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For $frac1a^2+x^2, exists K>0$ such that $|f(x)|+|f'(x)| leq fracK1+x^2 ; forall x in mathbbR$
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Given $a>0$, for $f(x) = frac1a^2+x^2, exists K>0$ such that $|f(x)|+|f'(x)| leq fracK1+x^2 ; forall x in mathbbR$
This is a property that my pde book used without proving. It is supposed to be true, but I´m failing to verify.
I got:
$f’(x) = displaystyle frac-2x(a^2+x^2)^2$, then $|f(x)|+|f'(x)| = frac1a^2+x^2 + frac(a^2+x^2)^2 = frac+x^2(a^2+x^2)^2 $. What can be this constant $K$?
Exercise photo from the book:
It says: Apply the above formula (i) to the function $f(x) = (a^2+x^2)^-1$ to obtain the series below. It works when I apply the function, but why does this hypothesis of (i) holds for this $f$?
calculus inequality
asked Jul 25 at 1:31
dude3221
11510
 |Â
show 3 more comments
up vote
0
down vote
favorite
Given $a>0$, for $f(x) = frac1a^2+x^2, exists K>0$ such that $|f(x)|+|f'(x)| leq fracK1+x^2 ; forall x in mathbbR$
This is a property that my pde book used without proving. It is supposed to be true, but I´m failing to verify.
I got:
$f’(x) = displaystyle frac-2x(a^2+x^2)^2$, then $|f(x)|+|f'(x)| = frac1a^2+x^2 + frac(a^2+x^2)^2 = frac+x^2(a^2+x^2)^2 $. What can be this constant $K$?
Exercise photo from the book:
It says: Apply the above formula (i) to the function $f(x) = (a^2+x^2)^-1$ to obtain the series below. It works when I apply the function, but why does this hypothesis of (i) holds for this $f$?
calculus inequality
asked Jul 25 at 1:31
dude3221
11510
1
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
1
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
1
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $a>0$, for $f(x) = frac1a^2+x^2, exists K>0$ such that $|f(x)|+|f'(x)| leq fracK1+x^2 ; forall x in mathbbR$
This is a property that my pde book used without proving. It is supposed to be true, but I´m failing to verify.
I got:
$f’(x) = displaystyle frac-2x(a^2+x^2)^2$, then $|f(x)|+|f'(x)| = frac1a^2+x^2 + frac(a^2+x^2)^2 = frac+x^2(a^2+x^2)^2 $. What can be this constant $K$?
Exercise photo from the book:
It says: Apply the above formula (i) to the function $f(x) = (a^2+x^2)^-1$ to obtain the series below. It works when I apply the function, but why does this hypothesis of (i) holds for this $f$?
calculus inequality
asked Jul 25 at 1:31
dude3221
11510
Given $a>0$, for $f(x) = frac1a^2+x^2, exists K>0$ such that $|f(x)|+|f'(x)| leq fracK1+x^2 ; forall x in mathbbR$
This is a property that my pde book used without proving. It is supposed to be true, but I´m failing to verify.
I got:
$f’(x) = displaystyle frac-2x(a^2+x^2)^2$, then $|f(x)|+|f'(x)| = frac1a^2+x^2 + frac(a^2+x^2)^2 = frac+x^2(a^2+x^2)^2 $. What can be this constant $K$?
Exercise photo from the book:
It says: Apply the above formula (i) to the function $f(x) = (a^2+x^2)^-1$ to obtain the series below. It works when I apply the function, but why does this hypothesis of (i) holds for this $f$?
calculus inequality
asked Jul 25 at 1:31
dude3221
11510
edited Jul 25 at 3:26
asked Jul 25 at 1:31
dude3221
11510
asked Jul 25 at 1:31
dude3221
11510
asked Jul 25 at 1:31
dude3221
11510
11510
1
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
1
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
1
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07
 |Â
show 3 more comments
1
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
1
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
1
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07
1
1
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
1
1
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
1
1
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Let $g(x):=(1+x^2)(|f(x)|+|f'(x)|)$. Since $lim_xtoinfty g(x)$ and $lim_xto-infty g(x)$ are finite and $g$ doesn't have any vertical asymptote we conclude that there exists $K=K(a) > 0$ such that
$$g(x) leq K quad forall x in mathbb R,$$ or equivalently
$$ |f(x)|+|f'(x)| leq fracK1+x^2 quad forall x in mathbb R. $$
answered Jul 25 at 4:17
Gonzalo Benavides
581317
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Let $g(x):=(1+x^2)(|f(x)|+|f'(x)|)$. Since $lim_xtoinfty g(x)$ and $lim_xto-infty g(x)$ are finite and $g$ doesn't have any vertical asymptote we conclude that there exists $K=K(a) > 0$ such that
$$g(x) leq K quad forall x in mathbb R,$$ or equivalently
$$ |f(x)|+|f'(x)| leq fracK1+x^2 quad forall x in mathbb R. $$
answered Jul 25 at 4:17
Gonzalo Benavides
581317
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
add a comment |Â
up vote
2
down vote
accepted
Let $g(x):=(1+x^2)(|f(x)|+|f'(x)|)$. Since $lim_xtoinfty g(x)$ and $lim_xto-infty g(x)$ are finite and $g$ doesn't have any vertical asymptote we conclude that there exists $K=K(a) > 0$ such that
$$g(x) leq K quad forall x in mathbb R,$$ or equivalently
$$ |f(x)|+|f'(x)| leq fracK1+x^2 quad forall x in mathbb R. $$
answered Jul 25 at 4:17
Gonzalo Benavides
581317
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Let $g(x):=(1+x^2)(|f(x)|+|f'(x)|)$. Since $lim_xtoinfty g(x)$ and $lim_xto-infty g(x)$ are finite and $g$ doesn't have any vertical asymptote we conclude that there exists $K=K(a) > 0$ such that
$$g(x) leq K quad forall x in mathbb R,$$ or equivalently
$$ |f(x)|+|f'(x)| leq fracK1+x^2 quad forall x in mathbb R. $$
answered Jul 25 at 4:17
Gonzalo Benavides
581317
Let $g(x):=(1+x^2)(|f(x)|+|f'(x)|)$. Since $lim_xtoinfty g(x)$ and $lim_xto-infty g(x)$ are finite and $g$ doesn't have any vertical asymptote we conclude that there exists $K=K(a) > 0$ such that
$$g(x) leq K quad forall x in mathbb R,$$ or equivalently
$$ |f(x)|+|f'(x)| leq fracK1+x^2 quad forall x in mathbb R. $$
answered Jul 25 at 4:17
Gonzalo Benavides
581317
answered Jul 25 at 4:17
Gonzalo Benavides
581317
answered Jul 25 at 4:17
Gonzalo Benavides
581317
answered Jul 25 at 4:17
Gonzalo Benavides
581317
581317
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
add a comment |Â
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
1
1
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
This is beautiful. Thanks a lot!
– dude3221
Jul 25 at 5:30
add a comment |Â
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1
If $a=0$, LHS of the inequality grows without bounds as $xto 0$ while RHS is bounded by $K$. Have you got the expressions correctly?
– Macavity
Jul 25 at 2:37
1
@Macavity I think we should assume $a neq 0$
– Gonzalo Benavides
Jul 25 at 2:56
1
@GonzaloBenavides Even if one assumes that, you’ll always have a nbd of $0$ where the inequality fails. Take any $ain (0, 1/sqrt K)$ for a counter example.
– Macavity
Jul 25 at 2:58
oh, sorry! yes, $a>0$
– dude3221
Jul 25 at 2:58
@Macavity, so the inequality isn´t valid for this function? That´s weird, it´s also in an exercise of my pde book...
– dude3221
Jul 25 at 3:07