Find the best upper bound of $varepsilon$ such that $g(x) = 1 + varepsilon f(x) > 0$, $f(x)$ bounded on a...
$begingroup$
Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:
$$g(x) = 1 + varepsilon f(x),$$
for $x in [0, 1]$ and some $varepsilon > 0$.
Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:
$$0 < varepsilon < M,$$
for some $M > 0$.
My attempt
Let $[0, 1] = I^- cup I^+$, where:
$f(x) < 0 ~forall x in I^-$;
$f(x) geq 0 ~forall x in I^+$.
For $x in I^+$, then $g(x) > 0$.
For $x in I^-$, since function $f$ is (smooth, then) bounded, then:
$$exists F > 0 : f(x)> -F ~forall x in I^-.$$
Therefore:
$$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$
If $varepsilon < frac{1}{F}$, then:
$$1 - varepsilon F > 0,$$
and hence
$$g(x) > 0 ~forall x in [0, 1].$$
Questions
Is there some easies way to achieve this kind of results?
Is there any better upper bound for $varepsilon$?
algebra-precalculus proof-verification upper-lower-bounds
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
$endgroup$
add a comment |
$begingroup$
Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:
$$g(x) = 1 + varepsilon f(x),$$
for $x in [0, 1]$ and some $varepsilon > 0$.
Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:
$$0 < varepsilon < M,$$
for some $M > 0$.
My attempt
Let $[0, 1] = I^- cup I^+$, where:
$f(x) < 0 ~forall x in I^-$;
$f(x) geq 0 ~forall x in I^+$.
For $x in I^+$, then $g(x) > 0$.
For $x in I^-$, since function $f$ is (smooth, then) bounded, then:
$$exists F > 0 : f(x)> -F ~forall x in I^-.$$
Therefore:
$$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$
If $varepsilon < frac{1}{F}$, then:
$$1 - varepsilon F > 0,$$
and hence
$$g(x) > 0 ~forall x in [0, 1].$$
Questions
Is there some easies way to achieve this kind of results?
Is there any better upper bound for $varepsilon$?
algebra-precalculus proof-verification upper-lower-bounds
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
$endgroup$
add a comment |
$begingroup$
Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:
$$g(x) = 1 + varepsilon f(x),$$
for $x in [0, 1]$ and some $varepsilon > 0$.
Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:
$$0 < varepsilon < M,$$
for some $M > 0$.
My attempt
Let $[0, 1] = I^- cup I^+$, where:
$f(x) < 0 ~forall x in I^-$;
$f(x) geq 0 ~forall x in I^+$.
For $x in I^+$, then $g(x) > 0$.
For $x in I^-$, since function $f$ is (smooth, then) bounded, then:
$$exists F > 0 : f(x)> -F ~forall x in I^-.$$
Therefore:
$$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$
If $varepsilon < frac{1}{F}$, then:
$$1 - varepsilon F > 0,$$
and hence
$$g(x) > 0 ~forall x in [0, 1].$$
Questions
Is there some easies way to achieve this kind of results?
Is there any better upper bound for $varepsilon$?
algebra-precalculus proof-verification upper-lower-bounds
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
$endgroup$
Consider a smooth function $f : [0, 1] to mathbb{R}$. Moreover, consider:
$$g(x) = 1 + varepsilon f(x),$$
for $x in [0, 1]$ and some $varepsilon > 0$.
Which are the conditions on $varepsilon$ which guarantee that $g(x) > 0$? I guess the answer is something in the form:
$$0 < varepsilon < M,$$
for some $M > 0$.
My attempt
Let $[0, 1] = I^- cup I^+$, where:
$f(x) < 0 ~forall x in I^-$;
$f(x) geq 0 ~forall x in I^+$.
For $x in I^+$, then $g(x) > 0$.
For $x in I^-$, since function $f$ is (smooth, then) bounded, then:
$$exists F > 0 : f(x)> -F ~forall x in I^-.$$
Therefore:
$$g(x) = 1 + varepsilon f(x) > 1 - varepsilon F ~forall x in I^-.$$
If $varepsilon < frac{1}{F}$, then:
$$1 - varepsilon F > 0,$$
and hence
$$g(x) > 0 ~forall x in [0, 1].$$
Questions
Is there some easies way to achieve this kind of results?
Is there any better upper bound for $varepsilon$?
algebra-precalculus proof-verification upper-lower-bounds
algebra-precalculus proof-verification upper-lower-bounds
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
asked Dec 2 '18 at 16:01
the_candymanthe_candyman
8,91632145
8,91632145
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)
$$
g(x)geq 0 quad forall xin [0,1] Leftrightarrow
1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
$$
If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
$varepsilongeq frac{-1}{m}$
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
$endgroup$
add a comment |
$begingroup$
If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.
If $f(x) < 0$, then
begin{align}
g(x) > 0
&iff 1 + epsilon f(x) > 0 \
&iff epsilon f(x) > -1 \
&iff epsilon < -dfrac{1}{f(x)}
end{align}
So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
$endgroup$
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)
$$
g(x)geq 0 quad forall xin [0,1] Leftrightarrow
1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
$$
If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
$varepsilongeq frac{-1}{m}$
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
$endgroup$
add a comment |
$begingroup$
It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)
$$
g(x)geq 0 quad forall xin [0,1] Leftrightarrow
1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
$$
If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
$varepsilongeq frac{-1}{m}$
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
$endgroup$
add a comment |
$begingroup$
It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)
$$
g(x)geq 0 quad forall xin [0,1] Leftrightarrow
1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
$$
If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
$varepsilongeq frac{-1}{m}$
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
$endgroup$
It's easier than you think. You just need that your function is continuous. Indeed, since $f$ is continuous, for the extreme value theorem $f$ must attain a maximum $M$ and a minimum $m$.
Then (SPOILER)
$$
g(x)geq 0 quad forall xin [0,1] Leftrightarrow
1+varepsilon f(x)geq 0 Leftrightarrow 1+varepsilon mgeq 0
$$
If $m=0$, it's obvious that $g(x)>0$, otherwise we select $varepsilon >0$ such that
$varepsilongeq frac{-1}{m}$
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
answered Dec 2 '18 at 16:34
Andrea CavaliereAndrea Cavaliere
1156
1156
add a comment |
add a comment |
$begingroup$
If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.
If $f(x) < 0$, then
begin{align}
g(x) > 0
&iff 1 + epsilon f(x) > 0 \
&iff epsilon f(x) > -1 \
&iff epsilon < -dfrac{1}{f(x)}
end{align}
So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
$endgroup$
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
add a comment |
$begingroup$
If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.
If $f(x) < 0$, then
begin{align}
g(x) > 0
&iff 1 + epsilon f(x) > 0 \
&iff epsilon f(x) > -1 \
&iff epsilon < -dfrac{1}{f(x)}
end{align}
So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
$endgroup$
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
add a comment |
$begingroup$
If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.
If $f(x) < 0$, then
begin{align}
g(x) > 0
&iff 1 + epsilon f(x) > 0 \
&iff epsilon f(x) > -1 \
&iff epsilon < -dfrac{1}{f(x)}
end{align}
So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
$endgroup$
If $f(x) ge 0$, then $1+epsilon f(x) > 0$ for all values of $epsilon$.
If $f(x) < 0$, then
begin{align}
g(x) > 0
&iff 1 + epsilon f(x) > 0 \
&iff epsilon f(x) > -1 \
&iff epsilon < -dfrac{1}{f(x)}
end{align}
So $displaystyle epsilon = -max_{f(x)<0} dfrac{1}{f(x)}$
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
edited Dec 2 '18 at 19:39
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
answered Dec 2 '18 at 19:27
steven gregorysteven gregory
18.2k32258
18.2k32258
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
add a comment |
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
Thanks a lot for the answer. Then, can I take the lowest between $$-min_{f(x)<0} dfrac{1}{f(x)}$$ and $$-frac{1}{min_{f(x)<0} f(x)},$$ right?
$endgroup$
– the_candyman
Dec 2 '18 at 19:30
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
I had to rethink my answer and change it. That should be $-frac{1}{max_{f(x)<0} f(x)},$
$endgroup$
– steven gregory
Dec 2 '18 at 19:40
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
In that case, your answer is the same of the one provided by Andrea Cavaliere.
$endgroup$
– the_candyman
Dec 2 '18 at 19:49
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
$begingroup$
Yes. I just did the logic differenly.
$endgroup$
– steven gregory
Dec 2 '18 at 20:06
add a comment |
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