
Consider a function $g$ such that $g in mathcal{K}^s(0,T)$ and such that there is $delta > 0$ such that $g(t) geq delta, forall t in [0,T]$then prove that $frac{1}{g} in mathcal{K}^s(0,T)$
I don’t know how to approach this problem. The assumption made on $g$ are difficult to understand intuitively. We understand that it means that $g$ is extremely regular but then how can I conclude that $1/g$ has the same property.
An idea is to use binomial theorem to expand the derivative but it doesn’t really work since the expression of $(1/g)^{(n)}$ depend on some of the derivative of $1/g$.
calculus real-analysis integration sequences-and-series functions
asked Nov 22 at 20:38
Interesting problems
13310
$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
1
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A friend of mine asked me the following problem :
Let $s in [1,infty)$ and $T > 0$.
We say that a function $f in C^{infty}([0,T], mathbb{R})$ is in $mathcal{K}^s(0,T)$ iff there is $M, R > 0$ such that :
$$mid f^{(n)}(t) mid leq frac{M(n!)^s}{R^n}, forall n in mathbb{N}, t in [0,T]$$
Consider a function $g$ such that $g in mathcal{K}^s(0,T)$ and such that there is $delta > 0$ such that $g(t) geq delta, forall t in [0,T]$then prove that $frac{1}{g} in mathcal{K}^s(0,T)$
I don’t know how to approach this problem. The assumption made on $g$ are difficult to understand intuitively. We understand that it means that $g$ is extremely regular but then how can I conclude that $1/g$ has the same property.
An idea is to use binomial theorem to expand the derivative but it doesn’t really work since the expression of $(1/g)^{(n)}$ depend on some of the derivative of $1/g$.
calculus real-analysis integration sequences-and-series functions
asked Nov 22 at 20:38
Interesting problems
13310
A friend of mine asked me the following problem :
Let $s in [1,infty)$ and $T > 0$.
We say that a function $f in C^{infty}([0,T], mathbb{R})$ is in $mathcal{K}^s(0,T)$ iff there is $M, R > 0$ such that :
$$mid f^{(n)}(t) mid leq frac{M(n!)^s}{R^n}, forall n in mathbb{N}, t in [0,T]$$
Consider a function $g$ such that $g in mathcal{K}^s(0,T)$ and such that there is $delta > 0$ such that $g(t) geq delta, forall t in [0,T]$then prove that $frac{1}{g} in mathcal{K}^s(0,T)$
I don’t know how to approach this problem. The assumption made on $g$ are difficult to understand intuitively. We understand that it means that $g$ is extremely regular but then how can I conclude that $1/g$ has the same property.
An idea is to use binomial theorem to expand the derivative but it doesn’t really work since the expression of $(1/g)^{(n)}$ depend on some of the derivative of $1/g$.
calculus real-analysis integration sequences-and-series functions
calculus real-analysis integration sequences-and-series functions
asked Nov 22 at 20:38
Interesting problems
13310
asked Nov 22 at 20:38
Interesting problems
13310
edited Nov 22 at 20:44
asked Nov 22 at 20:38
Interesting problems
13310
asked Nov 22 at 20:38
Interesting problems
13310
asked Nov 22 at 20:38
Interesting problems
13310
13310
$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
1
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47
add a comment |
$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
1
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47
$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
1
1
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47
add a comment |
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$1/g$ could be unbounded if $g=0$ at some point?
– MisterRiemann
Nov 22 at 20:41
@MisterRiemann Yes thank you for noticing, I forgot an assumption. Now it’s ok.
– Interesting problems
Nov 22 at 20:45
1
For a slightly more general version of this assertion, check out Theorem 2.9 (p.14) of this thesis.
– MisterRiemann
Nov 22 at 20:47