Alternative definiton of differentiability
$begingroup$
Is it correct to say that both of definitions of differentiability are equivalent?
$(1);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_oiffexists_{ainmathbb{C}}left(lim_{zrightarrow z_0}frac{f(z)-f(z_0)}{z-z_o}=aright)$.
$(2);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_o
qquad iffexists_{ainmathbb{C}};forall_{epsilon>0};exists_{delta>0};forall_{zin D}left(left(vert z-z_0vert<deltaright)Rightarrow left(|frac{f(z)-f(z_0)}{z-z_o}-a|right)<epsilon.right)$
real-analysis derivatives
edited Jan 19 at 19:11
jordan_glen
1
asked Jan 19 at 18:47
RM777RM777
39112
$endgroup$
|
show 1 more comment
$begingroup$
Is it correct to say that both of definitions of differentiability are equivalent?
$(1);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_oiffexists_{ainmathbb{C}}left(lim_{zrightarrow z_0}frac{f(z)-f(z_0)}{z-z_o}=aright)$.
$(2);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_o
qquad iffexists_{ainmathbb{C}};forall_{epsilon>0};exists_{delta>0};forall_{zin D}left(left(vert z-z_0vert<deltaright)Rightarrow left(|frac{f(z)-f(z_0)}{z-z_o}-a|right)<epsilon.right)$
real-analysis derivatives
edited Jan 19 at 19:11
jordan_glen
1
asked Jan 19 at 18:47
RM777RM777
39112
$endgroup$
1
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
1
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30
|
show 1 more comment
$begingroup$
Is it correct to say that both of definitions of differentiability are equivalent?
$(1);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_oiffexists_{ainmathbb{C}}left(lim_{zrightarrow z_0}frac{f(z)-f(z_0)}{z-z_o}=aright)$.
$(2);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_o
qquad iffexists_{ainmathbb{C}};forall_{epsilon>0};exists_{delta>0};forall_{zin D}left(left(vert z-z_0vert<deltaright)Rightarrow left(|frac{f(z)-f(z_0)}{z-z_o}-a|right)<epsilon.right)$
real-analysis derivatives
edited Jan 19 at 19:11
jordan_glen
1
asked Jan 19 at 18:47
RM777RM777
39112
$endgroup$
Is it correct to say that both of definitions of differentiability are equivalent?
$(1);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_oiffexists_{ainmathbb{C}}left(lim_{zrightarrow z_0}frac{f(z)-f(z_0)}{z-z_o}=aright)$.
$(2);;displaystyle f:Drightarrowmathbb{C},Dsubseteqmathbb{R}text{ is differentiable in }z_o
qquad iffexists_{ainmathbb{C}};forall_{epsilon>0};exists_{delta>0};forall_{zin D}left(left(vert z-z_0vert<deltaright)Rightarrow left(|frac{f(z)-f(z_0)}{z-z_o}-a|right)<epsilon.right)$
real-analysis derivatives
real-analysis derivatives
edited Jan 19 at 19:11
jordan_glen
1
asked Jan 19 at 18:47
RM777RM777
39112
edited Jan 19 at 19:11
jordan_glen
1
asked Jan 19 at 18:47
RM777RM777
39112
edited Jan 19 at 19:11
jordan_glen
1
edited Jan 19 at 19:11
jordan_glen
1
edited Jan 19 at 19:11
jordan_glen
1
1
asked Jan 19 at 18:47
RM777RM777
39112
asked Jan 19 at 18:47
RM777RM777
39112
asked Jan 19 at 18:47
RM777RM777
39112
39112
1
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
1
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30
|
show 1 more comment
1
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
1
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30
1
1
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
1
1
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30
|
show 1 more comment
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1
$begingroup$
If you unpack the definition of a limit in the first definition, what do you get?
$endgroup$
– Xander Henderson
Jan 19 at 18:55
$begingroup$
I am not sure whether I get what I wrote down or $exists_{ainmathbb{C}}forall_{epsilon>0}exists_{delta>0}forall_{zin D}|z-z_0|<deltaRightarrow |frac{f(z)-a}{z-z_o}|<epsilon$
$endgroup$
– RM777
Jan 19 at 19:01
$begingroup$
But this is exactly what you wrote in the question, no? (Apart from the fact that $|z-z_0|<delta$ should read $0<|z-z_0|<delta$.)
$endgroup$
– Did
Jan 19 at 19:20
$begingroup$
@Did the Right side is different. In my comment I wrote $|frac{f(z)-a}{z-z_0}|$But in my Question I wrote down $|frac{f(z)-f(z_0)}{z-z_0}-a|$
$endgroup$
– RM777
Jan 19 at 19:24
1
$begingroup$
Ah, I missed that. But then the comment is truly absurd. Would not work even for $f(z)=z$, whose derivative you probably know...
$endgroup$
– Did
Jan 19 at 19:30