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Is that sentence true?
How find this $f(overline{E})=overline{f(E)} Longleftrightarrow ?$True/False Real Analysis Qualifying Exam QuestionsSufficient condition for inflection pointDoes $max{x|h_n(x) = 0} to max{x|h(x) = 0}$ provided $h_n$ is continuous, convex and $h_n to h$ uniformly?Derive an $O(h^4)$ five point formula to approximate $f'(x_0)$ that uses $f(x_0 - h), f(x_0), f(x_0 + h), f(x_0 + 2h), f(x_0 + 3h)$True or false, prove or find a counterexample. If ${x_n}$ is a sequence such that ${x_n^2}$ converges, then ${x_n}$ convergesProve that $f(x)=begin{cases}1& text{if x is rational},\0 &text{if x is irrational}end{cases}$ is discontinuous at every real number.If $f$ is analytic on $(a, b)$ containing at point $x_{0}$ with $f^{(n)}(x_{0}) = 0$ for $n in mathbb{N}$, prove $f(x) = 0$ for all $x$.Writing proofs to show that a set is open, not closed and infinite
$begingroup$
Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.
Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)
real-analysis derivatives
edited Mar 16 at 12:01
daw
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
$endgroup$
add a comment |
$begingroup$
Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.
Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)
real-analysis derivatives
edited Mar 16 at 12:01
daw
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
$endgroup$
$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18
add a comment |
$begingroup$
Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.
Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)
real-analysis derivatives
edited Mar 16 at 12:01
daw
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
$endgroup$
Let $f$ to have in point $x_0inmathbb{R}$ two different (to each other) final one-sided derivations. Then $f(x_0)=0$ is urgent (necessary) and ample (satisfactory) condition for existence of $[f^2]'(x_0)$.
Does anyone know if it is true or false ? I can't go through it, don't know how ? And if it is so, can you prove it ?
Thank you for your help :)
real-analysis derivatives
real-analysis derivatives
edited Mar 16 at 12:01
daw
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
edited Mar 16 at 12:01
daw
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
edited Mar 16 at 12:01
daw
24.8k1745
edited Mar 16 at 12:01
daw
24.8k1745
edited Mar 16 at 12:01
daw
24.8k1745
24.8k1745
asked Mar 15 at 17:29
JankaJanka
105
asked Mar 15 at 17:29
JankaJanka
105
asked Mar 15 at 17:29
JankaJanka
105
105
$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18
add a comment |
$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18
$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18
$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$
This limit exists iff $f'(x)$ exists or $f(x)=0$
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
$endgroup$
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
|
show 2 more comments
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$
This limit exists iff $f'(x)$ exists or $f(x)=0$
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
$endgroup$
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
|
show 2 more comments
$begingroup$
If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$
This limit exists iff $f'(x)$ exists or $f(x)=0$
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
$endgroup$
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
|
show 2 more comments
$begingroup$
If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$
This limit exists iff $f'(x)$ exists or $f(x)=0$
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
$endgroup$
If the two one sided derivatives exists finite, and $f$ is continuous in $x$ then
$(f^2)'(x)=lim_{h to 0}frac{f^2(x+h)-f^2(x)}{h}=lim_{hto 0}frac{left(f(x+h)-f(x)right)left(f(x+h)+f(x)right)}{h}=lim_{hto 0}frac{f(x+h)-f(x)}{h}2f(x)$
This limit exists iff $f'(x)$ exists or $f(x)=0$
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
edited Mar 16 at 11:25
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
answered Mar 15 at 19:33
Gabriele CasseseGabriele Cassese
1,171316
1,171316
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
|
show 2 more comments
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
Thanks, but how did $2f(x)$ appear ? From what ?
$endgroup$
– Janka
Mar 15 at 19:49
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
@Janka $lim_{hto 0}f(x+h)+f(x)=f(x)+f(x)=2f(x)$
$endgroup$
– Gabriele Cassese
Mar 15 at 19:51
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
and what does $h$ mean ?
$endgroup$
– Janka
Mar 15 at 19:53
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
@Janka $h$ is the increment, that we take $hto 0$ in the definition of the derivative
$endgroup$
– Gabriele Cassese
Mar 15 at 19:55
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
$begingroup$
$f$ has to be continuous in $x_0$, too.
$endgroup$
– Michael Hoppe
Mar 16 at 11:16
|
show 2 more comments
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$begingroup$
It's false. Consider $f(x)=x$ for $xleq0$ and $f(x)=1-x$ for $x>0$ and $x_0=0$.
$endgroup$
– Michael Hoppe
Mar 16 at 11:18