Mixing
If function is positive and continuous at a point, then it is positive in a neighborhood of the point?
$begingroup$
Let $f:mathbb Rrightarrow mathbb R$ be defined as
$$
f(x) = left{ begin{array}{c}
x-x^2 & text{ if } xinmathbb Q \
x+x^2 & text{ if } xnotin mathbb Q.
end{array}
right.
$$
Show that $f'(0)=1$ and yet there is no neighborhood of the point $0$ on which $f$ is monotonically increasing.
For the first part, using the density of irrationals we have
$$f'(0)=text{lim}_{xrightarrow0^-}frac{f(x)-f(0)}{x-0}=text{lim}_{xrightarrow0^+}frac{f(x)-f(0)}{x-0}=frac{x+x^2-0}{x-0}=1.$$ Is it correct?
For the second part, I've read in the book that "if a function is positive and continuous at a point, then it is positive in a neighborhood of the point." So why is this not working here? And how should I prove that there is no neighborhood of the point $0$ on which $f$ is monotonically increasing?
real-analysis
edited Jan 27 at 23:28
DeepSea
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb Rrightarrow mathbb R$ be defined as
$$
f(x) = left{ begin{array}{c}
x-x^2 & text{ if } xinmathbb Q \
x+x^2 & text{ if } xnotin mathbb Q.
end{array}
right.
$$
Show that $f'(0)=1$ and yet there is no neighborhood of the point $0$ on which $f$ is monotonically increasing.
For the first part, using the density of irrationals we have
$$f'(0)=text{lim}_{xrightarrow0^-}frac{f(x)-f(0)}{x-0}=text{lim}_{xrightarrow0^+}frac{f(x)-f(0)}{x-0}=frac{x+x^2-0}{x-0}=1.$$ Is it correct?
For the second part, I've read in the book that "if a function is positive and continuous at a point, then it is positive in a neighborhood of the point." So why is this not working here? And how should I prove that there is no neighborhood of the point $0$ on which $f$ is monotonically increasing?
real-analysis
edited Jan 27 at 23:28
DeepSea
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
$endgroup$
1
$begingroup$
Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
$endgroup$
– Mindlack
Jan 26 at 23:24
$begingroup$
Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
$endgroup$
– Matt A Pelto
Jan 26 at 23:35
$begingroup$
f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
$endgroup$
– William Elliot
Jan 26 at 23:43
$begingroup$
@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
$endgroup$
– MetaColon
Jan 26 at 23:52
1
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14
add a comment |
$begingroup$
Let $f:mathbb Rrightarrow mathbb R$ be defined as
$$
f(x) = left{ begin{array}{c}
x-x^2 & text{ if } xinmathbb Q \
x+x^2 & text{ if } xnotin mathbb Q.
end{array}
right.
$$
Show that $f'(0)=1$ and yet there is no neighborhood of the point $0$ on which $f$ is monotonically increasing.
For the first part, using the density of irrationals we have
$$f'(0)=text{lim}_{xrightarrow0^-}frac{f(x)-f(0)}{x-0}=text{lim}_{xrightarrow0^+}frac{f(x)-f(0)}{x-0}=frac{x+x^2-0}{x-0}=1.$$ Is it correct?
For the second part, I've read in the book that "if a function is positive and continuous at a point, then it is positive in a neighborhood of the point." So why is this not working here? And how should I prove that there is no neighborhood of the point $0$ on which $f$ is monotonically increasing?
real-analysis
edited Jan 27 at 23:28
DeepSea
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
$endgroup$
Let $f:mathbb Rrightarrow mathbb R$ be defined as
$$
f(x) = left{ begin{array}{c}
x-x^2 & text{ if } xinmathbb Q \
x+x^2 & text{ if } xnotin mathbb Q.
end{array}
right.
$$
Show that $f'(0)=1$ and yet there is no neighborhood of the point $0$ on which $f$ is monotonically increasing.
For the first part, using the density of irrationals we have
$$f'(0)=text{lim}_{xrightarrow0^-}frac{f(x)-f(0)}{x-0}=text{lim}_{xrightarrow0^+}frac{f(x)-f(0)}{x-0}=frac{x+x^2-0}{x-0}=1.$$ Is it correct?
For the second part, I've read in the book that "if a function is positive and continuous at a point, then it is positive in a neighborhood of the point." So why is this not working here? And how should I prove that there is no neighborhood of the point $0$ on which $f$ is monotonically increasing?
real-analysis
real-analysis
edited Jan 27 at 23:28
DeepSea
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
edited Jan 27 at 23:28
DeepSea
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
edited Jan 27 at 23:28
DeepSea
71.3k54488
edited Jan 27 at 23:28
DeepSea
71.3k54488
edited Jan 27 at 23:28
DeepSea
71.3k54488
71.3k54488
asked Jan 26 at 23:20
dxdydzdxdydz
42110
asked Jan 26 at 23:20
dxdydzdxdydz
42110
asked Jan 26 at 23:20
dxdydzdxdydz
42110
42110
1
$begingroup$
Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
$endgroup$
– Mindlack
Jan 26 at 23:24
$begingroup$
Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
$endgroup$
– Matt A Pelto
Jan 26 at 23:35
$begingroup$
f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
$endgroup$
– William Elliot
Jan 26 at 23:43
$begingroup$
@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
$endgroup$
– MetaColon
Jan 26 at 23:52
1
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14
add a comment |
1
$begingroup$
Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
$endgroup$
– Mindlack
Jan 26 at 23:24
$begingroup$
Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
$endgroup$
– Matt A Pelto
Jan 26 at 23:35
$begingroup$
f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
$endgroup$
– William Elliot
Jan 26 at 23:43
$begingroup$
@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
$endgroup$
– MetaColon
Jan 26 at 23:52
1
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14
1
1
$begingroup$
Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
$endgroup$
– Mindlack
Jan 26 at 23:24
$begingroup$
Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
$endgroup$
– Mindlack
Jan 26 at 23:24
$begingroup$
Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
$endgroup$
– Matt A Pelto
Jan 26 at 23:35
$begingroup$
Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
$endgroup$
– Matt A Pelto
Jan 26 at 23:35
$begingroup$
f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
$endgroup$
– William Elliot
Jan 26 at 23:43
$begingroup$
f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
$endgroup$
– William Elliot
Jan 26 at 23:43
$begingroup$
@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
$endgroup$
– MetaColon
Jan 26 at 23:52
$begingroup$
@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
$endgroup$
– MetaColon
Jan 26 at 23:52
1
1
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$a)$: We can rewrite $f(x) - x = left { begin {array} {c} -x^2 & text{ if } x in mathbb{Q}\
x^2 & text{ if } x notin mathbb{Q}. end{array}\
right.$. Thus $displaystyle lim_{x to 0} left|dfrac{f(x) - f(0)}{x-0} - 1right|= displaystyle lim_{x to 0} left|dfrac{f(x)-x}{x}right|= displaystyle lim_{x to 0} left|dfrac{x^2}{x}right|= displaystyle lim_{x to 0} |x| = 0implies displaystyle lim_{x to 0} left(dfrac{f(x)-f(0)}{x-0} - 1right) = 0implies displaystyle lim_{x to 0} dfrac{f(x) - f(0)}{x-0} = 1implies f'(0) = 1$.
$b)$: Let $a > 0$ and consider $(-a,a)$ be a neighborhood of $0$. We might consider further that $a < dfrac{1}{2}, a in mathbb{Q}$. Observe that $dfrac{-1+sqrt{1+4a-4a^2}}{2}< aimplies $ if we take an irrational number $b in left(dfrac{-1+sqrt{1+4a-4a^2}}{2}, aright)$ then $b+b^2 > a-a^2implies f(b) > f(a)$. Next we choose a rational number $d in (0, b)implies d < b implies d-d^2 < b < b+b^2 implies d-d^2 < b+b^2implies f(d) < f(b)$. Thus we have: $d < b < a $ and $f(d) < f(b) > f(a)$, proving $f$ is not monotonically increasing.
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
add a comment |
$begingroup$
Keep in mind that a function can only be differentiable where it is continuous (and not always there, either).
The density of the rationals and irrationals in the reals means that this piecewise function will only be continuous (and, therefore, may only be differentiable) where the two definitional formulas agree. That is, for $f$ to be continuous, we require that $$x-x^2=x+x^2\0=2x^2\0=x^2\0=x.$$
Now, to prove that $f'(0)$ even exists, we have to show that $$lim_{xto 0}frac{x-x^2-f(0)}{x-0}=lim_{xto 0}frac{x+x^2-f(0)}{x-0},$$ but this is easily seen, since $f(0)=0,$ so that we need only show that $$lim_{xto 0}frac{x-x^2}{x}=lim_{xto 0}frac{x+x^2}{x},$$ or (equivalently) that $$lim_{xto 0}1-x=lim_{xto 0}1+x.$$ I leave it to you to show that both limits exist, and that both are equal to $1,$ so that $f'(0)=1.$
On the other hand, given any $x_0neq 0,$ we can readily show that $$lim_{xto x_0}frac{x-x^2}{x}neqlim_{xto x_0}frac{x+x^2}{x},$$ so that $f'$ is defined only at $x=0.$
What your book leaves out (apparently) is that the neighborhood of the point must be a relative neighborhood of the function's domain. In this case, this means that there must be some (more general) neighborhood (say $U$) of the point $0$ such that, for all $xinoperatorname{dom}(f')cap U,$ we have that $f(x)>0.$ But this is trivially true, regardless of the neighborhood $U$ we choose, since $f'$ is defined only at $x=0,$ so that $operatorname{dom}(f')cap U={0},$ whence $f'$ is positive in a relative neighborhood of $x=0,$ as desired.
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
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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
$begingroup$
$a)$: We can rewrite $f(x) - x = left { begin {array} {c} -x^2 & text{ if } x in mathbb{Q}\
x^2 & text{ if } x notin mathbb{Q}. end{array}\
right.$. Thus $displaystyle lim_{x to 0} left|dfrac{f(x) - f(0)}{x-0} - 1right|= displaystyle lim_{x to 0} left|dfrac{f(x)-x}{x}right|= displaystyle lim_{x to 0} left|dfrac{x^2}{x}right|= displaystyle lim_{x to 0} |x| = 0implies displaystyle lim_{x to 0} left(dfrac{f(x)-f(0)}{x-0} - 1right) = 0implies displaystyle lim_{x to 0} dfrac{f(x) - f(0)}{x-0} = 1implies f'(0) = 1$.
$b)$: Let $a > 0$ and consider $(-a,a)$ be a neighborhood of $0$. We might consider further that $a < dfrac{1}{2}, a in mathbb{Q}$. Observe that $dfrac{-1+sqrt{1+4a-4a^2}}{2}< aimplies $ if we take an irrational number $b in left(dfrac{-1+sqrt{1+4a-4a^2}}{2}, aright)$ then $b+b^2 > a-a^2implies f(b) > f(a)$. Next we choose a rational number $d in (0, b)implies d < b implies d-d^2 < b < b+b^2 implies d-d^2 < b+b^2implies f(d) < f(b)$. Thus we have: $d < b < a $ and $f(d) < f(b) > f(a)$, proving $f$ is not monotonically increasing.
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
add a comment |
$begingroup$
$a)$: We can rewrite $f(x) - x = left { begin {array} {c} -x^2 & text{ if } x in mathbb{Q}\
x^2 & text{ if } x notin mathbb{Q}. end{array}\
right.$. Thus $displaystyle lim_{x to 0} left|dfrac{f(x) - f(0)}{x-0} - 1right|= displaystyle lim_{x to 0} left|dfrac{f(x)-x}{x}right|= displaystyle lim_{x to 0} left|dfrac{x^2}{x}right|= displaystyle lim_{x to 0} |x| = 0implies displaystyle lim_{x to 0} left(dfrac{f(x)-f(0)}{x-0} - 1right) = 0implies displaystyle lim_{x to 0} dfrac{f(x) - f(0)}{x-0} = 1implies f'(0) = 1$.
$b)$: Let $a > 0$ and consider $(-a,a)$ be a neighborhood of $0$. We might consider further that $a < dfrac{1}{2}, a in mathbb{Q}$. Observe that $dfrac{-1+sqrt{1+4a-4a^2}}{2}< aimplies $ if we take an irrational number $b in left(dfrac{-1+sqrt{1+4a-4a^2}}{2}, aright)$ then $b+b^2 > a-a^2implies f(b) > f(a)$. Next we choose a rational number $d in (0, b)implies d < b implies d-d^2 < b < b+b^2 implies d-d^2 < b+b^2implies f(d) < f(b)$. Thus we have: $d < b < a $ and $f(d) < f(b) > f(a)$, proving $f$ is not monotonically increasing.
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
$endgroup$
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
add a comment |
$begingroup$
$a)$: We can rewrite $f(x) - x = left { begin {array} {c} -x^2 & text{ if } x in mathbb{Q}\
x^2 & text{ if } x notin mathbb{Q}. end{array}\
right.$. Thus $displaystyle lim_{x to 0} left|dfrac{f(x) - f(0)}{x-0} - 1right|= displaystyle lim_{x to 0} left|dfrac{f(x)-x}{x}right|= displaystyle lim_{x to 0} left|dfrac{x^2}{x}right|= displaystyle lim_{x to 0} |x| = 0implies displaystyle lim_{x to 0} left(dfrac{f(x)-f(0)}{x-0} - 1right) = 0implies displaystyle lim_{x to 0} dfrac{f(x) - f(0)}{x-0} = 1implies f'(0) = 1$.
$b)$: Let $a > 0$ and consider $(-a,a)$ be a neighborhood of $0$. We might consider further that $a < dfrac{1}{2}, a in mathbb{Q}$. Observe that $dfrac{-1+sqrt{1+4a-4a^2}}{2}< aimplies $ if we take an irrational number $b in left(dfrac{-1+sqrt{1+4a-4a^2}}{2}, aright)$ then $b+b^2 > a-a^2implies f(b) > f(a)$. Next we choose a rational number $d in (0, b)implies d < b implies d-d^2 < b < b+b^2 implies d-d^2 < b+b^2implies f(d) < f(b)$. Thus we have: $d < b < a $ and $f(d) < f(b) > f(a)$, proving $f$ is not monotonically increasing.
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
$endgroup$
$a)$: We can rewrite $f(x) - x = left { begin {array} {c} -x^2 & text{ if } x in mathbb{Q}\
x^2 & text{ if } x notin mathbb{Q}. end{array}\
right.$. Thus $displaystyle lim_{x to 0} left|dfrac{f(x) - f(0)}{x-0} - 1right|= displaystyle lim_{x to 0} left|dfrac{f(x)-x}{x}right|= displaystyle lim_{x to 0} left|dfrac{x^2}{x}right|= displaystyle lim_{x to 0} |x| = 0implies displaystyle lim_{x to 0} left(dfrac{f(x)-f(0)}{x-0} - 1right) = 0implies displaystyle lim_{x to 0} dfrac{f(x) - f(0)}{x-0} = 1implies f'(0) = 1$.
$b)$: Let $a > 0$ and consider $(-a,a)$ be a neighborhood of $0$. We might consider further that $a < dfrac{1}{2}, a in mathbb{Q}$. Observe that $dfrac{-1+sqrt{1+4a-4a^2}}{2}< aimplies $ if we take an irrational number $b in left(dfrac{-1+sqrt{1+4a-4a^2}}{2}, aright)$ then $b+b^2 > a-a^2implies f(b) > f(a)$. Next we choose a rational number $d in (0, b)implies d < b implies d-d^2 < b < b+b^2 implies d-d^2 < b+b^2implies f(d) < f(b)$. Thus we have: $d < b < a $ and $f(d) < f(b) > f(a)$, proving $f$ is not monotonically increasing.
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
edited Jan 27 at 22:45
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
answered Jan 27 at 0:13
DeepSeaDeepSea
71.3k54488
71.3k54488
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any ideas for the second part?
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– dxdydz
Jan 27 at 3:40
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I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
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– dxdydz
Jan 28 at 16:18
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It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
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– DeepSea
Jan 28 at 21:05
add a comment |
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
any ideas for the second part?
$endgroup$
– dxdydz
Jan 27 at 3:40
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
I dont understand, where did $frac{-1+sqrt{1+4a-4a^2}}{2}<a$ come from?
$endgroup$
– dxdydz
Jan 28 at 16:18
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
$begingroup$
It comes from solving for $b$ the solutions of the inequality: $b^2+b > a-a^2$ and the expression above is solution of $b$ using quadratic formula.
$endgroup$
– DeepSea
Jan 28 at 21:05
add a comment |
$begingroup$
Keep in mind that a function can only be differentiable where it is continuous (and not always there, either).
The density of the rationals and irrationals in the reals means that this piecewise function will only be continuous (and, therefore, may only be differentiable) where the two definitional formulas agree. That is, for $f$ to be continuous, we require that $$x-x^2=x+x^2\0=2x^2\0=x^2\0=x.$$
Now, to prove that $f'(0)$ even exists, we have to show that $$lim_{xto 0}frac{x-x^2-f(0)}{x-0}=lim_{xto 0}frac{x+x^2-f(0)}{x-0},$$ but this is easily seen, since $f(0)=0,$ so that we need only show that $$lim_{xto 0}frac{x-x^2}{x}=lim_{xto 0}frac{x+x^2}{x},$$ or (equivalently) that $$lim_{xto 0}1-x=lim_{xto 0}1+x.$$ I leave it to you to show that both limits exist, and that both are equal to $1,$ so that $f'(0)=1.$
On the other hand, given any $x_0neq 0,$ we can readily show that $$lim_{xto x_0}frac{x-x^2}{x}neqlim_{xto x_0}frac{x+x^2}{x},$$ so that $f'$ is defined only at $x=0.$
What your book leaves out (apparently) is that the neighborhood of the point must be a relative neighborhood of the function's domain. In this case, this means that there must be some (more general) neighborhood (say $U$) of the point $0$ such that, for all $xinoperatorname{dom}(f')cap U,$ we have that $f(x)>0.$ But this is trivially true, regardless of the neighborhood $U$ we choose, since $f'$ is defined only at $x=0,$ so that $operatorname{dom}(f')cap U={0},$ whence $f'$ is positive in a relative neighborhood of $x=0,$ as desired.
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
$endgroup$
add a comment |
$begingroup$
Keep in mind that a function can only be differentiable where it is continuous (and not always there, either).
The density of the rationals and irrationals in the reals means that this piecewise function will only be continuous (and, therefore, may only be differentiable) where the two definitional formulas agree. That is, for $f$ to be continuous, we require that $$x-x^2=x+x^2\0=2x^2\0=x^2\0=x.$$
Now, to prove that $f'(0)$ even exists, we have to show that $$lim_{xto 0}frac{x-x^2-f(0)}{x-0}=lim_{xto 0}frac{x+x^2-f(0)}{x-0},$$ but this is easily seen, since $f(0)=0,$ so that we need only show that $$lim_{xto 0}frac{x-x^2}{x}=lim_{xto 0}frac{x+x^2}{x},$$ or (equivalently) that $$lim_{xto 0}1-x=lim_{xto 0}1+x.$$ I leave it to you to show that both limits exist, and that both are equal to $1,$ so that $f'(0)=1.$
On the other hand, given any $x_0neq 0,$ we can readily show that $$lim_{xto x_0}frac{x-x^2}{x}neqlim_{xto x_0}frac{x+x^2}{x},$$ so that $f'$ is defined only at $x=0.$
What your book leaves out (apparently) is that the neighborhood of the point must be a relative neighborhood of the function's domain. In this case, this means that there must be some (more general) neighborhood (say $U$) of the point $0$ such that, for all $xinoperatorname{dom}(f')cap U,$ we have that $f(x)>0.$ But this is trivially true, regardless of the neighborhood $U$ we choose, since $f'$ is defined only at $x=0,$ so that $operatorname{dom}(f')cap U={0},$ whence $f'$ is positive in a relative neighborhood of $x=0,$ as desired.
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
$endgroup$
add a comment |
$begingroup$
Keep in mind that a function can only be differentiable where it is continuous (and not always there, either).
The density of the rationals and irrationals in the reals means that this piecewise function will only be continuous (and, therefore, may only be differentiable) where the two definitional formulas agree. That is, for $f$ to be continuous, we require that $$x-x^2=x+x^2\0=2x^2\0=x^2\0=x.$$
Now, to prove that $f'(0)$ even exists, we have to show that $$lim_{xto 0}frac{x-x^2-f(0)}{x-0}=lim_{xto 0}frac{x+x^2-f(0)}{x-0},$$ but this is easily seen, since $f(0)=0,$ so that we need only show that $$lim_{xto 0}frac{x-x^2}{x}=lim_{xto 0}frac{x+x^2}{x},$$ or (equivalently) that $$lim_{xto 0}1-x=lim_{xto 0}1+x.$$ I leave it to you to show that both limits exist, and that both are equal to $1,$ so that $f'(0)=1.$
On the other hand, given any $x_0neq 0,$ we can readily show that $$lim_{xto x_0}frac{x-x^2}{x}neqlim_{xto x_0}frac{x+x^2}{x},$$ so that $f'$ is defined only at $x=0.$
What your book leaves out (apparently) is that the neighborhood of the point must be a relative neighborhood of the function's domain. In this case, this means that there must be some (more general) neighborhood (say $U$) of the point $0$ such that, for all $xinoperatorname{dom}(f')cap U,$ we have that $f(x)>0.$ But this is trivially true, regardless of the neighborhood $U$ we choose, since $f'$ is defined only at $x=0,$ so that $operatorname{dom}(f')cap U={0},$ whence $f'$ is positive in a relative neighborhood of $x=0,$ as desired.
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
$endgroup$
Keep in mind that a function can only be differentiable where it is continuous (and not always there, either).
The density of the rationals and irrationals in the reals means that this piecewise function will only be continuous (and, therefore, may only be differentiable) where the two definitional formulas agree. That is, for $f$ to be continuous, we require that $$x-x^2=x+x^2\0=2x^2\0=x^2\0=x.$$
Now, to prove that $f'(0)$ even exists, we have to show that $$lim_{xto 0}frac{x-x^2-f(0)}{x-0}=lim_{xto 0}frac{x+x^2-f(0)}{x-0},$$ but this is easily seen, since $f(0)=0,$ so that we need only show that $$lim_{xto 0}frac{x-x^2}{x}=lim_{xto 0}frac{x+x^2}{x},$$ or (equivalently) that $$lim_{xto 0}1-x=lim_{xto 0}1+x.$$ I leave it to you to show that both limits exist, and that both are equal to $1,$ so that $f'(0)=1.$
On the other hand, given any $x_0neq 0,$ we can readily show that $$lim_{xto x_0}frac{x-x^2}{x}neqlim_{xto x_0}frac{x+x^2}{x},$$ so that $f'$ is defined only at $x=0.$
What your book leaves out (apparently) is that the neighborhood of the point must be a relative neighborhood of the function's domain. In this case, this means that there must be some (more general) neighborhood (say $U$) of the point $0$ such that, for all $xinoperatorname{dom}(f')cap U,$ we have that $f(x)>0.$ But this is trivially true, regardless of the neighborhood $U$ we choose, since $f'$ is defined only at $x=0,$ so that $operatorname{dom}(f')cap U={0},$ whence $f'$ is positive in a relative neighborhood of $x=0,$ as desired.
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
answered Jan 26 at 23:59
Cameron BuieCameron Buie
86k772161
86k772161
add a comment |
add a comment |
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Why did you choose to forget about the rationals in your limit computation? For the rest: where is $f$ differentiable (or even continuous)?
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– Mindlack
Jan 26 at 23:24
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Just because the irrationals are dense in $mathbb R$, you still have to show that the limit of interest is $1$ when restricted to rationals. The fact that $mathbb Q$ and $mathbb R setminus mathbb Q$ are dense in $mathbb R$ seems more useful for showing that $f$ is not monotone increasing on any neighborhood of $0 in mathbb R$. What you have read (the quote) is true but I am unsure how it is helpful here.
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– Matt A Pelto
Jan 26 at 23:35
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f is continuous at 0 but not positive at 0. f' is positive at 0 but not continuous at 0.
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– William Elliot
Jan 26 at 23:43
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@WilliamElliot why is $f'$ not continuous at $0$? Doesn't it always approach $1$ as well as $f(0)=1$?
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– MetaColon
Jan 26 at 23:52
1
$begingroup$
If $x ne 0$ then $f'(x) = lim_{hto 0} frac {f(x+h) - f(x)}h=lim frac {x pm x^2 pm 2hx pm h^2 - x pm x^2}h = lim frac {{pm 2x^2,0}}h + pm 2x$ does not have consistent values depending upon whether $x$ and $x + h$ are both rational, irrational or of different types. So $f'$ doesn't exist anywhere but at $x=1$. So $f'(x)$ is not continuous at $x=0$.
$endgroup$
– fleablood
Jan 27 at 0:14