Mixing
Implications among assertions
$begingroup$
Explain implications among these assertions:
a. The function $f:mathbb R^2rightarrow mathbb R$ is continuously differentiable.
b. The function $f:mathbb R^2rightarrow mathbb R$ has directional derivatives in all directions at each point in $mathbb R^2$.
c.The function $f:mathbb R^2rightarrow mathbb R$ has first order partial derivatives at each point in $mathbb R^2$.
(Hint: for $bnRightarrow a$ consider $f=begin {cases} sin(frac{y^2}{x})sqrt{x^2+y^2}, xneq0 \0,x=0 end {cases}$ ).
So, if $f$ is continuously differentiable, then, obviously, it has first order partial derivatives, so $a Rightarrow c$, however the converse is not true.
Following the hint consider $$frac{partial f}{partial vec{p}}(vec{0})=lim_{trightarrow 0}frac{f(tp_1,tp_2)-f(0,0)}{t}=lim_{trightarrow 0}frac{sin(frac{(tp_2)^2}{tp_1})sqrt{t^2(p_1^2+p_2^2)}}{t}=lim_{trightarrow 0}frac{sin(frac{tp_2^2}{p_1})|t|sqrt{(p_1^2+p_2^2)}}{t}=0$$
Also $frac{partial f}{partial x} (x,y)=frac{x^3 sin(frac{y^2}{x}) - y^2 (x^2 + y^2) cos(frac{y^2}{x})}{x^2sqrt{x^2 + y^2}}$, and $frac{partial f}{partial x} (0,0)=frac{f(t,0)-f(0,0)}{t}=0.$ But $$lim_{(x,y)rightarrow (0,0)}frac{partial f}{partial x} (x,y)=text{(restricted to the line } x=my^2)=lim_{yrightarrow 0} frac{m^3 y^6 sin(frac{1}{m}) - y^2 (m^2 y^4 + y^2) cos(frac{1}{m}))}{m^2 y^4 sqrt{m^2 y^4 + y^2}} = -infty neq 0.$$
So $frac{partial f}{partial x} (x,y)$ is not continuous at (0,0).
Are there any mistakes in my prove? And I don't know how to prove/disprove other implications, can somebody show how to prove or give counterexamples to these implications?
real-analysis partial-derivative
asked Feb 2 at 16:27
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
Explain implications among these assertions:
a. The function $f:mathbb R^2rightarrow mathbb R$ is continuously differentiable.
b. The function $f:mathbb R^2rightarrow mathbb R$ has directional derivatives in all directions at each point in $mathbb R^2$.
c.The function $f:mathbb R^2rightarrow mathbb R$ has first order partial derivatives at each point in $mathbb R^2$.
(Hint: for $bnRightarrow a$ consider $f=begin {cases} sin(frac{y^2}{x})sqrt{x^2+y^2}, xneq0 \0,x=0 end {cases}$ ).
So, if $f$ is continuously differentiable, then, obviously, it has first order partial derivatives, so $a Rightarrow c$, however the converse is not true.
Following the hint consider $$frac{partial f}{partial vec{p}}(vec{0})=lim_{trightarrow 0}frac{f(tp_1,tp_2)-f(0,0)}{t}=lim_{trightarrow 0}frac{sin(frac{(tp_2)^2}{tp_1})sqrt{t^2(p_1^2+p_2^2)}}{t}=lim_{trightarrow 0}frac{sin(frac{tp_2^2}{p_1})|t|sqrt{(p_1^2+p_2^2)}}{t}=0$$
Also $frac{partial f}{partial x} (x,y)=frac{x^3 sin(frac{y^2}{x}) - y^2 (x^2 + y^2) cos(frac{y^2}{x})}{x^2sqrt{x^2 + y^2}}$, and $frac{partial f}{partial x} (0,0)=frac{f(t,0)-f(0,0)}{t}=0.$ But $$lim_{(x,y)rightarrow (0,0)}frac{partial f}{partial x} (x,y)=text{(restricted to the line } x=my^2)=lim_{yrightarrow 0} frac{m^3 y^6 sin(frac{1}{m}) - y^2 (m^2 y^4 + y^2) cos(frac{1}{m}))}{m^2 y^4 sqrt{m^2 y^4 + y^2}} = -infty neq 0.$$
So $frac{partial f}{partial x} (x,y)$ is not continuous at (0,0).
Are there any mistakes in my prove? And I don't know how to prove/disprove other implications, can somebody show how to prove or give counterexamples to these implications?
real-analysis partial-derivative
asked Feb 2 at 16:27
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
Explain implications among these assertions:
a. The function $f:mathbb R^2rightarrow mathbb R$ is continuously differentiable.
b. The function $f:mathbb R^2rightarrow mathbb R$ has directional derivatives in all directions at each point in $mathbb R^2$.
c.The function $f:mathbb R^2rightarrow mathbb R$ has first order partial derivatives at each point in $mathbb R^2$.
(Hint: for $bnRightarrow a$ consider $f=begin {cases} sin(frac{y^2}{x})sqrt{x^2+y^2}, xneq0 \0,x=0 end {cases}$ ).
So, if $f$ is continuously differentiable, then, obviously, it has first order partial derivatives, so $a Rightarrow c$, however the converse is not true.
Following the hint consider $$frac{partial f}{partial vec{p}}(vec{0})=lim_{trightarrow 0}frac{f(tp_1,tp_2)-f(0,0)}{t}=lim_{trightarrow 0}frac{sin(frac{(tp_2)^2}{tp_1})sqrt{t^2(p_1^2+p_2^2)}}{t}=lim_{trightarrow 0}frac{sin(frac{tp_2^2}{p_1})|t|sqrt{(p_1^2+p_2^2)}}{t}=0$$
Also $frac{partial f}{partial x} (x,y)=frac{x^3 sin(frac{y^2}{x}) - y^2 (x^2 + y^2) cos(frac{y^2}{x})}{x^2sqrt{x^2 + y^2}}$, and $frac{partial f}{partial x} (0,0)=frac{f(t,0)-f(0,0)}{t}=0.$ But $$lim_{(x,y)rightarrow (0,0)}frac{partial f}{partial x} (x,y)=text{(restricted to the line } x=my^2)=lim_{yrightarrow 0} frac{m^3 y^6 sin(frac{1}{m}) - y^2 (m^2 y^4 + y^2) cos(frac{1}{m}))}{m^2 y^4 sqrt{m^2 y^4 + y^2}} = -infty neq 0.$$
So $frac{partial f}{partial x} (x,y)$ is not continuous at (0,0).
Are there any mistakes in my prove? And I don't know how to prove/disprove other implications, can somebody show how to prove or give counterexamples to these implications?
real-analysis partial-derivative
asked Feb 2 at 16:27
dxdydzdxdydz
49110
$endgroup$
Explain implications among these assertions:
a. The function $f:mathbb R^2rightarrow mathbb R$ is continuously differentiable.
b. The function $f:mathbb R^2rightarrow mathbb R$ has directional derivatives in all directions at each point in $mathbb R^2$.
c.The function $f:mathbb R^2rightarrow mathbb R$ has first order partial derivatives at each point in $mathbb R^2$.
(Hint: for $bnRightarrow a$ consider $f=begin {cases} sin(frac{y^2}{x})sqrt{x^2+y^2}, xneq0 \0,x=0 end {cases}$ ).
So, if $f$ is continuously differentiable, then, obviously, it has first order partial derivatives, so $a Rightarrow c$, however the converse is not true.
Following the hint consider $$frac{partial f}{partial vec{p}}(vec{0})=lim_{trightarrow 0}frac{f(tp_1,tp_2)-f(0,0)}{t}=lim_{trightarrow 0}frac{sin(frac{(tp_2)^2}{tp_1})sqrt{t^2(p_1^2+p_2^2)}}{t}=lim_{trightarrow 0}frac{sin(frac{tp_2^2}{p_1})|t|sqrt{(p_1^2+p_2^2)}}{t}=0$$
Also $frac{partial f}{partial x} (x,y)=frac{x^3 sin(frac{y^2}{x}) - y^2 (x^2 + y^2) cos(frac{y^2}{x})}{x^2sqrt{x^2 + y^2}}$, and $frac{partial f}{partial x} (0,0)=frac{f(t,0)-f(0,0)}{t}=0.$ But $$lim_{(x,y)rightarrow (0,0)}frac{partial f}{partial x} (x,y)=text{(restricted to the line } x=my^2)=lim_{yrightarrow 0} frac{m^3 y^6 sin(frac{1}{m}) - y^2 (m^2 y^4 + y^2) cos(frac{1}{m}))}{m^2 y^4 sqrt{m^2 y^4 + y^2}} = -infty neq 0.$$
So $frac{partial f}{partial x} (x,y)$ is not continuous at (0,0).
Are there any mistakes in my prove? And I don't know how to prove/disprove other implications, can somebody show how to prove or give counterexamples to these implications?
real-analysis partial-derivative
real-analysis partial-derivative
asked Feb 2 at 16:27
dxdydzdxdydz
49110
asked Feb 2 at 16:27
dxdydzdxdydz
49110
asked Feb 2 at 16:27
dxdydzdxdydz
49110
asked Feb 2 at 16:27
dxdydzdxdydz
49110
asked Feb 2 at 16:27
dxdydzdxdydz
49110
49110
add a comment |
add a comment |
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